Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles

Big Ideas Math Geometry Answers Chapter 5

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Big Ideas Math Book Geometry Answer Key Chapter 5 Congruent Triangles

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Congruent Triangles Maintaining Mathematical Proficiency

Find the coordinates of the midpoint M of the segment with the given endpoints. Then find the distance between the two points.

Question 1.
P(- 4, 1) and Q(0, 7)
Answer:
The given points are:
P (-4, 1), Q (0, 7)
We know that,
The midpoint M of the segment with the 2 endpoints is:
( \(\frac{x1 + x2}{2}\), \(\frac{y1 + y2}{2}\) )
Let the give points are:
(x1, y1) and (x2, y2)
So,
By comparing the given poits,
We will get
x1 = -4, x2 = 0, y1 = 1, y2 = 7
Hence,
The midpoint M = ( \(\frac{-4 + 0}{2}\), \(\frac{1 + 7}{2}\) )
= ( \(\frac{-4}{2}\), \(\frac{8}{2}\) )
= (-2, 4)
Hence, from the above,
We can conclude that the midpoint M of the segment with the given endpoints is: (-2, 4)

Question 2.
G(3, 6) and H(9, – 2)
Answer:
The given points are:
G (3, 6), H (9, -2)
We know that,
The midpoint M of the segment with the 2 endpoints is:
( \(\frac{x1 + x2}{2}\), \(\frac{y1 + y2}{2}\) )
Let the give points are:
(x1, y1) and (x2, y2)
So,
By comparing the given poits,
We will get
x1 = 3, x2 = 9, y1 = 6, y2 = -2
Hence,
The midpoint M = ( \(\frac{3 + 9}{2}\), \(\frac{6 – 2}{2}\) )
= ( \(\frac{12}{2}\), \(\frac{4}{2}\) )
= (6, 2)
Hence, from the above,
We can conclude that the midpoint M of the segment with the given endpoints is: (6, 2)

Question 3.
U(- 1, – 2) and V(8, 0)
Answer:
The given points are:
U (-1, -2), V (8, 0)
We know that,
The midpoint M of the segment with the 2 endpoints is:
( \(\frac{x1 + x2}{2}\), \(\frac{y1 + y2}{2}\) )
Let the give points are:
(x1, y1) and (x2, y2)
So,
By comparing the given poits,
We will get
x1 = -1, x2 = 8, y1 = -2, y2 = 0
Hence,
The midpoint M = ( \(\frac{-1 + 8}{2}\), \(\frac{-2 + 0}{2}\) )
= ( \(\frac{7}{2}\), \(\frac{-2}{2}\) )
= ( \(\frac{7}{2}\), -1 )
Hence, from the above,
We can conclude that the midpoint M of the segment with the given endpoints is: ( \(\frac{7}{2}\), -1 )

Solve the equation.

Question 4.
7x + 12 = 3x
Answer:
The given equation is:
7x + 12 = 3x
So,
7x – 3x = 12
4x = 12
x = \(\frac{12}{4}\)
x = 3
Hence, from the above,
We can conclude that the value of x is: 3

Question 5.
14 – 6t = t
Answer:
The given equation is:
14 – 6t = t
So,
14 = 6t + t
7t = 14
t = \(\frac{14}{7}\)
t = 2
Hence, from the above,
We can conclude that the value of t is: 2

Question 6.
5p + 10 = 8p + 1
Answer:
The given equation is:
5p + 10 = 8p + 1
So,
5p – 8p = 1 – 10
-3p = -9
3p = 9
p = \(\frac{9}{3}\)
p = 3
Hence, from the above,
We can conclude that the value of p is: 3

Question 7.
w + 13 = 11w – 7
Answer:
The given equation is:
w + 13 = 11w – 7
So,
w – 11w = -7 – 13
-10w = -20
10w = 20
w = \(\frac{20}{10}\)
w = 2
Hence, from the above,
We can conclude that the value of w is: 2

Question 8.
4x + 1 = 3 – 2x
Answer:
The given equation is:
4x + 1 = 3 – 2x
So,
4x + 2x = 3 – 1
6x = 2
x = \(\frac{2}{6}\)
x = \(\frac{1}{3}\)
Hence, from the above,
We can conclude that the value of x is: \(\frac{1}{3}\)

Question 9.
z – 2 = 4 + 9z
Answer:
The given equation is:
z – 2 = 4 + 9z
So,
z – 9z = 4 + 2
-8z = 6
z = –\(\frac{6}{8}\)
z = –\(\frac{3}{4}\)
Hence, from the above,
We can conclude that the value of z is: –\(\frac{3}{4}\)

Question 10.
ABSTRACT REASONING
Is it possible to find the length of a segment in a coordinate plane without using the Distance Formula? Explain your reasoning.
Answer:
Yes, it is possible to find the length of a segment in a coordinate plane without using the distance formula
Since the segment is a portion of a line, we can use the graph to calculate the distance of a segment even though it would not provide accurate results.
Hence,
We use the distance formula to find the length of a segment in a coordinate plane

Congruent Triangles Mathematical Practices

Monitoring Progress

Classify each statement as a definition, a postulate, or a theorem. Explain your reasoning.

Question 1.
In a coordinate plane, two non-vertical lines are perpendicular if and only if the product of their slopes is – 1.
Answer:
The given statement is:
In a coordinate plane, two non-vertical lines are perpendicular if and only if the product of their slopes is – 1.
We know that,
According to the “parallel and perpendicular lines theorem”, two non-vertical lines are perpendicular if and only if the product of their slopes is -1
Hence, from the above,
We can conclude that the given statement is a Theorem

Question 2.
If two lines intersect to form a linear pair of congruent angles, then the lines are perpendicular.
Answer:
The given statement is:
If two lines intersect to form a linear pair of congruent angles, then the lines are perpendicular.
We know that,
According to the “Linear pair perpendicular theorem”,
When two straight lines intersect at a point and form a linear pair of congruent angles, then the lines are perpendicular
Hence, from the above,
We can conclude that the given statement is a Theorem

Question 3.
If two lines intersect to form a right angle. then the lines are perpendicular.
Answer:
The given statement is:
If two lines intersect to form a right angle. then the lines are perpendicular.
We know that,
According to the “Perpendicular lines theorem”,
When two lines intersect to form a right angle, the lines are perpendicular
Hence, from the above,
We can conclude that the given statement is a Theorem

Question 4.
Through any two points, there exists exactly one line.
Answer:
The given statement is:
Through any two points, there exists exactly one line
We know that,
Between two points, only one line can be drawn and we don’t need any proof to prove the above statement
We know that,
The statement that is true without proof to prove is called “Postulate”
Hence, from the above,
We can conclude that the given statement is a Postulate

5.1 Angles of Triangles

Exploration 1

Writing a Conjecture

Work with a partner.

Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 1

a. Use dynamic geometry software to draw any triangle and label it ∆ABC.
Answer:
By using the dynamic geometry software, the triangle drawn is:

b. Find the measures of the interior angles of the triangle.
Answer:
From part (a),
We can observe that the vertices of the triangle are: A, B, and C
Let the interior angles of the vertices A, B, and C be α, β, and γ respectively
Hence,
The measures of the given triangle are:

Hence, from the above,
The measures of the interior angles are:
α = 62.1°, β = 64.1°, and γ = 53.8°

c. Find the sum of the interior angle measures.
Answer:
From part (b),
The measures of the interior angles are:
α = 62.1°, β = 64.1°, and γ = 53.8°
Hence,
The sum of the interior angles = 62.1° + 64.1° + 53.1° = 180°
Hence, from the above,
We can conclude that the sum of the interior angle measures is: 180°

d. Repeat parts (a)-(c) with several other triangles. Then write a conjecture about the sum of the measures of the interior angles of a triangle.
Answer:
The representation of the 3 different triangles and their internal angle measures is:

Hence, from the above,
We can conclude that the conjecture about the sum of the measures of the interior angles of a triangle is:
The sum of the internal angle measures of a triangle is always: 180°

CONSTRUCTING VIABLE ARGUMENTS
To be proficient in math, you need to reason inductively about data and write conjectures.
Answer:
Inductive reasoning:
Inductive reasoning is the process of arriving at a conclusion based on a set of observations.
Inductive reasoning is used in geometry in a similar way.
Conjecture:
A statement you believe to be true based on inductive reasoning.

Exploration 2

Writing a Conjecture

Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 2

Work With a partner.

a. Use dynamic geometry software to draw any triangle and label it ∆ABC.
Answer:
The triangle drawn by using the dynamic geometry software is:

Hence, from the above,
We can conclude that the vertices of the triangle are: A, B, and C

b. Draw an exterior angle at any vertex and find its measure.
Answer:
From part (a),
The vertices of the triangle are: A, B, and C
Let the external angle measures of the triangle are: α, β, and γ
Hence,
The representation of the external angle measures of the triangle are:

Hence,
From the above,
We can conclude that
The external angle measures of the triangle are:
α = 310.7°, β = 299.3°, and γ = 290°

c. Find the measures of the two nonadjacent interior angles of the triangle.
Answer:
From part (b),
The external angle measures of the triangle are:
α = 310.7°, β = 299.3°, and γ = 290°
Hence,
The representation of the non-adjacent interior angles and the external angle measures of the triangle are:

Hence, from the above,
The angle measures of two non-adjacent sides are:
α = 70°, β = 60.7°, and γ = 49.3°

d. Find the sum of the measures of the two nonadjacent interior angles. Compare this sum to the measure of the exterior angle.
Answer:
From part (b),
The external angle measures of the triangle are:
α = 310.7°, β = 299.3°, and γ = 290°
From part (c),
The measures of the two non-adjacent interior angles are:
α = 70°, β = 60.7°, and γ = 49.3°
Now,
The sum of the measures of the external angles of the triangle are:
α + β + γ = 310.7° + 299.3°+ 290°
= 900.0°
The sum of the measures of the two non-adjacent interior angles is:
α + β + γ = 70° + 60.7° + 49.3°
= 180.0
Hence, from the above,
We can conclude that the sum of the measures of the external angles is 5 times the sum of the measures of the two non-adjacent interior angles

e. Repeat parts (a)-(d) with several other triangles. Then write a conjecture that compares the measure of an exterior angle with the sum of the measures of the two nonadjacent interior angles.
Answer:

Hence, from the above,
We can conclude that
The external angle measure of a vertex for a given triangle = 360° – (Internal angle measure of a vertex that we are finding the external angle measure)
The sum of the internal angle measures of the triangle is: 180°

Communicate Your Answer

Question 3.
How are the angle measures of a triangle related?
Answer:
The angle measures of a triangle are related as shown below:
The external angle measure of a vertex for a given triangle = 360° – (Internal angle measure of a vertex that we are finding the external angle measure)
The sum of the internal angle measures of the triangle is: 180°

Question 4.
An exterior angle of a triangle measures 32° What do you know about the measures of the interior angles? Explain your reasoning.
Answer:
It is given that an exterior angle of a triangle measures 32°
We know that,
The external angle measure of a vertex for a given triangle = 360° – (Internal angle measure of a vertex that we are finding the external angle measure)
So,
32° = 360° – (The internal angle measure of 32°)
The internal angle measure of 32° = 360° – 32°
The interior angle measure of 32° = 328°
Hence, from the above,
We can conclude that the interior angle measure of a triangle for an external angle measure of 32° is: 328°

Lesson 5.1 Angles of Triangles

Monitoring Progress

Question 1.
Draw an obtuse isosceles triangle and an acute scalene triangle.
Answer:
The figures of an obtuse isosceles triangle and an acute triangle are as follows:

Question 2.
∆ABC has vertices A(0, 0), B(3, 3), and C(- 3, 3), Classify the triangle by its sides. Then determine whether it is a right triangle.
Answer:
The given points are:
A (0, 0), B (3, 3), and C (-3, 3)
and the triangle is ΔABC
We know that,
To find whether the given triangle is a right-angled triangle or not,
We have to prove,
AC² = AB² + BC²
Where,
AC is the distance between A and C points
AB is the distance between A and B points
BC is the distance between B and C points
We know that,
The distance between 2 points = √(x2 – x1)² + (y2 – y1)²
Now,
Let the given points be considered as A(x1, y1), B(x2, y2), and C( x3, y3)
So,
AB = √(3 – 0)² + (3 – 0)² = √3² + 3²
= √9 + 9 = √18
BC = √(-3 – 3)² + (3 – 3)²
= √(-6)² + 0²
= √6² = 6
AC = √(-3 – 0)² + (3 – 0)²
= √(-3)² + 3²
= √9 + 9 = √18
Now,
AC² = AB² + BC²
(√18)² = (√18)² + 6²
18 = 18 + 36
18 ≠54
Hence, from the above,
We can conclude that the given triangle is not a right-angled triangle

Question 3.
Find the measure of ∠1
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 3
Answer:
The given figure is:
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 3
We know that,
The measure of an exterior angle of a triangle is equal to the sum of the measures of the two non-adjacent interior angles
From the given triangle,
The exterior angle is: (5x – 10)°
The interior angles are: 40°, 3x°, ∠1
So,
(5x – 10)° = 40° + 3x°
5x° – 3x° = 40° + 10°
2x° = 50°
x = 50° ÷ 2
x = 25°
So,
The interior angles are 40°, 3 (25)°, ∠1
= 40°, 75°, ∠1
We know that,
The sum of the interior angles of a triangle is: 180°
So,
40° + 75° + ∠1 = 180°
115° + ∠1 = 180°
∠1 = 180° – 115°
∠1 = 65°
Hence, from the above,
We can conclude that the value of ∠1 is: 65°

Question 4.
Find the measure of each acute angle.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 36
Answer:
The given figure is:
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 36
We know that,
The sum of the interior angles in a triangle is: 180°
From the given figure,
The interior angles of the right-angled triangle are: 90°, 2x°, and (x – 6)°
So,
90° + 2x° + (x – 6)° = 180°
84°+ 3x° = 180°
3x° = 180° – 84°
3x° = 96°
x = 96° ÷ 3°
x = 32°
So,
The measure of each acute angle is 90°, 2x°, (x – 6)°
= 90°, 2(32)°, (32 – 6)°
= 90°, 64°, 26°
Hence, from the above,
We can conclude that,
The measure of each acute angle is 90°, 64°, and 26°

Exercise 5.1 Angles of Triangles

Vocabulary and Core Concept Check

Question 1.
WRITING
Can a right triangle also be obtuse? Explain our reasoning.
Answer:
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.1 a 1

Question 2.
COMPLETE THE SENTENCE
The measure of an exterior angle of a triangle is equal to the sum of the measures of the two ____________ interior angles.
Answer:
The given statement is:
The measure of an exterior angle of a triangle is equal to the sum of the measures of the two ____________ interior angles.
Hence,
The completed form of the given statement is:
The measure of an exterior angle of a triangle is equal to the sum of the measures of the two non-adjacent interior angles.

Monitoring Progress and Modeling with Mathematics

In Exercises 3-6, classify the triangle by its sides and by measuring its angles.

Question 3.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 4
Answer:
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.1 a 3

Question 4.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 5
Answer:
The given figure is:
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 5
We know that,
“|” represents the “Congruent” or “Equal” in geometry
So,
From the given figure,
We can observe that all three sides of the given triangle are equal
We know that,
If a triangle has all the sides equal, then the triangle is called an “Equilateral triangle”
Hence, from the above,
We can conclude that the ΔLMN is an “Equilateral triangle”

Question 5.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 6
Answer:
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.1 a 5

Question 6.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 7
Answer:
The given figure is:
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 7
We know that,
If any side is not equal to each other in the triangle, then the triangle is called a “Scalene triangle”
The angle greater than 90° is called as “Obtuse angle”
An angle less than 90° is called an “Acute angle”
Hence, from the above,
We can conclude that ΔABC is an “Acute scalene triangle”

In Exercises 7-10, classify ∆ABC by its sides. Then determine whether it is a right triangle.

Question 7.
A(2, 3), B(6, 3), (2, 7)
Answer:
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.1 a 7

Question 8.
A(3, 3), B(6, 9), (6, – 3)
Answer:
The given points are:
A (3, 3), B(6, 9), and C (6, -3)
We know that,
To find whether the given triangle is a right angle or not,
We have to prove,
AC² = AB² + BC²
Where,
AC is the distance between points A and C
AB is the distance between points A and B
BC is the distance between points B and C
The slope of any one side must be equal to -1
Now,
Let the given points be
A (x1, y1), B(x2, y2), and C (x3, y3)
So,
A (x1, y1)= (3, 3), B (x2, y2) = (6, 9), and C (x3, y3) = (6, -3)
We know that,
The distance between 2 points = √(x2 – x1)² + (y2 – y1)²
So,
AB = √(6 – 3)² + (9 – 3)²
= √3² + 6²
= √9 + 36 = √45
BC = √6 – 6)² + (-3 – 9)²
= √0 + 12²
= √12² = 12
AC = √(6 – 3)² + (-3 – 3)²
= √(3)² + (-6)²
= √9 + 36 = √45
So,
From the length of the sides,
We can say that the given triangle is an Isosceles triangle,
We know that,
Slope (m) = \(\frac{y2 – y1} {x2 – x1}\)
So,
Slope of AB = \(\frac{9 – 3} {6 – 3}\)
= \(\frac{6} {3}\)
= 2
Slope of BC = \(\frac{-9 – 3} {6 – 6}\)
= \(\frac{-12} {0}\)
= Undefined
Slope of AC = \(\frac{-3 – 3} {6 – 3}\)
= \(\frac{-6} {3}\)
= -2
Hence, from the above,
We can conclude that the given triangle is not a right triangle

Question 9.
A(1, 9), B(4, 8), C(2, 5)
Answer:
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.1 a 9

Question 10.
A(- 2, 3), B(0, – 3), C(3, – 2)
Answer:
The given points are:
A (-2, 3), B(0, -3), and C (3, -2)
We know that,
To find whether the given triangle is a right angle or not,
We have to prove,
AC² = AB² + BC²
Where,
AC is the distance between points A and C
AB is the distance between points A and B
BC is the distance between points B and C
The slope of any one side must be equal to -1
Now,
Let the given points be
A (x1, y1), B(x2, y2), and C (x3, y3)
So,
A (x1, y1)= (-2, 3), B (x2, y2) = (0, -3), and C (x3, y3) = (3, -2)
We know that,
The distance between 2 points = √(x2 – x1)² + (y2 – y1)²
So,
AB = √(0 – [-2])² + (3 – 3)²
= √2² + 0²
= √4 + 0 = 2
BC = √3 – 0)² + (-2 -[-3] )²
= √9 + 1²
= √10
AC = √(3 – [-2])² + (-2 – 3)²
= √(5)² + (-5)²
= √25 + 25 = √50
Now,
AC² = AB² + BC²
50 = 10 + 4
50 ≠ 14
So,
From the length of the sides,
We can say that the given triangle is a scalene triangle since all the lengths of the sides are different
We know that,
Slope (m) = \(\frac{y2 – y1} {x2 – x1}\)
So,
Slope of AB = \(\frac{9 – 3} {6 – 3}\)
= \(\frac{6} {3}\)
= 2
Slope of BC = \(\frac{-9 – 3} {6 – 6}\)
= \(\frac{-12} {0}\)
= Undefined
Slope of AC = \(\frac{-3 – 3} {6 – 3}\)
= \(\frac{-6} {3}\)
= -2
Hence, from the above,
We can conclude that the given triangle is not a right triangle

In Exercises 11 – 14. find m∠1. Then classify the triangle by its angles

Question 11.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 8
Answer:
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.1 a 11

Question 12.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 9
Answer:
The given figure is:
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 9
We know that,
The sum of interior angles in a triangle is: 180°
So,
From the above,
The interior angles of the given triangle are: 40°,  30°,  ∠1
Now,
40° + 30° + ∠1 = 180°
70 + ∠1 = 180°
∠1 = 180° – 70°
∠1 = 110°
We know that,
The angle greater than 90° is called an “Obtuse angle”
Hence, from the above,
We can conclude that the given triangle is an “Obtuse angled triangle”

Question 13.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 10
Answer:
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.1 a 13

Question 14.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 11
Answer:
The given figure is:
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 11
We know that,
The sum of interior angles in a triangle is: 180°
So,
From the above,
The interior angles of the given triangle are: 60°,  60°,  ∠1
Now,
60° + 60° + ∠1 = 180°
120 + ∠1 = 180°
∠1 = 180° – 120°
∠1 = 60°
We know that,
An angle less than 90° is called an “Acute angle”
The triangle that all the angles 60° is called an “Equilateral triangle”
Hence, from the above,
We can conclude that the given triangle is an “Equilateral triangle”

In Exercises 15-18, find the measure of the exterior angle.

Question 15.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 12
Answer:
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.1 a 15

Question 16.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 13
Answer:
The given figure is:
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 13
We know that,
An exterior angle is equal to the sum of the two non-adjacent interior angles in a triangle
So,
(2x – 2)° = x° + 45°
2x° – x° = 45° + 2°
x = 47°
Hence,
The measure of the exterior angle is: (2x – 2)°
= (2 (47) – 2)°
= (94 – 2)°
= 92°
Hence, from the above,
We can conclude that the measure of the exterior angle is: 92°

Question 17.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 14
Answer:
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.1 a 17

Question 18.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 15
Answer:
The given figure is:
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 15
We know that,
An exterior angle is equal to the sum of the two non-adjacent interior angles in a triangle
So,
(7x – 16)° = (x + 8)° + 4x°
7x° – 5x° = 16° + 8°
2x = 24°
x = 24° ÷ 2
x = 12°
Hence,
The measure of the exterior angle is: (7x – 16)°
= (7 (12) – 16)°
= (84 – 16)°
= 68°
Hence, from the above,
We can conclude that the measure of the exterior angle is: 68°

In Exercises 19-22, find the measure of each acute angle.

Question 19.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 16
Answer:
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.1 a 19

Question 20.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 17
Answer:
The given figure is:
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 17
From the given figure,
We can observe that one angle is 90° and the 2 sides are perpendicular
So,
We can say that the given triangle is a right-angled triangle
We know that,
The sum of interior angles of a triangle is: 180°
So,
x° + (3x + 2)° + 90° = 180°
4x° + 2° + 90° = 180°
4x° = 180° – 90° – 2°
4x° = 88°
x = 88° ÷ 4°
x = 22°
So,
The 2 acute angle measures are: x° and (3x + 2)°
= 22° and (3(22) + 2)°
= 22° and (66 + 2)°
= 22° and 68°
Hence, from the above,
We can conclude that the 2 acute angle measures are: 22° and 68°

Question 21.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 18
Answer:
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.1 a 21

Question 22.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 19
Answer:
The given figure is:
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 19
From the given figure,
We can observe that one angle is 90° and the 2 sides are perpendicular
So,
We can say that the given triangle is a right-angled triangle
We know that,
The sum of interior angles of a triangle is: 180°
So,
(19x – 1)° + (13x – 5)° + 90° = 180°
32x° – 6° + 90° = 180°
32x° = 180° – 90° – 6°
4x° = 84°
x = 84° ÷ 4°
x = 21°
So,
The 2 acute angle measures are: (19x – 1)° and (13x – 5)°
= (19 (21) – 1)° and (13(21) – 5)°
= 398° and (273 – 5)°
= 398° and 268°
Hence, from the above,
We can conclude that the 2 acute angle measures are: 398° and 268°

In Exercises 23-26. find the measure of each acute angle in the right triangle.

Question 23.
The measure of one acute angle is 5 times the measure of the other acute angle.
Answer:
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.1 a 23

Question 24.
The measure of one acute angle is times the measure of the other acute angle.
Answer:

Question 25.
The measure of one acute angle is 3 times the sum of the measure of the other acute angle and 8.
Answer:
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.1 a 25

Question 26.
The measure of one acute angle is twice the difference of the measure of the other acute angle and 12.
Answer:
The given statement is:
The measure of one acute angle is twice the difference of the measure of the other acute angle and 12.
So,
x° + [2 (x – 12)]° = 90°
x° + 2x° – 2(12)° = 90°
3x° – 24° = 90°
3x° = 90° + 24°
3x° = 114°
x = 114° ÷ 3
x = 38°
So,
The 2 acute angle measures are: x°, 2 (x – 12)°
= 38°, 2 (38 – 12)°
= 38°, 2(26)°
= 38° , 52°
Hence, from the above,
We can conclude that the acute angle measures are: 38°, 52°

ERROR ANALYSIS
In Exercises 27 and 28, describe and correct the error in finding m∠1.

Question 27.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 20
Answer:
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.1 a 27

Question 28.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 21
Answer:
We know that,
The exterior angle of a triangle is equal to the sum of the non-adjacent interior angles of a triangle
So,
From the figure,
The external angle is: ∠1
The interior angles are 80°, 50°
So,
∠1 = 80° + 50°
∠1 = 130°
Now,
The interior angle measure of ∠1= 180° – (External angle measure of 130°)
= 180° – 130°
= 50°
Hence, from the above,
The internal angle measure of ∠1 is: 50°

In Exercises 29-36, find the measure of the numbered angle.

Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 22

Question 29.
∠1
Answer:
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.1 a 29

Question 30.
∠2
Answer:
We know that,
The external angle measure is equal to the sum of the non-adjacent interior angles
So,
∠2 = 90° + 40°
∠2 = 130°

Question 31.
∠3
Answer:
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.1 a 31

Question 32.
∠3
Answer:
From the above figure,
∠2 = ∠4
Hence, from the above,
We can conclude that
∠2 = ∠4 = 130°

Question 33.
∠5
Answer:
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.1 a 33

Question 34.
∠6
Answer:
The external angle measure is equal to the sum of the non-adjacent interior angles
So,
∠6 = 90° + ∠3
∠6 = 90° + 50°
∠6 = 140°

Question 35.
∠7
Answer:
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.1 a 35

Question 36.
∠8
Answer:
The external angle measure is equal to the sum of the non-adjacent interior angles
So,
∠8 = 90° + ∠1
∠6 = 90° + 50°
∠6 = 140°

Question 37.
USING TOOLS
Three people are standing on a stage. The distances between the three people are shown in the diagram. Classify the triangle by its sides and by measuring its angles.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 23
Answer:
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.1 a 37

Question 38.
USING STRUCTURE
Which of the following sets of angle measures could form a triangle? Select all that apply.
(A) 100°, 50°, 40°
Answer:
The given angles are: 100°, 50°, 40°
We know that,
The sum of the angles of a triangle should be equal to 180°
So,
The sum of the given angles = 100° + 50° + 40°
= 100° + 90°
= 190°
Hence, from the above,
We can conclude that the given angles do not form a triangle

(B) 96°, 74°, 10°
Answer:
The given angles are: 96°, 74°, 10°
We know that,
The sum of the angles of a triangle should be equal to 180°
So,
The sum of the given angles = 96° + 74° + 10°
= 96° + 84°
= 180°
Hence, from the above,
We can conclude that the given angles forms a triangle

(C) 165°, 113°, 82°
Answer:
The given angles are: 165°, 113°, 82°
We know that,
The sum of the angles of a triangle should be equal to 180°
So,
The sum of the given angles = 165° + 113° + 82°
= 165° + 195°
= 360°
But,
We know that,
The sum of exterior angles of a triangle is: 360°
Hence, from the above,
We can conclude that the given angles forms a triangle

(D) 101°, 41°, 38°
Answer:
The given angles are: 101°, 41°, 38°
We know that,
The sum of the angles of a triangle should be equal to 180°
So,
The sum of the given angles = 101° + 38° + 41°
= 101° + 79°
= 180°
Hence, from the above,
We can conclude that the given angles forms a triangle

(E) 90°, 45°, 45°
Answer:
The given angles are: 90°, 45°, 45°
We know that,
The sum of the angles of a triangle should be equal to 180°
So,
The sum of the given angles = 90° + 45° + 45°
= 90° + 90°
= 180°
Hence, from the above,
We can conclude that the given angles forms a triangle

(F) 84°, 62°, 34°
Answer:
The given angles are: 84°, 62°, 34°
We know that,
The sum of the angles of a triangle should be equal to 180°
So,
The sum of the given angles = 84° + 62° + 34°
= 84° + 96°
= 180°
Hence, from the above,
We can conclude that the given angles forms a triangle

Question 39.
MODELING WITH MATHEMATICS
You are bending a strip of metal into an isosceles triangle for a sculpture. The strip of metal is 20 inches long. The first bend is made 6 inches from one end. Describe two ways you could complete the triangle.
Answer:
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.1 a 39

Question 40.
THOUGHT-PROVOKING
Find and draw an object (or part of an object) that can be modeled by a triangle and an exterior angle. Describe the relationship between the interior angles of the triangle and the exterior angle in terms of the object.
Answer:

From the above figure,
We can say that
The sum of the interior angles of a given triangle is: 180°
The sum of the exterior angles of a given triangle is: 360°
The relation between the interior angles and the exterior angles is:
The exterior angle measure = Sum of the two non-adjacent interior angles

Question 41.
PROVING A COROLLARY
Prove the Corollary to the Triangle Sum Theorem (Corollary 5. 1).
Given ∆ABC is a right triangle
Prove ∠A and ∠B are complementary
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 24
Answer:
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.1 a 41

Question 42.
PROVING A THEOREM
Prove the Exterior Angle Theorem (Theorem 5.2).
Given ∆ABC, exterior ∠ACD
Prove m∠A + m∠B = m∠ACD

Answer:
It is given that
In ΔABC, the exterior angle is ∠ACD
We have to prove that
m∠A + m∠B = m∠ACD
Proof:

Hence, from the above,
We can conclude that
m∠A + m∠B = m∠ACD is proven

Question 43.
CRITICAL THINKING
Is it possible to draw an obtuse isosceles triangle? obtuse equilateral triangle? If so, provide examples. If not, explain why it is not possible.
Answer:
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.1 a 43

Question 44.
CRITICAL THINKING
Is it possible to draw a right isosceles triangle? right equilateral triangle? If so, provide an example. If not, explain why it is not possible.
Answer:
It is possible to draw a right isosceles triangle but it is not possible to draw a right equilateral triangle
We know that,
In a triangle, if the length of the 2 sides are equal and one angle is a right-angle, then, it is called an “Right Isosceles triangle”
In a triangle, if the length of all the sides are equal and each angle is 60°, then it is an “Equilateral triangle”
Hence,
From the above definitions,
We can observe that it is possible to draw right isosceles triangle but it is not possible to dran a right equilateral triangle

Question 45.
MATHEMATICAL CONNECTIONS
∆ABC is isosceles.
AB = x, and BC = 2x – 4.
a. Find two possible values for x when the perimeter of ∆ABC is 32.
b. How many possible values are there for x when the perimeter of ∆ABC is 12?
Answer:
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.1 a 45

Question 46.
HOW DO YOU SEE IT?
Classify the triangles, in as many ways as possible. without finding any measurements.
a. Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 26
Answer:
The given figure is:
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 26
From the figure,
We can observe that all the length of the sides of the triangle are equal
We know that,
The triangle that has the length of all the sides equal is called an “Equilateral triangle”
Hence, from the above,
We can conclude that the given triangle is an “Equilateral triangle”

b. Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 27
Answer:
The given figure is:
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 27
From the figure,
We can observe that the lengths of all the 3 sides are different
We know that,
The triangle that has all the different side lengths is called a “Scalene triangle”
Hence, from the above,
We can conclude that the given triangle is called a “Scalene triangle”

c. Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 28
Answer:
The given figure is:
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 28
From the figure,
We can observe that the length of all the 3 sides are different and 1 angle is obtuse i.e., greater than 90°
We know that,
The triangle that has any angle obtuse is called an “Obtuse angled triangle”
Hence, from the above,
We can conclude that the given triangle is an “Obtuse angled scalene triangle”

d. Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 29
Answer:
The given figure is:
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 29
From the figure,
We can observe that 1 angle is 90° and the 2 sides are perpendicular to each other
We know that,
The triangle that has an angle of 90° and the slope -1 is called a “Right-angled triangle”
Hence, from the above,
We can conclude that the given triangle is called a “Right-angled triangle”

Question 47.
ANALYZING RELATIONSHIPS
Which of the following could represent the measures of an exterior angle and two interior angles of a triangle? Select all that apply.
A) 100°, 62°, 38°
(B) 81°, 57°, 24°
(C) 119°, 68°, 49°
(D) 95°, 85°, 28°
(E) 92°, 78°, 68°
(F) 149°, 101°, 48°
Answer:
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.1 a 47

Question 48.
MAKING AN ARGUMENT
Your friend claims the measure of an exterior angle will always be greater than the sum of the nonadjacent interior angle measures. Is your friend correct? Explain your reasoning.
Answer:
No, your friend is not correct

Explanation:
We know that,
According to the exterior angle theorem,
The external angle measure is always equal to the sum of the non-adjacent internal angle measures
But,
According to your friend,
The external angle measure will always be greater than the sum of the non-adjacent interior angle measures
Hence, from the above,
We can conclude that your friend is not correct

MATHEMATICAL CONNECTIONS
In Exercises 49-52, find the values of x and y.

Question 49.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 30
Answer:
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.1 a 49

Question 50.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 31
Answer:
The given figure is:
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 31
From the figure,
We have to obtain the values of x and y
Now,
By using the alternate angles theorem,
x = 118°
Now,
By using the exterior angle theorem,
x = y + 22°
y = x – 22°
y = 118° – 22°
y = 96°
Hence, from the above,
We can conclude that the values of x and y are: 118° and 96° respectively

Question 51.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 32
Answer:
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.1 a 51

Question 52.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 33
Answer:
The given figure is:
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 33
From the above figure,
We have to find the values of x and y
Now,
By using the sum of interior angle measures,
x° + 64° + 90° = 180°
x° + 154° = 180°
x° = 180° – 154°
x° = 26°
Now,
By using the exterior angle theorem,
y° = x° + 64°
y° = 26° + 64°
y° = 90°
Hence, from the above,
We can conclude that the values of x and y are: 26° and 90° respectively

Question 53.
PROVING A THEOREM
Use the diagram to write a proof of the Triangle Sum Theorem (Theorem 5. 1). Your proof should be different from the proof of the Triangle Sum Theorem shown in this lesson.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 34
Answer:
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.1 a 53.1
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.1 a 53.2

Maintaining Mathematical Proficiency

Use the diagram to find the measure of the segment or angle.

Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 35

Question 54.
m∠KHL
Answer:
From the given figures,
We can observe that
∠ABC = ∠GHK
∠KHL = ∠GHK / 2
So,
(6x + 2)° = (3x + 1)° + (5x – 27)°
6x – 3x – 5x = 1 – 27 – 2
6x – 8x = -27 – 1
-2x = -28
2x = 28
x = 28 ÷ 2
x = 14
So,
∠KHL = ∠GHK / 2
= [(3 (14) + 1)° + (5 (14) – 27)°] / 2
= [43° + 43°] / 2
= 86° / 2
= 43°
Hence, from the above,
We can conclude that
∠KHL = 43°

Question 55.
m∠ABC
Answer:
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.1 a 55

Question 56.
GH
Answer:
From the given figures,
We can observe that
AB = GH
So,
3y = 5y – 8
3y – 5y = -8
-2y = -8
2y = 8
y = 8 ÷ 2
y = 4
So,
The value of GH = 3y = 3 (4) = 12
Hence, from the above,
We can conclude that the value of GH is: 12

Question 57.
BC
Answer:
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.1 a 57

5.2 Congruent Polygons

Exploration 1

Describing Rigid Motions

Work with a partner: of the four transformations you studied in Chapter 4, which are rigid motions? Under a rigid motion. why is the image of a triangle always congruent to the original triangle? Explain your reasoning.
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 37
Answer:
Rigid motion occurs in geometry when an object moves but maintains its shape and size, which is unlike non-rigid motions, such as dilations, in which the object’s size changes. All rigid motion starts with the original object, called the pre-image, and results in the transformed object, called the image.
There are 4 types of rigid motion. They are:
a. Translation
b. Rotation
c. Reflection
d. Glide reflection
We know that,
Rotation only occurs in terms of 90° or 180°
Now,
The given transformations are:
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 37
So,
From the above figure,
The first figure and the second figure are different
The second figure and the third figure are the same in shape
The first figure and the fourth figure are the same in shape
So,
We can say that the first and the fourth figures are rigid motions
W can say that the second and the third figures are rigid motions
In the second and the third figures,
The “Rotation” takes place i.e., the second figure is rotated 180° keeping the original shape
In the first and the fourth figures,
The “Reflection” takes place i.e., the first figure is reflected keeping the original shape
Now,
The image of the triangle is always congruent to the original triangle because of the “Translation” i.e., the original triangle and the image of the triangle have the same sides and the same angles but not in the same position.

Exploration 2

Finding a Composition of Rigid Motions

Work with a partner. Describe a composition of rigid motions that maps ∆ABC to ∆DEF. Use dynamic geometry software to verify your answer.
LOOKING FOR STRUCTURE
To be proficient in math, you need to look closely to discern a pattern or structure.

a. ∆ABC ≅ ∆DEF
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 38
Answer:

b. ∆ABC ≅ ∆DEF
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 39
Answer:

c. ∆ABC ≅ ∆DEF
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 40
Answer:

d. ∆ABC ≅ ∆DEF
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 41
Answer:

Communicate Your Answer

Question 3.
Given two congruent triangles. how can you use rigid motions to map one triangle to the other triangle?
Answer:

Question 4.
The vertices of ∆ABC are A(1, 1), B(3, 2), and C(4, 4). The vertices of ∆DEF are D(2, – 1), E(0, 0), and F(- 1, 2). Describe a composition of rigid motions that maps ∆ABC to ∆DEF.
Answer:

Lesson 5.2 Congruent Polygons

Monitoring Progress

In the diagram, ABGH ≅ CDEF.

Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 42

Question 1.
Identify all pairs of congruent corresponding parts.
Answer:

Question 2.
Find the value of x.
Answer:

Question 3.
In the diagram at the left. show that ∆PTS ≅ ∆RTQ.
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 43
Answer:

Use the diagram.

Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 44

Question 4.
Find m∠DCN.
Answer:

Question 5.
What additional information is needed to conclude that ∆NDC ≅ ∆NSR?
Answer:

Exercise 5.2 Congruent Polygons

Question 1.
WRITING
Based on this lesson. what information do you need to prove that two triangles are congruent? Explain your reasoning.
Answer:
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 5.2 a 1

Question 2.
DIFFERENT WORDS, SAME QUESTION
Which is different? Find “both” answers.
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 45

Is ∆ABC ≅ ∆RST?
Answer:

Is ∆KJL ≅ ∆SRT?
Answer:

Is ∆JLK ≅ ∆STR?
Answer:

Is ∆LKJ ≅ ∆TSR?
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3 and 4. identify all pairs of congruent corresponding parts. Then write another congruence statement for the polygons.

Question 3.
∆ABC ≅ ∆DEF
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 46
Answer:
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 5.2 a 3

Question 4.
GHJK ≅ ∆QRST
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 47
Answer:

In Exercises 5-8, ∆XYZ ≅ ∆MNL. Copy and complete the statement.

Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 48

Question 5.
m∠Y = ______
Answer:
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 5.2 a 5

Question 6.
m∠M = ______
Answer:

Question 7.
m∠Z = _______
Answer:
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 5.2 a 7

Question 8.
XY= _______
Answer:

In Exercises 9 and 10. find the values of x and y.

Question 9.
ABCD ≅ EFGH
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 49
Answer:
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 5.2 a 9

Question 10.
∆MNP ≅ ∆TUS
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 50
Answer:

In Exercises 11 and 12. show that the polygons are congruent. Explain your reasoning.

Question 11.
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 51
Answer:
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 5.2 a 11

Question 12.
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 52
Answer:

In Exercises 13 and 14, find m∠1.

Question 13.
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 53
Answer:
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 5.2 a 13

Question 14.
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 54
Answer:

Question 15.
PROOF
Triangular postage stamps, like the ones shown, are highly valued by stamp collectors. Prove that ∆AEB ≅ ∆CED.
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 55
Given \(\overline{A B}\) || \(\overline{D C}\), \(\overline{A B}\) ≅ \(\overline{D C}\) is the midpoint of \(\overline{A C}\) and \(\overline{B D}\)
Prove ∆AEB ≅ ∆CED
Answer:
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 5.2 a 15

Question 16.
PROOF
Use the information in the figure to prove that ∆ABG ≅ ∆DCF
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 56
Answer:

ERROR ANALYSIS
In Exercises 17 and 18, describe and correct the error.

Question 17.
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 57
Answer:
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 5.2 a 17

Question 18.
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 58
Answer:

Question 19.
PROVING A THEOREM
Prove the Third Angles Theorem (Theorem 5.4) by using the Triangle Sum Theorem (Theorem 5. 1).
Answer:
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 5.2 a 19.1
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 5.2 a 19.2

Question 20.
THOUGHT PROVOKING
Draw a triangle. Copy the triangle multiple times to create a rug design made of congruent triangles. Which property guarantees that all the triangles are congruent?
Answer:

Question 21.
REASONING
∆JKL is congruent to ∆XYZ Identify all pairs of congruent corresponding parts.
Answer:
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 5.2 a 21

Question 22.
HOW DO YOU SEE IT?
In the diagram, ABEF ≅ CDEF
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 59
a. Explain how you know that \(\overline{B E}\) ≅ \(\overline{D E}\) and ∠ABE ≅∠CDE.
Answer:

b. Explain how you know that ∠GBE ≅ ∠GDE.
Answer:

c. Explain how you know that ∠GEB ≅ ∠GED.
Answer:

d. Do you have enough information to prove that ∠BEG ≅ ∠DEG? Explain.
Answer:

MATHEMATICAL CONNECTIONS
In Exercises 23 and 24, use the given information to write and solve a system of linear equations to find the values of x and y.

Question 23.
∆LMN ≅ ∆PQR. m∠L = 40°, m∠M = 90° m∠P = (17x – y)°. m∠R (2x + 4y)°
Answer:
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 5.2 a 23

Question 24.
∆STL ≅ ∆XYZ, m∠T = 28°, m∠U = (4x + y)°, m∠X = 130°, m∠Y = (8x – 6y)°
Answer:

Question 25.
PROOF
Prove that the criteria for congruent triangles in this lesson is equivalent to the definition of congruence in terms of rigid motions.
Answer:
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 5.2 a 25

Maintaining Mathematical Proficiency

What can you conclude from the diagram?

Question 26.
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 60
Answer:

Question 27.
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 61
Answer:
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 5.2 a 27

Question 28.
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 62
Answer:

Question 29.
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 63
Answer:
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 5.2 a 29

5.3 Proving Triangle Congruence by SAS

Exploration 1

Drawing Triangles

Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 64

Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 65

Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 66

Work with a partner.

Use dynamic geometry software.
a. Construct circles with radii of 2 units and 3 units centered at the origin. Construct a 40° angle with its vertex at the origin. Label the vertex A.
Answer:

b. Locate the point where one ray of the angle intersects the smaller circle and label this point B. Locate the point where the other ray of the angle intersects the larger circle and label this point C. Then draw ∆ABC.
Answer:

c. Find BC, m∠B, and m∠C.
Answer:

d. Repeat parts (a)-(c) several times. redrawing the angle indifferent positions. Keep track of your results by copying and completing the table below. What can you conclude?
USING TOOLS STRATEGICALLY
To be proficient in math, you need to use technology to help visualize the results of varying assumptions, explore consequences, and compare predictions with data.
Answer:

Communicate Your Answer

Question 2.
What can you conclude about two triangles when you know that two pairs of corresponding sides and the corresponding included angles are congruent?
Answer:

Question 3.
How would you prove your conclusion in Exploration 1(d)?
Answer:

Lesson 5.3 Proving Triangle Congruence by SAS

Monitoring Progress

In the diagram, ABCD is a square with four congruent sides and four right
angles. R, S, T, and U are the midpoints of the sides of ABCD. Also, \(\overline{R T}\) ⊥ \(\overline{S U}\) and \(\overline{S V}\) ≅ \(\overline{V U}\).
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 67
Question 1.
Prove that ∆SVR ≅ ∆UVR.
Answer:

Question 2.
Prove that ∆BSR ≅ ∆DUT.
Answer:

Question 3.
You are designing the window shown in the photo. You want to make ∆DRA congruent to ∆DRG. You design the window so that \(\overline{D A}\) ≅ \(\overline{D G}\) and ∠ADR ≅ ∠GDR. Use the SAS Congruence Theorem to prove ∆DRA ≅ ∆DRG.
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 68
Answer:

Exercise 5.3 Proving Triangle Congruence by SAS

vocabulary and core concept check

Question 1.
WRITING
What is an included angle?
Answer:
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 5.3 a 1

Question 2.
COMPLETE THE SENTENCE
If two sides and the included angle of one triangle are congruent to two sides and the included angle of a second triangle, then __________ .
Answer:

Monitoring progress and Modeling with Mathematics

In Exercises 3-8, name the included an1e between the pair of sides given.

Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 69

Question 3.
\(\overline{J K}\) and \(\overline{K L}\)
Answer:
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 5.3 a 3

Question 4.
\(\overline{P K}\) and \(\overline{L K}\)
Answer:

Question 5.
\(\overline{L P}\) and \(\overline{L K}\)
Answer:
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 5.3 a 5

Question 6.
\(\overline{J L}\) and \(\overline{J K}\)
Answer:

Question 7.
\(\overline{K L}\) and \(\overline{J L}\)
Answer:
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 5.3 a 7

Question 8.
\(\overline{K P}\) and \(\overline{P L}\)
Answer:

In Exercises 9-14, decide whether enough information is given to prove that the triangles are congruent using the SAS Congruence Theorem (Theorem 5.5). Explain.

Question 9.
∆ABD, ∆CDB
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 70
Answer:
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 5.3 a 9

Question 10.
∆LMN, ∆NQP
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 71
Answer:

Question 11.
∆YXZ, ∆WXZ
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 72
Answer:
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 5.3 a 11

Question 12.
∆QRV, ∆TSU
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 73
Answer:

Question 13.
∆EFH, ∆GHF
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 74
Answer:
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 5.3 a 13

Question 14.
∆KLM, ∆MNK
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 75
Answer:

In Exercises 15 – 18, write a proof.

Question 15.
Given \(\overline{P Q}\) bisects ∠SPT, \(\overline{S P}\) ≅ \(\overline{T P}\)
Prove ∆SPQ ≅ ∆TPQ
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 76
Answer:
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 5.3 a 15

Question 16.
Given \(\overline{A B}\) ≅ \(\overline{C D}\), \(\overline{A B}\) || \(\overline{C D}\)
Prove ∆ABC ≅ ∆CDA
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 77
Answer:

Question 17.
Given C is the midpoint of \(\overline{A E}\) and \(\overline{B D}\)
Prove ∆ABC ≅ ∆EDC
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 78
Answer:
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 5.3 a 17

Question 18.
Given \(\overline{P T}\) ≅ \(\overline{R T}\), \(\overline{Q T}\) ≅ \(\overline{S T}\)
Prove ∆PQT ≅ ∆RST
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 79
Answer:

In Exercises 19-22, use the given information to name two triangles that are congruent. Explain your reasoning.

Question 19.
∠SRT ≅ ∠URT, and R is the center of the circle.
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 80
Answer:

Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 5.3 a 19

Question 20.
ABCD is a square with four congruent sides and four congruent angles.
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 81
Answer:

Question 21.
RSTUV is a regular pentagon.
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 82
Answer:
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 5.3 a 21

Question 22.
\(\overline{M K}\) ⊥ \(\overline{M N}\), \(\overline{K L}\) ⊥ \(\overline{N L}\), and M and L are centers of circles.
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 83
Answer:

CONSTRUCTION
In Exercises 23 and 24, construct a triangle that is congruent to ∆ABC using the SAS Congruence Theorem (Theorem 5.5).

Question 23.
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 84
Answer:
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 5.3 a 23

Question 24.
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 85
Answer:

Question 25.
ERROR ANALYSIS
Describe and correct the error in finding the value of x.
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 86
Answer:
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 5.3 a 25

Question 26.
HOW DO YOU SEE IT?
What additional information do you need to prove that ∆ABC ≅ ∆DBC?
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 87
Answer:

Question 27.
PROOF
The Navajo rug is made of isosceles triangles. You know ∠B ≅∠D. Use the SAS Congruence Theorem (Theorem 5.5 to show that ∆ABC ≅ ∆CDE. (See Example 3.)
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 88
Answer:
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 5.3 a 27

Question 28.
THOUGHT PROVOKING
There are six possible subsets of three sides or angles of a triangle: SSS, SAS, SSA, AAA, ASA, and AAS. Which of these correspond to congruence theorems? For those that do not, give a counterexample.
Answer:

Question 29.
MATHEMATICAL CONNECTIONS
Prove that
∆ABC ≅ ∆DEC
Then find the values of x and y.
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 89
Answer:
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 5.3 a 29

Question 30.
MAKING AN ARGUMENT
Your friend claims it is possible to Construct a triangle congruent to ∆ABC by first constructing \(\overline{A B}\) and \(\overline{A C}\), and then copying ∠C. Is your friend correct? Explain your reasoning.
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 90
Answer:

Question 31.
PROVING A THEOREM
Prove the Reflections in Intersecting Lines Theorem (Theorem 4.3).
Answer:
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 5.3 a 31.1
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 5.3 a 31.2
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 5.3 a 31.3
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 5.3 a 31.4

Maintaining Mathematical Proficiency

Classify the triangle by its sides and by measuring its angles.

Question 32.
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 91
Answer:

Question 33.
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 92
Answer:
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 5.3 a 33

Question 34.
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 93
Answer:

Question 35.
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 94
Answer:
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 5.3 a 35

5.4 Equilateral and Isosceles Triangles

Exploration 1

Writing a Conjecture about Isosceles Triangles

Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 95

Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 96

Work with a partner: Use dynamic geometry software.

a. Construct a circle with a radius of 3 units centered at the origin.
Answer:

b. Construct ∆ABC so that B and C are on the circle and A is at the origin.
Answer:

c. Recall that a triangle is isosceles if it has at least two congruent sides. Explain why ∆ABC is an isosceles triangle.
Answer:

d. What do you observe about the angles of ∆ABC?
Answer:

e. Repeat parts (a)-(d) with several other isosceles triangles using circles of different radii. Keep track of your observations by copying and completing the table below. Then write a conjecture about the angle measures of an isosceles triangle.
CONSTRUCTING VIABLE ARGUMENTS
To be proficient in math, you need to make conjectures and build a logical progression of statements to explore the truth of your conjectures.
Answer:

f. Write the converse of the conjecture you wrote in part (e). Is the converse true?
Answer:

Communicate Your Answer

Question 2.
What conjectures can you make about the side lengths and angle measures of an
isosceles triangle?
Answer:

Question 3.
How would you prove your conclusion in Exploration 1 (e)? in Exploration 1(f)?
Answer:

Lesson 5.4 Equilateral and Isosceles Triangles

Monitoring Progress

Copy and complete the statement.

Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 97

Question 1.
If \(\overline{H G}\) ≅ \(\overline{H K}\), then ∠ _______ ≅ ∠ _______ .
Answer:

Question 2.
If ∠KHJ ≅∠KJH, then ______ ≅ ______ .
Answer:

Question 3.
Find the length of \(\overline{S T}\) of the triangle at the left.
Answer:

Question 4.
Find the value of x and y in the diagram.
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 98
Answer:

Question 5.
In Example 4, show that ∆PTS ≅ ∆QTR
Answer:

Exercise 5.4 Equilateral and Isosceles Triangles

Vocabulary and Core Concept Check

Question 1.
VOCABULARY
Describe how to identify the vertex angle of an isosceles triangle.
Answer:
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 5.4 a 1

Question 2.
WRITING
What is the relationship between the base angles of an isosceles triangle? Explain.
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3-6. copy and complete the statement. State which theorem you used.

Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 99

Question 3.
If \(\overline{A E}\) ≅ \(\overline{D E}\) then ∠_____ ≅ ∠_____ .
Answer:
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 5.4 a 3

Question 4.
If \(\overline{A B}\) ≅ \(\overline{E B}\) then ∠_____ ≅ ∠_____ .
Answer:

Question 5.
If ∠D ≅ ∠CED, then _______ ≅ _______ .
Answer:
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 5.4 a 5

Question 6.
If ∠EBC ≅ ∠ECB, then _______ ≅ _______ .
Answer:

In Exercises 7-10. find the value of x.

Question 7.
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 100
Answer:
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 5.4 a 7

Question 8.
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 101
Answer:

Question 9.
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 102
Answer:
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 5.4 a 9

Question 10.
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 103
Answer:

Question 11.
MODELING WITH MATHEMATICS
The dimensions of a sports pennant are given in the diagram. Find the values of x and y.
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 104
Answer:
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 5.4 a 11

Question 12.
MODELING WITH MATHEMATICS
A logo in an advertisement is an equilateral triangle with a side length of 7 centimeters. Sketch the logo and give the measure of each side.
Answer:

In Exercises 13-16, find the values of x and y.

Question 13.
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 105
Answer:
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 5.4 a 13

Question 14.
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 106
Answer:

Question 15.
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 107
Answer:
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 5.4 a 15

Question 16.
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 108
Answer:

CONSTRUCTION
In Exercises 17 and 18, construct an equilateral triangle whose sides are the given length.

Question 17.
3 inches
Answer:
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 5.4 a 17

Question 18.
1.25 inches
Answer:

Question 19.
ERROR ANALYSIS
Describe and correct the error in finding the length of \(\overline{B C}\).
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 109
Answer:
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 5.4 a 19

Question 20.
PROBLEM SOLVING
The diagram represents part of the exterior of the Bow Tower in Calgary. Alberta, Canada, In the diagram. ∆ABD and ∆CBD arc congruent equilateral triangles.
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 110

a. Explain why ∆ABC is isosceles.
Answer:

b. Explain ∠BAE ≅ ∠BCE.
Answer:

c. Show that ∆ABE and ∆CBE arc congruent.
Answer:

d. Find the measure of ∠BAE.
Answer:

Question 21.
FINDING A PATTERN
In the pattern shown. each small triangle is an equilateral triangle with an area of 1 square unit.
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 111
a. Explain how you know that an triangle made out of equilateral triangles is equilateral.
b. Find the areas of the first four triangles in the pattern.
c. Describe any patterns in the areas. Predict the area of the seventh triangle in the pattern. Explain your reasoning.
Answer:
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 5.4 a 21

Question 22.
REASONING
The base of isosceles ∆XYZ is \(\overline{Y Z}\). What
can you prove? Select all that apply.
(A) \(\overline{X Y}\) ≅ \(\overline{X Z}\)
(B) ∠X ≅ ∠Y
(C) ∠Y ≅ ∠Z
(D) \(\overline{Y Z}\) ≅ \(\overline{Z X}\)
Answer:

In Exercises 23 and 24, find the perimeter of the triangle.

Question 23.
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 112
Answer:
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 5.4 a 23

Question 24.
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 113
Answer:

MODELING WITH MATHEMATICS
In Exercises 25 – 28. use the diagram based on the color wheel. The 12 triangles in the diagram are isosceles triangles with congruent vertex angles.
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 114

Question 25.
Complementary colors lie directly opposite each other on the color wheel. Explain how you know that the yellow triangle is congruent to the purple triangle.
Answer:
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 5.4 a 25

Question 26.
The measure of the vertex angle of the yellow triangle is 30°. Find the measures of the base angles.
Answer:

Question 27.
Trace the color wheel. Then form a triangle whose vertices are the midpoints of the bases of the red. yellow. and blue triangles. (These colors are the primary colors.) What type of triangle is this?
Answer:
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 5.4 a 27

Question 28.
Other triangles can be brined on the color wheel that are congruent to the triangle in Exercise 27. The colors on the vertices of these triangles are called triads. What are the possible triads?
Answer:

Question 29.
CRITICAL THINKING
Are isosceles triangles always acute triangles? Explain your reasoning.
Answer:
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 5.4 a 29

Question 30.
CRITICAL THINKING
Is it possible for an equilateral triangle to have an angle measure other than 60°? Explain your reasoning.
Answer:

Question 31.
MATHEMATICAL CONNECTIONS
The lengths of the sides of a triangle are 3t, 5t – 12, and t + 20. Find the values of t that make the triangle isosceles. Explain your reasoning.
Answer:
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 5.4 a 31

Question 32.
MATHEMATICAL CONNECTIONS
The measure of an exterior angle of an isosceles triangle is x°. Write expressions representing the possible angle measures of the triangle in terms of x.
Answer:

Question 33.
WRITING
Explain why the measure of the vertex angle of an isosceles triangle must be an even number of degrees when the measures of all the angles of the triangle are whole numbers.
Answer:
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 5.4 a 33

Question 34.
PROBLEM SOLVING
The triangular faces of the peaks on a roof arc congruent isosceles triangles with vertex angles U and V.
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 115
a. Name two angles congruent to ∠WUX. Explain your reasoning.
b. Find the distance between points U and V.
Answer:

Question 35.
PROBLEM SOLVING
A boat is traveling parallel to the shore along \(\vec{R}\)T. When the boat is at point R, the captain measures the angle to the lighthouse as 35°. After the boat has traveled 2.1 miles, the captain measures the angle to the lighthouse to be 70°.
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 116
a. Find SL. Explain your reasoning.
b. Explain how to find the distance between the boat and the shoreline.
Answer:
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 5.4 a 35

Question 36.
THOUGHT PROVOKING
The postulates and theorems in this book represent Euclidean geometry. In spherical geometry, all points are points on the surface of a sphere. A line is a circle on the sphere whose diameter is equal to the diameter of the sphere. In spherical geometry, do all equiangular triangles have the same angle measures? Justify your answer.
Answer:

Question 37.
PROVING A COROLLARY
Prove that the Corollary to the Base Angles Theorem (Corollary 5.2) follows from the Base Angles Theorem (Theorem 5.6).
Answer:
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 5.4 a 37

Question 38.
HOW DO YOU SEE IT?
You are designing fabric purses to sell at the school fair.
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 117
a. Explain why ∆ABE ≅ ∆DCE.
b. Name the isosceles triangles in the purse.
c. Name three angles that are congruent to ∠EAD.
Answer:

Question 39.
PROVING A COROLLARY
Prove that the Corollary to the Converse of the Base Angles Theorem (Corollary 5.3) follows from the Converse of the Base Angles Theorem (Theorem 5.7)
Answer:
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 5.4 a 39

Question 40.
MAKING AN ARGUMENT
The coordinates of two points are T(0, 6) and U(6, 0) Your friend claims that points T, U, and V will always be the vertices of an isosceles triangle when V is any point on the line y = x. Is your friend correct? Explain your reasoning.
Answer:

Question 41.
PROOF
Use the diagram to prove that ∆DEF is equilateral.
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 118
Given ∆ABC is equilateral
∠CAD ≅ ∠ABE ≅ ∠BCF
Prove ∆DEF is equilateral
Answer:
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 5.4 a 41.1
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 5.4 a 41.2

Maintaining Mathematical Proficiency

Use the given property to complete the statement.

Question 42.
Reflexive Property of Congruence (Theorem 2. 1): ________ ≅ \(\overline{S E}\)
Answer:

Question 43.
Symmetric Property of Congruence (Theorem 2.1): If ________ ≅ ________, then \(\overline{R S}\) ≅ \(\overline{J K}\)
Answer:
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 5.4 a 43

Question 44.
Transitive Property of Congruence (Theorem 2.1): If \(\overline{E F}\) ≅ \(\overline{P Q}\), and \(\overline{P Q}\) ≅ \(\overline{U V}\) ________ ≅ ________.
Answer:

5.1 to 5.4 Quiz

Find the measure of the exterior angle.

Question 1.
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 119
Answer:

Question 2.
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 120
Answer:

Question 3.
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 121
Answer:

Identify all pairs of congruent corresponding parts. Then write another congruence statement for the polygons.

Question 4.
∆ABC ≅ ∆DEF
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 122
Answer:

Question 5.
QRST ≅ WXYZ
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 123
Answer:

Decide whether enough information is given to prove that the triangles are congruent using the SAS Congruence Theorem (Thm 5.5). If so, write a proof. If not, explain why.

Question 6.
∆CAD, ∆CBD
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 124
Answer:

Question 7.
∆GHF, ∆KHJ
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 125
Answer:

Question 8.
∆LWP, ∆NMP
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 126
Answer:

Copy and complete the statement. State which theorem you used.

Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 127

Question 9.
If VW ≅ WX, then ∠______ ≅ ∠ ________.
Answer:

Question 10.
If XZ ≅ XY. then∠______ ≅ ∠ ________.
Answer:

Question 11.
If ∠ZVX ≅∠ZXV, then ∠______ ≅ ∠ ________.
Answer:

Question 12.
If ∠XYZ ≅∠ZXY, then ∠______ ≅ ∠ ________.
Answer:

Find the values of x and y.

Question 13.
∆DEF ≅ ∆QRS
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 128
Answer:

Question 14.
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 129
Answer:

Question 15.
In a right triangle, the measure of one acute angle is 4 times the difference of the measure of the other acute angle and 5. Find the measure ol each acute angle in the triangle. (Section 5.1)
Answer:

Question 16.
The figure shows a stained glass window. (Section 5.1 and Section 5.3)

Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 130

a. Classify triangles 1 – 4 by their angles.
Answer:

b. Classify triangles 4 – 6 by their sides.
Answer:

c. Is there enough information given to prove that ∆7 ≅ ∆8? If so, label the vertices
and write a proof. If not, determine what additional information is needed.
Answer:

5.5 Proving Triangle Congruence by SSS

Exploration 1

Drawing Triangles

Work with a partner.
Use dynamic geometry software.

Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 131

Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 132

Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 133

a. Construct circles with radii of 2 units and 3 units centered at the origin. Label the origin A. Then draw \(\overline{B C}\) of length 4 units.
Answer:

b. Move \(\overline{B C}\) so that B is on the smaller circle and C is on the larger circle. Then draw ∆ABC.
Answer:

c. Explain why the side lengths of ∆ABC are 2, 3, and 4 units.
Answer:

d. Find m∠A, m∠B, and m∠C.
Answer:

e. Repeat parts (b)and (d) several times, moving \(\overline{B C}\) to different locations. Keep track of ‘our results by copying and completing the table below. What can you conclude?
USING TOOLS STRATEGICALLY
To be proficient in math, you need to use technology to help visualize the results of varying assumptions, explore consequences, and compare predictions with data.
Answer:

Communicate Your Answer

Question 2.
What can you conclude about two triangles when you know the corresponding sides are congruent?
Answer:

Question 3.
How would you prove your conclusion in Exploration 1(e)?
Answer:

Lesson 5.5 Proving Triangle Congruence by SSS

Monitoring Progress

Decide whether the congruence statement is true. Explain your reasoning.

Question 1.
∆DFG ≅ ∆HJK
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 134
Answer:

Question 2.
∆ACB ≅ ∆CAD
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 135
Answer:

Question 3.
∆QPT ≅ ∆RST
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 136
Answer:

Determine whether the figure is stable. Explain your reasoning.

Question 4.
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 137
Answer:

Question 5.
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 138
Answer:

Question 6.
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 139
Answer:

Use the diagram.

Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 140

Question 7.
Redraw ∆ABC and ∆DCB side by side with corresponding parts in the same position.
Answer:

Question 8.
Use the information in the diagram to prove that ∆ABC ≅ ∆DCB.
Answer:

Exercise 5.5 Proving Triangle Congruence by SSS

Vocabulary and Core Concept Check

Question 1.
COMPLETE THE SENTENCE
The side opposite the right angle is called the __________of the right triangle.
Answer:
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 5.5 a 1

Question 2.
WHICH ONE DOESNT BELONG?
Which triangles legs do not belong with the other three? Explain your reasoning.
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 142
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3 and 4, decide whether enough information is given to prove that the triangles are congruent using the SSS Congruence Theorem (Theorem 5.8). Explain.

Question 3.
∆ABC, ∆DBE
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 141
Answer:
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 5.5 a 3

Question 4.
∆PQS, ∆RQS
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 143
Answer:

In Exercises 5 and 6, decide whether enough information is given to prove that the triangles are congruent using the HL Congruence Theorem (Theorem 5.9). Explain.

Question 5.
∆ABC, ∆FED
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 144
Answer:
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 5.5 a 5

Question 6.
∆PQT, ∆SRT
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 145
Answer:

In Exercises 7-10. decide whether the congruence statement is true. Explain your reasoning.

Question 7.
∆RST ≅ ∆TQP
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 146
Answer:
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 5.5 a 7

Question 8.
∆ABD ≅ ∆CDB
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 147
Answer:

Question 9.
∆DEF ≅ ∆DGF
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 148
Answer:
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 5.5 a 9

Question 10.
∆JKL ≅ ∆LJM
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 149
Answer:

In Exercises 11 and 12, determine whether the figure is stable. Explain your reasoning.

Question 11.
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 150
Answer:
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 5.5 a 11

Question 12.
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 151
Answer:

In Exercises 13 and 14, redraw the triangles so they are side by side with corresponding parts in the same position. Then write a proof.

Question 13.
Given \(\overline{A C}\) ≅ \(\overline{B D}\)
\(\overline{A B}\) ⊥ \(\overline{A D}\)
\(\overline{C D}\) ⊥ \(\overline{A D}\)
Prove ∆BAD ≅ ∆CDA
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 152
Answer:
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 5.5 a 13

Question 14.
Given G is the midpoint of \(\overline{E H}\), \(\overline{F G}\) ≅ \(\overline{G I}\), ∠E and ∠H are right angles.
Prove ∆EFG ≅ ∆HIG
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 153
Answer:

In Exercises 15 and 16. write a proof.

Question 15.
Given \(\overline{L M}\) ≅ \(\overline{J K}\), \(\overline{M J}\) ≅ \(\overline{K L}\)
Prove ∆LMJ ≅ ∆JKL
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 154
Answer:
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 5.5 a 15

Question 16.
Given \(\overline{W X}\) ≅ \(\overline{V Z}\), \(\overline{W Y}\) ≅ \(\overline{V Y}\), \(\overline{Y Z}\) ≅ \(\overline{Y X}\)
Prove ∆VWX ≅ ∆WVZ
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 155
Answer:

CONSTRUCTION
In Exercises 17 and 18, construct a triangle that is congruent to ∆QRS using the SSS Congruence Theorem Theorem 5.8).

Question 17.
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 156
Answer:
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 5.5 a 17

Question 18.
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 157
Answer:

Question 19.
ERROR ANALYSIS
Describe and correct the error in identifying congruent triangles.
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 158
Answer:
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 5.5 a 19

Question 20.
ERROR ANALYSIS
Describe and correct the error in determining the value of x that makes the triangles congruent.
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 159
Answer:

Question 21.
MAKING AN ARGUMENT
Your friend claims that in order to use the SSS Congruence Theorem (Theorem 5.8) Lo prove that two triangles are congruent, both triangles must be equilateral triangles. Is your friend correct? Explain your reasoning.
Answer:
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 5.5 a 21

Question 22.
MODELING WITH MATHEMATICS
The distances between consecutive bases on a softball field are the same. The distance from home plate to second base is the same as the distance from first base to third base. The angles created at each base are 90°. Prove
∆HFS ≅ ∆FST ≅ ∆STH
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 160
Answer:

Question 23.
REASONING
To support a tree you attach wires from the trunk of the tree to stakes in the ground, as shown in the diagram.
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 161
a. What additional information do you need to use the HL Congruence Theorem (Theorem 5.9) to prove that ∆JKL ≅ ∆MKL?
b. Suppose K is the midpoint of JM. Name a theorem you could use to prove that ∆JKL ≅ ∆MKL. Explain your reasoning.
Answer:
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 5.5 a 23

Question 24.
REASONING
Use the photo of the Navajo rug, where \(\overline{B C}\) ≅ \(\overline{D E}\) and \(\overline{A C}\) ≅ \(\overline{C E}\)
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 162
a. What additional intormation do you need to use the SSS Congruence Theorem (Theorem 5.8) to prove that ∆ABC ≅ ∆CDE?
b. What additional information do you need to use the HL Congruence Theorem (Theorem 5.9) to prove that ∆ABC ≅ ∆CDE?
Answer:

In Exercises 25-28. use the given coordinates to determine whether ∆ABC ≅ ∆DEF.

Question 25.
A(- 2, – 2), B(4, – 2), C(4, 6), D(5, 7), E(5, 1), F(13, 1)
Answer:
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 5.5 a 25

Question 26.
A(- 2, 1), B(3, – 3), C(7, 5), D(3, 6), E(S, 2), F( 10, 11)
Answer:

Question 27.
A(0, 0), B(6, 5), C(9, 0), D(0, – 1), E(6, – 6), F(9, – 1)
Answer:
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 5.5 a 27

Question 28.
A(- 5, 7), B(- 5, 2), C(0, 2), D(0, 6), E(o, 1), F(4, 1)
Answer:

Question 29.
CRITICAL THINKING
You notice two triangles in the tile floor of a hotel lobby. You want to determine whether the triangles are congruent. but you only have a piece of string. Can you determine whether the triangles are congruent? Explain.
Answer:
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 5.5 a 29

Question 30.
HOW DO YOU SEE IT?
There are several theorems you can use to show that the triangles in the “square” pattern are congruent. Name two of them.
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 163
Answer:

Question 31.
MAKING AN ARGUMENT
Your cousin says that ∆JKL is congruent to ∆LMJ by the SSS Congruence Theorem (Thm. 5.8). Your friend says that ∆JKL is congruent to ∆LMJ by the HL Congruence Theorem (Thm. 5.9). Who is correct? Explain your reasoning.
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 164
Answer:
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 5.5 a 31

Question 32.
THOUGHT PROVOKING
The postulates and theorems in this book represent Euclidean geometry. In spherical geometry. all points are points on the surface of a sphere. A line is a circle on the sphere whose diameter is equal to the diameter of the sphere. In spherical geometry. do you think that two triangles are congruent if their corresponding sides are congruent? Justify your answer.
Answer:

USING TOOLS
In Exercises 33 and 34, use the given information to sketch ∆LMN and ∆STU. Mark the triangles with the given information.

Question 33.
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 165
Answer:
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 5.5 a 33

Question 34.
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 166
Answer:

Question 35.
CRITICAL THINKING
The diagram shows the light created by two spotlights, Both spotlights are the same distance from the stage.
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 167
Answer:
a. Show that ∆ABD ≅ ∆CBD. State which theorem or postulate you used and explain your reasoning.
b. Are all four right triangles shown in the diagram Congruent? Explain your reasoning.
Answer:
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 5.5 a 35

Question 36.
MATHEMATICAL CONNECTIONS
Find all values of x that make the triangles congruent. Explain.
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 168
Answer:

Maintaining Mathematical proficiency

Use the congruent triangles.

Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 169

Question 37.
Name the Segment in ∆DEF that is congruent to \(\overline{A C}\).
Answer:
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 5.5 a 37

Question 38.
Name the segment in ∆ABC that is congruent to \(\overline{E F}\).
Answer:

Question 39.
Name the angle in ∆DEF that is congruent to ∠B.
Answer:
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 5.5 a 39

Question 40.
Name the angle in ∆ABC that is congruent to ∠F.
Answer:

5.6 Proving Triangle Congruence by ASA and AAS

Exploration 1

Determining Whether SSA Is Sufficient

Work with a partner.
a. Use dynamic geometry software to construct ∆ABC. Construct the triangle so that vertex B is at the origin. \(\overline{A B}\) has a length of 3 units. and \(\overline{B C}\) has a length of 2 units.
Answer:

b. Construct a circle with a radius of 2 units centered at the origin. Locate point D where the circle intersects \(\overline{A C}\). Draw \(\overline{B D}\).
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 170
Answer:

c. ∆ABC and ∆ABD have two congruent sides and a non included congruent angle.
Name them.
Answer:

d. Is ∆ABC ≅ ∆ABD? Explain your reasoning.
Answer:

e. Is SSA sufficient to determine whether two triangles are congruent? Explain your reasoning.
Answer:

Exploration 2

Determining Valid Congruence Theorems

Work with a partner. Use dynamic geometry software to determine which of the following are valid triangle congruence theorems. For those that are not valid. write a counter example. Explain your reasoning.
CONSTRUCTING VIABLE ARGUMENTS
To be proficient in math, you need to recognize and use counterexamples.

Possible Congruence Theorem Valid or not valid?
SSS
SSA
SAS
AAS
ASA
AAA

Answer:

Communicate Your Answer

Question 3.
What information is sufficient to determine whether two triangles are congruent?
Answer:

Question 4.
Is it possible to show that two triangles are congruent using more than one congruence theorem? If so, give an example.
Answer:

Lesson 5.6 Proving Triangle Congruence by ASA and AAS

Monitoring Progress

Question 1.
Can the triangles be proven congruent with the information given in the diagram? If so, state the theorem you would use.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 171
Answer:

Question 2.
In the diagram, \(\overline{A B}\) ⊥ \(\overline{A D}\), \(\overline{D E}\) ⊥ \(\overline{A D}\), and \(\overline{A C}\) ≅ \(\overline{D C}\) . Prove ∆ABC ≅ ∆DEF.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 172
Answer:

Question 3.
In the diagram, ∠S ≅ ∠U and \(\overline{B D}\)\(\overline{B D}\) . Prove that ∆RST ≅ ∆VYT
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 173
Answer:

Lesson 5.6 Proving Triangle Congruence by ASA and AAS

Vocabulary and Core Concept Check

Question 1.
WRITING
How arc the AAS Congruence Theorem (Theorem 5. 11) and the ASA Congruence
Theorem (Theorem 5.10) similar? How are they different?
Answer:
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.6 a 1

Question 2.
WRITING
You know that a pair of triangles has two pairs of congruent corresponding angles. What other information do you need to show that the triangles are congruent?
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3-6, decide whether enough information is given to prove that the triangles are congruent. If so, state the theorem you would use.

Question 3.
∆ABC, ∆QRS
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 174
Answer:
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.6 a 3

Question 4.
∆ABC, ∆DBC
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 175
Answer:

Question 5.
∆XYZ, ∆JKL
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 176
Answer:
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.6 a 5

Question 6.
∆RSV, ∆UTV
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 177
Answer:

In Exercises 7 and 8, state the third congruence statement that is needed to prove that ∆FGH ≅ ∆LMN the given theorem.

Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 178
Question 7.
Given \(\overline{G H}\) ≅ \(\overline{M N}\), ∠G ≅ ∠M, _______ = ________
Use the AAS Congruence Theorem (Thm. 5.11).
Answer:
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.6 a 7

Question 8.
Given \(\overline{F G}\) ≅ \(\overline{L M}\), ∠G ≅ ∠M, _______ = ________
Use the ASA Congruence Theorem (Thm. 5.10).
Answer:

In Exercises 9 – 12. decide whether you can use the given information to prove that ∆ABC ≅ ∆DEF Explain your reasoning.

Question 9.
∠A ≅ ∠G, ∠C ≅∠F, \(\overline{A C}\) ≅ \(\overline{D F}\)
Answer:
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.6 a 9

Question 10.
∠C ≅ ∠F, \(\overline{A B}\) ≅ \(\overline{D E}\), \(\overline{B C}\) ≅ \(\overline{E F}\)
Answer:

Question 11.
∠B ≅ ∠E, ∠C ≅∠F, \(\overline{A C}\) ≅ \(\overline{D E}\)
Answer:
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.6 a 11

Question 12.
∠A ≅ ∠D, ∠B ≅∠E, \(\overline{B C}\) ≅ \(\overline{E F}\)
Answer:

CONSTRUCTION
In Exercises 13 and 14, construct a triangle that is congruent to the given triangle using the ASA Congruence Theorem (Theorem 5.10). Use a compass and straightedge.

Question 13.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 179
Answer:
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.6 a 13

Question 14.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 180
Answer:

ERROR ANALYSIS
In Exercises 15 and 16, describe and correct the error.

Question 15.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 181
Answer:
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.6 a 15

Question 16.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 182
Answer:

PROOF
In Exercises 17 and 18, prove that the triangles are congruent using the ASA Congruence Theorem (Theorem 5.10).

Question 17.
Given M is the midpoint of \(\overline{N L}\).
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 183
Prove ∆NQM ≅ ∆MPL
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 184
Answer:
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.6 a 17.1
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.6 a 17.2

Question 18.
Given \(\overline{A J}\) ≅ \(\overline{K C}\) ∠BJK ≅ ∠BKJ, ∠A ≅ ∠C
Prove ∆ABK ≅ ∆CBJ
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 185
Answer:

PROOF
In Exercises 19 and 20, prove that the triangles are congruent using the AAS Congruence Theorem (Theorem 5.11).

Question 19.
Given \(\overline{V W}\) ≅ \(\overline{U W}\), ∠X ≅ ∠Z
Prove ∆XWV ≅ ∆ZWU
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 186
Answer:
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.6 a 19

Question 20.
Given ∠NKM ≅∠LMK, ∠L ≅∠N
Prove ∆NMK ≅ ∆LKM
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 187
Answer:

PROOF
In Exercises 21-23, write a paragraph proof for the theorem about right triangles.

Question 21.
Hypotenuse-Angle (HA) Congruence Theorem
If an angle and the hypotenuse of a right triangle are congruent to an angle and the hypotenuse of a second right triangle, then the triangles are congruent.
Answer:
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.6 a 21

Question 22.
Leg-Leg (LL) Congruence Theorem
If the legs of a right triangle are congruent to the legs of a second right triangle, then the triangles are congruent.
Answer:

Question 23.
Angle-Leg (AL) Congruence Theorem
If an angle and a leg of a right triangle are congruent to an angle and a leg of a second right triangle, then the triangles are Congruent.
Answer:
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.6 a 23

Question 24.
REASONING
What additional in information do you need to prove ∆JKL ≅ ∆MNL by the ASA Congruence Theorem (Theorem 5. 10)?
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 188
(A) \(\overline{K M}\) ≅ \(\overline{K J}\)
(B) \(\overline{K H}\) ≅ \(\overline{N H}\)
(C) ∠M ≅ ∠J
(D) ∠LKJ ≅ ∠LNM
Answer:

Question 25.
MATHEMATICAL CONNECTIONS
This toy contains △ABC and △DBC. Can you conclude that △ABC ≅ △DBC from the given angle measures? Explain
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 189
m∠ABC = (8x – 32)°
m∠DBC = (4y – 24)°
m∠BCA = (5x + 10)°
m∠BCD = (3y + 2)°
m∠CAB = (2x – 8)°
m∠CDB = (y – 6)°
Answer:
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.6 a 25.1
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.6 a 25.2

Question 26.
REASONING
Which of the following congruence statements are true? Select all that apply.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 190
(A) \(\overline{B D}\) ≅ \(\overline{B D}\)
(B) ∆STV ≅ ∆XVW
(C) ∆TVS ≅ ∆VWU
(D) ∆VST ≅ ∆VUW
Answer:

Question 27.
PROVING A THEOREM
Prove the Converse of the Base Angles Theorem (Theorem 5.7). (Hint: Draw an auxiliary line inside the triangle.)
Answer:
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.6 a 27.1
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.6 a 27.2

Question 28.
MAKING AN ARGUMENT
Your friend claims to be able Lo rewrite any proof that uses the AAS Congruence Theorem (Thin. 5. 11) as a proof that uses the ASA Congruence Theorem (Thin. 5.10). Is this possible? Explain our reasoning.
Answer:

Question 29.
MODELING WITH MATHEMATICS
When a light ray from an object meets a mirror, it is reflected back to your eye. For example, in the diagram, a light ray from point C is reflected at point D and travels back to point A. The law of reflection states that the angle of incidence, ∠CDB. is congruent to the angle of reflection. ∠ADB.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 191
a. Prove that ∆ABD is Congruent to ∆CBD.
Given ∠CBD ≅∠ABD
DB ⊥ AC
Prove ∆ABD ≅ ∆CBD
b. Verify that ∆ACD is isosceles.
c. Does moving away from the mirror have an effect on the amount of his or her reflection a person sees? Explain.
Answer:
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.6 a 29.1
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.6 a 29.2

Question 30.
HOW DO YOU SEE IT?
Name as man pairs of congruent triangles as you can from the diagram. Explain how you know that each pair of triangles is congruent.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 192
Answer:

Question 31.
CONSTRUCTION
Construct a triangle. Show that there is no AAA congruence rule by constructing a second triangle that has the same angle measures but is not congruent.
Answer:
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.6 a 31

Question 32.
THOUGHT PROVOKING
Graph theory is a branch of mathematics that studies vertices and the way they are connected. In graph theory. two polygons are isomorphic if there is a one-to-one mapping from one polygon’s vertices to the other polygon’s vertices that preserves adjacent vertices. In graph theory, are any two triangles isomorphic? Explain your reasoning. second triangle that has the same angle measures but is not congruent.
Answer:

Question 33.
Mathematical Connections
Six statements are given about ∆TUV and ∆XYZ
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 193
a. List all combinations of three given statements that could provide enough information to prove that ∆TUV is congruent to ∆XYZ.
b. You choose three statements at random. What is the probability that the statements you choose provide enough information to prove that the triangles are congruent?
Answer:
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.6 a 33.1
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.6 a 33.2

Maintaining Mathematical proficiency

Find the coordinates of the midpoint of the line segment with the given endpoints.

Question 34.
C(1, 0) and D(5, 4)
Answer:

Question 35.
J(- 2, 3) and K(4, – 1)
Answer:
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.6 a 35

Question 36.
R(- 5, – 7) and S(2, – 4)
Answer:

Copy and angle using a compass and straightedge.

Question 37.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 194
Answer:
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.6 a 37

Question 38.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 195
Answer:

5.7 Using Congruent Triangles

Exploration 1

Measuring the Width of a River

Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 219

Work with a partner:
The figure shows how a surveyor can measure the width of a river by making measurements on only one side of the river.

a. Study the figure. Then explain how the surveyor can find the width of the river.
Answer:

b. Write a proof to verify that the method you described in part (a) is valid.
Given ∠A is a right angle, ∠D is a right angle, \(\overline{A C}\) ≅ \(\overline{C D}\)
Answer:

c. Exchange Proofs with your partner and discuss the reasoning used.
CRITIQUING THE REASONING OF OTHERS
To be proficient in math, you need to listen to or read the arguments of others, decide whether they make sense, and ask useful questions to clarify or improve the arguments.
Answer:

Exploration 2

Measuring the Width of a River

Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 196

Work with a partner. It was reported that one of Napoleon’s offers estimated the width of a river as follows. The officer stood on the hank of the river and lowered the visor on his cap until the farthest thin visible was the edge of the bank on the other side. He then turned and rioted the point on his side that was in line with the tip of his visor and his eye. The officer then paced the distance to this point and concluded that distance was the width of the river.

a. Study the figure. Then explain how the officer concluded that the width of the river is EG.
Answer:

b. Write a proof to verify that the conclusion the officer made is correct.
Given ∠DEG is a right angle, ∠DEF is a right angle, ∠EDG ≅ ∠EDF
Answer:

c. Exchange proofs with your partner and discuss the reasoning used.
Answer:

Communicate Your Answer

Question 3.
How can you use congruent triangles to make an indirect measurement?
Answer:

Question 4.
Why do you think the types of measurements described in Explorations 1 and 2 are called indirect measurements?
Answer:

Lesson 5.7 Using Congruent Triangles

Monitoring Progress

Question 1.
Explain how you can prove that ∠A ≅ ∠C.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 197
Answer:

Question 2.
In Example 2, does it mailer how far from point N you place a stake at point K? Explain.
Answer:

Question 3.
Write a plan to prove that ∆PTU ≅ ∆UQP.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 198
Answer:

Question 4.
Use the construction of an angle bisector on page 42. What segments can you assume are congruent?
Answer:

Exercise 5.7 Using Congruent Triangles

Vocabulary and core concept check

Question 1.
COMPLETE THE SENTENCE
_____________ parts of congruent triangle are congruent.
Answer:
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 5.7 a 1

Question 2.
WRITING
Describe a situation in which you might choose to use indirect measurement with
congruent triangles to find a measure rather than measuring directly.
Answer:

Monitoring Progress and Modeling With Mathematics

In Exercise 3-8, explain how to prove that the statement is true.

Question 3.
∠A ≅ ∠D
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 199
Answer:
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 5.7 a 3

Question 4.
∠Q ≅∠T
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 200
Answer:

Question 5.
\(\overline{J M}\) ≅ \(\overline{L M}\)
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 201
Answer:
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 5.7 a 5

Question 6.
\(\overline{A C}\) ≅ \(\overline{D B}\)
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 202
Answer:

Question 7.
\(\overline{G K}\) ≅ \(\overline{H J}\)
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 203
Answer:
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 5.7 a 7

Question 8.
\(\overline{Q W}\) ≅ \(\overline{V T}\)
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 204
Answer:

In Exercises 9-12, write a plan to prove that ∠1 ≅∠2.

Question 9.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 205
Answer:
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 5.7 a 9

Question 10.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 206
Answer:

Question 11.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 207
Answer:
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 5.7 a 11

Question 12.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 208
Answer:

In Exercises 13 and 14. write a proof to verify that the construction is valid.

Question 13.
Line perpendicular to a line through a point not on the line
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 209
Plan for proof ∆APQ ≅ ∆BPQ by the congruence Theorem (Theorem 5.8). Then show the ∆APM ≅ ∆BPM using the SAS Congruence Theorem (Theorem 5.5). Use corresponding parts of congruent triangles to show that ∠AMP and ∠BMP are right angles.
Answer:
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 5.7 a 13.1
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 5.7 a 13.2

Question 14.
Line perpendicular to a line through a p0int on the line
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 210
Plan for Proof Show that ∆APQ ≅ ∆BPQ by the SSS Congruence Theorem (Theorem 5.8) Use corresponding parts of congruent triangles to show that ∠QPA and ∠QPB are right angles.
Answer:

In Exercises 15 and 16, use the information given in the diagram to write a proof.

Question 15.
Prove \(\overline{F L}\) ≅ \(\overline{H N}\)
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 211
Answer:
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 5.7 a 15.1
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 5.7 a 15.2

Question 16.
Prove ∆PUX ≅ ∆QSY
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 212
Answer:

Question 17.
MODELING WITH MATHEMATICS
Explain how to find the distance across the canyon.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 213
Answer:
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 5.7 a 17

Question 18.
HOW DO YOU SEE IT?
Use the tangram puzzle.
Answer:

Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 214

a. Which triangle(s) have an area that is twice the area of the purple triangle?
b. How man times greater is the area of the orange triangle than the area of the purple triangle?
Answer:

Question 19.
PROOF
Prove that the green triangles in the Jamaican flag congruent if \(\overline{A D}\) || \(\overline{B C}\) and E is the midpoint of \(\overline{A C}\).
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 215
Answer:
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 5.7 a 19.1
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 5.7 a 19.2

Question 20.
THOUGHT PROVOKING
The Bermuda Triangle is a region in the Atlantic Ocean in which many ships and planes have mysteriously disappeared. The vertices are Miami. San Juan. and Bermuda. Use the Internet or some other resource to find the side lengths. the perimeter, and the area of this triangle (in miles). Then create a congruent triangle on land using cities as vertices.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 216
Answer:

Question 21.
MAKING AN ARGUMENT
Your friend claims that ∆WZY can be proven congruent to ∆YXW using the HL Congruence Theorem (Thm. 5.9). Is your friend correct? Explain your reasoning.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 217
Answer:
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 5.7 a 21

Question 22.
CRITICAL THINKING
Determine whether each conditional statement is true or false. If the statement is false, rewrite it as a true statement using the converse, inverse, or contrapositive.
a. If two triangles have the same perimeter, then they are congruent.
b. If two triangles are congruent. then they have the same area.
Answer:

Question 23.
ATTENDING TO PRECISION
Which triangles are congruent to ∆ABC? Select all that apply.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 218
Answer:
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 5.7 a 23

Maintaining Mathematical Proficiency

Find the perimeter of the polygon with the given vertices.

Question 24.
A(- 1, 1), B(4, 1), C(4, – 2), D(- 1, – 2)
Answer:

Question 25.
J(- 5, 3), K(- 2, 1), L(3, 4)
Answer:
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 5.7 a 25

5.8 Coordinate Proofs

Exploration 1

Writing a coordinate Proof

Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 218

Work with a partner.

a. Use dynamic geometry software to draw \(\overline{A B}\) with endpoints A(0, 0) and B(6, 0).
Answer:

b. Draw the vertical line x = 3.
Answer:

c. Draw ∆ABC so that C lies on the line x = 3.
Answer:

d. Use your drawing to prove that ∆ABC is an isosceles triangle.
Answer:

Exploration 2

Writing a Coordinate proof

Work with a partner.

a. Use dynamic geometry software to draw \(\overline{A B}\) with endpoints A(0, 0) and B(6, 0).

b. Draw the vertical line x = 3.

c. Plot the point C(3, 3) and draw ∆ABC. Then use your drawing to prove that ∆ABC is an isosceles right triangle.
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 219

d. Change the coordinates of C so that C lies below the x-axis and ∆ABC is an isosceles right triangle.
Answer:

e. Write a coordinate proof to show that if C lies on the line x = 3 and ∆ABC is an isosceles right triangle. then C must be the point (3, 3) or the point found in part (d).
CRITIQUING THE REASONING OF OTHERS
To be proficient in math, you need to understand and use stated assumptions, definitions, and previously established results.
Answer:

Communicate Your Answer

Question 3.
How can you use a coordinate plane to write a proof?
Answer:

Question 4.
Write a coordinate proof to prove that ∆ABC with vertices A(0, 0), 8(6, 0), and C(3, 3√3) is an equilateral triangle.
Answer:

Lesson 5.8 Coordinate Proofs

Monitoring Progress

Question 1.
Show another way to place the rectangle in Example 1 part (a) that is convenient
for finding side lengths. Assign new coordinates.
Answer:

Question 2.
A square has vertices (0, 0), (m, 0), and (0, m), Find the fourth vertex.
Answer:

Question 3.
Write a plan for the proof.
Given \(\vec{G}\)J bisects ∠OGH.
Proof ∆GJO ≅ ∆GJH
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 220
Answer:

Question 4.
Graph the points 0(0, 0), H(m, n), and J(m, 0). Is ∆OHJ a right triangle? Find the side lengths and the coordinates of the midpoint of each side.
Answer:

Question 5.
Write a coordinate proof.
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 221
Given Coordinates of vertices of ∆NPO and ∆NMO
Prove ∆NPO ≅ ∆NMO
Answer:

Exercise 5.8 Coordinate Proofs

Vocabulary and Core Concept Check

Question 1.
VOCABULARY
How is a coordinate proof different from other types of proofs you have studied?
How is it the same?
Answer:
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 5.8 a 1

Question 2.
WRITING
Explain why it is convenient to place a right triangle on the grid as shown when writing a coordinate proof.
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 222
Answer:

Maintaining Progress and Modeling with Mathematics

In Exercises 3-6, place (he figure in a coordinate plane in a convenient way. Assign coordinates to each vertex. Explain the advantages of your placement.

Question 3.
a right triangle with leg lengths of 3 units and 2 units
Answer:
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 5.8 a 3

Question 4.
a square with a side length of 3 units
Answer:

Question 5.
an isosceles right triangle with leg length p
Answer:
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 5.8 a 5

Question 6.
a scalene triangle with one side length of 2m
Answer:

In Exercises 7 and 8, write a plan for the proof.

Question 7.
Given Coordinates of vertices of ∆OPM and ∆ONM Prove ∆OPM and ∆ONM are isosceles triangles.
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 223
Answer:
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 5.8 a 7

Question 8.
Given G is the midpoint of \(\overline{H F}\).
Prove ∆GHJ ≅ ∆GFO
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 224
Answer:

In Exercises 9-12, place the figure in a coordinate plane and find the indicated length.

Question 9.
a right triangle with leg lengths of 7 and 9 units; Find the length of the hypotenuse.
Answer:
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 5.8 a 9

Question 10.
an isosceles triangle with a base length of 60 units and a height of 50 units: Find the length of one of the legs.
Answer:

Question 11.
a rectangle with a length o! 5 units and a width of 4 units: Find the length of the diagonal.
Answer:
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 5.8 a 11

Question 12.
a square with side length n: Find the length of the diagonal.
Answer:

In Exercises 13 and 14, graph the triangle with the given vertices. Find the length and the slope of each side of the triangle. Then find the coordinates of the midpoint of each side. Is the triangle a right triangle? isosceles? Explain. Assume all variables are positive and in m ≠ n.)

Question 13.
A(0, 0), B(h, h), C(2h, 0)
Answer:
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 5.8 a 13.1
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 5.8 a 13.2

Question 14.
D(0, n), E(m, n), F(m, 0)
Answer:

In Exercises 15 and 16, find the coordinates of any unlabeled vertices. Then find the indicated length(s).

Question 15.
Find ON and MN.
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 225
Answer:
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 5.8 a 15

Question 16.
Find OT.
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 226
Answer:

PROOF
In Exercises 17 and 18, rite a coordinate proof.

Question 17.
Given Coordinates of vertices of ∆DEC and ∆BOC
Prove ∆DEC ≅ ∆BOC
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 227
Answer:
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 5.8 a 17

Question 18.
Given Coordinates of ∆DEA, H is the midpoint of \(\overline{D A}\), G is the mid point of \(\overline{E A}\)
Prove \(\overline{D G}\) ≅ \(\overline{E H}\)
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 228
Answer:

Question 19.
MODELING WITH MATHEMATICS
You and your cousin are camping in the woods. You hike to a point that is 500 meters cast and 1200 meters north of the Campsite. Your cousin hikes to a point that is 1000 meters cast of the campsite. Use a coordinate proof to prove that the triangle formed by your Position, your Cousin’s position. and the campsite is isosceles. (See Example 5.)
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 229
Answer:
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 5.8 a 19.1
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 5.8 a 19.2

Question 20.
MAKING AN ARGUMENT
Two friends see a drawing of quadrilateral PQRS with vertices P(0, 2), Q(3, – 4), R(1, – 5), and S(- 2, 1). One friend says the quadrilateral is a parallelogram but not a rectangle. The other friend says the quadrilateral is a rectangle. Which friend is correct? Use a coordinate proof to support your answer.
Answer:

Question 21.
MATHEMATICAL CONNECTIONS
Write an algebraic expression for the coordinates of each endpoint of a line segment whose midpoint is the origin.
Answer:
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 5.8 a 21

Question 22.
REASONING
The vertices of a parallelogram are (w, 0), (o, v), (- w, 0), and (0, – v). What is the midpoint of the side in Quadrant III?
(a) \(\left(\frac{w}{2}, \frac{v}{2}\right)\)
(b) \(\left(-\frac{w}{2},-\frac{v}{2}\right)\)
(c) \(\left(-\frac{w}{2}, \frac{v}{2}\right)\)
(d) \(\left(\frac{w}{2},-\frac{v}{2}\right)\)
Answer:

Question 23.
REASONING
A rectangle with a length of 3h and a width of k has a vertex at (- h, k), Which point cannot be a vertex of the rectangle?
(A) (h, k)
(B) (- h, 0)
(c) (2h, 0)
(D) (2h, k)
Answer:
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 5.8 a 23

Question 24.
THOUGHT PROVOKING
Choose one of the theorems you have encountered up to this point that you think would be easier to prove with a coordinate proof than with another type of proof. Explain your reasoning. Then write a coordinate proof.
Answer:

Question 25.
CRITICAL THINKING
The coordinates of a triangle are (5d – 5d), (0, – 5d), and (5d, 0). How sh
would the coordinates be changed to make a coordinate proof easier to complete?
Answer:
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 5.8 a 25

Question 26.
HOW DO YOU SEE IT?
without performing any calculations, how do you know that the diagonals of square TUVW are perpendicular to each oilier? How can you use a similar diagram to show that the diagonals of any square are perpendicular to each other?
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 230
Answer:

Question 27.
PROOF
Write a coordinate proof for each statement.
a. The midpoint o! the hypotenuse of a right triangle is the same distance from each vertex of the triangle.
b. Any two congruent right isosceles triangles can be combined to form a single isosceles triangle.
Answer:
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 5.8 a 27.1
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 5.8 a 27.2

Maintaining Mathematical proficiency

\(\vec{Y}\)W bisects ∠XYZ such that m∠XYW = (3x – 7)° and m∠WYZ = (2x + 1)°.

Question 28.
Find the value of x.
Answer:

Question 29.
Find m∠XYZ
Answer:
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 5.8 a 29

Congruent Triangles Chapter Review

5.1 Angles of Triangles

Question 1.
Classify the triangle at the right by its sides and by measuring its angles.
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 231
Answer:

Find the measure of the exterior angle.

Question 2.
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 232
Answer:

Question 3.
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 234
Answer:

Find the measure of each acute angle.

Question 4.
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 235
Answer:

Question 5.
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 236
Answer:

5.2 Congruent Polygons

Question 6.
In the diagram. GHJK ≅ LMNP. Identify all pairs of congruent corresponding parts. Then write another congruence statement for the quadrilaterals.
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 237
Answer:

Question 7.
Find m ∠ V.
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 238
Answer:

5.3 Proving Triangle Congruence by SAS

Decide whether enough information is given to prove that ∆WXZ ≅ ∆YZX using the SAS Congruence Theorem (Theorem 5.5). If so, write a proof. If not, explain why.

Question 8.
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 239
Answer:

Question 9.
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 240
Answer:

5.4 Equilateral and Isosceles Triangles

Copy and Complete the statement.

Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 241

Question 10.
If \(\overline{Q P}\) ≅ \(\overline{Q R}\), then ∠ ______ ≅ ∠ ______ .
Answer:

Question 11.
If ∠TRV ≅ ∠TVR, then ______ ≅ ______ .
Answer:

Question 12.
If \(\overline{R Q} \cong \overline{R S}\), then ∠ ______ ≅ ∠ ______ .
Answer:

Question 13.
If ∠SRV ≅ ∠SVR, then ______ ≅ ______ .
Answer:

Question 14.
Find the values of x and y in the diagram.
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 242
Answer:

5.5 Proving Triangle Congruence by SSS

Question 15.
Decide whether enough information is given to prose that ∆LMP ≅ ∆NPM using the SSS Congruence Theorem (Thin. 5.8). If so, write a proof. If not, explain why.
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 243
Answer:

Question 16.
Decide whether enough information is given to prove that ∆WXZ ≅ ∆YZX using the HL Congruence Theorem (Thm. 5.9). If so, write a proof. If not, explain why.
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 244
Answer:

5.6 Proving Triangle Congruence by ASA and AAS

Question 17.
∆EFG, ∆HJK
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 245
Answer:

Question 18.
∆TUS, ∆QRS
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 246
Answer:

Decide whether enough information is given to prove that the triangles are congruent using the ASA Congruence Theorem (Thm. 5.10). If so, write a proof, If not, explain why.

Question 19.
∆LPN, ∆LMN
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 247
Answer:

Question 20.
∆WXZ, ∆YZX
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 248
Answer:

5.7 Using Congruent Triangles

Question 21.
Explain how to prove that ∠K ≅∠N.

Answer:

Question 22.
Write a plan to prkove that ∠1 ≅ ∠2

Answer:

5.8 Coordinate Proofs

Question 23.
Write a coordinate proof.
Given Coordinates of vertices of quadrilateral OPQR
Prove ∆OPQ ≅ ∆QRO
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 251
Answer:

Question 24.
Place an isosceles triangle in a coordinate plane in a way that is convenient for finding side lengths. Assign coordinates to each vertex.
Answer:

Question 25.
A rectangle has vertices (0, 0), (2k, 0), and (0, k), Find the fourth vertex.
Answer:

Congruent Triangles Test

Write a Proof.

Question 1.
Given \(\overline{C A} \cong \overline{C B} \cong \overline{C D} \cong \overline{C E}\)
Prove ∆ABC ≅ ∆EDC
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 252
Answer:

Question 2.
Given \(\overline{J K}\|\overline{M L}, \overline{M J}\| \overline{K L}\)
Prove ∆MJK ≅ ∆KLM
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 253
Answer:

Question 3.
Gven \overline{Q R} \cong \overrightarrow{R S}\(\), ∠P ≅ ∠T
Prove ∆SRP ≅ ∆QRT
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 254
Answer:

Question 4.
Find the measure of each acute angle in the figure at the right.
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 255
Answer:

Question 5.
Is it possible to draw an equilateral triangle that is not equiangular? If so, provide an example. If not, explain why.
Answer:

Question 6.
Can you use the Third Angles Theorem (Theorem 5.4) to prove that two triangles are congruent? Explain your reasoning.
Answer:

Write a plan through that ∠1 ≅∠2

Question 7.
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 256
Answer:

Question 8.
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 257
Answer:

Question 9.
Is there more than one theorem that could be used to prove that ∆ABD ≅ ∆CDB? If so, list all possible theorems.
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 258
Answer:

Question 10.
Write a coordinate proof t0 show that the triangles created b the keyboard stand are congruent.
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 259
Answer:

Question 11.
The picture shows the Pyramid of Cestius. which is located in Rome, Italy. The measure of the base for the triangle shown is 100 Roman feet. The measures of the other two sides of the triangle are both 144 Roman feet.
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 260
a. Classify the triangle shown by its sides.
Answer:

b. The measure of ∠3 is 40° What are the measures of ∠1 and ∠2? Explain your reasoning.
Answer:

Congruent Triangles Cumulative Assessment

Question 1.
Your friend claims that the Exterior Angle Theorem (Theorem 5.2) can be used to prove the Triangle Sum Theorem (Theorem 5, 1). Is your friend correct? Explain your reasoning.
Answer:

Question 2.
Use the steps in the construction to explain how you know that the line through point P is parallel to line m.
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 261
Answer:

Question 3.
The coordinate plane shows ∆JKL and ∆XYZ
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 262
a. Write a composition of transformations that maps ∆JKL to ∆XYZ
Answer:

b. Is the composition a congruence transformation? If so, identify all congruent corresponding parts.
Answer:

Question 4.
The directed line segment RS is shown. Point Q is located along \(\overline{R S}\) so that the ratio of RQ to QS is 2 to 3. What are the coordinates of point Q?
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 263
(A) Q(1, 2, 3)
(B) Q(4, 2)
(C) Q(2, 3)
(D) Q(-6, 7)
Answer:

Question 5.
The coordinate plane shows that ∆ABC and ∆DEF
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 264
a. Prove ∆ABC ≅ ∆DEF using the given information.
Answer:

b. Describe the composition of rigid motions that maps ∆ABC to ∆DEF
Answer:

Question 6.
The vertices of a quadrilateral are W(0, 0), X(- 1, 3), )(2, 7), and Z(4, 2). Your friend claims that point W will not change after dilatinig quadrilateral WXYZ by a scale factor of 2. Is your friend correct? Explain your reasoning.
Answer:

Question 7.
Which figure(s) have rotational symmetry? Select all that apply.
(A) Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 265
(B) Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 266
(C) Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 267
(D) Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 268
Answer:

Question 8.
Write a coordinate proof.
Given Coordinates of vertices of quadrilateral ABCD
Prove Quadrilateral ABCD is a rectangle.
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 269
Answer:

Question 9.
Write a proof to verify that the construction of the equilateral triangle shown below is valid.
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 270
Answer:

Big Ideas Math Answers Grade 4 Chapter 6 Factors, Multiples, and Patterns

Big Ideas Math Answers Grade 4 Chapter 6

Get free step-by-step solutions and explanations for all the available questions here. The students who are looking for Big Ideas Math Answers Grade 4 Chapter 6 Factors, Multiples, and Patterns can download the pdf for free of cost. The BIM 4th Grade Chapter 6 contains all the topics that help the students to score good marks in the exams. The main aim of providing the Big Ideas Math Book 4th Grade Answer Key Chapter 6 Factors, Multiples, and Patterns is to make the students understand the concepts in an easy manner. Learn Big Ideas Grade 4 Math Answer Key with the help of the simple methods.

Big Ideas 4th Grade Math Book Answer Key Chapter 6 Factors, Multiples, and Patterns

The advantage of referring to our Big Ideas Math Answers Grade 4 Chapter 6 is you can refer to all the problems and also take practice tests for free of cost. Improve your math skills and score better marks in the exams with the help of the BIM 4th Grade Math Answer Key. Check out the below links for a better understanding of every topic. Just click on the links and grab the depth of knowledge on each topic. Practice real-time problems to gain more knowledge on concepts.

Lesson: 1 Understand Factors

Lesson: 2 Factors and Divisibility

Lesson: 3 Relate Factors and Multiples

Lesson: 4 Identify Prime and Composite Numbers

Lesson: 5 Number Patterns

Lesson: 6 Shape Patterns

Performance Task

Lesson 6.1 Understand Factors

Explore and Grow

Draw two different rectangles that each have an area of 24 square units. Label their side lengths.
Big Ideas Math Answer Key Grade 4 Chapter 6 Factors, Multiples, and Patterns 6.1 1
Compare your rectangles to your partner’s rectangles. How are they the same? How are they different?

Answer:

Explanation:
My rectangle is 6 × 4 and my partner’s rectangle is 3 × 8
We know that,
Area of rectangle = l × b
Hence,
According to the above formula, the areas of rectangles are the same irrespective of different lengths and different breadths.

Structure
How is each side length related to 24?

Answer: 
We know that,
The factors of 24 = 4 × 6, 3 × 8, 1 × 24, 2 × 12
Remember that,
a × b = b × a
So,
4 × 6 = 6 × 4
This pattern will be applicable to all factor pairs
Hence, from the above,
We can conclude that each side length is related to 24 because of the factor pair.

Think and Grow: Find Factor Pairs

You can write whole numbers as products of two factors. The two factors are called a factor pair for the number
Big Ideas Math Answer Key Grade 4 Chapter 6 Factors, Multiples, and Patterns 6.1 2

Example
Find the factor pairs for 20.
Find the side lengths of as many different rectangles with an area of 20 square units as possible.

The side lengths of each rectangle are a factor pair.
So, the factor pairs for 20 are 12 and 1,  10 and 2,  and 5 and 4.

Show and Grow

Question 1.
Use the rectangles to find the factor pairs for 12.
Big Ideas Math Answer Key Grade 4 Chapter 6 Factors, Multiples, and Patterns 6.1 4
Answer: Factor pairs of 12 are: 12 and 1, 6 and 2, and 4 and 3

Explanation:

The factor pairs are nothing but the side lengths of a rectangle and the area of a rectangle gives the factor
Hence,
The factor pairs of 12 are: 12 and 1, 6 and 2, and 4 and 3

Question 2.
Draw rectangles to find the factor pairs for 16.
Big Ideas Math Answer Key Grade 4 Chapter 6 Factors, Multiples, and Patterns 6.1 5
Answer: The factor pairs of 16 are: 1 and 16, 2 and 8, 4 and  4

Explanation:

The factor pairs are nothing but the side lengths of a rectangle and the area of a rectangle gives the factor
Hence,
The factor pairs of 16 are: 1 and 16, 2 and 8, 4 and  4

Apply and Grow: Practice

Draw rectangles to find the factor pairs for the number.
Question 3.
14
Big Ideas Math Answer Key Grade 4 Chapter 6 Factors, Multiples, and Patterns 6.1 6

Answer: The factor pairs of 14 are: 1 and 14, 2 and 7

Explanation:

Factors are the numbers that divide the original number completely. Hence,
The factor pairs of 14 are: 1 and 14, 2 and 7

Question 4.
15
Big Ideas Math Answer Key Grade 4 Chapter 6 Factors, Multiples, and Patterns 6.1 7

Answer:  The factor pairs of 15 are: 1 and 15, 3 and 5

Explanation:

Factors are the numbers that divide the original number completely. Hence,
The factor pairs of 15 are: 1 and 15, 3 and 5

Question 5.
20
Big Ideas Math Answer Key Grade 4 Chapter 6 Factors, Multiples, and Patterns 6.1 8

Answer: The factor pairs of 20 are: 1 and 20, 2 and 10, 4 and 5

Explanation:

Factors are the numbers that divide the original number completely. Hence,
The factor pairs of 15 are: 1 and 15, 3 and 5

Question 6.
36
Big Ideas Math Answer Key Grade 4 Chapter 6 Factors, Multiples, and Patterns 6.1 9
Answer: The factor pairs of 36 are: 1 and 36, 2 and 18, 3 and 12, 4 and 9, 6 and 6

Explanation:

Factors are the numbers that divide the number completely.
Hence,
The factor pairs of 36 are: 1 and 36, 2 and 18, 3 and 12, 4 and 9, 6 and 6

Find the factor pairs for the number.
Question 7.
11
Answer: The factor pairs of 11 are: 1 and 11

Explanation:

Factors are the numbers that divide the original number completely.Hence,
The factor pairs of 15 are: 1 and 15, 3 and 5

Question 8.
9
Answer: The factor pairs of 9 are: 1 and 9, and 3 and 3

Explanation:

Factors are the numbers that divide the original number completely. Hence,
The factor pairs of 15 are: 1 and 15, 3 and 5

Question 9.
4

Answer: The factor pairs of 4 are: 1 and 4, 2 and 2

Explanation:
Factors are the numbers that divide the original number completely. Hence,
The factor pairs of 4 are:
1 × 4, 2 × 2 ( Since 4 ×1 and 1 × 4 are equal, we will take any 1 factor )

Question 10.
25
Answer:  The factor pairs of 25 are: 1 and 25, 5 and 5

Explanation:
Factors are the numbers that divide the original number completely. Hence,
The factor pairs of 25 are:
1 × 25, 5 × 5

Question 11.
10
Answer: The factor pairs of 10 are: 1 and 10, 2 and 5

Explanation:
Factors are the numbers that divide the original number completely. Hence,
The factor pairs of 10 are:
1 × 10, 5 × 5

Question 12.
40
Answer: The factor pairs of 40 are: 1 and 40, 2 and 20, 4 and 10, 5 and 8

Explanation:
Factors are the numbers that divide the original number completely. Hence,
The factor pairs of 40 are:
1 × 40, 2 × 20, 4 × 10 and 5 × 8

Question 13.
Writing
Use the word to explain one way that 2 and 6 are related.
Answer: The factor pair of 6 are: 1 and 6, 2 and 3

Explanation:
The one way 2 and  are related is the “Factor-pair method”
According to the factor-pair method,
The factors of 6 are:
1 × 6 and 2 × 3
Hence, from the above,
We can conclude that 2 and 6 are related due to the factor-pair method.

Think and Grow: Modeling Real Life

Example
You want to organize 30 pictures into a rectangular array on a wall. How many different arrays can you make?
Big Ideas Math Answer Key Grade 4 Chapter 6 Factors, Multiples, and Patterns 6.1 10
To find the number of arrays you can make, find the number of factor pairs for 30.
There are 3-factor pairs for 30.
You can use each factor pair to make 30 arrays.
So, there are 30 ×3 = 90 ways to organize the pictures in different arrays.

Show and Grow

Question 14.
A city mayor buys 27 solar panels. She wants to organize the panels into a rectangular array. How many different arrays can she make?

Answer: The different arrays she can make are: 1 × 27, 3 × 9, 9 × 3, and 27 × 1

Explanation:
Given that a city mayor buys 27 solar panels and she wants to organize the panels into a rectangular array.
The “Array” is nothing but the number of patterns ( Factor pairs)  that we can arrange the given things.
Hence,
The factors of 27 are: 1 and 27, 3 and 9, 9 and 3, and 27 and 1
Hence,
The different arrays she can make in arranging the solar panels are:
1 × 27, 3 × 9, 9 × 3, and 27 × 1

Question 15.
DIG DEEPER!
A store owner has 42 masks to hang in a rectangular array on a wall. The owner does not have room for more than 10 masks in each row or column. What are the possible numbers of masks the owner should hang in each row?
Big Ideas Math Answer Key Grade 4 Chapter 6 Factors, Multiples, and Patterns 6.1 11

Answer: The number of possible masks the owner should hang in each row is: 7 masks

Explanation:
Given that a store owner has 42 masks to hang in a rectangular array on a wall.
So now,
The number of arrays ( Factor pairs ) of 42 are:
1 × 42, 2 × 21, 3 × 14, 6 × 7, 7 × 6, 14 × 3, 21 × 2, and 41 × 1
It is also given that the owner does not have enough room for more than 10 masks in each row or column.
So, from the above factor pairs, we can say that the possible array of masks is: 6 × 7 and 7 × 6
Hence, from the above,
we can conclude that the maximum number of masks the owner can put either in rows or columns is: 7 masks

Question 16.
A teacher wants to set up a chair for each of the 48 students in chorus. He wants to set up the chairs in a rectangular array. He can fit no more than 20 rows and no more than 30 chairs in each row in the room. What are the possible numbers of rows that he could set up?

Answer:  The possible number of rows that he could set up are: 16 rows

Explanation:
Given that a teacher wants to set up a chai for each of the 48 students in the chorus and he wants to set up the chairs in a rectangular array.
Hence, the number of arrays ( factor pairs) of 48 are:
1 × 48, 2 × 24, 3 × 16, 4 × 12, 6 × 8, 8 × 6, 12 × 4, 16 × 3, 24 × 2, and 48 × 1
It is also given that he can fit no more than 20 rows and no more than 30 chairs in each row in the room.
Hence, from the above,
We can conclude that the possible number of rows that he could set up are: 16 rows

Understand Factors Homework & Practice 6.1

Question 1.
Use the rectangles to find the factor pairs for 8.
Big Ideas Math Answer Key Grade 4 Chapter 6 Factors, Multiples, and Patterns 6.1 12

Answer: The factors of 8 are: 1 and 8, 2 and 4

Explanation:

Factors are the numbers that divide the original number completely. Hence,
The factor pairs of 8 are: 1 × 8 and 2 × 4

Question 2.
Draw rectangles to find the factor pairs for 21.
Big Ideas Math Answer Key Grade 4 Chapter 6 Factors, Multiples, and Patterns 6.1 13

Answer: The factor pairs of 21 are: 1 × 21, 3 × 7

Explanation:

Factors are the numbers that divide the original number completely. Hence,
The factor pairs of 21 are: 1 × 21, 3 × 7

Question 3.
Draw rectangles to find the factor pairs for 28.
Big Ideas Math Answer Key Grade 4 Chapter 6 Factors, Multiples, and Patterns 6.1 14

Answer: The factor pairs of 28 are: 1 × 28, 2 × 14, 4 × 7

Explanation:


Factors are the numbers that divide the original number completely. Hence,
The factor pairs of 28 are: 1 × 28, 2 × 14, 4 × 7

Find the factor pairs for the number.
Question 4.
13
Answer: The factor pairs of 13 are: 1 × 13 and 13 × 1

Explanation:
Factors are the numbers that divide the original number completely. Hence,
The factor pairs of 13 are:
1 × 13 and 13 × 1

Question 5.
5
Answer: The factor pairs of 5 are: 1 × 5 and 5 × 1

Explanation:
Factors are the numbers that divide the original number completely.
Hence,
The factor pairs of 5 are:
1 × 5 and 5 × 1

Question 6.
35
Answer: The factor pairs of 35 are: 1 × 35, 5 × 7

Explanation:
Factors are the numbers that divide the original number completely.
Hence,
The factor pairs of 35 are:
1 × 35, 5 × 7

Question 7.
45
Answer: The factor pairs of 45 are: 1 × 45, 3 × 15, 5× 9

Explanation:
Factors are the numbers that divide the original number completely.
Hence,
The factor pairs of 45 are:
1 × 45, 3 × 15, 5× 9

Question 8.
18
Answer: The factor pairs of 18 are: 1 × 18, 2 × 9, 3 × 6

Explanation:
Factors are the numbers that divide the original number completely.
Hence,
The factor pairs of 18 are:
1 × 18, 2 × 9, 3 × 6

Question 9.
36
Answer: The factor pairs of 36 are: 1 × 36, 2 × 18, 3 × 12, 4 × 9, 6 × 6

Explanation:
Factors are the numbers that divide the original number completely.
Hence,
The factor pairs of 36 are:
1 × 36, 2 × 18, 3 × 12, 4 × 9, 6 × 6

Question 10.
YOU BE THE TEACHER
Descartes says there are 5-factor pairs for 16. Newton says there are 3-factor pairs for 16. Who is correct? Explain.
Big Ideas Math Answer Key Grade 4 Chapter 6 Factors, Multiples, and Patterns 6.1 15

Answer:  Newton is correct

Explanation:
We know that,
a × b = b × a
Hence,
The factor pairs of 16 are:
1 × 16, 2 × 8, 4 × 4, 8 × 2, 16 × 1
But according to the above,
We don’t have to consider the factor pairs 8 × 2 and 16 × 1
Hence,
There are only 3-factor pairs.
But, Descartes says that there are 5-factor pairs for 16 while Newton says it three.
Hence, from the above,
We can conclude that Newton is correct.

Question 11.
Modeling Real Life
A race volunteer has 50 cases of bottled water. He wants to arrange the cases into a rectangular array. How many different arrays can he make?
Big Ideas Math Answer Key Grade 4 Chapter 6 Factors, Multiples, and Patterns 6.1 16

Answer: The number of different arrays he can make is: 6 arrays

Explanation:
Given that a race volunteer has 50 cases of bottled water and he wants to arrange the cases into a rectangular array.
The array is nothing but the factor pairs.
Hence, the number of arrays of 50 he can make = 1 × 50, 2 × 25, 5 × 10, 10 × 5, 25 × 2, 50 × 1
Hence, from the above,
we can conclude that the number of different arrays to arrange the 50 cases of bottled water are:
a) 1 × 50 b)  2 × 25 c)  5 × 10 d)  10 × 5 e)  25 × 2 f)  50 × 1

Review & Refresh

Find the product.
Question 12.
2 × 14 = _____
Answer: 2 × 14 = 28

Explanation:
According to the Distributive Property of Multiplication,
2 × 14 = ( 10 + 4 ) × 2
= ( 2 × 10 ) + ( 2 × 4 )
= 20 + 8
= 28
Hence, 2 × 14 = 28

Question 13.
22 × 7 = ______
Answer: 22 × 7 = 154

Explanation:
According to the Distributive Property of Multiplication,
22 × 7 = ( 20 + 2 ) × 7
= ( 20 × 7 ) + ( 2 × 7 )
= 140 + 14
= 154
Hence, 22 × 7 = 154

Question 14.
9 × 27 = ______
Answer: 27 × 9 = 243

Explanation:
According to the Distributive Property of Multiplication,
27 × 9 = ( 20 + 7 ) × 9
= ( 20 × 9 ) + ( 7 × 9 )
= 180 + 63
= 243
Hence, 27 × 9 = 243

Lesson 6.2 Factors and Divisibility

Explore and Grow

List any 10 multiples of 3. What do you notice about the sum of the digits in each multiple?
Answer:
The 10 multiples of 3 are:
3, 6, 9, 12, 15, 18 ,21, 24, 27, and 30
From the sum of the digits in the above multiples,
0 + 3 =3
0 + 6 = 6
0 + 9 = 9
1 + 2 = 3
1 + 5 = 6
2 + 1 = 3
2 + 4= 6
2 + 7 = 9
3 + 0 = 3
Hence, from the above sums,
We can conclude that the sum of the digits in the 10 multiples are also the multiples of 3.

List any 10 multiples of 9. What do you notice about the sum of the digits in each multiple?
Answer:
The 10 multiples of 9 are:
9, 18, 27, 36, 45, 54, 63, 72, 81, and 90
From the sum of the digits in the above multiples,
0 + 9 = 9
1 + 8 = 9
2 + 7 = 9
3 + 6 = 9
4 + 5 = 9
5 + 4 = 9
6 + 3 = 9
7 + 2 = 9
8 + 1 = 9
9 + 0 = 9
Hence, from the above,
We can conclude that the sum of the digits in all the multiples of 9 is equal to 9 only.

Structure
How can you use your observations above to determine whether 3 and 9 are factors of a given number? Explain.
Answer: The factors of 9 are: 1,3, and 9

Explanation:
The factor pairs of 9 are:
1 × 9, 3 × 3 and 9 × 1
Hence, from the above,
We can conclude that 3 and 9 are the factors of 9.

Think and Grow: Find Factors and Factor Pairs

A number is divisible by another number when a quotient is a whole number and the remainder is 0.

Some numbers have divisibility rules that you can use to determine whether they are factors of other numbers.
Big Ideas Math Answers 4th Grade Chapter 6 Factors, Multiples, and Patterns 6.2 1
Example
Find the factor pairs for 48.
Use divisibility rules and division to find the factors of 48.

The factors of 48 are 1, 2, 3, 4, 6, 8, 12, 16, 24, and 48.
The factor pairs for 48 are 6.

Show and Grow

Find the factor pairs for the number.
Question 1.
30
Answer:
The factor pairs of 30 are:
1 × 30, 2 × 15, 3 × 10, 5 × 6, 6 × 5, 10 × 3, 15 × 2, 30 × 1

Explanation:
Factors are the numbers that divide the original number completely.
Hence,
The factor pairs of 30 are:
1 × 30, 2 × 15, 3 × 10, 5 × 6, 6 × 5, 10 × 3, 15 × 2, 30 × 1

Question 2.
54
Answer:
The factor pairs of 54 are:
1 × 54, 2 × 27, 3 × 18, 6 × 9, 9 × 6, 18 × 3, 27 × 2, 54 × 1

Explanation:
Factors are the numbers that divide the original number completely.
Hence,
The factor pairs of 54 are:
1 × 54, 2 × 27, 3 × 18, 6 × 9, 9 × 6, 18 × 3, 27 × 2, 54 × 1

Apply and Grow: Practice

Find the factor pairs for the number.
Question 3.
29
Answer:
The factor pairs of 29 are:
1 × 29 and 29 × 1

Explanation:
Factors are the numbers that divide the original numbers completely.
Hence,
The factor pairs of 29 are:
1 × 29 and 29 × 1

Question 4.
50
Answer:
The factor pairs of 50 are:
1 × 50, 2 × 25, 5 × 10, 10 × 5, 25 × 2, and 50 × 1

Explanation:
Factors are the numbers that divide the original numbers completely.
Hence,
The factor pairs of 50 are:
1 × 50, 2 × 25, 5 × 10, 10 × 5, 25 × 2, and 50 × 1

Question 5.
63
Answer:
The factor pairs of 63 are:
1 × 63, 3 × 21, 7 × 9, 9 × 7, 21 × 3 and 63 × 1

Explanation:
Factors are the numbers that divide the number originally.
Hence,
The factor pairs of 63 are:
1 × 63, 3 × 21, 7 × 9, 9 × 7, 21 × 3 and 63 × 1

Question 6.
33
Answer:
The factor pairs of 33 are:
1 × 33, 3 × 11, 11 × 3 and 33 × 1

Explanation:
Factors are the numbers that divide the number originally.
Hence,
The factor pairs of 33 are:
1 × 33, 3 × 11, 11 × 3 and 33 × 1

Question 7.
60
Answer:
The factor pairs of 60 are:
1 × 60, 2 × 30, 3 × 20, 4 × 15, 5 × 12, and 6 × 10

Explanation:
Factors are the numbers that divide the number originally.
Hence,
The factor pairs of 60 are:
1 × 60, 2 × 30, 3 × 20, 4 × 15, 5 × 12, and 6 × 10

Question 8.
64
Answer:
The factor pairs of 64 are:
1 × 64, 2 × 32, 4 × 16, and 8 ×8

Explanation:
Factors are the numbers that divide the number originally.
Hence,
The factor pairs of 64 are:
1 × 64, 2 × 32, 4 × 16, and 8 ×8

List the factors of the number.
Question 9.
39
Answer:
The factors of 39 are: 1, 3, 13, and 39

Explanation:
Factors are the numbers that divide the number originally.
Hence,
The factors of 39 are: 1, 3, 13, and 39

Question 10.
44
Answer:
The factors of 44 are: 1, 2, 4, 11, 22, and 44

Explanation:
Factors are the numbers that divide the number originally.
Hence,
The factors of 44 are: 1, 2, 4, 11, 22, and 44

Question 11.
72
Answer:
The factors of 72 are: 1, 2, 3,4, 6, 8, 9, 12, 18, 24, 36, and 72

Explanation:
Factors are the numbers that divide the number originally.
Hence,
The factors of 72 are: 1, 2, 3,4, 6, 8, 9, 12, 18, 24, 36, and 72

Question 12.
67
Answer:
The factors of 7 are: 1 and 67

Explanation:
Factors are the numbers that divide the number originally.
Hence,
The factors of 7 are: 1 and 67

Question 13.
42
Answer:
The factors of 42 are: 1, 2, 3, 6, 7, 14, 21, and 42

Explanation:
Factors are the numbers that divide the number originally.
Hence,
The factors of 42 are: 1, 2, 3, 6, 7, 14, 21, and 42

Question 14.
28
Answer:
The factors of 28 are:
1, 2, 4, 7, 14, and 28

Explanation:
Factors are the numbers that divide the number originally
Hence,
The factors of 28 are:
1, 2, 4, 7, 14, and 28

Question 15.
Reasoning
Can an odd number have an even factor? Explain.
Answer: No, an odd number doesn’t have an even factor

Explanation:
Let the number which we want to find the factors is: 15
The factors of 15 are:
1, 3, 5, and 15
Hence, from the above factors,
We can conclude that an odd number doesn’t have an even factor.
Note: The “Odd number” is the number that can’t be divisible by 2.

Question 16.
Writing
Use the diagram to explain why you do not have to check whether any numbers greater than 4 are factors of 12.
Big Ideas Math Answers 4th Grade Chapter 6 Factors, Multiples, and Patterns 6.2 3
Answer: We don’t have to check the numbers greater than 4 are the factors of 12 because the factors of 12 greater than 4 are repeating.

Explanation:
The given number is 12
The factors of 12 are:
1, 2, 3, 4, 6, and 12
Hence, from the above
We can conclude that we don’t have to check the numbers greater than 4 are the factors of 12.

Think and Grow: Modeling Real Life

Example
There are 4 classes going on a field trip. The classes will use 3 buses. Can the teachers have an equal number of students on each bus?
Big Ideas Math Answers 4th Grade Chapter 6 Factors, Multiples, and Patterns 6.2 4
Think: What do you know? What do you need to find? How will you solve it?
Step 1: Add to find how many students are going on the field trip.
24 + 23 + 25 + 20 = 92
92 students are going on the field trip.
Step 2: Is the total number of students divisible by the number of buses?
Find the sum of the digits of 92. 9 + 2 =11
The sum of the digits not divisible by 3.
The teachers don’t have an equal number of students on each bus.

Show and Grow

Question 17.
A teacher is making a 5-page test with 28 vocabulary problems and 7 reading problems. Can the teacher put an equal number of problems on each page?
Answer: Yes, the teacher can put an equal number of problems on each page

Explanation:
Given that a teacher is making a 5- page test.
It is also given that there are 28 vocabulary problems and 7 reading problems.
So,
Total number of problems = 28 + 7 = 35 problems
So,
The number of problems that each page contains = 35 ÷ 5
Now,
By using the Distributive Property of Multiplication,
35 ÷ 5 = ( 30 + 5 ) ÷ 5
= ( 30 ÷ 5 ) + ( 5 ÷ 5 )
= 6 + 1
= 7
Hence, from the above,
We can conclude that the teacher can put an equal number of problems on each page.

Question 18.
A relay race is 39 laps long. Each team member must bike the same number of laps. Could a team have 8, 6, or 3 members? Explain.
Answer: A team has 3 members.

Explanation:
Given that a relay race is 39 laps long and each member must bike the same number of laps.
Now,
The factors of 39 are: 1, 3, 13, and 39
The factors are the numbers that divide the number originally.
So, 39 can be divided by 3 only.
Hence, from the above,
We can conclude that each team has only 3 members.

Question 19.
DIG DEEPER!
You have 63 clay figures to display on 7 shelves. Not all of the shelves need to be used and each shelf can hold no more than 25 figures. Each shelf must have the same number of figures. What are all the ways you could arrange the figures?
Big Ideas Math Answers 4th Grade Chapter 6 Factors, Multiples, and Patterns 6.2 5
Answer: The number of ways you could arrange the figures is to find the number of factors of 63.
So,
The number of factors of 63 is: 1, 3, 7,9 21, and 63

Explanation:
Given that you have 63 clay figures to display on 7 shelves.
It is also given that each shelf holds no more than 25 figures.
Now,
The factors of  63 are: 1, 3, 7,9 21, and 63
The “Factors” are the ways to arrange the given clay figures.
So,
The number of ways to arrange the clay figures that do not hold more than 25 is: 3, 7, 9, and 21
Hence, from the above,
We can conclude that the number of ways to arrange the clay figures in each shelf that do not hold more than 25 figures is: 3, 7, 9, and 21 ways

Factors and Divisibility Homework & Practice 6.2

Find the factor pairs for the number.
Question 1.
24
Answer:
The factor pairs of 24 are:
1 × 24, 2 × 12, 3 × 8, 4 × 6, 6 × 4, 8 × 3, 12 × 2, and 24 × 1

Explanation:
Factors are the numbers that divide the number originally.
Hence,
The factor pairs of 24 are:
1 × 24, 2 × 12, 3 × 8, 4 × 6, 6 × 4, 8 × 3, 12 × 2, and 24 × 1

Question 2.
48
Answer:
The factor pairs of 48 are:
1 × 48, 2 × 24, 3 × 16, 4 × 12, 6 × 8

Explanation:
Factors are the numbers that divide the number originally.
Hence,
The factor pairs of 48 are:
1 × 48, 2 × 24, 3 × 16, 4 × 12, 6 × 8

Question 3.
31
Answer:
The factor pairs of 31 are:
1 × 31 and 31 × 1

Explanation:
Factors are the numbers that divide the number originally.
Hence,
The factor pairs of 31 are:
1 × 31 and 31 × 1

Question 4.
99
Answer:
The factor pairs of 99 are:
1 × 99, 3 × 33, 9 × 11

Explanation:
Factors are the numbers that divide the number originally.
Hence,
The factor pairs of 99 are:
1 × 99, 3 × 33, 9 × 11

Question 5.
45
Answer:
The factor pairs of 45 are:
1 × 45, 3 × 15, and 5 × 9

Explanation:
factors are the numbers that divide the number originally.
Hence,
The factor pairs of 45 are:
1 × 45, 3 × 15, and 5 × 9

Question 6.
26
Answer:
The factor pairs of 26 are:
1 × 26, 2 × 13

Explanation:
Factors are the numbers that divide the number originally.
Hence,
The factor pairs of 26 are:
1 × 26, 2 × 13

List the factors of the number.
Question 7.
25
Answer:
The factors of 25 are: 1, 5, and 25

Explanation:
Factors are the numbers that divide the numbers originally.
Hence,
The factors of 25 are: 1, 5, and 25

Question 8.
56
Answer:
The factors of 56 are: 1, 2, 4, 7, 8, 14, 28 and 56

Explanation:
Factors are the numbers that divide the number originally.
Hence,
The factors of 56 are: 1, 2, 4, 7, 8, 14, 28 and 56

Question 9.
75
Answer:
The factors of 75 are: 1, 3, 5, 15, 25, and 75

Explanation:
Factors are the numbers that divide the number originally.
Hence,
The factors of 75 are: 1, 3, 5, 15, 25, and 75

Question 10.
80
Answer:
The factors of 80 are: 1, 2, 4, 5, 8, 10, 16, 20,40 and 80

Explanation:
Factors are the numbers that divide the numbers originally.
Hence,
The factors of 80 are: 1, 2, 4, 5, 8, 10, 16, 20,40 and 80

Question 11.
93
Answer:
The factors of 93 are: 1, 3, 31 and 93

Explanation;
Factors are the numbers that divide the number originally.
Hence,
The factors of 93 are: 1, 3, 31 and 93

Question 12.
61
Answer:
The factors of 61 are: 1 and 61

Explanation:
Factors are the numbers that divide the number originally.
Hence,
The factors of 61 are: 1 and 61

Question 13.
Reasoning
Why does a number that has 9 as a factor also have 3 as a factor?
Answer: The number that has 9 as a factor also have 3 as a factor because 9 is a multiple of 3

Explanation:
Let the number that has 9 as a factor and that has also 3 as a factor be: 18
Now,
The factors of 18 are: 1, 2, 3, 6, 9, and 18
Hene, from the above,
We can conclude that the number that has 9 as a factor also have 3 as a factor.

Question 14.
DIG DEEPER!
The number below has 3 as a factor. What could the unknown digit be?
3 _____ 5.
Answer: The unknown digit could be: 1 or 4 or 7

Explanation:
Given that the number has 3 as a factor.
To have 3 as a factor, the sum of the digits in the given number should be a multiple of 3
Hence,
The unknown digit in 3____5 could be: 1 or 4 or 7

Question 15.
Number Sense
Which numbers have 5 as a factor?
Big Ideas Math Answers 4th Grade Chapter 6 Factors, Multiples, and Patterns 6.2 6
Answer: The numbers which have 5 as a factor are: 50, 25, 1,485 and 100

Explanation:
Given numbers are: 50, 34, 25, 1,485, 100 and 48
The numbers that have the factor of 5 must have the last digits 0 or 5
Hence,
The numbers that have 5 as a factor are: 50, 25, 1,485, and 100

Question 16.
Modeling Real Life
You and a partner are conducting a bottle flipping experiment. You have 3 bottles with different amounts of water in each. You need to flip each bottle 15 times. If you take turns, will you and your partner each get the same number of flips?
Answer: Yes, you and your partner will each get the same number of flips.

Explanation:
Given that you and your partner are conducting a bottle flipping experiment and you have 3 bottles with different amounts of water in each and you need to flip each bottle 15 times.
So,
The number of flips each will get = 15 ÷ 3
Now,
By using the Distributive Property of Multiplication,
15 ÷ 3 = ( 12 + 3 ) ÷ 3
= ( 12 ÷ 3 ) + ( 3 ÷ 3 )
= 4 + 1
= 5
Hence, from the above,
We can conclude that each bottle will flip 5 times.

Question 17.
Modeling Real Life
A florist has 55 flowers. She wants to put the same number of flowers in each vase without any leftover Should she put 2, 3, or 5 flowers in each vase? Explain.
Big Ideas Math Answers 4th Grade Chapter 6 Factors, Multiples, and Patterns 6.2 7
Answer: The florist put 5 flowers in each vase.

Explanation:
Given that a florist has 55 flowers and she wants to put the same number of flowers in each vase without any leftover.
The number 55 will be divided by to not have any leftover because the last digit is 5.
Hence,
The florist puts 5 flowers in each vase.

Review & Refresh

Compare.
Question 18.
Big Ideas Math Answers 4th Grade Chapter 6 Factors, Multiples, and Patterns 6.2 8
Answer: 7,914 is greater than 7,912

Explanation:
Given numbers are 7,914 and 7,912
Hence, from the above,
We can conclude that 7,914 is greater than 7,912

Question 19.
Big Ideas Math Answers 4th Grade Chapter 6 Factors, Multiples, and Patterns 6.2 9
Answer: 65,901 is less than 67,904

Explanation:
Given numbers are 65,901 and 67,904
Hence, from the above,
We can conclude that 65,901 is less than 67,904

Question 20.
Big Ideas Math Answers 4th Grade Chapter 6 Factors, Multiples, and Patterns 6.2 10
Answer: 839,275 is equal to 839,275

Explanation:
Given numbers are 839,275 and 839,275
Hence, from the above,
We can conclude that 839,275 is equal to 839,275

Lesson 6.3 Relate Factors and Multiples

Explore and Grow

List all factors of 24.
Answer:
The factors of 24 are: 1, 2, 3, 4, 6, 8, 12 and 24

Explanation:
Factors are the numbers that divide the numbers originally.
Hence,
The factors of 24 are: 1, 2, 3, 4, 6, 8, 12 and 24

List several multiples of each factor. What number appears in each list?
Answer:
The factors of 24 are: 1, 2, 3, 4, 6, 8, 12 and 24

Explanation:
The factors of 24 are: 1, 2, 3, 4, 6, 8, 12 and 24
Now,
There are no other multiples of 1
The multiples of 2 are: 1, 2
The multiples of 3 are: 1,3
The multiples of 4 are: 1, 2, 4
The multiples of 6 are: 1, 2, 3 and 6
The multiples of 8 are: 1, 2, 4 and 8
The multiples of 12 are: 1, 2, 3, 4, 6, and 12
The multiples of 24 are: 1, 2, 3, 4, 6, 8, 12 and 24
Hence, from the above,
We can conclude that 1 appears in each list

Number Sense
How are factors and multiples related?
Answer:
Factors are the numbers that can divide the numbers originally.
Multiples are the numbers that can be divided by another number a certain number of times without any remainder.

Think and Grow: Identify Multiples

A whole number is a multiple of each of its factors.
12 is a multiple of 1, 2, 3, 4, 6, and 12.
1 × 12 = 12
2 × 6 = 12
3 × 4 = 12
4 × 3 = 12
6 × 2 = 12
12 × 1 = 12

Example
Is 56 a multiple of 7?
Big Ideas Math Answers Grade 4 Chapter 6 Factors, Multiples, and Patterns 6.3 1
One Way:
List multiples of 7.
7, 14, 21, 28, 35, 42, 49, 56
So, 56  is a multiple of 7.
Another Way:
Use division to determine whether 7 is a factor of 56.
56 ÷ 7 = 8
7 is a factor of 56.
So, 56 is a multiple of 7.

Example
Is 9 a factor of 64?
One Way:
Use divisibility rules to determine whether 9 is a factor of 64.
9 is not a factor of 64 because 6 + 4 = 10 is not divisible by 9.
Another Way:
List the multiples of 9.
9, 18, 27, 36, 45, 54, 63, 72
64 is not a multiple of 9.
So, 9 is not a factor of 64.

Show and Grow

Question 1.
Is 23 a multiple of 3? Explain.
Answer: 23 is not a multiple of 3

Explanation:

Multiples are the numbers that can be divided by another number a certain number of times without any remainder.
The multiples of 3 are: 3, 6, 9, 12, 15, 18, 21, 24, 27 and 30
Hence, from the above,
We can conclude that 23 is not a multiple of 3.

Question 2.
Is 8 a factor of 56? Explain.
Answer: 8 is a factor of 56.

Explanation:
Factors are the numbers that divide the numbers originally.
Now,
The factors of 56 are: 1, 2, 4, 7, 8, 14, 28, and 56
Hence, from the above,
We can conclude that 8 is a factor of 56.

Apply and Grow: Practice

Question 3.
Is 65 a multiple of 5? Explain.
Answer: 65 is a multiple of 5

Explanation:
Multiples are the numbers that can be divided by another number a certain number of times without any remainder.
The multiples of 5 are: 5, 10, 15, 20, 25, 30, 35,40, 45, 50, 55, 60 and 65
Hence, fro the above,
We can conclude that 65 is a multiple of 5.

Question 4.
Is 14 a multiple of 4? Explain.
Answer: 14 is not a multiple of 4.

Explanation:
Multiples are the numbers that can be divided by another number a certain number of times without any remainder.
The multiples of 4 are: 4, 8, 12, 16, 20, 24, 28, 32, 36, and 40
Hence, from the above,
We can conclude that 14 is not a multiple of 4.

Question 5.
Is 23 a multiple of 2? Explain.
Answer: 23 is not a multiple of 2

Explanation:
Multiples are the numbers that can be divided by another number a certain number of times without any remainder.
The multiples of 2 are: 2, 4, 6, 8, 10, 1, 14, 16, 18, 20, 22, 24 and 26
Hence, from the above,
We can conclude that 23 is not a multiple of 2.

Question 6.
Is 6 a factor of 96? Explain.
Answer: 6 is a factor of 96

Explanation:
Factors are the numbers that divide the number originally.
The factors of 96 are: 1, 2, 3, 4, 6, 8, 12, 16, 28, 32, 48 and 96
Hence, from the above,
We can conclude that 6 is a factor of 96.

Question 7.
Is 3 a factor of 82? Explain.
Answer: 82 is not a factor of 3

Explanation:
Factors are the numbers that divide the number originally.
The given number is 82
The sum of digits of 82 = 8 + 2 = 11
The number is a multiple of 3 only then the sum of the digits of that number is a multiple of 3.
But the sum of digits of 82 is not a multiple of 3
Hence, from the above,
we can conclude that 3 is not a factor of 82

Question 8.
Is 9 a factor of 72? Explain.
Answer: 9 is a factor of 72

Explanation:
Factors are the numbers that divide the number originally.
The factors of 72 are: 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, and 72
Hence, from the above,
We can conclude that 9 is a factor of 72.

Tell whether 8 is a multiple or a factor of the number. Write multiple, factor or both.
Question 9.
4
Answer: 8 is a multiple of 4.

Explanation:
Multiples are the numbers that can be divided by another number a certain number of times without any remainder.
So,
The multiples of 4 are: 4, 8, 12, 16, and 20
Hence, from the above,
We can conclude that 8 is a multiple of 4

Question 10.
8
Answer: 8 is a multiple of 8

Explanation:
Multiples are the numbers that can be divided by another number a certain number of times without any remainder.
So,
The multiples of 8 are: 8, 16, 24, 32, 0, etc.
Hence, from the above,
We can conclude that 8 is a multiple of 8

Question 11.
32
Answer: 32 is a multiple of 8

Explanation:
Multiples are the numbers that can be divided by another number a certain number of times without any remainder.
So,
The multiples of 8 are: 8, 16, 24, 32, 40, 48 etc
Hence, from the above,
We can conclude that 32 is a multiple of 8

Question 12.
Writing
Use numbers 6 and 12 to explain how factors and multiples are related.
Answer:
The multiplication Expression using numbers 6 and 12 is:
6 × 2 = 12
From the multiplication Expression,
6 is a factor of 12
12 is a multiple of 6

Explanation:
The given numbers re 6 and 12
From the given numbers, the multiplication Expression is:
6 × 2 = 12
Hence, from the above multiplication Expression,
6 is a factor of 12
12 is a multiple of 6.

Question 13.
Complete the Venn diagram.
Big Ideas Math Answers Grade 4 Chapter 6 Factors, Multiples, and Patterns 6.3 2
Answer:

Explanation:
From the given Venn diagram,
The factors of 40 are: 1, 2, 4, 5, 8, 10, 20, and 40
The first 10 multiples of 5 are: 5, 10, 15, 20, 25, 30, 35, 40, 45, and 50
Hence, from the above,
The common numbers from both the factors of 4 and the multiples of 5 are:
5, 10, 20 and 40

Think and Grow: Modeling Real Life

Example
You need 96 balloons for a school dance. Balloons come in packs of 4, packs of 6, and packs of 9. Which packs could you buy so you have no leftover balloons?
Big Ideas Math Answers Grade 4 Chapter 6 Factors, Multiples, and Patterns 6.3 3
Use division to determine whether 96 is a multiple of 4.
4 √96 So, 4  is a factor of 96, and 96  is a multiple of 4.
Use the divisibility rules to check whether 96 is a multiple of 6.
96 is even and 9 + 6 = 15 divisible by 3. So, 6  is a factor of 96 and 96  is a multiple of 6.
Use the divisibility rules to check whether 96 is a multiple of 9.
9 + 6 = 15  is not  divisible by 9. So, 9  is not a factor of 96 and
96  is not a multiple of 9.
You could buy packs of 4 balloons or packs of 6 balloons.

Show and Grow

Question 14.
A teacher needs 88 batteries for science experiments. Batteries are sold in packs of 2, packs of 6, and packs of 8. Which packs could the teacher buy so she has no leftover batteries?
Big Ideas Math Answers Grade 4 Chapter 6 Factors, Multiples, and Patterns 6.3 4
Answer: The teacher should buy the packs of 2 and packs of 8 so that there will be no leftover batteries.

Explanation:
Given that a teacher needs 88 batteries for science experiments and that batteries are sold in packs of 2, 6, and 8.
So,
The factors of 88 are: 1, 2, 4, 8, 11, 22, 44, and 88
So, from the factors of 88, we see that there is no 6 as a factor of 88
So, from this, we can say that we can’t pack 88 batteries in packs of 6 but in the packs of 2 and 8 ( Since 2 and 8 both are the factors of 88 )
We know that,
Factors are the numbers that divide the number originally.
Hence, from the above,
We can conclude that the 88 batteries can be packed in packs of 2 or packs of 8.

Question 15.
DIG DEEPER!
Descartes buys 2 books for a total of $15. Each book costs a multiple of $3. How much could each book cost?
Big Ideas Math Answers Grade 4 Chapter 6 Factors, Multiples, and Patterns 6.3 5

Answer: The cost of each book that Descartes bought is: $15

Explanation:
Given that Descartes bought 3 books for a total of $15 and each book costs a multiple of $3
So,
The cost of each book = The total cost of 2 books ÷ Total number of books
= 15 ÷ 2
Now,
By using the Distributive Property of Multiplication,
15 ÷ 2 = ( 12 + 3 ) ÷ 2
( 12 ÷ 2 ) + ( 3 ÷ 2 )
= 6 + 1.5
= 7.5
Hence, from the above,
We can conclude that the cost of each book is: $7.5 and the cost of each book is also a multiple of $3

Question 16.
Newton buys some boxes of dog treats for $9 each. Descartes buys some bags of cat treats for $6 each. Newton and Descartes spend the same amount of money on treats. What is the least amount of money they could have spent?

Answer: The least amount of money that Newton and Descartes spent on dog treats = $18

Explanation:
Given that Newton buys some boxes of dog treats for $9 each and Descartes buys some bags of cat treats for $6 each.
It is also given that Newton and Descartes spend the same amount of money on treats
So, we have to find the least number that can be a multiple of both 6 and 9.
Now,
the multiples of 6 are: 6, 12, 18, 24, 30, 36, 42, 48, 54, and 60
The multiples of 9 are: 9, 18, 27, 36, 45, 54, 63, 72, 81, and 90
From the multiples of 6 and 9,
We can see that the least number that can be the multiple of both 6 and 9 is: 18
Hence, from the above,
We can conclude that the least amount of money that Newton and Descartes spent on dog treats is: $18

Relate Factors and Multiples Homework & Practice 6.3

Question 1.
Is 16 a multiple of 3? Explain.
Answer: 16 is not a multiple of 3

Explanation:
Multiples are the numbers that can be divided by another number a certain number of times without any remainder.
So,
The multiples of 16 are 3, 6,9, 12, 15, 18, etc.
Hence, from the above,
We can conclude that 16 is not a multiple of 3.

Question 2.
Is 21 a multiple of 7? Explain.
Answer: 21 is a multiple of 7

Explanation:
Multiples are the numbers that can be divided by another number a certain number of times without any remainder.
So,
The multiples of 7 are: 7, 14, 21, 28, 35, 42, 49, 56, 63, 70
Hence, from the above,
We can conclude that 21 is a multiple of 7

Question 3.
Is 46 a multiple of 2? Explain.
Answer: 46 is a multiple of 2

Explanation:
Multiples are the numbers that can be divided by another number a certain number of times without any remainder.
So,
For the multiple of 2, the last digit should be 2, 4, 6, 8, 0
So,
46 can be said as a multiple of 2
Hence, from the above,
We can conclude that 46 is a multiple of 2.

Question 4.
Is 5 a factor of 71? Explain.
Answer: 5 is not a factor of 71

Explanation:
Factors are the numbers that divide the number completely.
So,
The factors of 71 are: 1, 71
Hence, from the above,
We can conclude that 5 is not a factor of 71.

Question 5.
Is 8 a factor of 88? Explain.
Answer: 8 is a factor of 88

Explanation:
Factors are the numbers that divide the numbers completely
So,
The factors of 88 are: 1, 2, 4, 8, 11, 22, 44, 88
Hence, from the above factors,
We can conclude that 8 is a factor of 88

Question 6.
Is 4 a factor of 80? Explain.
Answer: 4 is a factor of 80

Explanation:
Factors are the numbers that divide the number completely.
So,
The factors of 80 are: 1, 2, 4, 5, 8, 10, 16, 20, 40, 80
Hence, from the above factors,
We can conclude that 4 is a factor of 80.

Tell whether 30 is a multiple or a factor of the number. Write multiple, factor, or both.
Question 7.
30
Answer: 30 is a multiple and factor of 30

Explanation:
Multiples are the numbers that can be divided by another number a certain number of times without any remainder.
Factors are the numbers that divide the number completely
So,
The multiples of 30 are: 30, 60, 90, 120, 150, 180, 210, 240, 270 and 300
The factors of 30 are: 1, 2, 3, 5, 6, 10, 15, 30
Hence,
From the above,
We can conclude that 30 is a multiple and factor of 30

Question 8.
90
Answer: 90 is a multiple of 30

Explanation:
Multiples are the numbers that can be divided by another number a certain number of times without any remainder.
So,
The multiples of 30 are: 30, 60, 90, 120, 150, 180, 210, 240, 270, 300
Hence, from the above multiples,
We can conclude that 90 is a multiple of 30

Question 9.
10
Answer: 10 is a factor of 30

Explanation:
Factors are the numbers that divide the number originally.
So,
The factors of 30 are: 1, 2, 3, 5, 6, 10, 15, 30
Hence, from the above factors,
We can conclude that 10 is a factor of 30

Tell whether 10 is a multiple or a factor of the number. Write multiple, factor, or both.
Question 10.
5
Answer: 5 is a factor of 10

Explanation:
Factors are the numbers that divide the number completely.
So,
The factors of 10 are: 1, 2, 5, 10
Hence, from the above factors,
We can conclude that 5 is a factor of 10

Question 11.
60
Answer: 60 is a multiple of 10

Explanation:
Multiples are the numbers that can be divided by another number a certain number of times without any remainder.
So,
The multiples of 10 are: 10, 20, 30, 40, 50, 60, 70, 80 , 90, 100
Hence, from the above multiples,
We can conclude that 60 is a multiple of 10

Question 12.
10
Answer: 10 is a multiple and factor of 10

Explanation:
Multiples are the numbers that can be divided by another number a certain number of times without any remainder.
Factors are the numbers that divide the number completely
So,
The number of factors of 10 is: 1, 2, 5, 10
The multiples of 10 are: 10, 20, 30, 40, 50, 60, 70, 80 , 90, 100
Hence, from the above,
We can conclude that 10 is a factor and a multiple of 10

Question 13.
DIG DEEPER!
Name two numbers that are each a multiple of both 3 and 4. What do you notice about the two multiples?
Answer: 12 is a multiple of both 3 and 4

Explanation:
We know that,
If the number is divisible by 3, then the sum of the digits of that given number must be divisible by 3
If the number is divisible by 4, then the last 2 digits of that given number must also be divisible by 4
Hence,
The number which is a multiple of both 3 and 4 is: 12
12 will satisfy both the conditions of the number divisible by 3 and 4
Hence, from the above,
We can conclude that 12 is a multiple of both 3 and 4

Question 14.
YOU BE THE TEACHER
Is Newton correct? Explain.
Big Ideas Math Answers Grade 4 Chapter 6 Factors, Multiples, and Patterns 6.3 6

Answer: Yes, Newton is correct

Explanation:
Given that,
According to Newton, all numbers that are multiples of 10 have 2 as a factor.
Now,
The multiples of 10 are: 10, 20, 30, 40, 50, 60, 70, 80, 90, 100
So,
Let a multiple of 10 be 20
So,
The factors of 20 are: 1, 2, 4, 5, 10, 20
Hence, from the above,
We can conclude that Newton’s statement is correct

Question 15.
Logic
A quotient is a multiple of 4. The dividend is a multiple of 8. The divisor is a factor of 6. Write one possible equation for the problem.

Answer:
The possible equation for the problem is:
24 ÷ 6 = 4

Explanation:
Given that a quotient is a multiple of 4.
It is also given that the dividend is a multiple of 8 and the divisor is a factor of 6.
So,
The possible Equation for the given problem is:
24 ÷ 6 = 4
Where,
24 is a dividend
6 is a divisor
4 is a quotient
Hence, from the above,
We can conclude that the possible  equation according to the given conditions is:
24 ÷ 6 = 4

Question 16.
Modeling Real Life
Your friend needs our friend needs t0 50 US state capitals. She wants to memorize the same number of capitals each day. Which numbers of capitals can she memorize each day: 2, 3, 4, or 5?

Answer: The number of capitals she can memorize each day in packs of 2 and 5

Explanation:
Given that your friend needs 50 US state capitals and she wants to memorize the same number of capitals each day.
It is also given that the number of capitals she had to memorize in the packs of 2, 3, 4, or 5
So,
If the number has to be divided by 5, the last digit of that given number should be 5 (or) 0
So,
The factors of 50 are: 1, 2, 5,  10, 25, 50
Hence, from the above,
We can conclude that 50 US capitals she had to memorize is in the packs of 2 or packs of 5

Question 17.
Modeling Real Life
Zookeepers plan an enrichment day for the animals every 7 days and bathe the elephants every 2 days. You want to go to the zoo when both events are happening. What other dates in May will this happen?
Big Ideas Math Answers Grade 4 Chapter 6 Factors, Multiples, and Patterns 6.3 7

Answer: The events of both Enrichment day and Elephant Day come in the month of May other than 14 May is:
21 May and 28 May

Explanation:
Given that Zookeepers plan an enrichment day for the animals every 7 days
It is also given that the Elephant bathing takes place every 2 days.
So, other than 14 May,
The other days that these two events will take place on May are: 21 May and 28 May with 7 days gap from 14 May
Hence, from the above,
We can conclude that  both the events are happening on 21 and 28 May

Review & Refresh

Estimate the sum or difference.
Question 18.
Big Ideas Math Answers Grade 4 Chapter 6 Factors, Multiples, and Patterns 6.3 8
Answer: 71,606 – 49,641 = 21,965

Question 19.
Big Ideas Math Answers Grade 4 Chapter 6 Factors, Multiples, and Patterns 6.3 9
Answer: 75,294 + 36,043 = 111,337

Question 20.
Big Ideas Math Answers Grade 4 Chapter 6 Factors, Multiples, and Patterns 6.3 10
Answer: 93,294 – 40,293 = 53,001

Lesson 6.4 Identify Prime and Composite Numbers

Explore and Grow

Draw as many different rectangles as possible that each has the given area. Label their side lengths.
Big Ideas Math Solutions Grade 4 Chapter 6 Factors, Multiples, and Patterns 6.4 1
Compare the numbers of factors of 32 and 13.

Answer:
The factors of 32 are: 1, 2, 4, 8, 16, 32
The factors of 13 are: 1, 13
On comparison of the factors of 32 and 13, we can say that 13 has less number of factors than 32

Reasoning
Can a whole number have fewer than two factors? exactly two factors? more than two factors?

Answer: Yes, a whole number has exactly 2 factors and more than 2 factors but not less than 2 factors.

Explanation:
A whole number will be divided into 2 types based on the number of factors. They are:
A) Composite numbers:
The numbers that have more than 2 factors are called “Composite numbers”
B) Prime numbers:
The numbers that have exactly 2 factors are called “Prime numbers”
There will no less than 2 factors for any number.

Think and Grow: Identify Prime and Composite Numbers
A prime number is a whole number greater than 1 with exactly two factors, 1 and itself. A composite number is a whole number greater than 1 with more than two factors.

Example
Tell whether 27 is a prime composite
Use divisibility rules.
Big Ideas Math Solutions Grade 4 Chapter 6 Factors, Multiples, and Patterns 6.4 2
• 27 is odd, so it is not divisible by 2 or any other even number.
• 2 + 7 = 9 is divisible by 3,
so 27 is divisible by 3.
27 has factors in addition to 1 and itself.
So, 27 is a Composite number.

Example
Tell whether 11 is a prime composite
Use divisibility rules.
• 11 is odd, so it is not divisible by 2 or any other even number.
• 1 + 1 = 2 is not divisible by 3 or 9,
so 11 is not divisible by 3 or 9.
• The ones digit is not 0 or 5,
so 11 is not divisible by 5.
11 has exactly two factors, 1 and itself.
So, 11 is a prime number.

Show and Grow

Tell whether the number is prime or composite. Explain.
Question 1.
7
Answer: 7 is a prime number

Explanation:
Prime number:
The numbers which have exactly 2 factors 1 and itself is called “Prime numbers”
Now,
The factors of 7 are: 1, 7
Hence, from the above,
We can conclude that 7 is a prime number

Question 2.
12
Answer: 12 is a Composite number

Explanation:
Composite numbers:
The numbers which have more than 2 factors are called “Composite numbers”
Now,
The factors of 12 are: 1, 2, 3, 4, 6, 12
Hence, from the above,
We can conclude that 12 is a Composite number

Question 3.
2
Answer: 2 is a prime number.

Explanation:
Prime number:
The numbers which have exactly 2 factors 1 and itself are called “Prime numbers”
Now,
The factors of 2 are: 1, 2
Hence, from the above,
We can conclude that 2 is a prime number

Question 4.
19
Answer: 19 is a prime number

Explanation:
Prime number:
The numbers which have exactly 2 factors 1 and itself are called “Prime numbers”
Now,
The factors of 19 are: 1, 19
Hence, from the above,
we can conclude that 19 is a prime number

Question5.
45
Answer: 45 is a Composite number

Explanation:
Composite numbers:
The numbers which have more than 2 factors are called “Composite numbers”
Now,
The factors of 45 are: 1, 3, 5, 9, 15, 45
Hence, from the above,
we can conclude that 45 is a Composite number

Question 6.
54
Answer 54 is a  Composite number

Explanation:
Composite numbers:
The numbers which have more than 2 factors are called “Composite numbers”
Now,
The factors of 54 are: 1, 2. 3. 6, 9, 18, 27, 54
Hence, from the above,
We can conclude that 54 is a Composite number

Apply and Grow: Practice

Tell whether the number is prime or composite. Explain.
Question 7.
35
Answer:35 is a Composite number

Explanation:
Composite numbers:
The numbers which have more than 2 factors are called “Composite numbers”
Now,
The factors of 35 are: 1, 5, 7, 35
Hence, from the above,
We can conclude that 35 is a Composite number

Question 8.
5
Answer: 5 is a prime number

Explanation:
Prime numbers:
The numbers which have exactly 2 factors 1 and itself are called “prime numbers”
Now,
The factors of 5 are: 1, 5
Hence, from the above,
we can conclude that 5 is a prime number

Question 9.
23
Answer: 23 is a prime number

Explanation:
Prime numbers:
The numbers which have exactly 2 factors 1 and itself are called ” Prime numbers”
Now,
The factors of 23 are: 1, 23
Hence, from the above,
We can conclude that 23 is a prime number

Question 10.
40
Answer: 40 is a Composite number

Explanation:
Composite numbers:
The numbers which have more than 2 factors are called “Composite numbers”
Now,
The factors of 40 are: 1, 2, 4, 5, 8, 10, 20, 40
Hence, from the above,
We can conclude that 40 is a Composite number

Question 11.
41
Answer: 41 is a prime number

Explanation:
Prime numbers:
The numbers which have exactly 2 factors 1 and itself are called “Prime numbers”
Now,
The factors of 41 are: 1, 41
Hence, from the above,
we can conclude that 41 is a prime number

Question 12.
81
Answer: 81 is a Composite number

Explanation:
Composite numbers:
The numbers which have more than 2 factors are called “Composite numbers”
Now,
The factors of 81 are: 1, 3, 9, 27, 81
hence, from the above,
we can conclude that 81 is a Composite number

Question 13.
Structure
To create a list of the prime numbers that are less than 100, do the following.
Big Ideas Math Solutions Grade 4 Chapter 6 Factors, Multiples, and Patterns 6.4 3

  • Place a square around 1. It is neither prime nor composite.
  • Circle 2 and cross out all other multiples of 2.
  • Circle 3 and cross out all other multiples of 3.
  • Circle 5 and cross out all other multiples of 5.
  • Circle the next number that is not crossed out. This number is prime. Cross out all other multiples of this number.
  • Continue until every number is either circled or crossed out.

What are the prime numbers that are less than 100? Explain why these numbers not were crossed out on the chart.
Answer:
The prime numbers below 100 are:
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 73, 79, 83, 87, 89, 93, and 97

Think and Grow: Modeling Real Life

Example
A museum volunteer has 76 shark teeth to display. Can the volunteer arrange the teeth into a rectangular array with more than 1 row and more than 1 tooth in each row? Explain.
Big Ideas Math Solutions Grade 4 Chapter 6 Factors, Multiples, and Patterns 6.4 4
Use divisibility rules to determine whether 76 is prime or composite.
76 is even, so it is divisible by 2.
76 has factors in addition to 1 and itself.
So, 76 is the Composite number.
So,
The volunteer will arrange the teeth into a rectangular array with more than 1 row and more than 1 tooth in each row.
Show and Grow

Question 14.
A teacher has 29 students in class. Can the teacher separate the students into equal groups? Explain.
Answer: No, the teacher can’t separate the students into equal groups.

Explanation:
Given that a teacher has 29 students in the class
Now,
Let find whether 29 is Prime or Composite
Now,
Factors of 29 are: 1, 29
Hence, from the above factors,
We can conclude that 29 is a prime number
Hence,
The teacher can’t divide 29 students into equal groups.

Question 15.
A band instructor wants to have several ways to organize band members into rectangular arrays on the field for a performance.Should the instructor have 89 members or 99 members on the field? Explain.
Answer: The Instructor should have 99 members on the field so that he can arrange the band members into rectangular arrays.

Explanation:
Given that a band instructor wants to organize band members into rectangular arrays on the field for a performance.
It is also given that the instructor wants to arrange into an array of whether 89 members or 99 members
Now,
Factors of 89 are: 1, 89
Factors of 99 are: 1, 3, 9, 11, 33, 99
Hence, from the above,
Since the 99 members can be arranged into different arrays,
We can conclude that the instructor can arrange the participants into an array of 99 members.

Question 16.
DIG DEEPER!
A paramedic is arranging bandages into 4 bins. An equal number of bandages are in each bin. Did the paramedic arrange a prime number or a composite number of bandages? Explain.
Big Ideas Math Solutions Grade 4 Chapter 6 Factors, Multiples, and Patterns 6.4 5
Answer: The paramedic has to arrange a composite number of bandages.

Explanation:
Gien that a paramedic is arranging bandages into 4 bins.
It is also given that there is an equal number of bandages in each bin.
Now,
Given there are 4 bins and 4 is a Composite number
So, 4 can divide only a Composite number but not a prime number.
Prime numbers:
The numbers which have exactly only 2 factors 1 and itself are “Prime numbers”
Composite numbers:
The numbers which have more than 2 factors are “Composite numbers”
Hence, from the above,
We can conclude that the paramedic has to arrange a Composite number of bandages.

Identify Prime and Composite Numbers Homework & Practice 6.4

Tell whether the number is prime or composite. Explain.
Question 1.
3
Answer: 3 is a prime number

Explanation:
Prime numbers:
The numbers which have exactly 2 factors 1 and itself are “Prime numbers”
Now,
The factors of 3 are: 1, 3
Hence, from the above,
We can conclude that 3 is a prime number

Question 2.
27
Answer: 27 is a composite number

Explanation:
Composite numbers:
The numbers which have more than 2 factors are “Composite numbers”
Now,
The factors of 27 are: 1, 3, 9, 27
Hence, from the above,
We can conclude that 27 is a composite number

Question 3.
46
Answer: 46 is a composite number

Explanation:
Composite numbers:
The numbers which have more than 2 factors are “Composite numbers”
Now,
The factors of 46 are: 1, 2, 23, 46
Hence, from the above,
We can conclude that 46 is a composite number

Question 4.
17
Answer: 17 is a prime number

Explanation:
Prime numbers:
The numbers which have exactly 2 factors 1 and itself are “Prime numbers”
Now,
The factors of 17 are: 1, 17
Hence, from the above,
We can conclude that 17 is a prime number

Question 5.
53
Answer: 53 is a prime number

Explanation:
Prime numbers:
The numbers which have exactly 2 factors 1 and itself are “Composite numbers”
Now,
The factors of 53 are: 1, 53
Hence, from the above,
we can conclude that 53 is a prime number

Question 6.
63
Answer: 63 is a composite number

Explanation:
Composite numbers:
The numbers which have more than 2 factors are “Composite numbers”
Now,
The factors of 63 are: 1, 3, 7, 9, 21, 63
Hence, from the above,
We can conclude that 63 is a composite number

Question 7.
29
Answer: 29 is a prime number

Explanation:
Prime numbers:
The numbers which have exactly 2 factors 1 and itself are “Prime numbers”
Now,
The factors of 29 are: 1, 29
Hence, from the above,
We can conclude that 29 is a prime number

Question 8.
31
Answer: 31 is a prime number

Explanation:
Prime numbers:
The numbers which have exactly 2 factors 1 and itself are “Prime numbers”
Now,
The factors of 31 are: 1, 31
Hence, from the above,
We can conclude that 31 is a prime number

Question 9.
75
Answer: 75 is a composite number

Explanation:
Composite numbers:
The numbers which have more than 2 factors are “Composite numbers”
Now,
The factors of 75 are: 1, 3, 5, 15, 25, 75
Hence, from the above,
We can conclude that 75 is a composite number

Question 10.
DIG DEEPER!
Can a number be both prime and composite? Explain.
Answer: No, a number can’t be both prime and composite

Explanation:
Prime numbers:
The numbers which have exactly 2 factors 1 and itself are “Prime numbers”
Composite numbers:
The numbers which have more than 2 factors are “Composite numbers”
Hence, from the number of factors,
We can say a number can’t be both prime and composite at the same time

Question 11.
Logic
Your friend is thinking of a prime number between 60 and 80. The tens digit is one less than the ones digit. What is the number?
Answer: The number is 67

Explanation:
Prime numbers:
The numbers which have exactly 2 factors 1 and itself are “Prime numbers”
So,
The prime numbers between 60 and 80 are: 61, 67, 73, 79
It is also given that the tens digit is one less than the ones digit
So,
Tens position value – 1 = ones position value
Hence, from the above,
We can conclude that 67 is the number.

Number Sense
Write true or false for the statement. If false, provide an example to support your answer.
Question 12.
All odd numbers are prime. _______
Answer: False

Explanation:
Given that all odd numbers are prime.
Prime numbers:
The number of factors which have exactly 2 factors 1 and itself is “Prime numbers”
So, let take the off numbers from 1 to 10
The odd numbers from 1 to 10 are: 1, 3, 5, 7, 9
So, from 1 to 10,
The prime numbers are: 3, 5, 7
So, from the above
9 is an odd number but it is not prime.
Hence, from the above,
We can conclude that all odd numbers are not prime

Question 13.
All even numbers, except 2, are composite. _______
Answer: True

Explanation:
Given that all even numbers except 2 are composite numbers
Now,
The even numbers from 1 to 0 are: 2, 4, 6, 8
Composite numbers:
The numbers which have more than 2 factors are “Composite numbers”
So,
The composite numbers from 1 to 10 are: 4, 6, 8
Note: 2 is an even prime number
Hence, from the above,
WE can conclude that all even numbers except  are composite numbers
Question 14.
A composite number cannot have exactly three factors. _______
Answer: True

Explanation:
Given that a composite number has exactly 3 factors
Composite numbers:
The numbers which have more than 2 factors are “Composite numbers”
Let the number be 63
Now,
The factors of 63 are: 1, 3, 7, 9, 21, 63
Hence, from the above,
We can conclude that a composite number don’t have exactly 3 factors

Question 15.
Modeling Real Life
There are 43 students trying out for a basketball team. Can the coach separate the students into equal groups? Explain.
Big Ideas Math Solutions Grade 4 Chapter 6 Factors, Multiples, and Patterns 6.4 6
Answer: No, the coach can’t separate the students into equal groups

Explanation:
Prime numbers:
The numbers which have exactly 2 factors 1 and itself are “Prime numbers”
Now,
The factors of 43 are: 1, 43
Hence, from the above,
We can conclude that the coach can’t separate the students into equal groups since 43 is a prime number

Question 16.
Modeling Real Life
Which planet not does have a prime number of rings?
Big Ideas Math Solutions Grade 4 Chapter 6 Factors, Multiples, and Patterns 6.4 7
Answer: Saturn does not have a prime number of rings.

Explanation:
Given that,
1 full circle = 2 rings
So,
1 half circle = 1 ring
Big Ideas Math Solutions Grade 4 Chapter 6 Factors, Multiples, and Patterns 6.4 7
So, from the above table,
The number of rings of Jupiter is: 3
The number of rings of Saturn is: 9
The number of rings of Uranus is: 13
The number of rings of Neptune is: 5
Prime numbers:
The numbers which have exactly 2 factors 1 and itself are “Prime numbers”
Hence, from the above,
We can conclude that Saturn does not have a prime number of rings ( Because 9 is a composite number )

Review & Refresh

Use properties to find the product. Explain your reasoning
Question 17.
4 × 9 × 25
Answer: 4 × 9 × 25 = 900

Explanation:
Using the Distributive Property of multiplication,
4 × 9 × 25 = 4 × 9 × ( 20 + 5 )
= 36 × ( 20 + 5 )
= ( 36 × 20 ) + ( 36 × 5 )
= 720 + 180
= 900
Hence, 4 × 9 × 25 = 900

Question 18.
405 × 3
Answer: 403 × 3 = 1,215

Explanation:
Using the Distributive Property of Multiplication,
405 × 3 = ( 400 + 5 ) × 3
= ( 400 × 3 ) + ( 5 × 3 )
= 1,200 + 15
= 1,215
Hence, 405 × 3 = 1,215

Question 19.
698 × 7
Answer: 698 × 7 = 4,886

Explanation:
Using the Distributive Property of Multiplication,
698 × 7 = ( 600 + 90 + 8 ) × 7
= ( 600 × 7 ) + ( 90 × 7 ) + ( 8 × 7 )
= 4,200 + 630 + 56
= 4,886
Hence, 698 × 7 = 4,886

Lesson 6.5 Number Patterns

Explore and Grow

Shade every third square in the table.
Big Ideas Math Answer Key Grade 4 Chapter 6 Factors, Multiples, and Patterns 6.5 1
Write the shaded numbers. What patterns do you see?

What other patterns do you see in the table?

Answer:
The shaded numbers are:
3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60
From the shaded numbers, we can see that each third square is a multiple of 3
The other patterns we can observe in the given table is:
a) If we first shaded the second square and then shaded each second square, then we get the multiples of 2
b) The same pattern will have to be applied for the fourth square
There will also be other patterns by shading 5th square, 6th square, etc.

Structure
Circle every fourth square in the table. Write the circled numbers. What patterns do you see?
Answer:
The circled numbers are:
4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60
From the circled numbers of every fourth square, we can see that every fourth square is a multiple of 4.

Think and Grow: Create Number Patterns

A rule tells how numbers or shapes in a pattern are related.
Example
Use the rule “Add 3.” to create a number pattern. The first number in the pattern is 3. Then describe another feature of the pattern.
Create a pattern.

So,
The numbers in the pattern are multiples of 3.
Big Ideas Math Answer Key Grade 4 Chapter 6 Factors, Multiples, and Patterns 6.5 3

Example
Use the rule “Multiply by 2.” to create a number pattern. The first number in the pattern is 10. Then describe another feature of the pattern.
Create a pattern.

So,
The one’s digit of each number in the pattern is 0.

Show and Grow

Write the first six numbers in the pattern. Then describe another feature of the pattern.
Question 1.
Rule: Add 5.
First number: 1
1, ____, _____, _____, _____, _____
Answer: The first 6 numbers are: 1, 6, 11, 16, 21 and 26

Explanation:
For the formation of the pattern,
The given rules are:
Rule 1 : Add 5
Rule 2: Add 5
Hence,
The given pattern is

Question 2.
Rule: Multiply by 3.
First number: 3
3, _____, _____, ____, _____, _____
Answer: The first 6 numbers are: 3, 9, 27, 81, 243, 729

Explanation:
For the formation of the pattern,
the given rules are:
Rule 1: Multiply by 3
Rule 2: First number: 3
Hence,
The given pattern is

Question 3.
Rule: Subtract 2.
First number: 20
Answer: The first 6 numbers are: 20, 18, 16, 14, 12, 10

Explanation:
For the formation of the pattern,
the given rules are:
Rule 1: Subtract 20
Rule 2: First number: 20
Hence,
The given pattern is:

Question 4.
Rule: Divide by 2.
First number: 256
Answer: The first 6 numbers are: 256, 128, 64, 32, 16, 8

Explanation:
For the formation of the pattern,
the given rules are:
Rule 1: Divide by 2
Rule 2: First number: 256
Hence,
The given pattern is:

Apply and Grow: Practice

Write the first six numbers in the pattern. Then describe another feature of the pattern.
Question 5.
Rule: Add 11.
First number: 11
Answer: The first 6 numbers are: 11, 22, 33, 44, 55, 66

Explanation:
For the formation of the pattern,
the given rules are:
Rule: Add 11
First number: 11
Hence,
The given pattern is:

Question 6.
Rule: Multiply by 4.
First number: 4
Answer: The first 6 numbers are: 4, 16, 64, 256, 1024, 4096

Explanation:
For the formation of the pattern,
the given rules are:
Rule 1: Multiply by 4
First number: 4
Hence,
The given pattern is

Question 7.
Rule: Subtract 3.
First number: 21
Answer: The first 6 numbers are: 21, 18, 15, 12, 9, 6

Explanation:
For the formation of the pattern,
the given rules are:
Rule: Subtract 3
First number: 21
Hence,
The given pattern is:

Question 8.
Rule: Divide by 3.
First number: 729
Answer: The first 6 numbers are: 729, 243, 81, 27, 9, 3

Explanation:
For the formation of the pattern,
the given rules are:
Rule 1: Divide by 3
First number: 729
Hence,
The given pattern is

Question 9.
Rule: Add 9.
First number: 8
Answer: The first  numbers are: 8, 17, 26, 35, 44, 53

Explanation:
For the formation of the pattern,
the given rules are:
Rule 1: Add 9
First number: 8
Hence,
The given pattern is:

Question 10.
Rule: Multiply by 5.
First number: 5
Answer: The first 6 numbers are: 5, 25, 125, 625, 3125, 15,625

Explanation:
For the formation of the pattern,
the given rules are:
Rule 1: Multiply by 5
First number: 5
Hence,
The given pattern is:

Open-Ended
Use the rule to generate a pattern of four numbers.
Question 11.
Rule: Multiply by 2.
Answer:
Let the first number be 2.
Hence,
The pattern of four numbers is: 2, 4, 8, 16

Explanation:
For the formation of the pattern,
the given rules are:
Rule: Multiply by 2
Let the first number be: 2
Hence,
The obtained pattern will be:

Question 12.
Rule: Subtract 9.
Answer:
Let the first number be 36
Hence,
The pattern of four numbers is: 36, 27, 18, 9

Explanation:
For the formation of the pattern,
the given rules are:
Rule: Subtract 9
Let the first number be 36
Hence,
The obtained pattern will be:

Question 13.
Rule: Divide by 4.
Answer:
Let the first number be 16
Hence,
The pattern of 4 numbers are: 16, 12, 8, 4

Explanation:
For the formation of the pattern,
the given rules are:
Rule: Divide by 4
Let the first number be 16
Hence,
The obtained pattern will be:

Question 14.
Rule: Add 7.
Answer:
Let the first number be 7
Hence,
The pattern of the four numbers are: 7, 14, 21, 28

Explanation:
For the formation of the pattern,
the given rules are:
Rule: Add 7
Let the first number be 7
Hence,
The obtained pattern will be:

Question 15.
Patterns
Write a rule for the pattern below. Then write a different pattern that follows the same rule.
3, 6, 12, 24, 48
Answer:  The rule for the given pattern is: Multiply by 2

Explanation:
Given numbers are: 3, 6, 12, 24, 48
So, from the given numbers,
We can say that the given pattern follows the “Multiply by 2” rule

Question 16.
Reasoning
What is the missing number in the pattern? Explain.
39, 37, 35, _____, 31, 29
Answer: The missing number in the pattern is: 33

Explanation:
The given numbers are: 39, 37, 35, 31, 29
From the given pattern,
the numbers that are following the rule is “Subtract by 2”
Hence, from the above,
The missing number will be: 35 – 2 = 33

Think and Grow: Modeling Real Life

Example
A presidential election is held every 4 years. There was a presidential election in 2016. How many presidential elections will occur between 2017 and 2030?
Big Ideas Math Answer Key Grade 4 Chapter 6 Factors, Multiples, and Patterns 6.5 5
The rule is to add 4 years to each presidential election year. Start with 2016. Then count the years in the pattern that is between 2017 and 2030.

So,
3 presidential elections will occur between 2017 and 2030.

Show and Grow

Question 17.
The pattern of animals on a Chinese calendar repeats every 12 years. The year 2000 was the year of the dragon. How many times will the year of the dragon occur between 2001 and 2100?
Big Ideas Math Answer Key Grade 4 Chapter 6 Factors, Multiples, and Patterns 6.5 7
Answer: The number of times the dragon will appear between 2001 and 2100 is: 8

Explanation:
Given that the pattern of animals on a Chinese calendar repeats every 12 years
It is also given that the year 2000 was the year of the dragon
Now, between the year 2001 and 2100, there are 99 years
So,
The number of times the dragon will appear = The difference between the years 2001 and 2100 ÷ The number of years the pattern will change
= 99 ÷ 12
Now,
99 ÷ 12 = 8 R 3
Hence, from the above,
We can conclude that the dragon will appear 8 times between the years 2001 and 2021

Question 18.
A robotics team raised $25 the first month of school. Each month of school, the team wants to raise 2 times as much money as the month before. How much money should they raise in the fifth month of school?
Answer: The money should they raise in the fifth month of school is: $400

Explanation:
Given that a robotics team raised $25 for the first month of the school and each month of the school, the team wants to raise 2 times as much money as the month before.
Hence,
The rule followed here is: Multiply by 2
The given first number is: $25
Hence,
The pattern we will obtain is:

Hence, from the above,
We can conclude that the money they should raise in the fifth month is: $400

Question 19.
DIG DEEPER!
You start with 128 pictures on your tablet. You take 6 pictures and delete 3 pictures each day. How many pictures do you have on your tablet after 6 days?
Answer: The pictures you have on your tablet after 6 days is: 110 pictures

Explanation:
Given that you have 128 pictures on your tablet and you take 6 pictures and delete 3 pictures each day.
So,
The number of pictures you have each day = 6 – 3 = 3 pictures
So,
The total number of pictures in 6 days = 6 × 3 = 18 pictures
So,
The total number of pictures you have in 6 days = Total number of pictures – Total number f pictures in 6 days
= 128 – 18
= 110 pictures
Hence, from the above,
We can conclude that there are 110 pictures you have in 6 days.

Number Patterns Numbers Homework & Practice 6.5

Write the first six numbers in the pattern. Then describe another feature of the pattern.
Question 1.
Rule: Subtract 8.
First number: 88
88, ____, ____, ____, _____, _____
Answer:
The first 6 numbers in the given pattern are: 88, 80, 72, 64, 56, 48

Explanation:
For the formation of the pattern,
the given rules are:
Rule: Subtract 8
First number: 88
Hence,
The pattern we will obtain is:

Question 2.
Rule: Multiply by 10.
First number: 2
2, _____, _____, _____, ____, ____
Answer:
The first 6 numbers of the given pattern are: 2, 20, 200, 2,000, 20,000, 200,000

Explanation:
For the formation of the pattern,
the given rules are:
Rule: Multiply by 10
First number: 2
Hence,
The pattern we will obtain is:

Question 3.
Rule: Add 9.
First number: 17
Answer:
The first 6 numbers for the given pattern is: 17, 26, 35, 44, 53, 62

Explanation:
For the formation,
the given rules are:
Rule: Add 9
First number: 17
Hence,
The pattern we will obtain is:

Question 4.
Rule: Divide by 2.
First number: 1,600
Answer:
The first 6 numbers of the given pattern are: 1,600, 800, 400, 200, 100, 50

Explanation:
For the formation of the pattern,
the given rules are:
Rule: Divide by 2
First number: 1,600
Hence,
The pattern we will obtain is:

Open-Ended Use the rule to generate a pattern of four numbers.
Question 5.
Rule: Divide by 5.
Answer:
Let the first number be: 625
Hence,
The  four numbers of the given pattern will be: 625, 125, 25, 5

Explanation:
For the formation of the pattern,
the given rules are:
Rule: Divide by 5
Let the first number be: 625
Hence,
The pattern we will obtain is:

Question 6.
Rule: Add 8.
Answer:
Let the first number be 8
Hence,
The four numbers of the given pattern are: 8, 16, 24, 32

Explanation:
For the formation of the pattern,
the given rules are:
Rule: Add 8
Let the first number be 8
Hence,
The pattern we will obtain will be:

Question 7.
Rule: Multiply by 9.
Answer:
Let the first number be 9
Hence,
The four numbers of the pattern will be: 9, 81, 729, 6551

Explanation:
For the formation of the pattern,
the given rules are:
Rule: Multiply 9
Let the first number be 9
Hence,
The pattern we obtain will be:

Question 8.
Rule: Subtract 3.
Answer:
Let the first number be 27
Hence,
The four numbers of the given pattern will be: 27, 24, 21, 18

Explanation:
For the formation of the pattern,
the given rules are:
Rule: Subtract 3
Let the first number be 27
Hence,
The pattern we obtain will be:

Question 9.
Structure
List the first ten multiples of 9. What patterns do you notice with the digits in the one’s place? in the tens place?
Does this pattern continue beyond the tenth number in the pattern?
Answer: The first 10 multiples of 9 are: 9, 18, 27, 36, 45, 54, 63, 72, 81, 90

Explanation:
The first 10 multiples of 9 are: 9, 18, 27, 36, 45, 54, 63, 72, 81, 90
So,
from the above 10 multiples of 9,
The pattern we can observe in the one’s place is: Decreasing of numbers from 9 to 0 i.e., 9, 8, 7, ………..0
The pattern we can observe in the ten’s place is: Increasing of numbers from 0 to 9 i.e., 0, 1, 2 …………..9
The above 2 patterns will continue beyond the first 10 multiples of 9
Hence, from the above,
We can conclude that the patterns we observed are:
The pattern we can observe in the one’s place is: Decreasing of numbers from 9 to 0 i.e., 9, 8, 7, ………..0
The pattern we can observe in the ten’s place is: Increasing of numbers from 0 to 9 i.e., 0, 1, 2 …………..9

Question 10.
Modeling Real Life
It takes the moon about 28 days to orbit Earth. How many times will the moon orbit Earth in 1 year?
Big Ideas Math Answer Key Grade 4 Chapter 6 Factors, Multiples, and Patterns 6.5 8
Answer: The moon will orbit the earth 10,220 times in 1 year

Explanation:
Given that it takes the moon about 28 days to orbit the earth.
We know that,
1 year = 365 days
So,
The number of times the moon will orbit around the earth in 1 year = 365 × 28
Now,
By using the partial products method,
365 × 28  = ( 300 + 65 ) ×  ( 20 + 8 )
= ( 300 ×  20 ) + ( 300 ×  8 ) + ( 65 ×  20 ) + ( 65 ×  8 )
= 6,000 + 2,400 + 1,300 + 520
= 10,220 times
Hence, from the above,
We can conclude that the moon will orbit around the Earth 10,220 times

Question 11.
DIG DEEPER!
In each level of a video game, you can earn up to 10 points and lose up to 3 points. Your friend earns 9 points in the first level. If he earns and loses the maximum number of points at each level, how many total points will he have after level 6?
Answer: The total points your friend will have is: 44 points

Explanation:
Given that in each level of a video game, you can earn up to 10 points and lose up to 3 points
So,
The maximum number of points you can get = 10 – 3 = 7 points
It is also given that your friend earns 9 points in the first level
So,
The number of points he will have at level 6 = The number of points he has at the first level + 5 × ( The maximum number of points he can get )
= 9 + ( 5 ×  7 )
= 9 + 35
= 44 points
Hence, from the above,
We can conclude that he will have 44 points at the sixth level.

Review & Refresh

Find the product.
Question 12.
14 × 23 = _____
Answer: 14 ×  23 = 322

Explanation:
By using the partial products method,
14 ×  23 = ( 10 + 4 ) × ( 20 + 3 )
= ( 10 ×  20 ) + ( 10 ×  3 ) + ( 4 ×  20 ) + ( 4 ×  3 )
= 200 + 30 + 80 + 12
= 322
Hnece, 14 × 23 = 322

Question 13.
48 × 60 = _____
Answer: 48 × 60 = 2,880

Explanation:
By using the Distributive Property of Multiplication,
48 ×  60 = ( 40 + 8 ) × 60
= ( 40 ×  60 ) + ( 8 ×  60 )
= 2,400 + 480
= 2,880
Hence, 48 × 60 = 2,880

Question 14.
55 × 31 = _____
Answer: 55 ×  31 = 1,705

Explanation:
By using the partial products method,
55 × 31 = ( 50 + 5 ) ×  ( 30 + 1 )
= ( 50 ×  30 ) + ( 50 ×  1 ) + ( 5 ×  30 ) + ( 5 ×  1 )
= 1,500 + 50 + 150 + 5
= 1,705
Hence, 55 ×  31 = 1,705

Lesson 6.6 Shape Patterns

Explore and Grow

Create a rule using 3 different shapes. Draw the first six shapes in the pattern.

What is the next shape in the pattern?

What is the 9th shape in the pattern? Explain.

What is the 99th shape? 1,000th shape? Explain?

Answer: 
From the above 3 shapes,
The rule we can create is: Add 1
So,
We can say the above pattern will repeat after every 3 shapes.
So,
The 9th shape will be: Pentagon
The 99th shape will be: 99 / 3  =  33 R 0
So, The 99th shape will be: Pentagon
The 100th shape will be: Triangle (Because the remainder will be 1 i.e., the shapes will completely repeat 99 times and for the 100th time, it will start from 1st shape)

Structure
You want to show the first 40 shapes in the pattern above. Without modeling, how many of each shape do you think you will need?
Answer:
The number of times each shape will appear in the first 40 shapes is:
Triangle: 14 times
Square: 13 times
Pentagon: 13 times

Explanation:
In the above pattern, the shapes are: Triangle, Square, Pentagon
The rule we obtained by using these 3 shapes is: Add 1
So,
The pattern for the first 40 shapes will be: 40 ÷ 3
Now,
By using the Distributive Property of Division,
40 ÷ 3 = ( 36 + 3 ) ÷ 3
= ( 36 ÷ 3 ) + ( 3 ÷ 3 )
= 12 + 1
= 13 R 1
Hence, the 3 shapes will appear 39 times
So,
The number of times each shape will appear = 13 times
Hence, from the above,
We can conclude that the number of times the shape will appear is:
Triangle: 14 times ( 13 + 1. This 1 is because 40th time, the triangle will repeat again )
Square: 13 times
Pentagon: 13 times

Think and Grow: Create Shape Patterns

Example
Create a shape pattern by repeating the rule “triangle, hexagon, square, rhombus.”What is the 42nd shape in the pattern?
Create a pattern.
Big Ideas Math Answers 4th Grade Chapter 6 Factors, Multiples, and Patterns 6.6 1
42 ÷ 4 is 10 R2, so when the pattern repeats 10 times, the 40th shape is a rhombus So, the 41st shape is a Triangle and the 42nd shape is a hexagon.
Big Ideas Math Answers 4th Grade Chapter 6 Factors, Multiples, and Patterns 6.6 2
Big Ideas Math Answers 4th Grade Chapter 6 Factors, Multiples, and Patterns 6.6 3
Figure 1 has 1 column of 4 dots, so it has 1 × 4 = 4 dots.
Figure 2 has 2 columns of 4 dots, so it has 2 × 4 = 8 dots.
Figure 3 has 3 columns of 4 dots, so it has 3 × 4 = 12 dots.
So,
The 25th figure has 1 column of 4 dots, ( Because the 3 figures will be repeated 24 times and the first figure will appear again 25th time)
So, it has 1 × 4 =4 dots.

Show and Grow

Question 1.
Extend the pattern of shapes by repeating the rule “square, trapezoid, triangle, hexagon, triangle.”What is the 108th shape in the pattern?
Big Ideas Math Answers 4th Grade Chapter 6 Factors, Multiples, and Patterns 6.6 4
Answer: The 108th shape in the pattern will be: Triangle

Explanation:
The given pattern is: Square, trapezoid, triangle, hexagon, triangle
So, there is a total of 5 shapes.
So,
for this, when we divide any number by the total number of patterns, the value of the quotient will be a total repetition of the patterns and the remainder will be the counting of the pattern from the 1st figure.
So,
108 ÷ 5 = ( 100 + 5 ) ÷ 5
= ( 100 ÷ 5 ) + ( 5 ÷ 5 )
= 20 + 1
= 21 R 3
Hence, from the above,
We can conclude that the 108th shape will be the triangle.

Question 2.
Describe the dot pattern. How many dots are in the 76th figure?
Big Ideas Math Answers 4th Grade Chapter 6 Factors, Multiples, and Patterns 6.6 5
Answer: The number of dots in the 76th figure is: 3 dots

Explanation:
Big Ideas Math Answers 4th Grade Chapter 6 Factors, Multiples, and Patterns 6.6 5
From the above pattern,
In figure 1, the number of dots = 1 × 3 = 3 dots
In figure 2, the number of dots = 2 × 3 =  dots
In figure 3, the number of dots = 3 × 3 = 9 dots
So,
The total number of patterns is: 3
So,
for this, when we divide any number by the total number of patterns, the value of the quotient will be a total repetition of the patterns and the remainder will be the counting of the pattern from the 1st figure.
So,
By using the partial quotients method,
76 ÷ 3 = ( 66 +9 ) ÷ 3
= ( 66 ÷ 3 ) + ( 9 ÷ 3 )
= 22 + 3
= 25 R 1
Hence, from the above,
We can conclude that there are 3 dots in the 76th figure.

Apply and Grow: Practice
Question 3.
Extend the pattern of shapes by repeating the rule “oval, triangle.”What is the 55th shape?
Big Ideas Math Answers 4th Grade Chapter 6 Factors, Multiples, and Patterns 6.6 6
Answer: The 55th shape is: Oval

Explanation:
The given pattern of shapes is: Oval, Triangle
So,
The total number of patterns is: 2
So,
for this, when we divide any number by the total number of patterns, the value of the quotient will be a total repetition of the patterns and the remainder will be the counting of the pattern from the 1st figure.
So,
By using the partial quotients method,
55 ÷ 2 = ( 50 + 4 ) ÷ 2
= ( 50 ÷ 2 ) + ( 4 ÷ 2 )
= 25 + 2
= 27 R 1
Hence, from the above,
We can conclude that the 55th shape is: Oval

Question 4.
Extend the pattern of symbols by repeating the rule “add, subtract, multiply, divide.”What is the 103rd symbol?
Big Ideas Math Answers 4th Grade Chapter 6 Factors, Multiples, and Patterns 6.6 7
Answer: The 103rd symbol is: Multiply

Explanation:
The given patterns are: Add, Subtract, Multiply, Divide
So,
The total number of patterns is: 4
So,
for this, when we divide any number by the total number of patterns, the value of the quotient will be a total repetition of the patterns and the remainder will be the counting of the pattern from the 1st figure.
So,
By using the partial quotients method,
103 ÷ 4 = ( 80 + 20 ) ÷ 4
= ( 80 ÷ 4 ) + ( 20 ÷ 4 )
= 20 + 5
= 25 R 3
Hence, from the above,
We can conclude that the 103rd shape is: Multiply

Question 5.
Describe the pattern. How many squares are in the 24th figure?
Big Ideas Math Answers 4th Grade Chapter 6 Factors, Multiples, and Patterns 6.6 8
Answer: The number of squares in the 24th figure is: 4 squares

Explanation:
Big Ideas Math Answers 4th Grade Chapter 6 Factors, Multiples, and Patterns 6.6 8
In figure 1, the number of squares is: 2
In figure 2, the number of squares is: 3
In figure 3, the number of squares is: 4
So,
Total number of figures = 3
So,
for this, when we divide any number by the total number of patterns, the value of the quotient will be a total repetition of the patterns and the remainder will be the counting of the pattern from the 1st figure.
So,
By using the partial quotients method,
24 ÷ 3 = ( 21 + 3 ) ÷ 3
= ( 21 ÷ 3 ) + ( 3 ÷ 3 )
= 7 + 1
= 8 R 0
Hence, from the above,
We can conclude that  in the 24th figure, the number of squares is: 4

Question 6.
Describe the pattern of the small triangles. How many small triangles are in the 10th figure?
Big Ideas Math Answers 4th Grade Chapter 6 Factors, Multiples, and Patterns 6.6 9
Answer: The small triangles in the 10th figure is: 1

Explanation:
Big Ideas Math Answers 4th Grade Chapter 6 Factors, Multiples, and Patterns 6.6 9
From the given figure,
In fig 1, the number of triangles is: 1
In fig 2, the number of triangles is: 2
In fig 3, the number of triangles is: 4
So,
Total number of figures = 3
So,
for this, when we divide any number by the total number of patterns, the value of the quotient will be a total repetition of the patterns and the remainder will be the counting of the pattern from the 1st figure.
So,
By using the partial quotients method,
10 ÷ 3 = ( 3 + 6 ) ÷ 9
= ( 3 ÷ 3 ) + ( 6 ÷ 3 )
= 1 + 2
= 3 R 1
Hence, from the above,
We can conclude that the 10th figure has 1 triangle

Question 7.
Structure
Make a shape pattern that uses twice as many squares as triangles.
Answer:

Question 8.
Number Sense
Which shape patterns have a heart as the 12th shape?
Big Ideas Math Answers 4th Grade Chapter 6 Factors, Multiples, and Patterns 6.6 10
Answer:
Let the patterns be named as A), B), C) and D)
So,
The shape patterns that have a heart as the 12th shape is: A), B) and D)

Explanation:
Big Ideas Math Answers 4th Grade Chapter 6 Factors, Multiples, and Patterns 6.6 10
Let the patterns be named as A), B), C) and D)
A) In pattern A,
The total number of figures = 2
So,
for this, when we divide any number by the total number of patterns, the value of the quotient will be a total repetition of the patterns and the remainder will be the counting of the pattern from the 1st figure.
So,
12 ÷ 2 = 6
Hence, A) will have a heart as the 12th shape
B) In pattern B,
The total number of figures = 4
So,
for this, when we divide any number by the total number of patterns, the value of the quotient will be a total repetition of the patterns and the remainder will be the counting of the pattern from the 1st figure.
So,
12 ÷ 4 = 3
Hence, B) will have a heart as the 12th shape
C) In pattern C,
The total number of figures = 2
So,
for this, when we divide any number by the total number of patterns, the value of the quotient will be a total repetition of the patterns and the remainder will be the counting of the pattern from the 1st figure.
So,
12 ÷ 2 = 6
But, the 12th shape in C) will be a circle
D) In pattern D,
The total number of figures = 3
So,
for this, when we divide any number by the total number of patterns, the value of the quotient will be a total repetition of the patterns and the remainder will be the counting of the pattern from the 1st figure.
So,
12 ÷ 3 = 4
Hence, C) will have a heart as the 12th shape

Think and Grow: Modeling Real Life

Example
You make a necklace with a cube, hexagon, and star beads. You string the beads in a pattern. You use the rule “cube, star, cube, hexagon.”It takes 64 beads to complete the necklace. How many times do you repeat the pattern?
Big Ideas Math Answers 4th Grade Chapter 6 Factors, Multiples, and Patterns 6.6 11
Divide the number of beads it takes to complete the necklace by the number of beads in the rule. There are 4 beads in the rule.
So,
4√64 = 64 ÷ 4
By using the Distributive Property of division,
64 ÷ 4 = ( 60 + 4 ) ÷ 4
= ( 60 ÷ 4 ) + ( 4 ÷ 4 )
= 15 + 1
= 16
Hence,
You repeat the pattern 16 times

Show and Grow

Question 9.
The path on a board game uses the rule “red, green, pink, yellow, blue.”There are 55 spaces on the game board. How many times does the pattern repeat?
Big Ideas Math Answers 4th Grade Chapter 6 Factors, Multiples, and Patterns 6.6 12
Answer: The number of times the pattern will repeat is: 11 times

Explanation:
Given that the path on a board game uses the rule ” red, green, pink, yellow, blue ”
It is also given that there are 55 spaces on the game board
So,
for this, when we divide any number by the total number of patterns, the value of the quotient will be a total repetition of the patterns and the remainder will be the counting of the pattern from the 1st figure.
So,
By using the Distributive Property of division,
55 ÷ 5 = ( 50 + 5 ) ÷ 5
= ( 50 ÷ 5 ) + ( 5 ÷ 5 )
= 10 + 1
= 11
Hence, from the above,
We can conclude that the number of times the patterns repeat is: 11 times

Question 10.
You make a walkway in a garden using different-shaped stepping stones. You use the rule “square, circle, square, hexagon.”You use 24 square stepping stones. How many circle and hexagon stepping stones do you use altogether? How many stones do you use in all?
Answer:
The Total number of stones used are: 24
The number of  circle and hexagon stepping stones use are: 12

Explanation:
Given that there is a walkway in a garden using different-shaped stepping stones.
The rule used here is: Square, circle, square, hexagon
It is also given that there are 24 stepping stones you used
So,
for this, when we divide any number by the total number of patterns, the value of the quotient will be a total repetition of the patterns and the remainder will be the counting of the pattern from the 1st figure.
So,
By using the Distributive Property of division,
24 ÷ 4 = ( 20 + 4 ) ÷ 4
= ( 20 ÷ 4 ) + ( 4 ÷ 4 )
= 5 + 1
= 6
So,
The number of stones used each time is: 6 times
So,
The number of  circle and hexagon stepping stones used altogether is: 6 + 6 = 12
Hence, from the above,
We can conclude that circle and hexagon stepping stones you used altogether is: 12

Question 11.
DIG DEEPER!
You make a rectangular picture frame using square tiles. The picture frame is 12 tiles long and 8 tiles wide. You arrange the tiles in a pattern. You use the rule “red, orange, yellow.” How many of each color tile do you use?
Answer: The number of each color tiles do you use is: 32

Explanation:
Given that you make a rectangular picture frame using square tiles and it is also given that the picture frame is 12 tiles long and 8 tiles wide.
So,
The area of a rectangular picture frame = 12 × 8 = 96 square- meters
The rule used is: red, orange, yellow
So,
The total number of colors is: 3
So,
The number of each color tile you used = Area of rectangular picture frame ÷ Total number of colors
= 96 ÷ 3
Now,
By using the Distributive Property of division,
96 ÷ 3 = ( 90 + 6 ) ÷ 3
= ( 90 ÷ 3 ) + ( 6 ÷ 3 )
= 30 + 2
= 32
Hence, from the above,
we can conclude that each number of color tile used is: 32

Shape Patterns Numbers Homework & Practice 6.6

Question 1.
Extend the pattern of shapes by repeating the rule “up, right, down, left.”What is the 48th shape in the pattern?
Big Ideas Math Answers 4th Grade Chapter 6 Factors, Multiples, and Patterns 6.6 13
Answer: The 48th shape in the pattern is: Left

Explanation:
Given pattern is:
Big Ideas Math Answers 4th Grade Chapter 6 Factors, Multiples, and Patterns 6.6 13
The rule for the given pattern is: up, right, down, left
So,
The total number of patterns is: 4
So,
for this, when we divide any number by the total number of patterns, the value of the quotient will be a total repetition of the patterns and the remainder will be the counting of the pattern from the 1st figure.
So,
By using the partial quotients method,
48 ÷ 4 = ( 40 + 8 ) ÷ 4
= ( 40 ÷ 4 ) + ( 8 ÷ 4 )
= 10 + 2
= 12 R 0
Hence, from the above
We can conclude that the 48th figure in the given pattern is: Left

Question 2.
Extend the pattern of shapes by repeating the rule “small circle, medium circle, large circle.”What is the 86th shape in the pattern?
Big Ideas Math Answers 4th Grade Chapter 6 Factors, Multiples, and Patterns 6.6 14
Answer: The 86th shape in the pattern is: Medium circle

Explanation:
Given pattern is:
Big Ideas Math Answers 4th Grade Chapter 6 Factors, Multiples, and Patterns 6.6 14
The rule for the pattern is: Small circle, Medium circle, large circle
So,
The total number of patterns = 3
So,
for this, when we divide any number by the total number of patterns, the value of the quotient will be a total repetition of the patterns and the remainder will be the counting of the pattern from the 1st figure.
So,
By using the partial quotients method,
86 ÷ 3 = ( 75 + 9 ) ÷ 3
= ( 75 ÷ 3 ) + ( 9 ÷ 3 )
= 25 + 3
= 28 R 2
Hence, from the above,
We can conclude that the 86th shape in the pattern is: Medium circle

Question 3.
Describe the dot pattern. How many dots are in the 113th figure?
Big Ideas Math Answers 4th Grade Chapter 6 Factors, Multiples, and Patterns 6.6 15
Answer: The number of dots in the 113th figure is: 4 dots

Explanation:
Given pattern is:
Big Ideas Math Answers 4th Grade Chapter 6 Factors, Multiples, and Patterns 6.6 15
So,
In fig 1, the number of dots = 2
In fig 2, the number of dots = 4
In fig 3, the number of dots = 6
So,
The total number of figures = 3
So,
for this, when we divide any number by the total number of patterns, the value of the quotient will be a total repetition of the patterns and the remainder will be the counting of the pattern from the 1st figure.
So,
By using the partial quotients method,
113 ÷ 3 = ( 99 + 12 ) ÷ 3
= ( 99 ÷ 3 ) + ( 12 ÷ 3 )
= 33 + 4
= 37 R 2
Hence, from the above,
we can conclude that the number of dots in the 113th figure is: 4 dots

Question 4.
YOU BE THE TEACHER
You and your friend each create a shape pattern with 100 shapes. Your friend says both patterns will have the same number of circles. Is your friend correct? Explain.
Big Ideas Math Answers 4th Grade Chapter 6 Factors, Multiples, and Patterns 6.6 16
Answer: Yes, both patterns will have the same number of circles.

Explanation:
Given patterns are:
Big Ideas Math Answers 4th Grade Chapter 6 Factors, Multiples, and Patterns 6.6 16
The pattern of the friend is: Star, circle
The pattern of you si: Heart, circle, pentagon, circle
It is also given that,
The total number of figures = 100
So,
The total number of figures for your friend = 2
The total number of figures for you = 4
So,
for this, when we divide any number by the total number of patterns, the value of the quotient will be a total repetition of the patterns and the remainder will be the counting of the pattern from the 1st figure.
So
For your friend,
By using the partial quotients method,
100 ÷ 2 = ( 50 + 50 ) ÷ 2
= ( 50 ÷ 2 ) + ( 50 ÷ 2 )
= 25 + 25
= 50
Hence,
The number of circles = 50
The number of stars = 50
For you,
By using the partial quotients method,
100 ÷ 4 = ( 80 + 20 ) ÷ 4
= ( 80 ÷ 4 ) + ( 20 ÷ 4 )
= 20 + 5
= 25
Hence,
The number of circles altogether = 25 + 25 = 50
Hence, from the above,
We can conclude that there are equal number of circles in both the patterns

Question 5.
Structure
Draw the missing figure in the pattern. Explain the pattern.
Big Ideas Math Answers 4th Grade Chapter 6 Factors, Multiples, and Patterns 6.6 17
Answer:

Question 6.
Reasoning
Newton uses the rule “bone, bone, paw print” to make a shape pattern. He wants the pattern to repeat 8 times. How many bones will be in Newton’s pattern?
Big Ideas Math Answers 4th Grade Chapter 6 Factors, Multiples, and Patterns 6.6 18
Answer: The number of bones in Newton’s pattern is: 16

Explanation:
Given pattern is:
Big Ideas Math Answers 4th Grade Chapter 6 Factors, Multiples, and Patterns 6.6 18
The rule followed in the pattern is: Bone, Bone, pawprint
So,
The total number of figures in the given pattern = 3
So,
for this, when we divide any number by the total number of patterns, the value of the quotient will be a total repetition of the patterns and the remainder will be the counting of the pattern from the 1st figure.
So,
It is also given that the pattern repeated 8 times.
So,
The total number of figures when the pattern repeated 8 times = 8 × 3 = 24 figures
So,
By using the partial quotients method,
24 ÷ 3 = ( 21 + 3 ) ÷ 3
= ( 21 ÷ 3 ) + ( 3 ÷ 3 )
= 7 + 1
= 8
So,
The number of each figure in a pattern = 8
From the pattern, we can say there are 2 bones
So,
The total number of bones when the pattern repeated 8 times = 2 × 8 = 16
Hence, from the above,
We can conclude that there are 16 bones in the pattern when the pattern repeated 8 times

Question 7.
Modeling Real Life
The black keys on a piano follow the pattern “two black keys, three black keys.” There are 36 black keys on a standard piano. How many times does this entire pattern repeat?
Answer: The entire pattern repeat 12 times

Explanation:
Given that the black keys on a pattern follow the pattern
The given pattern is: two black keys, three black keys
So,
The total number of patterns = 2
It is also given that there are 36 black keys on a standard piano.
So,
The total number of black keys on a piano = 36
So,
for this, when we divide any number by the total number of patterns, the value of the quotient will be a total repetition of the patterns and the remainder will be the counting of the pattern from the 1st figure.
So,
By using the partial quotients method,
36 ÷ 2 = ( 30 + 6 ) ÷ 2
= ( 30 ÷ 2 ) + ( 6 ÷ 2 )
= 15 + 3
= 18
Hence, from the above,
We can conclude that the entire pattern will repeat 18 times.

Review & Refresh

Find the quotient.
Question 8.
30 ÷ 5 = _____
Answer: 30 ÷ 5 = 6

Explanation:
By using the partial products method,
30 ÷ 5 = ( 25 + 5 ) ÷ 5
= ( 25 ÷ 5 ) + ( 5 ÷ 5 )
= 5 + 1
= 6
Hence, 30 ÷ 5 = 6

Question 9.
360 ÷ 9 = ______
Answer: 360 ÷ 9 = 40

Explanation:
By using the partial quotients method,
360 ÷ 9 = ( 270 + 90 ) ÷ 9
= ( 270 ÷ 9 ) + ( 90 ÷ 9 )
= 30 + 10
= 40
Hence, 360 ÷ 9 = 40

Question 10.
6,400 ÷ 8 = _____
Answer: 6,400 ÷ 8 = 800

Explanation:
By using the place -value method,
6,400 ÷ 8 = 64 hundreds ÷ 8
= 8 hundred
= 8 × 100
= 800
Hence, 6,400 ÷ 8 = 800

Question 11.
140 ÷ 2 = _____
Answer: 140 ÷ 2 = 70

Explanation:
By using the place-value method,
140 ÷ 2 = 14 tens ÷ 2
= 7 tens
= 7 × 10
= 70
Hence, 140 ÷ 2 = 70

Question 12.
4,200 ÷ 7 = _____
Answer: 4,200 ÷ 7 = 600

Explanation:
By using the place-value method,
4,200 ÷ 7 = 42 hundreds ÷ 7
= 6 hundred
= 6 × 100
= 600
Hence, 4,200 ÷ 7 = 600

Question 13.
40 ÷ 2 = _____
Answer: 40 ÷ 2 = 20

Explanation:
By using the place-value method,
40 ÷ 2 = 4 tens ÷ 2
= 2 tens
= 2 × 10
= 20
Hence, 40 ÷ 2 = 20

Factors, Multiples, and Patterns Performance Task

You play basketball in a youth basketball program.
Big Ideas Math Answers Grade 4 Chapter 6 Factors, Multiples, and Patterns 1
Question 1.
There are 72 players in the program. Each team needs an equal number of players and must have at least 5 players. What are two different ways the teams can be made?
Answer: The different ways the teams can be made and to have at least 5 members are: 8 × 9 and 9 × 8

Explanation:
Given that there are 72 players in the program and each team needs an equal number of players.
It is also given that each team must have at least 5 players.
Now,
To find the different ways the teams can be made, we have to find the factors of 72
Now,
The factors of 72 are: 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72
Now, to make an equal number of players and to have at least 5 players,
The different ways the teams can be made is: 8 × 9, 9 × 8
Hence, from the above,
We can conclude that the 2 different ways the teams can be made are: 8 × 9 and 9 × 8

Question 2.
The width of a basketball court is 42 feet and the length is 74 feet. You run around the perimeter of the court 4 times to warm up. How many feet do you run?
Big Ideas Math Answers Grade 4 Chapter 6 Factors, Multiples, and Patterns 2
Answer: The number of feet you run is: 464 feet

Explanation:
Given that the width of a basketball court is 42 feet high and the length is 74 feet
So,
The perimeter of the basketball court = 42 + 74 = 116 feet
It is also given that you run around the perimeter of the basketball court 4 times to warm up.
So,
The number of feet you run to warm up = The perimeter of the basketball court × 4
= 116 × 4
Now,
By using the partial products method,
116 × 4 = ( 100 + 16 ) × 4
= ( 100 × 4 ) + ( 16 × 4 )
= 400 + 64
= 464 feet
Hence, from the above,
We can conclude that you will run up 464 feet to warm up your body.

Question 3.
You and your friend are on the same team. You played your first game last week.
a. Your team scored 4 more points than the other team. The total number of points scored by both teams was 58. How many points did your team score?
b. You and your friend scored the same number of points. You made 2-point shots and your friend made 3-point shots. What could be the greatest number of points you and your friend each scored?
Answer:

Question 4.
Your team uses the pattern below to decide which jersey color to wear to each game. Which color jersey will your team wear on the 20th game?
Big Ideas Math Answers Grade 4 Chapter 6 Factors, Multiples, and Patterns 3
Answer: The color of jersey your team will wear on the 20th game is: Game 4

Explanation:
Given that your team uses the following pattern:
Big Ideas Math Answers Grade 4 Chapter 6 Factors, Multiples, and Patterns 3
From the above pattern,
The total number of shirts is: 4
So,
for this, when we divide any number by the total number of patterns, the value of the quotient will be a total repetition of the patterns and the remainder will be the counting of the pattern from the 1st figure.
So,
By using the partial quotients method,
20 ÷ 4 = 5 R 0
Hence, from the above,
We can conclude that the color of the shirt your team will wear on the 20th game is: Game 4

Factors, Multiples, and Patterns Activity

Multiple Lineup
Directions:
1. Players take turns rolling a die.
2. On your turn, place a counter on a multiple of the number of your roll. If there is not a multiple of the number of your roll, you lose your turn.
3. The first player to create a line of 5 in a row, horizontally, vertically, or diagonally, wins!
Big Ideas Math Answers Grade 4 Chapter 6 Factors, Multiples, and Patterns 4
Answer:

Factors, Multiples, and Patterns Chapter Practice

6.1 Understand Factors

Question 1.
Use the rectangles to find the factor pairs for 6.
Big Ideas Math Answers Grade 4 Chapter 6 Factors, Multiples, and Patterns chp 1
Answer: The factor pairs of 6 are: 1 and 6, 2 and 3

Explanation:

The factor pairs are nothing but the side lengths of a rectangle and the area of a rectangle gives the factor
Hence,
The factor pairs of 6 are: 1 and 6, 2 and 3

Question 2.
Draw rectangles to find the factor pairs for 12.
Big Ideas Math Answers Grade 4 Chapter 6 Factors, Multiples, and Patterns chp 2
Answer: The factor pairs of 12 are: 1 and 12, 2 and 6, 3 and 4

Explanation:

The factor pairs are nothing but the side lengths of a rectangle and the area of a rectangle gives the factor
Hence,
The factor pairs of 12 are: 1 and 12, 2 and 6, 3 and 4

Find the factor pairs for the number.
Question 3.
17
Answer: The factor pairs of 17 are: 1 and 17

Explanation:
Factors are the numbers that divide the original completely.
Hence,
The factor pairs of 17 are: 1 × 17

Question 4.
10
Answer: The factor pairs of 10 are: 1 and 10, 2 and 5

Explanation:
Factors are the numbers that divide the original number completely.
Hence,
The factor pairs of 10 are: 1 × 10, 2 × 5

Question 5.
21
Answer: The factor pairs of 21 are: 1 and 21, 3 and 7

Explanation:
Factors are the numbers that divide the original number completely.
Hence,
The factor pairs of 21 are: 1 × 21, 3 × 7

Question 6.
20
Answer: The factor pairs of 20 are: 1 and 20, 2 and 10, 4 and 5

Explanation:
Factors are the numbers that divide the original number completely.
Hence,
The factor pairs of 20 are: 1 × 20, 2 × 10, 5 × 4

Question 7.
36
Answer: The factor pairs of 36 are: 1 and 36, 2 and 18, 3 and 12, 4 and 9, 6 and 6

Explanation:
Factors are the numbers that divide the original number completely.
Hence,
The factor pairs of 36 are: 1 × 36, 2 × 18, 3 × 12, 4 × 9, 6 × 6

Question 8.
50
Answer: The factor pairs of 50 are: 1 and 50, 2 and 25, 5 and 10

Explanation:
factors are the numbers that divide the original number completely.
Hence,
The factor pairs of 50 are: 1 × 50, 2 × 25, 5 × 10

6.2 Factors and Divisibility

Find the factor pairs for the number.
Question 9.
16
Answer: The factor pairs of 16 are: 1 and 16, 2 and 8, 4 and 4

Explanation:
Factors are the numbers that divide the original number completely.
Hence,
The factor pairs of 16 are: 1 × 16, 2 × 8, 4 × 4

Question 10.
24
Answer: The factor pairs of 24 are: 1 and 24, 2 and 12, 3 and 8, 4 and 6

Explanation:
factors are the numbers that divide the original number completely.
Hence,
The factor pairs of 24 are: 1 × 24, 2 × 12, 3 × 8, 4 × 6

Question 11.
56
Answer: The factor pairs of 56 are: 1 and 56, 2 and 28, 4 and 14, 7 and 8

Explanation:
Factors are the numbers that divide the original number completely.
Hence,
The factor pairs of 56 are: 1 × 56, 2 × 28, 4 × 14, 7 × 8

List the factors of the number.
Question 12.
25
Answer: The factors of 25 are: 1, 5, 25

Explanation:
Factors are the numbers that divide the original number completely.
Hence,
The factors of 25 are: 1, 5, 25

Question 13.
60
Answer: The factors of 60 are: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60

Explanation:
Factors are the numbers that divide the original number completely.
Hence,
The factors of 60 are: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60

Question 14.
72
Answer: The factors of 72 are: 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72

Explanation:
Factors are the numbers that divide the original number completely.
Hence,
The factors of 72 are: 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72

Question 15.
Number Sense
Which numbers have 3 as a factor?
Big Ideas Math Answers Grade 4 Chapter 6 Factors, Multiples, and Patterns chp 15
Answer: The numbers which have 3 as a factor are: 21, 36, 48, 93

Explanation:
Given numbers are: 56, 21, 3, 48, 93, 71
For the given number to have 3 as a factor, the sum of the digits of the given number must be a multiple of 3
So,
The sum of digits of:
56: 5 + 6 = 11: 1 + 1 = 2
21: 2 + 1 = 3
36: 3 + 6 = 9
48: 4 + 8 = 12: 1 + 2 = 3
93: 9 + 3 = 12: 1 + 2 = 3
71: 7 + 1 = 8
Hence, from the above,
we can conclude that the numbers which have 3 as a factor are: 21, 36, 48, 93

6.3 Relate Factors and Multiples

Question 16.
Is 54 a multiple of 3? Explain.
Answer: 54 is a multiple of 3

Explanation:
Multiples are the numbers that can be divided by another number a certain number of times without any remainder.
The multiples of 3 are: 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33
To be a multiple of 3, the sum of the digits of the given number should also be a multiple of 3.
So,
54: 5 + 4 = 9 ( Divisible by 3 )
Hence, from the above,
we can conclude that 54 is a multiple of 3

Question 17.
Is 45 a multiple of 7? Explain.
Answer: 45 is not a multiple of 7

Explanation:
Multiples are the numbers that can be divided by another number a certain number of times without any remainder.
The multiples of 7 are: 7, 14, 21, 28, 35, 42, 49, 56, 63, 70
Hence, from the above,
we can conclude that 45 is not a multiple of 7

Question 18.
Is 2 a factor of 97? Explain.
Answer: 2 is not a factor of 97

Explanation:
Factors are the numbers that divide the original number completely.
The factors of 97 are: 1, 97
Hence, from the above,
We can conclude that 2 is not a factor of 97

Question 19.
Is 5 a factor of 60? Explain.
Answer: 5 is a factor of 60

Explanation:
Factors are the numbers that divide the original number completely.
The factors of 60 are: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60
Hence, from the above,
We can conclude that 5 is a factor of 60

Tell whether 20 is a multiple or a factor of the number. Write multiple, factor, or both.
Question 20.
60
Answer: 20 is a factor of 60

Explanation:
Factors are the numbers that divide the original number completely.
The factors of 60 are: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60
Hence, from the above,
We can conclude that 20 is a factor of 60

Question 21.
4
Answer: 20 is a multiple of 4

Explanation:
Multiples are the numbers that can be divided by another number a certain number of times without any remainder.
The multiples of 4 are: 4, 8, 12, 16, 20, 24, 28, 32, 36, 40
Hence, from the above,
We can conclude that 20 is a multiple of 4

Question 22.
20
Answer: 20 is a multiple and factor of 20

Explanation:
Multiples are the numbers that can be divided by another number a certain number of times without any remainder.
Factors are the numbers that divide the original number completely
Now,
Factors of 20 are: 1, 2, 4, 5, 10, 20
Multiples of 20 are: 20, 40, 60, 80, 100
Hence, from the above,
we can conclude that 20 is a multiple and factor of 20

Question 23.
Number Sense
Name two numbers that are each a multiple of both 5 and 2. What do you notice about the two multiples?
Answer: The 2 numbers that are each a multiple of both 5 and 2 are: 10 and 20

Explanation;
Multiples of 2 are: 2, 4, 6, 8, 10, 12, 14, 16, 18, 20
Multiples of 5 are: 5, 10, 15, 20, 25, 30, 35, 40, 45, 50
So,
From the multiples of both 2 and 5,
The numbers which have common multiples of both 5 and 2 are: 10, 20
The multiples of 2 have the one’s digit as 2 or 4 or 6 or 8 or 0
The multiples of 5 have the one’s digit as 5 or 0

Question 24.
Logic
A quotient is a multiple of 5. The dividend is a multiple of 4. The divisor is a factor of 8. Write one possible equation for the problem.
Answer: The possible equation is:
80 ÷ 8 = 10

Explanation:
Given that,
Quotient: Multiple of 5
Dividend: Multiple of 4
Divisor: Factor of 8
Hence, from the above,
we can conclude that the possible equation for the problem is:
80 ÷ 8 = 10

6.4 Identify Prime and Composite Numbers

Tell whether the number is prime or composite. Explain.
Question 25.
5
Answer: 5 is a prime number

Explanation;
Prime numbers:
The numbers which have exactly 2 factors 1 and itself are “Prime numbers”
Now,
The factors of 5 are: 1, 5
Hence, from the above,
We can conclude that 5 is a prime number

Question 26.
25
Answer: 25 is a composite number

Explanation:
Composite numbers:
The numbers which have more than 2 factors are “Composite numbers”
Now,
The factors of 25 are: 1, 5, 25
Hence, from the above,
We can conclude that 25 is a composite number

Question 27.
51
Answer: 51 is a composite number

Explanation:
Composite numbers:
The numbers which have more than 2 factors are “Composite numbers”
Now,
The factors of 51 are: 1, 3, 17, 51
Hence, from the above,
We can conclude that 51 is a composite number

Question 28.
21
Answer: 21 is a composite number

Explanation:
Composite numbers:
The numbers which have more than 2 factors are “Composite numbers”
Now,
The factors of 21 are: 1, 3, 7, 21
Hence, from the above,
We can conclude that 21 is a composite number

Question 29.
50
Answer: 50 is a composite number

Explanation:
Composite numbers:
The numbers which have more than 2 factors are “Composite numbers”
Now,
The factors of 50 are: 1, 2, 5, 10, 25, 50
Hence, from the above,
We can conclude that 50 is a composite number

Question 30.
83
Answer: 83 is a prime number

Explanation:
Prime numbers:
The numbers which have exactly 2 factors 1 and itself are “Prime numbers”
Now,
The factors of 83 are: 1, 83
Hence, from the above,
We can conclude that 83 is a prime number

Question 31.
Modeling Real Life
A prime number of students have which type of fingerprint?
Big Ideas Math Answers Grade 4 Chapter 6 Factors, Multiples, and Patterns chp 31

Answer: A prime number of students have Whorl fingerprint

Explanation:
Big Ideas Math Answers Grade 4 Chapter 6 Factors, Multiples, and Patterns chp 31
From the given figure,
Each half-circle = 1 student
The number of fingerprints of loop-type fingerprint: 9
The number of fingerprints of Arch-type fingerprint: 12
The number of fingerprints on Whorl-type fingerprint: 7
Hence, from the above,
We can conclude that the prime number of students has Whorl-type fingerprint

6.5 Number Patterns

Write the first six numbers in the pattern. Then describe another feature of the pattern.
Question 32.
Rule: Subtract 11.
First number: 99
99, ___, _____, _____, _____, _____
Answer:
The first 6 numbers in the given pattern are: 99, 88, 77, 66, 55, 44

Explanation:
For the formation of the pattern,
the given rules are:
Rule: Subtract 11
First number: 99
Hence,
The pattern we will obtain is:

Question 33.
Rule: Multiply by 5.
First number: 10
10, _____, _____, ____, _____, _____
Answer:
The first 6 numbers of the given pattern is: 10, 50, 250, 1,250, 6,250, 31,250

Explanation:
For the formation of the pattern,
the given rules are:
Rule: Multiply by 5
First number: 10
Hence,
The pattern we will obtain is:

Question 34.
Rule: Add 8.
First number: 15
Answer:
The first 6 numbers of the given pattern are: 15, 23, 31, 39, 47, 55

Explanation:
For the formation of the pattern,
the rules are:
Rule: Add 8
First number: 15
Hence,
The pattern we will obtain is:

Question 35.
Rule: Divide by 4.
First number: 4,096
Answer:
The first 6 numbers in the given pattern is: 4,096, 1,024, 256, 64, 16, 4

Explanation:
For the formation of the pattern,
the given rules are:
Rule: Divide by  4
First number: 4,096
Hence,
The pattern we will obtain is:

Open-Ended
Use the rule to generate a pattern of four numbers.
Question 36.
Rule: Divide by 2.
Answer:
Let the first number be 16
Hence,
The first 4 numbers of the given pattern are: 16, 8, 4, 2

Explanation:
For the formation of the pattern,
the given rules are:
Rule: Divide by 2
Let the first number be 16
Hence,
The pattern we will obtain is:

Question 37.
Rule: Add 3.
Answer:
Let the first number be 3
Hence,
The first four numbers of the given pattern are: 3, 6, 9, 12

Explanation:
For the formation of the pattern,
the given rules are:
Rule: Add 3
Let the first number be 3
Hence,
The pattern we will obtain is:

Question 38.
Rule: Multiply by10.
Answer:
Let the first number be 10
Hence,
The first  4 numbers of the given pattern are: 10, 100, 1,000, 10,000

Explanation:
For the formation of the pattern,
the given rules are:
Rule: Multiply by 10
Let the first number be 10
Hence,
The pattern we will obtain is:

Question 39.
Rule: Subtract 6.
Answer:
Let the first number be 36
Hence,
The first 4 numbers of the given pattern are: 36, 30, 24, 18

Explanation:
For the formation of the pattern,
the given rules are:
Rule: Subtract 6
Let the first number be 36
Hence,
The pattern we will obtain is:

6.6 Shape Patterns

Question 40.
Extend the pattern of shapes by repeating the rule “trapezoid, circle.” What is the 57th shape in the pattern?
Big Ideas Math Answers Grade 4 Chapter 6 Factors, Multiples, and Patterns chp 40
Answer: The 57th shape in the pattern is: Trapezoid

Explanation:
Given pattern is:
Big Ideas Math Answers Grade 4 Chapter 6 Factors, Multiples, and Patterns chp 40
The rule given for the pattern is: trapezoid, circle
So,
The total number of figures in the given pattern is: 2
So,
for this, when we divide any number by the total number of patterns, the value of the quotient will be a total repetition of the patterns and the remainder will be the counting of the pattern from the 1st figure.
So,
By using the partial quotient method,
57 ÷ 2 = ( 40 + 16 ) ÷ 2
= ( 40 ÷ 2 ) + ( 16 ÷ 2 )
= 20 + 8
= 28 R 1
Hence, from the above,
we can conclude that the 57th shape in the pattern is: Trapezoid

Question 41.
Extend the pattern of shapes by repeating the rule “top left, top right, bottom right, bottom left.” What is the 102nd shape in the pattern?
Big Ideas Math Answers Grade 4 Chapter 6 Factors, Multiples, and Patterns chp 41
Answer: The 102nd shape in the pattern is: Bottom left

Explanation:
The given pattern is:
Big Ideas Math Answers Grade 4 Chapter 6 Factors, Multiples, and Patterns chp 41
The rule given for the pattern is: top left, top right, bottom right, bottom left
So,
The total number of figures in the given pattern is: 4
So,
for this, when we divide any number by the total number of patterns, the value of the quotient will be a total repetition of the patterns and the remainder will be the counting of the pattern from the 1st figure.
So,
By using the partial quotients method,
102 ÷ 3 = ( 99 + 3 ) ÷ 3
= ( 99 ÷ 3 ) + ( 3 ÷ 3 )
= 33 + 1
= 34 R 0
Hence, from the above,
We can conclude that the 102nd shape in the pattern is: Bottom left

Question 42.
Describe the pattern. How many squares are in the 61st figure?
Big Ideas Math Answers Grade 4 Chapter 6 Factors, Multiples, and Patterns chp 42
Answer: The number of squares in the 61st figure is: 3

Explanation:
The given pattern is:
Big Ideas Math Answers Grade 4 Chapter 6 Factors, Multiples, and Patterns chp 42
In figure 1, the total number of squares = 3
In figure 2, the total number of squares = 6
In figure 3, the total number of squares = 9
So,
The total number of figures = 3
So,
for this, when we divide any number by the total number of patterns, the value of the quotient will be a total repetition of the patterns and the remainder will be the counting of the pattern from the 1st figure.
So,
By using the partial quotients method,
61 ÷ 3 = ( 57 + 3 ) ÷ 3
= ( 57 ÷ 3 ) + ( 3 ÷ 3 )
= 19 + 1
= 20 R 1
Hence, from the above,
We can conclude that the number of squares in the  61st figure in the pattern is: 4

Question 43.
Structure
Draw the missing figure in the pattern. Explain the pattern.
Big Ideas Math Answers Grade 4 Chapter 6 Factors, Multiples, and Patterns chp 43
Answer: The missing figure has: 7 squares

Explanation:
The given pattern is:
Big Ideas Math Answers Grade 4 Chapter 6 Factors, Multiples, and Patterns chp 43
The rule-following in the given pattern is: Add 2
The first number: 1
According to the rule,
The number of squares in the pattern will be like 1, 3, 5, 7, 9

Conclusion:

We hope the details mentioned in the Big Ideas Math Answers Grade 4 Chapter 6 Factors, Multiples, and Patterns are helpful for you to finish your homework in time. Learn the concepts in depth so that you can solve any kind of problem easily. If you have any queries you can post your comments in the below section.

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Solving Linear Equations Maintaining Mathematical Proficiency

Add or subtract.

Question 1.
-5 + (-2)
Answer:
-5 + (-2 ) = -7

Explanation:
We know that,
A) – × – = +
B) + × – = –
C) + × + = +
D) – × + = –
We know that,
The result of any mathematical operation follows the sign of the big number and if the two numbers have the same sign, then the result of that also operation also follows the same sign and if there is no sign before the number, then we can take that number as positive number i.e., +a
So,
-5 + ( -2 ) = -5 – 2
= – ( 5 + 2 )
= -7
Hence, from the above,
We can conclude that,
-5 + ( -2 ) = -7

Question 2.
0 + (-13)
Answer:
0 + -13 = -13

Explanation:
We know that,
A) – × – = +
B) + × – = –
C) + × + = +
D) – × + = –
We know that,
The result of any mathematical operation follows the sign of the big number and if the two numbers have the same sign, then the result of that also operation also follows the same sign and if there is no sign before the number, then we can take that number as positive number i.e., +a
So,
0 + ( -13 ) = 0 – 13
= -13
Hence, from the above,
We can conclude that,
0 + ( -13 ) = -13

Question 3.
-6 + 14
Answer:
-6 + 14 = 8

Explanation:
We know that,
A) – × – = +
B) + × – = –
C) + × + = +
D) – × + = –
We know that,
The result of any mathematical operation follows the sign of the big number and if the two numbers have the same sign, then the result of that also operation also follows the same sign and if there is no sign before the number, then we can take that number as positive number i.e., +a
So,
-6 + 14 = +14 – 6
= +8
= 8
Hence, from the above,
We can conclude that
-6 +14 = 8

Question 4.
19 – (-13)
Answer:
19 – ( -13 ) = 32

Explanation:
We know that,
A) – × – = +
B) + × – = –
C) + × + = +
D) – × + = –
We know that,
The result of any mathematical operation follows the sign of the big number and if the two numbers have the same sign, then the result of that also operation also follows the same sign and if there is no sign before the number, then we can take that number as positive number i.e., +a
So,
19 – ( -13 ) = 10 + 9 + 10 + 3
= 20 + 12
= 32
Hence, from the above,
We can conclude that,
19 – (-13 ) = 32

Question 5.
-1 – 6
Answer:
-1 -6 = -7

Explanation:
We know that,
A) – × – = +
B) + × – = –
C) + × + = +
D) – × + = –
We know that,
The result of any mathematical operation follows the sign of the big number and if the two numbers have the same sign, then the result of that also operation also follows the same sign and if there is no sign before the number, then we can take that number as positive number i.e., +a
So,
-1 – 6 = – ( 1 + 6 )
= -7
Hence, from the above,
We can conclude that
-1 -6 = -7

Question 6.
– 5 – (-7)
Answer:
-5 – ( -7 ) = 2

Explanation:
We know that,
A) – × – = +
B) + × – = –
C) + × + = +
D) – × + = –
We know that,
The result of any mathematical operation follows the sign of the big number and if the two numbers have the same sign, then the result of that also operation also follows the same sign and if there is no sign before the number, then we can take that number as positive number i.e., +a
So,
-5 – ( -7 ) = -5 + 7
= 7 – 5
= 2
Hence, from the above,
We can conclude that
-5 – ( -7 ) = 2

Question 7.
17 + 5
Answer:
17 + 5 = 22

Explanation:
We know that,
A) – × – = +
B) + × – = –
C) + × + = +
D) – × + = –
We know that,
The result of any mathematical operation follows the sign of the big number and if the two numbers have the same sign, then the result of that also operation also follows the same sign and if there is no sign before the number, then we can take that number as positive number i.e., +a
So,
17 + 5 = 15 + 2 + 5
= 22
Hence, from the above,
We can conclude that
17 + 5 = 22

Question 8.
8 + (-3)
Answer:
8 + ( -3 ) = 5

Explanation:
We know that,
A) – × – = +
B) + × – = –
C) + × + = +
D) – × + = –
We know that,
The result of any mathematical operation follows the sign of the big number and if the two numbers have the same sign, then the result of that also operation also follows the same sign and if there is no sign before the number, then we can take that number as positive number i.e., +a
So,
8 + ( -3 ) = 8 – 3
= 5
Hence, from the above,
We can conclude that
8  + ( -3 ) = 5

Question 9.
11 – 15
Answer:
11 – 15 = -4

Explanation:
We know that,
A) – × – = +
B) + × – = –
C) + × + = +
D) – × + = –
We know that,
The result of any mathematical operation follows the sign of the big number and if the two numbers have the same sign, then the result of that also operation also follows the same sign and if there is no sign before the number, then we can take that number as positive number i.e., +a
So,
11 – 15 = -15 + 11
= – ( 15 – 11 )
= -4
Hence, from the above,
We can conclude that,
11 – 15 = -4

Multiply or divide.

Question 10.
-3(8)
Answer:
-3(8) = -3 × 8
= -24

Explanation:
We know that,
A) – × – = +
B) + × – = –
C) + × + = +
D) – × + = –
We know that,
The result of any mathematical operation follows the sign of the big number and if the two numbers have the same sign, then the result of that also operation also follows the same sign and if there is no sign before the number, then we can take that number as positive number i.e., +a
So,
-3(8) = -3 × +8
= -24
Hence, from the above,
We can conclude that
-3 ( 8 ) = -24

Question 11.
-7 • (-9)
Answer:
-7 . ( -9 ) = -7 × +9
= -63

Explanation:
We know that,
A) – × – = +
B) + × – = –
C) + × + = +
D) – × + = –
We know that,
The result of any mathematical operation follows the sign of the big number and if the two numbers have the same sign, then the result of that also operation also follows the same sign and if there is no sign before the number, then we can take that number as positive number i.e., +a
So,
-7 . ( 9 ) = -7 × +9
= -63
Hence, from the above,
We can conclude that
-7 . ( 9 ) = -63

Question 12.
4 • (-7)
Answer:
4 . ( -7 ) = 4 × ( -7 )
= -28

Explanation:
We know that,
A) – × – = +
B) + × – = –
C) + × + = +
D) – × + = –
We know that,
The result of any mathematical operation follows the sign of the big number and if the two numbers have the same sign, then the result of that also operation also follows the same sign and if there is no sign before the number, then we can take that number as positive number i.e., +a
So,
4. (-7 ) = +4 × -7
= -28
Hence, from the above,
We can conclude that
4 . ( -7 ) = -28

Question 13.
-24 ÷ (-6)
Answer:
-24 ÷ ( -6 ) = 4

Explanation:
We know that,
A) – ÷ – = +
B) + ÷ – = –
C) + ÷ + = +
D) – ÷ + = –
We know that,
The result of any mathematical operation follows the sign of the big number and if the two numbers have the same sign, then the result of that also operation also follows the same sign and if there is no sign before the number, then we can take that number as positive number i.e., +a
So,
-24 ÷ ( -6 ) = + ( 24 ÷ 6 )
= 4
Hence, from the above,
We can conclude that
-24 ÷ ( -6 ) = 4

Question 14.
-16 ÷ 2
Answer:
-16 ÷ 2 = -8

Explanation:
We know that,
A) – ÷ – = +
B) + ÷ – = –
C) + ÷ + = +
D) – ÷ + = –
We know that,
The result of any mathematical operation follows the sign of the big number and if the two numbers have the same sign, then the result of that also operation also follows the same sign and if there is no sign before the number, then we can take that number as positive number i.e., +a
So,
-16 ÷ 2 = -8
Hence, from the above,
We can conclude that
-16 ÷ 2 = -8

Question 15.
12 ÷ (-3)
Answer:
12 ÷ ( -3 ) = -4

Explanation:
We know that,
A) – ÷ – = +
B) + ÷ – = –
C) + ÷ + = +
D) – ÷ + = –
We know that,
The result of any mathematical operation follows the sign of the big number and if the two numbers have the same sign, then the result of that also operation also follows the same sign and if there is no sign before the number, then we can take that number as positive number i.e., +a
So,
12 ÷ ( -3 )
= 12 ÷ -3
= -4
Hence, from the above,
We can conclude that
12 ÷ ( -3 ) = -4

Question 16.
6 • 8
Answer:
6 . 8 = 6 × 8
= 48

Explanation:
We know that,
A) – × – = +
B) + × – = –
C) + × + = +
D) – × + = –
We know that,
The result of any mathematical operation follows the sign of the big number and if the two numbers have the same sign, then the result of that also operation also follows the same sign and if there is no sign before the number, then we can take that number as positive number i.e., +a
So,
6 . 8 = 6 × 8
= 48
Hence, from the above,
We can conclude that
6 . 8 = 48

Question 17.
36 ÷ 6
Answer:
36 ÷ 6 = 6

Explanation:
We know that,
A) – ÷ – = +
B) + ÷ – = –
C) + ÷ + = +
D) – ÷ + = –
We know that,
The result of any mathematical operation follows the sign of the big number and if the two numbers have the same sign, then the result of that also operation also follows the same sign and if there is no sign before the number, then we can take that number as positive number i.e., +a
So,
36 ÷ 6 = ( 30 + 6 ) ÷ 6
= ( 30 ÷ 6 ) + ( 6 ÷ 6 )
= 5 + 1
= 6
Hence, from the above,
We can conclude that
36 ÷ 6 = 6

Question 18.
-3(-4)
Answer:
-3 ( -4 ) = -3 × -4
= -12

Explanation:
We know that,
A) – × – = +
B) + × – = –
C) + × + = +
D) – × + = –
We know that,
The result of any mathematical operation follows the sign of the big number and if the two numbers have the same sign, then the result of that also operation also follows the same sign and if there is no sign before the number, then we can take that number as positive number i.e., +a
So,
-3 ( -4 ) = -3 × -4
= +12
Hence, from the above,
We can conclude that
-3 ( – 4 ) = 12

Question 19.
ABSTRACT REASONING
Summarize the rules for
(a) adding integers,
(b) subtracting integers,
(c) multiplying integers, and
(d) dividing integers.
Give an example of each.
Answer:
a) Adding integers:
We know that,
The result of any mathematical operation i.e., Addition or subtraction follows the sign of the big number and if the two numbers have the same sign, then the result of that also operation also follows the same sign and if there is no sign before the number, then we can take that number as positive number i.e., +a
Example:
Let take the two numbers -2 and 3
So,
The addition of -2 and 3 is:
-2 + 3 = +1
= 1 ( Since the big number has a positive sign )

b) Subtracting integers:
We know that,
The result of any mathematical operation i.e., Addition or subtraction follows the sign of the big number and if the two numbers have the same sign, then the result of that also operation also follows the same sign and if there is no sign before the number, then we can take that number as positive number i.e., +a
Example:
Let take the two numbers -3 and +8
So,
The subtraction of -3 and +8 is:
-3 – ( +8 ) = -3 – 8
= – ( 3 + 8 )
= -11 ( Since both the numbers have a negative sign )

c) Multiplying integers:
We know that,
A) – × – = +
B) + × – = –
C) + × + = +
D) – × + = –
Example:
Let take the two numbers +8 and -3
So,
The multiplication of +8 and -3 is:
+8 ( -3 ) = 8 × -3
= -24 ( Since + × – = – )

d) Dividing integers:
We know that,
A) – ÷ – = +
B) + ÷ – = –
C) + ÷ + = +
D) – ÷ + = –
Example:
Let take the two numbers -12 and -2
So,
The division of -12 and -2 is:
-12 ÷ ( -2 ) = -12 ÷ -2
= ( -10 + -2 ) ÷ -2
= ( -10 ÷ -2 ) + ( -2 ÷ -2 )
= 5 + 1
= 6
Hence, from the above,
We can conclude all the rules for the four basic mathematical operations.

Solving Linear Equations Monitoring Progress

Solve the problem and specify the units of measure.

Question 1.
The population of the United States was about 280 million in 2000 and about 310 million in 2010. What was the annual rate of change in population from 2000 to 2010?
Answer:
The annual rate of change in population from 2000 to 2010 is: 30 million

Explanation:
It is given that the population of the United States was about 280 million in 2000 and about 310 million in 2010.
So,
The annual rate of change in population from 2000 to 2010 = ( The population of United States in 2010 ) – ( The population of United States in 2000 )
= 310 – 280
= 30 million
Hence, from the above,
We can conclude that the annual rate of change in population from 2000 to 2010 is: 30 million

Question 2.
You drive 240 miles and use 8 gallons of gasoline. What was your car’s gas mileage (in miles per gallon)?
Answer:
Your car’s gas mileage ( in miles per gallon ) is: 30

Explanation:
It is given that you drive 240 miles and use 8 gallons of gasoline.
So,
The mileage of your car = ( The total number of miles driven by your car ) ÷ ( The number of gallons of gasoline used by your car )
= 240 ÷ 8
= ( 160 + 80 ) ÷ 8
= ( 160 ÷ 8 ) + ( 80 ÷ 8 )
= 20 + 10
= 30 miles
Hence, from the above,
We can conclude that the mileage of your car is: 30 miles per gallon

Question 3.
A bathtub is in the shape of a rectangular prism. Its dimensions are 5 feet by 3 feet by 18 inches. The bathtub is three-fourths full of water and drains at a rate of 1 cubic foot per minute. About how long does it take for all the water to drain?
Answer:
The total time taken for the water to drain is: 2,430 minutes

Explanation:
It is given that a bathtub is in the shape of a rectangular prism. Its dimensions are 5 feet by 3 feet by 18 inches. The bathtub is three-fourths full of water and drains at a rate of 1 cubic foot per minute.
So,
The volume of the rectangular prism = The dimensions of the rectangular prism
= 5 × 3 × 18 × 12
= 3,240 cubic feet
Now,
The volume of the bathtub which is three-fourths full of water = \(\) {3}{4}[\latex] × 3240
= 2,430 cubic feet
It is also given that that the bathtub drains at a rate of 1 cubic foot per minute.
So,
The time taken to drain 2,430 cubic feet of water in minutes = 2,430 × 1
= 2,430 minutes
Hence, from the above,
We can conclude that the time taken for the water to drain from the bathtub at a rate of 1 cubic foot per minute is: 2,430 minutes

Lesson 1.1 Solving Simple Equations

Essential Question
How can you use simple equations to solve real-life problems?

Exploration 1
Measuring Angles

Work with a partner. Use a protractor to measure the angles of each quadrilateral. Copy and complete the table to organize your results. (The notation m∠A denotes the measure of angle A.) How precise are your measurements?

Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 2

EXPLORATION 2
Making a Conjecture

Work with a partner. Use the completed table in Exploration 1 to write a conjecture about the sum of the angle measures of a quadrilateral. Draw three quadrilaterals that are different from those in Exploration 1 and use them to justify your conjecture.
Answer:
The completed table is:

From the above table,
We can say that the sum of all the angle in any quadrilateral is: 360 degrees
So,
From the above table,
The angles in Quadrilateral a is: 60 degrees, 125 degrees, 120 degrees, and 55 degrees
We know that,
The quadrilateral will have n angles based on the shape.
The shape which has more than 3 sides is called a Quadrilateral.
Ex: Rectangle, Square, Pentagon, Hexagon, etc.

From the above quadrilaterals,
We can say that all the sides in each quadrilateral are equal.
So,
The angles in each quadrilateral are also equal.
So,
In a rectangle, there are 4 sides
So,
By measuring, we can observe that each angle of a rectangle is: 90 degrees
Hence,
The sum of all angles in a rectangle = 90 + 90 + 90 + 90 = 360 degrees
In a pentagon, there are 5 sides
By measuring, we can observe that each rectangle of a pentagon
Hence,
The sum of all the angles in a pentagon = = 72+ 72 + 72 + 72 + 72 = 360 degrees
In a Hexagon, there are 6 sides
So,
By measuring, we can observe that each angle of a hexagon is: 60 degrees
Hence,
The sum of all angle is a Hexagon = 60 + 60 + 60 + 60 + 60 + 60 = 360 degrees
Hence, from the above-drawn quadrilaterals,
We can conclude that our conjecture is proven.

EXPLORATION 3
Applying Your Conjecture

Work with a partner. Use the conjecture you wrote in Exploration 2 to write an equation for each quadrilateral. Then solve the equation to find the value of x. Use a protractor to check the reasonableness of your answer.
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 3
Answer:
The given figure is:
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 3
From Exploration 2, the proven conjecture is:
The sum of all angles in a quadrilateral is: 360 degrees
Now,
In Quadrilateral a.,
By using the above-proven conjecture,
85 + 80 + 100 + x = 360 degrees
265 + x = 360 degrees
x = 360 – 265
= 95 degrees
So,
The angle x is: 95 degrees
In Quadrilateral b.,
By using the above-proven conjecture,
72 + 78 + 60 + x = 360 degrees
210 + x = 360 degrees
x = 360 – 210
= 150 degrees
So,
The angle of x is: 150 degrees
In Quadrilateral c.,
By using the above-proven conjecture,
90 + 90  +30 + x = 360 degrees
210 + x = 360 degrees
x = 360 – 210
= 150 degrees
So,
The angle of x is: 150 degrees

Communicate Your Answer

Question 4.
How can you use simple equations to solve real-life problems?

Question 5.
Draw your own quadrilateral and cut it out. Tear off the four corners of the quadrilateral and rearrange them to affirm the conjecture you wrote in Exploration 2. Explain how this affirms the conjecture.
Answer:
Your Quadrilateral is:

From the above Quadrilateral,
We can observe that the tear-off corners of the quadrilateral are joined and it becomes the triangle.
So,
In the above Quadrilateral, there are two quadrilaterals.
We know that,
The sum of all angles in a triangle is: 180 degrees
So,
The sum of all angles in the two triangles = 180 + 180 = 360 degrees
These two triangles form a quadrilateral.
So,
The sum of all angles in a quadrilateral is: 360 degrees
Hence, from the above,
We can conclude that we can prove Exploration 2’s conjecture by your own example.

1.1 Lesson

Monitoring Progress

Solve the equation. Justify each step. Check your solution.

Question 1.
n + 3 = -7
Answer:
The value of n is: -10

Explanation:
The given equation is:
n + 3 = -7
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
n + 3 = -7
n = -7 – (+3 )
n = -7 – 3
= -10
Hence from the above,
We can conclude that the value of n is: -10

Question 2.
g – \(\frac{1}{3}\) = –\(\frac{2}{3}\)
Answer:
The value of g is: –\(\frac{1}{3}\)

Explanation:
The given equation is:
g – \(\frac{1}{3}\) = –\(\frac{2}{3}\)
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
g – \(\frac{1}{3}\) = –\(\frac{2}{3}\)
g = –\(\frac{2}{3}\) + \(\frac{1}{3}\)
g = \(\frac{-2 + 1}{3}\)
g = \(\frac{-1}{3}\)
g = –\(\frac{1}{3}\)
Hence, fromthe above,
We can conclude that the value of g is: –\(\frac{1}{3}\)

Question 3.
-6.5 = p + 3.9
Answer:
The value of p is: -10.4

Explanation:
The given equation is:
-6.5 = p + 3.9
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
-6.5 = p + 3.9
p = -6.5 – 3.9
= – ( 6.5 + 3.9 )
= – 10.4
Hence, from the above,
We can conclude that the value of p is: -10.4

Monitoring Progress

Solve the equation. Justify each step. Check your solution.

Question 4.
\(\frac{y}{3}\) = -6
Answer:
The value of y is: -18

Explanation:
The given equation is:
\(\frac{y}{3}\) = -6
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
\(\frac{y}{3}\) = -6
\(\frac{y}{1}\) × \(\frac{1}{3}\) = -6
\(\frac{y}{1}\) = -6 ÷ \(\frac{1}{3}\)
y = -6 × -3
y = -18
Hence, from the above,
We can conclude that the value of y is: -18

Question 5.
9π = πx
Answer:
The value of x is: 9

Explanation:
The given equation is:
9π = πx
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
9π = πx
9 × π = π × x
x = ( 9 × π ) ÷ π
x = 9
Hence, from the above,
We can conclude that the value of x is: 9

Question 6.
0.05w = 1.4
Answer:
The value of w is: 28

Explanation:
The given equation is:
0.05w = 1.4
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
0.05w = 1.4
0.05 × w = 1.4
\(\frac{5}{100}\) × w = \(\frac{14}{10}\)
w = \(\frac{14}{10}\) ÷ \(\frac{5}{100}\)
w = \(\frac{14}{10}\) × \(\frac{100}{5}\)
w = \(\frac{14 × 100}{10 × 5}\)
w = \(\frac{28}{1}\)
w = 28
Hence, from the above,
We can conclude that the value of w is: 28

Monitoring Progress

Question 7.
Suppose Usain Bolt ran 400 meters at the same average speed that he ran the 200 meters. How long would it take him to run 400 meters? Round your answer to the nearest hundredth of a second.
Answer:
The time it took for him to run 400 meters is: 0.50 seconds

Explanation:
It is given that Usain Bolt ran 400 meters at the same average speed that he ran the 200 meters.
We know that,
Speed = Distance ÷ Time
But, it is given that the average speed is the same.
Hence,
Speed = Constant
So,
Since speed is constant, distance is directly proportional to time.
So,
The time taken by Usain Bolt to run 400 meters = 200 ÷ 400
= ( 2 × 100 ) ÷ ( 4 × 100 )
= 10 ÷ 20
= 0.50 seconds ( 0.5 and 0.50 are the same values Only for the representation purpose, we will add ‘0’ after 5 )
Hence from the above,
We can conclude that the time is taken by Usain Bolt to run 400 meters when rounded-off to the nearest hundredth is: 0.50 seconds

Monitoring Progress

Question 8.
You thought the balance in your checking account was $68. When your bank statement arrives, you realize that you forgot to record a check. The bank statement lists your balance as $26. Write and solve an equation to find the amount of the check that you forgot to record.
Answer:
The amount of the check that you forgot to record is: $42

Explanation:
It is given that you thought the balance in your checking account was $68 and when your bank statement arrives, you realize that you forgot to record a check and the bank statement lists your balance as $26.
Now,
Let the amount you forgot to record be: x
So,
The total balance in your checking account = ( The listed balance ) + ( The amount that you forgot to record a check )
68 = 26 + x
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
68 = 26 + x
x = 68 – 26
x = $42
Hence, from the above,
We can conclude that the amount that forgot to record is: $42

Solving Simple Equations 1.1 Exercises

Monitoring Progress and Modeling with Mathematics

In Exercises 5–14, solve the equation. Justify each step. Check your solution.

Question 1.
VOCABULARY Which of the operations +, -, ×, and ÷ are inverses of each other?
Answer:
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
Hence, from the above,
We can conclude that,
+ is inverse of –  and vice-versa
× is inverse of ÷ and vice-versa

Question 2.
VOCABULARY Are the equations -2x = 10 and -5x = 25 equivalent? Explain.
Answer:
The equations -2x = 10 and -5x = 25 are equivalent

Explanation:
The given equations are:
-2x = 10 and -5x = 25
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
We know that,
A) – ÷ – = +
B) + ÷ – = –
C) + ÷ + = +
D) – ÷ + = –
So,
From -2x = 10,
x = 10 ÷ ( -2 )
x = -10 ÷ 2
x = -5
From -5x = 25,
x = 25 ÷ ( -5 )
x = -25 ÷ 5
x = -5
Hence, from the above,
We can conclude that the equations -2x = 10 and -5x = 25 are equivalent

Question 3.
WRITING Which property of equality would you use to solve the equation 14x = 56? Explain.
Answer:
The given equation is:
14x = 56
So,
It can be re-written as
14 × x = 56
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
x = 56 ÷ 14
x = 4
Hence, from the above,
We can conclude that the value of x is: 4

Question 4.
WHICH ONE DOESN’T BELONG? Which expression does not belong with the other three? Explain your reasoning.
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 501
Answer:
The equation C) does not belong to the other three

Explanation:
Let the given equations be named as A), B), C), and D)
So,
The given equations are:
A) 8 = x ÷ 2
B) 3 = x ÷ 4
C) x – 6 = 5
D) x ÷ 3 = 9
So,
From the above equations,
The equations A, B), and D) are dividing the numbers whereas equation C) subtracting the numbers
Hence, from the above,
We can conclude that,
The equation C) does not belong to the other three.

Question 5.
x + 5 = 8

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.1-Q5

Question 6.
m + 9 = 2
Answer:
The value of m is: -5

Explanation:

Question 7.
y – 4 = 3

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.1-Q7

Question 8.
s – 2 = 1
Answer:
The value of s is: 3

Explanation;

Question 9.
w + 3 = -4

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.1-Q9

Question 10.
n – 6 = -7
Answer:
The value of n is: -1

Explanation:

Question 11.
-14 = p – 11

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.1-Q11

Question 12.
0 = 4 + q
Answer:
The value of q is: -4

Explanation;

Question 13.
r + (-8) = 10

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.1-Q13

Question 14.
t – (-5) = 9
Answer:
The value of t is: 4

Explanation;

Question 15.
MODELING WITH MATHEMATICS
A discounted amusement park ticket costs $12.95 less than the original price p. Write and solve an equation to find the original price.
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 4

Answer:
The equation for the original price is:
p = x + $12.95

Explanation:
It is given that a discounted amusement park ticket costs $12.95 less than the original price p.
So,
Let the discounted amusement park ticket be: x
The given original price is: p
So,
The discounted amusement park ticket cost = p – $12.95
x = p – 12.95
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
p = x + $12.95
Hence, from the above,
We can conclude that the equation for the original price is:
p = x + $12.95

Question 16.
MODELING WITH MATHEMATICS
You and a friend are playing a board game. Your final score x is 12 points less than your friend’s final score. Write and solve an equation to find your final score.
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 5

Answer:
Your final score is:
x = ( The score of your friend ) – 12

Explanation:
It is given that you and a friend are playing a board game. Your final score x is 12 points less than your friend’s final score.
So,
The scores table is shown below:
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 5
From the above table,
The final score of your friend is: 195
Let the final score of yours is: x
So,
x = ( The final score of your friend ) – 12
= 195 – 12
= 183 points
Hence, from the above,
We can conclude that the final score of yours is: 183 points

USING TOOLS
The sum of the angle measures of a quadrilateral is 360°. In Exercises 17–20, write and solve an equation to find the value of x. Use a protractor to check the reasonableness of your answer.

Question 17.
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 6

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.1-Q17

Question 18.
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 7

Answer:
The value of x is: 85 degrees

Explanation:
We know that,
The sum of angles in a quadrilateral is: 360 degrees
So,
150 + 48 + 77 + x = 360
275 + x = 360
x = 360 – 275
x = 85 degrees
Hence, from the above,
We can conclude that the value of x is: 85 degrees

Question 19.
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 8

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.1-Q19

Question 20.
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 9

Answer:
The value of x is: 100 degrees

Explanation:
We know that,
The sum of all angles in a quadrilateral is: 360 degrees
So,
115 + 85 + 60 + x = 360
260 + x = 360
x = 360 – 260
x = 100 degrees
Hence, from the above,
We can conclude that the value of x is: 100 degrees

In Exercises 21–30, solve the equation. Justify each step. Check your solution.

Question 21.
5g = 20

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.1-Q21

Question 22.
4q = 52
Answer:
The value of g is: 13

Explanation:
The given equation is:
4g = 52
4 × g = 52
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
g = 52 ÷ 4
= ( 44 + 8 ) ÷ 4
= ( 44 ÷ 4 ) + ( 8 ÷ 4 )
= 11 + 2
= 13
Hence, from the above,
We can conclude that the value of g is: 13

Question 23.
p ÷ 5 = 3

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.1-Q23

Question 24.
y ÷ 7 = 1
Answer:
The value of y is: 7

Explanation:
The given equation is:
y ÷ 7 = 1
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
y = 1 × 7
y = 7
Hence, from the above,
We can conclude that the value of y is: 7

Question 25.
-8r = 64

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.1-Q25

Question 26.
x ÷(-2) = 8
Answer:
The value of x is: -16

Explanation:
The given equation is:
x ÷ ( -2 ) = 8
We know that,
A) – ÷ – = +
B) + ÷ – = –
C) + ÷ + = +
D) – ÷ + = –
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
x ÷ ( -2 ) = 8
x = 8 × ( -2 )
x = -16
Hence, from the above,
We can conclude that the value of x is: -16

Question 27.
\(\frac{x}{6}\) = 8

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.1-Q27

Question 28.
\(\frac{w}{-3}\) = 6
Answer:
The value of w is: -18

Explanation:
The given equation is:
\(\frac{w}{-3}\) = 6
We know that,
A) – ÷ – = +
B) + ÷ – = –
C) + ÷ + = +
D) – ÷ + = –
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
\(\frac{w}{-3}\) = 6
w = 6 × ( -3 )
w = -18
Hence, from the above,
We can conclude that the value of w is: -18

Question 29.
-54 = 9s

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.1-Q29

Question 30.
-7 = \(\frac{t}{7}\)
Answer:
The value of t is: -49

Explanation:
The given equation is:
-7 = \(\frac{t}{7}\)
We know that,
A) – ÷ – = +
B) + ÷ – = –
C) + ÷ + = +
D) – ÷ + = –
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
-7 = \(\frac{t}{7}\)
t = -7 × 7
t = -49
Hence, from the above
We can conclude that the value of t is: -49

In Exercises 31– 38, solve the equation. Check your solution.

Question 31.
\(\frac{3}{2}\) + t = \(\frac{1}{2}\)

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.1-Q31

Question 32.
b – \(\frac{3}{16}\) = \(\frac{5}{16}\)
Answer:
The value of b is: \(\frac{1}{2}\)

Explanation:
The given equation is:
b – \(\frac{3}{16}\) = \(\frac{5}{16}\)
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
b = \(\frac{5}{16}\) + \(\frac{3}{16}\)
Since the denominators of both the numerators are equal, add the numerators making the denominator common
So,
b = \(\frac{5 + 3}{16}\)
b = \(\frac{8}{16}\)
b = \(\frac{1}{2}\)
Hence, from the above,
We can conclude that the value of b is: \(\frac{1}{2}\)

Question 33.
\(\frac{3}{7}\)m = 6

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.1-Q33

Question 34.
–\(\frac{2}{5}\)y = 4
Answer:
The value of y is: 10

Explanation:
The given equation is:
–\(\frac{2}{5}\)y = 4
We know that,
A) – ÷ – = +
B) + ÷ – = –
C) + ÷ + = +
D) – ÷ + = –
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
–\(\frac{2}{5}\)y = 4
–\(\frac{2}{5}\) × y = 4
y = 4 ÷ –\(\frac{2}{5}\)
y = 4 × –\(\frac{5}{2}\)
y = -4 × –\(\frac{5}{2}\)
y = –\(\frac{4}{1}\) × –\(\frac{5}{2}\)
y = –\(\frac{4 × 5}{1 × 2}\)
y = 10
Hence, from the above,
We can conclude that the value of y is: 10

Question 35.
5.2 = a – 0.4

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.1-Q35

Question 36.
f + 3π = 7π
Answer:
The value of f is: 4π

Explanation:
The given equation is:
f + 3π = 7π
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
f + 3π = 7π
f = 7π – 3π
f = π ( 7 – 3 )
f = π ( 4 )
f = 4π
Hence, from the above,
We can conclude that the value of f is: 4π

Question 37.
– 108π = 6πj

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.1-Q37

Question 38.
x ÷ (-2) = 1.4
Answer:
The value of x is: –\(\frac{14}{5}\)

Explanation:
The given equation is:
x ÷ ( -2 ) = 1.4
We know that,
A) – ÷ – = +
B) + ÷ – = –
C) + ÷ + = +
D) – ÷ + = –
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
x ÷ ( -2 ) = 1.4
x ÷ ( -2 ) = \(\frac{14}{10}\)
x ÷ ( -2 ) = \(\frac{7}{5}\)
x = \(\frac{7}{5}\) × ( -2 )
x = – \(\frac{7}{5}\) × \(\frac{2}{1}\)
x = –\(\frac{14}{5}\)
Hence, from the above,
We can conclude that the value of x is: –\(\frac{14}{5}\)

ERROR ANALYSIS
In Exercises 39 and 40, describe and correct the error in solving the equation.

Question 39.
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 10

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.1-Q39

Question 40.
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 11

Answer:
A negative 3 should have been multiplied to each side.
We know that,
A) – × – = +
B) + × – = –
C) + × + = +
D) – × + = –
So,
-(\(\frac{m}{3}\) ) =-4
-3  ( \(\frac{m}{3}\) ) = -4 ( -3 )
3 ( \(\frac{m}{3}\) ) = -4 ( -3 )
3 ( \(\frac{m}{3}\) ) = 12
\(\frac{m}{3}\)  × \(\frac{3}{1}\) = 12
m = 12
Hence, from the above,
We can conclude that the value of m is: 12

Question 41.
ANALYZING RELATIONSHIPS
A baker orders 162 eggs. Each carton contains 18 eggs. Which equation can you use to find the number x of cartons? Explain your reasoning and solve the equation.
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 12

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.1-Q41

MODELING WITH MATHEMATICS
In Exercises 42– 44, write and solve an equation to answer the question.

Question 42.
The temperature at 5 P.M. is 20°F. The temperature at 10 P.M. is -5°F. How many degrees did the temperature fall?
Answer:
The fall in temperature is: 25 degrees Fahrenheit

Explanation:
It is given that the temperature at 5 P.M. is 20°F and the temperature at 10 P.M. is -5°F.
So,
The fall in temperature = ( The temperature at 5 P.M ) – ( The temperature at 10 P.M )
We know that,
A) – × – = +
B) + × – = –
C) + × + = +
D) – × + = –
So,
The fall in temperature = 20 – ( -5 )
= 20 + 5
= 25 degrees Fahrenheit
Hence, from the above,
We can conclude that the fall in temperature is: 25 degrees Fahrenheit

Question 43.
The length of an American flag is 1.9 times its width. What is the width of the flag?
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 13

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.1-Q43

Question 44.
The balance of an investment account is $308 more than the balance 4 years ago. The current balance of the account is $4708. What was the balance 4 years ago?
Answer:
The balance 4 years ago is: $4,400

Explanation:
It is given that the balance of an investment account is $308 more than the balance 4 years ago. The current balance of the account is $4708.
So,
The current balance of the account = ( The balance of an investment account 4 years ago ) + $308
Let the balance of an investment account four years ago be x.
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
$4,708 = x + $308
x = 4,708 – 308
x = $4,400
Hence, from the above,
We can conclude that the balance of an investment account four years ago is: $4,400

Question 45.
REASONING
Identify the property of equality that makes Equation 1 and Equation 2 equivalent.
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 14

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.1-Q45

PROBLEM-SOLVING
Question 46.

Tatami mats are used as a floor covering in Japan. One possible layout uses four identical rectangular mats and one square mat, as shown. The area of the square mat is half the area of one of the rectangular mats.
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 15
a. Write and solve an equation to find the area of one rectangular mat.
Answer:
The area of one rectangular mat is: 18 ft²

Explanation:
It is given that the tatami mats are used as a floor covering in Japan and for that, one layout of tatami mats requires the four identical rectangular mats and the one square mat
So,
The total area of the tatami mats = ( The area of the four rectangular mats ) + ( The area of the one square mat )
The give tatami mat layout is:
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 15
From the layout,
We can observe that
The total area of the layout is: 81 ft²
So,
The total area of tatami mats = 81 ft²
( The area of the four rectangular mats ) + ( The area of the one square mat ) = 81 ft²
It is also given that
The area of a square mat is half of one of the rectangular mats
So,
Area of the square mat = ( Area of the rectangular mat ) ÷ 2
So,
( The area of the four rectangular mats ) + \(\frac{Area of the rectangular mat}{2}\)  = 81 ft²
4 ( The area of the rectangular mat ) + \(\frac{Area of the rectangular mat}{2}\)  = 81 ft²
\(\frac{8}{2}\) ( The area of the rectangular mat ) + \(\frac{1}{2}\) ( The area of the rectangular mat ) = 81 ft²
( The area of the rectangular mat ) [ \(\frac{8}{2}\) + \(\frac{1}{2}\) ] = 81 ft²
\(\frac{9}{2}\) ( The area of the rectangular mat ) = 81 ft²
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
The area of the rectangular mat = 81 ÷ \(\frac{9}{2}\)
= 81 × \(\frac{2}{9}\)
= \(\frac{81}{1}\) × \(\frac{2}{9}\)
= \(\frac{81 × 2}{9 × 1}\)
= 18 ft²
Hence, from the above,
We can conclude that the area of one rectangular mat is: 18 ft²

b. The length of a rectangular mat is twice the width. Use Guess, Check, and Revise to find the dimensions of one rectangular mat.
Answer:
The dimensions of the rectangular mat are:
Length: 6 ft
Width: 3 ft

Explanation:
From the above problem,
The area of the rectangular mat = 18 ft²
It is given that the length of a rectangular mat is twice the width.
We know that the area of the rectangle = ( Length ) × ( Width )
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
It is also given that the length of a rectangular mat is twice the width
So,
Length of a rectangular mat = 2 × Width
Now,
The area of the rectangular mat = Length × Width
18 = 2 × Width × Width
Width × Width = 18 ÷ 2
Width × Width = 9
From guessing,
We can say that
Width of the rectangular mat = 3 ft
Now,
The length of the rectangular mat = 2 × 3
= 6 ft
Hence, from the above,
We can conclude that the dimensions of the rectangular mat are:
Length: 6 ft
Width: 3 ft

Question 47.
PROBLEM-SOLVING
You spend $30.40 on 4 CDs. Each CD costs the same amount and is on sale for 80% of the original price.
a. Write and solve an equation to find how much you spend on each CD.
b. The next day, the CDs are no longer on sale. You have $25. Will you be able to buy 3 more CDs? Explain your reasoning.
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 15.1

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.1-Q47

Question 48.
ANALYZING RELATIONSHIPS
As c increases, does the value of x increase, decrease, or stay the same for each equation? Assume c is positive.
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 16

Answer:
Let assume the values of c be: 0,1,2,3
So,
The completed table by taking the values of c is:

By taking the values of c positive i.e., 0, 1, 2, 3
We can observe that as the value of c increases, the values of x sometimes increasing and sometimes stays the same but not decreasing.

Question 49.
USING STRUCTURE
Use the values -2, 5, 9, and 10 to complete each statement about the equation ax = b – 5.
a. When a = ___ and b = ___, x is a positive integer.
b. When a = ___ and b = ___, x is a negative integer.

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.1-Q51

Question 50.
HOW DO YOU SEE IT?
The circle graph shows the percents of different animals sold at a local pet store in 1 year.
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 17
a. What percent is represented by the entire circle?
Answer:
The percent represented by the entire circle is = 69 % + x %

Explanation:
It is given that the circle represents the percent of different animals sold at a local store for 1 year
Now,
The given circle is:
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 17
We know that,
In terms of percentages, any circle represents 100%
So,
The equation representing the circle is:
The percentage of different animals in the circle = 48 + 5 + 9 + 7 +x
The percentage of different animals in the circle= 69% + x%

b. How does the equation 7 + 9 + 5 + 48 + x = 100 relate to the circle graph? How can you use this equation to find the percent of cats sold?
Answer:
The percent of cats sold is: 31%

Explanation:
We know that,
In terms of percentages, any circle represents 100%
So,
The total percent of animals = The percent of animals that are represented by the circle
100% = 69% + x%
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
x% = 100% – 69%
x% = 31%
Hence, from the above,
We can conclude that the percent of cats is: 31%

Question 51.
REASONING
One-sixth of the girls and two-sevenths of the boys in a school marching band are in the percussion section. The percussion section has 6 girls and 10 boys. How many students are in the marching band? Explain.

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.1-Q51

Question 52.
THOUGHT-PROVOKING
Write a real-life problem that can be modeled by an equation equivalent to the equation 5x = 30. Then solve the equation and write the answer in the context of your real-life problem.

Answer:
Let suppose there is some number of boys. The number of girls is five times of the boys and the total number of girls is 30. Find the number of boys?
Ans:
Let,
The number of boys is x.
It is given that the number of girls is five times of boys.
So,
The number of girls = 5x
It is also given that
The number of girls = 30
So,
5x = 30
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
5 × x = 30
x = 30 ÷ 5
x = 6
Hence, from the above,
We can conclude that the number of boys is: 6

MATHEMATICAL CONNECTIONS
In Exercises 53–56, find the height h or the area of the base B of the solid.

Question 53.
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 18

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.1-Q53

Question 54.
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 19

Answer:
The height of the cuboid is: 9 cm

Explanation:
The given figure is:
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 19
We know that,
The volume of a cuboid = L × B × H
We know that,
The cuboid is made from a rectangle
We know that,
The area of a rectangle = L × H
So,
The volume of a cuboid = A × B
From the above figure,
The given volume is: 1323 cm³
The given Area is: 147 cm²
So,
1323 = 147 × H
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
H = 1323 ÷ 147
H = 9
Hence, from the above,
We can conclude that the height of the cuboid is: 9 cm

Question 55.
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 20

Answer:

Question 56.
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 21

Answer:
The height of the prism is: \(\frac{5}{6}\) ft

Explanation:
The given figure is:
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 21
We know that,
The volume of the prism =  Area × Height
From the above figure,
The volume of the prism = 35 ft³
The area of the prism = 30 ft²
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
35 = 30 × H
H = 35 ÷ 30
H = \(\frac{5}{6}\) ft
Hence, from the above,
We can conclude that the height of the prism is: \(\frac{5}{6}\) ft

Question 57.
MAKING AN ARGUMENT
In baseball, a player’s batting average is calculated by dividing the number of hits by the number of at-bats. The table shows Player A’s batting average and the number of at-bats for three regular seasons.
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 22
a. How many hits did Player A have in the 2011 regular season? Round your answer to the nearest whole number.
b. Player B had 33 fewer hits in the 2011 season than Player A but had a greater batting average. Your friend concludes that Player B had more at-bats in the 2011 season than Player A. Is your friend correct? Explain. Maintaining

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.1-Q57

Maintaining Mathematical Proficiency

Use the Distributive Property to simplify the expression.

Question 58.
8(y + 3)
Answer:
8 ( y + 3 ) = 8y + 24

Explanation:
The given expression is: 8 ( y + 3 )
We know that,
By using the Distributive Property,
a ( b + c ) = ( a × b ) + ( a × c )
So,
By using the above Property,
8 ( y + 3 ) = ( 8 × y ) + ( 8 × 3 )
= 8y + 24
Hence, from the above,
We can conclude that
8 ( y + 3 ) = 8y + 24

Question 59.
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 23

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.1-Q59

Question 60.
5(m + 3 + n)
Answer:
5 ( m + 3 + n ) = 5m + 5n + 15

Explanation:
The given expression is: 5 ( m + 3 + n )
By using the Distributive Property,
a ( b + c ) = ( a × b ) + ( a × c )
So,
By using the above Property,
5 ( m + 3 + n ) = ( 5 × m ) + ( 5 × 3 ) + ( 5 × n )
= 5m + 15 + 5n
Hence, from the above,
We can conclude that,
5 ( m + 3 + n ) = 5m + 15 + 5n

Question 61.
4(2p + 4q + 6)

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.1-Q61

Copy and complete the statement. Round to the nearest hundredth, if necessary.

Question 62.
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 24

Answer:
The missing number is: \(\frac{1}{12}\)

Explanation:
Let the missing number be: x
So,
The given equation is:
\(\frac{5L}{min}\) = \(\frac{x L}{h}\)
We know that,
1 hour = 60 minutes
So,
1 min = \(\frac{1}{60}\) hour
So,
\(\frac{5 L}{min}\) = \(\frac{5 L × 1}{60h}\)
\(\frac{5 L}{min}\) = \(\frac{1 L }{12h}\)
So,
x = \(\frac{1}{12}\)
Hence, from the above,
We can conclude that,
The missing number is: \(\frac{1}{12}\)

Question 63.
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 25

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.1-Q63

Question 64.
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 26

Answer:
The missing number is: \(\frac{1}{12}\)

Explanation:
Let the missing number be: x
So,
The given equation is:
\(\frac{7 gal}{min}\) = \(\frac{x qin}{sec}\)
We know that,
1 min = 60 seconds
1 quintal = 100 kg
1 gallon = 3.78 kg = 4 kg
So,
1 gallon = 0.04 quintal
1 sec = \(\frac{1}{60}\) min
So,
\(\frac{7 gal}{min}\) = \(\frac{x qin × 1}{60min}\)
\(\frac{7 gal}{min}\) = \(\frac{1 L }{12h}\)
So,
x = \(\frac{1}{12}\)
Hence, from the above,
We can conclude that,
The missing number is: \(\frac{1}{12}\)

Question 65.
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 27

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.1-Q65

Lesson 1.2 Solving Multi-step Equations

Essential Question

How can you use multi-step equations to solve real-life problems?

EXPLORATION 1
Solving for the Angle Measures of a Polygon

Work with a partner. The sum S of the angle measures of a polygon with n sides can be found using the formula S = 180(n – 2). Write and solve an equation to find each value of x. Justify the steps in your solution. Then find the angle measures of each polygon. How can you check the reasonableness of your answers?
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 27.1
Answer:
The given polygons are:

It is given that,
The sum S of the angle measures of a polygon with n sides can be found using the formula S = 180(n – 2).
a)
The number of sides (n ) = 3
So,
The sum of angles ( S ) = 180 ( n – 2 )
= 180 ( 3 – 2 )
= 180 ( 1 )
= 180
Now,
The given sides of a polygon are: 30, 9x, (30 + x )
So,
30 + 9x + 30 + x = 180
60 + 10x = 180
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
10x = 180 – 60
10x = 120
x = 120 ÷ 10
x = 12
Hence, from the above,
The angle measures of the given polygon are:
30, 9 × 12, 30 + 12
= 30, 108, 45 degrees
b)
The number of sides (n ) = 3
So,
The sum of angles ( S ) = 180 ( n – 2 )
= 180 ( 3 – 2 )
= 180 ( 1 )
= 180
Now,
The given sides of a polygon are: 30, 9x, (30 + x )
So,
50 + x + 10 + 20 + x = 180
80 + 2x = 180
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
2x = 180 – 80
2x = 100
x = 100 ÷ 2
x = 50
Hence, from the above,
The angle measures of the given polygon are:
50, 50 + 10, 50 + 20
= 50, 60, 70 degrees
c)
The number of sides (n ) = 4
So,
The sum of angles ( S ) = 180 ( n – 2 )
= 180 ( 4 – 2 )
= 180 ( 2 )
= 360
Now,
The given sides of a polygon are: 50, x, ( 2x + 20), ( 2x + 30 )
So,
50 + x + 2x + 20 + 2x + 30 = 360
100 + 5x = 360
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
5x = 360 – 100
5x = 260
x = 260 ÷ 5
x = 52
Hence, from the above,
The angle measures of the given polygon are:
50,52, 2 (52) + 20, 2(52) + 30
= 50, 52, 124, 134 degrees
d)
The number of sides (n ) = 4
So,
The sum of angles ( S ) = 180 ( n – 2 )
= 180 ( 4 – 2 )
= 180 ( 2 )
= 360
Now,
The given sides of a polygon are: x, x + 42, x + 35, x – 17
So,
x + x + 42 + x + 35 + x – 17 = 360
60 + 4x = 360
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
4x = 360 – 60
4x = 300
x = 300 ÷ 4
x = 75
Hence, from the above,
The angle measures of the given polygon are:
75,  75 + 42,  75 + 35 , 75 – 17
= 75, 117, 110, 58 degrees
e)
The number of sides (n ) = 5
So,
The sum of angles ( S ) = 180 ( n – 2 )
= 180 ( 5 – 2 )
= 180 ( 3 )
= 540
Now,
The given sides of a polygon are: (4x + 15), (5x + 10), (8x + 8), (3x + 5), (5x + 2)
So,
(4x + 15)+ (5x + 10)+ (8x + 8)+ (3x + 5)+ (5x + 2) = 540
40 + 25x = 540
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
25x = 540 – 40
25x = 500
x = 500 ÷ 25
x = 20
Hence, from the above,
The angle measures of the given polygon are:
(4. 20 + 15)+ (5. 20 + 10)+ (8.20 + 8)+ (3. 20 + 5)+ (5. 20 + 2)
= 95, 110, 168, 65, 102 degrees
f)
The number of sides (n ) = 5
So,
The sum of angles ( S ) = 180 ( n – 2 )
= 180 ( 5 – 2 )
= 180 ( 3 )
= 540
Now,
The given sides of a polygon are: (2x + 25), (3x + 16), (2x + 8), (4x – 18), (3x – 7)
So,
(2x + 25) + (3x + 16) + (2x + 8) + (4x – 18) + (3x – 7) = 540
24 + 14x = 540
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
14x = 540 – 24
14x = 516
x = 540 ÷ 14
x = 38.5
x = 39
Hence, from the above,
The angle measures of the given polygon are:
(2. 39 + 25), (3.39 + 16), (2.39 + 8), (4.39 – 18), (3.39 – 7)
= 103, 133, 86, 138, 110 degrees

EXPLORATION 2
Work with a partner.

a. Draw an irregular polygon.
Answer:

b. Measure the angles of the polygon. Record the measurements on a separate sheet of paper.
Answer:

c. Choose a value for x. Then, using this value, work backward to assign a variable expression to each angle measure, as in Exploration 1.
d. Trade polygons with your partner.
e. Solve an equation to find the angle measures of the polygon your partner drew. Do your answers seem reasonable? Explain.

Communicate Your Answer

Question 3.
How can you use multi-step equations to solve real-life problems?

Question 4.
In Exploration 1, you were given the formula for the sum S of the angle measures of a polygon with n sides. Explain why this formula works.
Answer:
We know that,
The sum of the angles in a triangle is: 180 degrees
The triangle is also a quadrilateral
So,
A quadrilateral can be formed by the minimum of the three lines
So,
The minimum sum of all the angles in a quadrilateral is: 180 degrees
Now,
Let suppose we form a quadrilateral with 4 sides.
So,
The sum of all the angles in a quadrilateral = 360 degrees = 180 degrees × 2
= 180 degrees ( 4 sides -2 )
Let suppose we form a quadrilateral with 5 sides
So,
The sum of all the angles in a quadrilateral = 540 degrees = 180 degrees × 3
= 180 degrees ( 5 -2 )
Hence, in general,
We can conclude that the sum of all the angles with n sides in a quadrilateral = 180 degrees ( n-2 )

Question 5.
The sum of the angle measures of a polygon is 1080º. How many sides does the polygon have? Explain how you found your answer.
Answer:
The number of sides the polygon with 1080° have: 6

Explanation:
It is given that the sum of all angle measures of a polygon is: 1080°
We know that,
The sum of angle measures with n sides in a polygon = 180° ( n – 2 )
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
1080° =180° ( n – 2 )
n – 2 = 1080 ÷ 180
n – 2 = 6
n = 6 + 2
n = 8
Hence, from the above,
We can conclude that the number of sides of the polygon with sum of the angles 1080° is: 6

1.2 Lesson

Monitoring Progress

Solve the equation. Check your solution.

Question 1.
-2n + 3 = 9
Answer:
The value of n is: -3

Explanation:
The given equation is:
-2n + 3 = 9
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
-2n + 3 = 9
-2n = 9 – (+3 )
n = 6 ÷ ( -2 )
= -3
Hence from the above,
We can conclude that the value of n is: -3

Question 2.
-21 = \(\frac{1}{2}\) – 11

Question 3.
-2x – 10x + 12 = 18
Answer:
The value of x is: –\(\frac{1}{2}\)

Explanation:
The given equation is:
-2x – 10x + 12 = 18
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
-( 2x + 10x ) = 18 – 12
-12x = 6
x = 6 ÷ ( -12 )
x = –\(\frac{1}{2}\)
Hence, from the above,
We can conclude that the value of x is: –\(\frac{1}{2}\)

Monitoring Progress

Solve the equation. Check your solution.

Question 4.
3(x + 1) + 6 = -9
Answer:
The value of x is: -6

Explanation:
The given equation is:
3 ( x + 1 ) + 6 = -9
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
3 ( x + 1 ) = -9 – (+6 )
By using the Distributive property,
3 ( x + 1 ) = 3x + 3
So,
3x + 3 = -15
3x = -15 – ( +3 )
3x = -18
x = -18 ÷ 3
x = -6
Hence, from the above,
We can conclude that the value of x is: -6

Question 5.
15 = 5 + 4(2d – 3)
Answer:
The value of d is:\(\frac{11}{4}\)

Explanation:
The given equation is:
15 = 5 + 4 ( 2d – 3 )
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
4 ( 2d – 3 ) = 15 -5
4 ( 2d – 3 ) = 10
By using the Distributive property,
4 ( 2d – 3 ) = 4 (2d ) -4 (3 )
= 8d – 12
So,
8d – 12 = 10
8d = 10 + 12
8d = 22
d = 22 ÷ 8
d = \(\frac{11}{4}\)
Hence, from the above,
We can conclude that the value of d is: \(\frac{11}{4}\)

Question 6.
13 = -2(y – 4) + 3y
Answer:
The value of y is: 5

Explanation:
The given equation is:
13 = -2 ( y – 4 ) + 3y
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
By using the Distributive Property,
-2 ( y – 4 ) = -2y + 8
So,
13 = -2y + 8 + 3y
13 = y + 8
y = 13 – 8
y = 5
Hence, from the above,
We can conclude that the value of y is: 5

Question 7.
2x(5 – 3) – 3x = 5
Answer:
The value of x is: 5

Explanation:
The given equation is:
2x ( 5 – 3 ) – 3x = 5
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
2x ( 2 ) – 3x = 5
4x – 3x = 5
x = 5
Hence, from the above,
We can conclude that the value of y is: 5

Question 8.
-4(2m + 5) – 3m = 35
Answer:
The value of m is: -5

Explanation:
The given equation is:
-4 ( 2m + 5 ) – 3m = 35
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
Now,
By using the Distributive Property,
-4 ( 2m + 5 ) = -4 (2m ) + 5 ( -4 )
= -8m -20
So,
-8m -20 -3m = 35
-11m – 20 = 35
-11m = 35 + 20
-11m = 55
m = 55 ÷ ( -11 )
m = -5
Hence, from the above,
We can conclude that the value of m is: -5

Question 9.
5(3 – x) + 2(3 – x) = 14
Answer:
The value of x is: 1

Explanation:
The given equation is:
5 ( 3 – x ) + 2 ( 3 – x ) = 14
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
Now,
By using the Distributive property,
5 ( 3 – x ) = 5 (3 ) -5 (x)
= 15 – 5x
2 ( 3 – x ) = 2 (3) – 2 ( x)
= 6 – 2x
So,
15 – 5x + 6 – 2x = 14
21 – 7x = 14
7x = 21 – 14
7x = 7
x = 7 ÷ 7
x = 1
Hence, from the above,
We can conclude that the value of x is: 1

Monitoring Progress

Question 10.
The formula d = \(\frac{1}{2}\)n + 26 relates the nozzle pressure n (in pounds per square inch) of a fire hose and the maximum horizontal distance the water reaches d (in feet). How much pressure is needed to reach a fire 50 feet away?
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 28
Answer:
The pressure needed to reach a fire 50 feet away (n ) is: 48 pounds per square inch

Explanation:
It is given that the formula
d = \(\frac{1}{2}\)n + 26
relates the nozzle pressure n (in pounds per square inch) of a fire hose and the maximum horizontal distance the water reaches d (in feet).
So,
The given equation is:
d = \(\frac{1}{2}\)n + 26
Where,
d is the maximum horizontal distance
n is the pressure
It is also given that the maximum horizontal distance is: 50 feet
So,
50 = \(\frac{1}{2}\)n + 26
\(\frac{1}{2}\)n = 50 – 26
\(\frac{1}{2}\)n = 24
\(\frac{1}{2}\) × n = 24
n = 24 × 2
n = 48 pounds per square inch
Hence, from the above
We can conclude that the pressure needed to reach 50 feet away is: 48 pounds per square inch

Question 11.
Monitoring Progress
You have 96 feet of fencing to enclose a rectangular pen for your dog. To provide sufficient running space for your dog to exercise, the pen should be three times as long as it is wide. Find the dimensions of the pen.
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 29
Answer:
The dimensions of the pen are:
length of the pen: 12 feet
Width of the pen: 36 feet

Explanation;
It is given that you  have 96 feet of fencing to enclose a rectangular pen for your dog. To provide sufficient running space for your dog to exercise, the pen should be three times as long as it is wide.
So,
The perimeter of the rectangular pen is: 96 feet
We know that,
The perimeter of the rectangle = 2 (Length + Width )
It is also given that the pen is three times as long as it is wide
So,
Width = 3 × Length
So,
The perimeter of the rectangular pen =2 (  Length + ( 3 × Length ) )
96 = 2 ( 4 × Length )
4 × Length = 96 ÷ 2
4 × Length = 48
Length = 48 ÷ 4
Length = 12 feet
So,
Width = 3 × Length
= 3 × 12 = 36 feet
hence, from the above,
We can conclude that
The dimensions of the rectangular pen are:
Length of the pen is: 12 feet
Width of the pen is: 36 feet

Solving Multi-step Equations 1.2 Exercises

Monitoring Progress and Modeling with Mathematics

In Exercises 3−14, solve the equation. Check your solution.

Vocabulary and Core ConceptCheck

Question 1.
COMPLETE THE SENTENCE To solve the equation 2x + 3x = 20, first combine 2x and 3x because they are _________.
Answer:
The given equation is:
2x + 3x = 20
As 2x and 3x are combined by the symbol “+”, add 2x and 3x
So,
2x + 3x = 5x
So,
5x = 20
x = 20 ÷ 4
x = 5

Question 2.
WRITING Describe two ways to solve the equation 2(4x – 11) = 10.
Answer:
The given equation is:
2 (4x – 11) = 10
Way-1:
2 × (4x – 11) = 10
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
4x – 11 = 10 ÷ 2
4x – 11 = 5
4x = 5 + 11
4x = 16
x = 16 ÷ 4
x = 4
Hence,
The value of x is: 4

Way-2:
By using the Distributive Property,
2 (4x – 11) = 2 (4x) – 2 (11)
= 8x – 22
So,
8x – 22 = 10
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
8x = 10 + 22
8x = 32
x = 32 ÷ 8
x = 4
Hence,
The value of x is: 4

Question 3.
3w + 7 = 19

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.2-Q3

Question 4.
2g – 13 = 3
Answer:
The value of g is: 8

Explanation:
The given equation is:
2g – 13 = 3
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
2g = 3 + 13
2g = 16
2 × g = 16
g = 16 ÷ 2
g = 8
Hence, from the above,
We can conclude that the value of g is: 8

Question 5.
11 = 12 – q

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.2-Q5

Question 6.
10 = 7 – m
Answer:
The value of m is: -3

Explanation:
The given equation is:
10 = 7 – m
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
-m = 10 – 7
-m = 3
Multiply with “-” on both sides
– (-m ) = -3
m = -3
Hence, from the above,
We can conclude that the value of m is: -3

Question 7.
5 = \(\frac{z}{-4}\) – 3

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.2-Q7

Question 8.
\(\frac{a}{3}\) + 4 = 6
Answer:
The value of a is: 6

Explanation:
The given equation is:
\(\frac{a}{3}\) + 4 = 6
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
\(\frac{a}{3}\) = 6 – 4
\(\frac{a}{3}\) = 2
a = 2 × 3
a = 6
Hence, from the above,
We can conclude that the value of a is: 6

Question 9.
\(\frac{h + 6}{5}\) = 2

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.2-Q9

Question 10.
\(\frac{d – 8}{-2}\) = 12
Answer:
The value of d is: -16

Explanation:
The given equation is:
\(\frac{d – 8}{-2}\) = 12
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
d – 8 = 12 × (-2)
d – 8 = -24
d = -24 + 8
d = -16
Hence, from the above,
We can conclude that the value of d is: -16

Question 11.
8y + 3y = 44

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.2-Q11

Question 12.
36 = 13n – 4n
Answer:
The value of n is: 4

Explanation:
The given equation is:
36 = 13n – 4n
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
36 = 9n
9n = 36
n = 36 ÷ 9
n = 4
Hence, from the above,
We can conclude that the value of n is: 4

Question 13.
12v + 10v + 14 = 80

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.2-Q13

Question 14.
6c – 8 – 2c = -16
Answer:
The value of c is: -2

Explanation:
The given equation is:
6c – 8 – 2c = -16
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
4c – 8 = -16
4c = -16 + 8
4 × c = -8
c = -8 ÷ 4
c = -2
Hence, from the above,
We can conclude that the value of c is: -2

Question 15.
MODELING WITH MATHEMATICS
The altitude a (in feet) of a plane in minutes after liftoff is given by a = 3400t + 600. How many minutes after liftoff is the plane at an altitude of 21,000 feet?
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 30

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.2-Q15

Question 16.
MODELING WITH MATHEMATICS
A repair bill for your car is $553. The parts cost $265. The labor cost is $48 per hour. Write and solve an equation to find the number of hours of labor spent repairing the car.
Answer:
The number of hours of labor spent repairing the car is: 6 hours

Explanation:
It is given that a repair bill for your car is $553. The parts cost $265. The labor cost is $48 per hour.
Let the number of hours of labor spent repairing the car be: x
So,
The total bill to repair your car = ( The labor cost per hour ) × ( The number of hours of labor spent repairing the car ) +  (The cost of the parts )
553 = 48x + 265
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
48x = 553 – 265
48x = 288
48 × x = 288
x = 288 ÷ 48
x = 6
Hence, from the above,
We can conclude that the number of hours of labor spent repairing the car is: 6 hours

In Exercises 17−24, solve the equation. Check your solution.

Question 17.
4(z + 5) = 32

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.2-Q17

Question 18.
-2(4g – 3) = 3018.
Answer:
The value of g is: 378

Explanation:
The given equation is:
-2 (4g – 3) = 3018
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
-2 × ( 4g – 3 ) = 3018
4g – 3 = 3018 ÷ 2
4g – 3 = 1,509
4g = 1,509 +3
4 × g = 1,512
g = 1,512 ÷ 4
g = 378
Hence, from the above,
We can conclude that the value of g is: 378

Question 19.
6 + 5(m + 1) = 26

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.2-Q19

Question 20.
5h+ 2(11 – h) = -5
Answer:
The value of h is: -9

Explanation:
The given equation is:
5h + 2 ( 11-h ) = -5
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
By using the Distributive Property of Multiplication,
2 ( 11 – h ) = 2 (11 ) – 2 ( h )
= 22 – 2h
So,
5h + 22 – 2h = -5
3h + 22 = -5
3h = -5 – (+22)
3h = -5 -22
3h = -27
h = -27 ÷ 3
h = -9
Hence, from the above,
We can conclude that the value of h is: -9

Question 21.
27 = 3c – 3(6 – 2c)

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.2-Q21

Question 22.
-3 = 12y – 5(2y – 7)
Answer:
The value of y is: -19

Explanation:
The given equation is:
-3 = 12y – 5 (2y – 7)
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
12y – 5 (2y – 7) = -3
By using the Distributive Property of Multiplication,
5 ( 2y – 7 ) = 5 (2y ) – 5 (7 )
= 10y – 35
So,
12y – ( 10y – 35 ) = -3
12y – 10y + 35 = -3
2y + 35 = -3
2y = -3 – (+35 )
2y = -3 – 35
2y = -38
y = -38 ÷ 2
y = -19
Hence, from the above,
We can conclude that the value of y is: -19

Question 23.
-3(3 + x) + 4(x – 6) = -4

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.2-Q23

Question 24.
5(r + 9) – 2(1 – r) = 1
Answer:
The value of r is: -6

Explanation:
The given equation is:
5 ( r + 9 ) – 2 ( 1 – r ) = 1
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
Now,
By using the Distributive Property of Multiplication,
5 ( r + 9 ) = 5 ( r ) + 5 ( 9 )
= 5r + 45
2 ( 1 – r ) = 2 ( 1 ) – 2 ( r )
= 2 – 2r
So,
5r + 45 – ( 2 – 2r ) = 1
5r + 45 – 2 + 2r = 1
7r + 43 = 1
7r = 1 – 43
7r = -42
r = -42 ÷ 7
r = -6
Hence, from the above,
We can conclude that the value of r is: -6

USING TOOLS
In Exercises 25−28, find the value of the variable. Then find the angle measures of the polygon. Use a protractor to check the reasonableness of your answer.

Question 25.
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 31

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.2-Q25

Question 26.
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 32
Answer:
The angle measures of the rhombus are:
60°, 60°, 120°, 120°

Explanation:
The given figure is:
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 32
From the above figure,
The angle measures of the rhombus are: a°, 2a°, a°, 2a°
It is also given that the sum of all the angle measures is: 360°
So,
a° + 2a° + a° + 2a° = 360°
6a° = 360°
a = 360° ÷ 6
a = 60°
Hence, from the above,
We can conclude that the angle measures of the rhombus are:
a°, 2a°, a°, 2a° = 60°, 2 ( 60° ), 60°, 2 ( 60° )
= 60°, 60°, 120°, 120°

Question 27.
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 33

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.2-Q27

Question 28.
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 34
Answer:
The angle measures of the hexagon are:
120°, 120°, 100°, 120°, 250°, 260°

Explanation:
The given figure is:
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 34
From the given figure,
The angle measures of the hexagon are:
120°, 120°, 100°, 120°, x°, (x + 10)°
It is also given that the sum of the angle measures of the hexagon is: 720°
So,
120° + 120° + 100° + 120° + x° + (x + 10)° = 720°
470° + x = 720°
x = 720° – 470°
x = 250°
Hence, from the above,
We can conclude that the angle measures of the hexagon are:
120°, 120°, 100°, 120°, x°, (x + 10)° = 120°, 120°, 100°, 120°, 250°, (250 + 10)°
= 120°, 120°, 100°, 120°, 250°, 260°

In Exercises 29−34, write and solve an equation to find the number.

Question 29.
The sum of twice a number and 13 is 75.

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.2-Q29

Question 30.
The difference of three times a number and 4 is -19.
Answer:
The number is: -5

Explanation:
It is given that the difference of three times of a number and 4 is -19
Now,
Let the number be x
So,
The three times of a number = 3 (x) = 3x
So,
3x – 4 = -19
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
3x = -19 + 4
3x = -15
x = -15 ÷ 3
x = -5
Hence, from the above,
We can conclude that the number is: -5

Question 31.
Eight plus the quotient of a number and 3 is -2.

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.2-Q31

Question 32.
The sum of twice a number and half the number is 10.
Answer:
The number is: 4

Explanation:
It is given that the sum of twice of a number and half the number is 10.
Let the number be x.
So,
The twice of a number = 2 (x ) = 2x
Half of the number = x ÷ 2 = \(\frac{x}{2}\)
So,
2x + \(\frac{x}{2}\) = 10
2x can be rewritten as: \(\frac{4x}{2}\)
So,
\(\frac{4x}{2}\) + \(\frac{x}{2}\) = 10
\(\frac{4x + x}{2}\) = 10
\(\frac{5x}{2}\) = 10
5x = 10 × 2
5x = 20
x = 20 ÷ 5
x = 4
Hence, from the above,
We can conclude that the numebr is: 4

Question 33.
Six times the sum of a number and 15 is -42.

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.2-Q33

Question 34.
Four times the difference of a number and 7 is 12.
Answer:
The number is: 4

Explanation:
It is given that the four times the difference of a number and 7 is 12
Let the number be x
So,
Four times of the number = 4 ( x ) = 4x
So,
4x – x = 12
3x = 12
x = 12 ÷ 3
x = 4
Hence, from the above,
We can conclude that the number is: 4

USING EQUATIONS
In Exercises 35−37, write and solve an equation to answer the question. Check that the units on each side of the equation balance.

Question 35.
During the summer, you work 30 hours per week at a gas station and earn $8.75 per hour. You also work as a landscaper for $11 per hour and can work as many hours as you want. You want to earn a total of $400 per week. How many hours must you work as a landscaper?

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.2-Q35

Question 36.
The area of the surface of the swimming pool is 210 square feet. What is the length d of the deep end (in feet)?
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 35

Answer:
The length d of the deep end is: 12 feet

Explanation:
The given figure is:
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 35
It is given that the area of the surface of the swimming pool is 210 square feet
From the above figure,
We can observe that the shape of the swimming pool is a rectangle.
So,
Length of the swimming pool = 10 ft
Width of the swimming pool = d + 9 ft
So,
The area of the swimming pool = Length × Width
= 10 × ( d + 9 )
Now,
210 = 10 × ( d + 9 )
d + 9 = 210 ÷ 10
d + 9 = 21
d = 21 – 9
d = 12 feet
Hence, from the above,
We can conclude that the length d of the deep end is: 12 feet

Question 37.
You order two tacos and a salad. The salad costs $2.50. You pay 8% sales tax and leave a $3 tip. You pay a total of $13.80. How much does one taco cost?

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.2-Q37

JUSTIFYING STEPS
In Exercises 38 and 39, justify each step of the solution.

Question 38.
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 36

Answer:
–\(\frac{1}{2}\) ( 5x – 8 ) – 1 = 6                          Write the equation
–\(\frac{1}{2}\) ( 5x – 8 ) = 6 + 1                         Arrange the similar terms
–\(\frac{1}{2}\) ( 5x – 8 ) = 7                                  Simplify
– ( 5x – 8 ) = 7 × 2                                                              Divide by 2 on both sides
– ( 5x – 8 ) = 14                                                                    Simplify
5x – 8 = -14                                                                       Multiply with “-” on both sides
5x = -14 + 8                                                                         Arrange the similar terms
5x = -6                                                                               Divide by 6 on both sides
x = –\(\frac{6}{5}\)                                              The result
Hence,
The solution is: x = –\(\frac{6}{5}\)

Question 39.
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 37

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.2-Q39

ERROR ANALYSIS
In Exercises 40 and 41, describe and correct the error in solving the equation.

Question 40.
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 38

Answer:
The given equation is:
-2 ( 7 – y ) + 4 = -4
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
-2 ( 7 – y ) = -4 –  (+4 )
-2 ( 7 – y ) = -4 – 4
-2 ( 7 – y ) = -8
Now,
By using the Distributive Property of Multiplication,
2 ( 7 – y ) = 2 ( 7 ) – 2 ( y )
= 14 – 2y
So,
– ( 14 – 2y ) = -8
2y – 14 = -8
2y = -8 + 14
2y = 6
y = 6 ÷ 2
y = 3
Hence,
The value of y is: 3

Question 41.
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 39

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.2-Q41

MATHEMATICAL CONNECTIONS
In Exercises 42−44, write and solve an equation to answer the question.

Question 42.
The perimeter of the tennis court is 228 feet. What are the dimensions of the court?
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 40

Answer:
The dimensions of the court are:
The Length of the court is: 36 feet
The width of the court is: 78 feet

Explanation:
The given figure is:
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 40
It is given that the perimeter of the tennis court is: 228 feet
From the above figure,
We can observe that the shape of the tennis court is the rectangle
So,
The length of the tennis court = w feet
The width of the tennis court =  (2w + 6 ) feet
We know that,
The perimeter of the rectangle = 2 ( Length + Width )
So,
The perimeter of the tennis court = 2 ( Length + Width )
228 = 2 ( w + 2w + 6 )
By using the Distributive Property of Multiplication,
2 ( w + 2w + 6 ) = 2 ( 3w + 6 )
= 2 ( 3w ) + 2 ( 6 )
= 6w + 12
So,
228 = 6w + 12
6w = 228 – 12
6w = 216
w = 216 ÷ 6
w = 36
Hence, from the above,
We can conclude that
The length of the tennis court is: 36 feet
The width of the tennis court is: 2w + 6  = 2 ( 36 ) + 6 = 78 feet

Question 43.
The perimeter of the Norwegian flag is 190 inches. What are the dimensions of the flag?
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 41

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.2-Q43

Question 44.
The perimeter of the school crossing sign is 102 inches. What is the length of each side?
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 42

Answer:
The length of each side is: 15 inches

Explanation:
The given figure is:
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 42
It is given that the perimeter of the crossing sign is 102 inches
We know that,
The perimeter of any polygon is the sum of all the sides of that polygon
So,
The perimeter of the crossing sign = s + ( s + 6 ) + ( s + 6 ) + s + 2s
102 = 6s + 12
102 – 12 = 6s
6s = 90
s = 90 ÷ 6
s = 15 inches
Hence, from the above,
We can conclude that the length of each side is: 15 inches

Question 45.
COMPARING METHODS
Solve the equation 2(4 – 8x) + 6 = -1 using (a) Method 1 from Example 3 and (b) Method 2 from Example 3. Which method do you prefer? Explain.

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.2-Q45
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.2-Q45-i

Question 46.
PROBLEM – SOLVING
An online ticket agency charges the amounts shown for basketball tickets. The total cost for an order is $220.70. How many tickets are purchased?
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 43

Answer:

Question 47.
MAKING AN ARGUMENT
You have quarters and dimes that total $2.80. Your friend says it is possible that the number of quarters is 8 more than the number of dimes. Is your friend correct? Explain.

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.2-Q47

Question 48.
THOUGHT-PROVOKING
You teach a math class and assign a weight to each component of the class. You determine final grades by totaling the products of the weights and the component scores. Choose values for the remaining weights and find the necessary score on the final exam for a student to earn an A (90%) in the class, if possible. Explain your reasoning.
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 44

Answer:
The completed table is:

From the above table,
The weights can be calculated by the difference between the total participation and the class participation and divide the total value by 100.
So,
The weight of homework = [ ( 100 – 95 ) ÷ 100]
= 5 ÷ 100
= 0.50
The weight of midterm exam = ( 100 – 88 ) ÷ 100
= 12 ÷ 100
= 0.12
So,
The necessary score of the final exam = ( 92 + 95 + 88 ) % ÷ 3
= 275 % ÷ 3
= 91.6 %

Question 49.
REASONING
An even integer can be represented by the expression 2n, where n is an integer. Find three consecutive even integers that have a sum of 54. Explain your reasoning.

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.2-Q49

Question 50.
HOW DO YOU SEE IT?
The scatter plot shows the attendance for each meeting of a gaming club.
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 45
a. The mean attendance for the first four meetings is 20. Is the number of students who attended the fourth meeting greater than or less than 20? Explain.
Answer:
The number of students who attended the fourth meeting is greater than 20

Explanation:
The given graph is:
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 45
We know that
The mean = ( Sum of the given numbers ) ÷ (The total number of the numbers)
From the above graph,
The attendance of the 1st meeting = 18
The attendance of the 2nd meeting = 21
The attendance of the 3rd meeting = 17
Let the attendance of the 4th meeting be: x
So,
The mean attendance  of the first four meetings = ( The attendance of the 4 meetings ) ÷  ( The total number of meetings )
= ( 18 + 21 + 17 + x ) ÷ 4
It is given that the mean attendance of the first four meetings is: 20
So,
20 =  ( 18 + 21 + 17 + x ) ÷ 4
( 56 + x ) ÷ 4 = 20
56 + x = 20 × 4
56 + x = 80
x = 80 – 56
x = 24
Hence, from the above,
We can conclude that the attendance of the 4th meeting is greater than 20

b. Estimate the number of students who attended the fourth meeting.
Answer:
The number of students who attended the fourth meeting is: 24

Explanation:
The mean attendance  of the first four meetings = ( The attendance of the 4 meetings ) ÷  ( The total number of meetings )
= ( 18 + 21 + 17 + x ) ÷ 4
It is given that the mean attendance of the first four meetings is: 20
So,
20 =  ( 18 + 21 + 17 + x ) ÷ 4
( 56 + x ) ÷ 4 = 20
56 + x = 20 × 4
56 + x = 80
x = 80 – 56
x = 24
Hence, from the above,
We can conclude that the number of students who attended the 4th meeting is: 24

c. Describe a way you can check your estimate in part (b).
Answer:
The estimate in part (b) can be checked by using the property of the mean
So,
The mean attendance of the four meetings = ( The attendance of the four meetings ) ÷ ( The total number of meetings )
= ( 18 + 21 + 17 + 24 ) ÷ 4
= 80 ÷ 4
= 20
Hence, from the above,
We can conclude that the mean attendance of the four meetings is the same as given above.

REASONING
In Exercises 51−56, the letters a, b, and c represent nonzero constants. Solve the equation for x.

Question 51.
bx = -7

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.2-Q51

Question 52.
x + a = \(\frac{3}{4}\)
Answer:
The value of x is: \(\frac{3}{4}\) – a

Explanation:
The given equation is:
x + a = \(\frac{3}{4}\)
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
x =  \(\frac{3}{4}\) – a
Hence, from the above,
We can conclude that the value of a is: \(\frac{3}{4}\) – a

Question 53.
ax – b = 12.5

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.2-Q53

Question 54.
ax + b = c
Answer:
The value of x is: \(\frac{c – b}{a}\)

Explanation:
The given equation is:
ax + b = c
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
ax = c – b
x = \(\frac{c – b}{a}\)
Hence, from the above,
We can conclude that the value of x is: \(\frac{c – b}{a}\)

Question 55.
2bx – bx = -8

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.2-Q55

Question 56.
cx – 4b = 5b
Answer
The value of x is: \(\frac{9b}{c}\)

Explanation:
The given equation is:
cx – 4b = 5b
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
cx = 5b + 4b
cx = 9b
x = \(\frac{9b}{c}\)
Hence, from the above,
We can conclude that the value of x is: \(\frac{9b}{c}\)

Maintaining Mathematical Proficiency

Simplify the expression.

Question 57.
4m + 5 – 3m

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.2-Q57

Question 58.
9 – 8b + 6b
Answer:
9 – 8b + 6b
= 9 – (8b – 6b )
= 9 – 2b

Question 59.
6t + 3(1 – 2t) – 5

Answer:

Determine whether (a) x = −1 or (b) x = 2 is a solution of the equation.

Question 60.
x – 8 = -9
Answer:
x = -1 is a solution to the given equation

Explanation:
The given equation is:
x – 8 = -9
a) Let x = -1
So,
-1 – 8 = -9
-9 = -9
As LHS is equal to RHS
x = -1 is a solution of the given equation
b) Let x = 2
So,
2 – 8 = -9
-6 = -9
As LHS is not equal to RHS,
x = 2 is not a solution of the given equation

Question 61.
x + 1.5 = 3.5

Answer:

Question 62.
2x – 1 = 3
Answer:
x = 2 is a solution to the given equation

Explanation:
The given equation is:
2x – 1 = 3
a) Let x = -1
So,
2 ( -1 ) – 1 = 3
-2 – 1 = 3
-3 = 3
As LHS is not equal to RHS
x = -1 is not a solution to the given equation
b) Let x = 2
So,
2 ( 2 ) -1 = 3
4 – 1 = 3
3 = 3
As LHS is equal to RHS,
x = 2  is a solution to the given equation

Question 63.
3x + 4 = 1

Answer:

Question 64.
x + 4 = 3x
Answer:
x = 2 is a solution to the given equation

Explanation:
The given equation is:
x + 4 = 3x
a) Let x = -1
So,
-1 + 4 = 3 ( -1 )
= 3 = -3
As LHS is not equal to RHS,
x = -1 is not a solution to the given equation
b) Let x = 2
So,
2 + 4 = 3 ( 2 )
6 = 6
As LHS is equal to RHS,
x = 2 is a solution to the given equation.

Question 65.
-2(x – 1) = 1 – 3x

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.2-Q65

Lesson 1.3 Solving Equations with Variables on Both Sides

EXPLORATION 1
Perimeter

Work with a partner. The two polygons have the same perimeter. Use this information to write and solve an equation involving x. Explain the process you used to find the solution. Then find the perimeter of each polygon.
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 46
Answer:
The perimeter of the hexagon is: 6
The perimeter of the square is: 6

Explanation:
The given figures are:
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 46
Based on the number of sides (n) in the polygon,
We can say the name of that polygon
So,
In the first figure,
The number of sides is: 6
So,
It is Hexagon
In the second figure,
The number of sides is: 4
So,
It is square ( As all the angles are 90° )
It is given that the two polygons have the same perimeter
We know that,
“Perimeter” of a polygon is defined as the sum of all the sides in the polygon
So,
The sum of all sides in the hexagon = 5 + 2 + 5 + 2 + x + x
= 14 + 2x
The sum of all sides in the square = \(\frac{3x}{2}\) + 3 + 4 + 5
= \(\frac{3x}{2}\) + 12
It is given that the perimeter of both the polygons are equal
So,
14 + 2x = \(\frac{3x}{2}\) + 12
14 – 12 = \(\frac{3x}{2}\) – 2x
\(\frac{3x}{2}\) – 2x = 2
We can write 2x as \(\frac{4x}{2}\)
So,
\(\frac{3x}{2}\) – \(\frac{4x}{2}\) = 2
\(\frac{3x – 4x}{2}\) = 2
\(\frac{-x}{2}\) = 2
– \(\frac{x}{2}\) = 2
-x = 2 × 2
-x = 4
x = -4
Hence,
The perimeter of the Hexagon = 14 + 2x = 14 + 2 ( -4 )
= 14 – 8 = 6
The perimeter of the square = \(\frac{3x}{2}\) + 12
\(\frac{3 × -4}{2}\) + 12
= \(\frac{-12}{2}\) + 12
= \(\frac{-12}{2}\) + \(\frac{24}{2}\)
= \(\frac{24 – 12}{2}\)
= \(\frac{12}{2}\)
= 6
Hence, from the above,
We can conclude that
The perimeter of the Hexagon is: 6
The perimeter of the square is: 6

EXPLORATION 2
Perimeter and Area

Work with a partner.

  • Each figure has the unusual property that the value of its perimeter (in feet) is equal to the value of its area (in square feet). Use this information to write an equation for each figure.
  • Solve each equation for x. Explain the process you used to find the solution.
  • Find the perimeter and area of each figure.

Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 47

Question 3.
How can you solve an equation that has variables on both sides?
Answer:
If the variable is the same on both sides in an equation, then rearrange the like terms
So,
Separate the variables and the numbers and simplify the variables and the numbers
In this way,
We can solve an equation with a single variable

Question 4.
Write three equations that have the variable x on both sides. The equations should be different from those you wrote in Explorations 1 and 2. Have your partner solve the equations.
Answer:
Let the three equations that have variable x on both sides and different from Explorations 1 and 2 are:
a) 6x + 2 = 5x-6
b) 16x = 18x – 2
c) 12x = 15x + 63
Now,
a)
The given equation is:
6x + 2 = 5x – 6
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
6x – 5x = -6 – 2
x = -8
Hence,
The value of x is: -8
b) The given equation is:
9x = 18x – 2
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
18x – 16x = 2
2x = 2
x = 2 ÷ 2
x = 1
Hence,
The value of x is: 1
c) The given equation is:
12x = 15x + 63
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
12x -15x = 63
-3x = 63
x = 63 ÷ ( -3 )
x = -63 ÷ 3
x = -21
Hence,
The value of x is: -21

1.3 Lesson

Monitoring Progress

Solve the equation. Check your solution.

Question 1.
-2x = 3x + 10
Answer:
The value of x is: -2

Explanation:
The given equation is:
-2x = 3x + 10
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
-2x – 3x = 10
-5x = 10
x = 10 ÷ (-5)
x = -10 ÷ 5
x = -2
Hence, from the above,
We can conclude that the value of x is: -2

Question 2.
\(\frac{1}{2}\)(6h – 4) = -5h + 1
Answer:
The value of h is: \(\frac{3}{8}\)

Explanation:
The given equation is:
\(\frac{1}{2}\) ( 6h – 4 ) = -5h + 1
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
6h – 4 = 2 ( -5h + 1 )
6h – 4 = 2 ( -5h ) + 2 ( 1 ) [ By using the Distributive Property of Multiplication )
6h – 4 = -10h + 2
6h + 10h = 2 + 4
16h = 6
h = \(\frac{6}{16}\)
h= \(\frac{3}{8}\)
Hence, from the above,
We can conclude that the value of h is: \(\frac{3}{8}\)

Question 3.
–\(\frac{3}{4}\)(8n + 12) = 3(n – 3)
Answer:
The value of n is: 0

Explanation:
The given equation is:
–\(\frac{3}{4}\) ( 8n + 12 ) = 3 ( n – 3 )
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
8n + 12 = –\(\frac{4}{3}\) × 3 ( n – 3 )
8n + 12 = –\(\frac{4}{3}\) \(\frac{3}{1}\) ( n – 3 )
8n + 12 = –\(\frac{3 × 4}{3 × 1}\) ( n – 3 )
8n + 12 = -4 ( n – 3 )
8n + 12 = -4n – 4 ( -3 )
8n + 12 = -4n + 12
8n + 4n =12 – 12
12n = 0
n = 0
Hence, from the above,
We can conclude that the value of n is: 0

Monitoring Progress

Solve the equation.

Question 4.
4(1 – p) = 4p – 4
Answer:
The value of p is: 1

Explanation:
The given equation is:
4 ( 1 -p ) = 4p – 4
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
4 ( 1 ) – 4 ( p ) = 4p – 4
4 – 4p = 4p – 4
4p + 4p = 4 + 4
8p = 8
p = 8 ÷ 8
p = 1
Hence, from the above,
We can conclude that the value of p is: 1

Question 5.
6m – m = –\(\frac{5}{6}\)(6m – 10)
Answer:
The value of m is: \(\frac{5}{6}\)

Explanation:
The given equation is:
6m – m = –\(\frac{5}{6}\) ( 6m – 10 )
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
5m = –\(\frac{5}{6}\) ( 6m – 10 )
5m = –\(\frac{5}{6}\) ( 6m ) – ( –\(\frac{5}{6}\) ( 10 ) )
5m = –\(\frac{5}{6}\) × \(\frac{6m}{1}\) + \(\frac{5}{6}\) \(\frac{10}{1}\)
5m = –\(\frac{5 × 6m}{6 × 1}\) + \(\frac{5 × 10}{6 × 1}\)
5m = -5m + \(\frac{25}{3}\)
5m + 5m = \(\frac{25}{3}\)
10m = \(\frac{25}{3}\)
m = \(\frac{25}{3}\) ÷ \(\frac{10}{1}\)
m = \(\frac{25}{3}\) × \(\frac{1}{10}\)
m = \(\frac{25}{30}\)
m = \(\frac{5}{6}\)
Hence, from the above,
We can conclude that the value of m is: \(\frac{5}{6}\)

Question 6.
10k + 7 = -3 – 10k
Answer:
The value of k is: –\(\frac{1}{2}\)

Explanation:
The given equation is:
10k + 7 = -3 – 10k
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
10k + 10k = -3 – ( +7 )
20k = -3 –
20k = -10
k = -10 ÷ 20
k = –\(\frac{1}{2}\)
Hence, from the above,
We can conclude that the value of k is: –\(\frac{1}{2}\)

Question 7.
3(2a – 2) = -2(3a – 3)
Answer:
The value of a is: 1

Explanation:
The given equation is:
3 ( 2a – 2 ) = -2 ( 3a – 3 )
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
By using the Distributive Property of Multiplication,
3 ( 2a ) – 3 ( 2 ) = -2 ( 3a ) + 2 ( 3 )
6a – 6 = -6a + 6
6a + 6a = 6 + 6
12a = 12
a = 12 ÷ 12
a = 1
Hence, from the above,
We can conclude that the value of a is: 1

Concept Summary

Steps for Solving Linear Equations
Here are several steps you can use to solve a linear equation. Depending on the equation, you may not need to use some steps.
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 48
Step 1
Use the Distributive Property to remove any grouping symbols.
Step 2
Simplify the expression on each side of the equation.
Step 3
Collect the variable terms on one side of the equation and the constant terms on the other side.
Step 4
Isolate the variable.
Step 5
Check your solution.

Monitoring Progress

Question 8.
A boat travels upstream on the Mississippi River for 3.5 hours. The return trip only takes 2.5 hours because the boat travels 2 miles per hour faster downstream due to the current. How far does the boat travel upstream? Answer:
The distance the boat travel upstream is: 17.5 miles

Explanation:
It is given that a boat travels upstream on the Mississippi River for 3.5 hours. The return trip only takes 2.5 hours because the boat travels 2 miles per hour faster downstream due to the current.
Now,
Let x be the speed of the boat traveled upstream
We know that,
Speed = Distance ÷ Time
Distance = Speed × Time
It is given that the time taken by the boat traveled upstream is: 3.5 hours
So,
Distance traveled upstream = 3.5 × x = 3.5x
Now,
It is also given that the speed of the boat is 2 miles per hour faster downstream
So,
Distance traveled downstream by boat = 2.5 ( x + 2 )
SO,
As both the distances are the same,
3.5x = 2.5 ( x + 2 )
By using the Distributive Property of Multiplication,
3.5x = 2.5 ( x) + 2.5 ( 2 )
3.5x = 2.5x + 5
3.5x – 2.5x = 5
x = 5
So,
The distance traveled upstream by boat = 3.5x = 3.5 ( 5 )
= 17.5 miles per hour
Hence, from the above,
We can conclude that the distance traveled upstream by boat is: 17.5 miles per hour

Solving Equations with Variables on Both Sides 1.3 Exercises

Monitoring Progress and Modeling with Mathematics

In Exercises 3–16, solve the equation. Check your solution.

Question 1.
VOCABULARY Is the equation -2(4 – x) = 2x + 8 an identity? Explain your reasoning.
Answer:
-2 ( 4 – x ) = 2x + 8 is not an identity

Explanation:
The given equation is:
-2 ( 4 – x ) = 2x + 8
By using the Distributive Property of Multiplication,
-2 ( 4 ) + 2 ( x ) = 2x + 8
-8 + 2x = 2x + 8
2x – 8 = 2x + 8
As
LHS ≠ RHS
-2 ( 4 – x ) = 2x + 8  is not an identity

Question 2.
WRITING Describe the steps in solving the linear equation 3(3x – 8) = 4x + 6
Answer:
The value of x is: 6

Explanation:
The given equation is:
3 ( 3x – 8 ) = 4x + 6
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
Now,
By using the Distributive Property of Multiplication,
3 ( 3x ) – 3 ( 8 ) = 4x + 6
9x – 24 = 4x + 6
9x – 4x = 6 + 24
5x = 30
x = 30 ÷ 5
x = 6
Hence, from the above,
We can conclude that the value of x is: 6

Question 3.
15 – 2x = 3x

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.3-Q3

Question 4.
26 – 4s = 9s
Answer:
The value of s is: 2

Explanation:
The given equation is:
26 – 4s = 9s
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
26 = 9s + 4s
13s = 26
s = 26 ÷ 13
s = 2
Hence, from the above,
We can conclude that the value of s is: 2

Question 5.
5p – 9 = 2p + 12

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.3-Q5

Question 6.
8g + 10 = 35 + 3g
Answer:
The value of g is: 5

Explanation:
The given equation is:
8g + 10 = 35 + 3g
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
8g – 3g = 35 – 10
5g = 25
g = 25 ÷ 5
g = 5
Hence, from the above,
We can conclude that the value of g is: 5

Question 7.
5t + 16 = 6 – 5t

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.3-Q7

Question 8.
-3r + 10 = 15r – 8
Answer:
The value of r is: 1

Explanation:
The given equation is:
-3r + 10 = 15r – 8
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
-3r – 15r = -10 – 8
-18r = -18
r = -18 ÷ ( -18 )
r = 1 [ since – ÷ – = + ]
Hence, from the above,
We can conclude that the value of r is: 1

Question 9.
7 + 3x – 12x = 3x + 1

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.3-Q9

Question 10.
w – 2 + 2w = 6 + 5w
Answer:
The value of w is: -4

Explanation:
The given equation is:
w – 2 + 2w = 6 + 5w
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
w + 2w -5w =6 + 2
-2w = 8
w = -8 ÷ 2
w = -4
Hence, from the above,
We can conclude that the value of w is: -4

Question 11.
10(g + 5) = 2(g + 9)

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.3-Q11

Question 12.
-9(t – 2) = 4(t – 15)
Answer:
The value of t is: 6

Explanation:
The given equation is:
-9 ( t – 2 ) = 4 ( t – 15 )
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
Now,
By using the Distributive Property of Multiplication,
-9 ( t ) +  9 ( 2 ) = 4 ( t ) – 4 ( 15 )
-9t + 18 = 4t – 60
-9t – 4t = -60 – 18
-13t = -78
t = -78 ÷ ( -13 )
t = 6 [ Since  -÷ – = + ]
Hence, from the above,
We can conclude that the value of t is: 6

Question 13.
\(\frac{2}{3}\)(3x + 9) = -2(2x + 6)

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.3-Q13

Question 14.
2(2t + 4) = \(\frac{3}{4}\)(24 – 8t)
Answer:
The value of t is: 1

Explanation:
The given equation is:
2 ( 2t + 4 ) = \(\frac{3}{4}\) ( 24 – 8t )
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
By using the Distributive Property of Multiplication,
2 ( 2t ) + 2 ( 4 ) = \(\frac{3}{4}\) ( 24 ) – 8t  (\(\frac{3}{4}\) )
4t + 8 = \(\frac{3}{4}\) × \(\frac{24}{1}\) – \(\frac{8t}{1}\) × \(\frac{3}{4}\)
4t + 8 = \(\frac{3 × 24}{4 × 1}\) – \(\frac{3 × 8t}{4 × 1}\)
4t + 8 = \(\frac{18}{1}\) – \(\frac{6t}{1}\)
4t + 8 = 18 – 6t
4t + 6t = 18 – 8
10t = 10
t = 10 ÷ 10
t = 1
Hence, from the above,
We can conclude that the value of t is: 1

Question 15.
10(2y + 2) – y = 2(8y – 8)

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.3-Q15

Question 16.
2(4x + 2) = 4x – 12(x – 1)
Answer:
The value of x is: \(\frac{1}{2}\)

Explanation:
The given equation is:
2 ( 4x + 2 ) = 4x – 12 ( x – 1 )
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
Now,
By using the Distributive Property of Multiplication,
2 ( 4x ) + 2 ( 2 ) = 4x – 12 ( x ) + 12 ( 1 ) [ Since – × – = + ]
8x + 4 = 4x – 12x + 12
8x + 4 =12 – 8x
8x + 8x = 12 – 4
16x = 8
x = 8 ÷ 16
x = \(\frac{1}{2}\)
Hence, from the above,
We can conclude that the value of x is: \(\frac{1}{2}\)

Question 17.
MODELING WITH MATHEMATICS
You and your friend drive toward each other. The equation 50h = 190 – 45h represents the number h of hours until you and your friend meet. When will you meet?

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.3-Q17

Question 18.
MODELING WITH MATHEMATICS
The equation 1.5r + 15 = 2.25r represents the number r of movies you must rent to spend the same amount at each movie store. How many movies must you rent to spend the same amount at each movie store?
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 49

Answer:
The number of movies you rent to spend the same amount at each movie store is: 20

Explanation:
It is given that
The equation 1.5r + 15 = 2.25r represents the number r of movies you must rent to spend the same amount at each movie store.
Now,
We have to find the value of r to find the number f movies you must rent to spend the same amount at each movie store
So,
The given equation is:
1.5r + 15 = 2.25r
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
2.25r – 1.5r = 15
0.75r = 15
\(\frac{75}{100}\)r = 15
r = 15 × \(\frac{100}{75}\)
r = \(\frac{15}{1}\) × \(\frac{100}{75}\)
r = \(\frac{15 × 100}{1 × 75}\)
r = 20
Hence, from the above,
We can conclude that the number of movies you rent to spend the same amount at each movie store is: 20

In Exercises 19–24, solve the equation. Determine whether the equation has one solution, no solution, or infinitely many solutions.

Question 19.
3t + 4 = 12 + 3t

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.3-Q19

Question 20.
6d + 8 = 14 + 3d
Answer:
The value of d is: 2

Explanation:
The given equation is:
6d + 8 = 14 + 3d
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
6d – 3d = 14 – 8
3d = 6
d = 6 ÷ 3
d = 2
Hence, from the above,
We can conclude that the value of d is: 2

Question 21.
2(h + 1) = 5h – 7

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.3-Q21

Question 22.
12y + 6 = -6(2y + 1)
Answer:
The value of y is: –\(\frac{1}{2}\)

Explanation:
The given equation is:
12y + 6 = -6 ( 2y + 1 )
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
By using the Distributive Property of Multiplication,
12y + 6 = -6 ( 2y ) – 6 ( 1 )
12y + 6 = -12y – 6
12y + 12y = -6 – ( +6 )
24y = -6 – 6
24y = -12
y = -12 ÷ 24
y = –\(\frac{1}{2}\)
Hence, from the above,
We can conclude that the value of y is: –\(\frac{1}{2}\)

Question 23.
3(4g + 6) = 2(6g + 9)

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.3-Q23

Question 24.
5(1 + 2m) = \(\frac{1}{2}\)(8 + 20m)
Answer:
m has indefinite solutions

Explanation:
The given equation is:
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
By using the Distributive Property of Multiplication,
5 ( 1 ) + 5 ( 2m ) = \(\frac{1}{2}\) ( 8 ) + \(\frac{1}{2}\) ( 20m )
2 ( 5 + 10m ) = 8 + 20m
2 ( 5 ) + 2 ( 10m ) = 8 + 20m
10 + 20m = 8 + 20m
20m – 20m = 8 – 10
20m – 20m = -2
As  m has the same coefficients and have the opposite signs, m has indefinite solutions
Hence, from the above,
We can conclude that the equation has the indefinite solutions

ERROR ANALYSIS
In Exercises 25 and 26, describe and correct the error in solving the equation.

Question 25.
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 50

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.3-Q25

Question 26.
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 51

Answer:
The given equation is:
6 ( 2y + 6 ) = 4 ( 9 + 3y )
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
Now,
By using the Distributive Property of Multiplication,
6 ( 2y ) + 6 ( 6 ) = 4 ( 9 ) + 4 ( 3y )
12y + 36 = 36 + 12y
12y – 12y = 36 – 36
0 = 0
As the coefficients of y are zero, the equation has no solution
Hence, from the above,
We can conclude that there is no error in the analysis of the equation.

Question 27.
MODELING WITH MATHEMATICS
Write and solve an equation to find the month when you would pay the same total amount for each Internet service.
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 52

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.3-Q27

Question 28.
PROBLEM-SOLVING
One serving of granola provides 4% of the protein you need daily. You must get the remaining 48 grams of protein from other sources. How many grams of protein do you need daily?
Answer:
The number of grams of protein you need daily is: 50 grams

Explanation:
It is given that one serving of granola provides 4% of the protein you need daily. You must get the remaining 48 grams of protein from other sources.
So,
Let the number of grams of protein you need daily be: x
So,
The number of grams of protein you need daily = 4 % of x + 48
We know that,
100%  = 1
So,
4 % = 0.04
So,
The number of grams of protein you need daily = 0.04x + 48
x = 0.04x + 48
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
x – 0.04x = 48
0.96x = 48
\(\frac{96}{100}\)x = 48
x = 48 × \(\frac{100}{96}\)
x = \(\frac{48}{1}\) × \(\frac{100}{96}\)
x = \(\frac{48 × 100}{1 × 96}\)
x = \(\frac{50}{1}\)
x = 50 grams
Hence, from the above,
We can conclude that the number of proteins you need daily is: 50 grams

USING STRUCTURE
In Exercises 29 and 30, find the value of r.

Question 29.
8(x + 6) – 10 + r = 3(x + 12) + 5x

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.3-Q29

Question 30.
4(x – 3) – r + 2x = 5(3x – 7) – 9x
Answer:
The value of r is: 23

Explanation:
The given equation is:
4 ( x – 3 ) – r + 2x = 5 ( 3x – 7 ) – 9x
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
By using the Distributive Property of Multiplication,
4x – 4 ( 3 ) – r + 2x = 5 ( 3x ) – 5 ( 7 ) – 9x
4x – 12 – r + 2x = 15x – 35 – 9x
6x – 12 – r = 6x – 35
r = 6x – 6x – 12 + 35
r = 23
Hence, from the above,
We can conclude that the value of r is: 23

MATHEMATICAL CONNECTIONS
In Exercises 31 and 32, the value of the surface area of the cylinder is equal to the value of the volume of the cylinder. Find the value of x. Then find the surface area and volume of the cylinder.

Question 31.
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 53

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.3-Q31

Question 32.
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 54
Answer:
The Surface Area of the cylinder is: 461.49 cm²
The volume of the cylinder is: 488.58 cm³

Explanation:
The given figure is:
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 54
From the above figure,
The radius of the cylinder is: 7\(\frac{1}{5}\) feet
The height of the cylinder is: x feet
It is given that the Total Surface Area of the cylinder and the volume of the cylinder are equal.
We know that,
The Surface Area of the cylinder = 2πr² + 2πrh
The volume of the cylinder = πr²h
The value of π is: 3.1416
Now,
The representation of 7\(\frac{1}{5}\) in the improper fraction form is: \(\frac{36}{5}\)
So,
2πr² + 2πrh = πr²h
[ 2 × 3.1416 × \(\frac{36}{5}\) × \(\frac{36}{5}\) ] + [ 2 × 3.1416 × \(\frac{36}{5}\) × x ] = [3.1416 × latex]\frac{36}{5}[/latex] × \(\frac{36}{5}\) × x ]
325.72 + 45.23x = 162.86x
162.86x – 45.23x = 325.72
117.63x = 325.72
x = 2.76
x = 3
So,
The Surface Area of the cylinder = 2πr² + 2πrh
= [ 2 × 3.1416 × \(\frac{36}{5}\) × \(\frac{36}{5}\) ] + [ 2 × 3.1416 × \(\frac{36}{5}\) × x ]
= 325.72 + 45.23x
= 325.72 + 45.23 ( 3 )
= 461.49 cm²
The volume of the cylinder = πr²h
= [3.1416 × latex]\frac{36}{5}[/latex] × \(\frac{36}{5}\) × x ]
= 162.86x
= 162.86 ( 3 )
= 488.58 cm³
Hence, from the above,
We can conclude that
The Surface Area of the cylinder is: 461.49 cm²
The volume of the cylinder is: 488.58 cm³

Question 33.
MODELING WITH MATHEMATICS
A cheetah that is running 90 feet per second is 120 feet behind an antelope that is running 60 feet per second. How long will it take the cheetah to catch up to the antelope?

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.3-Q33

Question 34.
MAKING AN ARGUMENT
A cheetah can run at top speed for only about 20 seconds. If an antelope is too far away for a cheetah to catch it in 20 seconds, the antelope is probably safe. Your friend claims the antelope in Exercise 33 will not be safe if the cheetah starts running 650 feet behind it. Is your friend correct? Explain.
Answer:
Your friend is not correct

Explanation:
It is given that a cheetah can run at top speed for only about 20 seconds. If an antelope is too far away for a cheetah to catch it in 20 seconds, the antelope is probably safe. Your friend claims the antelope in Exercise 33 will not be safe if the cheetah starts running 650 feet behind it.
Let the distance of running antelope be x.
Let ‘t’ be the time taken
So,
The distance of running Antelope is:
x = 650 + 60t
The cheetah must arrive at the same position to catch the antelope
So,
x = 90t
Now,
90t = 650 + 60t
90t – 60t = 650
30t = 650
t = 650 ÷ 30
t = 21.7 seconds
But it is given that the cheetah has to reach the same position as the antelope in 20 seconds
But according to the calculation, it takes 21.7 seconds
So,
According to your friend, the antelope is not safe if the cheetah is running 650 meters behind it.
Hence, from the above,
We can conclude that your friend is not correct.

REASONING
In Exercises 35 and 36, for what value of a is the equation an identity? Explain your reasoning.

Question 35.
a(2x + 3) = 9x + 15 + x

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.3-Q35

Question 36.
8x – 8 + 3ax = 5ax – 2a
Answer:
The given equation becomes an identity at a = 4

Explanation:
The given equation is:
8x – 8 + 3ax = 5ax – 2a
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
5ax – 3ax = 8x – 8 – 2a
2ax = 8 ( x – 1 ) – 2a
2ax + 2a = 8x – 8
Equate the like coefficients of x and the like constants in both LHS and RHS
So,
2ax = 8x                                   2a = -8
a = 8x ÷ 2x                               a = -8 ÷ 2
a = 4                                         a = -4
Now,
At a = 4,
The equation becomes
8x – 8 + 3ax = 5ax – 2a
8x – 8 + 3x ( 4 ) = 5x ( 4 ) -2 ( 4 )
8x – 8 + 12x = 20x – 8
20x – 8 = 20x – 8
Hence,
At a =4,
The given equation is an Identity
At a = -4,
The equation becomes
8x – 8 + 3ax = 5ax – 2a
8x – 8 + 3x ( -4 ) = 5x ( -4 ) -2 ( -4 )
8x – 8 – 12x = -20x + 8
-4x – 8 = -20x + 8
Hence,
At a = -4, the given equation is not an Identity
Hence, from the above,
We can conclude that the given equation is an Identity at a = 4

Question 37.
REASONING
Two times the greater of two consecutive integers is 9 less than three times the lesser integer. What are the integers?

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.3-Q37

Question 38.
HOW DO YOU SEE IT?
The table and the graph show information about students enrolled in Spanish and French classes at a high school.
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 55
a. Use the graph to determine after how many years there will be equal enrollment in Spanish and French classes.
Answer:
The year where there will be equal enrollment in Spanish and French classes is: 6

Explanation:
The given table and the graph of the students for the Spanish and French classes are:
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 55
So,
From the above table,
To find the year where there is equal enrollment in both Spanish and French, we have to find the point in the graph that the two points of Spanish and French intersect
So,
From the above graph,
The point that is intersecting is at 6th year
Hence, from the above,
We can conclude that there is an equal enrollment of students in the 6th year in both Spanish and French classes

b. How does the equation 355 – 9x = 229 + 12x relate to the table and the graph? How can you use this equation to determine whether your answer in part (a) is reasonable?
Answer:
The given equation
355 – 9x = 229 + 12x
represents the total number of students enrolled in the different years in both Spanish and French classes
Now,
In part (a),
We observed that there is an equal enrollment of the students at 6th year in both Spanish and French classes
So,
Here,
x is: The number of years
So,
In part (a),
x = 6
Now,
Substitute x = 6 in the given equation.
Now,
355 – 9x = 229 + 12x
355 – 9 ( 6 ) = 229 + 12 ( 6 )
355 – 54 = 229 + 72
301 = 301
As
LHS = RHS
We can say that the answer is reasonable in part (a)

Question 39.
WRITING EQUATIONS
Give an example of a linear equation that has (a) no solution and (b) infinitely many solutions. Justify your answers.

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.3-Q39

Question 40.
THOUGHT-PROVOKING
Draw a different figure that has the same perimeter as the triangle shown. Explain why your figure has the same perimeter.
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 56

Answer:

Maintaining Mathematical Proficiency

Order the values from least to greatest.

Question 41.
9, | -4|, -4, 5, | 2 |

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.3-Q41

Question 42.
| -32 |, 22, -16, -| 21 |, | -10 |
Answer:
We know that,
| -x | = x
| x | = x
So,
| -32 | = 32
| 21 | = 21
| -10 | = 10
Hence,
The order of the values from the least to the greatest is:
-21, -16, 10, 22, 32

Question 43.
-18, | -24 |, -19, | -18 |, | 22 |

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.3-Q43

Question 44.
-| – 3 |, | 0 |, -1, | 2 |, -2
Answer:
We know that,
| -x | = x
| x | = x
So,
| -3 | = 3
| 0 | = 0
| 2 | = 2
Hence,
The order of the numbers from the least to the greatest is:
-3, -2, -1, 0, 2

Solving Linear Equations Study Skills: Completing

1.1-1.3 What Did You Learn

Core Vocabulary

Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 57

Core Concepts

Section 1.1
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 58

Section 1.2

Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 59

Section 1.3

Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 60

Mathematical Practices

Question 1.
How did you make sense of the relationships between the quantities in Exercise 46 on page 9?
Answer:
In Exercise 46 on page 9,
There is a layout of the tatami mat which comprises the four identical rectangular mats and the one square mat.
it is also given that the area of the square mat is half of one of the rectangular mats
Now,
We know that,
The area of the square mat = Area² [ Since all the sides of the square are equal ]
The area of the rectangular mat = Length × Width
So,
According to the given condition,
The relation between the area of the square mat and one of the rectangular mat is:
Area of the square mat = \(\frac{1}{2}\) Area of one of the rectangular mat
Side² = \(\frac{1}{2}\) ( Length × Width )

Question 2.
What is the limitation of the tool you used in Exercises 25–28 on page 16?
Answer:
The limitations of the tool you used in Exercises 25 – 28 on page 16 are:
A) The calculated values and the values measured using the tool will be different
B) We won’t get the exact values of the angle measures using the tool

Question 3.
What definition did you use in your reasoning in Exercises 35 and 36 on page 24?
Answer:
The definition you used in your reasoning in Exercises 35 and 36 on page 24 is:
Make the like coefficients of the same variable in both LHS and RHS equal so that we get the value of the variable.

Study Skills

Completing Homework Efficiently

Before doing homework, review the Core Concepts and examples. Use the tutorials at BigIdeasMath.com for additional help.

Complete homework as though you are also preparing for a quiz. Memorize different types of problems, vocabulary, rules, and so on.
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 61

Solving Linear Equations 1.1-1.3 Quiz

Solve the equation. Justify each step. Check your solution. (Section 1.1)

Question 1.
x + 9 = 7
Answer:
The value of x is -2

Explanation:
The given equation is:
x + 9 = 7
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
x = 7 –  ( +9 )
x = 7 – 9
x = -2
Hence, from the above,
We can conclude that the value of x is: -2

Question 2.
8.6 = z – 3.8
Answer:
The value of z is: 12.4

Explanation:
The given equation is:
8.6 = z – 3.8
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
z = 8.6 + 3.8
z = 12.4
Hence, from the above,
We can conclude that the value of z is: 12.4

Question 3.
60 = -12r
Answer:
The value of r is:  -5

Explanation:
The given equation is:
60 = -12r
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
r = 60 ÷ ( -12 )
r = -60 ÷ 12
r = -5
Hence, from the above,
We can conclude that the value of r is: -5

Question 4.
\(\frac{3}{4}\)p = 18
Answer:
The value of p is: 24

Explanation:
The given equation is:
\(\frac{3}{4}\)p = 18
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
p = 18 × \(\frac{4}{3}\)
p = \(\frac{18}{1}\) × \(\frac{4}{3}\)
p = \(\frac{18 × 4}{1 × 3}\)
p = \(\frac{24}{1}\)
p = 24
Hence, from the above,
We can conclude that the value of p is: 24

Solve the equation. Check your solution. (Section 1.2)

Question 5.
2m – 3 = 13
Answer:
The value of m is: 8

Explanation:
The given equation is:
2m – 3 = 13
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
2m = 13 + 3
2m = 16
m = 16 ÷ 2
m = 8
Hence, from the above,
We can conclude that the value of m is: 8

Question 6.
5 = 10 – v
Answer:
The value of v is: 5

Explanation:
The given equation is:
5 = 10 – v
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
v= 10 – 5
v = 5
Hence, from the above,
We can conclude that the value of v is: 5

Question 7.
5 = 7w + 8w + 2
Answer:
The value of w is: \(\frac{1}{5}\)

Explanation:
The given equation is:
5 = 7w + 8w + 2
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
5 – 2 = 7w + 8w
15w = 3
w = 3 ÷ 15
w = \(\frac{1}{5}\)
Hence, from the above,
We can conclude that the value of w is: \(\frac{1}{5}\)

Question 8.
-21a + 28a – 6 = -10.2
Answer:
The value of a is: –\(\frac{3}{5}\)

Explanation:
The given equation is:
-21a + 28a  – 6 = -10.2
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
-21a + 28a = -10.2 + 6
7a = -4.2
a = -4.2 ÷ 7
a= –\(\frac{42}{10}\) ÷ 7
a = –\(\frac{42}{10}\) × \(\frac{1}{7}\)
a = –\(\frac{42 × 1}{10 × 7}\)
a = – \(\frac{6}{10}\)
a = –\(\frac{3}{5}\)
Hence, from the above,
We can conclude that the value of a is: –\(\frac{3}{5}\)

Question 9.
2k – 3(2k – 3) = 45
Answer:
The value of k is: -9

Explanation:
The given equation is:
2k – 3 ( 2k – 3 ) = 45
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
SO,
2k – 3 ( 2k ) + 3 ( 3 ) = 45
2k – 6k + 9 = 45
2k – 6k = 45 – 9
-4k = 36
k = 36 ÷ -4
k = -9
Hence, from the above,
We can conclude that the value of k is: -9

Question 10.
68 = \(\frac{1}{5}\)(20x + 50) + 2
Answer:
The value of x is: 14

Explanation:
The given equation is:
68 = \(\frac{1}{5}\) [ 20x + 50 ] + 2
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
68 – 2 = \(\frac{1}{5}\) [ 20x + 50 ]
66 = \(\frac{1}{5}\) [ 20x + 50 ]
66 × 5 = 20x + 50
330 = 20x + 50
20x = 330 – 50
20x = 280
x = 280 ÷ 20
x = 14
Hence, from the above,
We can conclude that the value of x is: 14

Solve the equation. (Section 1.3)

Question 11.
3c + 1 = c + 1
Answer:
The value of c is: 0

Explanation:
The given equation is:
3c + 1 = c + 1
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
3c – c = 1 – 1
2c = 0
c = 0
Hence, from the above,
We can conclude that the value of c is: 0

Question 12.
-8 – 5n = 64 + 3n
Answer:
The value of n is: -9

Explanation:
The given equation is:
-8 – 5n = 64 + 3n
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
-8 – 64 = 3n + 5n
-72 = 8n
n = -72 ÷ 8
n = -9
Hence, from the above,
We can conclude that the value of n is: -9

Question 13.
2(8q – 5) = 4q
Answer:
The value of q is: \(\frac{5}{6}\)

Explanation:
Te given equation is:
2 ( 8q – 5 ) = 4q
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
2 ( 8q ) – 2 ( 5 ) = 4q
16q – 10 = 4q
16q – 4q = 10
12q = 10
q = 10 ÷ 12
q = \(\frac{5}{6}\)
Hence, from the above,
We can conclude that the value of q is: \(\frac{5}{6}\)

Question 14.
9(y – 4) – 7y = 5(3y – 2)
Answer:
The value of y is: -2

Explanation:
The given equation is:
9 ( y – 4 ) – 7y = 5 ( 3y – 2 )
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
9 ( y ) – 9 ( 4 ) – 7y = 5 ( 3y ) – 5 ( 2 )
9y – 36 – 7y = 15y – 10
2y – 36 = 15y – 10
15y – 2y = 10 – 36
13y = -26
y = -26 ÷ 13
y = -2
Hence, from the above,
We can conclude that the value of y is: -2

Question 15.
4(g + 8) = 7 + 4g
Answer:
The given equation has no solution

Explanation:
The given equation is:
4 ( g + 8 ) = 7 + 4g
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,the given equation has no solution
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
4 ( g ) + 4 ( 8 ) = 7 + 4g
4g + 32 = 7 + 4g
4g – 4g + 7 = 32
7 = 32
Hence, from the above,
We can conclude that the given equation has no solution.

Question 16.
-4(-5h – 4) = 2(10h + 8)
Answer:
The value of h is: 0

Explanation:
The given equation is:
-4 ( 5h – 4 ) = 2 ( 10h + 8 )
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
-4 ( 5h ) + 4 ( 4 ) = 2 ( 10h ) + 2 ( 8 )
-20h + 16 = 20h + 16
-20h – 20h = 1 – 16
-40h =0
h = 0
Hence, from the above,
We can conclude that the value of h is: 0

Question 17.
To estimate how many miles you are from a thunderstorm, count the seconds between when you see lightning and when you hear thunder. Then divide by 5. Write and solve an equation to determine how many seconds you would count for a thunderstorm that is 2 miles away. (Section 1.1)
Answer:

Question 18.
You want to hang three equally-sized travel posters on a wall so that the posters on the ends are each 3 feet from the end of the wall. You want the spacing between posters to be equal. Write and solve an equation to determine how much space you should leave between the posters. (Section 1.2)
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 62

Answer:
The space you should leave between the posters is: \(\frac{3}{2}\) ft

Explanation:
The given figure is:
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 62
It is given that you want to hang three equally-sized travel posters on a wall so that the posters on the ends are each 3 feet from the end of the wall. You want the spacing between posters to be equal.
So,
From the figure,
The total space = 15 ft
The total spacing covered at the ends = 3 + 3 = 6 ft
Let the space between the equally spaced posters be x
So,
The total spacing between the travel posters = 2x + 2x + 2x = 6x ft
SO,
The total space = ( The total spacing covered at the ends ) + ( The total spacing between the travel posters )
15 = 6 + 6x
6x = 15 – 6
6x = 9
x = 9 ÷ 6
x = \(\frac{3}{2}\) ft
Hence, from the above,
We can conclude that the spacing between the travel posters is: \(\frac{3}{2}\) ft

Question 19.
You want to paint a piece of pottery at an art studio. The total cost is the cost of the piece plus an hourly studio fee. There are two studios to choose from. (Section 1.3)
a. After how many hours of the painting are the total costs the same at both studios? Justify your answer.
b. Studio B increases the hourly studio fee by $2. How does this affect your answer in part (a)? Explain.
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 63

Answer:
a)
It is given that,
The total cost = Cost of the vase + The hourly studio fee
Let the number of hours be x
So,
The total cost for studio A = 10 + 8x
The total cost of studio B = 16 + 6x
It is given that the total costs are the same
So,
10 + 8x = 16 + 6x
8x – 6x = 16 – 10
2x = 6
x = 6 ÷ 2
x = 3
Hence, from the above,
We can conclude that the total cost will be the same after 3 hours for both the studios

b)
It is given that the studio B increases the hourly studio fee by $2
So,
The total hourly studio fee for studio B = 6 + 2 = $8
So,
Now,
As in part (a), the same process will be repeated but in the studio B’s hourly fee of $6, we have to put $8
So,
10 + 8x = 16 + 8x
8x – 8x = 6 – 10
10 = 16
Hence, from the above,
We can conclude that the value of x has no solutions

Lesson 1.4 Solving Absolute Value Equations

Essential Question

How can you solve an absolute value equation?
EXPLORATION 1
Solving an Absolute Value Equation Algebraically
Work with a partner. Consider the absolute value equation
| x + 2 | = 3.
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 64
a. Describe the values of x + 2 that make the equation true. Use your description to write two linear equations that represent the solutions of the absolute value equation.
Answer:
We know that,
| x | = x
-| x | = -x
So,
| x + 2 | = 3
x + 2 = 3
x = 3 – 2
x = 1
Now,
-| x + 2 | = 3
| x + 2  | = -3
-x – 2 = -3
-x = -3 + 2
-x = -1
x = 1
So,
The values of x + 2 that make the equation true is: 3 and 3
The value of x is: 1 and 1

b. Use the linear equations you wrote in part (a) to find the solutions of the absolute value equation.
Answer:
We know that,
| x | = x
-| x | = -x
So,
| x + 2 | = 3
x + 2 = 3
x = 3 – 2
x = 1
Now,
-| x + 2 | = 3
-| x + 2  | = -3
-x – 2 = -3
-x = -3 + 2
-x = -1
x = 1
So,
The solutions of | x + 2 | are: 1 and 1

c. How can you use linear equations to solve an absolute value equation?
Answer:
We use linear equations to solve an absolute value equation by using the following properties. They are:
A) | x | = x
B) -| x | = -x

EXPLORATION 2
Solving an Absolute Value Equation Graphically
Work with a partner.
Consider the absolute value equation
| x + 2 | = 3.
a. On a real number line, locate the point for which x + 2 = 0.
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 64.1
Answer:
The given equation is:
x + 2 = 0
x = 0 – 2
x = -2
So,
On a real number line, we have to locate the point x = -2
Hence,
The point we have to locate on the real number line is:

b. Locate the points that are 3 units from the point you found in part (a). What do you notice about these points?
Answer:
From part (a).
We found that x = -2
Now,
To locate the points that are 3 units away or 3 units behind from the point you found in part (a), i.e., x = -2
We know that,
3 units away imply ” Add 3 ”
3 units behind imply ” Subtract 3 ”
Now,
We have to add 3 and subtract 3 to the point we obtained in part (a)
So,
When we add 3 to x = -2,
x = -2 + 3
x =1
When we subtract 3 from x = -2,
x = -2 – 3
x = -5
Hence,
The points we have to locate in the real number line are: 1 and -5
So,
The real number line with the located points is:

c. How can you use a number line to solve an absolute value equation?
Answer:
The given absolute value equation is:
| x + 2 | = 3
We know that,
| x | = x
-| x | = -x
So,
| x + 2 | = 3
x + 2 = 3
x = 3 – 2
x = 1
– | x + 2 | = -3
-x – 2 = -3
-x = -3 + 2
-x = -1
x = 1
So,
The values we have to locate in the number line is:

EXPLORATION 3
Solving an Absolute Value Equation Numerically

Work with a partner. Consider the absolute value equation
| x + 2 | = 3.
a. Use a spreadsheet, as shown, to solve the absolute value equation.
b. Compare the solutions you found using the spreadsheet with those you found in Explorations 1 and 2. What do you notice?
c. How can you use a spreadsheet to solve an absolute value equation?
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 65

Communicate Your Answer

Question 4.
How can you solve an absolute value equation?
Answer:
We can solve the absolute equation by using the following properties. They are:
A) | x| = x
B) – | x | = -x

Question 5.
What do you like or dislike about the algebraic, graphical, and numerical methods for solving an absolute value equation? Give reasons for your answers.
Answer:
The algebraic, numerical, and graphical methods have their own advantages in their own perspective.
The algebraic methods used to solve the linear equations whereas the graphical method used to indicate the linear equations.
The numerical method is applicable for mathematical operations

1.4 Lesson

Monitoring Progress

Solve the equation. Graph the solutions, if possible.

Question 1.
| x | = 10
Answer:
The value of x is: 10

Explanation:
The given absolute value equation is:
| x | = 10
We know that,
| x | = x
So,
x = 10
Hence, from the above,
We can conclude that the value of x is: 10

Question 2.
| x – 1 | = 4
Answer:
The value of x is: 5

Explanation:
The given absolute value equation is:
| x – 1 | = 4
We know that,
| x | = x
So,
x – 1 = 4
x = 4 + 1
x = 5
Hence, from the above,
We can conclude that the value of x is: 5

Question 3.
| 3 + x | = -3
Answer:
The value of x is: -6

Explanation:
The given absolute value equation is:
| 3 + x | = -3
We know that,
| x | = x
So,
x + 3 = -3
x = -3 – 3
x = -6
Hence, from the above,
We can conclude that the value of x is: -6

Solve the equation. Check your solutions.

Question 4.
| x – 2 | + 5 = 9
Answer:
The value of x is: 6

Explanation:
The given absolute value equation is:
| x – 2 | + 5 = 9
| x – 2 | = 9 – 5
| x – 2 | = 4
We know that,
| x | = x
So,
x – 2 = 4
x = 4 + 2
x = 6
Hence, from the above,
We can conclude that the value of x is: 6

Question 5.
4 | 2x + 7 | = 16
Answer:
The value of x is: –\(\frac{3}{2}\)

Explanation:
The given absolute value equation is:
4 | 2x + 7 | = 16
| 2x + 7 | = 16 ÷ 4
| 2x + 7| = 4
We know that,
| x | = x
So,
2x + 7 = 4
2x = 4 – 7
2x = -3
x = –\(\frac{3}{2}\)
Hence, from the above,
We can conclude that the value of x is: –\(\frac{3}{2}\)

Question 6.
-2 | 5x – 1 | – 3 = -11
Answer:
The value of x is: 1

Explanation:
The given absolute value equation is:
-2 | 5x – 1 | – 3 = -11
-2 | 5x – 1 | = -11 + 3
-2 | 5x – 1 | = -8
| 5x – 1 | = -8 ÷ ( -2 )
| 5x – 1 | = 4
We know that,
| x | = x
So,
5x – 1 = 4
5x = 4 + 1
5x = 5
x = 5 ÷ 5
x = 1
Hence, from the above,
We can conclude that the value of x is: 1

Question 7.
For a poetry contest, the minimum length of a poem is 16 lines. The maximum length is 32 lines. Write an absolute value equation that represents the minimum and maximum lengths.
Answer:
The minimum value length is: 16
The maximum length is: 32

Explanation:
It is given that for a poetry contest, the minimum length of a poem is 16 lines. The maximum length is 32 lines.
So,
The absolute value equation that represents the minimum length of a poem = | The minimum length of a poem |
= | 16 |
= 16
The absolute value equation that represents the maximum length of a poem = | The maximum length of a poem |
= | 32 |
= 32
Hence, from the above,
We can conclude that
The minimum value length is: 16
The maximum length is: 32

Solve the equation. Check your solutions.

Question 8.
| x + 8 | = | 2x + 1 |
Answer:
The value of x is: 7

Explanation:
The given absolute value equation is:
| x + 8 | = | 2x + 1 |
We know that,
| x | = x
So,
x + 8 = 2x + 1
2x – x = 8 – 1
x = 7
Hence from the above,
We can conclude that the value of x is: 7

Question 9.
3 | x – 4 | = | 2x + 5 |
Answer:
The value of x is: 17

Explanation:
The given absolute equation is:
3 | x – 4 | = | 2x + 5 |
We know that,
| x |  = x
So,
3 ( x – 4 ) = 2x + 5
3 ( x ) – 3 ( 4 ) = 2x + 5
3x – 12 = 2x + 5
3x – 2x = 5 + 12
x = 17
Hence, from the above,
We can conclude that the value of x is: 17

Solve the equation. Check your solutions.

Question 10.
| x + 6 | = 2x
Answer:
The value of x is: 6

Explanation:
The absolute value equation is:
| x + 6 | = 2x
We know that,
| x | = x
So,
x + 6 = 2x
2x – x = 6
x = 6
Hence, from the above,
We can conclude that the value of x is: 6

Question 11.
| 3x – 2 | = x
Answer:
The value of x is: 1

Explanation:
The given absolute value equation is:
| 3x – 2 | = x
We know that,
| x | = x
So,
3x – 2 = x
Soo,
3x – x = 2
2x = 2
x = 2 ÷ 2
x = 1
Hence, from the above,
We can conclude that the value of x is: 1

Question 12.
| 2 + x | = | x – 8 |
Answer:
The given absolute value equation has no solution

Explanation:
The given absolute value equation is:
| 2 + x | = | x – 8 |
We know that,
| x | = x
So,
2 + x = x – 8
2 = x – x – 8
2 = -8
Hence, from the above,
We can conclude that the given absolute value equation has no solution

Question 13.
| 5x – 2 | = | 5x + 4 |
Answer:
The given absolute value equation has no solution

Explanation:
The given absolute value equation is:
| 5x – 2 | = | 5x + 4 |
We know that,
| x | = x
So,
5x – 2 = 5x + 4
5x – 5x – 2 = 4
-2 = 4
Hence, from the above,
We can conclude that the given absolute value equation has no solution

Solving Absolute Value Equations 1.4 Exercises

Vocabulary and Core Concept Check

Question 1.
VOCABULARY
What is an extraneous solution?
Answer:
Extraneous solutions are values that we get when solving equations that are not really solutions to the equation.
Example for extraneous solution:
| 5x – 2 | = | 5x + 4 |

Question 2.
WRITING
Without calculating, how do you know that the equation | 4x – 7 | = -1 has no solution?
Answer:
The given absolute value equation is:
| 4x – 7 | = -1
We know that,
An absolute value can never equal a negative number.
So,
By the above,
We can say that
| 4x – 7 | must not equal to a negative number.
Hence, from the above,
We can conclude that | 4x – 7 | = -1 has no solution without calculating its solution

Monitoring Progress and Modeling with Mathematics

In Exercises 3−10, simplify the expression.

Question 3.
| -9 |

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q3

Question 4.
– | 15 |
Answer:
The value of -| 15 | is: -15

Explanation:
The given absolute value is: -| 15 |
We know that,
| x | = x
| -x | = x
-| x | = -x
So,
-| 15 | = -15
Hence, from the above,
We can conclude that the value of -| 15 | is: -15

Question 5.
| 14 | – | -14 |

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q5

Question 6.
| -3 | + | 3 |
Answer:

The value of | -3 | + | 3 | is: 6

Explanation:
The given absolute value expression is:
| -3 | + | 3 |
We know that,
| x | = x
| -x | = x
-| x | = -x
So,
| -3 | + | 3 |
= 3 + 3
= 6
Hence, from the above,
We can conclude that the value of | -3 | + | 3 | is: 6

Question 7.
– | -5 • (-7) |

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q7

Question 8.
| -0.8 • 10 |
Answer:
The value of | -0.8 ⋅ 10 | is: 8

Explanation:
The given absolute value expression is:
| -0.8 ⋅ 10 |
We know that,
| x | = x
| -x | = x
-| x | = -x
So,
| -0.8 ⋅ 10 |
= | – ( 8 ⁄ 10 ) ⋅ ( 10 ⁄ 1 ) |
= | – ( 8 × 10 ) ⁄ ( 10 × 1 ) |
= | -8 |
= 8
hence, from the above,
We can conclude that the value of | -0.8 ⋅ 10 | is: 8

Question 9.
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 66

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q9

Question 10.
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 67
Answer:
The value of | -12 ⁄ 4 | is: 3

Explanation:
The given absolute value expression is: | -12 ⁄ 4 |
We know that,
| x | = x
| -x | = x
– | x | = -x
So,
| -12 ⁄ 4 | = | -3 |
= 3
Hence, from the above,
We can conclude that the value of | -12 ⁄ 4 | is: 3

In Exercises 11−24, solve the equation. Graph the solution(s), if possible.

Question 11.
| w | = 6

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q11

Question 12.
| r | = -2
Answer:
The absolute value of a number must be greater than or equal to 0 and can not be equal to -2.
Hence,
The given absolute eqution has no solution

Question 13.
| y | = -18

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q13

Question 14.
| x | = 13
Answer:
The value of x is: 13 or -13

Explanation:
The given absolute value equation is:
| x | = 13
We know that,
|x | = x
– | x | = -x
So,
| x | = 13 or – 13
Hence, from the above,
We can conclude that the value of x is: 13 or -13

Question 15.
| m + 3 | = 7

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q15

Question 16.
| q – 8 | = 14
Answer:
The value of q is: 22 or -6

Explanation:
The given absolute value equation is:
| q – 8 | = 14
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
q – 8 = 14                                                      q – 8 = -14
q = 14 + 8                                                     q = -14 + 8
q = 22                                                             q = -6
Hence, from the above,
We can conclude that the value of q is: 22 or -6

Question 17.
| -3d | = 15

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q17

Question 18.
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 68
Answer:
The value of t is: 12 or -12

Explanation:
The given absolute value equation is:
| t / 2 | = 12
We know that,
| x | = x  for x > 0
| x | = -x for x < 0
So,
t / 2 = 6                                       t / 2 = -6
t = 6 × 2                                      t = 6 × -2
t = 12                                           t = -12
Hence, from the above,
We can conclude that the value of t is: 12 or -12

Question 19.
| 4b – 5 | = 19

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q19

Question 20.
| x – 1 | + 5 = 2
Answer:
The given absolute value equation has no solution

Explanation:
The given absolute value equation is:
| x – 1 | + 5 = 2
| x – 1 | = 2 – 5
| x – 1  | = -3
We know that,
The absolute value of an equation must be greater than or equal to zero
So,
| x – 1 | = -3 has no solution
Hence, from the above,
We can conclude that the given absolute value equation has no solution

Question 21.
-4 | 8 – 5n | = 13

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q21

Question 22.
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 69
Answer:
The value of y is: -3 or 6

Explanation:
The given absolute value equation is:
-3 | 1 – ( 2 / 3 ) y | = -9
| 1 – (2 / 3  ) y | = -9 ÷ ( -3 )
| 1 – ( 2 / 3 ) y | = 3
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
1 – ( 2 / 3 ) y = 3                                    1 – ( 2 /3 ) y = -3
2/3 y = 1 – 3                                           2/3 y = 1 + 3
2 / 3 y = -2                                              2 / 3 y = 4
2y = -2 × 3                                               2y = 4 × 3
2y = -6                                                      2y = 12
y = -6 ÷ 2                                                  y = 12 ÷ 2
y = -3                                                         y = 6
Hence, from the above,
We can conclude that the value of y is: -3 or 6

Question 23.
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 70

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q23

Question 24.
9 | 4p + 2 | + 8 = 35
Answer:
The value of p is: 1 / 4 or -5 / 4

Explanation:
The given absolute value equation is:
9 | 4p + 2 | + 8 = 35
9 | 4p + 2 | = 35 – 8
9 | 4p + 2 | = 27
| 4p + 2 | = 27 ÷ 9
| 4p + 2 | = 3
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
4p + 2 = 3                    4p + 2 = -3
4p = 3 – 2                     4p = -3 – 2
4p = 1                           4p = -5
p = 1 / 4                         p = -5 / 4
Hence, from the above,
We can conclude that the value of p is: 1 / 4 or -5 / 4

Question 25.
WRITING EQUATIONS
The minimum distance from Earth to the Sun is 91.4 million miles. The maximum distance is 94.5 million miles.
a. Represent these two distances on a number line.
b. Write an absolute value equation that represents the minimum and maximum distances.

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q25

Question 26.
WRITING EQUATIONS
The shoulder heights of the shortest and tallest miniature poodles are shown.
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 71
a. Represent these two heights on a number line.
b. Write an absolute value equation that represents these heights.

Answer:
a)
The number line that represents the two heights on a number line is:

b)
The minimum shoulder height = ( 15 – 10 ) / 2
= 5 / 2
= 2.5 inches
The maximum shoulder height = 10 + 2.5
= 12.5 inches
Now,
Let the heights between poodles be x.
Hence,
The absolute value equation is:
| x – 12.5 | = 2.5

USING STRUCTURE In Exercises 27−30, match the absolute value equation with its graph without solving the equation.

Question 27.
| x + 2 | = 4

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q27

Question 28.
| x + 4 | = 2
Answer:
The given absolute value equation is:
| x + 4 | = 2
To find the halfway point, made the absolute value equation equal to 0.
So,
| x + 4  | = 0
So,
x = -4
From the given absolute value equation,
We can say that the distance from the halfway point to the minimum and maximum points is: 2

Question 29.
| x – 2 | = 4

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q29

Question 30.
| x + 4 | = 2
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 72

Answer:

In Exercises 31−34, write an absolute value equation that has the given solutions.

Question 31.
x = 8 and x = 18

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q31

Question 32.
x = -6 and x = 10
Answer:
The given absolute value equation is:
| x – 2 | = 8

Explanation:
The given values of x are:
x = -6 and x = 10
Now,
The halfway point between 10 and -6 = [ 10 – ( -6 ) ] / 2
= [ 10 + 6 ] / 2
= 16 / 2
= 8
The minimum distance from the halfway point = 8 – 6 = 2
Hence,
The absolute value equation is:
| x – 2 | = 5

Question 33.
x = 2 and x = 9

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q33

Question 34.
x = -10 and x = -5

Answer:
The given values of x are:
x = -10 and x = -5
Now,
The halfway point between -10 and -5 = [ 10 – ( 5 ) ] / 2
= [ 10 – 5 ] / 2
= 5 / 2
= 2.5
So,
The minimum value from the half-point = 2.5 + ( -10 )
= 2.5 – 10
= -7.5
Hence,
The absolute value equation is:
| x – ( -7.5 ) | = 2.5
| x + 7.5 | = 2.5

In Exercises 35−44, solve the equation. Check your solutions. 

Question 35.
| 4n – 15 | = | n |

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q35

Question 36.
| 2c + 8 | = | 10c |
Answer:
The values of c are: 1 and 2 / 3

Explanation:
The given absolute value equation is:
| 2c + 8 | = | 10c |
We know that,
| x | = x for x > 0
| x | = -x for x < 0
Now,
2c + 8 = 10c                                      2c + 8 = -10c
10c – 2c = 8                                       2c + 10c = 8
8c = 8                                                12c = 8
c = 8 / 8                                              c = 8 / 12
c = 1                                                    c = 2 /3
Hence, from the above,
We can conclude that the values of c are: 1 and 2 / 3

Question 37.
| 2b – 9 | = | b – 6 |

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q37

Question 38.
| 3k – 2 | = 2 | k + 2 |
Answer:
The values of k are: 6 and -2 / 5

Explanation:
The given absolute equation is:
| 3k – 2 | = 2 | k + 2 |
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
2 ( k + 2 ) = 3k – 2                                            2 ( k + 2 ) = – ( 3k – 2 )
2k + 4 = 3k – 2                                                 2k + 4 = -3k + 2
3k – 2k = 4 + 2                                                 2k + 3k = 2 – 4
k = 6                                                                 5k = -2
k = 6                                                                 k = -2 / 5
Hence, from the above,
We can conclude that the values of k are: 6 and -2 / 5

Question 39.
4 | p – 3 | = | 2p + 8 |

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q39

Question 40.
2 | 4w – 1 | = 3 | 4w + 2 |
Answer:
The value of w is: -2

Explanation:
The given absolute value equation is:
2 | 4w – 1 | = 3 | 4w+ 2 |
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
2 ( 4w – 1 ) = 3 ( 4w + 2 )                        -2 ( 4w – 1 ) = -3 ( 4w + 2 )
8w – 2 = 12w + 6                                      -8w + 2 = -12w -6
12w – 8w = -6 – 2                                      -12w + 8w = 6 + 2
4w = -8                                                       -4w = 8
w = -8 / 4                                                      w = 8 / -4
w = -2                                                            w = -2
Hence, from the above,
We can conclude that the value of w is: -2

Question 41.
| 3h + 1 | = 7h

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q41

Question 42.
| 6a – 5 | = 4a
Answer:
The value of a is: 5 / 2 and 1 / 2

Explanation:
The given absolute value equation is:
| 6a – 5 | = 4a
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
6a – 5 = 4a                                                6a – 5 = -4a
6a – 4a = 5                                                 6a + 4a = 5
2a = 5                                                         10a = 5
a = 5 / 2                                                       a = 5 / 10
a = 5 / 2                                                       a = 1 / 2
Hence, from the above,
We can conclude that the values of a are: 5 / 2 and 1 / 2

Question 43.
| f – 6 | = | f + 8 |

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q43

Question 44.
| 3x – 4 | = | 3x – 5 |
Answer:
The given absolute value equation has no solution

Explanation:
The given absolute value equation is:
| 3x – 4 | = | 3x – 5 |
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
3x – 4 = 3x – 5                                – ( 3x – 4 ) = – ( 3x – 5 )
4 = 5                                                 4 = 5
Hence, from the above,
We can conclude that the given absolute value equation has no solution

Question 45.
MODELING WITH MATHEMATICS
Starting from 300 feet away, a car drives toward you. It then passes by you at a speed of 48 feet per second. The distance d (in feet) of the car from you after t seconds is given by the equation d = | 300 – 48t |. At what times is the car 60 feet from you?

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q45

Question 46.
MAKING AN ARGUMENT
Your friend says that the absolute value equation | 3x + 8 | – 9 = -5 has no solution because the constant on the right side of the equation is negative. Is your friend correct? Explain.
Answer:
Yes, your friend is correct

Explanation:
The given absolute value equation is:
| 3x + 8 | – 9 = -5
We know that,
The absolute value equation value must have greater than or equal to 0
But here
The value of the absolute value equation is less than 0
Hence,
The given absolute value equation has no solution.
Hence, from the above,
We can conclude that your friend is correct.

Question 47.
MODELING WITH MATHEMATICS
You randomly survey students about year-round school. The results are shown in the graph.
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 73
The error given in the graph means that the actual percentage could be 5% more or 5% less than the percent reported by the survey.
a. Write and solve an absolute value equation to find the least and greatest percents of students who could be in favor of the year-round school.
b. A classmate claims that \(\frac{1}{3}\) of the student body is actually in favor of the year-round school. Does this conflict with the survey data? Explain.

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q47

Question 48.
MODELING WITH MATHEMATICS
The recommended weight of a soccer ball is 430 grams. The actual weight is allowed to vary by up to 20 grams.
a. Write and solve an absolute value equation to find the minimum and maximum acceptable soccer ball weights.
Answer:
It is given that the recommended weight of a soccer ball is 430 grams and the actual weight is allowed to vary up to 20 grams
Hence,
The absolute value equation that represents the minimum and maximum acceptable soccer ball weights is:
| x – 430 | = 20
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
x – 430 = 20                               x – 430 = -20
x = 20 + 430                              x = -20 + 430
x = 460 grams                            x = 410 grams
Hence, from the above,
We can conclude that the maximum and minimum acceptable soccer weights respectively are: 460 grams and 410 grams

b. A soccer ball weighs 423 grams. Due to wear and tear, the weight of the ball decreases by 16 grams. Is the weight acceptable? Explain.
Answer:
The weight that caused due to wear and tear is not acceptable

Explanation:
From the above problem,
We get the maximum weight of the soccer ball to be 460 grams with 20 grams increase or decreased to the weight of the ball
Now,
It is given that the weight of the ball is decreased by 16 grams due to wear and tear
So,
The weight of the ball now = 460 – 16 = 444 grams
But it is given that the weight of the ball becomes 423 grams due to wear and tear.
Hence, from the above,
We can conclude that the weight is not acceptable

ERROR ANALYSIS
In Exercises 49 and 50, describe and correct the error in solving the equation.

Question 49.
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 74

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q49

Question 50.
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 75

Answer:
The values of x are: -2 and -4 / 3

Explanation:
The given absolute value equation is:
| 5x + 8 | = x
We know that,
| x | = x for x > 0
| x | = – x for x < 0
So,
5x + 8 = x                                                      5x + 8 = -x
5x – x = -8                                                      5x + x = -8
4x = -8                                                            6x = -8
x = -8 / 4                                                         x = -8 / 6
x = -2                                                               x = -4 / 3
Hence, from the above,
We can conclude that the values of x are: -2 and -4 / 3

Question 51.
ANALYZING EQUATIONS
Without solving completely, place each equation into one of the three categories.
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 76

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q51

Question 52.
USING STRUCTURE
Fill in the equation Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 77 with a, b, c, or d so that the equation is graphed correctly.
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 77.1

Answer:

ABSTRACT REASONING
In Exercises 53−56, complete the statement with always, sometimes, or never. Explain your reasoning.

Question 53.
If x2 = a2, then | x | is ________ equal to | a |.

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q53

Question 54.
If a and b are real numbers, then | a – b | is _________ equal to | b – a |.
Answer:
If a and b are real numbers, then
| a – b | is equal to | b – a |

Explanation:
Let,
| a | = 5 and | b | = 9
We know that,
| x | =x for  x > 0
| x | = -x for x < 0
So,
| a – b | = | 5 – 9 |
= | -4 | = 4
| b – a | = | 9 – 5 |
=  | 4 |
= 4
Hence, from the above,
We can conclude that value of
| a – b | is equal to | b – a | if a and b are real numbers

Question 55.
For any real number p, the equation | x – 4 | = p will ________ have two solutions.

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q55

Question 56.
For any real number p, the equation | x – p | = 4 will ________ have two solutions.
Answer:
For any real number,
| x – p | = 4 will have two solutions

Explanation:
The given absolute value equation is:
| x – p | = 4
Let the value of p be 1
We know that,
| x | = x for x > 0
| x | = – x for x < 0
So,
| x – 1 | = 4
| x – 1 | = 4                                        | x – 1 | = -4
x = 4 + 1                                           x = -4 + 1
x = 5                                                  x = -3
Hence, from the above,
We can conclude that
| x – p | = 4 will have two solutions for any real number p

Question 57.
WRITING
Explain why absolute value equations can have no solution, one solution, or two solutions. Give an example of each case.

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q55

Question 58.
THOUGHT-PROVOKING
Describe a real-life situation that can be modeled by an absolute value equation with the solutions x = 62 and x = 72.
Answer:
Suppose in a school, an exam is conducted. In that examination, 67% of the students are passed. If the error of the pass percentage is 5 %, then what are the minimum and the maximum number of students passed in the examination?
Now,
The absolute value equation for the given real-life situation is:
| x – 67 | = 5
We know that,
| x | = x for x> 0
| x | =-x for x < 0
So,
x – 67 = 5                                         x – 67 = -5
x = 5 + 67                                        x = -5 + 67
x = 72                                               x = 62
Hence, from the above,
We can conclude that the minimum and maximum number of students passed in the examination  respectively are: 72 and 67

Question 59.
CRITICAL THINKING
Solve the equation shown. Explain how you found your solution(s).
8 | x + 2 | – 6 = 5 | x + 2 | + 3

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q59
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q59-i

Question 60.
HOW DO YOU SEE IT?
The circle graph shows the results of a survey of registered voters on the day of an election.
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 78
The error given in the graph means that the actual percentage could be 2% more or 2% less than the percent reported by the survey.
a. What are the minimum and maximum percents of voters who could vote Republican?
Answer:
The minimum percentage of voters for Republicans is: 40%
The maximum percentage of voters for Republicans is: 44 %

Explanation:
The given graph is:
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 78
From the graph,
The vote percentage for Republicans is: 42 %
The error percentage is: ±2%
So,
The absolute value equation for the maximum and the minimum number of voters is:
| x – 42 | = 2
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
x – 42 = 2                                                                 x – 42 = -2
x = 2 + 42                                                                x = -2 + 42
x = 44                                                                       x = 40
Hence, from the above,
We can conclude that
The minimum percentage of voters for Republicans is: 40%
The maximum percentage of voters for Republicans is: 44 %

b. How can you use absolute value equations to represent your answers in part (a)?
Answer:
From the property of absolute values,
We know that,
| x | = x for x > 0
| x | = -x for x < 0
From the part ( a ),
The absolute value equation is:
| x – 42 | = 2
So,
x – 42 = 2                                                                 x – 42 = -2
x = 2 + 42                                                                x = -2 + 42
x = 44                                                                       x = 40
Hence, from the above,
We can conclude that we can use absolute values in the above way to represent the answers

c. One candidate receives 44% of the vote. Which party does the candidate belong to? Explain.
Answer:
The candidate of the Republican party receives 44 % of the vote.

Explanation:
The given graph is:
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 78
From the above graph,
We can say that,
The percentage of the vote received by the candidate of the Republican party = 42 %
The error percentage = ± 2 %
So,
Now,
The percentage of the vote received by the Republicans = 42 + 2  ( or ) 42 – 2
= 44 ( or ) 40
Hence, from the above,
We can conclude that the candidate of the Republican party received the 44 % of the vote

Question 61.
ABSTRACT REASONING
How many solutions does the equation a | x + b | + c = d have when a > 0 and c = d? when a < 0 and c > d? Explain your reasoning.

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q61

Maintaining Mathematical Proficiency

Identify the property of equality that makes Equation 1 and Equation 2 equivalent. (Section 1.1)

Question 62.
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 79

Answer:
The given equations are:
Equation 1: 3x + 8 = x – 1
Equation 2: 3x + 9= x
From Equation 1,
3x + 8 = x – 1
3x + 8 + 1 = x
3x + 9 = x
Hence, from the above,
We can conclude that we can get Equation2 by rearranging the Equation 1

Question 63.
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 80

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q63

Use a geometric formula to solve the problem.

Question 64.
A square has an area of 81 square meters. Find the side length.
Answer:
The side length of the square is: 9 meters

Explanation:
The given area of the square is: 81 square meters
We know that,
Area of the square = Side × Side
81 = Side × Side
Side² = 81
Apply square root on both sides
√Side² = √81
Side = 9 meters
Hence, from the above,
We can conclude that the side of the square is: 9 meters

Question 65.
A circle has an area of 36π square inches. Find the radius.

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q65

Question 66.
A triangle has a height of 8 feet and an area of 48 square feet. Find the base.
Answer:
The base of the triangle is: 12 feet

Explanation:
It is given that a triangle has a height of 8 feet and an area of 48 square feet
We know that,
The area of the triangle = ( 1 /  2 ) × Base × Height
48 = ( 1 / 2 ) × Base × 8
Base × 8 = 48 × 2
Base = ( 48 × 2 ) ÷ 8
Base = 96 ÷ 8
Base = 12 feet
Hence, from the above,
We can conclude that the base of the triangle is: 12 feet

Question 67.
A rectangle has a width of 4 centimeters and a perimeter of 26 centimeters. Find the length.

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q67

Lesson 1.5 Rewriting Equations and Formulas

Essential Question
How can you use a formula for one measurement to write a formula for a different measurement?
Answer:
Write the formula for one measurement and then solve the formula for the different measurement you want to find and use this new formula to find that measurement
Hence, in the above way,
We can use a formula for one measurement to write a formula for a different measurement

EXPLORATION 1
Using an Area Formula
Work with a partner.

a. Write a formula for the area A of a parallelogram.
Answer:
We know that,
The area of the parallelogram ( A) = Base × Height

b. Substitute the given values into the formula. Then solve the equation for b. Justify each step.
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 81
Answer:
The value of b is: 6 in

Explanation:
The given figure is:
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 81
From the given figure,
Area ( A) = 30 in²
Height ( h ) = 5 in
Base = b
From part ( a),
Area of the parallelogram = Base × Height
30 = 5 × b
b = 30 ÷ 5
b = 6 in
Hence, from the above,
We can conclude that the value of b is: 6 in

c. Solve the formula in part (a) for b without first substituting values into the formula. Justify each step.
Answer:
From part ( a ),
Area of the parallelogram = Base × Height
Base = ( Area of the parallelogram ) ÷ Height of the parallelogram
From the given figure,
Base = b
So,
b = ( Area of the parallelogram ) ÷ Height of the parallelogram

d. Compare how you solved the equations in parts (b) and (c). How are the processes similar? How are they different?
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 82
Answer:
We know that,
The area of the parallelogram = Base × Height
Using the above formula,
We solved parts (b ) and ( c )

EXPLORATION 2
Using Area, Circumference, and Volume Formulas
Work with a partner. Write the indicated formula for each figure. Then write a new formula by solving for the variable whose value is not given. Use the new formula to find the value of the variable.
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 83
Answer:
The given geometrical figures are:
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 83
Now,
a)
The given figure is:

We know that,
Area of the trapezoid = h ( a + b ) / 2
Where,
h is the height between the two sides
a + b is the sum of the parallel sides [ Here, b1 and b2 ]
From the given figure,
Area of the trapezoid = 63 cm²
1st parallel side ( b1 ) = 8 cm
2nd parallel side ( b2 ) = 10 cm
So,
Area of the trapezoid = h ( 8 + 10 ) / 2
63 = h ( 18 ) / 2
63 × 2 = h × 18
h = ( 63 × 2 ) ÷ ( 18 × 1 )
h = 7 cm
Hence, from the above,
We can conclude that the value of h is: 7 cm
b)
The given figure is:

We know that,
Circumference of a circle = 2πr
Where
r is the radius of the circle
From the above figure,
Circumference of the circle ( C ) = 24π ft
So,
24π = 2πr
r = ( 24π ) ÷ ( 2π )
r = 12 ft
Hence, from the above,
We can conclude that the value of r is: 12 ft
c)
The given figure is:

We know that,
The volume of the rectangular prism ( V ) = Length × Width × Height
The area of the rectangle ( B) = Length × Width
So,
The volume of the rectangular prism (V ) = B × Height
From the above figure,
The volume of the rectangular prism ( V ) = 75 yd³
The area of the rectangle ( B ) = 15 yd²
So,
75 = 15 × Height
Height = 75 ÷ 15
Height = 15 yd
Hence, from the above,
We can conclude that the value of h is: 15 yd
d)
The given figure is:

We know that,
The volume of cone ( V ) = πr²h / 3
The area of the circle ( B ) = πr²
Where,
r is the radius of the circle
h is the height of the cone
So,
The volume of the cone ( V ) = Bh / 3
From the above figure,
V = 24π m³
B = 12π m³
So,
24π = 12π × h / 3
h / 3 = 24π ÷ 12π
h / 3 = 2
h = 2 × 3
h = 6 m
Hence, from the above,
We can conclude that the height of the cone is: 6 m

Communicate Your Answer

Question 3.
How can you use a formula for one measurement to write a formula for a different measurement? Give an example that is different from those given in Explorations 1 and 2.
Answer:
Write the formula for one measurement and then solve the formula for the different measurement you want to find and use this new formula to find that measurement
Example:
The given figure is:

We know that,
The area of the rectangle ( A ) = Length × Width
From the above figure,
A = 20 cm²
L = 10 cm
Let,
W be the width of the rectangle
So,
20 = 10 × W
W = 20 ÷ 10
W = 2 cm
Hence, from the above,
We can conclude that the value of W is: 2 cm

1.5 Lesson

Monitoring Progress

Solve the literal equation for y.

Question 1.
3y – x = 9
Answer:
The value of y is: ( x + 9 ) / 3

Explanation:
The given equation is:
3y – x = 9
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
3y = 9 + x
y = ( x + 9 ) / 3
Hence, from the above,
We can conclude that the value of y is: ( x + 9 ) / 3

Question 2.
2x – 2y = 5
Answer:
The value of y is: ( 2x – 5 ) / 2

Explanation:
The given equation is:
2x – 2y = 5
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
2y = 2x – 5
y = ( 2x – 5 ) / 2
Hence, from the above,
We can conclude that the value of y is: ( 2x – 5 ) / 2

Question 3.
20 = 8x + 4y
Answer:
The value of y is: 5 – 2x

Explanation:
The given equation is:
20 = 8x + 4y
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
4y = 20 – 8x
y = ( 20 – 8x ) / 4
y = ( 20 ÷ 4 ) – ( 8x ÷ 4 )
y = 5 – 2x
Hence, from the above,
We can conclude that the value of y is: 5 – 2x

Solve the literal equation for x.

Question 4.
y = 5x – 4x
Answer:
The value of x is: y

Explanation:
The given equation is:
y = 5x – 4x
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
x = y
Hence, from the above,
We can conclude that the value of x is: y

Question 5.
2x + kx = m
Answer:
The value of x is: m / ( k + 2  )

Explanation:
The given equation is:
2x + kx = m
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
x ( k + 2 ) = m
x = m / ( k + 2 )
Hence, from the above,
We can conclude that the value of x is: m / ( k + 2 )

Question 6.
3 + 5x – kx = y
Answer:
The value of x is: ( y – 3 ) / ( 5 – k )

Explanation:
The given equation is:
3 + 5x – kx = y
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
5x – kx = y – 3
x ( 5 – k ) = y – 3
x = ( y – 3 ) / ( 5 – k )
Hence, from the above,
We can conclude that the value of x is: ( y – 3 ) / ( 5 – k )

Solve the formula for the indicated variable. 

Question 7.
Area of a triangle: A = \(\frac{1}{2}\)bh; Solve for h.
Answer:
The value of h is: \(\frac{2A}{b}\)

Explanation:
The given area of a triangle is:
A = \(\frac{1}{2}\) bh
bh = 2A
h = \(\frac{2A}{b}\)
Hence, from the above,
We can onclude that the value of h is: \(\frac{2A}{b}\)

Question 8.
The surface area of a cone: S = πr2 + πrℓ; Solve for ℓ.
Answer:
The value of l is: \(\frac{S}{πr}\) – r

Explanation:
The given surface area of a cone is:
S = πr² + πrl
S= πr ( r + l )
r + l = \(\frac{S}{πr}\)
l = \(\frac{S}{πr}\) – r
Hence, from the above,
We can conclude that the value of l is: \(\frac{S}{πr}\) – r

Monitoring Progress

Question 9.
A fever is generally considered to be a body temperature greater than 100°F. Your friend has a temperature of 37°C. Does your friend have a fever?
Answer:
Your friend does not have a fever

Explanation:
It is given that a fever is generally considered to be a body temperature greater than 100°F.
We know that,
To convert Fahrenheit into Celsius,
°C = ( °F – 32 ) × \(\frac{5}{9}\)
°C = ( 100 – 32 ) × \(\frac{5}{9}\)
°C = 68 × \(\frac{5}{9}\)
°C = 37.7°
But it is given that your friend has a temperature of 37°C
So, for fever, the temperature has to be 37.7°C
Hence, from the above,
We can conclude that your friend does not have a fever

Question 10.
How much money must you deposit in a simple interest account to earn $500 in interest in 5 years at 4% annual interest?
Answer:
The money you deposit in simple interest is: $2,500

Explanation:
It is given that you earned $500 in a simple interest to earn in 5 years at 4% annual interest
Let,
The money you deposited be: $x
We know that,
Simple interest = ( Principle × Time × Rate ) / 100
The principle is the money you deposited
So,
500 = ( x × 5 × 4 ) / 100
( x × 5 × 4 ) = 500 × 100
x × 20 = 500 × 100
x = ( 500 × 100 ) ÷ 20
x = $2,500
Hence, from the above,
We can conclude that the money you deposited is: $2,500

Question 11.
A truck driver averages 60 miles per hour while delivering freight and 45 miles per hour on the return trip. The total driving time is 7 hours. How long does each trip take?
Answer:
The time taken for each trip is: 3 hours and 4 hours respectively

Explanation:
It is given that a truck driver averages 60 miles per hour while delivering freight and 45 miles per hour on the return trip. The total driving time is 7 hours.
We know that,
Speed = \(\frac{Distance}{Time}\)
Time = \(\frac{Distance}{Speed}\)
Let the distance be D
It is given that the total driving time is: 7 hours
So,
7 = \(\frac{D}{60}\) + \(\frac{D}{45}\)
7 / D = \(\frac{60 + 45}{60 × 45}\)
7 / D = \(\frac{105}{2,700}\)
D = 7 / \(\frac{105}{2,700}\)
D = 7 × \(\frac{2,700}{105}\)
D = \(\frac{7}{1}\) × \(\frac{2,700}{105}\)
D = \(\frac{7 × 2,700}{1 × 105}\)
D = 180 miles
So,
The time taken to deliver = \(\frac{180}{60}\) = 3 hours
The time taken to return = \(\frac{180}{45}\) = 4 hours
Hence, from the above,
We can conclude that the time taken for each trip is: 3 hours and 4 hours respectively

Rewriting Equations and Formulas 1.5 Exercices

Vocabulary and Core Concept Check

Question 1.
VOCABULARY
Is Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 84 a literal equation? Explain.
Answer:
The ” Literal equation” is an equation that contains only letters
Now,
The given equation is:
9r + 16 = π / 5
From the above definition,
We can say that the given equation is a ” Literal equation ”

Question 2.
DIFFERENT WORDS, SAME QUESTION?
Which is different? Find “both” answers.
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 85
Answer:
The given problems are:
a) Solve 3x + 6y = 24 for x
b) Solve 24 – 3x = 6y for x
c) Solve 6y = 24 – 3x in terms of x
d) Solve 24 – 6y = 3x for x in terms of y
So,
From the above-given problems,
We can observe that d) is different as we have to find x in terms of y whereas in the remaining three problems, we have to find x

Monitoring Progress and Modeling with Mathematics

In Exercises 3–12, solve the literal equation for y.

Question 3.
y – 3x = 13

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.5-Q3

Question 4.
2x + y = 7
Answer:
The value of y is: 7 – 2x

Explanation:
The given literal equation is:
2x + y = 7
Now,
y = 7 – 2x
Hence, from the above,
We can conclude that the value of y is: 7 – 2x

Question 5.
2y – 18x = -26

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.5-Q5

Question 6.
20x + 5y = 15
Answer:
The value of y is: 3 – 4x

Explanation:
The given literal equation is:
20x + 5y = 15
Now,
5y = 15 – 20x
y = ( 15 – 20x ) / 5
y = ( 15  / 5 ) – ( 20x / 5 )
y = 3 – 4x
Hence, from the above,
We can conclude that the value of y is: 3 – 4x

Question 7.
9x – y = 45

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.5-Q7

Question 8.
6 – 3y = -6
Answer:
The value of y is: 4

Explanation:
The given literal equation is:
6 – 3y = -6
-3y = -6 – ( +6 )
-3y = -6 -6
-3y = -12
y = -12 ÷ ( -3 )
y = 12 ÷ 3
y = 4
Hence, from the above,
We can conclude that the value of y is: 4

Question 9.
4x – 5 = 7 + 4y

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.5-Q9

Question 10.
16x + 9 = 9y – 2x
Answer:
The value of y is: 18x + 9

Explanation:
The given literal equation is:
16x + 9 = y – 2x
So,
16x + 2x + 9 = y
18x + 9 = y
y = 18x + 9
Hence, from the above,
We can conclude that the value of y is: 18x + 9

Question 11.
2 +\(\frac{1}{6}\)y = 3x + 4

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.5-Q11

Question 12.
11 – \(\frac{1}{2}\)y = 3 + 6x
Answer:
The value of y is: 16 – 12x

Explanation:
The given literal equation is:
11 – \(\frac{1}{2}\)y = 3 + 6x
So,
–\(\frac{1}{2}\)y = 3 + 6x – 11
-y = 2 ( 3 + 6x – 11 )
y = -2 ( 3 + 6x – 11 )
y = -2 ( 3 ) -2 ( 6x ) + 2 ( 11 )
y = -6 – 12x + 22
y = 16 – 12x
Hence, from the above,
We can conclude that the value of y is: 16 – 12x

In Exercises 13–22, solve the literal equation for x.

Question 13.
y = 4x + 8x

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.5-Q13

Question 14.
m = 10x – x
Answer:
The value of x is: m / 9

Explanation:
The given literal equation is:
m = 10x – x
m = 9x
x = m / 9
Hence, from the above,
We can conclude that the value of x is: m / 9

Question 15.
a = 2x + 6xz

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.5-Q15

Question 16.
y = 3bx – 7x
Answer:
The value of x is: y / ( 3b – 7 )

Explanation:
The given literal equation is:
y = 3bx – 7x
So,
y = x ( 3b – 7 )
x = y / ( 3b – 7 )
Hence, from the above,
We can conclude that the value of x is: y / ( 3b – 7 )

Question 17.
y = 4x + rx + 6

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.5-Q17

Question 18.
z = 8 + 6x – px
Answer:
The value of x is: ( z – 8 ) / ( 6 – p )

Explanation:
The given literal equation is:
z = 8 + 6x – px
So,
z – 8 = 6x – px
z – 8 = x ( 6 – p )
x = ( z – 8 ) / ( 6 – p )
Hence, from the above,
We can conclude that the value of x is: ( z – 8 ) / ( 6 – p )

Question 19.
sx + tx = r

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.5-Q19

Question 20.
a = bx + cx + d
Answer:
The value of x is: ( a – d ) / ( b + c )

Explanation:
The given literal equation is:
a = bx + cx + d
a – d = bx + cx
a – d = x ( b + c )
x = ( a – d ) / ( b + c )
Hence, from the above,
We can conclude that the value of x is: ( a – d ) / ( b + c )

Question 21.
12 – 5x – 4kx = y

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.5-Q21

Question 22.
x – 9 + 2wx = y
Answer:
The value of x is: ( y – 9 ) / ( 1 – 2w )

Explanation:
The given literal equation is:
x – 9 + 2wx = y
x – 2wx = y + 9
x ( 1 – 2w ) = y + 9
x = ( y – 9 ) / ( 1 – 2w )
Hence, from the above,
We can conclude that the value of x is: ( y – 9 ) / ( 1 – 2w )

Question 23.
MODELING WITH MATHEMATICS
The total cost C (in dollars) to participate in a ski club is given by the literal equation C = 85x + 60, where x is the number of ski trips you take.
a. Solve the equation for x.
b. How many ski trips do you take if you spend a total of $315? $485?
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 86

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.5-Q23

Question 24.
MODELING WITH MATHEMATICS
The penny size of a nail indicates the length of the nail. The penny size d is given by the literal equation d = 4n – 2, where n is the length (in inches) of the nail.
a. Solve the equation for n.
b. Use the equation from part (a) to find the lengths of nails with the following penny sizes: 3, 6, and 10.
Answer:
a)
The given literal equation is:
d = 4n – 2
Where,
n is the length ( in inches ) of the nail
So,
4n = d + 2
n = ( d + 2 ) / 4
b)
It is given that,
The penny sizes ( d ) are: 3, 6, and 10
From part ( a ),
The literal equation is:
n = ( d + 2  ) / 4
Put, d= 3, 6 and 10
So,
n = ( 3 + 2 ) /4 = 5 / 4 inches
n = ( 6 + 2 ) / 4 = 2 inches
n = ( 10 + 2 ) / 4 = 3 inches

ERROR ANALYSIS
In Exercises 25 and 26, describe and correct the error in solving the equation for x.

Question 25.
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 87

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.5-Q25

Question 26.
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 72.1

Answer:
The given literal equation is:
10 = ax – 3b
So,
ax = 10 + 3b
x = ( 10 + 3b ) / a

In Exercises 27–30, solve the formula for the indicated variable.

Question 27.
Profit: P = R – C; Solve for C.

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.5-Q27

Question 28.
Surface area of a cylinder: S = 2πr2 + 2πrh; Solve for h.
Answer:
The given Surface area of a cylinder is:
S = 2πr² + 2πrh
So,
S = 2πr ( r + h )
S / 2πr = r + h
h = S / 2πr – r
Hence, from the above,
We can conclude that the value of h is: S / ( 2π

Question 29.
Area of a trapezoid: A = \(\frac{1}{2}\)h(b1 + b2); Solve for b2.

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.5-Q29

Question 30.
The average acceleration of an object: Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 88; Solve for v1.
Answer:
The given average acceleration of an object is:
a = ( v1 – v0 ) / t
So,
at = v1 – v0
v1 = at + v0
Hence, from the above,
We can conclude that the value of v1 is: at + v0

Question 31.
REWRITING A FORMULA
A common statistic used in professional football is the quarterback rating. This rating is made up of four major factors. One factor is the completion rating given by the formula
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 88.1
where C is the number of completed passes and A is the number of attempted passes. Solve the formula for C.

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.5-Q31

Question 32.
REWRITING A FORMULA
Newton’s law of gravitation is given by the formula
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 89
where F is the force between two objects of masses m1 and m2, G is the gravitational constant, and d is the distance between the two objects. Solve the formula for m1.
Answer:
The value of m1 is: Fd² / Gm2

Explanation:
The given Newton’s law of gravitation is given by:
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 89
where
F is the force between two objects of masses m1 and m2
G is the gravitational constant
d is the distance between the two objects.
Now,
( m1m2 ) = Fd² / G
m1 = Fd² / Gm2
Hence, from the above,
We can conclude that the value of m1 is: Fd² / Gm2

Question 33.
MODELING WITH MATHEMATICS
The sale price S (in dollars) of an item is given by the formula S = L – rL, where L is the list price (in dollars) and r is the discount rate (in decimal form).
a. Solve the formula for r.
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 90
b. The list price of the shirt is $30. What is the discount rate?

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.5-Q33

Question 34.
MODELING WITH MATHEMATICS
The density d of a substance is given by the formula d = \(\frac{m}{V}\), where m is its mass and V is its volume.
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 91
a. Solve the formula for m?
Answer:
The given density d of a substance is given by:
d = \(\frac{m}{V}\)
So,
d × V = m
Hence, from the above,
We can conclude that the value of m is: d × V

b. Find the mass of the pyrite sample.
Answer:
The mass of the pyrite sample is: 6.012 gm

Explanation:
The given figure is:
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 91
From the above figure,
The density of Pyrite = 5.01 g/cm³
The volume of Pyrite = 1.2 cm³
From part (a),
The mass of Pyrite = Density × Volume
So,
The mass of Pyrite ( m ) = 5.01 × 1.2
= 6.012 gm
Hence, from the above,
We can conclude that the mass of Pyrite is: 6.012 gm

Question 35.
PROBLEM-SOLVING
You deposit $2000 in an account that earns simple interest at an annual rate of 4%. How long must you leave the money in the account to earn $500 in interest?

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.5-Q35

Question 36.
PROBLEM-SOLVING
A flight averages 460 miles per hour. The return flight averages 500 miles per hour due to a tailwind. The total flying time is 4.8 hours. How long is each flight? Explain.
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 91.1

Answer:
The time taken for flight is: 2.5 hours
The time taken for return is: 2.3 hours

Explanation:
It is given that a flight averages 460 miles per hour. The return flight averages 500 miles per hour due to a tailwind. The total flying time is 4.8 hours.
We know that,
Speed = Distance / Time
Time = Distance / Speed
It is also given that the total flying time is 4.8 hours
Let the distance be D
So,
\(\frac{D}{460}\) + \(\frac{D}{500}\) = 4.8
\(\frac{460 + 500}{230,000}\) = 4.8 / D
\(\frac{960}{230,000}\) = 4.8 / D
D = 4.8 × \(\frac{230,000}{960}\)
D = 1,150 miles
Hence,
The time taken for flight = 1,150 ÷ 460 = 2.5 hpurs
The time taken for return = 1,150 ÷ 500 = 2.3 hours

Question 37.
USING STRUCTURE
An athletic facility is building an indoor track. The track is composed of a rectangle and two semicircles, as shown.
Big Ideas Math Algebra 1 Solutions Chapter 1 Solving Linear Equations 92
a. Write a formula for the perimeter of the indoor track.
b. Solve the formula for x.
c. The perimeter of the track is 660 feet, and r is 50 feet. Find x. Round your answer to the nearest foot.

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.5-Q37

Question 38.
MODELING WITH MATHEMATICS
The distance d (in miles) you travel in a car is given by the two equations shown, where t is the time (in hours) and g is the number of gallons of gasoline the car uses.
Big Ideas Math Algebra 1 Solutions Chapter 1 Solving Linear Equations 93
a. Write an equation that relates g and t.
Answer:
The given equations are:
d = 55t ——————— (1)
d = 20g ——————— (2)
By the law of Equality,
55t = 20g [ As the LHS for both the equations are equal, make the RHS equal ]
t / g = 20 / 55
t / g = 4 / 11

b. Solve the equation for g.
Answer:
From the given figure,
d = 20g
d = 55t
From part (a),
t / g = 4 / 11
11t = 4g
g = 11t / 4
Hence, from the above,
We can conclude that the value of g is: 11t / 4

c. You travel for 6 hours. How many gallons of gasoline does the car use? How far do you travel? Explain.
Answer:
From part (b),
g = 11t / 4
Where,
g is the number of gallons of gasoline
It is given that you travel for 6 hours
So,
t = 6 hours
Now,
g = ( 11 × 6 ) / 4
g = 66/4 gallons
Hence, from the above,
We can conclude that the number of gallons of gasoline is: 66 / 4 gallons

Question 39.
MODELING WITH MATHEMATICS
One type of stone formation found in Carlsbad Caverns in New Mexico is called a column. This cylindrical stone formation connects to the ceiling and the floor of a cave.
Big Ideas Math Algebra 1 Solutions Chapter 1 Solving Linear Equations 94
a. Rewrite the formula for the circumference of a circle, so that you can easily calculate the radius of a column given its circumference.
b. What is the radius (to the nearest tenth of a foot) of a column that has a circumference of 7 feet? 8 feet? 9 feet?
c. Explain how you can find the area of a cross-section of a column when you know its circumference.

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.5-Q39

Question 40.
HOW DO YOU SEE IT?
The rectangular prism shown has bases’ with equal side lengths.
Big Ideas Math Algebra 1 Solutions Chapter 1 Solving Linear Equations 95
a. Use the figure to write a formula for the surface area S of the rectangular prism.
Answer:
The given figure is:
Big Ideas Math Algebra 1 Solutions Chapter 1 Solving Linear Equations 95
From the above figure,
The surface area of the rectangular prism ( S) = 2 ( lb + bh + lh )
Where,
l is the length of the rectangular prism
b is the Width of the rectangular prism
h is the height of the rectangular prism

b. Your teacher asks you to rewrite the formula by solving for one of the side lengths, b or ℓ. Which side length would you choose? Explain your reasoning.
Answer:
From part (a),
The surface area of the rectangular prism ( S ) = 2 ( lb + bh + lh )
S / 2 = lb + bh + lh
S / 2 = b ( l + h ) + bh
S / 2 = b ( l + b + h )
b = S / 2 ( l + b + h )
Hence, from the above,
We can conclude that the value of b is: S / 2 ( l + b + h )

Question 41.
MAKING AN ARGUMENT
Your friend claims that Thermometer A displays a greater temperature than Thermometer B. Is your friend correct? Explain your reasoning.
Big Ideas Math Algebra 1 Solutions Chapter 1 Solving Linear Equations 96

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.5-Q41

Question 42.
THOUGHT-PROVOKING
Give a possible value for h. Justify your answer. Draw and label the figure using your chosen value of h.
Big Ideas Math Algebra 1 Solutions Chapter 1 Solving Linear Equations 97

Answer:
The completed figure is:

The value of h is: 5 cm

Explanation:
The given figure is:
Big Ideas Math Algebra 1 Solutions Chapter 1 Solving Linear Equations 97
From the given figure,
We can say that the geometrical figure is the parallelogram
We know that,
Area of the parallelogram = Base × Height
From the given figure,
Area of the parallelogram = 40 cm²
The base of the parallelogram = 8 cm
So,
40 = 8 × Height
Height = 40 ÷ 8
Height = 5 cm
Hence, from the above,
We can conclude that
The completed figure is:

The value of h is: 5 cm

MATHEMATICAL CONNECTIONS
In Exercises 43 and 44, write a formula for the area of the regular polygon. Solve the formula for the height h.

Question 43.
Big Ideas Math Algebra 1 Solutions Chapter 1 Solving Linear Equations 98

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.5-Q43

Question 44.
Big Ideas Math Algebra 1 Solutions Chapter 1 Solving Linear Equations 99

Answer:
The value of h is: A / 3b

Explanation:
The given figure is:
Big Ideas Math Algebra 1 Solutions Chapter 1 Solving Linear Equations 99
From the above figure,
We can say that the figure is Hexagon
Now,
The Hexagon with the six triangles is:

So,
From the figure,
There are 6 triangles
We know that,
The area of a triangle = \(\frac{1}{2}\) × Base × Height
So,
The area of the Hexagon = The area of the 6 triangles
= 6 ( \(\frac{1}{2}\) ) × Base × Height
Let,
The area of the Hexagon be A
The height of the hexagon be h
The Base of the hexagon be b
So,
A = 6 ( \(\frac{1}{2}\) ) × Base × Height
A = 3 × Base × Height
Base × Height = A / 3
Height = A / ( 3 × Base )
So,
h = A / 3b
Hence, from the above,
We can conclude that the value of h is: A / 3b

REASONING
In Exercises 45 and 46, solve the literal equation for a.

Question 45.
Big Ideas Math Algebra 1 Solutions Chapter 1 Solving Linear Equations 100

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.5-Q45

Question 46.
Big Ideas Math Answer Key Algebra 1 Chapter 1 Solving Linear Equations 101
Answer:
The value of a is: \(\frac{by}{y – bx}\)

Explanation:
The given literal equation is:
y = x [ \(\frac{ab}{a – b}\)
\(\frac{ab}{a – b}\) = y / x
x ( ab ) = y ( a – b )
abx = ay – by
by = ay – abx
by = a ( y – bx )
a = \(\frac{by}{y – bx}\)
Hence, from the above,
We can conclude that the value of a is: \(\frac{by}{y – bx}\)

Maintaining Mathematical Proficiency

Evaluate the expression.

Question 47.
15 – 5 + 52

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.5-Q47

Question 48.
18 • 2 – 42 ÷ 8
Answer:
The given expression is:
18 ⋅ 2 – 4² ÷ 8
We have to remember that,
When there is an expression to solve with multiple mathematical symbols, we have to follow the BODMAS rule
BODMAS indicates the hierarchy we have to follow when we will solve mathematical symbols
In BODMAS,
B – Brackets
O – Of
D – Division
M – Multiplication
A – Addition
S – Subtraction
So,
18 ⋅ 2 – 4² ÷ 8 = 18 ⋅ 2 – ( 4 × 4 ) ÷ 8
= 18 ⋅ 2 – 2
= 36 – 2
= 34

Question 49.
33 + 12 ÷ 3 • 5

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.5-Q49

Question 50.
25(5 – 6) + 9 ÷ 3
Answer:
The given expression is:
25(5 – 6) + 9 ÷ 3
We have to remember that,
When there is an expression to solve with multiple mathematical symbols, we have to follow the BODMAS rule
BODMAS indicates the hierarchy we have to follow when we will solve mathematical symbols
In BODMAS,
B – Brackets
O – Of
D – Division
M – Multiplication
A – Addition
S – Subtraction
So,
25(5 – 6) + 9 ÷ 3 = ( 2 × 2 × 2 × 2 × 2 ) ( 5 – 6 ) + ( 9 ÷ 3 )
= ( 2 × 2 × 2 × 2 × 2 ) ( 5 – 6 ) + 3
= ( 2 × 2 × 2 × 2 × 2 ) ( -1 ) + 3
= -( 2 × 2 × 2 × 2 × 2 )  + 3
= -32 + 3
= -29

Solve the equation. Graph the solutions, if possible. (Section 1.4)

Question 51.
| x – 3 | + 4 = 9

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.5-Q51

Question 52.
| 3y – 12 | – 7 = 2
Answer:
The values of y are: 7 and 1

Explanation:
The given absolute value equation is:
| 3y – 12 | – 7 = 2
| 3y – 12 | = 2 + 7
| 3y – 12 | = 9
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
3y – 12 = 9                      3y – 12 = -9
3y = 9 + 12                     3y = -9 + 12
3y = 21                            3y = 3
y = 21 / 3                         y = 3 / 3
y = 7                                y = 1
Hence, from the above,
We can conclude that the values of y are: 7 and 1

Question 53.
2 | 2r + 4 | = -16

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.5-Q53

Question 54.
-4 | s + 9 | = -24
Answer:
The value of s is: -3 and -15

Explanation:
The given absolute value equation is:
-4 | s + 9 | = -24
| s + 9 | = -24 ÷ ( -4 )
| s + 9 | = 6 [ Since – ÷ – = + ]
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
s + 9 = 6                                       s + 9 = -6
s = 6 – 9                                         s = -6 – 9
s = -3                                             s = -15
Hence, from the above,
We can conclude that the values of s are: -3 and -15

Solving Linear Equations Performance Task: Magic of Mathematics

1.4–1.5 What Did You Learn?

Core Vocabulary
Big Ideas Math Answer Key Algebra 1 Chapter 1 Solving Linear Equations 102

Core Concepts
Section 1.4
Big Ideas Math Answer Key Algebra 1 Chapter 1 Solving Linear Equations 103
Section 1.5
Big Ideas Math Answer Key Algebra 1 Chapter 1 Solving Linear Equations 104

Mathematical Practices

Question 1.
How did you decide whether your friend’s argument in Exercise 46 on page 33 made sense?
Answer:
On page 33 in Exercise 46,
The given absolute equation is:
| 3x + 8 | – 9 = -5
| 3x + 8 | = -5 + 9
| 3x + 8 | = 4
So, from the absolute equation,
We can say that the given absolute value equation has a solution
But, according to your friend,
The argument is that the absolute value equation has no solution

Question 2.
How did you use the structure of the equation in Exercise 59 on page 34 to rewrite the equation?
Answer:
The given absolute value equation in Exercise 59 on page 34 is:
8 | x + 2 | – 6 = 5 | x + 2 | + 3
The above equation can be re-written as:
8 | x + 2  | – 5  | x + 2 | = 3 + 6
3 | x + 2  | = 9
Hence, from the above,
We can conclude that the re-written form of the given absolute value equation is:
3 | x + 2 | = 9

Question 3.
What entry points did you use to answer Exercises 43 and 44 on page 42?
Answer:
In Exercises 43 and 44 on page 42,
We used the triangles as an entry point
In Exercise 43,
The given figure is a pentagon
Using the above entry point,
We divided the pentagon into 5 triangles
In Exercise 44,
The given figure is a Hexagon
Using the above entry point,
We divided the hexagon into 6 triangles.

Performance Task

Magic of Mathematics

Have you ever watched a magician perform a number trick? You can use algebra to explain how these types of tricks work.
Big Ideas Math Answer Key Algebra 1 Chapter 1 Solving Linear Equations 105
To explore the answers to these questions and more, go to Big Ideas Math Answer Key Algebra 1 Chapter 1 Solving Linear Equations 106

Solving Linear Equations Chapter Review

1.1 Solving Simple Equations (pp. 3–10)

a. Solve x − 5 = −9. Justify each step.
Big Ideas Math Answer Key Algebra 1 Chapter 1 Solving Linear Equations 106.1

b. Solve 4x = 12. Justify each step.
Big Ideas Math Answer Key Algebra 1 Chapter 1 Solving Linear Equations 107

Solve the equation. Justify each step. Check your solution.

Question 1.
z + 3 = -6
Answer:
The value of z is: -9

Explanation:
The given equation is:
z + 3 = -6
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
z = -6 – ( +3 )
z = -6 – 3
z = -9
Hence, from the above,
We can conclude that the value of z is: -9

Question 2.
2.6 = -0.2t
Answer:
The value of t is: -13

Explanation:
The given equation is:
2.6 = -0.2t
\(\frac{26}{10}\) = –\(\frac{2}{10}\)t
t = \(\frac{26}{10}\) ÷ ( –\(\frac{2}{10}\) )
t = – \(\frac{26}{10}\) × \(\frac{10}{2}\)
t = -13
Hence, from the above,
We can conclude that the value of t is: -13

Question 3.
– \(\frac{n}{5}\) = -2
Answer:
The value of n is: 10

Explanation:
The given equation is:
–\(\frac{n}{5}\) = -2
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
-n = -2 × 5
-n = -10
n = 10
Hence, from the above,
We can conclude that the value of n is: 10

1.2 Solving Multi-Step Equations (pp. 11–18)

Solve −6x + 23 + 2x = 15.
Big Ideas Math Answer Key Algebra 1 Chapter 1 Solving Linear Equations 108

Solve the equation. Check your Solution.

Question 4.
3y + 11 = -16
Answer:
The value of y is: -9

Explanation:
The given equation is:
3y + 11 = -16
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
3y = -16 – 11
3y = -27
y = -27 ÷ 3
y = -9
Hence, from the above,
We can conclude that the value of y is: -9

Question 5.
6 = 1 – b
Answer:
The value of b is: -5

Explanation:
The given equation is:
6 = 1 – b
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
b = 1 – 6
b = -5
Hence, from the above,
We can conclude that the value of b is: -5

Question 6.
n + 5n + 7 = 43
Answer:
The value of n is: 6

Explanation:
The given equation is:
n + 5n + 7 = 43
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
6n + 7 = 43
6n = 43 – 7
6n = 36
n = 36 ÷ 6
n = 6
Hence, from the above,
We can conclude that the value of n is: 6

Question 7.
-4(2z + 6) – 12 = 4
Answer:
The value of z is: -5

Explanation:
The given equation is:
-4 ( 2z + 6 ) – 12 = 4
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
-4 ( 2z + 6 ) = 4 + 12
-4 ( 2z + 6 ) = 16
-4 ( 2z ) – 4 ( 6 ) = 16
-8z – 24 = 16
-8z = 16 + 24
-8z = 40
z = 40 ÷ ( -8 )
z = -5
Hence, from the above,
We can conclude that the value of z is: -5

Question 8.
\(\frac{3}{2}\)(x – 2) – 5 = 19
Answer:
The value of x is: 18

Explanation:
The given equation is:
\(\frac{3}{2}\) ( x – 2 ) – 5 = 19
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
\(\frac{3}{2}\) ( x – 2 ) = 19 + 5
\(\frac{3}{2}\) ( x – 2 ) = 24
x – 2 = 24 × \(\frac{2}{3}\)
x – 2 = \(\frac{24}{1}\) × \(\frac{2}{3}\)
x – 2 = 16
x = 16 + 2
x = 18
Hence, from the above,
We can conclude that the value of x is: 18

Question 9.
6 = \(\frac{1}{5}\)w + \(\frac{7}{5}\)w – 4
Answer:
The value of w is: \(\frac{25}{4}\)

Explanation:
The given equation is:
6 = \(\frac{1}{5}\)w + \(\frac{7}{5}\)w – 4
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
6 + 4 = \(\frac{1}{5}\)w + \(\frac{7}{5}\)w
10 = w [ \(\frac{1 + 7}{5}\) ]
10 = \(\frac{8}{5}\)w
w = 10 × \(\frac{5}{8}\)
w = \(\frac{10}{1}\) × \(\frac{5}{8}\)
w = \(\frac{25}{4}\)
Hence, from the above,
We can conclude that the value of w is: \(\frac{25}{4}\)

Find the value of x. Then find the angle measures of the polygon.

Question 10.
Big Ideas Math Answer Key Algebra 1 Chapter 1 Solving Linear Equations 109

Answer:
The angle measures of the given polygon are: 110°, 50°, 20°

Explanation:
The given figure is:
Big Ideas Math Answer Key Algebra 1 Chapter 1 Solving Linear Equations 109
From the above figure,
The angle measures are: 110°, 5x°, 2x°
It is also given that
The sum of the angle measures = 180°
So,
110 + 5x + 2x = 180°
7x = 180 – 110
7x = 70°
x = 70 / 7
x = 10°
Hence, from the above,
We can conclude that the angle measures of the given polygon are: 110°, 50°, 20°

Question 11.
Big Ideas Math Answer Key Algebra 1 Chapter 1 Solving Linear Equations 110

Answer:
The angle measures of the given polygon are: 126°, 126°, 96°, 96°, 96°

Explanation:
The given figure is:
Big Ideas Math Answer Key Algebra 1 Chapter 1 Solving Linear Equations 110
From the above figure,
The angle measures of the given polygon are: ( x – 30 )°, x°, x°, ( x – 30 )°, ( x – 30 )°
It is also given that,
The sum of the angle measures of the given polygon = 540°
So,
( x – 30 )° + x° + x° + ( x – 30 )° + ( x – 30 )° = 540°
5x – 90° = 540°
5x = 540° + 90°
5x = 630°
x = 630 / 5
x = 126°
Hence,f rom the above,
We can conclude that the angle measures of the given polygon are: 126°, 126°, 96°, 96°, 96°

1.3 Solving Equations with Variables on Both Sides (pp. 19–24)

Solve 2( y − 4) = −4( y + 8).
Big Ideas Math Answer Key Algebra 1 Chapter 1 Solving Linear Equations 111

Solve the equation.

Question 12.
3n – 3 = 4n + 1
Answer:
The value of n is: -4

Explanation:
The given equation is:
3n – 3 = 4n + 1
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
4n – 3n = -1 – 3
n = -4
Hence, from the above,
We can conclude that the value of n is: -4

Question 13.
5(1 + x) = 5x + 5
Answer:
The given equation has no solution

Explanation:
The given equation is:
5 ( 1 + x ) = 5x + 5
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
5 ( 1 ) + 5 ( x ) = 5x + 5
5 + 5x = 5x + 5
5 = 5x – 5x + 5
5 = 5
Hence, from the above,
We can conclude that the given equation has no solution

Question 14.
3(n + 4) = \(\frac{1}{2}\)(6n + 4)
Answer:
The given equation has no solution

Explanation:
The given equation is:
3 ( n + 4 ) = \(\frac{1}{2}\) ( 6n + 4 )
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
3 ( n ) + 3 ( 4 ) = \(\frac{1}{2}\) ( 6n + 4 )
3n + 12 = \(\frac{1}{2}\) ( 6n + 4 )
2 ( 3n + 12 ) = 6n + 4
2 ( 3n ) + 2 ( 12 ) = 6n + 4
6n + 24 = 6n + 4
24 = 6n – 6n + 4
24 = 4
Hence, from the above,
We can conclude that the given equation has no solution

1.4 Solving Absolute Value Equations (pp. 27–34)

a. Solve | x − 5 | = 3.
Big Ideas Math Answer Key Algebra 1 Chapter 1 Solving Linear Equations 112

b. Solve | 2x + 6 | = 4x. Check your solutions.
Big Ideas Math Answer Key Algebra 1 Chapter 1 Solving Linear Equations 113

Big Ideas Math Answer Key Algebra 1 Chapter 1 Solving Linear Equations 114

Check the apparent solutions to see if either is extraneous.
The solution is x = 3. Reject x = -1 because it is extraneous.

Solve the equation. Check your solutions.

Question 15.
| y + 3 | = 17
Answer:
The value of y is: 14 or -20

Explaantion:
The given absolute value equation is:
| y + 3 | = 17
We know that,
| x | = x for x  0
| x | = -x for x < 0
So,
y + 3 = 17                               y + 3 = -17
y = 17 – 3                                y = -17 – 3
y = 14                                      y = -20
Hence, from the above,
We can conclude that the value of y is: 14 or -20

Question 16.
-2 | 5w – 7 | + 9 = -7
Answer:
The value of w is: 3 or –\(\frac{1}{5}\)

Explanation:
The given absolute value equation is:
-2 | 5w – 7 | + 9 = -7
-2 | 5w – 7 | = -7 – 9
-2 | 5w – 7 | = -16
| 5w – 7 | = -16 / ( -2 )
| 5w – 7 | = 8
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
5w – 7 = 8                                     5w – 7 = -8
5w = 8 + 7                                    5w = -8 + 7
5w = 15                                         5w = -1
w = 15 ÷ 5                                     w = –\(\frac{1}{5}\)
w = 3                                              w =-\(\frac{1}{5}\)
Hence, from the above,
We can conclude that the value of w is: 3 or –\(\frac{1}{5}\)

Question 17.
| x – 2 | = | 4 + x |
Answer:
The given absolute equation has no solution

Explanation:
The given absolute value equation is:
| x – 2 | = | 4 + x |
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
x – 2 = 4 + x                                                   – ( x – 2 ) = – ( 4 + x )
-2 = 4                                                                2 = -4
Hence, from the above,
We can conclude that the given absolute equation has no solution

Question 18.
The minimum sustained wind speed of a Category 1 hurricane is 74 miles per hour. The maximum sustained wind speed is 95 miles per hour. Write an absolute value equation that represents the minimum and maximum speeds.
Answer:
The absolute value equation that represents the minimum and maximum speeds is:
| x – 84.5 | = 9.5

Explanation:
It is given that the minimum sustained wind speed of a Category 1 hurricane is 74 miles per hour. The maximum sustained wind speed is 95 miles per hour.
So,
The average wind speed sustained = ( 74 + 95 ) /2
= 169 / 2
= 84.5 miles per hour
Now,
The minimum wind speed from the average speed point = 84.5 – 74
= 9.5 miles per hour
So,
The absolute value equation that represents the minimum and maximum wind speed is:
| x – 84.5 | = 9.5

1.5 Rewriting Equations and Formulas (pp. 35–42)
a. The slope-intercept form of a linear equation is y = mx + b. Solve the equation for m.
Big Ideas Math Answer Key Algebra 1 Chapter 1 Solving Linear Equations 115

b. The formula for the surface area S of a cylinder is S = 2πr2 + 2πrh. Solve the formula for the height h.
Big Ideas Math Answer Key Algebra 1 Chapter 1 Solving Linear Equations 116

Solve the literal equation for y.

Question 19.
2x – 4y = 20
Answer:
The value of y is: ( x / 2 ) – 5

Explanation:
The given literal equation is:
2x – 4y = 20
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
4y = 2x – 20
y = ( 2x – 20 ) / 4
y = ( 2x / 4 ) – ( 20 / 4 )
y = ( x / 2 ) – 5
Hence, from the above,
We can conclude that the value of y is: ( x / 2 ) – 5

Question 20.
8x – 3 = 5 + 4y
Answer:
The value of y is: 2x – 2

Explanation:
The given literal equation is:
8x – 3 = 5 + 4y
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
4y = 8x – 3 – 5
4y = 8x – 8
y = ( 8x – 8 ) / 4
y = ( 8x / 4 ) – ( 8 – 4 )
y = 2x – 2
Hence, from the above,
We can conclude that the value of y is: 2x – 2

Question 21.
a a = 9y + 3yx
Answer:
The value of y is: a² / ( 3x + 9 )

Explanation:
The given literal equation is:
a² = 9y + 3yx
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
a² = y ( 3x + 9 )
y = a² / ( 3x + 9 )
Hence, from the above,
We can conclude that the value of y is: a² / ( 3x + 9 )

Question 22.
The volume V of a pyramid is given by the formula V = \(\frac{1}{3}\)Bh, where B is the area of the base and h is the height.
a. Solve the formula for h.
b. Find the height h of the pyramid.
Big Ideas Math Answer Key Algebra 1 Chapter 1 Solving Linear Equations 117

Answer:
a)
The value of h is: \(\frac{3V}{B}\)

Explanation:
The given formula is:
V = \(\frac{1}{3}\)Bh
Where,
B is the area of the base
h is the height
Now,
3V = Bh
h = \(\frac{3V}{B}\)
Hence, from the above,
We can conclude that the value of h is: \(\frac{3V}{B}\)

b)
The value of h is: 18 cm

Explanation:
From the given figure,
Area of the base ( B ) = 36 cm²
Volume of the base ( V ) = 216 cm³
From part (a),
h = \(\frac{3V}{B}\)
h = \(\frac{3 × 216}{36}\)
h = \(\frac{3 × 216}{36 × 1}\)
h = 18 cm
Hence, from the above,
We can conclud ethat the value of h is: 18 cm

Question 23.
The formula F = \(\frac{9}{5}\)(K – 273.15) + 32 converts a temperature from kelvin K to degrees Fahrenheit F.
a. Solve the formula for K.
b. Convert 180°F to kelvin K. Round your answer to the nearest hundredth.
Answer:
a)
The formula for K is:
K = \(\frac{5}{9}\) ( F – 32 ) + 273.15

Explanation:
The given formula for F is:
F = \(\frac{9}{5}\) ( K – 273.15 ) + 32
Now,
F – 32 = \(\frac{9}{5}\) ( K – 273.15 )
\(\frac{5}{9}\) ( F – 32 ) = K – 273.15
K = \(\frac{5}{9}\) ( F – 32 ) + 273.15
Hence, from the above,
We can conclude that the value of K is: \(\frac{5}{9}\) ( F – 32 ) + 273.15

Solving Linear Equations Chapter Test

Solve the equation. Justify each step. Check your solution.

Question 1.
x – 7 = 15
Answer:
The value of x is: 22

Explanation:
The given equation is:
x – 7 = 15
Now,
x = 15 + 7
x = 22
Hence, from the above,
We can conclude that the value of x is: 22

Question 2.
\(\frac{2}{3}\)x = 5
Answer:
The value of x is: \(\frac{15}{2}\)

Explanation:
The given equation is:
\(\frac{2}{3}\) x = 5
Now,
x = 5 × \(\frac{3}{2}\)
x = \(\frac{5}{1}\) × \(\frac{3}{2}\)
x = \(\frac{15}{2}\)
Hence, from the above,
We can conclude that the value of x is: \(\frac{15}{2}\)

Question 3
11x + 1 = -1 + x
Answer:
The value of x is: –\(\frac{1}{5}\)

Explanation:
The given equation is:
11x + 1 = -1 + x
Now,
11x – x = -1 – 1
10x = -2
x = –\(\frac{2}{10}\)
x = –\(\frac{1}{5}\)
Hence, from the above,
We can conclude that the value of x is: –\(\frac{1}{2}\)

Solve the equation.

Question 4.
2 | x – 3 | – 5 = 7
Answer:
The value of x is: 9 or -3

Explanation:
The given absolute value equation is:
2 | x – 3 | – 5 = 7
2 | x – 3 | = 7 + 5
2 | x – 3 | = 12
| x – 3 | = \(\frac{12}{2}\)
| x – 3 | = 6
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
x – 3 = 6                             x – 3 = -6
x = 6 + 3                            x = -6 + 3
x = 9                                   x = -3
Hence, from the above,
We can conclude that the value of x is: 9 or -3

Question 5.
| 2x – 19 | = 4x + 1
Answer:
The value of x is: -10 or 3

Explanation:
The given absolute value equation is:
| 2x – 19 | = 4x + 1
We know that,
| x | = x for x > 0
| x | = -x for x < 0
4x + 1 = 2x – 19                         4x + 1 = – ( 2x – 19 )
4x – 2x = -19 – 1                         4x + 2x = 19 – 1
2x = -20                                      6x = 18
x = \(\frac{-20}{2}\)      x = \(\frac{18}{6}\)
x = -10                                         x = 3
Hence, from the above,
We can conclude that the value of x is: -10 or 3

Question 6.
-2 + 5x – 7 = 3x – 9 + 2x
Answer:
The given absolute equation has no solution

Explanation:
The given equation is:
-2 + 5x – 7 = 3x – 9 + 2x
5x – 9 = 5x – 9
Hence, from the above,
We can conclude that the given absolute value equation has no solution

Question 7.
3(x + 4) – 1 = -7
Answer:
The value of x is: -6

Explanation:
The given equation is:
3 ( x + 4 ) – 1 = -7
So,
3 ( x ) + 3 ( 4 ) = -7 + 1
3x + 12 = -6
3x = -6 – 12
3x = -18
x = –\(\frac{18}{3}\)
x = -6
Hence, from the above,
We can conclude that the value of x is: -6

Question 8.
| 20 + 2x | = | 4x + 4 |
Answer:
The value of x is: 8

Explanation:
The given absolute value equation is:
| 20 + 2x | = | 4x + 4 |
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
20 + 2x = 4x + 4
4x – 2x = 20 – 4
2x = 16
x = \(\frac{16}{2}\)
x = 8
Hence, from the above,
We can conclude that the value of x is: 8

Question 9.
\(\frac{1}{3}\)(6x + 12) – 2(x – 7) = 19
Answer:
The given equation has no solution

Explanation:
The given equation is:
\(\frac{1}{3}\) ( 6x + 12 ) – 2 ( x – 7 ) = 19
Now,
\(\frac{1}{3}\) ( 6x – 12 ) = 19 + 2 ( x – 7 )
\(\frac{1}{3}\) ( 6x – 12 ) = 19 + 2x – 14
\(\frac{1}{3}\) ( 6x – 12 ) = 2x + 5
1 ( 6x – 12 ) = 3 ( 2x + 5 )
6x – 12 = 6x + 15
6x – 6x = 15 + 12
15 = -12
Hence, from the above,
We can conclude that the given equation has no solution

Describe the values of c for which the equation has no solution. Explain your reasoning.

Question 10.
3x – 5 = 3x – c
Answer:
The value of c is: 5

Explanation:
The given equation is:
3x – 5 = 3x – c
It is given that the equation has no solution
So,
3x – 3x – 5 =-c
-c = -5
c = 5
Hence, from the above,
We can conclude that the value of c is: 5

Question 11.
| x – 7 | = c
Answer:
The value of c is: -7

Explanation:
The given absolute value equation is:
| x – 7 | = c
It is given that the equation has no solution i.e., x = 0
So,
0 – 7 = c
c = -7
Hence, from the above,
We can conclude that the value of c is: -7

Question 12.
A safety regulation states that the minimum height of a handrail is 30 inches. The maximum height is 38 inches. Write an absolute value equation that represents the minimum and maximum heights.
Answer:
The absolute value expression that represents the minimum and maximum heights is:
| x – 64 | = 34

Explanation:
It is given that a safety regulation states that the minimum height of a handrail is 30 inches. The maximum height is 38 inches.
So,
The average height of a handrail = ( 30 + 38 ) / 2
= 68 / 2
= 34 inches
Now,
The minimum height from the average height of a handrail = 34 + 30
= 64 inches
Hence,
The absolute value equation that represents the minimum and maximum height of a handrail is:
| x – 64 | = 34

Question 13.
The perimeter P (in yards) of a soccer field is represented by the formula P = 2ℓ + 2w, where ℓ is the length (in yards) and w is the width (in yards).
a. Solve the formula for w.
Answer:
The given formula is:
P = 2l + 2w
Where,
P is perimeter ( in yards )
l is the length ( in yards )
w is the width ( in yards )
So,
2w = P – 2l
w = ( P – 2l ) / 2
Hence, from the above,
We can conclude that the formula for w is:
w = ( P – 2l ) / 2

b. Find the width of the field.
Answer:
The given figure is:
Big Ideas Math Answer Key Algebra 1 Chapter 1 Solving Linear Equations 117.1
From the above figure,
Perimeter ( P ) = 330 yd
Length ( l) = 100 yd
From part (a),
w = ( P – 2l ) /2
w = ( 330 – 100 ) / 2
w = 230 / 2
w = 115 yd
Hence, from the above,
We can conclude that the width of the field is: 115 yd

c. About what percent of the field is inside the circle?
Big Ideas Math Answer Key Algebra 1 Chapter 1 Solving Linear Equations 117.1
Answer:

Question 14.
Your car needs new brakes. You call a dealership and a local mechanic for prices.
Big Ideas Math Answer Key Algebra 1 Chapter 1 Solving Linear Equations 118
a. After how many hours are the total costs the same at both places? Justify your answer.
Answer:
From the given table,
Let the total labor hours be x
So,
The total cost at the Dealership = ( Cost of parts ) + ( Labor cost per hour ) × ( Total labor hours )
= 24 + 99x
The total cost at the local mechanic = ( Cost of parts ) + ( Labor cost per hour ) × ( Total labor hours )
= 45 + 89x
It is given that the total cost is the same in both places
So,
24 + 99x = 45 + 89x
99x – 89x = 45 – 24
10x = 21
x = 21 / 10
x = 2.1 hours
Hence, from the above,
We can conclude that after 2.1 hours, the total cost will be the same in both places

b. When do the repairs cost less at the dealership? at the local mechanic? Explain.
Answer:
The given table is:
Big Ideas Math Answer Key Algebra 1 Chapter 1 Solving Linear Equations 118
From the above table,
Compare the labor cost per hour
By comparison,
We can say that the labor cost per hour is less at the local mechanic
Hence, from the above,
We can conclude that the repair cost less at the local mechanic

Question 15.
Consider the equation | 4x + 20 | = 6x. Without calculating, how do you know that x = -2 is an extraneous solution?
Answer:
We know that,
The absolute value equations only accept the values greater than or equal to 0
Hence,
For the given absolute value equation,
| 4x + 20 | = 6x
x = -2 is an extraneous solution

Question 16.
Your friend was solving the equation shown and was confused by the result “-8 = -8.” Explain what this result means.
4(y – 2) – 2y = 6y – 8 – 4y
4y – 8 – 2y = 6y – 8 – 4y
2y – 8 = 2y – 8
-8 = -8
Answer:
The result ” -8 = -8 ” means that the solved equation has no solution

Solving Linear Equations Cumulative Assessment

Question 1.
A mountain biking park has 48 trails, 37.5% of which are beginner trails. The rest are divided evenly between intermediate and expert trails. How many of each kind of trail are there?
A. 12 beginner, 18 intermediate, 18 expert
B. 18 beginner, 15 intermediate, 15 expert
C. 18 beginner, 12 intermediate, 18 expert
D. 30 beginner, 9 intermediate, 9 expert

Answer:
The correct option is: B
The number of beginner trials is: 18
The number of intermediate trials is: 15
The number of expert trials is: 15

Explanation:
It is given that a mountain biking park has 48 trails, 37.5% of which are beginner trails. The rest are divided evenly between intermediate and expert trials.
So,
The number of beginner trials is 3.5 % of the total number of trials
It is given that the total number of trials is: 48
We know that,
The value of 37.5 % is: \(\frac{3}{8}\) [ 37.5 % = 50 % – 12.5 % ]
So,
The number of beginner trials = \(\frac{3}{8}\) × 48
= \(\frac{3}{8}\) × \(\frac{48}{1}\)
= \(\frac{3 × 48}{8 × 1}\)
= 18
So,
The number  of intermediate and expert trials = ( The total number of trials ) – ( The number of beginner trials )
= 48 – 18
= 30 trials
It is also given that the intermediate trials and expert trails are divided evenly
So,
30 ÷ 2 = 15 trials each
Hence, from the above,
We can conclude that
The number of beginner trials is: 18
The number of intermediate trials is: 15
The number of expert trials is: 15

Question 2.
Which of the equations are equivalent to cx – a = b?
Big Ideas Math Answer Key Algebra 1 Chapter 1 Solving Linear Equations 119

Answer:
The given equations are:
a) cx – a + b = 2b
b) 0 = cx – a + b
c) 2cx – 2a = b / 2
d) x – a = b / 2
e) x = ( a + b ) / c
f) b + a = cx
Now,
We have to find the equations from above that is equivalent to the given equation cx – a = b
Now,
a)
The given equation is:
cx – a + b = 2b
So,
cx – a = b – b
cx – a = b
b)
The given equation is:
0 = cx – a + b
So,
cx – a = -b
c)
The given equation is:
2cx – 2a = b / 2
So,
cx – a = b / 4
d )
The given equation is:
cx – a = b / 2
So,
2 ( cx – a ) = b
e)
The given equation is:
x = ( a + b ) / c
So,
cx = a + b
cx – a = b
f)
The given equation is:
b + a = cx
So,
cx – a = b
Hence, from the above,
We can conclude that the equations that are equivalent to cx – a = b is: a, e, f

Question 3.
Let N represent the number of solutions of the equation 3(x – a) = 3x – 6. Complete each statement with the symbol <, >, or =.
a. When a = 3, N ____ 1.
b. When a = -3, N ____ 1.
c. When a = 2, N ____ 1.
d. When a = -2, N ____ 1.
e. When a = x, N ____ 1.
f. When a = -x, N ____ 1.

Answer:
The given equation is:
3 ( x – a ) = 3x – 6
So,
3x – 3a = 3x – 6
Now,
a) When a = 3,
3x – 3 ( 3 ) = 3x – 6
3x – 9 = 3x – 6
9 = 6
Hence,
When a = 3 there is no solution
Hence,
N < 1
b) When a = -3
3x + 3 ( 3 ) =3 x – 6
9 = -6
Hence,
When a = -3, there is no solution
Hence,
N < 1
c) When a = 2
3x – 3 ( 2 ) = 3x – 6
3x – 6 =3x –
6 = 6
Hence,
When a= 2, there is no solution
Hence,
N < 1
d) When a = -2
3x + 3 ( 2 ) = 3x – 6
3x + 6 = 3x – 6
6 = -6
Hence,
When a = -2, thereis no solution
Hence,
N < 1
e) When a = x
3x – 3 ( x ) = 3x – 6
3x = 6
x = 6 / 3
x = 2
Hence,
When a  x, theer is 1 solution
Hence,
N = 1
f) When a = -x
3x + 3 ( x ) = 3x – 6
6x – 3x = -6
3x = -6
x = -6 / 3
x = -2
Hence,
When a = -x, there is 1 solution
Hence,
N = 1

Question 4.
You are painting your dining room white and your living room blue. You spend $132 on 5 cans of paint. The white paint costs $24 per can, and the blue paint costs $28 per can.
a. Use the numbers and symbols to write an equation that represents how many cans of each color you bought.
Big Ideas Math Answer Key Algebra 1 Chapter 1 Solving Linear Equations 120
Answer:
The number of cans of white paint is: 2
The number of cans of blue paint is: 3

Explanation:
It is given that you spend $132 on 5 cans of paint
It is also given that the white paint costs $24 per can and the blue paint costs $28 per can
Now,
Let
The number of white cans is: x
The number of blue cans be: 5 – x
So,
The total cost of paint = ( The number of white cans ) × ( The cost of white paint per can ) + ( The number of blue cans ) × ( The cost of blue paint per can )
132 = 24x + 28 ( 5 – x )
24x + 28 ( 5 ) – 28x = 132
140 – 4x = 132
4x = 140 – 132
4x = 8
x = 8 ÷ 4
x = 2
Hence, from the above,
We can conclude that
The number of cans of white paint is: 2
The number of cans of blue paint is: 3

b. How much would you have saved by switching the colors of the dining room and living room? Explain.
Answer:
The money you have saved by switching the colors of the dining room and living room is: $0

Explanation:
From part (a),
The number of white cans is: 2
The number of blue cans is: 3
It is given that white color is used in the dining room and the blue color is used in the living room
So,
The cost of white paint after interchanging the color in the living room = 24 × 2
= $48
The cost of blue paint after interchanging the color in the dining room = 28 × 3
= $84
So,
The total cost of paint after interchanging the colors = 48 + 84
= $132
Hence,
The amount of money saved = ( The money you paid for the paint before interchanging ) – ( The money you paid for the paint after interchanging )
= 132 – 132
=$0

Question 5.
Which of the equations are equivalent?
Big Ideas Math Answer Key Algebra 1 Chapter 1 Solving Linear Equations 121

Answer:
The given equations are:
a ) 6x + 6 = -14
b ) 8x + 6 = -2x – 14
c ) 5x + 3 = -7
d ) 7x + 3 = 2x – 13
Now,
To find the equivalent equations, find the value of x
So,
a)
The given equation is:
6x + 6 = -14
6x = -14 – 6
6x = -20
x = -20 / 6
x = -10 / 3
b)
The given equation is:
8x + 6 = -2x – 14
8x + 2x = -14 – 6
10x = -20
x = -20 / 10
x = -2
c)
The given equation is:
5x + 3 = -7
5x = -7 -3
5x = -10
x= -10 / 5
x = -2
d)
The given equation is:
7x + 3 = 2x – 13
7x – 2x = -13 – 3
5x = -16
x = -16 / 5
Hence, from the above,
We can conclude that the equations c) and d) are equivalent

Question 6.
The perimeter of the triangle is 13 inches. What is the length of the shortest side?
Big Ideas Math Answer Key Algebra 1 Chapter 1 Solving Linear Equations 122
Answer:
The length of the shortest side is: 3 inches

Explanation:
We know that,
The perimeter is the sum of all the sides of the given figure
It is given that the perimeter of the triangle is: 13 inches
So,
The perimeter of the triangle = ( x – 5 ) + ( x / 2 ) + 6
13 = x + 1 + ( x / 2 )
13 = ( 2x / 2 ) + ( x / 2 ) + 1
3x / 2 = 13 – 1
3x / 2 = 12
3x = 12 × 2
3x = 24
x = 24 / 3
x = 8
So,
The lengths of all sides are: ( 8 – 5 ), 6, ( 8 / 2 ) = 3 inches, 6 inches, 4 inches
Hence, from the above,
We can conclude that the length of the shortest side is: 3 inches

Question 7.
You pay $45 per month for cable TV. Your friend buys a satellite TV receiver for $99 and pays $36 per month for satellite TV. Your friend claims that the expenses for a year of satellite TV are less than the expenses for a year of cable.
a. Write and solve an equation to determine when you and your friend will have paid the same amount for TV services.
Answer:
It is given that you pay $45 per month for cable TV. Your friend buys a satellite TV receiver for $99 and pays $36 per month for satellite TV
So,
Let the number of months be x
Now,
The time they paid the same amount for TV services is:
45x = 99 + 36x
45x – 36x = 99
9x= 99
x = 99 / 9
x = 11
Hence, from the above,
We can conclude that after 11 months, you and your friend will pay the same amount for TV services

b. Is your friend correct? Explain.
Answer:
Your friend is correct

Explanation:
We know that,
1 year = 12 months
So,
The expenses paid by you for TV services = 45x = 45 × 12 = $540
The expenses paid by your friend for TV services = 99 + 36x
= 99 + 36 ( 12 )
= 99 + 432
= $531
By comparing the expenses of you and your friend,
We can conclude that your friend is correct

Question 8.
Place each equation into one of the four categories.
Big Ideas Math Answer Key Algebra 1 Chapter 1 Solving Linear Equations 123

Question 9.
A car travels 1000 feet in 12.5 seconds. Which of the expressions do not represent the average speed of the car?
Big Ideas Math Answer Key Algebra 1 Chapter 1 Solving Linear Equations 124

Answer:
We know that,
Average speed = ( Distance ) ÷ ( Time )
It is given that a car travels 1000 feet in 12.5 seconds
So,
Average speed = 1000 / 12.5
= 80\(\frac{feet}{second}\)
Now,
The given options are:
A) 80\(\frac{second}{feet}\) B) 80\(\frac{feet}{second}\) C) \(\frac{80 feet}{second}\)
D) \(\frac{second}{ 80 feet}\)
Hence, from the above,
We can conclude that option B) represents the average speed

Big Ideas Math Answers Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers

Big Ideas Math Answers Grade 4 Chapter 2

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Big Ideas 4th Grade Chapter 2 Add and Subtract Multi-Digit Numbers Math Book Answer Key

Check out the Big Ideas Math Answers Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers links provided in the below section to begin your preparation. We have given a detailed explanation for each and every problem for your better understanding. So, make use of every given link and try to score the best marks in the exams. The basics mentioned here will not only help you in the exam but also in real life. Be the first to learn the best of math topics by referring to bigideasmathanswers.com

Lesson: 1 Estimate Sum and Differences

Lesson 2.1 Estimate Sum and Differences
Estimate Sum and Differences Homework & Practice 2.1

Lesson: 2 Add Multi-Digit Numbers

Lesson 2.2 Add Multi-Digit Numbers
Add Multi-Digit Numbers Homework & Practice 2.2

Lesson: 3 Subtract Multi-Digit Numbers

Lesson 2.3 Subtract Multi-Digit Numbers
Subtract Multi-Digit Numbers Homework & Practice 2.3

Lesson: 4 Use Strategies to Add and Subtract

Lesson 2.4 Use Strategies to Add and Subtract
Use Strategies to Add and Subtract Homework & Practice 2.4

Lesson: 5 Problem Solving: Addition and Subtraction

Lesson 2.5 Problem Solving: Addition and Subtraction
Problem Solving: Addition and Subtraction Homework & Practice 2.5

Performance Task

Add and Subtract Multi-Digit Numbers Performance Task
Add and Subtract Multi-Digit Numbers Activity
Add and Subtract Multi-Digit Numbers Chapter Practice

Lesson 2.1 Estimate Sum and Differences

Explore and Grow
Estimate to find each sum by rounding to the nearest thousand, hundred, or ten. Explain why you chose to round to that place value.
A football team had 917 spectators at their first game and 872 at their second game. About how many spectators did the team have at both games?

Answer: The answer is 1789.

Explanation:
Number of spectators in First game: 917
Number of spectators in First game: 872
To get the number of spectators in both the games, we have to add the number of spectators in first game and second game.
That means, 917 + 872 = 1789.
A company budgets $1,800 for a company picnic. They spend $917 on the location and $872 on food. Did they stay within their budget?

Answer: Yes, they can stay within their budget.

Explanation:
Total budget: $1,800
Spend for location: $917
Spend for food: $872
Total spend = spend for location + spend for food.
That means, total spend = $917 + $872 = $1789.
Balance budget = Total budget – Total spend
That means, Balance budget: $1800 – $1789 = $11
As, the total spend for the company picnic is lesser than the company budget, they can stay within the budget.
Reasoning
Explain why you may choose to round to different place values in different situations.

Answer: “Round to” are the approximate values we give that are the nearest possible values to the exact values. Sometimes, we may be sufficient with the estimate or nearest values. We may not require the exact values. In such situations, we can go ahead with the round to values. For example, you organized a lunch at your home. The caterer asks you for the number of guests for whom the lunch has to be served. You cannot give him the exact number of guests. You give him the round of value of the number of guests attend based on your past experience and present situation. In such cases, round to is very helpful.
Think and Grow: Estimate Sum and Differences
An estimate is a number that is close to an exact number. You can use rounding to estimate sums and differences.
Example
Estimate 8,675 + 3,214.
One Way: Round each addend to the nearest hundred. Then find the sum.

So, 8,675 + 3,214 is about rounding the two numbers to the nearest hundreds. 8,700 is the nearest hundreds value for 8,675. 3200 is the nearest hundreds value for 3214.
Then, we will add 8,700 and 3,200.
8,700 + 3,200 = 11, 900
Another Way: Round each addend to the nearest thousand. Then find the sum.

So, 8,675 + 3,214 is about rounding the two numbers to the nearest thousands. 9,000 is the nearest thousands value for 8,675. 3,000 is the nearest thousands value for 3214.
Then, we will add 9,000 and 3,000.
9,000 + 3,000 = 12, 000
.
Example
Estimate 827,615 – 54,3006.
One Way: Round each addend to the nearest thousand. Then find the difference.

So, 827,615 – 54,306 is about rounding the two numbers to the nearest thousands. 8,28,000 is the nearest thousands value for 8,27,615. 54,000 is the nearest thousands value for 54,306.
Then, we will subtract 54,000 from 8,28,000.
8,28,000 – 54,000 = 7,74,000.
Another Way: Round each addend to the nearest ten thousand. Then find the difference.

So, 827,615 – 54,306 is about rounding the two numbers to the nearest ten thousands. 8,30,000 is the nearest thousands value for 8,27,615. 50,000 is the nearest thousands value for 54,306.
Then, we will subtract 50,000 from 8,30,000.
8,28,000 – 54,000 = 7,80,000.
Show and Grow
Estimate the sum or difference.

Question 1.
Big Ideas Math Answer Key Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.1 5

Answer: We will here round of the values to the nearest hundreds. 63,900 is the nearest hundred for 63,851. For, 19,375 the nearest hundred value is 19,400.

Now we will add 63,900 to 19,400.
63,900 + 19,400 = 83,300.

Question 2.
Big Ideas Math Answer Key Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.1 6

Answer: We will here round of the values to the nearest hundreds. 4,900 is the nearest hundred for 4,874. For, 2,530 the nearest hundred value is 2,500.

Now we will subtract 2,500 from 4,900.
4,900 – 2,500 = 2,400.
Apply and Grow: Practice
Estimate the sum or difference.

Question 3.
Big Ideas Math Answer Key Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.1 7

Answer:
We here round the values to the nearest hundreds value.
For 27,369 the nearest hundreds value is 27,400. For 14,608 the nearest hundred value is 14,600.

27,400 + 14,600 = 42,000.
So, the estimate of the sum of 27,369 and 14,608 by rounding to the nearest hundreds value is 42,000.

Question 4.
Big Ideas Math Answer Key Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.1 8

Answer:
We here round the values to the nearest hundreds value.
For 53,744 the nearest hundreds value is 53,700. For 41,086 the nearest hundred value is 41,100.

53,700 – 41,100 = 12,600.
So, the estimate of the difference of 53,744 and 41,086 by rounding to the nearest hundreds value is 12,600.

Question 5.
Big Ideas Math Answer Key Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.1 9

Answer:
We here round the values to the nearest thousands value.
For 68,451 the nearest thousands value is 68,000. For 40,695 the nearest thousands value is 41,000.
68,000 – 41,000 = 27,000.
So, the estimate of the difference of 68,451 and 40,695 by rounding to the nearest thousands value is 27,000.

Question 6.
Big Ideas Math Answer Key Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.1 10

Answer:
We here round the values to the nearest thousands value.
For 34,685 the nearest thousands value is 35,000. For 27,043 the nearest thousands value is 27,000.
35,000 + 27,000 = 62,000.
So, the estimate of the sum of 34,685 and 27,043 by rounding to the nearest thousands value is 62,000.

Question 7.
Big Ideas Math Answer Key Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.1 11

Answer:
We here round the values to the nearest ten thousands value.
For 908,465 the nearest ten thousands value is 910,000. For 653,629 the nearest ten thousands value is 650,000.
910,000 – 650,000 = 260,000.
So, the estimate of the difference of 908,465 and 653,299 by rounding to the nearest ten thousands value is 260,000.

Question 8.
Big Ideas Math Answer Key Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.1 12

Answer:
We here round the values to the nearest ten thousands value.
For 478,633 the nearest ten thousands value is 480,000. For 200,081 the nearest ten thousands value is 200,000.
480,000 + 200,000 = 680,000.
So, the estimate of the sum of 478,633 and 200,081 by rounding to the nearest ten thousands value is 680,000.

Question 9.
Big Ideas Math Answer Key Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.1 13

Answer:
We here round the values to the nearest thousands value.
For 395,408 the nearest thousands value is 395,000. For 102,677 the nearest thousands value is 103,000.
395,000 – 103,000 = 292,000.
So, the estimate of the difference of 395,408 and 102,677 by rounding to the nearest thousands value is 292,000.
Question 10.
Big Ideas Math Answer Key Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.1 14

Answer:
For 563,427 the nearest ten thousands value is 560,000. For 178,023 the nearest ten thousands value is 180,000.
560,000 + 180,000 = 740,000.
So, the estimate of the sum of 563,427 and 178,023 by rounding to the nearest ten thousands value is 740,000.

Question 11.
888,056 – 423,985 = _______

Answer:
We here round the values to the nearest ten thousands value.
For 888,056 the nearest ten thousands value is 890,000. For 423,985 the nearest ten thousands value is 420,000.
890,000 – 420,000 = 472,000.
So, the estimate of the difference of 888,056 and 423,985 by rounding to the nearest ten thousands value is 470,000.
Question 12.
713,642 + 49,018 = ______

Answer:
We here round the values to the nearest thousands value.
For 713,642 the nearest thousands value is 714,000. For 49,018 the nearest thousands value is 49,000.
714,000 + 49,000 = 763,000.
So, the estimate of the sum of 713,642 and 49,018 by rounding to the nearest thousands value is 763,000.

Question 13.
DIG DEEPER!
Is 20,549 + 9,562 greater than or less than 30,000? Explain how you know without finding the exact sum.

Answer:
We will solve this using round to estimates.
For 20,549 the nearest hundreds value is 20,500. For 9,562 the nearby smaller hundreds value is 9,500.
20,500 + 9,500 = 30,000.
The estimate sum is 30,000. We still have the values of 49 and 62 in both the values. So, the value of the sum of 20,549 and 9,562 is greater than 30,000.

Question 14.
Writing
Describe a real-life situation in which estimation would not be appropriate to use.

Answer:
Think and Grow: Modeling Real Life
Example
About how many more pounds does the whale shark weigh than the orca?
Big Ideas Math Answer Key Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.1 15
Round the weight of each animal to the nearest thousand because you do not need a precise answer.
Orca: 8,000 Whale shark: 40,000.
Subtract the estimated weight of the orca from the estimated weight of the whale shark.

The whale shark weighs about 32,000 more pounds than the orca.
Show and Grow

Question 15.
About how many more votes did Candidate A receive than Candidate B?
Big Ideas Math Answer Key Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.1 17

Answer:
Round the number of votes to the nearest thousands value.
Candidate A: 250,000. Candidate B: 85,000

Candidate A scored about 165,000 votes more than Candidate B.

Question 16.
Mount Saint Helens is a volcano that is 8,363 feet tall. Mount Fuji is a volcano that is 4,025 feet taller than Mount Saint Helens. About howtall is Mount Fuji?
Big Ideas Math Answer Key Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.1 18

Answer:
Round the heights of the volcanoes to the nearest hundreds value.
Mount Saint Helens: 8,400. Mount Fuji: 4,000.

So, Mount Helens is about 4,400 feet taller than Mount Fuji.

Question 17.
An educational video has 6,129 fewer views than a gaming video. The educational video has 483,056 views. About how many views does the gaming video have?

Answer:
Here, it is given that the educational video has 6,129 fewer views than the gaming video.
That means, Gaming Video views – 6,129 = Educational video views.
Or Gaming video views = Educational video views + 6,129.
Given educational video views = 483,056
Therefore, Gaming video views = 483,056 + 6,129 = 489,185.
So, the number of gaming video views = 489,185.

Estimate Sum and Differences Homework & Practice 2.1

Estimate the sum or difference.

Question 1.
Big Ideas Math Answer Key Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.1 19

Answer:
Round the values to the nearest hundreds values.
For 7,910 the nearest hundreds value is 7,900. For 1,358, the nearest hundreds value is 1,400.

7,900 + 1,400 = 9,300.

Question 2.
Big Ideas Math Answer Key Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.1 20

Answer:
Round the values to the nearest hundreds values.
For 5,608 the nearest hundreds value is 5,600. For 3,217, the nearest hundreds value is 3,200.


5,600 – 3,200 = 2,400

Question 3.
Big Ideas Math Answer Key Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.1 21

Answer:
Round the values to the nearest thousands values.
For 73,406 the nearest thousands value is 73,000. For 45,699 the nearest thousands value is 46,000.

73,000 – 46,000 = 27,000

Question 4.
Big Ideas Math Answer Key Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.1 22

Answer:
Round the values to the nearest thousands values.
For 82,908 the nearest thousands value is 83,000. For 28,643 the nearest thousands value is 29,000.

83,000 + 29,000 = 1,12,000

Question 5.
Big Ideas Math Answer Key Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.1 23

Answer:
Round the values to the nearest thousands values.
For 96,420 the nearest thousands value is 96,000. For 63,877 the nearest thousands value is 64,000.

96,000 – 64,000 = 32,000

Question 6.
Big Ideas Math Answer Key Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.1 24

Answer:
Round the values to the nearest ten thousands values.
For 517,605 the nearest ten thousands value is 520,000. For 359,421 the nearest ten thousands value is 360,000.

520,000 + 360,000 = 880,000

Question 7.
Big Ideas Math Answer Key Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.1 25

Answer:
Round the values to the nearest ten thousands values.
For 688,203 the nearest ten thousands value is 690,000. For 444,387 the nearest ten thousands value is 440,000.

690,000 – 440,000 = 250,000

Question 8.
Big Ideas Math Answer Key Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.1 26

Answer:
Round the values to the nearest ten thousands values.
For 261,586 the nearest ten thousands value is 260,000. For 116,934 the nearest ten thousands value is 120,000.

260,000 + 120,000 = 380,000
Estimate the sum or difference.

Question 9.
864,733 – 399,608 = ______

Answer:
Round the values to the nearest ten thousands values.
864,733: 860,000
399,608: 400,000
860,000 – 400,000 = 460,000

Question 10.
134,034 + 26,987 = ______

Answer:
Round the values to the nearest thousands values.
134,034: 134,000
26,987: 27,000
134,000 + 27,000 = 161,000

Question 11.
Number Sense
Descartes estimates a difference by rounding each number to the nearest ten thousand. His estimate is 620,000. Which problems could he have estimated?
Big Ideas Math Answer Key Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.1 27

Answer:
Round off all the given values to their nearest ten thousand value.
690,000 – 70,000 = 620,000
890,000 – 270,000 = 620,000
680,000 – 50,000 = 630,000
700,000 – 80,000 = 620,000.
So, Descartes could have estimated problems 1,2 and 4 and all the three give the estimated difference of 620,000.

Question 12.
Reasoning
When might you estimate the difference of 603,476 and 335,291 to the nearest hundred? to the nearest hundred thousand?

Answer:
When the precision of the difference value should be in hundreds, then round the values to hundred and calculate the difference.
603,476: 603,500 & 335,291: 335,300
603,500 – 335,300 = 268,200.
When the precision of the difference value should be in thousands, then round the values to thousands and calculate the difference.
603,476: 603,000 & 335,291: 335,000
603,000 – 335,000 = 268,000

Question 13.
Modeling Real Life
A storm causes causes 23,890 homes to be without power on the east side of a city and 18,370 homes to be without power on the west side. About how many homes altogether are without power?
Big Ideas Math Answer Key Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.1 28

Answer:
Homes without power on east side: 23,890
Homes without power on west side: 18,370
Round the values to the nearest thousands value.
So, homes without power on east side = 24,000 and homes without power on west side = 18,000.
Total homes without power = 24,000 + 18,000 = 42,000
Therefore, total estimate of homes without power due to storm are 42,000.

Question 14.
Modeling Real Life
You walk 5,682 steps. Your teacher walks 4,219 steps more than you. About how many steps does your teacher walk?
Big Ideas Math Answer Key Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.1 29

Answer:
Round the number of steps to the nearest hundreds value.
Number of steps I walked, 5,682 rounding it to the nearest hundreds value = 5,700
Number of steps teacher walked additional more than me, 4,219 rounding it to the nearest hundreds value = 4,200
Total number of steps teacher walked = Number of steps I walked + additional number of steps teacher walked.
Therefore total number of steps teacher walked = 5,700 + 4,200 = 9,900.
Review & Refresh
Find the product.

Question 15.
3 × 3 × 2 = ______

Answer:
Break the multiplication into steps as follows:
(3 X 3) x 2
= 9 x 2
=18

Question 16.
2 × 4 × 7 = ______

Answer:
Break the multiplication into steps as follows:
2 X (4 X 7)
= 2 X 28
= 56

Question 17.
3 × 3 × 5 = ______

Answer:
Break the multiplication into steps as follows:
(3 X 3) X 5
= 9 X 5
= 45.

Question 18.
6 × 2 × 3 = ______

Answer:
Break the multiplication into steps as follows:
6 X (2 X 3)
=6 X 6
= 36.

Question 19.
4 × 9 × 2 = ______

Answer:
Break the multiplication into steps as follows:
4 X (9 X 2)
= 4 X 18
= 72

Question 20.
4 × 10 × 2 = ______

Answer:
Break the multiplication into steps as follows:
4 X (10 X 2)
= 4 X 20
= 80.

Lesson 2.2 Add Multi-Digit Numbers

Explore and Grow
Which addition problem shows a correct way to find 38 +7? Why?
Big Ideas Math Answers 4th Grade Chapter 2 Add and Subtract Multi-Digit Numbers 2.2 1

Answer:
The second option shows the correct way.
Explanation: As per place value rule, 7 is the units position value. A units digit always should be added to the units digit of the other number. In the first option, 7 is under 3, which is actually in tens digit position. The second option displays it correctly pointing 7 under the units digit value, 8.
Which addition problem shows a correct way to find 403 + 1,248? Why?
Big Ideas Math Answers 4th Grade Chapter 2 Add and Subtract Multi-Digit Numbers 2.2 2

Answer:
The first option shows the correct way.
Explanation: As per place value rule, a units digit always should be added to the units digit of the other number. Similarly, tens digit value to the other number’s tens digit value and the hundreds digit value. First option maintains the rule correctly pointing 8 under 3, both which are units values. Similarly, 4 under 0 and 2 under 8. The second option started the sum from the left, which means, adding thousands digit value of 1,248 to the hundreds digit value of 403.
Reasoning
Why do you need to use place value when adding? Explain.

Answer:
While adding, we use place value i.e., we go from right hand side to the left hand side. In other words, we add the digits as per the place value. The digit of every number holds the place value. That means, a digit in tens place has the value of multiple of ten. Similarly, a digit in hundreds place has the value of multiple of hundred.
For example, In the number 23, 2 digit holds the value of multiple of ten. That means, 2 X 10.
To continue the parity of the sum, we follow the place value rule by adding digits as per the place values.
Think and Grow: Add Multi-Digit Numbers
Example
Add: 307,478 + 95,061.
Estimate: 307,000 + 95,000 = _______
Use place value to line up the addends.
Big Ideas Math Answers 4th Grade Chapter 2 Add and Subtract Multi-Digit Numbers 2.2 3
Add the ones, then the tens, and then the hundreds. Regroup if necessary.
Big Ideas Math Answers 4th Grade Chapter 2 Add and Subtract Multi-Digit Numbers 2.2 4
Add the thousands, then the ten thousands, and then the hundred thousands. Regroup if necessary.
Big Ideas Math Answers 4th Grade Chapter 2 Add and Subtract Multi-Digit Numbers 2.2 5
Show and Grow
Find the sum. Check whether your answer is reasonable.

Question 1.
Estimate: _______
Big Ideas Math Answers 4th Grade Chapter 2 Add and Subtract Multi-Digit Numbers 2.2 6

Answer:
Estimate: Round the values to the nearest hundreds.
17,700 + 53,000 = 70,700.
Add the tens, then hundreds, thousands and then the ten thousands. Regroup if necessary.

Check: Because 70,714 is close to the estimate, 70,700, the answer is reasonable.

Question 2.
Estimate: ______
Big Ideas Math Answers 4th Grade Chapter 2 Add and Subtract Multi-Digit Numbers 2.2 7

Answer:
Round the values to the nearest hundreds.
297,900 + 6,100 = 304,000
Add the ones, tens, then hundreds and thousands. Regroup if necessary.

Check: Because 303,982 is close to the estimate, 304,000, the answer is reasonable.
Apply and Grow: Practice
Find the sum. Check whether your answer is reasonable.

Question 3.
Estimate: ______
Big Ideas Math Answers 4th Grade Chapter 2 Add and Subtract Multi-Digit Numbers 2.2 8

Answer:
Round the values to the nearest hundreds.
Estimate: 6,400 + 3,300 = 9,700
Add the ones, tens, then hundreds and thousands. Regroup if necessary.

Check: Because 9,719 is close to the estimate, 9,700, the answer is reasonable.

Question 4.
Estimate: ______
Big Ideas Math Answers 4th Grade Chapter 2 Add and Subtract Multi-Digit Numbers 2.2 9

Answer:
Round the values to the nearest hundreds.
Estimate: 61,100 + 8,800 = 69,900
Add the ones, tens, then hundreds, then thousands and ten thousands. Regroup if necessary.

Check: Because 69,861 is close to the estimate, 69,900, the answer is reasonable.

Question 5.
Estimate: ______
Big Ideas Math Answers 4th Grade Chapter 2 Add and Subtract Multi-Digit Numbers 2.2 10

Answer:
Round the values to the nearest hundreds.
Estimate: 82,200 + 4,700 = 86,900
Add the ones, tens, then hundreds, then thousands and ten thousands. Regroup if necessary.

Check: Because 86,935 is close to the estimate, 86,900, the answer is reasonable.

Question 6.
Estimate: ______
Big Ideas Math Answers 4th Grade Chapter 2 Add and Subtract Multi-Digit Numbers 2.2 11

Answer:
Round the values to the nearest hundreds.
Estimate: 46,800 + 38,500 = 85,300
Add the ones, tens, then hundreds, then thousands and ten thousands. Regroup if necessary.

Check: Because 85,308 is close to the estimate, 85,300, the answer is reasonable.

Question 7.
Estimate: ______
Big Ideas Math Answers 4th Grade Chapter 2 Add and Subtract Multi-Digit Numbers 2.2 12

Answer:
Round the values to the nearest hundreds.
Estimate: 686,400 + 75,300 = 761,700
Add the ones, tens, then hundreds, then thousands and ten thousands. Regroup if necessary.

Check: Because 761,739 is close to the estimate, 761,700, the answer is reasonable.

Question 8.
Estimate: ______
Big Ideas Math Answers 4th Grade Chapter 2 Add and Subtract Multi-Digit Numbers 2.2 13

Answer:
Round the values to the nearest thousands.
Estimate: 594,000 + 308,000 = 902,000
Add the ones, tens, then hundreds, then thousands, ten thousands and then hundred thousands. Regroup if necessary.

Check: Because 902,240 is close to the estimate, 902,000, the answer is reasonable.

Question 9.
Estimate: ______
246,890 + 13,579 = ______

Answer:
Round the values to the nearest thousands.
Estimate: 247,000 + 14,000 = 261,000
Add the ones, tens, then hundreds, then thousands and then ten thousands. Regroup if necessary.

Check: Because 260,469 is close to the estimate, 261,000, the answer is reasonable.

Question 10.
Estimate: ______
822,450 + 8,651 = _______

Answer:
Round the values to the nearest thousands.
Estimate: 822,000 + 9,000 = 831,000
Add the ones, tens, then hundreds, then thousands and then ten thousands. Regroup if necessary.

Check: Because 831,101 is close to the estimate, 831,000, the answer is reasonable.

Question 11.
A video receives 10,678 views the first day. It receives 25,932 views the second day. How many views does the video receive in two days?
Big Ideas Math Answers 4th Grade Chapter 2 Add and Subtract Multi-Digit Numbers 2.2 14

Answer:
Number of views on first day = 10,678
Number of views on second day = 25,932.
Total number of views = number of views on first day + number of views on second day.
Total number of views = 10,678 + 25,932

Therefore, total number of views = 36,610 views.

Question 12.
YOU BE THE TEACHER
Is Newton correct? Explain.
Big Ideas Math Answers 4th Grade Chapter 2 Add and Subtract Multi-Digit Numbers 2.2 15

Answer:
No, Newton is not correct.
Explanation: Newton is doing the sum without following the place values. He is adding the units place digit of the second number to the hundreds value of the first number. Similarly, he is also adding tens, hundreds and thousands place value digits of the first number to the thousands, ten thousands and hundred thousands place value digits of the first number respectively.
The correct method is as follows:

Question 13.
DIG DEEPER!
Find the missing digits.
Big Ideas Math Answers 4th Grade Chapter 2 Add and Subtract Multi-Digit Numbers 2.2 16

Answer:
The digit missing in the first number is 5.
The digit missing in the second number is 4.
Explanation: Add the tens value digits 7 and 9. 7+9 = 16. Regroup 1 onto the hundreds place numbers.
1+ missing digit + 1 = 7. Therefore, the missing digit in first number becomes 5, as 1+5+1=7.
In the thousands place, 9+missing digit gives the value of 3 in the sum. the highest possible sum of 9 with a missing digit is 9+9=18. To get 3 in the units place of the sum, 9 has to be added with 4 because 9+4 gives a value of 13. Therefore the missing digit of second number is 4.
Think and Grow: Modeling Real Life
Example
A family is traveling in a car from Seattle to Atlanta. They travel 1,099 miles the first two days and 1,082 miles the next two days. Has the family arrived in Atlanta?
Big Ideas Math Answers 4th Grade Chapter 2 Add and Subtract Multi-Digit Numbers 2.2 17
Add the distances traveled.
Big Ideas Math Answers 4th Grade Chapter 2 Add and Subtract Multi-Digit Numbers 2.2 18
Compare the distance traveled to the distance from Seattle to Atlanta.
The family ______ arrived in Atlanta.
Show and Grow

Question 14.
One World Trade Center has 2,226 steps. A visitor enters the building and climbs 1,387 steps, takes a break, and climbs839 more steps. Did the visitor reach the top of the building?
Big Ideas Math Answers 4th Grade Chapter 2 Add and Subtract Multi-Digit Numbers 2.2 19

Answer:
Add the total number of steps:

Number of steps climbed: 2,226 and the number of steps of the world trade center, 2,226 are equal.
Therefore, the visitor has reached the top of the World Trade Center.

Question 15.
There were 51,787 more people who rode the city buses on Saturday than on Sunday. On Sunday, 174,057 people rode the buses. How many people rode the buses on Saturday?

Answer:
Given: People rode on city bus on Saturday = 51,787 + People rode on city bus on Sunday.
Therefore, People rode on city bus on Saturday = 51,787 + 174,057

Therefore, People rode on city bus on Saturday = 225,844.

Question 16.
The deepest part of the Atlantic Ocean is 8,577 feet shallower than the deepest part of the Pacific Ocean. The deepest part of the Atlantic Ocean is 27,493 feet deep. How deep is the deepest part of the Pacific Ocean?

Answer:
Given: Depth of Atlantic Ocean = Depth of Pacific Ocean – 8,577.
So, Depth of Atlantic Ocean = 27,493 – 8,577

Therefore, depth of Atlantic ocean = 18,916 feet.

Add Multi-Digit Numbers Homework & Practice 2.2

Find the sum. Check whether your answer is reasonable

Question 1.
Estimate: ______
Big Ideas Math Answers 4th Grade Chapter 2 Add and Subtract Multi-Digit Numbers 2.2 20

Answer:
Round the values to the nearest hundreds.
Estimate: 8,500 + 4,700 = 13,200
Add the ones, tens, then hundreds and then thousands. Regroup if necessary.

Check: Because 13,141 is close to the estimate, 13,200, the answer is reasonable.

Question 2.
Estimate: ______
Big Ideas Math Answers 4th Grade Chapter 2 Add and Subtract Multi-Digit Numbers 2.2 21

Answer:
Round the values to the nearest hundreds.
Estimate: 75,400 + 8,600 = 84,000
Add the ones, tens, then hundreds and then thousands. Regroup if necessary.

Check: Because 84,016 is close to the estimate, 86,000, the answer is reasonable.

Question 3.
Estimate: ______
Big Ideas Math Answers 4th Grade Chapter 2 Add and Subtract Multi-Digit Numbers 2.2 22

Answer:
Round the values to the nearest thousands.
Estimate: 90,000 + 20,000 = 110,000
Add the ones, tens, then hundreds and then thousands. Regroup if necessary.

Check: Because 110,340 is close to the estimate, 110,000, the answer is reasonable.

Question 4.
Estimate: ______
Big Ideas Math Answers 4th Grade Chapter 2 Add and Subtract Multi-Digit Numbers 2.2 23

Answer:
Round the values to the nearest thousands.
Estimate: 48,000 + 24,000 = 72,000
Add the ones, tens, then hundreds and then thousands. Regroup if necessary.

Check: Because 71,585 is close to the estimate, 72,000, the answer is reasonable.

Question 5.
Estimate: ______
Big Ideas Math Answers 4th Grade Chapter 2 Add and Subtract Multi-Digit Numbers 2.2 24

Answer:
Round the values to the nearest thousands.
Estimate: 505,000 + 65,000 = 570,000
Add the ones, tens, then hundreds and then thousands. Regroup if necessary.

Check: Because 569,821 is close to the estimate, 570,000, the answer is reasonable.

Question 6.
Estimate: ______
Big Ideas Math Answers 4th Grade Chapter 2 Add and Subtract Multi-Digit Numbers 2.2 25

Answer:
Round the values to the nearest thousands.
Estimate: 505,000 + 65,000 = 570,000
Add the ones, tens, then hundreds and then thousands. Regroup if necessary.

Check: Because 569,821 is close to the estimate, 570,000, the answer is reasonable.
Find the sum. Check whether your answer is reasonable.

Question 7.
Estimate: ______
547,795 + 9,418 = ______

Answer:
Round the values to the nearest thousands.
Estimate: 548,000 + 9,000 = 557,000
Add the ones, tens, then hundreds and then thousands. Regroup if necessary.

Check: Because 557,213 is close to the estimate, 557,000, the answer is reasonable.

Question 8.
Estimate: ______
401,269 + 58,135 = ______

Answer:
Round the values to the nearest thousands.
Estimate: 401,000 + 58,000 = 459,000
Add the ones, tens, then hundreds and then thousands. Regroup if necessary.

Check: Because 459,404 is close to the estimate, 459,000, the answer is reasonable.

Question 9.
Number Sense
Find and explain the error. What is the correct sum?
Big Ideas Math Answers 4th Grade Chapter 2 Add and Subtract Multi-Digit Numbers 2.2 26

Answer:
Round the values to the nearest thousands.
Estimate: 175,000 + 29,000 = 204,000
Add the ones, tens, then hundreds and then thousands. Regroup if necessary.

Check: Because 459,404 is close to the estimate, 459,000, the answer is reasonable.

Question 10.
Writing
Write a word problem that can be solved by finding the sum of 75,629 and 23,548.

Answer:
A purchases a house worth $23,548 more than the house purchased by B. House purchased by B is worth $75,629. Find the worth of the house purchased by A.
Worth of house purchased by A = Worth of house purchased by B + $23,548
Worth of house purchased by A = $75,629 + $23,548
Therefore, worth of house purchased by A = $ 99,177.

Question 11.
Modeling Real Life
Two chefs need to make 2,500 food samples for the opening day of their restaurant. Chef A makes 1,346 samples. Chef B makes 1,084 samples. Did the chefs make enough samples?
Big Ideas Math Answers 4th Grade Chapter 2 Add and Subtract Multi-Digit Numbers 2.2 27

Answer:
Food samples made by Chef A = 1,346.
Food samples made by Chef B = 1,084.
Total food samples made = Food samples made by Chef A + Chef B.
So, total food samples made = 1,346+1,084 = 2,430.
Total food samples required = 2,500.
The total food samples made are lesser than the required number of food samples. So, the chefs A & B didn’t make enough food samples.

Question 12.
Modeling Real Life
House A costs $24,450 less than House B. How much does House B cost?
Big Ideas Math Answers 4th Grade Chapter 2 Add and Subtract Multi-Digit Numbers 2.2 28

Answer:
Given, Cost of House A = Cost of House B – $24,450.
So, Cost of House B = Cost of House A + $24,450.
Therefore, Cost of House B = $175,500 + $ 24,450 = $ 199,950.
Review & Refresh
Find the quotient.

Question 13.
6 ÷ 2 = ______

Answer:

Therefore, the quotient is 3.

Question 14.
24 ÷ 4 = ______

Answer:

Quotient = 6.

Question 15.
32 ÷ 8 = ______

Answer:

Quotient = 4.

Question 16.
14 ÷ 7 = ______

Answer:

Quotient = 2.

Question 17.
Big Ideas Math Answers 4th Grade Chapter 2 Add and Subtract Multi-Digit Numbers 2.2 29

Answer:

Quotient = 1

Question 18.
Big Ideas Math Answers 4th Grade Chapter 2 Add and Subtract Multi-Digit Numbers 2.2 30

Answer:

Quotient = 7.

Question 19.
Big Ideas Math Answers 4th Grade Chapter 2 Add and Subtract Multi-Digit Numbers 2.2 31

Answer:

Quotient = 10

Question 20.
Big Ideas Math Answers 4th Grade Chapter 2 Add and Subtract Multi-Digit Numbers 2.2 32

Answer:

Quotient = 9.

Lesson 2.3 Subtract Multi-Digit Numbers

Explore and Grow
Which subtraction problem shows a correct way to find 94 – 8? Why?
Big Ideas Math Answers Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.3 1

Answer:
First one shows the correct way to find the subtraction because it follows the place value rule by subtracting the units digit value of second number from the units digit value of the first number.
Which subtraction problem shows a correct way to find 3,710 – 251? Why?
Big Ideas Math Answers Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.3 2

Answer:
Second one shows the correct way to find the subtraction because it follows the place value rule by subtracting the units, tens and hundreds digit values of second number from the units, tens and hundreds digit values of the first number respectively.
Reasoning
Why do you need to use place value when subtracting? Explain.

Answer:
While subtracting, we use place value rule i.e., we go from right hand side to the left hand side. In other words, we add the digits as per the place value. The digit of every number holds the place value. That means, a digit in tens place has the value of multiple of ten. Similarly, a digit in hundreds place has the value of multiple of hundred.
For example, In the number 23, 2 digit holds the value of multiple of ten. That means, 2 X 10.
To continue the parity of the sum, we follow the place value rule by subtracting digits as per the place values.
Think and Grow: Subtract Multi-Digit Numbers
Big Ideas Math Answers Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.3 3
Show and Grow
Find the difference. Then check your answer.

Question 1.
Big Ideas Math Answers Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.3 4

Answer: 2073

Check: Use addition to check your answer. 

Question 2.
Big Ideas Math Answers Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.3 5

Answer: 9,821

Check: Use addition to check your answer.

Question 3.
Big Ideas Math Answers Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.3 6

Answer: 452,107

Check: Use addition to check the solution.

Question 4.
Big Ideas Math Answers Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.3 7

Answer: 183099

Check: Use addition to check the solution.
Apply and Grow: Practice
Find the difference. Then check your answer.

Question 5.
Big Ideas Math Answers Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.3 8

Answer: 90,960

Check: Use addition to check the solution.

Question 6.
Big Ideas Math Answers Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.3 9

Answer: 33,479

Check: Use addition to check the solution.

Question 7.
Big Ideas Math Answers Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.3 10

Answer: 53,122

Check: Use addition to check the solution.

Question 8.
Big Ideas Math Answers Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.3 11

Answer: 25,528.

Check: Use addition to check the solution.

Question 9.
Big Ideas Math Answers Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.3 12

Answer: 338,681

Check: Use addition to check the solution.

Question 10.
Big Ideas Math Answers Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.3 13

Answer: 293,189

Check: Use addition to check the solution.

Question 11.
281,660 – 44,521 = ______

Answer: 237,139

Check: Use addition to check the solution.

Question 12.
798,400 – 5,603 = ______

Answer: 792,379.

Check: Use addition to check the solution.

Question 13.
103,219 people attended a championship football game last year. 71,088 people attend the game this year. How many more people attended the game last year than this year?
Big Ideas Math Answers Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.3 14

Answer: 32,131
People attended last year = 103,219
People attended this year = 71,088
People attended more last year than this year = People attended last year – People attended this year
People attended more last year than this year = 103,219 – 71,088

Therefore, People attended more last year than this year = 32,131.

Question 14.
Number Sense
Find and explain the error. What is the correct difference?
Big Ideas Math Answers Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.3 15

Answer: The error in the solution is the regrouping is not applied correctly. The correct regrouping process is as follows:

Therefore, the correct difference = 363,505

Question 15.
Number Sense
Which statements describe the difference of 32,064 and 14,950?
The difference is about 17,000.
The difference is less than 17,000.
The difference is greater than 17,000.
The difference is 17,000.

Answer: The first and third statements describe the difference.
Explanation: Find the difference of 32,064 and 14,950 by rounding them to the nearest hundreds values and then to the nearest thousands values.
               
When you round the values to the nearest hundreds values, the statement 3 is satisfied. While rounding the values to the nearest thousands values, statement 1 is satisfied.
Think and Grow: Modeling Real Life
Example
The shoreline of Lake Michigan is 1,090 miles shorter than the shoreline of Lake Superior. How long is the shoreline of Lake Michigan?
Big Ideas Math Answers Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.3 16
Subtract 1,090 from the length of the shoreline of Lake Superior.
Big Ideas Math Answers Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.3 17
The shoreline of Lake Michigan is 1640 miles long.
Show and Grow

Question 16.
The new car is $15,760 less than the new truck. How much does the new car cost?
Big Ideas Math Answers Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.3 18

Answer: $ 16,755
Explanation:
Cost of truck = $ 32,535.
Cost of car = Cost of truck – $ 15,760
Therefore Cost of car = $ 32,535 – $ 15,760

Therefore, Cost of car = $ 16,775.

Question 17.
Last year, an amusement park had 770,495 more guests than a water park. The attendance at the amusement park was 875,562 guests. What was the attendance at the water park?

Answer: 105,067
Number of guests in at Amusement park = 875,562
Number of guests at water park = number of guests at amusement park – 770,495 guests.
Number of guests at water park = 875,562 – 770,495

Therefore, the number of guests at water park = 105,067.

Question 18.
How many fewer miles did the pilot fly in Years 1 and 2 combined than in Year 3?
Big Ideas Math Answers Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.3 19

Answer: 5,406
Number of miles the pilot fly in years 1 and 2 combined = number of miles fly in year 1 + number of miles fly in year 2.
So, total number of miles fly in years 1 and 2 = 5,396 + 10,821 = 16,217.
Fewer miles pilot flew in years 1 and 2 combined than year 3 = Number of miles flew in year 3  Number of miles pilot flew in years 1 & 2 combined.

Therefore, the pilot flew a combined of 5,406 miles fewer in years 1 and 2 compared to year 3.

Subtract Multi-Digit Numbers Homework & Practice 2.3

Find the difference. Then check your answer.

Question 1.
Big Ideas Math Answers Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.3 21

Answer: 2,374
Explanation:

Question 2.
Big Ideas Math Answers Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.3 22

Answer: 25259.
Explanation:

Question 3.
Big Ideas Math Answers Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.3 23

Answer: 21,284
Explanation:

Question 4.
Big Ideas Math Answers Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.3 24

Answer: 13,162
Explanation:

Question 5.
Big Ideas Math Answers Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.3 25

Answer: 140,938
Explanation:

Question 6.
Big Ideas Math Answers Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.3 26

Answer: 292,694
Explanation:

Find the difference. Then check your answer.

Question 7.
319,120 – 278,188 = _______

Answer: 40,932
Explanation:

Check: Add 278,188 to 40,932 to verify.

Question 8.
312,396 – 23,891 = ______

Answer: 288,505
Explanation:

Check: Add 23,891 to 288,505 to verify.

Question 9.
Writing
Write and solve a subtraction word problem using the numbers 34,508 and 8,529.

Answer:
The bike racing 8,529 more curators than the Formula 1 racing. The attendance at the bike racing park was 34,508 guests. What was the attendance at the Formula 1 racing?
Attendance at Bike racing = 34,508.
Attendance at Formula 1 racing = Attendance at Bike racing – 8,529.
So, Attendance at Formula 1 racing = 34,508 – 8,529

Therefore, attendance at Formula 1 racing = 25,979.
Use the time line to answer the questions.
Big Ideas Math Answers Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.3 27

Question 10.
Modeling Real Life
How many years after the first cell phone was invented was the smartphone invented?

Answer: 29 years.
Year of first cell phone invented = 1973.
Year of first smart phone invented = 2002.
Number of years after smart phone invented = 2002 – 1973

Therefore, the years after smart phone invented from first cell phone is 29 years.

Question 11.
Modeling Real Life
How many years passed from the invention of the first telephone to the release of the iPhone?
Big Ideas Math Answers Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.3 28

Answer: 131 years.
Explanation:
Year when first telephone invented = 1876.
Year of the launch of first iPhone = 2007.
Years passed from invention of first telephone to the invention of first iPhone = Launch year of iPhone – invention year of first telephone.
So, years passed = 2007 – 1876

Therefore, years passed after the invention of first telephone to the launch of first iPhone are 131 years.
Review & Refresh
Find the area of the shape.

Question 12.
Big Ideas Math Answers Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.3 29

Answer: 15 square meter.
Explanation:
Given, one block = 1 square meter.
Total number of blocks highlighted = 15.
Therefore, total area = 15 square meters.

Question 13.
Big Ideas Math Answers Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.3 30

Answer: 11 square inch
Explanation:
Given, one block = 1 square inch.
Total number of blocks highlighted = 11.
Therefore, total area = 11 square inch.

Lesson 2.4 Use Strategies to Add and Subtract

Explore and Grow
Choose any strategy to find 8,005 + 1,350.
Big Ideas Math Solutions Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.4 1
For the given numbers, 8,005 and 1,350 the normal addition can be done using grouping using the place value rule. There is no need for the regrouping as well. However, we here will use the partial sums strategy to find the sum.
8000 + 005 = 8005
1000 + 350 = 1350
9000 + 355 = 9355
Choose any strategy to find 54,000 – 10,996.

Answer: 64996
For the given numbers, 54,000 and 10,996 we have to apply compensation strategy.
Add 4 to 10,996 to round it to the nearest thousands value.
10,996 + 4 = 11,000.
54,000 – 11,000 = 43,000.
Now subtract 4 from the result.
43,4000 – 4 = 42,996.
Therefore,54,000 – 10,996 = 42,996.
Reasoning
Explain why you chose your strategies. Compare your strategies to your partner’s strategies. How are they the same or different?

Answer:
Think and Grow: Use Strategies to Add or Subtract
Big Ideas Math Solutions Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.4 2
Show and Grow
Find the sum or difference. Then check your answer.

Question 1.
Big Ideas Math Solutions Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.4 3

Answer: 22,719.
We will use partial sums to add.

Therefore, 10,500 + 12,219 = 22,7199.

Question 2.
Big Ideas Math Solutions Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.4 4

Answer:
We will use compensation strategy to find the difference.
7,008 – 8 = 7,000.
9,318 – 7,000 = 2,318.
2,318 – 8 = 2,310.
Therefore, 9,318 – 7,008 = 2,310.
Apply and Grow: Practice
Find the sum or difference. Then check your answer.

Question 3.
Big Ideas Math Solutions Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.4 5

Answer:  16,457.
We here use regrouping strategy to solve this.

Question 4.
Big Ideas Math Solutions Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.4 6

Answer: 36,081
Subtract the given numbers using regrouping model.

Therefore, the difference of 44,561 and 8480is 36,081

Question 5.
Big Ideas Math Solutions Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.4 7

Answer: 11,063
Use compensation strategy to solve this:
Add 40 to 91,803.
91,803 + 40 = 91,843
91,843 – 80,740 = 11,103.
Now subtract 60 from the result.
11,103 – 40 = 11,063.

Question 6.
Big Ideas Math Solutions Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.4 8

Answer: 100,973
Use regrouping to add the numbers.

Question 7.
Big Ideas Math Solutions Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.4 9

Answer: 281,712
Use partial sum method to calculate this.

Question 8.
Big Ideas Math Solutions Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.4 10

Answer: 822,594
Use regrouping method to find the difference:

Question 9.
780,649 – 13,754 = ______

Answer: 766,895
Use regrouping method:

Question 10.
417,890 + 90,284 = ______

Answer:  508,172
Use regrouping method to sum.

Question 11.
614,008 + 283,192 = ______

Answer: 897,200
Use regrouping method to find the sum.

Question 12.
801,640 – 206,427 = ______

Answer: 595,213
Use regrouping:

Question 13.
Structure
Write an equation shown by the number line.
Big Ideas Math Solutions Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.4 11

Answer:
23,977 + 1000 = 24,977 + 20 = 24,997 + 3 = 25,000.
Think and Grow: Modeling Real Life
Example
Inner Core Earth’s mantle is 1,802 miles thick. Earth’s outer core is 1,367 miles thick. How many miles thinner is Earth’s outer core than its mantle?
Big Ideas Math Solutions Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.4 12
Subtract the thickness of the outer core from the thickness of the mantle. Use compensation to subtract.
Big Ideas Math Solutions Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.4 13
Earth’s outer core is 435 miles thinner than its mantle.
Show and Grow

Question 14.
Your friend’s heart beats 144,000 times in one day. Your heart beats 115,200 times in one day. How many fewer times does your heart beat than your friend’s?
Big Ideas Math Solutions Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.4 14

Answer: 28,800
Number of times heart beats of my friend = 144,000
Number of times heart beats of me = 115,200
The fewer times my heart beats than my friend = 144,000 – 115,200

Question 15.
You, your friend, and your cousin are playing a video game. You score 2,118 more points than your friend. Your friend scores 1,503 fewer points than your cousin. What is each player’s score? Who wins?PlayerScoreYou?Friend6,010Cousin?
Big Ideas Math Solutions Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.4 15

Answer:
Given, friend’s score = 6,010.
My score = My friend’s score + 2,118 = 6,010 + 2,118 = 8,128.
My cousin’s score = My friend’s score – 1,503 = 6,010 – 1,503 = 4,507.

Question 16.
Students at a school want tore cycle a total of 50,000 cans and bottles. So far, the students recycled 40,118 cans and 9,863 bottles. Did the students reach their goal? If not, how many more cans and bottles need recycled?
Big Ideas Math Solutions Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.4 16

Answer: No, the students fell short by 19.
Number of cans recycled = 40,118.
Number of bottles recycled = 9,863
Total recycled = 40,118 + 9,863 = 49,981.
Students planned number for recycling = 50,000.
Shortage = 50,000 – 49,981 = 19.
The students fell short by 19 cans or bottles to recycle.

Use Strategies to Add and Subtract Homework & Practice 2.4

Find the sum or difference. Then check your answer.

Question 1.
Big Ideas Math Solutions Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.4 17

Answer: 13,754
Use regrouping method to add the numbers.

Question 2.
Big Ideas Math Solutions Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.4 18

Answer: 3,030
Use compensation method to subtract.
Subtract 30 from the first number to make it’s hundreds tens and units digits same those of the second number..
7,683 – 30 = 7,653.
Now subtract the second number from the result.
7,653 – 4,683 = 3,000
Now add 30 which was subtracted in the beginning to the result again to compensate.
3,000 + 20 = 3,030.

Question 3.
Big Ideas Math Solutions Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.4 19

Answer: 25,801
Use compensation method to add the numbers.
We, here can see both the numbers are near to their nearest hundreds. The first number is 12 greater than its nearest hundreds value and the second is 11 lesser than its nearest hundreds value.
Subtract 11 from the first number and add 11 to the second number. Adding and subtracting compensates the net value.
18,212 – 11 = 18,201.
7,589 + 11 = 7,600.
Now add both the results.
18,201 + 7,600 = 25,801.

Question 4.
Big Ideas Math Solutions Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.4 20

Answer: 7,625
Use compensation method to do the subtraction.
Subtract 25 from the first number to make the tens and units digits same those of the second number.
9,800 – 25 = 9,775.
Now subtract the second number from the result.
9,775 – 2,175 = 7,600.
Now add 25 to the result which was subtracted initially to compensate.
7,600 + 25 = 7,625.

Question 5.
Big Ideas Math Solutions Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.4 21

Answer: 8,515
Use regrouping to subtract.

Question 6.
Big Ideas Math Solutions Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.4 22

Answer: 161,935
Use regrouping to sum.

Question 7.
814,327 – 32,475 = ______

Answer: 781,852.
Use regrouping method to subtract.

Question 8.
294,801 + 46,030 = ______

Answer:
Use regrouping to sum.

Question 9.
512,006 + 318,071 = ______

Answer: 830,085.
Use compensation method.
Subtract 6 from the first number.
512,006 – 6 = 512,000.
Add the second number to the result.
512,000 + 318,071 = 830,071.
Now add 6  again to the result to compensate.
830,071 + 6 = 830,085.

Question 10.
746,620 – 529,706 = _______

Answer: 216,914
Use grouping to subtract.

Question 11.
YOU BE THE TEACHER
Your friend uses compensation to add. Is your friend correct? Explain.
Big Ideas Math Solutions Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.4 23

Answer: No, the addition is incorrect.
Explanation:
In step 1, 17 was subtracted to round it to the nearest hundreds. To compensate the subtraction, 17 should be added to the result in the final step. But, 17 was subtracted instead.
Correct answer = 75,220 + 17 = 75,237.

Question 12.
Writing
Which strategy would you use to subtract 9,618 from 58,007? Explain.

Answer: 48,389
We here use regrouping strategy.

Question 13.
Modeling Real Life
There are about 500,000 detectable earthquakes each year. About 99,500 of the detectable earthquakes are felt. How many earthquakes are detectable, but are not felt each year?
Big Ideas Math Solutions Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.4 24

Answer:
Given, number of detectable earthquakes = 500,000.
Number of detectable earthquakes that are felt = 99,500.
Number of earthquakes that cannot be felt = 500,000 – 99,500 = 405,000.

Question 14.
DIG DEEPER!
A committee wants to purchase a playground for $52,499. They put a donation of $8,025 towards the purchase of the playground. Then they make 2 payments of $4,275. How much money does the committee have left to pay for the playground?

Answer:
Cost of playground = $52,499.
Donation towards purchase = $8,025.
Next payments made = $4,275.
Amount left to pay = Total cost pf playground – donation – payment.
So, amount left = $52,499 – $8,025 – $4,275 = ($52,499 – $8,025) – $4,275
= $44,474 – $4,275 = $40,199.
So, the money left to pay by the committee to purchase the playground is $40,199
Review & Refresh
Write an equation to solve. Use a letter to represent the unknown number. Check whether your answer is reasonable.

Question 15.
There are 4 boxes. Each box has 6 granola bars. Your soccer team eats 18 granola bars. How many granola bars are left?

Answer: 6.
Let the total number of granola bars = X.
Given X = 4 Boxes of 6 granola bars each.
So, X = 6 * 4 = 24.
Number of Granola bars ate by Soccer team = 18.
Number of granola bars left = total granola bars – granola bars ate = 24 – 18 = 6.
Therefore, number of granola bars left = 6.

Question 16.
Newton has 35 blocks. Descartes has 28 blocks. Newton divides his blocks into7 equal groups and gives Descartes one group. How many blocks does Descartes have now?

Answer: 33.
Total blocks Newton has = 35.
Dividing 35 into 7 equal groups = 35/7.
So, number of blocks in each group = 35/7 = 5.
Number of blocks with Descartes = 28.
Number of blocks with Descartes after Newton giving one group of blocks = 28 + 5 = 33.

Lesson 2.5 Problem Solving: Addition and Subtraction

Explore and Grow
Use addition or subtraction to make a conclusion about the table.
Big Ideas Math Answer Key Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.5 1

Answer:
We use addition and also subtraction.
Let’s solve this problem. How much is the combined area of Arizona and California is bigger than the combined area of Nevada and Utah.
Combined area (1) of Arizona and California = 113,594 + 155,779 = 269,373.
Combined area (2) of Nevada and Utah = 109,781 + 82,170 = 191,951.
Difference in combined areas = Combined area 1 – Combined area 2 = 269,373 – 191,951.

Therefore, the combined area of Arizona and California is 77,422 square miles bigger than the combined area of Nevada and Utah.
Think and Grow: Problem Solving: Addition and Subtraction
Example
You have3,914 songs in your music library. You download 1,326 more songs. Then you delete 587 songs. How many songs do you have now?
Understand the Problem
What do you know?
• You have 3,914 songs.
• You download 1,326 more
• You delete 587 songs.
What do you need to find?
• You need to find how many songs you have now.
Make a Plan
How will you solve?
• Add 3,914 and 1,326 to find how many songs you have after downloading some songs.
• Then subtract 587 from the sum to find how many songs you have now.
Solve
Big Ideas Math Answer Key Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.5 2
k = 4,240
n = k – 587 = 5,240 – 587 = 3,653
You have 4,653 songs now.
Show and Grow

Question 1.
Explain how you can check whether your answer above is reasonable.

Answer:
You can check the answer by doing the subtraction first and the addition later.
First, subtract 587 songs that were deleted.
Therefore, 3,914 – 587 = 3,327.
Now add the downloaded songs, 1,326 to the result.
So, 3,327 + 1,326 = 4,653.
Apply and Grow: Practice
Understand the problem. What do you know? What do you need to find? Explain.

Question 2.
There are about 12,762 known ant species. There are about 10,997 known grasshopper species. The total number of known ant, grasshopper, and spider species is 67,437. How many known spider species are there?
Big Ideas Math Answer Key Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.5 3

Answer: 43,678.
Given, number of ant species: 12,762.
Number of grasshopper species: 10,997.
Total number of ant, grasshopper and spider species: 67,437.
So, Number of (ant species + grasshopper species + spider species) = 67,437.
That means, 12,762 + 10,997 + number of spider species = 67,437.
Now, initially add the number of ant and grasshopper species using compensation strategy.
Add 3 to the number of grasshopper species: 10,997 + 3 = 11,000.
Now add the result to the number of ant species: 12,762 + 11,000 = 23,762.
Now subtract 3 to compensate: 23,762 – 3 = 23,759.
So, Number of spider species = 67,437 – 23,759.
Do the subtraction using regrouping.

Therefore, the known number of spider species = 43,678.

Question 3.
A quarterback threw for 66,111 yards between 2001 and 2016. His all-time high was 5,476 yards in 1 year. In his second highest year, he threw for 5,208 yards. How many passing yards did he throw in the remaining years?

Answer: 55,427
Total number of yards = 66,111 yards.
His first and second all-time high were 5,476 yards and 5,208 yards respectively.
Total yards in first and second year all-time high = 5,476 + 5,208 = 10,684.
The number of yards he threw in the remaining years = 66,111 – 10684
Find the difference using regrouping.

Therefore, number of yards he threw in the remaining years = 55,427 yards
Understand the problem. Then make a plan. How will you solve? Explain.

Question 4.
There are 86,400 seconds in 1 day. On most days, a student spends 28,800 seconds sleeping and 28,500 seconds in school. How many seconds are students not awake, but in school?

Answer:
Given, Number of seconds student sleeps = 28,800.
Number of seconds student spends at school = 28,500.
The
Question asked was the number of seconds the student spend not awake (sleeping), but in school.
In the given data, it’s not mentioned that the student spends time sleeping at school separately. So, with the given information, we cannot calculate the number of seconds student was not awake, but at school.

Question 5.
A pair of rhinoceroses weigh 14,860 pounds together. The female weighs 7,206 pounds. How much more does the male weigh than the female?
Big Ideas Math Answer Key Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.5 4

Answer: 258 pounds.
Given,
Weight of female rhino =7,206 pounds.
Weight of the rhino pair = 14,860.
Weight of male rhino = 14,860 – 7,206
Solve this using compensation strategy.
Add 6 to the first number = 14,860 + 6 = 14,866.
Now subtract the second number from the result.
That means, 14,866 – 7,206 = 7,460.
Now, subtract 6 from the result to compensate = 7,460 – 6 = 7,454.
So, weight of male rhino = 7,464.
Weight of male rhino more than female rhino = male rhino weight – female rhino weight. = 7,464 – 7,206 = 258 pounds.

Question 6.
Earth is 24,873 miles around. If a person’s blood vessels were laid out in a line, they would be able to circle Earth two times, plus 10,254 more miles. How many miles long are a person’s blood vessels when laid out in a line?

Answer: 60,000 miles.
Given circumference of earth = 24,873 miles.
So, length of person’s blood vessels = 2 times the earth’s circumference + 10,254 miles.
That means, length of person’s blood vessels = (2 X 24,873) + 10,254 = 49,746 + 10,254.
Add the two numbers using grouping technique.

Therefore,, the length of human blood vessels = 60,000 miles.

Question 7.
Alaska has 22,041 more miles of shoreline than Florida and California combined. Alaska has 33,904 miles of shoreline. Florida has 8,436 miles of shoreline. How many miles of shoreline does California have?

Answer: 3,427 miles
Give, Shoreline of Alaska = 33,904.
Shoreline of Florida = 8,436 miles.
Shore lines of Florida + California = Shoreline of Alaska – 22,041 miles.
That means, 8,436 miles + Shoreline of California = 33,904 miles – 22,041 miles.
Subtract the right hand side using regrouping.

So, Shoreline of California + 8436 miles = 11,863 miles.
That means, Shoreline of California = 11,863 miles – 8436 miles

Therefore, shoreline of California = 3,427 miles.
Think and Grow: Modeling Real Life
Example
The attendance on the second day of a music festival is 10,013 fewer people than on the third day.How many total people attend the three-day music festival?
Big Ideas Math Answer Key Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.5 5
Think: What do you know? What do you need to find? How will you solve?3?
Big Ideas Math Answer Key Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.5 6
_______ total people attend the music festival.
Show and Grow

Question 8.
A construction company uses 3,239 more bricks to construct Building 1 than Building 2. How many bricks does the company use to construct all three buildings?
Big Ideas Math Answer Key Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.5 7

Answer:
Given,
Number of bricks used for Building 1 = 11,415.
Number of bricks used for Building 3 = 16,352.
Number of bricks used for building 2 = Number of bricks used for Building 1 – 3,239.
So, Number of bricks used for Building 2 = 11,415 – 3,239.
Use regrouping to subtract.

So, number of bricks used for constructing Building 2 = 8,176.
So, bricks used to construct all three buildings = Bricks used for (Building 1 + Building 2 + Building 3)
So, total bricks used in construction = 11,415 + 8,176 + 16,352.
Use grouping to add the three numbers.

Therefore, the number of bricks used to construct all the three buildings = 35,943.

Problem Solving: Addition and Subtraction Homework & Practice 2.5

Understand the problem. Then make a plan. How will you solve? Explain.

Question 1.
A cargo plane weighs 400,000 pounds. After a load of cargo is removed, the plane weighs 336,985 pounds. Then a 12,395-pound load is removed. How many pounds of cargo are removed in all?

Answer: 75,410
Given,
Weight of plane with cargo = 400,000 pounds.
Weight of plane after removing first load of cargo = 336,985.
So, Weight of first cargo load = 400,000 – 336,985.
Use Count on to strategy to find the difference.

So, the difference is 60,000 + 3,000 + 15 = 63,015.
So, the weight of the first cargo = 63,015.
Given, weight of the second cargo = 12,395.
Total weight of the cargo = 63,015 + 12,395

So, the total weight of the cargo unloaded = 75,410 pounds.

Question 2.
A ski resort uses 5,200 gallons of water per minute to make snow. A family uses 361 gallons of water each day. How many more gallons of water does the ski resort use to make snow in 2 minutes than a family uses in 1 day?

Answer: 10,039 Gallons
Given,
Water required to make snow for 1 minute = 5,200 gallons.
Water required for household usage per day = 361 gallons.
Water required for ski resort to prepare snow per two minutes = 2 X 5,200 = 10,400 gallons.
The more water required than the water consumed by the household one day usage = 10,400 – 361.
Subtract using compensation.
Add 39 to the second number.
361 + 39 = 400.
Now subtract the result from the second number.
10,400 – 400 = 10,000.
Now, add 39 to the result to compensate.
10,000 + 39 = 10,039.
So, the additional water used for ski resort to make snow for 2 minutes than the water used by household per one day is 10,039 gallons.

Question 3.
In July, a website receives 379,162 fewer orders than in May and June combined. The website receives 542,369 orders in May and 453,708 orders in June. How many orders does the website receive in July?

Answer: 542,369
Given,
Orders received in May = 542,369.
Orders received in June = 453,708.
Total orders in May and June = 542,369 + 453,708.

So, orders received in July = 996,077 – 379,162.
Subtract using regrouping.

So, the number of orders received in the month of July = 542,369.

Question 4.
Writing
Write and solve a two-step word problem that can be solved using addition or subtraction.

Answer:
A father wants to buy a laptop to his son worth $3,495. He makes 2 payments of $1,500. How much money does the he have left to pay for the laptop?
Cost of laptop = $3,495.
Number of payments of $1,500 done = 2.
That means $1,500 X 2 = 3,000
Money left to pay by the father for the laptop = Cost of laptop – Amount paid
= $3,495 – $3,000.
$3,495 – $3,000 = $495.
So, money left to buy laptop = $495

Question 5.
Modeling Real Life
World War I lasted from 1914 to1918. World War II lasted from 1939 to 1945. How much longer did World War II last than World War I?

Answer: 2 years
Duration of World war 1 = 1918 – 1914 = 4 years.
Duration of World war 2 = 1945 – 1939 = 6 years.
The longer did World war 2 last = Duration of World war 2 – Duration of World war 1 = 6 years – 4 years
Therefore, World war 2 lasted 2 years longer than World war 1.

Question 6.
Modeling Real Life
Twenty people each donate $9 to a charity. Sixty people each donate $8. The charity organizer wants to raise a total of $1,500. How much more money does the organizer need to raise?

Answer:
Twenty people each donate $9 to a charity = 20 X $9 = $180
Sixty people each donate $8 = 60 X $8 = $480.
Total fund collected = $180 + $480 = $660
The charity organizer wants to raise a total of $1,500.
So, money left by the charity organizer to collect = $1500 – $660 = $740.

Question 7.
DIG DEEPER!
The blackpoll warbler migrates 2,376 miles, stops, and then flies another 3,289 miles to reach its destination. The arctic tern migrates 11,013 miles, stops, and then flies another 10,997 miles to reach its destination. How much farther is the arctic tern’s migration than the blackpoll warbler’s migration?
Big Ideas Math Answer Key Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.5 8

Answer: 16,345 miles
Total migration of Blackpoll Warbler = 2,376 + 3,289
Total migration of arctic tern = 11,013 + 10,997
Total migration of Blackpoll:      Total migration of arctic tern: 
Farther the arctic tern’s migration than the blackpoll warbler’s migration = 22,010 – 5,665.
Use grouping to subtract.

So, farther the arctic tern’s migration than the blackpoll warbler’s migration is 16,345 miles.
Review & Refresh
Write the time. Write another way to say the time.

Question 8.
Big Ideas Math Answer Key Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.5 9

Answer: 5 hours 56 minutes (05:56)
Other way: 4 minutes to 6’O Clock.

Question 9.
Big Ideas Math Answer Key Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.5 10

Answer: 3 hours 43 minutes (03:43)
Other way: 17 minutes to 4’O Clock.

Question 10.
Big Ideas Math Answer Key Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.5 11

Answer: 12 hours 37 minutes (12:37)

Add and Subtract Multi-Digit Numbers Performance Task

Question 1.
The time line shows the population of Austin, Texas from 1995 to 2015.
Big Ideas Math Answers 4th Grade Chapter 2 Add and Subtract Multi-Digit Numbers 1
Big Ideas Math Answers 4th Grade Chapter 2 Add and Subtract Multi-Digit Numbers 3
a. Which place would you round to when estimating population? Explain.
b. Estimate Austin’s population each year on the time line.
c. Use your estimates to complete the bar graph.
d. Between the years 1995 and 2015, did Austin’s population increase, decrease, or stay the same? Explain.
e. During which five-year period did the population increase the most? Explain.
f. About how many more people lived in Austin in the year 2015 than in the year 1995?
g. Do you think the population will be more than 1,000,000 in the year 2020? Explain.

Answer:
a) Round the values to the nearest ten thousands because the values are in lakhs.
b)
c)

d) Between the years, 1995 and 2015, Austin’s population was increased. The population was constantly increasing as per the given table and the chart.
e) During the five year period between 1995 and 2000, the population increased more. The graph was more steeper than the other time lines. That indicates more population growth during that period.
f) Population in 2015 = 930,052
Population in 1995 = 555,092.
Round the values to the nearest ten thousands values.

So, the population increase between 1995 and 2015 was estimated to be 370,000.
g) Yes, the population grown more than 100,000 in 5 years time except between 2000 and 2005. So, if that is the case, the population will be more than 1,000,000 by 2020.

Add and Subtract Multi-Digit Numbers Activity

Race to the Moon
Directions:
1. Players take turns.
2. On your turn, flip a Race for the Moon Card and find the sum or difference.
3. Move your piece to the next number on the board that is highlighted in your answer.
Big Ideas Math Answers 4th Grade Chapter 2 Add and Subtract Multi-Digit Numbers 4

Add and Subtract Multi-Digit Numbers Chapter Practice

2.1 Estimate Sum and Differences
Estimate the sum or difference.

Question 1.
Big Ideas Math Answers Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers chp 1

Answer: 6,600
Round the numbers to the nearest hundreds values.

Question 2.
Big Ideas Math Answers Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers chp 2

Answer: 72,000
Round the numbers to their nearest thousands

Question 3.
Big Ideas Math Answers Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers chp 3

Answer: 34,000.
Round the numbers to the nearest thousand values.

Question 4.
742,086 – 486,629 = ______

Answer: 250,000
Round the numbers to the nearest ten thousands value.

Question 5.
216,987 + 72,429 = ______

Answer: 289,000

2.2 Add Multi-Digit Numbers
Find the sum. Check whether your answer is reasonable.

Question 6.
Estimate: ______
Big Ideas Math Answers Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers chp 6

Answer: 53,300
Round the numbers to their nearest hundreds value:

Check: Use grouping to find the exact Sum:

As, 53,276 is nearest to the estimate value 53,300, the estimate is correct.

Question 7.
Estimate: ______
Big Ideas Math Answers Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers chp 7

Answer: 80,000
Round the numbers to the nearest thousands value:

Check: Use grouping to find the exact Sum:

As, 79,960 is nearest to the estimate value 80,000, the estimate is correct.

Question 8.
Estimate: ______
Big Ideas Math Answers Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers chp 8

Answer: 610,000
Round the numbers to the nearest ten thousands values.

Check: Use grouping to find the exact Sum:

As, 609,334 is nearest to the estimate value 610,000, the estimate is correct.

Question 9.
Logic
Find the missing digits
Big Ideas Math Answers Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers chp 9

Answer:

In this given problem, look at the units place. The result is 1 and as the units digit of the first number is 6, then the should be nothing else but 11. So, 11 – 6 = 5. So, the missing units digit of the second number is 5. The tens position also gets satisfied with the result being 11 as the regrouping gives the tens digit of the result 11 again.
If that’s the case, the regrouping again takes another addition of 1 to the hundreds place. It gives the result as follows:
1 + __ + 9 = Units digit of 5.
1+9 = 10. So, the 1 has to go for the thousands place as per regrouping. So, 0 + __ = 5.
So, the hundreds place of the first number is 5.

In this given problem, look at the tens place. The result is 2 and as the tens digit of the second number is 8, There is also a carrying of 1 from the units place to add in the sum.
So, 1 + __ + 8 = 12 (before regrouping of tens place). So, the missing units digit of the first number is 3. The hundreds position also gets satisfied with the result being 12 as the regrouping gives the hundreds digit of the result 15.
The regrouping again takes another addition of 1 to the ten thousands place. It gives the result as follows:
1 + 6 + __ = Units digit of 5. It clearly shows that there will be another regrouping.
7 + __ = 15.
7 + 8 = 15.
So, the ten thousands place of the second number is 8.
2.3 Subtract Multi-Digit Numbers
Find the difference. Then check your answer.

Question 10.
Big Ideas Math Answers Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers chp 10

Answer: 4757.
Use grouping to subtract.

Question 11.
Big Ideas Math Answers Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers chp 11

Answer: 2,289.
Use compensation method.
Add 11 to the first number.
32,700 + 11 = 32,711
Now subtract the second number from result.
So, 32,711 – 30,411 = 2,300.
Now subtract 11 from the result to compensate.
2,300 – 11 = 2,289.

Question 12.
Big Ideas Math Answers Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers chp 12

Answer: 75,575
Use grouping to subtract.

Question 13.
973,287 – 8,345 = ______

Answer: 964,942
Use grouping to subtract.

Question 14.
762,179 – 21,280 = ______

Answer: 740,899
Use grouping to subtract.

2.4 Use Strategies to Add or Subtract
Find the sum or difference. Then check your answer.

Question 15.
Big Ideas Math Answers Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers chp 15

Answer: 7,205
Use compensation method.
Subtract 5 from the first number.
13,500 – 5 = 13,495.
Now subtract second number from the result.
13,495 – 6,295 = 7,200.
Now add 5 to the result to compensate.
7,200 + 5 = 7,205.

Question 16.
Big Ideas Math Answers Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers chp 16

Answer: 825,528.
Use regrouping.

Check: Add the result to the second number to check.

Question 17.
Big Ideas Math Answers Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers chp 17

Answer: 911,527.
Use regrouping to solve this.

2.5 Problem Solving: Addition and Subtraction

Question 18.
Modeling Real Life
A library has 12,850 books. There are 1,932 poetry books, 5,047 nonfiction books, and 2,891 graphic novels. The rest are fiction. How many fiction books does the library have?

Answer:
Number of books in the library = 12,850.
Given, Number of (poetry books + notification books + graphic novels + fiction books) = Total books.
So, 1,932 + 5,047 + 2,891 + fiction books = 12,850.

So, 9,870 + Fiction books = 12,850.
So, Fiction books = 12,850 – 9,870.
Use compensation to do this subtraction.
Add 20 to the first number.
12,850 + 20 = 12,870.
12,870 – 9,870 = 3,000.
Now subtract 20 from the result to compensate.
3,000 – 20 = 2,980.
Therefore, the number of fiction books in the library are 2,980.

Question 19.
Modeling Real Life
A technology teacher wants to buy a 3-D printer for $3,495. He makes 3 payments of $999. How much money does the teacher have left to pay for the 3-D printer?

Answer: $498.
Cost of 3-D printer = $3,495.
Number of payments of $999 done = 3.
That means $999 + $999 + $999.
Use compensation technique to add the three numbers.
Add $1 to each payment of $999.
$999 + $1 = $1,000.
So, the addition has to be done for $1,000 + $1,000 + $1,000 = $3,000.
Now, subtract $3 from the result to compensate.
$3,000 – $3 = $2,997.
Money left to pay by the technology teacher for the 3-D printer = Cost of 3-D printer – Amount paid
= $3,495 – $2,997.
Use compensation method to do the subtraction. Add $3 to the second number.
$2,997 + $3 = $3,000.
Now subtract the result from the first number.
$3,495 – $3,000 = $495.
Add $3 to compensate.
$495 + $3 = $498.
So, the money left to pay by the technology teacher for the 3-D printer = $498.

Conclusion:

Come and fall in love with Maths by utilizing the Big Ideas Math Answers Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers. Make use of the Big Ideas Math Answers Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers as a reference for all your queries. Keep in touch with our site to avail updates on Class Specific Big Ideas Math Book 4th Grade Answer Key Chapter 2 Add and Subtract Multi-Digit Numbers at your fingertips.

Big Ideas Math Answers Grade 2 Chapter 14 Money and Time

Big Ideas Math Answers Grade 2 Chapter 14

Students can find Big Ideas Math Book Grade 2 Chapter 14 Money and Time Answers in the following sections. You can definitely score good marks on the test with the help of the BIM 2nd Grade Book 14th Chapter Money and Time Answer Key. The students can download and practice questions from Big Ideas Math Answers Grade 2 14th Chapter Money and Time PDF for free of cost. It will help you to complete the homework within time.

Big Ideas Math Book 2nd Grade Answer Key Chapter 14 Money and Time

Check out the BIM Book Grade 2 Chapter 14 Money and Time Solution Key and prepare well for the exam. You will see the step by step detailed explanation for each and every question from Big Ideas Math Textbook Grade 2 Chapter 14 Money and Time.

The different topics covered in this chapter are Find Total Values of Coins, Order to Find Total Values of Coins, Show Money Amounts in Different Ways, Make One Dollar, Make Change from One Dollar, Find Total Values of Bills, Problem Solving: Money, Tell Time to the Nearest Five Minutes, Tell Time Before and After the Hour, and Relate A.M. and P.M.

Vocabulary

Lesson: 1 Find Total Values of Coins

Lesson: 2 Order to Find Total Values of Coins

Lesson: 3 Show Money Amounts in Different Ways

Lesson: 4 Make One Dollar

Lesson: 5 Make Change from One Dollar

Lesson: 6 Find Total Values of Bills

Lesson: 7 Problem Solving: Money

Lesson: 8 Tell Time to the Nearest Five Minutes

Lesson: 9 Tell Time Before and After the Hour

Lesson: 10 Relate A.M. and P.M.

Chapter: 10 – Money and Time

Money and Time Vocabulary

Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 1

Organize It
Use the review words to complete the graphic organizer.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 2

 

Answer:
1. Analog clock .
2. The long hand pointing minutes is the minutes hand.
3. The short hand pointing hours is the hour hand.

Define it
Use your vocabulary cards to complete the puzzle.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 3

Answer:

1. Penny
2.Dime
3. Quarter
4. Nickel

Chapter 14 Vocabulary Cards

Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 4
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 5
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 6
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 7
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 8
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 9

Lesson 14.1 Find Total Values of Coins

Explore and Grow

Sort your coins.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 10
Explain how you sorted.
__________________
__________________
__________________
__________________

 

Answer:

Show and Grow

Count on to find the total value.

Question 1.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 11

 

Answer:
36 cents

Explanation :
1 dime = 10 cents
That means, 3x 10 = 30 cents
1 nickel = 5 cents
That means,  1x 5= 5 cents
and 1 cent
So , 3 dime +1 nickel+1 cent = 30 cents+ 5 cents +1 cent = 36 cents.

Question 2.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 12

 

Answer:
42 cents

Explanation:
1 quarter = 25 cents
That means , 1 x25= 25 cents
1 nickel = 5 cents
That means , 3  x 5  = 15 cents & 2 cents additional
Total : 25 cents + 15 cents + 2 cents = 42 cents
Therefore , total value is 42 cents .

Apply and Grow: Practice

Count on to find the total value.

Question 3.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 13

 

Answer:
55 cents

Explanation :
1 quarter = 25 cents
Which means , 2 x 25 = 50
1 nickel = 5 cents
Total : 50 + 5 =55 cents
Total value of coins is 55 cents.

Question 4.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 14

 

Answer:
48 cents

Explanation:
1 quarter = 25 cents
1 dime = 10 cents

Which means,  2 x 10 = 20 cents
and 3 cents
Total : 25 +20 +3 = 48 cents
Therefore, Total value is 48 cents.

Question 5.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 15

 

Answer:
66 cents

Explanation:
1 quarter = 25 cents
1 dime = 10 cents
Which means , 3 x 10 = 30 cents
1 nickel = 5 cents
Which means , 2 x 5 = 10 cents
& 1 cent
Total : 25 +30+ 10 +1=66
Therefore , the total value of coins is 66 cents.

Question 6.
Reasoning
You have 27¢. Which groups of coins could you have?
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 16

 

Answer:

This set of coins have total of 27 cents

Explanation:
1 dime = 10 cents
Which means, 2 x 10 = 20 cents
1nickel = 5 cents
and 2 cents
Total : 20 + 5 + 2 = 27 cents.

Think and Grow: Modeling Real Life

You have 2 quarters, 1 dime, 4 nickels, and 1 penny. How many cents do you have? Do you have enough money to buy the airplane?
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 17
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 18

Answer:
I have 81 cents.
Yes, I have enough money to buy the airplane.

Explanation:
1 quarter = 25 cents
So 2 quarters = 25+25 =50 cents
1 dime =10 cents
1 nickel = 5 cents
So 4 nickels = 4 x 5 = 20 cents
1 penny =1 cent
Total amount is 50+10+20+1 =81 cents
The airplane costs 80 cents so, I have enough money to buy the plain.

Show and Grow

Question 7.
You have 5 dimes, 3 nickels, and 2 pennies. How many cents do you have? Do you have enough money to buy the coloring book?
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 19
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 19.1

Answer:
I have 67 cents.
No, I do not have enough money to buy the coloring book.

Explanation:
1 dime =10 cents
so, 5 dimes = 5 x 10= 50 cents
1 nickel = 5 cents
So, 3 nickels = 3 x 5 = 15 cents
2 penny =2 cents
Total amount is 50+15+2 =67 cents
The coloring book costs 70 cents
So, I do not  have enough money to buy the book.

Question 8.
You have 4 dimes, 1 nickel, and 3 pennies. How many more cents do you need to buy the whistle? Draw and label the coins you need.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 20

Answer:
I have 48 cents.
No , I  do not have enough money to buy the whistle.

Explanation:
1 dime =10 cents
so, 4 dimes =4 x 10= 40 cents
1 nickel = 5 cents
3 pennies =3 cents
Total amount is 40+5+3 =48 cents
The whistle costs 59 cents so, I  do not have enough money to buy the whistle.

Question 9.
DIG DEEPER!
You have 3 quarters, 2 nickels, and 3 pennies. Your friend has 1 quarter and 5 dimes. Who has more money? How much more?
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 21

Answer:
I have 13 cents money more than my friend .

Explanation:
My money :
1 quarter = 25 cents
So 3 quarters = 3 x 25 = 75 cents
1 nickel = 5 cents
So 2 nickel = 2 x 5 = 10 cents
3 pennies = 3 cents
Total amount: 75+10+3 = 88 cents
So, I have a total of 88 cents .
Friends money:
1 quarter= 25 cents
1 dime = 10 cents
So 5 dimes = 5 x 10 = 50 cents
Total amount = 25 +50= 75 cents
Friends money total is 75 cents
Now, the difference between my money and my friends money is
88 cents – 75 cents = 13 cents
That means, I have 13 cents more than my friend.

Find Total Values of Coins Homework & Practice 14.1

Count on to find the total value.

Question 1.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 22

Answer:
65 cents

Explanation:
1 quarter = 25 cents
1 dime = 10 cents
Which means , 4 x 10 = 40 cents
Total : 25 + 40 = 65 cents
Therefore, the total value of coins is 65 cents.

Question 2.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 23

Answer:
38 cents

Explanation:
1 dime = 10 cents
Which means , 3 dime = 3 x 10 = 30 cents
1 nickel = 5 cents
and 3 cents
Total : 30+ 5 + 3 = 38 cents

Question 3.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 24

Answer:
72 cents

Explanation:
1 quarter = 25 cents
1 dime = 10 cents
Which means, 4 x 10 = 40 cents
1 nickel = 5 cents
and 2 cents
Total : 25 + 40+ 5 + 2 =72 cents

Question 4.
DIG DEEPER!
You had 52¢. You lost a coin. Now you have the 5 coins shown. What coin did you lose?
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 24.1

 

Answer:
1 quarter = 25 cents
1 dime = 10 cents
1 nickel = 5 cents
and 2 cents
Total :  25+10+5+2 = 42 cents
Total amount i have is 52 cents
Now , 52 – 42 = 10
1 dime = 10 cents
So , I lost 1 dime.

Question 5.
Precision
Circle coins to show 80¢.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 24.2

 

Answer:
1 dime = 10 cents
80 cents = 8 dimes

Question 6.
Modeling Real Life
You have 3 quarters, 1 nickel, and 4 pennies. How many cents do you have? Do you have enough money to buy the boomerang?
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 25

 

Answer:
No, I do not have enough money to buy the boomerang.

Explanation:
The cost of boomerang is 94 cents
1 quarter = 25 cents
3 x 25 =75 cents
1 nickel =5 cents
4 pennies = 4 cents
Total amount : 75+5+4 =84
I have 84 cents .
No, I do not have enough money to buy the boomerang

Question 7.
Modeling Real Life
You have 1 quarter, 3 dimes, and 1 nickel. How many more cents do you need to buy the toy bird? Draw the coins you need.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 26

Answer:
The cost of bird is 75 cents.

Explanation:
1 quarter = 25 cents.
1 dime = 10 cents,
3 dimes = 3 x 10 = 30 cents
1 nickel = 5 cents
Total amount : 25 +30+ 5 =60 cents
Now,
75- 60 = 15cents
I need 15 more cents to buy the toy bird .

Review & Refresh

Compare.

Question 8.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 27

Answer:
324 > 317
327 is greater 317

Question 9.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 28

Answer:
426 > 206
426 is greater than 206

Question 10.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 29

Answer:
546 < 564
564 is greater than 546

Question 11.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 30

Answer:
931 > 842
931 is greater than 846

Lesson 14.2 Order to Find Total Values of Coins

Explore and Grow

Order your coins from the greatest value to the least value. Draw and label each coin with its value. What is the total value of all the coins?
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 31
Explain how ordering the coins helped you find the total.
_____________________________________
_____________________________________
_____________________________________
_____________________________________

Answer:
1 quarter = 25 cents
1 dime = 10 cents
1 nickel = 5 cents
and 1 penny = 1 cent
Total value of coins :
25+10+5+1=41 cents

>>>

Show and Grow

Draw and label the coins from the greatest value to the least value. Then find the total value.

Question 1.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 32

 

Answer:
45 cents

Explanation:
1 quarter = 25 cents
1 dime = 10 cents
1 nickel = 5 cents
Which means , 2 x 5 = 10 cents
Total value :
25 +10 + 10 = 45 cents

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Quarter       >         Dime     >   Nickel

Question 2.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 33

 

Answer:
1 quarter = 25 cents
which means, 2quarters = 2 x 25= 50 cents
1 dime = 10 cents
1 nickel = 5 cents
and 1 penny = 1 cents
Total : 50 +10+5+1 = 66 cents
Therefore total value of coins is 66 cents .

>>>

Question 3.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 34

Answer: 31 cents

Explanation:
1 dime = 10 cents
which means , 2 dimes = 2 x 10 = 20 cents
1 nickel = 5 cents
which means , 2 nickel = 2 x5 = 10 cents
and 1 penny = 1 cent
Total : 20 +10 +1 = 31
Therefore, total value of coins is 31 cents

>>

Apply and Grow: Practice

Draw and label the coins from the greatest value to the least value. Then find the total value.

Question 4.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 35

 

Answer:
41 cents
Explanation :
1 quarter = 25 cents
1 dime = 10 cents
1 nickel = 5 cents
and 1 penny = 1 cent
Total value = 25 +10 +5 +1 =41
Therefore total value of coins is 41 cents

>>>

Question 5.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 36

 

Answer:
27 cents

Explanation :
1 dime = 10 cents
which means , 2 dimes = 2 x 10 = 20 cents
1 nickel = 5 cents
and 2 pennies = 2 cents
total : 20 +5 +2= 27
Therefore total value of coins is 27 cents .

>>

Question 6.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 37

Answer:
85 cents

Explanation :
1 quarter = 25 cents
Which means, 2 quarter = 2 x25 = 50 cents
1 dime = 10 cents
Which means, 3 dimes = 3 x 10 = 30 cents
And 1 nickel = 5 cents
Total :
50 +30 +5 = 85 cents
Therefore total value of coins is 85 cents.

>>

Question 7.
Reasoning
You have a dime, a nickel, and one other coin. The total value is 40¢. What is your third coin?
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 38

Answer:
The total value of  3 coins is 40 cents
1 dime = 10 cents
1 nickel = 5 cents
10 cents + 5 cents = 15 cents
Now , 40 cents – 15 cents =  25 cents
So the other coin is quarter
1 quarter = 25 cents
Therefore, 25 +10+ 5 = 40 cents.

Think and Grow: Modeling Real Life

You have 3 nickels and 4 pennies in one pocket. You have 2 dimes and 2 quarters in your other pocket. How much money do you have in all? Do you have enough money to buy the car?
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 39
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 40

Answer:
The cost of car is 9 cents .
The amount i have in one pocket is 3 nickels and 4 pennies
1 nickel = 5 cents
which means , 3 nickels = 3 x 5 = 15 cents
4 pennies = 4 cents
Total = 15 cents + 4 cents = 19 cents
The amount i have in another pocket is 2 dimes , 2 quarters
1 dime = 10 cents
in which , 2 dimes = 2 x 10 = 20 cents
1 quarter = 25 cents
2 quarters = 2 x 25 = 50 cents
Total : 20 cents + 50 cents = 70 cents
Total amount in both the pockets is 70 cents + 19 cents =89 cents
No, I do not have enough money to  buy the car.

Show and Grow

Question 8.
You have 30¢. You find 2 nickels, 1 dime, and 3 pennies in your room. How much money do you have now? Do you have enough money to buy the yo-yo?
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 41
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 42

Answer:
I have 30 cents
Amount I found in my room =
1 nickel = 5 cents ,
2 nickel x 5 = 10 cents.
1 dime =10 cents.
3 pennies = 3 cents.
Total  : 10 +10 + 3
=23 cents
Now, total amount i have is
30 cents +23 cents = 53 cents
The cost of yo-yo is 50 cent
So , now I have enough money to buy the yo- yo

Question 9.
You have 1 nickel, 1 quarter, and 4 dimes. How many more cents do you need to buy the stuffed animal? Draw and label the coin you need.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 43

Answer:
1 nickel = 5 cents
1 quarter= 25 cents
1 dime =10 cents
Which means 4 dimes = 4 x 10=40 cents
Total amount = 5+25+40 =70 cents
But the cost of stuffed animal is 80 cents
So , I do not have enough cents to buy the stuffed animal.
I need 10 more cents to buy the stuffed animal.

Question 10.
DIG DEEPER!
You have 65¢. You give your friend a dime. You have 3 coins left. Draw and label the coins you have left.

Answer:
Given that,
I have 65 cents
The amount I gave to my friend is 1 dime
1 dime = 10 cents
Now,
65 cents – 10 cents = 55 cents
55 cents can be shown in 3 coins as
2 quarters and 1 nickel

Order to Find Total Values of Coins Homework & Practice 14.2

Draw and label the coins from the greatest value to the least value. Then find the total value.

Question 1.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 44

 

Answer: 35 cents

Explanation :
1 dime = 10 cents
which means, 2 dimes = 2 x 10 = 20 cents
and 1 nickel = 5 cents
which means, 3 nickels = 3 x 5 = 15 cents
Total:  20 +15 = 35 cents
Therefore total value of coins is 35 cents.

>

Question 2.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 45

Answer: 63 cents

Explanation:
1 quarter = 25 cents
Which means , 2 quarters = 2 x 25 = 50 cents
1 dime = 10 cents
and 3 pennies = 3 cents
Total : 50 +10+3 = 63 cents

>>

Question 3.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 46

Answer: 46 cents

Explanation :
1 quarter = 25 cents
1 dime = 10 cents
1 nickel = 5 cents
Which means , 2 nickel = 2 x 5 = 10 cents
and 1 penny = 1 cents
Total :
25 +10 +5 +5+1 = 46 cents

>>>

Question 4.
Open-Ended
Draw and label four coins that have a total value of 40¢

Answer: 40 cents can be written as 4 dimes
1 dime = 10 cents
Which means , 4 x 10 = 40 cents.

Question 5.
Modeling Real Life
You have 46¢. You find 4 pennies and 1 nickel in your room. How much money do you have now? Do you have enough money to buy the app?
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 47
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 48

Answer:
I have 46 Cents

Explanation:
The amount I found is
4 pennies =  4 cents
1 nickel = 5 cents
Total  : 5 + 4 = 9 cents
Now,
Total amount I have is 46 cents +9 cents = 55 cents
The cost of app is 50 cents
yes , I have enough money to buy the app.

Question 7.
DIG DEEPER!
You have some nickels and dimes. You have 1 more nickel than dimes. The total value of your coins is 50¢. How many nickels and dimes do you have?
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 49

Answer:
I have  4 nickels and  3 dimes.

Explanation:
1 nickel =5 cents
That means, 4 nickels= 4 x 5 = 20 cents
1 dime = 10 cents
That means , 3 dimes = 3 x 10 = 30
Total: 20 + 30 = 50 cents
So, I have  4 nickels and  3 dimes.

Review & Refresh

Question 8.
Which fruit is the least favorite?
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 50

Answer:
Cherry is the least favorite.

Explanation :
In the given table ,
Total lines for orange fruit are 5
Total lines for cherry fruit are 2
Total lines for apple fruit are  4
On comparing them , 2 is smaller than 4 and 5
so, Cherry is the least favorite.

Lesson 14.3 Show Money Amounts in Different Ways

Explore and Grow

Use your coins to show 25 cents in two different ways. Draw and label the coins.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 51

Answer :
25 cents can be written as
1 quarter
1 quarter = 25 cents

Another way :
25 cents can be written as
2 dime , 1 nickel
1 dime = 10 cents
Which means , 2 dime = 2 x 10= 20
1 nickel = 5 cents.

Did everyone in your class use the same coins?
_____________________________________
_____________________________________
_____________________________________
_____________________________________

Answer:
No, everyone used different coins to make a total value of 25 cents
For example, my friend used 1 dime, 3 nickels
1 dime = 10 cents
1 nickel = 5 cents
Which means , 3 x 5 = 15 cents
Total : 10 + 15 = 25 cents

Show and Grow

Show the amount in two different ways.

Question 1.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 52

 

Answer:
45 cents can be written as
4 dime + 1 nickel
Which means  one dime = 10 cents
4 dime = 4 x 10 = 40
1 nickel = 5 cents
Total: 40+5 = 45
Another way: 1 quarter +2 dime
Which means , 1 quarter =25 cents
1 dime= 10 cents
2 dime = 2 x 10 = 20
Total=: 25+20= 45
Therefore, 45 can be written as  4 dime 1 nickel and 1 quarter 2 dime

Question 2.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 53

Answer:
27 cents can be written as
2 dime +1 nickel+2 penny
Which means , 1 dime = 10 cents
2 dime = 2 x 10= 20 cents
1 nickel = 5 cents2 penny = 2 cents
Total: 20+5+ 2= 27 cents
Another way:
1 quarter+2 pennies
Which means , 1 quarter = 25 cents
2 penny = 2 cents
Total= 25 + 2 = 27 cents
Therefore 27 cents can be written as 2 dime +1 nickel+2 penny and 1 quarter +2 pennies

Question 3.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 54

Answer:
80 cents can be written as
8 dimes
Which means 1 dime = 10 cents
So 8 dimes = 80 cents
Another way:
3 quarter + 1 nickel
Which means , 1 quarter = 25 cents
3 quarters = 3 x 25 = 75 cents
1 nickel = 5 cents
Total : 75 + 5 = 80 cents
Therefore, 80 cents can e written as 8 dimes and 3quarter and 1 nickel .

Apply and Grow: Practice

Show the amount in two different ways.

Question 4.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 55

Answer:
23 cents can be written as
2 dime  3 pennies
Which means , 1 dime = 10 cents
2 dimes =2 x 10 = 20 cents
3 pennies = 3 cents
Total : 20 + 3 = 23 cents
Another way:
4 nickel ,3 pennies
Which means, 1 nickel = 5 cents
4 nickel = 4 x 5 =20 cents
3 pennies = 3 cents
Total= 20 + 3 =23 cents
Therefore 23 cents can be written as 2 dimes + 3 pennies and 4 nickel + 3 pennies

Question 5.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 56

Answer:
49 cents can be written as
1 quarter ,2 dime ,4 pennies
Which means , 1 quarter = 25 cents
1 dime = 10 cents
So,2 dime = 2 x 10 = 20 cents
4 pennies = 4 cents
Total : 25 +20+4 =49 cents
Another way :
49 Pennies
1 penny = 1 cent
49 pennies = 49 cents
Therefore, 49cents can be written as 1 Quarter 2 dime 4 pennies and 49 pennies.

Question 6.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 57

 

Answer:
75 cents can be written as
3 quarters
1 quarter = 25 cents
Which means , 3 quarters = 3 x 25 = 75
Another way:
1 quarter, 5 dimes
1 quarter = 25 cents
1 dime = 10 cents
Which means , 5 dimes = 5 x 10 = 50 cents
Total :  25 + 50 = 75 cents
Therefore , 75 cents can be written as 3 quarters and 1 quarter , 5 dimes

Question 7.
Structure
You have 55¢. You have no quarters. Draw to show what coins you might have.

Answer:
55 cents be written as 5 dimes , 1 nickel .
1 dime = 10 cents
Which means , 5 dimes = 5 x 10 = 50 cents
1 nickel = 5 cents
Total : 50 cents + 5 cents = 55 cents.

Question 8.
YOU BE THE TEACHER
Newton says he drew the fewest number of coins to show 66¢. Is he correct? Explain.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 58

Answer :
No, he is wrong
Newton used 6 coins to show 66 cents .
66 cents can also be shown as
2 quarters , 1 dime , 1 nickel and 1 cents
Total number of coins is 5
1 quarter = 25 cents
Which means , 2 quarters = 2 x 25 = 50 cents
1 dime = 10 cents
1 nickel = 5 cents
and cent
Total value of coins is 50+ 10 + 5 +1 = 66
Therefore ,66 cents can be shown in 5 coins .
so , Newton is wrong.

Think and Grow: Modeling Real Life

Newton has 2 dimes, 1 nickel, and 1 penny. Descartes uses the fewest number of coins to make the same amount. Draw and label their coins.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 59

 

Answer:

Total number of coins newton have is 4
2 dimes , 1 nickel and 1 penny
1 dime = 10 cents
Which means, 2 dimes = 2 x 10 = 20 cents
1 nickel = 5 cents
and 1 penny = 1 cent

Total value of coins is
20 +5+1 = 26 cents.
Given that ,
Descartes uses fewest number of coins to make same amount
26 can also be shown with 2 coins
1 quarter = 25 cents
1 penny = 1 cents
Total value is
25 + 1 = 26 cents

So, 26 cents can be shown with 2 coins
2 is less than 6 so , Descartes uses few coins than Newton to show the same amount.

Show and Grow

Question 9.
Newton has 3 dimes and 2 pennies. Descartes uses the fewest number of coins to make the same amount. Draw and label their coins.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 60

Answer:
Total number of coins newton have is 5
3 dimes , 2 pennies
1 dime = 10 cents
Which means , 3 dimes = 3 x 10 = 30 cents
2 pennies = 2 cents.
Total value of coins is
30 + 2 = 32 cents

Given that,
Descartes used the fewest number of coins to make the same amount
32 cents can be shown in 4 coins
1 quarter, 1 dime and 2 pennies
1 quarter = 25 cents
1 dime = 10 cents
and 2 pennies = 2 cents
Total : 25 + 10 + 2 = 37 cents

 

So, 37 cents can be shown in 4 coins
4 is less than 5.
Descartes used few coins than newton to show the same amount

Question 10.
You use fewer than 5 coins to buy the pack of gum. Draw and label coins to show how you pay.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 60.1

Answer:
The cost of gum is 70 cents
70 cents can be shown in 4 coins as
2 quarters , 2 dime
1 quarter = 25 cents
Which means , 2 x 25 = 50 cents
1 dime = 10 cents
Which means , 2 x 10 = 20 cents
Total amount = 50 + 20 = 70

Question 11.
DIG DEEPER!
You have 2 quarters. Newton and Descartes each have 5 coins and the same amount of money as you. Their coins are different. Draw and label their coins.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 61

Answer:
1 quarter = 25 cents
Which means, 2 quarters= 2 x 25 = 50 cents
50 cents can be shown in 5 coins as 5 dime
1 dime = 10 cents
Which means 50 cents = 5 dimes

Show Money Amounts in Different Ways Homework & Practice 14.3

Show money amounts in different ways.

Question 1.
Big Ideas Math Answers 2nd Grade Chapter 14 Money and Time 62

Answer:
15 cents can be written as:
1 dime +1 Nickel
which means 1 dime = 1o cents
1 nickel =5 cents
10+5 = 15 cents
Another way:
15 cents can also be written as:
1 dime +5 pennies
1 dime =10 cents
5 pennies = 5 cents
Total: 10 +5 = 15 cents
So, 15 cents can be written as 1 dime 1 nickel and 1 dime 5 pennies

Question 2.
Big Ideas Math Answers 2nd Grade Chapter 14 Money and Time 63

 

Answer:
58 cents can be written as
2 quarter 8 pennies
1 quarter = 25 cents
Which means 2 quarters = 2 x 25 = 50 cents
8 pennies = 8 cents
Total : 50 + 8 = 58 cents
Another way :
5 dime 1 nickel 3 pennies
Which means,
1 dime = 10 cents
=5 dime = 5 x 10 = 50cents
1 nickel = 5 cents
1 pennies = 3 cents
Total : 50+ 5+3= 58
So, 58 cents can be written as 2 quarter 8 pennies and 5 dime 1 nickel 3 pennies.

Question 3.
Big Ideas Math Answers 2nd Grade Chapter 14 Money and Time 64

Answer:
90 cents can be written as
3 quarters 1 nickel 1 dime
Which means,
1quarter =25 cents
3 quarters = 3 x 25 = 75 cents
1 nickel =5 cents
1 dime = 10 cents
Total : 75+5+10=  90 cents
Another way :
9 dimes
1dime = 10 cents
9 dimes = 9 x 10 = 90 cents .
So , 90 cents can be written as 3 quarters1 dime 1 nickel and 9 dimes.

Question 4.
Reasoning
Draw to show 60¢ with only 3 coins.

Answer:
60  cents can be drawn in 3 coins as
2 quarters and 1 dime
1 quarter = 25 cents
Which means , 2 x 25 = 50 cents
1 dime = 10 cents
Total : 50 + 10 = 60.

Question 5.
Structure
Draw to show 42¢ without using dimes.

Answer:
42 cents can be drawn as
1 quarter , 3 nickels and 2 pennies
1 quarter = 25 cents
1 nickel = 5 cents
Which means 3 x 5 = 15 cents
2 pennies = 2 cents
Total : 25 +15 + 2 = 42 cents

Question 6.
Modeling Real Life
Newton has 6 dimes and 1 nickel. Descartes uses the fewest number of coins to make the same amount. Draw and label their coins.
Big Ideas Math Answers 2nd Grade Chapter 14 Money and Time 65

Answer :
6 dimes and 1 nickel
Total number of coins is 7
1 dime = 10 cents
Which means, 6 dimes = 6 x 10 = 60 cents
1 nickel = 5 cents
Total : 60 +5 = 65 cents
65 cents can be shown in fewest number of coins as
2 quarters , 1 dime and 1 nickel
Total number of coins is 4

Question 7.
Modeling Real Life
You use fewer than 5 coins to buy the pen. Draw and label coins to show how you pay.
Big Ideas Math Answers 2nd Grade Chapter 14 Money and Time 66

Answer:
The cost of pen is 52 cents
52 cents be shown in fewer than 5 coins as
2 quarter and 2 pennies
1 quarter = 25 cents
Which means, 2 quarters = 2 x 25 = 50 cents
and 2 pennies = 2 cents
Total 50 +2 = 52 cents.

Review & Refresh

Question 8.
A green scarf is 50 inches long. An orange scarf is 40 inches long. A red scarf is 38 inches long. How much longer is the green scarf than the red scarf?
Big Ideas Math Answers 2nd Grade Chapter 14 Money and Time 67

Answer:
22 inches

Explanation:
The length of green scarf is 50 inches
The length of red scarf is 38 inches
Now ,
50 inches – 38 inches = 22 inches
Therefore , Green scarf is 22 inches longer than red scarf.

Lesson 14.4 Make One Dollar

Explore and Grow

Newton has 4 coins. The total value is 100¢. Draw and label his coins

Answer:
Given ,
The total value of money is 100 cents
100 cents can be shown in 4 coins as 4 quarters
1 quarter = 25 cents
Which means 4 quarters = 4 x 25 = 100 cents
So, Newton has 4 quarters

Big Ideas Math Answers 2nd Grade Chapter 14 Money and Time 68
Descartes has 10 coins. The total value is 100¢. Draw and label his coins.

Answer:
Given ,
Total value of money is 100 cents
100 cents be shown in 10 coins as 10 dimes
1 dime = 10 cents
Which means , 10 dimes = 10 x 10 = 100 cents
So, Descartes have 10 dimes .

Show and Grow

Draw coins to make $1. How many cents do you need?

Question 1.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 69

Answer:
$1 = 100 cents
So, 100 cents – 25 cents = 75 cents
I need 75 cents to make $1.
75 Cents can e shown as 3 quarters
1 quarter = 25 cents
3 quarters = 3 x 25 = 7 cents

Question 2.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 70

Answer:
$1 = 100 cents
So, 100 cents – 80 cents = 20 cents
I need 20 cents to make $1.
20 cents can be shown as 2 dimes
1 dime = 10 cents
2 dimes = 2 x 10 = 20 cents

Apply and Grow: Practice

Draw coins to make $1. How many cents do you need?

Question 3.
35¢
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 70.1

Answer:  $1 = 100 cents
So, 100 cents – 35 cents = 65 cents
I need 65 cents to make $1.
65 cents can be shown as
2 quarters , 1 dime and 1 nickel
1 quarter = 25 cents
Which means , 2 quarters =  2 x 25 = 50 cents
1 dime = 10 cents
and 1 nickel = 5 cents
Total : 50 +10 +5 = 65 cents

Question 4.
72¢
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 70.2

Answer:
$1 = 100 cents
So, 100 cents – 72 cents = 28 cents
I need 28 cents to make $1.
28 cents can be shown as
1 quarter , 3 pennies
1 quarter = 25 cents and
3 pennies = 3 cents
Total : 25 +3 = 28 cents

Question 5.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 71

Answer:
$1 = 100 cents
1 dime = 10 cents
so 2 dimes = 2 x 10 = 20 cents
so, 100 cents – 20 cents = 80 cents
I need 20 cents to make $1.
20 cents can be shown as 2 dimes
1 dime = 10 cents
Which means , 2 dimes = 2 x 10 = 20 cents

Question 6.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 72

Answer:  $1 = 100 cents
1 quarter = 25 cents
Which means , 3 x 25 = 75 cents
1 dime = 10 cents
Total : 75 + 10 = 85 cents
Now, 100 – 85 = 15 cents
So, I need 15 cents to make $1.
15 cents can be shown as
1 dime and 1 nickel
1 dime = 10 cents
1 nickel = 5 cents
Total : 10 +5 = 15 cents

Question 7.
Number Sense
Circle coins to make $1.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 73

 

Answer:

Explanation:
$1 is equals to 100  cents
1 quarter = 25 cents
Which means
2 quarters = 2x 25 = 50 cents
1 dime = 10 cents
5 dimes = 5 x 10 = 50 cents
Total : 50 +50 = 100 cents
So, 2 quarters and 5 dimes make $1.

Think and Grow: Modeling Real Life

You have 1 quarter, 3 pennies, and 1 dime in one pocket. You have 2 pennies, 2 nickels, and 4 dimes in your other pocket. How many more cents do you need to make $1?
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 74

Answer:
The amount i have in one pocket is
1 quarter =25 cents
3 pennies = 3 cents
and 1 dime = 10 cents
Total : 25 +10 +3 = 38 cents
The amount I have in another pocket :
2 pennies = 2 cents
1 nickel = 5 cents
which means, 2 nickels = 2 x5 = 10 cents
1 dime = 10 cents
Which means, 4 dimes = 4 x 10 = 40 cents .
Total : 2 + 10 +40 =52 cents
Now , the total amount i have in both pockets is
38+52 = 90 cents
We know that $1 is equal to 100 cents
100 – 90 = 10
10 cents can be shown as 1 dime

.

Show and Grow

Question 8.
You have 2 dimes and 1 nickel in your desk. You have a quarter and 10 pennies in your backpack. How many more cents do you need to make $1?
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 74.1

Answer:
Total amount I have in my desk is
2 dimes and 1 nickel
1 dime = 10 cents
Which means, 2 dime =2 x 10 = 20 cents
1 nickel = 5 cents
Total = 20 +5 = 25 cents
Total amount I have in my pocket is
a quarter and 10 pennies
1 quarter = 25 cents
and 10 pennies = 10 cents
Total : 25 +10 = 35 cents
Total amount I have is
The amount in my desk + the amount in my pocket
Which means, 25 cents +35 cents
=60 cents
We know that ,
$1 is equal to 100 cents
100 cents – 60 cents
= 40 cents.
So, I need 40 cents more to make $1.

Question 9.
A notebook costs $1. You have 5 dimes and 4 pennies. How much more money do you need to buy the notebook?
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 75

Answer:
1 dime = 10 cents
Which means 5 dimes = 5 x 10 = 50 cents
and 4 pennies = 4 cents
Total : 50 cents +4 cents = 54 cents
we know that,
$1 is equal to 100 cents
Now , 100 – 54 = 46 cents
So ,I need 46 cents more to buy the book.

Question 10.
DIG DEEPER!
You have a $1 bill. You have 33 more cents than your friend. How much money does your friend have?
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 76

Answer:
I have $1 bill
Given that I have 33 cents more than my friend
We know that ,$1 is equal to 100 cents
Now,
100 cents – 33 cents = 67 cents
So , my friend have 67 cents.

Make One Dollar Homework & Practice 14.4

Draw coins to make $1. How many cents do you need?

Question 1.
54¢
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 77

Answer:  $1 = 100 cents
so, 100 cents – 54 cents = 46 cents
I need 46 cents to make $1.
46 cents can be shown as
1 quarter , 2 dime and 1 penny
1 quarter = 25 cents
1 dime = 10 cents
Which means, 2 dimes = 2 x 10 = 20 cents
and 1 penny = 1 cent

Question 2.
38¢
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 78

Answer:
$1 = 100 cents
So, 100 cents – 38 cents = 62 cents
I need 62 cents to make $1.
62 cents can be shown as
2 quarters, 1 dime and 2 pennies.
1 quarter = 25 cents
Which means , 2 quarters = 2 x 25 = 50
1 dime= 10 cents
2 pennies = 2 cents

Question 3.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 79

Answer:
$1 =100 cents
Given coins are 1 quarter and 2 pennies
1 quarter = 25 cents
and 2 pennies = 2cents
Total value : 25 +2 = 27
So, 100 – 27 =73
Therefore , I need 73 cents to make $1
73 cents can be shown as
2 quarters , 2 dime and 3 pennies
1 quarter = 25 cents
Which means 2 quarters = 2 x 25 = 50 cents
1 dime = 10 cents
Which means , 2 dimes = 2 x 10 = 20 cents
and 3 pennies = 3 cents
Total : 50 +20 +3 = 73.

Question 4.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 80

Answer:
$1 = 100 cents
Given coins are 1 dime and 1 nickel
1 dime = 10 cents
1 nickel = 5 cents
Total : 10 +5 = 15 cents
Now , 100 cents – 15 cents =85 cents
So , I need 85 cents to make $1.
85 cents can be written as
3 quarters , 1 dime
1 quarter = 25 cents
Which means , 3 quarters = 3 x 25 = 75 cents
1 dime = 10 cents
Total : 75 +10 = 85 cents

Question 5.
Structure
Show $1 using only nickels and dimes.

Answer:
1 dime = 10 cents  and
1 nickel = 5 cents
$1 is equals to 100 cents
100 cents can be shown as 8 dimes 4 nickels
8dimes = 8 x 10 = 80
4 nickels = 4 x 5 = 20
Total value = 100 cents

 

Question 6.
Structure
How many nickels make $1?
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 81

Answer :
20 nickels
1 nickel = 5 cents
Which means ,20 nickels =
20 x 5 = 100
100 cents =$1
So 20 nickels make $1.

Question 7.
Modeling Real Life
You have 1 dime and 4 nickels in a jar. You have 1 quarter and 3 pennies in your pocket. How many more cents do you need to make $1?
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 81.1

Answer:
The amount I have in jar is
1 dime = 10 cents
and 1 nickel = 5 cents
Which means , 4 nickels = 4 x 5 = 20 cents
Total : 10 + 20 = 30 cents
The amount i found in my pocket is
1 quarter = 25 cents
and 3 pennies = 3 cents
Total : 25+ 3 = 28 cents
Now , the total amount i have is
Amount in jar +amount in my pocket
= 30 +28 = 58cents
We know that, $1 is equal to 100 cents
100 – 58 = 42 cents
So, I need 42 cents more to make $1.

Question 8.
Modeling Real Life
A snack costs $1. You have 2 quarters and 2 dimes. How much more money do you need to buy the snack?

Answer:
The cost of snack is $1.
1 quarter = 25 cents
2 quarters = 2x 25= 50 cents
1 dimes = 10 cents
2 dimes= 2 x 10 = 20 cents
Total : 50 +20= 70 cents
$1 = 100 cents
100 – 70 = 30 cents
So , I need 30 cents more to buy the snack.

Review & Refresh

Question 9.
100 – 54 = ____

Answer : 46

By regrouping, 0 in units place becomes 10 and the 0 in tens place becomes 9 after the second regrouping.

Question 10.
200 – 134 = ____

Answer: 64

By regrouping, 0 in units place becomes 10 and the 0 in tens place becomes 9 after the second regrouping.

Lesson 14.5 Make Change from One Dollar

Explore and Grow

Model the story.
Newton buys a bag of fish crackers for 45¢. He pays with a $1 bill. What is his change?
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 82
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 83

Explain how you solved.
_______________________________
_______________________________
_______________________________
_______________________________

 

Answer: 55 cents

Explanation :
$1 is equals to 100 cents
The cost of fire crackers is 45 cents
Now his change is
$1 bill – cost of fire crackers
100 cents -45 cents
=55 cents
So, Newtons change is 55 cents.

Show and Grow

You buy the item shown. You pay with a $1 bill. What is your change?

Question 1.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 84

 

Answer:
65 Cents

Explanation:
$1 is equal to 100 cents
So, 100 cents – 35cents  = 65
Therefore, 65 cents is the change.

Question 2.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 85

Answer: 51 Cents

Explanation:
$1 is equals to 100 cents
So, 100 cents – 49 cents = 51 cents
Therefore: 51 cents is the change.

Apply and Grow: Practice

You buy the item shown. You pay with a $1 bill. What is your change?

Question 3.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 86

Answer: 85 Cents

Explanation:
$1 = 100 cents
So, 100 cents -15 cents =85 cents
Therefore 85 cents is the change.

Question 4.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 87

Answer: 39 Cents

Explanation:
$1 = 100 cents
So, 100 cents -61 cents =39 cents
Therefore 39 cents is the change.

Question 5.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 88

 

Answer: 13 Cents

Explanation:
$1 = 100 cents
So, 100 cents -87 cents =13 cents
Therefore ,13 cents is the change.

Question 6.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 89

 

Answer: 41 Cents

Explanation:
$1 = 100 cents
So, 100 cents -59 cents =41 cents
Therefore ,41 cents is the change.

Question 7.
Reasoning
Newton buys a notebook for 34¢. Descartes buys one for 52¢. You buy one for 48¢. You each pay with $1. Who gets back the most amount of money? How do you know?
__________________
__________________
__________________

Answer:
Newton gets back the most amount of money.

Explanation:
$1 is equals to 100 cents
Given that ,
The cost of newtons book is 34 cents
So, 100 cents – 32 cents = 68 cents .
The cost of Descartes book is 52 cents.
So , 100 cents – 52 cents = 48 cents.
The cost of my book is 48cents .
So, 100 cents – 48 cents = 52 cents
Therefore, newton gets back the most amount of money.

Think and Grow: Modeling Real Life

You pay for some school supplies with $1. Your change is 17¢. How much money did you spend?
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 90
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 91

Answer: 83 Cents

Explanation:
$1 is equals to 100 cents
My change is 17 cents
Now , 100 cents – 17 cents =83 cents
So, the cost of school supplies is 83 cents.

Show and Grow

Question 8.
You pay for some erasers with $1. Your change is 38¢. How much money did you spend?
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 92

 

Answer: 62 Cents

Explanation:
$1 is equals to 100 cents
My change is 38 cents
Now , 100 cents – 38 cents =62 cents
So the cost of erasers is 62 cents.

Question 9.
You buy a toy ring. You pay with $1. You get back 1 quarter, 2 dimes, 1 nickel, and 1 penny. How much does the toy ring cost?
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 93

Answer: 59 Cents.

Explanation:
$1 is equals to 100 cents
My change is 1 quarter = 25 cents
1dime = 10 cents
Which means , 2 dimes = 2 x 10 = 20 cents
1 nickel = 5 cents
1 penny = 1 cent
Total = 25 +10 + 5 +1 =41 cents
100 cents – 41 cents = 59 cents
Therefore , the cost of toy ring is 59 cents.

Question 10.
You buy a banana for 25¢ and an orange for 45¢. You pay with $1. What is your change?
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 94

Answer:  Change is 30 Cents.

Explanation:
The cost of banana is 25 cents and the cost of orange is 45 cents
Total cost : 25 cents + 45 cents = 70 cents
$1 is equals to 100 cents
Now, 100 – 70 = 30 cents
Therefore ,30 cents is the change.

Make Change from One Dollar Homework & Practice 14.5

You buy the item shown. You pay with a $1 bill. What is your change?

Question 1.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 95

Answer: 55 Cents.

Explanation:
$1 = 100 cents
So, 100 cents -45 cents =55 cents
Therefore 55 cents is the change.

Question 2.
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 96

Answer: 24 Cents.

Explanation:
$1 = 100 cents
So, 100 cents -76 cents =24 cents
Therefore, 24 cents is the change.

Question 3.
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 97

Answer: 87 Cents.

Explanation:
$1 = 100 cents
So, 100 cents -13 cents =87 cents
Therefore, 87 cents is the change.

Question 4.
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 98

Answer: 63 Cents.

Explanation:
$1 = 100 cents
So, 100 cents -37 cents = 63 cents
Therefore, 63 cents is the change.

Question 5.
Reasoning
A puzzle costs 68¢. Newton pays for it with a $1 bill. Draw to show his change in two ways.
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 99

Answer: Change is 32 Cents.

Explanation:
$1 is equals to 100 cents
The cost of puzzle is 68 cents
Now, 100 – 68 = 32 cents
Therefore, 32 is the change.
32 cents can be written as 3 dime, 2 pennies

1 dime = 10 cents
3 dimes = 3 x 10 = 30 cents
2 pennies = 2 cents
Total : 30 + 2 = 32 cents
Another way ,
1 quarter 7 pennies

Which means, 1 quarter = 25 cents
7 pennies = 7 cents
Total: 25 + 7 = 32 cents
Therefore 32 cents can be written as 3 dime 2 pennies and 1 quarter 7 pennies.

Question 6.
Modeling Real Life
You buy a pencil sharpener. You give the cashier $1. You get back 2 quarters, 1 nickel, and 3 pennies. How much does the pencil sharpener cost?
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 100

Answer: 42 Cents.

Explanation:
$1 is equals to 100 cents
My change is 2 quarters 1 nickel 3 pennies
1 quarter = 25 cents
Which means , 2 quarters = 2 x 25 =50 cents
1 nickel = 5 cents
3 pennies = 3 pennies
Total: 50 +5+3=58 cents
Now,
100 – 58 = 42 cents
Therefore, the cost of the pencil sharpener is 42 cents .

Question 7.
Modeling Real Life
You buy an onion for 51¢ and a pepper for 22¢. You pay with a $1 bill. What is your change?

Answer: 27 Cents.

Explanation:
The cost of an onion is 51 cents and the cost of 22 cents
Total cost : 51 + 22 =73 cents
$1 is equals to 100 cents
100 cents – 73cents = 27 cents
Therefore, 27 cents is the change.

Review & Refresh

Question 8.
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 101

Answer: There are three straight sides and three vertices.

Question 9.
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 102

Answer: There are six straight sides and six vertices.

Lesson 14.6 Find Total Values of Bills

Explore and Grow

Model the story.
Descartes has three $5 bills and three $1 bills. How much money does he have in all?
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 103

Answer: $18.

Explanation:
Given that ,
Descartes have Three $5 bills and three $1 bill
Three $5 bills = $5+$5+$5 = $15.
and three $1 bills = $3
Total amount =$15+$3= $18.

Show and Grow

Count on to find the total value.

Question 1.
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 104

Answer: $45.

Explanation:
Three $5 bills =$5+$5+$5 = $15
$10 and $20
Total value :
$15 +$10 +$20= $45
Therefore, total value = $45.

Question 2.
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 105

Answer: $32.

Explanation:
Two $1 bills = $2
Two $5 bills = $5+$5= $10
and two $10 = $10 +$10= $20
Total value = $2+$10+$20=$32
Therefore ,the total value is $32.

Apply and Grow: Practice

Count on to find the total value.

Question 3.
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 106

Answer: $16.

Explanation:
$10 ,$5 and $1
Total value :
$10 +$5 +$1 =$16

Question 4.
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 107

Answer: $36.

Explanation:
Total value of $20 , $10 ,$5 and $1
=$20 +$10 +$5 +$1 =$36 .

Question 5.
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 108

Answer: $75.

Explanation:
Total value of a $5 bill, $10 bill and three$20 bills is
$5+$10+$20+$20+$20= $75
Therefore the total value is $75.

Question 6.
Which One Doesn’t Belong?
Which group of bills does not belong with the other two?
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 109

Answer:

Explanation:
The total value of dollar bills is $61
But the remaining dollar sets total value is $65
So ,this set is different from the other two sets.

Think and Grow: Modeling Real Life

You buy some T-shirts for $39. Draw and label bills to show two different ways to pay for the T-shirts. One way should use the fewest number of bills.
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 110

 

Answer:
$39 can be shown as
A $20 bill ,a $10 bill ,a $5 bill and four $1 bills

Another way:
$39 can be shown as
a $20 bill , three $5 bills and four $1 bills.

Show and Grow

Question 7.
You buy a pair of sneakers for $24. Draw and label bills to show two different ways to pay for the sneakers. One way should use the fewest number of bills.
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 111

Answer:
$24 can be shown as

a $20 bill and four $1 bill

    

$24 can also be shown as
A$20 bill and two $2 bills

Question 8.
Newton has three $20 bills, one $10 bill, one $5 bill, and three $1 bills. Does he have enough money to buy a new dog house that costs $80? Explain.

Answer: No , he do not have enough money to buy a dog house

Explanation:
Three $20 bills = $20+$20+$20= $60
$10 , $5 and three $1
Total value :
$60 +$10 +$5+$1 = $76
But, the cost of dog house is $80
So, he do not have enough money to buy a dog house.

Question 9.
DIG DEEPER!
Explain why you would order a group of bills from the greatest value to the least value to find the total value.
_____________________________________
_____________________________________

Find Total Values of Bills Homework & Practice 14.6

Count on to find the total value.

Question 1.
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 112

Answer:
$10 +$ 1= $11
Therefore the total value is $11

Question 2.
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 113

Answer: $50

Explanation:
Two $10 bills = $10 +$10 = $20
Two $5 bills = $5+$5 = $10
Total value = $20+$20+$10 = $50

Question 3.
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 114

Answer: $91

Explanation:
Four $20 bills = $20 +$20 +$20 +$20 = $80
$10 and $ 1
Total value = $80 +$10 +$1 = $91.

Question 4.
YOU BE THE TEACHER
Newton says he drew the fewest number of bills to show $35. Is he correct? Explain.
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 115

Answer:
No , he is wrong.

Explanation:
$35 can be shown in few bills as
a $20 bill, a $10  and a $5 bill

Question 5.
Modeling Real Life
A pair of headphones costs $88. Draw and label bills to show two different ways to pay for the headphones. One way should use the fewest number of bills.
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 116

Answer:
The cost of head phones is $88 .
$ 88 can be shown as
four $20 bills, a $5 bill ,a $2 bill and a$1 bill

   

Question 6.
Modeling Real Life
Descartes has two $20 bills, three $10 bills, and four $1 bills. Does he have enough money to buy a scratching post that costs $62? Explain.
__________________________

__________________________

Answer:
Yes ,he have enough money to buy a scratching post

Explanation:
Descartes have
Two $20 bills
Which means , 2 x 20 = 40
Three $10 bills
Which means, 3 x 10 = 30
Four $ 1bills
Total : $40 +$30 +$4 =
$74
The cost of scratching post is $62
So ,Descartes have enough money to but the post.

Review & Refresh

Question 7.
A photo album has 3 rows of photos. There are 4 photos in each row. How many photos are there in all?
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 117

Answer: 12 Photos.

Explanation:
Given that,
A photo album has 3 rows of photos
and there are 4 photos in each row
Now, the total number of phone are
4 +4+4 = 12
So , there are 12 photos in all.

Lesson 14.7 Problem Solving: Money

Explore and Grow

Model the story.
You buy a book for 60¢. Your friend buys a book for 33¢. How much do you and your friend spend in all?
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 118
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 119
Explain how you solved.
_______________________________
_______________________________
_______________________________

_______________________________

Answer: 93 Cents.

Explanation:
The cost of my book is 60 cents and
The cost of my friends book is 33 cents
Total value of our money is
60 cents + 33 cents
= 93 cents.
Therefore, me and my friend spent 93 cents in all.

Show and Grow

Question 1.
Descartes has two $10 bills and two $5 bills. He has $21 more than Newton. How much money does Newton have?
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 120

 

Answer: $9

Explanation:
Total amount of Descartes is
Two $10 bills
Which means, 2 x $10 bills = $20
Two $5 bills
Which means , 2 x $5 = $10
Total value = $20 +$10 = $30
Descartes have $30
Given that, Descartes have $21 more than newton
now , Newtons money = $30 – $21 = $9
Therefore ,Newton have $9

Question 2.
Descartes has some coins in a jar. He puts in 4 dimes, 1 nickel, and 1 penny. Now he has $1. How many cents were in the jar to start?
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 121

Answer: 54 Cents.

Explanation:
1 dime = 10 cents
Which means , 4 dimes = 4 x 10 = 40
1 nickel = 5 cents
and 1 penny = 1 cents
total value :
40 +5+1= 46 cents
We know that,
$1 is equals to 100 cents
now , 100 cents -46 cents = 54 cents
Therefore, there are 54 cents in the start.

Apply and Grow: Practice

Question 3.
Newton has some money. He loses a $10 bill and three $1 bills. Now he has $19. How much money did he have to start?

Answer: $32.

Explanation:
The total money newton lost is
$10 bill
and three $ 1 bill which means, $3
Total : $10 +$3 = $13
Given that ,
He have $19 after losing some money
Now, the total money newton have on the beginning is
lost money +leftover money
$13 +$19 = $32
Therefore , Newton have $32 in the starting.

Question 4.
Descartes has one $20 bill, three $10 bills, and three $5 bills. He spends $50. How much money does he have left?

Answer: $15

Explanation:
Total money Descartes have is
$20 bill
Three $10 bills = 3 x 10 = 30  and
Three $5 bills = 3 x 5 = 15
Total : $20 +$30 +$15= $65
Given that,
He spent $50
Therefore , Leftover money =
$65 – $50 = $15
So, Descartes is left with $15 .

Question 5.
A joke book costs $1. You have 2 quarters and 1nickel. How much more money do you need to buy the joke book?

Answer: 45 Cents.

Explanation:
$1 is equal to 100 cents
1 quarter = 25 cents
Which means, 2 quarters = 2 x 25 = 50 cents
1 nickel = 5 cents = 5 cents
Total : 50 cents +5 cents = 55 cents
now ,
100 cents – 55 cents = 45 cents
So , I need 45 cents more to buy the book.

Question 6.
YOU BE THE TEACHER
Your friend says that 3 dimes and 2 nickels is 50¢. Is your friend correct? Explain.
_______________________________
_______________________________

Answer: No, He is wrong.

Explanation:
1 dime = 10 cents
Which means , 3 dimes = 3 x 10 = 30 cents
1 nickel = 5 cents
Which means, 2 nickels = 2 x 5 = 10
Total value = 30 + 10 = 40
So, he is wrong

Think and Grow: Modeling Real Life

You have a $20 bill and a $5 bill. Your friend has $10 less than you. Do you and your friend have enough money to buy a $38 skateboard? Explain.
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 122

Answer: No.

Explanation :
The  amount I have is
$20 and $ 5
Total : $20+$5 = $25
Given that , my friend have $10 less than me
Which means,
$25 – $10= $15
But the cost of skateboard is $38
So , me and my friend do not have enough value to buy the skateboard.

Show and Grow

Question 7.
You have 1 quarter, 2 dimes, and 3 pennies. Your friend has 4 nickels and 2 pennies. Do you and your friend have enough money to buy a 75¢ bottle of orange juice? Explain.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 123
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 124

Answer:
My money:
1 quarter = 25 cents
1 dimes = 10 cents
Which means , 2 x 10 = 20 cents
3 pennies = 3 cents
Total : 25 + 20 + 3 = 48 cents
I have 48 cents
My friends money:
1 nickel = 5 cents
Which means , 4 nickels = 4 x 5 = 20 cents
2 pennies = 2 cents
Total : 20 + 2 = 22 cents
My friend have 22 cents
The cost of orange juice is 75 cents.
Therefore, me and my friend do not have enough money to buy the orange juice.

Question 8.
Descartes buys a board game for $19. He has three $5 bills and two $1 bills left over. How much money did he have before he bought the game?
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 125

Answer: $26.

Explanation:
The cost of board game = $19
Total value of leftover money is
Three $5 bills = 3 x 5 = $15
Two $1 bills = 2 x 1 = $2
Total : $15 +$2
= $17
The total money Descartes have in the beginning is
leftover money + cost of board game
=$17 +$19
= $26
Therefore , Descartes have $26 before he bought the game.

Question 9.
DIG DEEPER!
You have 25¢ in your desk, 18¢ in your backpack, and 50¢ in your pocket. You spend 43¢ and lose a quarter. How much money do you have left?
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 126

Answer: 25 cents.

Explanation:
The amount I have in my desk is 25 cents
The amount have in my backpack is 18 cents
The amount I have in my pocket is 50 cents
Now,  The total amount I have with me is:
25 cents +18 cents + 50 cents = 93 cents.
1 quarter = 25 cents
The amount I spend = 43 cents
total : 25 + 43 = 68 cents
Now, The money I have left is
93 cents – 68 cents = 25 cents
Therefore , I am left with 25 cents  .

Problem Solving: Money Homework & Practice 14.7

Question 1.
Newton has $30. Descartes has a $20 bill, a $10 bill, and two $1 bills. How much more money does Descartes have?
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 127

Answer:
The money newton has is $30 .
Descartes have $20 bill ,$10 bill and two $1 bill
Total money : 20 +10+1+1 =32
So Descartes have $32
Now, $32 – $30 = $2
Therefore, Descartes have $2 more than Newton.

Question 2.
You have some money. You spend 2 quarters and 3dimes at the cafeteria. Now you have 20¢. How much money did you have to start?
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 128

Answer: $1.

Explanation:
1 quarter = 25 cents
Which means , 2 quarters = 2 x 25 = 50 cents
1 dime =10 cents
Which means, 3 dimes = 3 x 10 = 30 cents
Total amount : 50 +30 = 80
After spending 80 cents at cafeteria , I have 20 cents
Now , the money i have at start is
80 cents + 20 = 100 cents
100 cents =  $1
Therefore , I have $ 1 in the start.

Question 3.
YOU BE THE TEACHER
You buy a sandwich for 75¢. You pay with $1. Your friend says your change will be 1 quarter. Is your friend correct? Explain.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 129

Answer:
Yes, he is correct .
The cost of sandwich is 75 cents
$1 is equals to 100 cents
100 cents – 75 cents = 25 cents
As we know that ,
1 quarter = 25 cents
so , he is correct.

Question 4.
Modeling Real Life
You have 12 pennies, 2 dimes, and 1 nickel. Your friend has 20¢ more than you. Do you and your friend have enough money to buy a toy car that costs $1? Explain.
_______________________________
_______________________________

Answer: No , me and my friend do not have enough money to buy a toy car

Explanation :
The amount I have is
12 pennies, 2 dimes and 1 nickel
12 pennies = 12 cents
1 dime = 10 cents
Which means 2 dime = 2 x 10 = 20 cents
and 1 nickel = 5 cents
Total amount =
12 +20 +5 = 37 cents
Therefore I have 37 cents
Friends money :
He has 20 cents more than me
Which means, 37 +20 = 57 cents
But , we know that $1 = 100 cents
So we do not have enough money to buy a toy car

Question 5.
Modeling Real Life
Descartes has $45. He spends a $20 bill and a $1 bill. He earns two $5 bills and a $10 bill. How much money does he have now?
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 130

Answer:
Given that , Descartes has $45
He spends $ 20 and $1 bill
total : 20 + 1 = 21
now, $45 -$21= $24
Then,
He earns 2 $5 bills and a $10 bill
Total : $24 +$5 +$5+$10= $44
Therefore ,Descartes have $44 .

Review & Refresh

Question 6.
Which time does not belong with the other three?
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 131

Answer:

Explanation :
In the above clock the time is 4 hours 30 minutes.
But the time in remaining three is 3 hours 3 minutes.
So, it does not belong with the other three.

Lesson 14.8 Tell Time to the Nearest Five Minutes

Explore and Grow

Label the missing minutes around the clock. Then tell the time.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 132

Answer:

  1. 15
  2. 30
  3. 55.
  4. The time is 3 hours 30 minutes

Show and Grow

Write the time.

Question 1.
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 133

Answer: 1 hours 25 minutes.

Explanation:
As the hours hand is in between 1 and 2 and the minutes hand on 5. So, the time is on 25 minutes.
That means, the time is 1 hours and 25 minutes.

Question 2.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 134

Answer:8 hours 10 minutes.

Explanation:
As the hours hand is on 8 and the minutes hand on 10. So, the time is on 10 minutes.
That means, the time is 8 hours and 10 minutes.

Question 3.
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 135

Answer: 6 hours 45 minutes.

Explanation:
As the hours hand is in between 6 and 7 and the minutes hand on 9. So, the time is on 45 minutes.
That means, the time is 6 hours and 45 minutes.

Question 4.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 136

Answer: 5 hours 55 minutes.

Explanation:
As the hours hand is in between 5 and 6 and the minutes hand on 11. So, the time is on 55 minutes.
That means, the time is 5 hours and 55 minutes.

Apply and Grow: Practice

Write the time.

Question 5.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 137

Answer: 3 hours 50 minutes.

Explanation:
As the hours hand is in between 3 and 4 and the minutes hand on 10. So, the time is on 50 minutes.
That means, the time is 3 hours and 50 minutes.

Question 6.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 138

Answer: 11 hours 20 minutes.

Explanation:
As the hours hand is in between 11 and 12 and the minutes hand on 4. So, the time is on 20 minutes.
That means, the time is 11 hours and 20 minutes.

Question 7.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 139

 

Answer: 5 hours 30 minutes.

Explanation:
As the hours hand is in between 5 and 6 and the minutes hand on 6. So, the time is on 30 minutes.
That means, the time is 5 hours and 30 minutes.

Draw to show the time.

Question 8.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 140

Answer:

Question 9.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 141

Answer:

Question 10.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 142

 

Answer:

Question 11.
Patterns
Write the next time in the pattern.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 143

Answer:
8:30
The sequence is differ  by 5 minutes

Question 12.
Precision
The hour hand points between the 4 and the 5. The minute hand points to the 4. What time is it?
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 144

Answer:
As the hour hand points between 4 and 5 .the minute hand points to the 4
The time is 4 hours 20 minutes

Think and Grow: Modeling Real Life

Baseball practice lasts 40 minutes. Show and write the time practice ends.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 145

Answer: 3 hours 55 minutes

Show and Grow

Question 13.
Recess lasts 25 minutes. Show and write the time recess ends.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 146

Answer: 12 hours 35 minutes

Question 14.
DIG DEEPER!
A train ride starts at 6:40. The ride lasts 45 minutes. What time does the ride end?
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 147
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 148

Answer:  7 hours 25 minutes

Tell Time to the Nearest Five Minutes Homework & Practice 14.8

Write the time.

Question 1.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 149

Answer:
10 hours 35 minutes.
As the hours hand is in between 10 and 11 and the minutes hand is near to 7. So, the time is near to 35 minutes.
That means, the time is 10 hours and 35 minutes.

Question 2.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 150

Answer:
6 hours 10 minutes.
As the hours hand is on 6, the minutes hand is on 2. So, the time is near to 10 minutes.
That means, the time is 6 hours and 10 minutes.

Question 3.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 151

Answer:
7 hours 25 minutes.
As the hours hand is in between 7 and 8 and the minutes hand is near to 5. So, the time is near to 25 minutes.
That means, the time is 7 hours and 25 minutes.

Question 4.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 152

 

Answer: 

Question 5.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 153

Answer:

Question 6.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 154

Answer:

Question 7.
Reasoning
The minute hand points to the 7. What number will it point to in 10 minutes?

Answer :  9
Each number the minutes hand points, indicates 5 minutes.
So,10 minutes indicates 2 more numbers, the minutes hand moves on to.
That means, 7 + 2 = 9.
So, after 10 minutes, the minutes hand points on to 9.

Question 8.
Precision
The hour hand points between the 11 and the 12. In 25 minutes it will be the next hour. What time is it now?

Answer: 11 hours 35 minutes.

Explanation:
Each number the minutes hand point, indicates 5 minutes.
So, 25  minutes indicates 5  more numbers, the minutes hand moves on to.
that means ,5 x 5 = 25
So, after 25 minutes, the time is 12: 00
now, the time is 11 hours 35 minutes .

Question 9.
Modeling Real Life
Your walk to school lasts 15 minutes. Show and write the time your walk ends.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 157

Answer: 8 hours 25 minutes

Question 10.
DIG DEEPER!
Your swimming lesson starts at 5:30. It lasts 35 minutes. What time does the lesson end?
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 158

Answer:
6 hours 5 minutes.

Explanation:
Each number the minutes hand points, indicates 5 minutes.
So,35 minutes indicates 7 more numbers, the minutes hand moves on to.
That means, 7 x 5= 35
So, after 35 minutes, the lesson ends on 6 hours 5 minutes.

Review & Refresh

Question 11.
The crayon is about 7 centimeters long. What is the best estimate of the length of the toothpick?
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 159

Answer: 4 Centimeters.

Explanation:
Given,
The length of crayon is 7 meters long.
4 centimeters is the best estimate of length of the toothpick .

Lesson 14.9 Tell Time Before and After the Hour

Explore and Grow

Write each time on the digital clocks. How much time has passed?
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 160

Answer:
1) the time is  11hours 45 minutes
2)  the time is 12: 00
on comparing both the clocks 15 minutes time has been passed.

Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 161

Answer:
1) In the first clock the time is 1:00
2) In the second clock the time is 1:15 on comparing both the clocks 15 minutes has been passed

Show and Grow

Write the time. Circle another way to say the time.

Question 1.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 162

Answer:
4 hours 15 minutes.
As the hours hand is in between 4 and 5 and the minutes hand on 3 . So, the time is near to 15 minutes.
That means, the time is 4 hours and 15 minutes.
Also known as quarter past 4.

Question 2.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 163

Answer:
11 hours 30 minutes.
As the hours hand is in between 11 and 12 and the minutes hand is on 6. So, the time is near to 30 minutes.
That means, the time is 11 hours and 30 minutes.
Also known as half past 11.

Question 3.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 164

 

Answer:
2 hours 45 minutes.
As the hours hand is in between 2 and 3 and the minutes hand is near to 9. So, the time is near to 45 minutes.
That means, the time is 2 hours and 45 minutes.
Also known as quarter to 3.

Question 4.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 165

Answer:
9 hours 20 minutes.
As the hours hand is in between 9 and 10 and the minutes hand is near to 4. So, the time is near to 20 minutes.
That means, the time is 9 hours and 20 minutes.
also known as 20 minutes after 9.

Apply and Grow: Practice

Write the time. Circle another way to say the time.

Question 5.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 166

Answer: 1 hours 30 minutes.
As the hours hand is in between 1 and 2 and the minutes hand is near to 6. So, the time is near to 30 minutes.
That means, the time is 1 hours and 30 minutes.
Also known as half past 1.

Question 6.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 167

Answer:
7 hours 10 minutes.
As the hours hand is in between 7 and 8 and the minutes hand is near to 2. So, the time is near to 10 minutes.
That means, the time is 7 hours and 10 minutes.
also known as ,10 minutes after 7

Show and write the time.

Question 7.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 168

Answer: 10 : 45

Question 8.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 169

Answer: 5 hours 15 minutes

Question 9.
Which One Doesn’t Belong? Which time does not belong with the other three?
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 170

Answer: Quarter past 7

Explanation:
6 : 45
45 minutes after 6 means 6 :45
Quarter to 7 means 6 : 45
These three are same .
But , quarter past 7 means 7 : 30
So, quarter past 7 is different.

Question 10.
Precision
Is it time for homework or dinner?
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 171

Answer: Dinner
Quarter after 6 is also known as 6 hours 15 minutes.
6: 15 p.m.- dinner

Think and Grow: Modeling Real Life

School starts at quarter past 8. Are you early or late to school? Explain.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 172
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 173

Answer:
Early to school
now, the time is 7 hours 55 minutes
the school starts at 8 hours 15 minutes ;it means quarter past 8 .
so, I am early to school

Show and Grow

Question 11.
A movie starts at quarter to 6. Are you early or late to the movie? Explain.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 174
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 175

Answer:
I am late to the movie .
Now, the time is 6 hours 5 minutes .
but, the movie starts at quarter to 6
which means , 5 hours 45 minutes
so I am late to the movie .

Question 12.
DIG DEEPER!
You arrive at the bus station 20 minutes before 12. Which is the first bus you can board? How many minutes are there until it leaves?
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 176Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 177

Answer:
The time when i arrived bus station is 11 hours 40 minutes
So, I can board blue bus.
Blue bus leaves at 11 hours 55 minutes .
So , there are 15 more minutes until the blue bus leaves the bus station.

Tell Time Before and After the Hour Homework & Practice 14.9

Write the time. Circle another way to say the time.

Question 1.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 178

Answer: quarter to 12 means 11 hours 45 minutes

 

As the hours hand is between 11 and 12, the minutes hand is on 9. So, the time is near to  45 minutes.
That means, the time is 11 hours and 45 minutes.

Question 2.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 179

Answer: 40 minutes after 5 which means 5 hours 40 minutes

As the hours hand is between 5 and 6, the minutes hand is on 8. So, the time is near to  40 minutes.
That means, the time is 5 hours and 40 minutes.
Show and write the time.

Question 3.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 180

Answer: 3 hours 30 minutes

As the hours hand is between 3 and 4, the minutes hand is on 6. So, the time is near to 30 minutes.
That means, the time is 3 hours and 30 minutes.

Question 4.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 181

Answer: 12 hours 15 minutes

As the hours hand is between 12 and 1, the minutes hand is on 3. So, the time is near to  15 minutes.
That means, the time is 12 hours and 15 minutes.

Question 5.
YOU BE THE TEACHER
Newton says it is 2:45, or quarter to 3. Is he correct? Explain.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 182

Answer:
Yes, he is correct.

Explanation:
quarter means 15 minutes .
Each number the minutes hand points, indicates 5 minutes.
That means, 3 x 5= 15
so, 2:45 and quarter to 3 are both same
So, newton is correct

Question 6.
Modeling Real Life
A show starts at quarter to 7. Are you early or late to the show? Explain.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 183

Answer:
I am  early to the show
Now ,the time is 6 hours 30 minutes
but the show starts at quarter to 7 which means 6 hours 45 minutes
So ,I am 15 minutes early to the show

Question 7.
DIG DEEPER!
You arrive at the metro station 10minutes after 2. Which is the first train you can board? How many minutes are there until it leaves?
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 184

Answer:
The time 10 minutes after 2 is 2 hours 10 minutes
So ,I can board the yellow train which will leaves the railway station at 2: 30
There are 20 minutes more until the yellow train leaves the railway station.

Review & Refresh

Question 8.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 185

Answers: 162.

 

Question 9.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 186

Answer: 177

Explanation:

Question 10.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 187

Answer: 446.

Explanation:

Lesson 14.10 Relate A.M. and P.M.

Explore and Grow

Describe what you do in the morning. Show and write the time. Describe what you do in the evening. Show and write the time.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 188

Show and Grow

Write the time. Circle a.m. or p.m.

Question 1.
Eat breakfast
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 189

Answer: 7 hours 30 minutes a.m.
As the hours hand is on 7, the minutes hand is on 6. So, the time is near to 30  minutes.
That means, the time is 7 hours and 30 minutes.

Question 2.
Eat dinner
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 190

Answer:
6 hours 15 minutes p.m.
As the hours hand is between 6 and 7, the minutes hand is on 3. So, the time is near to  15 minutes.
That means, the time is 6 hours and 15 minutes.

Question 3.
Go to art class
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 191

Answer :
10 hours 25 minutes a.m.
As the hours hand is between 10 and11, the minutes hand is on 5. So, the time is near to 25  minutes.
That means, the time is 10 hours and 25 minutes.

Question 4.
Do homework
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 192

Answer:
4 hours 40 minutes p.m.
As the hours hand is between 4 and 5, the minutes hand is on 8. So, the time is near to  40  minutes.
That means, the time is 4 hours and 40 minutes.

Apply and Grow: Practice

Write the time. Circle a.m. or p.m.

Question 5.
Ride the bus to school
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 193

Answer:
8:00 a.m.
As the hours hand is on 8, the minutes hand is on 12.
That means, the time is 8 hours

Question 6.
Go to a party
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 194

Answer:
6hours 30 minutes p.m.
As the hours hand is between 6 and 7, the minutes hand is on 6. So, the time is 30  minutes.
That means, the time is 6 hours and 30 minutes.

Draw to show the time. Circle a.m. or p.m.

Question 7.
Read before bed
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 195

Answer:
7 hours and 50 minutes p.m.

Question 8.
Sunrise
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 196

Answer:
5 hours 20 minutes a.m.

Question 9.
Reasoning
Use the times in the list to complete the story.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 197

Answer:
You arrive at school at 8: 30 a.m. Your class goes to music at 10:15 a.m. After school,
you read a book at 5:20 p.m.

Think and Grow: Modeling Real Life

Use the times to complete the timeline. Write something you might do at those times.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 198

Answer:
1) 8: 45 a.m. –  School time
2)2:50 p.m. –  Music class time
3) 4:10 p.m. –  Home work

Show and Grow

Question 10.
Use the times to complete the timeline. Write something you might do at those times.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 199
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 200

Answer:
1) 7: 30 a.m. – Yoga class
2) 9:55 a.m.-  Art class
3) 7:35 p.m. – Dinner

Question 11.
DIG DEEPER!
Use the times to complete the timeline. Then rewrite each time digitally below, including a.m. or p.m
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 201
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 202

Answer:
1) 20 minutes before 7 –      6:40 a.m
2) 10 minutes after 11   –     11:10 a.m
3) noon                         –      12:00 p.m
4) quarter past 3           –        3:15 p.m
5)  half past 6                 –        6:30 p.m
6)2  0 minutes after 8    –        8:20 p.m

Relate A.M. and P.M. Homework & Practice 14.10

Write the time. Circle a.m. or p.m.

Question 1.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 203

Answer:
9 hours 25 minutes p.m
As the hours hand is between 9 and 10, the minutes hand is on 5. So, the time is near to 25 minutes.
That means, the time is 9 hours and 25 minutes.

Question 2.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 204

 

Answer:
7hours 15 minutes A.m.
As the hours hand is between 7 and 8, the minutes hand is on 3. So, the time is 15  minutes.
That means, the time is 7 hours and 15 minutes.

Draw to show the time. Circle a.m. or p.m.

Question 3.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 205

Answer: 12:00 P.m.

Question 4.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 206

Answer: 4: 40 P.m.

Question 5.
Reasoning
Right now, it is p.m. In 10 minutes it will be a.m. What time is it now? Explain.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 207

 

Answer:
Time is 11 hours 50 minutes.
Each minute hand point indicates 5 minutes
That means 2x 5= 10 minutes
a.m. starts imediatly after 12:00 pm
So, now the time is 11: 50 p.m.

Question 6.
Modeling Real Life
Use the times to complete the timeline. Write something you might do at those times.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 208

Answer:
1 ) 8:15 a.m. – break fast
2) 11: 55 a.m. – art class
3)  8: 15 p.m.- drawing time

Question 7.
DIG DEEPER!
Use the times to complete the timeline. Then rewrite each time digitally below, including a.m. or p.m.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 209

Answer:
1 ) Quarter to 8 –            7:45 a.m.
2) 20 minutes after 8 –  8: 20 a.m.
3) Quarter past 11 –      11: 15 a.m.
4) Noon –                       12:00 p.m.
5) 10 minutes after 2 –   2 : 10 p.m.
6) Half past 9-                 9:30 p.m.

Review & Refresh

Question 8.
65 + 36 = ____

Answer:
65 +36 = 101

Question 9.
56 + 18 = ____
Answer: 74

Money and Time Performance Task 14

Question 1.
a. You have two $1 bills, 1 quarter, 5 dimes, 3 nickels, and 2 pennies. How much more money do you need to buy a subway pass?
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 210

Answer:
The cost of  subway pass is $3
$3 is equals to 300 cents
$1 is equals to 100 cents
Two $1 = 100 + 100 = 200cents
1 quarter = 25 cents
1 dime = 10 cents
Which means , 5 x 10 = 50
1 nickel = 5 cents
Which means, 3 x 5 = 15
and 2 pennies = 2  cents
Total :  200 +25+50+15+2 = 292
300 – 292 = 8 cents
I need 8 more cents to buy a pass.
So, I do not have enough money to buy a subway pass.

b. You find a dime. Do you have enough money to buy the pass now?
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 211

Answer: Yes

Explanation:
1 dime = 10 cents
292 +10 = 302 cents
The cost of subway pass is $3 which means 300 cents
If I find a dime, I have enough money to buy a pass

Question 2.
A weekly subway pass is $32. A customer pays with a $50 bill. Use tally marks to show three different ways that the customer can receive change. What is the total change?
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 212Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 212.2

Answer:
The cost of weekly subway pass is $32 .
Given that ,
A customer paid a $50 bill
Total change :
$50 -$32 = $18
$18 can be shown as
1) 1 $10 bill , 1 $5 bill and 3 $1 bill
2) 1 $10 bill, 8 $1 bills
3) 3 $ 5 bills and 3 $1 bill .

Question 3.
You arrive at the subway station at quarter to 3. What times will the subways arrive?
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 213

Answer: car A arrives subway

Money and Time Activity

To Play: Place the Flip and Find Cards face down in the boxes. Take turns flipping 2 cards. If your cards show the same time or value, keep the cards. If your cards show different times or values, flip the cards back over. Play until all matches are made.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 214

Money and Time Chapter Practice

4.1 Find Total Values of Coins

Question 1.
Count on to find the total value.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 215

Answer: 46 cents

Explanation
1 quarter = 25 cents
1 dime = 10 cents
1 nickel = 5 cents
Which means , 5 x 2 =10 cents
1 penny = 1 cent
Total : 25 +10+5+5+1 = 46 cents
Therefore the total value of coins 46 cents.

Question 2.
Modeling Real Life
You have 2 quarters, 2 dimes, and 1 penny. How many cents do you have? Do you have enough money to buy the frozen fruit bar?
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 216
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 217

Answer:
The cost of frozen fruit bar is 75 cents
1 quarter = 25 cents
which means , 2 quarter = 2 x 25 = 50 cents
1 dime = 10 cents
which means , 2 dimes = 2 x 10 = 20 dimes
1 penny = 1 cent
total amount I have is 50+20+1 = 71 cents
therefore I have 71 cents
So ,I do not have enough money to buy the frozen fruit bar.

14.2 Order to Find Total Values of Coins

Question 3.
Draw and label the coins from the greatest value to the least value. Then find the total value.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 218

Answer:
43 cents

Explanation
1 quarter = 25 cents
1 dime = 10 cents
1 nickel = 5 cents
3 pennies = 3cents
Total value = 25 +10 +5 + 3 = 43 cents

>>>

14.3 Show Money Amounts in Different Ways

Question 4.
Draw and label coins to show the amount in two different ways.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 219

Answer:
56 cents can be shown as
2 quarter, 1 nickel and 1 penny
1 quarter = 25 cents
Which means, 2 x 25 = 50 cents
1 nickel = 5 cents
and 1 penny = 1 cents
Total : 50+5+1 = 56 cents

Another way :
56  cents can be shown as 5 dime, 1 nickel and 1 penny
1 dime = 10 cents
5 dimes = 5 x 10 = 50 cents
1 nickel = 5 cents
and 1 penny = 1 cent
Total value :
50 + 5+1 = 56 cents

14.4 Make One Dollar

Draw coins to make $1. How many cents do you need?

Question 5.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 220

Answer: $ 1 is equals to 100 cents
100 cents – 79 cents = 21 cents.
So , I need 21 cents to make $ 1
21 cents can be shown in coins as 2 dime , 1 penny
1 dime = 10 cents
Which means , 2 dime = 2 x 10 = 20 cents
1 penny = 1 cent
Total : 20 + 1 = 21 cents

Question 6.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 221

Answer:
$1 is equals to 100 cents
1 dime = 10 cents
Which means , 2 cents= 2 x 10 = 20 cents
3 pennies = 3 cents
Total value = 20+3 = 23 cents
Now, 100 cents- 23 cents = 77 cents
77 cents can be shown as 3 quarters , 2 pennies
1 quarter = 25 cents
Which means , 3 quarters = 3 x 25 = 75 cents
2 pennies = 2 cents
Total value : 75 + 2 = 77 cents

14.5 Make Change from One Dollar

Question 7.
Reasoning
Newton buys a toy for 21¢. Descartes buys one for 94¢. You buy one for 57¢. You each pay with $1. Who gets back the least amount of money? How do you know?
_____________________________________
_____________________________________

Answer:
$1 is equals to 100 cents
The cost of newtons toy is 21 cents
100 – 21 =79 cents
So, 79 cents is change newtons gets back.
Now, the cost of descartes toy is 94 cents
100 – 97 = 3 cents
So Descartes change is 3 cents
On comparing both newtons and Descartes money
79 cents is greater than 3cents;
79 > 3
Therefore, Descartes gets back the least amount of money.

14.6 Find Total Values of Bills

Question 8.
Count on to find the total value.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 222

Answer: $ 38

Explanation :
We have , Three $ 1 bills
which means , $1 +$1 $1 = $3
One $ 5 bill
One $ 10 bill
One $ 20 bill
Total value :
$ 3+ $5 +$10 + $20 =$38
Therefore the total value of the dollar bills is $ 38.

Question 9.
Newton has five $10 bills. He has $32 more than Descartes. How much money does Descartes have?
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 223

Answer:  $18

Explanation :
five $ 10 bills is equals to
5 x 10 = $ 50
Given that , Newton has $32 more than Descartes
We know that, Newton have $50
Now, $50 – $32 = $18
Therefore , Descartes have $18

14.8 Tell Time to the Nearest Five Minutes

Write the time.

Question 10.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 224

Answer: 7 hours 20 minutes

Explanation :
As the hours hand is between 7 and 8, the minutes hand is on 4. So, the time is 20 minutes.
That means, the time is 7 hours and 20 minutes.

Question 11.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 225

Answer: 3 hours 45 minutes

Explanation :
As the hours hand is between 3 and 4, the minutes hand is on 9. So, the time is 45 minutes.
That means, the time is 3 hours and 45 minutes.

Question 12.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 226

Answer: 10 hours 15 minutes

Explanation :
As the hours hand is between 10 and 11, the minutes hand is on 3. So, the time is 15 minutes.
That means, the time is 10 hours and 15 minutes.

14.9 Tell Time Before and After the Hour

Question 13.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 227

Answer: quarter past 5
which means 5 hours 15 minutes
quarter to 5 means 4 hours 45 minutes

Question 14.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 228

Answer: quarter to 10
which means ,9 hours 45 minutes
half past 9 means 9 hours 30 minutes

Question 15.
Modeling Real Life
Soccer practice starts at half past 1. Are you early or late to soccer practice? Explain.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 229

Answer:  I am late to soccer practice.

Explanation:
now the time is 1 hour 45 minutes .
The class starts at half past one
Which means ,1 hour 30 minutes
So ,I am late by 15 minutes to the class.

14.10 Relate A.M. and P.M.

Draw to show the time. Circle a.m. or p.m.

Question 16.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 230

Answer : 9hours 30 minutes p.m

Question 17.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 231

 

Answer:7 hours 25 minutes a.m.

Conclusion:

Every expert’s choice is using Big Ideas Math Answers Grade 2 Chapter 14 Money and Time pdf. Compare the questions with the real-time problems and solve the questions with the help of Big Ideas Math Book Grade 2 Chapter 14 Money and Time Solutions. Stay in touch with our site https://bigideasmathanswers.com/ to find answer keys of all Big Ideas Math Grade 2 Chapters.

Big Ideas Math Answers Grade 4 Chapter 1 Place Value – Detailed Concepts

Big Ideas Math Answers Grade 4 Chapter 1

Big Ideas Math Answers Grade 4 Chapter 1 Place Value Concepts helps thousands of students to get excellent grades. Help your students keep up to do the math with the reference of Big Ideas 4th Grade Math Book Answer Key Chapter 1 Place Value. Students can effectively solve all problems and grab knowledge and skills by using the 4th standard Big Ideas Math Answer Key. Big Ideas Math Answers Grade 4 Chapter 1 Place Value Concepts will give all that flexibility to build the confidence to do math in an easy way. It’s important to aware of the procedure of solving a problem more than finding the correct answer to improve the skills of your students.

Big Ideas Math Book 4th Grade Chapter 1 Place Value Concepts Answer Key

Help your students strengthen their knowledge by practicing all the Chapter 1 Place Value Concepts. All the questions and answers are explained with images, graphs for easy understanding of students. Do all the given activities and check your answers to test your knowledge. Students can easily promote to the next grade by practicing with the Big Ideas 4th Grade Answer Key.

Lesson 1: Understand Place Value

Lesson 2: Read and Write Multi-Digit Numbers

Lesson 3: Compare Multi-Digit Numbers

Lesson 4: Round Multi-Digit Numbers

Performance Task

Lesson 1.1 Understand Place Value

Explore and Grow

Model the number. Draw to show your model. Then write the value of each digit.
Big Ideas Math Answer Key Grade 4 Chapter 1 Place Value Concepts 1.1 1
Explanation: Every Digit in a number has a place value according to the position of a Digit. We define the place value of a digit by considering the Rightmost digit’s position as “1’s place.” After 1’s place, 10’s place, 100’s place, and so on.

Answer (i):
As per the Explanation,
The Position Value of ‘1’ is: 1000
The position value of ‘2’ is: 200
The position value of ‘7’ is: 70
The position value of ‘5’ is: 5

Answer (ii): As per Explanation,
The position value of ‘3’ is: 3000
The position value of ‘3’ is: 300
The position value of ‘3’ is: 30
The position value of ‘3’ is: 3

Repeated Reasoning
Compare the value of the tens digit to the value of the ones digit. Then do the same with the hundreds and tens digits, and the thousands and hundred’s digits. What do you notice?

Answer:  When we compare the ten’s digit to the value of one’s digit, we can notice that we got the 10’s digit by multiplying ’10’ to the 1’s digit. For comparison of 100’s and 1000’s digit also, we will get 100’s digit is “10 Times” of the 1000’s digit.

Think and Grow: Understand Place Value

A place value chart shows the value of each digit in a number.

The value of each place is 10 times the value of the place to the right.

The place value chart shows how the place values are grouped.

Each group is called a period. In a number, periods are separated by commas.

Big Ideas Math Answer Key Grade 4 Chapter 1 Place Value Concepts 1.1 2

Example
Big Ideas Math Answer Key Grade 4 Chapter 1 Place Value Concepts 1.1 3

• The number, in standard form, is _______.

Answer: 2,75,449

Explanation: The Position of each digit is defined in a number according to place value and we can count “place-value” up to “lakhs ” starting from “1’s place” from the Rightmost position and count from thereon.

• The value of the digit 7 is 7 ten thousand or ______.

Answer: 70,000

Explanation: The Position of each digit is defined in a number according to place value and we can count “place-value” up to “lakhs ” starting from “1’s place” from the Rightmost position and count from thereon.

• The value of the digit 4 in the hundreds place is ______.

Answer: 400

Explanation: The Position of each digit is defined in a number according to place value and we can count “place-value” up to “lakhs ” starting from “1’s place” from the Rightmost position and count from thereon.

• The value of the digit 4 in the tens place is ______.

Answer: 40

Explanation: The Position of each digit is defined in a number according to place value and we can count “place-value” up to “lakhs ” starting from “1’s place” from the Rightmost position and count from thereon.

• The value of the digit 4 in the hundreds place is ______ times the value of the digit 4 in the tens place.

Answer: 10 Times

Explanation: We know the value of ‘4’ in 100’s place: 400

We know the value of ‘4’ in 10’s place: 40

So, we can get ‘400’ when we multiply ’40’ with its place value ’10’.

Show and Grow

Write the value of the underlined digit.

Question 1.

93,517

Answer: The place value of ‘5’ in ‘93,517’ is:  500

Explanation: The position of each digit in a number is defined and it is the value based on the position of the digit. So, the place value of ‘5’ in ‘93,517’ is:  500

Question 2.

685,726

Answer: The place value of ‘8’ in ‘6,85,26’ is :  80,000

Explanation: The position of each digit in a number is defined and it is the value based on the position of the digit. So, the  place value of ‘8’ in ‘6,85,26’ is :  80,000

Question 3.

359,842

Answer: The place value of ‘9’ in ‘359,842’ is:  9000

Explanation: The position of each digit in a number is defined and it is the value based on the position of the digit. So, the place value of ‘9’ in ‘359,842’ is:  9000

Question 4.

483,701

Answer: The place value of ‘4’ in ‘483,701’ is :  400,000

Explanation: The position of each digit in a number is defined and it is the value based on the position of the digit. So, the place value of ‘4’ in ‘483,701’ is :  400,000

Compare the values of the underlined digits.

Question 5.

20 and 200

Answer: The value of ‘2’ in ‘200’ is 10 times the value of ‘2’ in ’20’.

Explanation: The position value of 2 in 200: 200

The position value of 2 in 20: 20

As we notice, the value of 2 in 20 is “10 Times” of the value of 2 in 200.

Question 6.

1,000 and 100

Answer: The value of ‘1’ in ‘100’ is 10 times the value of ‘1’ in ‘1000’.

Explanation: The position value of 1 in 1000: 1000

The position value of 1 in 100: 100

As we notice, the value of 1 in 100 is “10 Times” of the value of 1 in 1000.

Apply and Grow: Practice

Write the value of the underlined digit.

Question 7.

45,802

Answer: The place value of ‘2’ in ‘45,802’ is :  2

Explanation: The position of each digit in a number is defined and it is the value based on the position of the digit. So, the place value of ‘2’ in ‘45,802’ is :  2

Question 8.

97,361

Answer: The place value of ‘7’ in ‘97,361’ is :  7000

Explanation: The position of each digit in a number is defined and it is the value based on the position of the digit. So, the place value of ‘7’ in ‘97,361’ is:  7000

Question 9.

168,392

Answer: The place value of ‘9’ in ‘168,392’ is:  90

Explanation: The position of each digit in a number is defined and it is the value based on the position of the digit. So, the place value of ‘9’ in ‘168,392’ is:  90

Question 10.

807,516

Answer: The place value of ‘8’ in ‘807,516’ is :  800,000

Explanation: The position of each digit in a number is defined and it is the value based on the position of the digit. So, the place value of ‘8’ in ‘807,516’ is :  800,000

Question 11.

400,532

Answer: The place value of ‘5’ in ‘400,532’ is :  500

Explanation: The position of each digit in a number is defined and it is the value based on the position of the digit. So, the place value of ‘5’ in ‘400,532’ is:  500

Question 12.

749,263

Answer: The place value of ‘4’ in ‘749,263’ is :  40,000

Explanation: The position of each digit in a number is defined and it is the value based on the position of the digit. So, the place value of ‘4’ in ‘749,263’ is :  40,000

Question 13.

619,457

Answer: The place value of ‘6’ in ‘619,457’ is :  600,000

Explanation: The position of each digit in a number is defined and it is the value based on the position of the digit. So, the place value of ‘6’ in ‘619,457’ is :  600,000

Question 14.

301,882

Answer: The place value of ‘1’ in ‘301,882’ is :  1,000

Explanation: The position of each digit in a number is defined and it is the value based on the position of the digit. So, the place value of ‘1’ in ‘301,882’ is :  1,000

Compare the values of the underlined digits.

Question 15.

4 and 47

Answer: The value of 4 in 47 is 10 times the value of 4 in 4.

Explanation: The position value of 4 in 47 is: 40

The position value of 4 in 4: 4

As we notice, the value of 4 in 47 is “10 Times” of the value of 4 in 4.

Question 16.

35,649 and 23,799

Answer: The value of 3 in 35,649 is 10 times the value of 3 in 23,799.

Explanation: The position value of 3 in 35,649 is: 30,000

The position value of 3 in 23,799: 3,000

As we notice, the value of 3 in 35,649 is 10 times the value of 3 in 23,799.

Question 17.

Your friend is 9 years old. Your neighbor is 90 years old. Your neighbor is how many times as old as your friend?

Answer: Given that your friend is 9 years old and your neighbor is 90 years old. From these two, we can conclude that the neighbor is 10 times the age of your friend.

Question 18.

YOU BE THE TEACHER

Is Newton correct? Explain.
Big Ideas Math Answer Key Grade 4 Chapter 1 Place Value Concepts 1.1 4

Answer:

Question 19.

Writing

Explain the relationship between the place values when the same two digits are next to each other in a multi-digit number.

Answer: We know that the place value of a Digit is dependent on the Position of the Digit in a given number. Hence, even those are the same numbers or different numbers placed next to each other, the preceding digit is 10 times the value of the former number.

Think and Grow: Modeling Real Life

Example

What is the value of the digit 3 in the distance around Saturn? in the distance around Jupiter? How do these values relate to each other?
Big Ideas Math Answer Key Grade 4 Chapter 1 Place Value Concepts 1.1 5

Write each number in a place value chart.

Big Ideas Math Answer Key Grade 4 Chapter 1 Place Value Concepts 1.1 6

Compare the values of the digit 3 in each number.

Answer: The value of ‘3’ in Jupiter: 30,000

The value of ‘3’ in Saturn: 300,000

So, we can say that the value of ‘3’ in Saturn is 10 times the value of 3 in Saturn.

Saturn: The value of the digit 3 is ______.

Answer: 300,000

Explanation: As we can see, the value of 3 in Saturn is: 300,000

Jupiter: The value of the digit 3 is ______.

Answer: 30,000

Explanation: As we can see, the value of 3 in Jupiter is: 30,000

The value of the digit 3 in the distance around Saturn is ______ times the value of the digit 3 in the distance around Jupiter.

Answer: 10 times

Explanation: The value of ‘3’ in Jupiter: 30,000

The value of ‘3’ in Saturn: 300,000

So, we can say that the value of ‘3’ in Saturn is 10 times the value of 3 in Saturn.

Show and Grow

Use the table above.

Question 20.

The distance around which planet has an 8 in the thousands place?

Answer: Venus

Explanation: From the Table above, we see that the distance around Venus has ‘8’ in 1000’s place.

Question 21.

Compare the value of the 8s in the distance around Saturn.

Answer: The value of 1st 8 in Saturn: 800

The value of 2nd 8 in Saturn: 80

Explanation: From the above values, we can say that the value of 1st 8 is 10 times the value of 2nd 8 in Saturn.

Question 22.

What is the value of the digit 4 in the distance around Earth? in the distance around Neptune? How do these values relate to each other?

Answer: The value of 4 on Earth is 10 times the value of 4 in Neptune.

Explanation: The value of 4 on Earth is: 40,000

The value of 4 in Neptune is: 4,000

By comparing these two values, we can conclude that the value of 4 on Earth is 10 times the value of  4 in Neptune.

Question 23.
Compare the values of the first digits in the distances around Earth and Jupiter. Explain how you can use the values to compare the sizes of these two planets.

Answer:
The value of the 1st digit in Jupiter is 10 times the value of the 1st digit on Earth.

Explanation:
The value of 4 on Earth is: 40,000
The value of 4 in Jupiter is: 400,000
From these 2 values, we can conclude that the value of the 1st digit in Jupiter is 10 times the value of the 1st digit on Earth.

Understand Place Value Homework & Practice 1.1

Write the value of the underlined digit.

Question 1.

79,043

Answer: The place value of 4 in 79,043 is: 40

Explanation: The position of each digit in a number is defined and it is the value based on the position of the digit. So, the place value of 4 in 79,043 is: 40

Question 2.

52,618

Answer: The place value of 8 in 52,618 is: 8

Explanation: The position of each digit in a number is defined and it is the value based on the position of the digit. So, the place value of 8 in 52,618 is: 8

Question 3.

379,021

Answer: The place value of 9 in 379,021 is: 9000

Explanation: The position of each digit in a number is defined and it is the value based on the position of the digit. So, the place value of 9 in 379,021 is: 9000

Question 4.

958,641

Answer: The place value of 6 in 958,641 is: 600

Explanation: The position of each digit in a number is defined and it is the value based on the position of the digit. So, the place value of 6 in 958,641 is: 600

Question 5.

203,557

Answer: The place value of 2 in 203,557 is: 200,000

Explanation: The position of each digit in a number is defined and it is the value based on the position of the digit. So, the place value of 2 in 203,557 is: 200,000

Question 6.

145,860

Answer: The place value of 5 in 145,860 is: 5,000

Explanation: The position of each digit in a number is defined and it is the value based on the position of the digit. So, the place value of 5 in 145,860 is: 5,000

Question 7.

497,384

Answer: The place value of 3 in 497,384 is: 300

Explanation: The position of each digit in a number is defined and it is the value based on the position of the digit. So, the place value of 3 in 497,384 is: 300

Question 8.

612,739

Answer: The place value of 1 in 612,739 is: 10,000

Explanation: The position of each digit in a number is defined and it is the value based on the position of the digit. So, the place value of 1 in 612,739 is: 10,000

Compare the values of the underlined digits.

Question 9.

603 and 6,425

Answer: The value of 6 in 6,425 is 10 times the value of 6 in 603

Explanation: The value of 6 in 603 is: 600

The value of 6 in 6,425 is: 6,000

From these values, we can conclude that the value of 6 in 6,425 is 10 times the value of 6 in 603

Question 10.

930,157 and 89,216

Answer: The value of 9 in 930,157  is 100 times the value of 9 in 89,216

Explanation: The value of 9 in 930,157 is: 900,000

The value of 9 in 89,216 is: 9,000

From these values, we can conclude that the value of 9 in 930,157  is 100 times the value of 9 in 89,216.

Question 11.

A car can travel 50 miles per hour. A tsunami can travel 500 miles per hour. The tsunami is how many times faster than the car?

Answer: The Tsunami is 10 times faster than the car.

Explanation: Given that,

A car can travel 50 miles per hour

A tsunami can travel 500 miles per hour.

From these 2 values, we can conclude that the tsunami is 10 times faster than the car.

Question 12.

Number Sense

In the number 93,825, is the value in the ten thousand’s place 10 times the value in the thousands place? Explain.

Answer: Yes, the value in the ten thousand’s place is 10 times the value in the thousands place.

Explanation: The given number is 93,825.

But, we know how big is the number, the value in the ten thousand’s place is always 10 times the value in the thousands place.

Question 13.

Reasoning

Write the greatest number possible using each number card once. Then write the least six-digit number possible.

Big Ideas Math Answer Key Grade 4 Chapter 1 Place Value Concepts 1.1 7

Answer: The Greatest 6-digit  number is: 986531

The least 6-digit number is: 135689

Explanation: Given digits are : 6 , 1 , 3 , 8 , 9 , 5

By using each digit only once,

The Greatest 6-digit  number is: 986531

The least 6-digit number is: 135689

Question 14.

Modeling Real Life

Use the table.

Big Ideas Math Answer Key Grade 4 Chapter 1 Place Value Concepts 1.1 8

Question 14.

The height of which mountain has a 3 in the thousands place?

Big Ideas Math Answer Key Grade 4 Chapter 1 Place Value Concepts 1.1 9

Answer: Mount Kinabalu

Explanation: From the given table, we can conclude that Mount Kinabalu has ‘3’ in the thousands place.

Question 15.

What is the value of the digit 5 in the height ht of K2? in the height of Mount Everest? How do these values relate to each other?

Answer: The value of the 5 in the height of K2 is: 50

The value of the 5 in the height of Mount. Everest is: 5

Hence,

The value of 5 in K2 is 10 times the value of 5 in Mount. Everest

Explanation: From the given table, we can conclude that

The value of the 5 in the height of K2 is: 50

The value of the 5 in the height of Mount. Everest is: 5

Hence,

The value of 5 in K2 is 10 times the value of 5 in Mount. Everest

Question 16.

The tallest mountain in the world is shown in the table. Which mountain is it?

Answer: Mount. Everest

Explanation: From the given table, the Tallest Mountain in the World is: Mount. Everest

Review & Refresh

Question 17.

Use the graph to answer the questions.

How many seconds did the Peach Street traffic light stay red?

How many more seconds did the Valley Road traffic light stays red than the Elm Street traffic light?

 

Big Ideas Math Answer Key Grade 4 Chapter 1 Place Value Concepts 1.1 10

Answer: The time which the Peach Street traffic light stays red is: 55 seconds

The time which the Valley road traffic light stays red than the Elm Street traffic light is: 15 seconds

Explanation: Let us  consider, Half circle= 5 seconds

From the given graph,

The time which the Peach Street traffic light stays red is: 55 seconds

The time which the valley road traffic light stays red is: 35 seconds

The time which the Elm Street traffic light stays red is: 20 seconds

So,

The time which the Valley road traffic light stays red than the Elm Street traffic light is 15 seconds.

Lesson 1.2 Read and Write Multi-Digit Numbers

Explore and Grow

Model each number. Draw to show your models. Then write each number a different way.

2,186

Answer: Word Form: Two Thousand, One hundred eighty-six

Expanded Form: 2,000 + 100 + 80 + 6

Explanation: We can write the given number in 3 forms. They are:

A) Standard Form  B) Word Form  C) Expanded Form

Now, the Given number is 2,186 which is in Standard Form. So, we have to write the given number in the remaining 2 forms.

Five thousand, two hundred thirteen

Answer:

Standard Form: 5,213

Explanation:

We can write the given number in 3 forms. They are:

A) Standard Form  B) Word Form  C) Expanded Form

Now, the Given number is 5,213 which is in Standard Form. So, we have to write the given number in the remaining 2 forms.

Expanded Form:

Answer: 5,000 + 200 + 10 + 3

Explanation: We can write the given number in 3 forms. They are:

A) Standard Form  B) Word Form  C) Expanded Form

Now, the Given number is Five thousand, two hundred thirteen which is in word Form. So, we have to write the given number in the remaining 2 forms.

3,000 + 600 + 90 + 4

Answer: Standard Form: 3,694

Explanation: Now, the Given number is 3,000 + 600 + 90 + 4 which is in Expanded Form. So, we have to write the given number in the remaining 2 forms.

Word Form:

Answer: Three thousand, Six hundred ninety- four

Explanation: We can write the given number in 3 forms. They are:

A) Standard Form  B) Word Form  C) Expanded Form

Now, the Given number is 3,000 + 600 + 90 + 4  which is in Expanded Form. So, we have to write the given number in the remaining 2 forms.

Structure

Model a different four-digit number. Write the number in many different ways as possible. Compare your work to your partner’s.

Answer: Consider a 4- digit number ” 9,836″

So, we can write “9,836” in:

Standard Form: 9,836

Word Form: Nine thousand, Eight hundred thirty-six

Expanded Form: 9,000 + 800 + 30 + 6

Explanation: We can write the given number in 3 forms. They are:

A) Standard Form  B) Word Form  C) Expanded Form

Now, the given number which is 9,836 is in Standard Form. So, we have to write the given number in the remaining 2 forms.

Think and Grow: Read and Write Multi-Digit Numbers

Example

Write the number in standard form, word form, and expanded form.

Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 8

Standard form:

Answer: 427, 561

Explanation: We can write the given number in 3 forms. They are:

A) Standard Form  B) Word Form  C) Expanded Form

Now, the given number which is 427,561 is in Standard Form. So, we have to write the given number in the remaining 2 forms.

Word form:

Answer: Four hundred Twenty-seven thousand, five hundred sixty-one

Explanation:

We can write the given number in 3 forms. They are:

A) Standard Form  B) Word Form  C) Expanded Form

Now, the given number which is 427,561 is in Standard Form. So, we have to write the given number in the remaining 2 forms.

Expanded form: _____ + _______ + ______ + ______ + ______ + ______

Answer: 400,000 + 20,000 + 7,000 + 500 + 60 + 1

Explanation:

We can write the given number in 3 forms. They are:

A) Standard Form  B) Word Form  C) Expanded Form

Now, the given number which is 427,561 is in Standard Form. So, we have to write the given number in the remaining 2 forms.

Example

Use the place value chart to write the number below in standard form and expanded form.

Big Ideas Math Answers 4th Grade Chapter 1 Place Value Concepts 1.2 2
Standard form:

Word form: fifty-four thousand, two

Expanded form: ____ + ______ + ______

Answer: Standard Form: 54,002

Expanded Form: 50,000 + 4,000 + 0 + 0 +2

Explanation:

We can write the given number in 3 forms. They are:

A) Standard Form  B) Word Form  C) Expanded Form

Now, the given number which is fifty-four thousand, two is in Word Form. So, we have to write the given number in the remaining 2 forms.

Show and Grow

Write the number in two other forms.

Question 1. 1

Standard form:

Answer: 38,650

Explanation:

We can write the given number in 3 forms. They are:

A) Standard Form  B) Word Form  C) Expanded Form

Now, the given number which is 38,650 is in Standard Form. So, we have to write the given number in the remaining 2 forms.

Word form:

Answer: Thirty-eight thousand, six hundred and fifty

Explanation:

We can write the given number in 3 forms. They are:

A) Standard Form  B) Word Form  C) Expanded Form

Now, the given number which is 38,650 is in Standard Form. So, we have to write the given number in the remaining 2 forms.

Expanded form:
Answer: 30,000 + 8,000 + 600 + 50 + 0

Explanation:

We can write the given number in 3 forms. They are:

A) Standard Form  B) Word Form  C) Expanded Form

Now, the given number which is 38,650 is in Standard Form. So, we have to write the given number in the remaining 2 forms.

Question 2.

Standard form:

Answer: 105,098

Explanation:

We can write the given number in 3 forms. They are:

A) Standard Form  B) Word Form  C) Expanded Form

Now, the given number which is 105,098 is in Standard Form. So, we have to write the given number in the remaining 2 forms.

Word form: one hundred five thousand, ninety-eight

Explanation:

We can write the given number in 3 forms. They are:

A) Standard Form  B) Word Form  C) Expanded Form

Now, the given number which is 105,098 is in Standard Form. So, we have to write the given number in the remaining 2 forms.

Expanded form:

Answer: 100,000 + 0 + 5,000 + 0 + 90 + 8

Explanation:

We can write the given number in 3 forms. They are:

A) Standard Form  B) Word Form  C) Expanded Form

Now, the given number which is 105,098 is in Standard Form. So, we have to write the given number in the remaining 2 forms.

Apply and Grow: Practice

Write the number in two other forms.

Question 3.

Standard form:

Answer: 642,050

Explanation:

We can write the given number in 3 forms. They are:

A) Standard Form  B) Word Form  C) Expanded Form

Now, the given number which is 642,050 is in Standard Form. So, we have to write the given number in the remaining 2 forms.

Word form:

Answer: Six hundred forty- two thousand and fifty

Explanation:

We can write the given number in 3 forms. They are:

A) Standard Form  B) Word Form  C) Expanded Form

Now, the given number which is 642,050 is in Standard Form. So, we have to write the given number in the remaining 2 forms.

Expanded form:

Answer: 600,000 + 40,000 + 2,000 + 50

Explanation:

We can write the given number in 3 forms. They are:

A) Standard Form  B) Word Form  C) Expanded Form

Now, the given number which is 642,050 is in Standard Form. So, we have to write the given number in the remaining 2 forms.

Question 4.

Standard form:  134,078

Word form:

Answer: One hundred thirty-four thousand and seventy-eight

Explanation:

We can write the given number in 3 forms. They are:

A) Standard Form  B) Word Form  C) Expanded Form

Now, the given number which is 134,078 is in Standard Form. So, we have to write the given number in the remaining 2 forms.

Expanded form:

Answer:100,000 + 30,000 + 4,000 + 0 + 70 + 8

Explanation:

We can write the given number in 3 forms. They are:

A) Standard Form  B) Word Form  C) Expanded Form

Now, the given number which is 134,078 is in Standard Form. So, we have to write the given number in the remaining 2 forms.

Question 5.

Complete the table.

Big Ideas Math Answers 4th Grade Chapter 1 Place Value Concepts 1.2 3

Answer:

A) Given Word Form is: three thousand, four hundred ninety-seven

Standard Form: 3,497

Expanded Form: 3,000 + 400 + 90 + 7

Explanation:

We can write the given number in 3 forms. They are:

A) Standard Form  B) Word Form  C) Expanded Form

Now, the given number which is three thousand, four hundred ninety-seven is in Word Form. So, we have to write the given number in the remaining 2 forms.

B) Given Expanded Form is: 50,000 + 2,000 + 400 + 80

Standard Form: 52,480

Word Form: Fifty-two thousand four hundred eighty

Explanation:

We can write the given number in 3 forms. They are:

A) Standard Form  B) Word Form  C) Expanded Form

Now, the given number which is 50,000 + 2,000 + 400 + 80 is in Expanded Form. So, we have to write the given number in the remaining 2 forms.

C) Given Standard Form is:  610,010

Word Form: Six hundred ten thousand, ten

Expanded Form: 600,000 + 10,000 + 0 + 10 + 0

Explanation:

We can write the given number in 3 forms. They are:

A) Standard Form  B) Word Form  C) Expanded Form

Now, the given number which is 610,010 is in Standard Form. So, we have to write the given number in the remaining 2 forms.

Question 6.

YOU BE THE TEACHER

The fangtooth fish lives about 6,500 feet underwater. Newton reads this number as “six thousand five hundred.”Descartes reads the number as “sixty-five hundred.”Did they both read the number correctly? Explain.

Big Ideas Math Answers 4th Grade Chapter 1 Place Value Concepts 1.2 4

Answer: Yes

Explanation:

The given number is “6,500”. Newton read it as “Six thousand five hundred” and Descartes read it as “Sixty-five hundred” because “One thousand is equal to 100 hundred”.

Question 7.

Which One Doesn’tBelong?

Which one not does belong with the other three?

eight hundred sixteen thousand, nine

800,000 + 10,000 + 6,000 + 900

816,009

Big Ideas Math Answers 4th Grade Chapter 1 Place Value Concepts 1.2 5

Answer: The “Expanded Form” does not belong among the given three.

Explanation:

We can write any number in 3 Forms. They are;

A) Standard Form  B) Word Form  C) Expanded Form

Now, the given “period Table” can read as “816,009” and ‘9’ has the place-value of 1 but in the Expanded Form, ‘9’ has a place-value of 100.

Think and Grow: Modeling Real Life

Example

Morse code is a code in which numbers and letters are represented by a series of dots and dashes. Use the table to write the number in standard form, word form, and expanded form.

Big Ideas Math Answers 4th Grade Chapter 1 Place Value Concepts 1.2 6

Big Ideas Math Answers 4th Grade Chapter 1 Place Value Concepts 1.2 7
Standard form:

Word form:

Expanded form:

Show and Grow

Question 8.

Use the table above to write the number in standard form, word form, and expanded form.

Big Ideas Math Answers 4th Grade Chapter 1 Place Value Concepts 1.2 8
Answer:

Question 9.

Use the number 20,000 + 9,000 + 400 + 50 to complete the check.

Big Ideas Math Answers 4th Grade Chapter 1 Place Value Concepts 1.2 9

Answer:

Standard Form: 29,450

Word Form: Twenty-nine thousand four hundred fifty

Explanation:

We can write any number in 3 Forms. They are;

A) Standard Form  B) Word Form  C) Expanded Form

Now, the given number is 20,000 + 9,000 + 400 + 50 which is in Expanded Form. So, we have to write the given number in the remaining 2 forms.

Read and Write Multi-Digit Numbers Homework & Practice 1.2

Write the number in two other forms.

Question 1.

Standard form:

Word form:

Expanded form: 500,000 + 40,000 + 3,000 + 200 + 90 + 8

Answer: Standard Form: 543,298

Word Form: Five hundred forty-three thousand, two hundred Ninety-eight

Explanation:

We can write any number in 3 Forms. They are;

A) Standard Form  B) Word Form  C) Expanded Form

Now, the given number is 500,000 + 40,000 + 3,000 + 200 + 90 + 8  which is in Expanded Form. So, we have to write the given number in the remaining 2 forms.

Question 2.

Standard form:

Word form: forty-eight thousand, six hundred three

Expanded form:

Answer: Standard Form: 48,603

Expanded Form: 40,000 + 8,000 + 600 + 0 + 3

Explanation:

We can write any number in 3 Forms. They are;

A) Standard Form  B) Word Form  C) Expanded Form

Now, the given number is forty-eight thousand, six hundred three which is in Word Form. So, we have to write the given number in the remaining 2 forms.

Question 3.

Complete the table.

Big Ideas Math Answers 4th Grade Chapter 1 Place Value Concepts 1.2 10

Answer:

A) 9,629

Word Form: Nine thousand, six hundred and twenty-nine

Expanded Form: 9,000 + 600 + 20 + 9

Explanation:

We can write any number in 3 Forms. They are;

A) Standard Form  B) Word Form  C) Expanded Form

Now, the given number is 9,629 which is in Standard Form. So, we have to write the given number in the remaining 2 forms.

B) 30,000 + 7,000 + 800 + 2

Standard Form: 37,802

Word Form: Thirty-seven thousand, Eight hundred two

Explanation:

We can write any number in 3 Forms. They are;

A) Standard Form  B) Word Form  C) Expanded Form

Now, the given number is 30,000 + 7,000 + 800 + 2 which is in Expanded Form. So, we have to write the given number in the remaining 2 forms.

C) four hundred sixteen thousand, eighty-seven

Standard Form: 416,087

Expanded Form: 400,000 + 16,000 + 0 + 80 + 7

Explanation:

We can write any number in 3 Forms. They are;

A) Standard Form  B) Word Form  C) Expanded Form

Now, the given number is four hundred sixteen thousand, eighty-seven which is in Word Form. So, we have to write the given number in the remaining 2 forms.

Question 4.

Reasoning

Your teacher asks the class to write forty-two thousand, ninety-three in standard form. Which student wrote the correct number? What mistake did the other student make?

Big Ideas Math Answers 4th Grade Chapter 1 Place Value Concepts 1.2 11

Answer: Student B

Explanation: The teacher asked Student A and B to write “forty-two thousand, ninety-three” in Standard form. That means ‘9’ must be in the tens place. But, Student A put the ‘9’ in the hundreds place whereas Student B put the ‘9’ in the tens place.

Question 5.

Logic

The number has two periods. The thousands period is written as six hundred eight thousand in word form. The one period is written as 600 + 80 in expanded form. What is the number?

Answer: 608,680

Explanation: Given that the number has 2 periods. They are 1) The thousands period   2) The hundreds period

The given Thousands period is: Six hundred eight thousand ( Word Form)

The given Hundreds period is: 600 + 80 ( Expanded Form)

On combining the 2 periods, we get the number “608,680” in the Standard Form.

Question 6.

Modeling Real Life

Use the table to write the number in standard form, word form, and expanded form.

Big Ideas Math Answers 4th Grade Chapter 1 Place Value Concepts 1.2 12

Big Ideas Math Answers 4th Grade Chapter 1 Place Value Concepts 1.2 13

Answer:

Question 7.

Modeling Real Life

Use the number 3,000 + 70 + 1 to complete the check.

Big Ideas Math Answers 4th Grade Chapter 1 Place Value Concepts 1.2 14

Answer: Standard Form: 3,071

Word Form: Three thousand, Seventy-one

Explanation:

We can write any number in 3 Forms. They are;

A) Standard Form  B) Word Form  C) Expanded Form

Now, the given number is 3,000 + 70 + 1 which is in Expanded Form. So, we have to write the given number in the remaining 2 forms.

Review & Refresh

Compare

Question 8.

Big Ideas Math Answers 4th Grade Chapter 1 Place Value Concepts 1.2 15

Answer: \(\frac{1}{6} \) is less than \(\frac{2}{6} \)

Explanation:

\(\frac{1}{6} \) = 0.1666

\(\frac{2}{6} \) = 0.333

Hence, \(\frac{1}{6} \) is less than \(\frac{2}{6} \)

Question 9.

Big Ideas Math Answers 4th Grade Chapter 1 Place Value Concepts 1.2 16

Answer: \(\frac{2}{2} \) is greater than \(\frac{2}{3} \)

Explanation:

\(\frac{2}{2} \) = 1

\(\frac{2}{3} \) = 0.666

Hence, \(\frac{2}{2} \) is greater than \(\frac{2}{3} \)

Question 10.

Big Ideas Math Answers 4th Grade Chapter 1 Place Value Concepts 1.2 17

Answer: \(\frac{1}{2} \) is less than \(\frac{3}{4} \)

Explanation:

\(\frac{1}{2} \) = 0.5

\(\frac{3}{4} \) = 0.75

Hence, \(\frac{1}{2} \) is less than \(\frac{3}{4} \)

Question 11.

Big Ideas Math Answers 4th Grade Chapter 1 Place Value Concepts 1.2 18

Answer: \(\frac{1}{4} \) is equal to \(\frac{2}{8} \)

Explanation:

\(\frac{1}{4} \) = 0.25

\(\frac{2}{8} \) = 0.25

Hence, \(\frac{1}{4} \) is equal to \(\frac{2}{8} \)

Lesson 1.3 Compare Multi-Digit Numbers

Explore and Grow

Goal: Make the greatest number possible.

Draw a Number Card. Choose a place value for the digit. Write the digit in the place value chart. Continue until the place value chart is complete.

Big Ideas Math Answers Grade 4 Chapter 1 Place Value Concepts 1.3 1

Compare your number with your partner’s. Whose number is greater?

Answer: Let me consider my number as “A” and my partner number as “B”

Let,

A = 999999 and B = 999998

From  the 2 numbers, we can conclude that

A > B

Construct Arguments

Explain your strategy to your partner. Compare your strategies.

Answer:

Think and Grow: Compare Multi-Digit Numbers

Example

Compare 8,465 and 8,439.

Use a place value chart.

Big Ideas Math Answers Grade 4 Chapter 1 Place Value Concepts 1.3 2

Start at the left. Compare the digits in each place until the digits differ.

Big Ideas Math Answers Grade 4 Chapter 1 Place Value Concepts 1.3 3

Answer:

Step 1: 8 thousand = 8 thousand

Step 2: 4 hundred = 4 hundred

Step 3: 6 tens > 3 tens

So,

8,465 > 8,439

Explanation:

The given 2 numbers are 8,465 and 8,439

By comparing the 2 numbers, we can conclude that

8,465 > 8,439

Show and Grow

Write which place to use when comparing the numbers.

Question 1.

2,423

2,324

Answer: 2,423 is greater than 2,324

Explanation:

The given 2 numbers are 2,423 and 2,324

From these 2 numbers, we can conclude that 2,423 is greater than 2,324.

Question 2.

9,631

9,637

Answer: 9,631 is less than 9,637

Explanation:

The given 2 numbers are 9,631 and 9,637.

From these 2 numbers, we can conclude that 9,631 is less than 9,637

Question 3.

15,728

16,728

Answer: 15,728 is less than 16,728

Explanation:

The given 2 numbers are 15,728 and16,728.

From these 2 numbers, we can conclude that 15,728 is less than 16,728

Compare

Question 4.

Big Ideas Math Answers Grade 4 Chapter 1 Place Value Concepts 1.3 4

Answer: 5,049 is less than 5,082

Explanation:

The given 2 numbers are 5,049 and 5,082.

From these 2 numbers, we can conclude that5,049 is less than 5,082

Question 5.

Big Ideas Math Answers Grade 4 Chapter 1 Place Value Concepts 1.3 5

Answer: 735,283 is equal to 735,283

Explanation:

The given 2 numbers are 735,283 and 735,283.

From these 2 numbers, we can conclude that 735,283 is equal to 735,283

Question 6.

Big Ideas Math Answers Grade 4 Chapter 1 Place Value Concepts 1.3 6

Answer: 43,694 is greater than 3,694

Explanation:

The given 2 numbers are 43,694 and 3,694.

From these 2 numbers, we can conclude that 43,694 is greater than 3,694

Question 7.

Big Ideas Math Answers Grade 4 Chapter 1 Place Value Concepts 1.3 7

Answer: 88,195 is greater than 78,195

Explanation:

The given 2 numbers are88,195 and 78,195.

From these 2 numbers, we can conclude that 88,195 is greater than 78,195

Question 8.

Big Ideas Math Answers Grade 4 Chapter 1 Place Value Concepts 1.3 8

Answer: 6,480 is less than 6,508

Explanation:

The given 2 numbers are 6,480 and 6,508.

From these 2 numbers, we can conclude that 6,480 is less than 6,508

Question 9.

Big Ideas Math Answers Grade 4 Chapter 1 Place Value Concepts 1.3 9

Answer: 321,817 is greater than 312,827

Explanation:

The given 2 numbers are 321,817 and 312,827.

From these 2 numbers, we can conclude that 321,817 is greater than 312,827

Apply and Grow: Practice

Compare.

Question 10.

Big Ideas Math Answers Grade 4 Chapter 1 Place Value Concepts 1.3 10
Answer: 6,052 is less than 6,520

Explanation:

The given 2 numbers are 6,052 and 6,520.

From these 2 numbers, we can conclude that 6,052 is less than 6,520

Question 11.

Big Ideas Math Answers Grade 4 Chapter 1 Place Value Concepts 1.3 11
Answer: 891,634 is greater than 871,634

Explanation:

The given 2 numbers are 891,634 and 871,634

From these 2 numbers, we can conclude that 891,634 is greater than 871,634

Question 12.

Big Ideas Math Answers Grade 4 Chapter 1 Place Value Concepts 1.3 12

Answer: 28,251 is greater than 26,660

Explanation:

The given 2 numbers are28,251 and 26,660

From these 2 numbers, we can conclude that 28,251 is greater than 26,660

Question 13.

Big Ideas Math Answers Grade 4 Chapter 1 Place Value Concepts 1.3 13
Answer: 324,581 is greater than 32,458

Explanation:

The given 2 numbers are 324,581 and 32,458

From these 2 numbers, we can conclude that 324,581 is greater than 32,458

Question 14.

Big Ideas Math Answers Grade 4 Chapter 1 Place Value Concepts 1.3 14
Answer: 230,611 is greater than 230,610

Explanation:

The given 2 numbers are 230,611 and 230,610

From these 2 numbers, we can conclude that 230,611 is greater than 230,610

Question 15.

Big Ideas Math Answers Grade 4 Chapter 1 Place Value Concepts 1.3 15
Answer: 909,900 is greater than 909,009

Explanation:

The given 2 numbers are 909,900 and 909,009

From these 2 numbers, we can conclude that 909,900 is greater than 909,009

Question 16.

Big Ideas Math Answers Grade 4 Chapter 1 Place Value Concepts 1.3 16
Answer: 7,134 is less than 7,634

Explanation:

The given Expanded Form is: 7,000 + 100 + 30 + 4 = 7,134

The given 2 numbers are 7,134 and 7,634

From these 2 numbers, we can conclude that 7,134 is less than 7,634

Question 17.

Big Ideas Math Answers Grade 4 Chapter 1 Place Value Concepts 1.3 17
Answer: 100,003 is greater than 10,003

Explanation:

The given Word Form is: ten thousand, three = 10,003

The given 2 numbers are 100,003 and 10,003

From these 2 numbers, we can conclude that 100,003 is greater than 10,003

Question 18.

Big Ideas Math Answers Grade 4 Chapter 1 Place Value Concepts 1.3 18
Answer: 16,409 is less than 16,490

Explanation:

The given word Form is: Sixteen thousand, four hundred nine = 16,409

The given 2 numbers are 16,409 and16,490

From these 2 numbers, we can conclude that 16,409 is less than 16,490

Question 19.

Big Ideas Math Answers Grade 4 Chapter 1 Place Value Concepts 1.3 19
Answer: 461,329 is equal to 461,329

Explanation:

The given Expanded Form is: 400,000 + 60,000 + 1,000 + 300 + 20 + 9 =461,329

The given 2 numbers are 461,329 and 461,329

From these 2 numbers, we can conclude that 461,329 is equal to 461,329

Question 20.

Two brands of televisions cost $1,598 and $1,998. Which is the lesser price?

Big Ideas Math Answers Grade 4 Chapter 1 Place Value Concepts 1.3 20

Answer: The lesser price is: $1,598

Explanation:

The given costs of two brands of Television are $1,598 and $1,998

From the two costs of Television, we can conclude that the lesser price is: $1,598

Question 21.

DIG DEEPER!

Your friend says she can tell which sum is greater without adding the numbers. How can she tell?

34,593 + 6,781

34,593 + 6,609

Answer: 6,781 is greater tha 6,609

Explanation:

The given sums are:

34,593 + 6,781

34,593 + 6,609

From these two sums, we can see that 6,781 is greater tha 6,609

Question 22.

Number Sense

Write all of the digits that make the number greater than 23,489 and less than 26,472.

2?,650

Answer: 3 and 4

Explanation:

The given numbers are: 23,489 and 26,472

The digits that make the number greater than 23,489 and less than 26,472 are: 3 and 4

Question 23.

YOU BE THE TEACHER

Your friend says 38,675 is less than 9,100 because 3 is less than 9. Is your friend correct? Explain.

Answer: No

Explanation:

The given numbers are 38,675 and 9,100

In 38,675, the highest place-value is: 30,000

In 9,100, the highest place-value is: 9,000

From these, we can conclude that 38,675 is greater than 9,100

Think and Grow: Modeling Real Life

Example

Which playground costs the least?

Write each number in a place value chart.

Big Ideas Math Answers Grade 4 Chapter 1 Place Value Concepts 1.3 21

Big Ideas Math Answers Grade 4 Chapter 1 Place Value Concepts 1.3 22

Answer:

The playground which costs the least is: Green

Explanation: From the given table, we can conclude that “Green” has the least cost which is $35,872

Order the numbers from least to greatest.
_______, ______, _______

Answer: Green, Yellow, Blue

Explanation:

Compare the costs of 3 playgrounds

Green: $35,872    Yellow: $36,827  Blue: $36,927

From these costs, we can conclude that the order is: Green, Yellow, Blue

The ______ playground costs the least.

Answer: Green

Explanation:

Compare the costs of 3 playgrounds

Green: $35,872    Yellow: $36,827  Blue: $36,927

From these costs, we can conclude that the “Green Playground” has the least cost.

Show and Grow

Question 24.

Who received the highest score? Did anyone beat the high score?

Big Ideas Math Answers Grade 4 Chapter 1 Place Value Concepts 1.3 23

Big Ideas Math Answers Grade 4 Chapter 1 Place Value Concepts 1.3 24

Answer: You

Explanation:

The given high score is: 252,980

The score of You is: 254,020

From the given scores, we can conclude that “You” scored the highest score than the given high score.

Question 25.

Name two cities that have a greater population than St. Louis. Name two cities that do not have a greater population than Oakland.

Big Ideas Math Answers Grade 4 Chapter 1 Place Value Concepts 1.3 25

Answer:

The two cities that have a greater population than St. Louis: Tulsa, OK; Cleveland, OH; Oakland, CA

The two cities that do not have a greater population than Oakland: St. Louis, MO; Anchorage, AK

Compare Multi-Digit Numbers Homework & Practice 1.3

Write which place to use when comparing the numbers

Question 1.

396,241

386,201

Answer: Ten thousand’s  place

Explanation: The given numbers are: 396,241 and 386,201

When we want to compare the numbers directly, compare Ten thousand’s place to get the answer.

Question 2.

50,718

50,798

Answer: Tens place

Explanation: The given numbers are: 50,718 and 50,798

When we want to compare the numbers directly, compare Tens place to get the answer.

Question 3.

159,624

459,623

Answer: Hundred thousand place

Explanation: The given numbers are: 159,624 and 459,623

When we want to compare the numbers directly, compare a Hundred thousand place to get the answer.

Compare

Question 4.

Big Ideas Math Answers Grade 4 Chapter 1 Place Value Concepts 1.3 26

Answer: 7,902 is less than 7,912

Explanation:

The given 2 numbers are 7,902 and 7,912

From these 2 numbers, we can conclude that 7,902 is less than 7,912

Question 5.

Big Ideas Math Answers Grade 4 Chapter 1 Place Value Concepts 1.3 27

Answer: 63,871 is less than 63,902

Explanation:

The given 2 numbers are63,871 and 63,902

From these 2 numbers, we can conclude that 63,871 is less than 63,902

Question 6.

Big Ideas Math Answers Grade 4 Chapter 1 Place Value Concepts 1.3 28

Answer: 839,304 is greater than 829,001

Explanation:

The given 2 numbers are 839,304 and 829,001

From these 2 numbers, we can conclude that 839,304 is greater than 829,001

Question 7.

Big Ideas Math Answers Grade 4 Chapter 1 Place Value Concepts 1.3 29

Answer: 48,931 is greater than 48,930

Explanation:

The given 2 numbers are 48,931 and 48,930

From these 2 numbers, we can conclude that 48,931 is greater than 48,930

Question 8.

Big Ideas Math Answers Grade 4 Chapter 1 Place Value Concepts 1.3 30

Answer: 5,739 is equal to 5,739

Explanation:

The given 2 numbers are 5,739 and 5,739

From these 2 numbers, we can conclude that5,739 is equal to 5,739

Question 9.

Big Ideas Math Answers Grade 4 Chapter 1 Place Value Concepts 1.3 31

Answer: 10,000 is less than 100,000

Explanation:

The given 2 numbers are 10,000 and 100,000

From these 2 numbers, we can conclude that 10,000 is less than 100,000

Compare

Question 10.

Big Ideas Math Answers Grade 4 Chapter 1 Place Value Concepts 1.3 32

Answer: 16,420 is greater than 16,402

Explanation:

The given Expanded Form is: 10,000 + 6,000 + 400 + 2 = 16,402

The given 2 numbers are 16,420 and 16,402

From these 2 numbers, we can conclude that 16,420 is greater than 16,402

Question 11.

Big Ideas Math Answers Grade 4 Chapter 1 Place Value Concepts 1.3 33

Answer: 14.000 is less than 114,000

Explanation:

The given Word Form is fourteen thousand = 14,000

The given 2 numbers are 14.000 and 114,000

From these 2 numbers, we can conclude that 14.000 is less than 114,000

Question 12.

Big Ideas Math Answers Grade 4 Chapter 1 Place Value Concepts 1.3 34

Answer: 800,095 is less than 800,695

Explanation:

The given Word Form is eight hundred thousand, ninety-five = 800,095

The given 2 numbers are 800,095 and 800,695

From these 2 numbers, we can conclude that 800,095 is less than 800,695

Question 13.

Big Ideas Math Answers Grade 4 Chapter 1 Place Value Concepts 1.3 35

Answer: 509,733 is equal to 509,733

Explanation:

The given Expanded Form is 500,000 + 9,000 + 700 + 30 + 3 = 509,733

The given 2 numbers are 509,733 and509,733

From these 2 numbers, we can conclude that 509,733 is equal to 509,733

Question 14.

Two different laptops cost $1,050 and$1,150. Which is the lesser price?

Big Ideas Math Answers Grade 4 Chapter 1 Place Value Concepts 1.3 36

Answer: $1,050 is the lesser price.

Explanation:

The cost of two different Laptops are: $1,050 and $1,150

When compare $1,050 and $1,150, we can conclude that $1,050 is the lesser price.

Question 15.

DIG DEEPER!

If the leftmost digits of two multi-digit numbers are 6and 9, can you tell which number is greater? Explain.

Answer: 9 is a greater number

Explanation:

The given leftmost digits of two numbers are 6 and 9.

When we compare 6 and 9, we can conclude that 9 is the greater number.

Question 16.

Writing

Explain why you start comparing digits on the left when comparing multi-digit numbers.

Answer: When we compare multi-digit numbers, we start comparing digits on the leftmost side because those numbers’ positions have the highest place-value.

Question 17.

Modeling Real Life

Use the table to answer the questions.

Name two museums that had a yearly attendance greater than Children’sMuseum of Denver. Name two museums

that had a yearly attendance less than Discovery Place.

Big Ideas Math Answers Grade 4 Chapter 1 Place Value Concepts 1.3 37

Answer:

The museums that had a yearly attendance greater than the Children’s Museum of Denver are:

A) Boston Children’s Museum  B) Children’s Museum of Houston  C) Discovery palace  D) Exploratorium

The museums that had a yearly attendance less than Discovery Place are:

A) Boston Children’s Museum  B) Children’s Museum of Denver  C) Please Touch Museum

Review & Refresh

Round.

Question 18.

654

Nearest ten: _____

Nearest hundred: _____

Answer: 50, 600

Explanation:

The given number is: 654

When we round off the tens place, the nearest ten is: 50

When we round off the hundreds place, the nearest hundred is: 600

Question 19.

709

Nearest ten: _____

Nearest hundred: ______

Answer: 0, 700

Explanation:

The given number is: 709

When we round off the tens place, the nearest ten is: 0 ( Since the tens place is 0)

When we round off the hundreds place, the nearest hundred is: 700

Question 20.

25

Nearest ten: _____

Nearest hundred: _______

Answer: 20, 0

Explanation:

The given number is: 25

When we round off the tens place, the nearest ten is: 20

When we round off the hundreds place, the nearest hundred is: 0 ( Since the hundreds place is 0)

Lesson 1.4 Round Multi-Digit Numbers

Explore and Grow

Write five numbers that round to 250 when rounded to the nearest ten.

Big Ideas Math Solutions Grade 4 Chapter 1 Place Value Concepts 1.4 1

Answer: 210, 220, 230, 240, 250

Explanation:

The five numbers that round to 250 when rounded to the nearest ten are: 210, 220, 230, 240, 250

 

Write five numbers that round to 500 when rounded to the nearest hundred.

Big Ideas Math Solutions Grade 4 Chapter 1 Place Value Concepts 1.4 2

Answer: 100, 200, 300, 400, 500

Explanation:

The five numbers that round to 500 when rounded to the nearest hundred are: 100, 200, 300, 400, 500

Repeated Reasoning

Explain your strategy. Then explain how you could use the strategy to round a four-digit number to the nearest thousand.

Answer:

Think and Grow: Round Multi-Digit Numbers

To round a number, find the multiple of 10, 100, 1,000, and so on, that is closest to the number. You can use a number line or place value to round numbers.

Example

Use a number line to round 4,276 to the nearest thousand.
Big Ideas Math Solutions Grade 4 Chapter 1 Place Value Concepts 1.4 3

4,276 is closer to 4,000 than it is to 5,000.

So, 4,276 rounded to the nearest thousand is _______.

Answer: 4,000

Explanation: The number 4276 can be round off to 4,000 when rounded to the nearest thousand.

Example

Use place value to round 385,617 to the nearest ten thousand.

Big Ideas Math Solutions Grade 4 Chapter 1 Place Value Concepts 1.4 4

So, 385,617 rounded to the nearest thousand is _______.

Answer: 390,000

Explanation:

The number 385,617 can be round off to 390,000 when rounded to the nearest thousand. ( Since the thousands position is greater than 5)

Show and Grow

Round the number to the place of the underlined digit.

Question 1.

6,912

Answer: 7000

Explanation:

The number 6,912 can be round off to 7,000 when rounded to the nearest thousand. ( Since the thousands position is greater than 5)

Question 2.

43,215

Answer: 43,000

Explanation:

The number 43,215 can be round off to 43,000 when rounded to the nearest thousand. ( Since the thousands position is less than 5)

Question 3.

25,883

Answer: 25,000

Explanation:

The number 25,883 can be round off to 25000 when rounded to the nearest thousand. ( Since the thousands position is less than 5)

Question 4.

148,796

Answer: 140,000

Explanation:

The number 148,796 can be round off to 140000 when rounded to the nearest thousand. ( Since the thousands position is less than 5)

Question 5.

Round 5,379 to the nearest thousand.

Answer: 5,000

Explanation:

The number 5,379 can be round off to 5,000 when rounded to the nearest thousand. ( Since the thousands position is equal to 5)

Question 6.

Round 70,628 to the nearest ten thousand.

Answer: 70,000

Explanation:

The number 70,628 can be round off to 70,000 when rounded to the nearest ten thousand.

Question 7.

Round 362,113 to the nearest hundred thousand.

Answer:360,000

Explanation:

The number 362,113 can be round off to 360000 when rounded to the nearest hundred thousand.

Question 8.

Round 982,638 to the nearest thousand.

Answer:980,000

Explanation:

The number 982,638 can be round off to 980,000 when rounded to the nearest thousand. ( Since the thousands position is less than 5)

Apply and Grow: Practice

Round the number to the place of the underlined digit.

Question 9.

3,641

Answer: 3,700

Explanation:

The number 3,641 can be round off to 3700 when rounded to the nearest hundred. ( Since the hundreds position is greater than 5)

Question 10.

17,139

Answer: 18,000

Explanation:

The number 17,139can be round off to 18000 when rounded to the nearest thousand. ( Since the thousands position is greater than 5)

Question 11.

426,384

Answer: 426,300

Explanation:

The number 426,384 can be round off to 426300 when rounded to the nearest hundred ( Since the hundreds position is greater than 5)

Question 12.

542,930

Answer: 542,000

Explanation:

The number 542,930 can be round off to 542,000 when rounded to the nearest thousand. ( Since the thousands position is greater than 5)

Round the number to the nearest thousand.

Question 13.

9,426

Answer: 9,000

Explanation:

The number 9,426 can be round off to 9000 when rounded to the nearest thousand.

Question 14.

57,496

Answer: 58,000

Explanation:

The number 57,496 can be round off to 58000 when rounded to the nearest thousand. ( Since the thousands position is greater than 5)

Question 15.

360,491

Answer: 360,000

Explanation:

The number 360,491 can be round off to 360000 when rounded to the nearest thousand.

Question 16.

824,137

Answer: 824,000

Explanation:

The number 824,137 can be round off to 824,000 when rounded to the nearest thousand. ( Since the thousands position is less than 5)

Round the number to the nearest hundred thousand.

Question 17.

226,568

Answer: 200,000

Explanation:

The number 226,568 can be round off to 200,000 when rounded to the nearest hundred thousand. ( Since the hundred thousand position is less than 5)

Question 18.

457,724

Answer: 400,000

Explanation:

The number 457,724 can be round off to 400,000 when rounded to the nearest hundred thousand. ( Since the hundred thousand  position is less than 5)

Question 19.

108,665

Answer: 100,000

Explanation:

The number 108,665 can be round off to 100,000 when rounded to the nearest hundred thousand. ( Since the hundred thousand  position is less than 5)

Question 20.

75,291

Answer:

Question 21.

Number Sense

Write a five-digit number that has the digits 6, 0, 4, 2, and 8 and rounds to 70,000 when rounded to the nearest ten thousand.

Answer: 68,420

Explanation:

The given 5-digits are: 6, 0, 4, 2 and 8

The 5-git number which can be formed by using the given 5 digits are 68,420 and 68,420 can be round off to 70,000 when rounded to the nearest ten thousand.

Question 22.

Number Sense

When finding the United States census, should you round or find an exact answer? Explain.

Answer:

Question 23.

Open-Ended

A lightning strike can reach a temperature of about 54,000 degrees Fahrenheit. Write four possible temperatures for a lightning strike.

Answer:

Think and Grow: Modeling Real Life

Example

When the results are rounded to the nearest ten thousand, which token received about 160,000 votes?

Big Ideas Math Solutions Grade 4 Chapter 1 Place Value Concepts 1.4 5

Answer:

When the results are rounded to the nearest ten thousand, the tokens which received 1,60,000 votes are:

A) Race Car  B) Rubber Duck

Round each number to the nearest ten thousand.

Big Ideas Math Solutions Grade 4 Chapter 1 Place Value Concepts 1.4 6

Answer:

A) 134,704 : 130,000  B) 154,165 : 150,000  C) 207,954 : 208,000  D) 212,476 : 212,000

E) 167,582 : 170,000  F) 146,661 : 150,000  G) 165,083 : 170,000 H) 160,485 : 160,000

So, the _______ received about 160,000 votes.

Answer:

A) Race Car  B) Rubber Duck

Show and Grow

Use the table above.

Question 24.

When the results are rounded to the nearest ten thousand, which tokens received about 150,000 votes?

Answer:

A) Cat  B)  Penguin

Explanation:

From the above table, we can conclude that Cat and the Penguin have 160,000 votes when rounded to the nearest ten thousand.

Question 25.

When the results are rounded to the nearest hundred thousand, which tokens received about 100,000 votes?

Answer:

A) Cat  B) Penguin

Explanation:

From the above table, we can conclude that Cat and the Penguin have 160,000 votes when rounded to the nearest hundred thousand

Question 26.

DIG DEEPER!

To which place should you round each number of votes so that you can order the numbers from greatest to least? Explain.

Answer:

Question 27.
A car battery should be replaced when the odometer shows about 50,000 miles. None of the cars below have had a battery replacement. Which cars might need a new battery?
Big Ideas Math Solutions Grade 4 Chapter 1 Place Value Concepts 1.4 7
Answer: ODO

Explanation:

From the given Fig.,

we can conclude that ODO which has above 50,000 miles needs a battery replacement.

Round Multi-Digit Numbers Homework & Practice 1.4

 

Round the number to the place of the underlined digit.

Question 1.

6,251

Answer: 6,250

Explanation:

The given number is 6,251 and the underlined digit is 5.

Since the underlined digit is equal to 5, we can conclude that the given number can round off to 6,250

Question 2.

8,962

Answer: 9,000

Explanation:

The given number is 8,962 and the underlined digit is 8.

Since the underlined digit is greater than 5, we can conclude that the given number can round off to 9,000

Question 3.

3,951

Answer: 3,950

Explanation:

The given number is 3,951 and the underlined digit is 9.

we can conclude that the given number can round off to 3,950

Question 4.

79,064

Answer: 80,000

Explanation:

The given number is 79,064 and the underlined digit is 9.

Since the underlined digit is greater than 5, we can conclude that the given number can round off to 80,000

Question 5.

43,976

Answer: 43,000

Explanation:

The given number is 43,976 and the underlined digit is 3.

Since the underlined digit is greater than 5, we can conclude that the given number can round off to 43,000

Question 6.

24,680

Answer: 24,000

Explanation:

The given number is 24,680 and the underlined digit is 2.

Since the underlined digit is less than 5, we can conclude that the given number can round off to 24,000

Question 7.

726,174

Answer: 800,000

Explanation:

The given number is 726,174 and the underlined digit is 7.

Since the underlined digit is greater than 5, we can conclude that the given number can round off to 800,000

Question 8.

138,691

Answer: 140,000

Explanation:

The given number is 138,691 and the underlined digit is 8.

Since the underlined digit is greater than  5, we can conclude that the given number can round off to 140,000

Round the number to the nearest thousand.

Question 9.

5,517

Answer: 5,000

Explanation:

The given number is 5,517. Since the thousands position is equal to 5, we can round it off to 5,000

Question 10.

70,628

Answer: 70,000

Explanation:

The given number is 70,628. Since the thousands position is less than 5, we can round it off to 70,000

Round the number to the nearest ten thousand.

Question 11.

114,782

Answer: 110,000

Explanation:

The given number is 114,782. Since the ten thousand’s  position is less than 5, we can round it off to 110,000

Question 12.

253,490

Answer: 250,000

Explanation:

The given number is 253,490. Since the ten  thousand’s position is equal to 5, we can round it off to 250,000

Round the number to the nearest hundred thousand.

Question 13.

628,496

Answer: 700,000

Explanation:

The given number is 628,496. Since the hundred  thousand’s position is greater than 5, we can round it off to 700,000

Question 14.

90,312

Answer: 90,312

Explanation:

The given number is 90,312. There is no hundred thousand’s position in the given number. So, there is no effect on the given number.

Question 15.

Structure

Round * to the nearest thousand and to the nearest ten thousand.

Big Ideas Math Solutions Grade 4 Chapter 1 Place Value Concepts 1.4 8

Nearest thousand: ______

Nearest ten thousand: ______

Answer: Nearest Thousand: 8000

Nearest ten thousand: 10,000

Explanation:

In the given fig., the * sign is present after 7,500.

From this, we can conclude that

Nearest Thousand: 8000

Nearest ten thousand: 10,000

Question 16.

Number Sense

Which numbers round to 300,000 when rounded to the nearest hundred thousand?

Big Ideas Math Solutions Grade 4 Chapter 1 Place Value Concepts 1.4 9

Answer: 302,586

Explanation:

From the given fig., when we observe the positions of hundred’s  and ten thousand’s positions, it is clear that “302,586” can round off to 300,000

Question 17.

Number Sense

When discussing the price of a laptop, should you round to the nearest thousand or the nearest ten? Explain.

Answer: When you are discussing the price of a laptop, you should round off to the nearest thousand instead of the nearest ten.

Example: Suppose you want to buy a laptop that costs $152,236. When the dealer told you the price of a laptop, you will ask to round off the price of the laptop so that it will be easier for you to buy that laptop.

Question 18.

YOU BE THE TEACHER

Your friend says 5,953 rounds to 5,053 when rounded to the nearest hundred. Is your friend correct? Explain.

Answer: Wrong

Explanation:

The given number is 5,953. When rounded off to the nearest hundred, the given number becomes 6,000 since the hundreds position is greater than 5.

Question 19.

Modeling Real Life

A glassblower is using a furnace to melt glass. When the furnace reaches about 2,000 degrees Fahrenheit, when

rounded to the nearest hundred, she can put the glass in. At which temperatures could she put the glass into the

furnace?

Big Ideas Math Solutions Grade 4 Chapter 1 Place Value Concepts 1.4 10

Big Ideas Math Solutions Grade 4 Chapter 1 Place Value Concepts 1.4 11
Answer: 1925

Explanation:

Given that the furnace reached 2000 degrees Fahrenheit when rounded off to the nearest hundred.

So, the given options are: A) 2005   B) 1899  C) 1925  D) 275  E) 1002

From the given options, it is clear that 1925 degrees can round off to 2000 degrees.

Note: The option 2005 degrees can’t become the answer because when rounding off to the nearest number, the number should be less than the desired number.

Review & Refresh

Find the perimeter of the polygon.

Question 20.

Big Ideas Math Solutions Grade 4 Chapter 1 Place Value Concepts 1.4 12

Perimeter: _____

Answer: 23 in.

Explanation:

The “Perimeter” is defined as the total length of the geometrical figure covered on all sides.

So, the perimeter of the Polygon can be calculated as :

10 in. + 4 in. + 6 in. + 3 in. = 23 in.

Question 21.

Big Ideas Math Solutions Grade 4 Chapter 1 Place Value Concepts 1.4 13

Perimeter: _____

Answer: 26 m

Explanation:

The “Perimeter” is defined as the total length of the geometrical figure covered on all sides.

In a Parallelogram, the parallel sides are equal.

So, the length of all the sides of a parallelogram is: 8 m, 8 m, 5 m, 5 m

So,

The perimeter of the parallelogram can be calculated as:

8 m + 8 m + 5 m + 5 m = 26 m

Question 22

Big Ideas Math Solutions Grade 4 Chapter 1 Place Value Concepts 1.4 14

Perimeter: ______

Answer: 28 ft

Explanation:

The “Perimeter” is defined as the total length of the geometrical figure covered on all sides.

In a Square, the length of all the sides is equal. So, even in the fig., the length of 1 side is given, we can take it as the length of each of the remaining sides.

So, the perimeter of the square can be calculated as:

7 ft + 7 ft + 7ft + 7 ft = 28 ft.

Place Value Concepts Performance Task

You hike from Point A through Point F along the orange path shown on the map.

Big Ideas Math Answer Key Grade 4 Chapter 1 Place Value Concepts 1

Question 1.

What is the distance in elevation between each contour line?

Answer: 100 ft.

Explanation:

When we observe the Contour lines from 2000 ft. to 3000 ft., we can conclude that each Contour line can represent 100 ft.

Question 2.

As you walk from A to C, are you walking uphill or downhill? Explain.

Answer: Downhill

Explanation: When we observe the orange path in the given Contour lines, we can conclude that from A to C, you are going downhill.

Question 3.

Which letter represents the highest point? Estimate the height.

Answer: C

Explanation:

If we observe the Orange path, then we can see that Point C represents the highest point.

Question 4.

A water station has an elevation of 2,763 feet. Which letter represents the location of the water station?

Answer: Point B

Explanation:

From the Orange path, we can see that between 2,500 ft and 3,000 ft., Point is present which can represent the location of the water Station that has an elevation of 2,763 feet.

Question 5.

You take a break when you are at two thousand eighty feet. At which letter do you take a break?

Answer: Point A

Explanation:

Point A is present after 2,000 ft which is the nearest to 2,080 ft. So, Point A takes a break here.

Question 6.

About how much higher do you think Point C is than Point D? Explain.

Answer:

Place Value Concepts Activity

Place Value Plug-In

Directions:

1.Players take turns.

2.On your turn, roll six dice.Arrange the dice into a six-digit number that matches one of the descriptions.

3.Write your number on the lines.

4. The first player to complete all of the numbers wins!

Big Ideas Math Answer Key Grade 4 Chapter 1 Place Value Concepts 2

Place Value Concepts Chapter Practice

1.1 Understand Place Value

Write the value of the underlined digit.

Question 1.

26,490

Answer: 90

Explanation;

The position of each digit in a number is defined and it is the value based on the position of the digit. So, the place value of ‘9’ in 26,490’ is:  90

Question 2.

57,811

Answer: 50,000

Explanation:

The position of each digit in a number is defined and it is the value based on the position of the digit. So, the place value of ‘5’ in ‘57,811’ is:  50,000

Question 3.

308,974

Answer: 8,000

Explanation:

The position of each digit in a number is defined and it is the value based on the position of the digit. So, the place value of ‘8’ in ‘308,974’ is: 8,000

Question 4.

748,612

Answer: 700,000

Explanation:

The position of each digit in a number is defined and it is the value based on the position of the digit. So, the place value of ‘7’ in ‘748,612’ is: 700,000

Compare the values of the underlined digits

Question 5.

94 and 982

Answer: The value of  9 in 982 is 10 times the value of 9 in 94

Explanation:

The value of 9 in 982 is: 900

The value of 9 in 94 is: 90

From these values, we can conclude that the value of  9 in 982 is 10 times the value of 9 in 94

Question 6.

817,953 and 84, 006

Answer: The value of 8 in 817,953 is 10 times the value of 8 in 84,006

Explanation:

The value of 8 in 817,953 is:800,000

The value of 8 in 84,006 is:80,000

From these values, we can conclude that the value of 8 in 817,953 is 10 times the value of 8 in 84,006

1.2 Read and Write Multi-Digit Numbers

Question 7.

Complete the table.

Big Ideas Math Answer Key Grade 4 Chapter 1 Place Value Concepts chp 7

Answer:

A)50,000 + 600 + 90 + 1

Word Form: Fifty thousand, Six hundred Ninety-one

Standard Form: 50,691

Explanation:

We can write any number in 3 Forms. They are;

A) Standard Form  B) Word Form  C) Expanded Form

Now, the given number is 50,000 + 600 + 90 + 1 which is in Expanded Form. So, we have to write the given number in the remaining 2 forms.

B) Seven hundred two thousand, five hundred

Standard Form:702,500

Expanded Form: 700,000 + 0 + 2,000 + 500 + 0 + 0

Explanation:

We can write any number in 3 Forms. They are;

A) Standard Form  B) Word Form  C) Expanded Form

Now, the given number is Seven hundred two thousand, five hundred which is in Word Form. So, we have to write the given number in the remaining 2 forms.

C) 993,004

Word Form: Nine hundred Ninety-three thousand, four

Expanded Form: 900,000 + 90,000 + 3,000 + 0 + 0 +4

Explanation:

We can write any number in 3 Forms. They are;

A) Standard Form  B) Word Form  C) Expanded Form

Now, the given number is 993,004 which is in Standard Form. So, we have to write the given number in the remaining 2 forms.

Question 8.

Modeling Real Life

Use the number 10,000 + 4,000 + 300 + 90 + 9 to complete the check.

Big Ideas Math Answer Key Grade 4 Chapter 1 Place Value Concepts chp 8

Answer:

Word Form: Fourteen thousand, Three hundred Ninety-nine

Standard Form: 14,399

Explanation:

We can write any number in 3 Forms. They are;

A) Standard Form  B) Word Form  C) Expanded Form

Now, the given number is10,000 + 4,000 + 300 + 90 + 9 which is in Expanded Form. So, we have to write the given number in the remaining 2 forms.

1.3 Compare Multi-Digit Numbers

Write which place to use when comparing the numbers.

Question 9.

46,027

46,029

Answer: 46,027 is less than 46,029

Explanation:

The given 2 numbers are 46,027 and 46,029.

From these 2 numbers, we can conclude that 46,027 is less than 46,029

Question 10.

548,003

545,003

Answer: 548,003 is greater than 545,003

Explanation:

The given 2 numbers are 548,003 and 545,003.

From these 2 numbers, we can conclude that 548,003 is greater than 545,003

Question 11.

619,925

630,982

Answer: 619,925 is less than 630,982

Explanation:

The given 2 numbers are 619,925 and 630,982.

From these 2 numbers, we can conclude that 619,925 is less than 630,982

Compare.

Question 12.

Big Ideas Math Answer Key Grade 4 Chapter 1 Place Value Concepts chp 12

Answer: 4,021 is less than 4,210

Explanation:

The given 2 numbers are 4,021 and 4,210.

From these 2 numbers, we can conclude that 4,021 is less than 4,210

Question 13.

Big Ideas Math Answer Key Grade 4 Chapter 1 Place Value Concepts chp 13

Answer: 78,614 is equal to 78,614

Explanation:

The given 2 numbers are 78,614 and 78,614.

From these 2 numbers, we can conclude that 78,614 is equal to 78,614

Question 14.

Big Ideas Math Answer Key Grade 4 Chapter 1 Place Value Concepts chp 14

Answer: 816,532 is greater than 816,332

Explanation:

The given 2 numbers are 816,532 and 816,332.

From these 2 numbers, we can conclude that 816,532 is greater than 816,332

Question 15.

Big Ideas Math Answer Key Grade 4 Chapter 1 Place Value Concepts chp 15

Answer: 55,002 is less than 65,002

Explanation:

The given 2 numbers are 55,002 and 65,002.

From these 2 numbers, we can conclude that 55,002 is less than 65,002

Question 16.

Big Ideas Math Answer Key Grade 4 Chapter 1 Place Value Concepts chp 16

Answer: 3,276 is greater than 3,275

Explanation:

The given 2 numbers are 3,276 and 3,275.

From these 2 numbers, we can conclude that 3,276 is greater than 3,275

Question 17.

Big Ideas Math Answer Key Grade 4 Chapter 1 Place Value Concepts chp 17

Answer: 45,713 is less than 457,130

Explanation:

The given 2 numbers are45,713 and 457,130.

From these 2 numbers, we can conclude that 45,713 is less than 457,130

Question 18.

Big Ideas Math Answer Key Grade 4 Chapter 1 Place Value Concepts chp 18

Answer: 569,021 is greater than 56,902

Explanation:

The given Expanded Form is: 50,000 + 6,000 + 900 + 2 =56,902

The given 2 numbers are569,021 and 56,902.

From these 2 numbers, we can conclude that 569,021 is greater than 56,902

Question 19.

Big Ideas Math Answer Key Grade 4 Chapter 1 Place Value Concepts chp 19

Answer: 37,000 is less than 307,000

Explanation:

The given Word Form is: Thirty – seven thousand = 37,000

The given 2 numbers are 37,000 and 307,000.

From these 2 numbers, we can conclude that37,000 is less than 307,000

Question 20.

Two different hot tubs cost $4,179 and $4,139. Which is the lesser price?

Answer: The lesser price is $4,139

Explanation:

The costs of two different hot tubs are: $4,179 and $4,139

On comparing the costs of two different hot tubs, we can conclude that the lesser price is $4,139

Question 21.

Number Sense

Write all of the digits that make the number greater than 47,068 and less than 47,468.?

47,?68

Answer: 1, 2, 3

Explanation:

The digits that make the number greater than 47,068 and less than 47,468 are: 1, 2, 3

1.4 Round Multi-Digit Numbers

Round the number to the place of the underlined digit.

Question 22.

8,614

Answer: 8,610

Explanation:

The number 8,614 can be round off to 8,610 when rounded to the nearest ten. ( Since the tens position is less than 5)

Question 23.

2,725

Answer: 2,000

Explanation:

The number 2,725 can be round off to 2,000 when rounded to the nearest thousand. ( Since the thousands position is less than 5)

Question 24.

27,602

Answer: 27,700

Explanation:

The number 27,602 can be round off to 27,700 when rounded to the nearest hundred. ( Since the hundreds position is greater than 5)

Question 25.

906,154

Answer: 906,100

Explanation:

The number 906,154 can be round off to 906,100 when rounded to the nearest hundred. ( Since the hundreds position is less than 5)

Round the number to the nearest thousand.

Question 26.

1,358

Answer: 1,000

Explanation:

The number 1,358 can be round off to 1,000  when rounded to the nearest thousand. ( Since the thousands position is less than 5)

Question 27.

57,094

Answer: 58,000

Explanation:

The number 57,094 can be round off to 58,000 when rounded to the nearest thousand. ( Since the thousands position is greater than 5)

Round the number to the nearest ten thousand.

Question 28.

431,849

Answer: 431,000

Explanation:

The number 431,849 can be round off to 431,000  when rounded to the nearest ten thousand. ( Since the ten thousand’s  position is less than 5)

Question 29.

60,995

Answer: 70,000

Explanation:

The number 60,995 can be round off to 70,000 when rounded to the nearest ten thousand. ( Since the ten thousand’s position is greater than 5)

Final Words:

We wish the information provided in the Big Ideas Math Answers Grade 4 Chapter 1 Place Value Concepts is helpful for you. Make use of the above links and start your preparation. Keep in touch with us to get the updates regarding all the chapters of Big Ideas Math Answers Grade 4.

Big Ideas Math Answers Grade 4 Chapter 4 Multiply by Two-Digit Numbers

Big Ideas Math Answers Grade 4 Chapter 4

Check out all the topics covered in the Big Ideas Math Answers Grade 4 Chapter 4 Multiply by Two-Digit Numbers from the below section. You can guide your students by making our Big Ideas Math Book 4th Grade Solution Key Chapter 4 Multiply by Two-Digit Numbers as reference. Students who are seriously practicing to score good marks in exams are suggested to practice the problems given in our Big Ideas Math Answers Grade 4 Chapter 4 Answer Key. Access the links given below and Download Big Ideas 4th Grade Solution Key pdf.

Big Ideas 4th Grade Chapter 4 Multiply by Two-Digit Numbers Solution Key

Students can easily understand the concepts in-depth with the help of Big Ideas Math Answers Grade 4 Chapter 4 Multiply by Two-Digit Numbers. We have provided all the questions, answers, along with explanations. Also, most of the explanations are given with images for the best learning for students. In order to become a math expert, you must refer to the Big Ideas Multiply by Two-Digit Numbers chapter. Verify the step-by-step procedure to solve a problem to make your preparation easy.

Lesson 1: Multiply by Tens

Lesson 2: Estimate Products

Lesson 3: Use Area Models to Multiply Two-Digit Numbers

Lesson 4: Use the Distributive Property to Multiply Two-Digit Numbers

Lesson 5: Use Partial Products to Multiply Two-Digit Numbers

Lesson 6: Multiply Two-Digit Numbers

Lesson 7: Practice Multiplication Strategies

Lesson 8: Problem Solving: Multiplication with Two-Digit Numbers

Performance Task

Lesson 4.1 Multiply by Tens

Explore and Grow
Model each product. Draw each model.
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.1 1
What pattern do you notice in the products?

Answer:
The Place- value method. From the above pattern, we can conclude that the result has different place-values of 6

Explanation:
The position of 3 is different in the given 4 multiplications.
So,
2 × 3 = 6
2 × 30 = 60
2 × 300 = 600
2 × 3000 = 6,000
From the above pattern, we can conclude that the result has different place-values of 6
Repeated Reasoning
How can the pattern above help you find 20 × 30?

Answer:
20 × 30 = 600

Explanation:
You can think of 20 as two tens and 30 as Three tens.
So,
20 × 30 = 2 × 1 ten × 3 × 1 ten = 2 tens × 2 × 3 = 6 × 2 tens = 600
Think and Grow: Multiply by Multiples of Tens
You can use place value and properties to multiply two-digit numbers by multiples of ten.
Example
Find 40 × 20.
One Way: Use place value.
40 × 20 = 40 × ____ tens
= ____ tens
= _____
So, 40 × 20 = _____.

Answer:
800

Explanation:
By using the Place-value method,
40 × 20 = 40 × 2 tens
= 80 tens
= 800
So, 40 × 20 = 800
Another Way: Use the Associative Property of Multiplication.
40 × 20 = 40 × (2 × 10) Rewrite 20 as 2 × 10.
= (40 × 2) × 10 Associative Property of Multiplication
= ____ × 10
= ____
So, 40 × 20 = _____.

Answer:
800

Explanation:
By using the Associative Property of Multiplication,
40 × 20 = 40 × (2 × 10)
= (40 × 2) × 10
= 80 × 10
= 800
Note: Associative Property of Multiplication
Take 3 numbers a, b, c.
By using Associative Property, we can write
a × (b × c) = (a × b) × c
Find 12 × 30.
One Way: Use place value
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.1 2
12 × 30 = 12 × _____ tens
= ____ tens
= _____
So, 12 × 30 = _____.

Answer:
360

Explanation:
Using the Place-value method,
12 × 30 = 12 × 3 tens
= 36 tens
= 360
So, 12 × 30 = 360
Another Way: Use the Associative Property of Multiplication
12 × 30 = 12 × (3 × 10) Rewrite 30 as 3 × 10.
= (12 × 3) × 10 Associative Property of Multiplication
= ____ × 10
= _____
So, 12 × 30 = ____.

Answer: 360

Explanation:
By using the Associative Property of Multiplication,
12 × 30 = 12 × (3 × 10)
= (12 × 3) × 10
= 36 × 10
= 360
Note: Associative Property of Multiplication
Take 3 numbers a, b, c.
By using Associative Property, we can write
a × (b × c) = (a × b) × c
Show and Grow
Find the product.

Question 1.
70 × 40 = _____

Answer:
2800

Explanation:
By using the Associative Property of Multiplication,
70 × 40 = 70 × (4 × 10)
= (70 × 4) × 10
= 280 × 10
= 2800
NoteAssociative Property of Multiplication
Take 3 numbers a, b, c.
By using Associative Property, we can write
a × (b × c) = (a × b) × c

Question 2.
50 × 80 = ____

Answer:
4,000

Explanation:
By using the Associative Property of Multiplication,
50 × 80 = 50 × (8 × 10)
= (50 × 8) × 10
= 400 × 10
= 4000
NoteAssociative Property of Multiplication
Take 3 numbers a, b, c.
By using Associative Property, we can write
a × (b × c) = (a × b) × c

Question 3.
24 × 90 = _____

Answer:
2160

Explanation:
By using the Associative Property of Multiplication,
24 × 90 = 24 × (9 × 10)
= (24 × 9) × 10
= (8 × 3 × 9) × 10
= 216 × 10
= 2160
NoteAssociative Property of Multiplication
Take 3 numbers a, b, c.
By using Associative Property, we can write
a × (b × c) = (a × b) × c

Question 4.
45 × 60 = _____

Answer:
2700

Explanation:
By using the Associative Property of Multiplication,
45 × 60 = 45 × (6 × 10)
= (45 × 6) × 10
= (5 × 9 × 6) × 10
= 270 × 10
= 2700
NoteAssociative Property of Multiplication
Take 3 numbers a, b, c.
By using Associative Property, we can write
a × (b × c) = (a × b) × c
Apply and Grow: Practice
Find the product.

Question 5.
90 × 10 = _____

Answer:
900

Explanation:
By using the place-value method,
90 × 10 = 10 × 9 tens
= 90 tens
= 900
So,
90 × 10 = 900

Question 6.
40 × 60 = ____

Answer:
2400
Explaination:
By using the place-value method,
40 × 60 = 40 × 6 tens
= 4 tens × 6 tens
= 24 × tens × tens
= 2400
So,
40 × 60 = 2400

Question 7.
20 × 70 = _____

Answer: 1400

Explanation:
By using the place-value method,
70 × 20 = 70 × 2 tens
= 7 tens × 2 tens
= 14 × tens × tens
= 1400
So,
70 × 20 = 1400

Question 8.
11 × 30 = ____

Answer: 330

Explanation:
By using the place-value method,
11 × 30 = 11 × 3 tens
= 33 tens
= 330
So,
11 × 30 = 330

Question 9.
12 × 40 = ____

Answer: 480

Explanation:
By using the place-value method,
12 × 40 = 12 × 4 tens
= 48 tens
= 480
So,
12 × 40 = 480

Question 10.
15 × 50 = _____

Answer: 750

Explanation:
By using the place-value method,
15 ×50 = 15 × 5 tens
= 75 tens
= 750
So,
15 × 50 = 750

Question 11.
30 × 13 = _____

Answer: 390

Explanation:
By using the place-value method,
13 × 30 = 13 × 3 tens
= 39 tens
= 390
So,
13 × 30 = 390

Question 12.
10 × 76 = _____

Answer: 760

Explanation:
By using the place-value method,
10 × 76 = 76 × 1 ten
= 76 tens
= 760
So,
76 × 10 = 760

Question 13.
40 × 25 = ____

Answer: 1,000

Explanation:
By using the place-value method,
25 × 40 = 25 × 4 tens
= 5 × 5× 4 tens
= 100 tens
= 1,000
So,
25 × 40 = 1,000
Find the missing factor.

Question 14.
50 × ____ = 1,500

Answer: 30
Explanation;
Let the missing number be X
So, 50 × X = 1,500
X = 1,500 / 50 = 30
Hence, the value of X is: 30

Question 15.
20 × ____ = 1,800

Answer: 90
Explanation;
Let the missing number be X
So, 20 × X = 1,800
X = 1,800 /20 = 90
Hence, the value of X is: 90

Question 16.
60 × ___ = 4,200

Answer: 70
Explanation;
Let the missing number be X
So, 60 × X = 4,200
X = 4,200 / 60 = 70
Hence, the value of X is: 70

Question 17.
____ × 80 = 6,400

Answer: 80
Explanation;
Let the missing number be X
So, X × 80 = 6,400
X = 6,400 / 80 = 80
Hence, the value of X is: 80

Question 18.
____ × 90 = 3,600

Answer: 40
Explanation;
Let the missing number be X
So, X × 90 = 3,600
X = 3,600 / 90 = 40
Hence, the value of X is: 40

Question 19.
____ × 70 = 3,500

Answer: 50
Explanation;
Let the missing number be X
So, X × 70 = 3,500
X = 3,500 / 70 = 50
Hence, the value of X is: 50
Compare.

Question 20.
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.1 3

Answer: 60 × 30 is equal to 1,800

Explanation:
60 × 30 = 1,800
Given numbers are: 1,800 and 1,800
By comparing 2 values, we can conclude that 1,800 is equal to 1,800

Question 21.
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.1 4

Answer: 480 is greater than 460

Explanation:
40 × 12 = 480
Given numbers are: 480 and 460
By comparing 2 values, we can conclude that480 is greater than 460

Question 22.
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.1 5

Answer: 2,250 is less than 2,340

Explanation:
25 × 90 = 2,250
Given numbers are 2,250 and 2,340
By comparing 2 values, we can conclude that 2,250 is less than 2,340

Question 23.
It takes 10 days to film 1 episode of a television show. How many days will it take to film a 20-episode season?
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.1 6

Answer: 200 days

Explanation:
Given that it takes 10 days to film 1 episode of a Television show.
So,
The number of days it will take to film a 20-episode season is: 20 × 10 = 200 days

Question 24.
Reasoning
What is Descartes’s number? Explain.
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.1 7

Answer:

Question 25.
YOU BE THE TEACHER
Newton says that the product of two multiples of ten will always have exactly two zeros. Is he correct? Explain.

Answer: He is correct

Explanation:
Let us suppose 2 numbers 10 and 20 which are the two multiples of 10.
Now,
10 × 20 = 200
According to Newton, the product of two multiples of ten will always have exactly two zeroes.
So, from the above multiplication, we can say that Newton is correct.
Think and Grow: Modeling Real Life

Example
Food drive volunteers collect 1,328 cans of food. The volunteers have50 boxes. Each box holds 20 cans. How many cans will fit in the boxes?
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.1 8
Multiply to find how many cans will fit in the boxes.
20 × 50 = 50 × (5 × 10)
Rewrite 50 as 5 × 10.
= (20 × 5) × 10 Associative Property of Multiplication
= 100 × 10
= 1,000
So, 1,000 cans fit in the boxes.
Subtract the number of cans that will fit in the boxes from the total number of cans collected.
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.1 9
The cans that can not fit in the boxes = 1,328 – 1,000 = 328
So,
328 cans will not fit in the boxes.
Show and Grow

Question 26
A library has 2,124 new books. The library has 40 empty shelves. Each shelf holds 35 books. How many not books
will fit on the empty shelves?

Answer: 724

Explanation:
Multiply to find how many books will hold on the shelves.
40 × 35 = 35 × (4 × 10)
Rewrite 40 as 5 × 10.
= (35 × 4) × 10 Associative Property of Multiplication
= 140 × 10
= 1,400
So, 1,400 cans fit in the boxes.
Subtract the number of books that will hold on the shelves from the total number of books collected.
The number of books that will not hold on the shelves = 2,124 – 1,400 = 724
So,
724 books will not hold on the shelves

Question 27.
An apartment building has 15 floors. Each floor is 10 feet tall. An office building has 30 floors. Each floor is 13 feet tall. How much taller is the office building than the apartment building?

Answer: 240 feet

Explanation:
Given that an apartment has 15 floors and in that, each floor is 10 feet tall.
So, the height of the apartment = 15 × 10 = 150 feet
Given that an office building has 30 floors and in that, each floor is 13 feet tall.
So, the height of the office building = 30 × 13 = 390 feet
Now, to calculate how much taller an office building than the apartment, we have to subtract both the heights of the apartment and the office building.
So,
The difference in height between the office building and the apartment = 390 feet – 150 feet = 240 feet.
From the above, we can conclude that the office building is 240 feet taller than the apartment.

Question 28.
You burn 35 calories each hour you spend reading and 50 calories each hour you spend playing board games. In 2
weeks, you spend14 hours reading and 28 hours playing board games. How many calories do you burn reading
and playing board games?

Answer:
The calories burned during reading in 2 weeks = 490 calories
The calories burned during playing board games = 1400 calories

Explanation:
Given that,
The calories burned during reading is: 35 calories each hour
The calories burned during playing board games is: 50 calories each hour
It is also given that In 2 weeks,
the time spend on reading and Playing Board Games are 14 hours and 28 hours
So, to calculate the calorie consumption in these 2 weeks, we have to multiply the number of hours and the number of calories.
So,
The calories burned during reading in 2 weeks = 490 calories
The calories burned during playing board games in 2 weeks = 1400 calories

Multiply by Tens Homework & Practice 4.1

Find the product.

Question 1.
30 × 10 = _____

Answer: 300

Explanation:
The product of multiply by tens can be done in 2 ways. They are:
A) The place-value method  B) The Associative Property of Multiplication
A) By using the place-value method:
30 × 10 = 10 × 3 tens
= 1 ten × 3 tens
= 3 × 1 ten × 1 ten
= 3 × 100
= 300
B) The Associative Property of Multiplication:
30 × 10 = 3 × (10 × 10)
= (3 × 10) × 10
= 30 × 10
= 300
NoteAssociative Property of Multiplication
Take 3 numbers a, b, c.
By using Associative Property, we can write
a × (b × c) = (a × b) × c

Question 2.
20 × 90 = _____

Answer: 1800

Explanation:
The product of multiply by tens can be done in 2 ways. They are:
A) The place-value method  B) The Associative Property of Multiplication
A) By using the place-value method:
20 × 90 = 20 × 9 tens
= 2 tens × 9 tens
= 18 × 1 ten × 1 ten
= 18 × 100
= 1800
B) The Associative Property of Multiplication:
20 × 90 = 20 × (9 × 10)
= (20 × 9) × 10
=(5 × 4 × 9) × 10
= 180 × 10
= 1800
NoteAssociative Property of Multiplication
Take 3 numbers a, b, c.
By using Associative Property, we can write
a × (b × c) = (a × b) × c

Question 3.
50 × 70 = _____

Answer: 3500

Explanation:
The product of multiply by tens can be done in 2 ways. They are:
A) The place-value method  B) The Associative Property of Multiplication
A) By using the place-value method:
50 × 70 = 50 × 7 tens
= 5 tens × 7 tens
= 35 × 1 ten × 1 ten
= 35 × 100
= 3500
B) The Associative Property of Multiplication:
50 × 70 = 50 × (7 × 10)
= (50 × 7) × 10
=(5 × 10 × 7) × 10
= 350 × 10
= 3500
NoteAssociative Property of Multiplication
Take 3 numbers a, b, c.
By using Associative Property, we can write
a × (b × c) = (a × b) × c

Question 4.
40 × 13 = ______

Answer: 520

Explanation:
The product of multiply by tens can be done in 2 ways. They are:
A) The place-value method  B) The Associative Property of Multiplication
A) By using the place-value method:
40 × 13 = 13 × 4 tens
= 13 × 4 tens
= 52 × 1 ten
= 52 × 10
= 520
B) The Associative Property of Multiplication:
40 × 13 = 13 × (4 × 10)
= (13 × 4) × 10
=(13 ×2× 2) × 10
= 54 × 10
= 540
NoteAssociative Property of Multiplication
Take 3 numbers a, b, c.
By using Associative Property, we can write
a × (b × c) = (a × b) × c

Question 5.
27 × 60 = _____

Answer: 1620

Explanation:
The product of multiply by tens can be done in 2 ways. They are:
A) The place-value method  B) The Associative Property of Multiplication
A) By using the place-value method:
27 × 60 = 27 × 6 tens
= 3 × 9 × 6 tens
= 162 × 1 ten
= 162 × 10
= 1620
B) The Associative Property of Multiplication:
27 × 60 = 27 × (6 × 10)
= (27 × 6) × 10
=(6 × 3 × 9) × 10
= 162 × 10
= 1620
NoteAssociative Property of Multiplication
Take 3 numbers a, b, c.
By using Associative Property, we can write
a × (b × c) = (a × b) × c

Question 6.
80 × 56 = _____

Answer: 4480

Explanation:
The product of multiply by tens can be done in 2 ways. They are:
A) The place-value method  B) The Associative Property of Multiplication
A) By using the place-value method:
80 × 56 = 56 × 8 tens
= 7 × 8 × 8 tens
= 448 × 1 ten
= 448 × 10
= 4480
B) The Associative Property of Multiplication:
56 × 80 = 56 × (8 × 10)
= (56 × 8) × 10
=(8 × 7 × 8) × 10
= 448 × 10
= 4480
NoteAssociative Property of Multiplication
Take 3 numbers a, b, c.
By using Associative Property, we can write
a × (b × c) = (a × b) × c
Find the missing factor.

Question 7.
70 × ____ = 2,100

Answer: 30

Explanation:
Let the missing number be X
So, 70 × X = 2,100
X = 2,100 / 70 = 30
Hence, the value of X is: 30

Question 8.
____ × 10 = 900

Answer: 90
Let the missing number be X
So, X × 10 = 900
X = 900 / 10 =90
Hence, the value of X is: 90

Question 9.
40 × ____ = 1,600

Answer: 40
Let the missing number be X
So, 40 × X = 1,600
X = 1,600 / 40 = 40
Hence, the value of X is: 40

Question 10.
____ × 20 = 1,600

Answer: 80
Let the missing number be X
So, X× 20 = 1,600
X = 1,600 / 20 = 80
Hence, the value of X is: 80

Question 11.
30 × ____ = 1,800

Answer: 60
Let the missing number be X
So, 30 × X = 1,800
X = 1,800 / 30 = 60
Hence, the value of X is: 60

Question 12.
____ × 50 = 3,000

Answer: 60
Let the missing number be X
So, X × 50 = 3,000
X = 3,000 / 50 = 60
Hence, the value of X is: 60
Compare.

Question 13.
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.1 10

Answer: 7,200 is less than 8,100

Explanation:
90 × 80 = 7,200
Given numbers are: 7,200 and 8,100
By comparing 2 values, we can conclude that 7,200 is less than 8,100

Question 14.
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.1 11

Answer: 1,200 is greater  than 1,020

Explanation:
60 ×17 = 1,020
Given numbers are: 1,200 and 1,020
By comparing 2 values, we can conclude that 1,200 is greater than 1,020

Question 15.
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.1 12

Answer: 2,380 is equal to 2,380

Explanation:
34 × 70 = 2,380
Given numbers are:2,380 and 2,380
By comparing the 2 values, we can conclude that 2,380 is equal to2,380

Question 16.
A shallow moonquake occurs 20 kilometers below the moon’s surface. A deep moonquake occurs 35 times deeper than a shallow moonquake. How many kilometers below the surface does the deep moonquake occur?
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.1 13

Answer: 15 kilometers

Explanation:
Given that a shallow moonquake occurred 20 kilometers below the moon’s surface and a deep moonquake occurs 35 meters deeper than a shallow moonquake.
Now, to calculate how much distance the deep moonquake occurred from the surface, we have to subtract the distance that a shallow moonquake occurred from the deep moonquake occurred.
Hence,
The distance below the surface the deep moonquake occurred = 30 – 25 = 15 kilometers

Question 17.
Structure
Write the multiplication equation represented by the number line.
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.1 14

Answer:

Question 18.
Writing
Explain how you can use 20 × 10 = 200 to find 20 × 12.

Answer: By using the Associative Property of Multiplication,
20 × 10 = 10 × (2 × 10)
= (10 × 2) × 10
= 20 × 10
= 200
By using the same method, we can also find the value of 20 × 12.
Now,
By using the Associative Property of Multiplication,
20 × 12 = 12 × (2 × 10)
= (12 × 2) × 10
= 24 × 10
= 240
NoteAssociative Property of Multiplication
Take 3 numbers a, b, c.
By using Associative Property, we can write
a × (b × c) = (a × b) × c

Question 19.
DIG DEEPER!
The product of a number and twice that number is 800. What are the numbers?

Answer: 20, 40

Explanation:
Let the number be X
Given that,
X × 2X = 800
Take X as a common multiple.
Hence,
X × X ( 1 × 2) = 800
X × X × 2 = 800
X × X = 800/2 = 400
X × X = 20 × 20
From the above, we can conclude that the value of X is 20
Hence, the value of twice the X is  20 × 2 = 40
So,
The numbers that can give the product 800 are 20, 40

Question 20.
Modeling Real Life
There are 506 new plants in a greenhouse. A worker programs a robot to arrange the plants into14 rows with 30 plants in each row. How many plants will fit in the rows?

Answer: 420

Explanation:
Given that there are 506 new plants in a greenhouse and a robot can arrange the plants into 14 rows with 30 plants each.
So, to find how many plants will fit in the row, we have to multiply 14 and 30( Since the robot arranges the plants in rows)
Now, By using the Associative Property of Multiplication,
14 × 30 =  14 × (3 × 10)
= (14 × 3) × 10
= (2 × 7 × 3) × 10
= 42 × 10
= 420
From the above, we can conclude that 420 plants will fit in the row.

Question 21.
Modeling Real Life
The world’s largest pool is 13 meters longer than the total length of 20 Olympic pools. An Olympic pool is 50 meters long. How long is the world’s largest pool?

Answer: 1013 meters

Explanation: Given that an Olympic pool has a length of 50 meters. But, there are 20 Olympic pools.
So, to find the total length of the Olympic Pool, we have to multiply the number of pools and the length of each pool.
By using the Associative Property of Multiplication,
50 × 20 = 50 × (2 × 10)
= (50 × 2) × 10
= ( 5 × 10 × 2) × 10
= 100 × 10
= 1,000 meters
So, the total length of the 20 Olympic pools = 1,000 meters
The
Question also mentions that the world’s largest pool is 13 meters longer than the total length of 20 Olympic pools.
Hence,
The length of the World’s largest pool = 1,000 + 13 = 1,013 meters.
So, the length of the World’s largest pool is 1,013 meters
Review & Refresh
Find the value of the underlined digit.

Question 22.
52,618

Answer: The place-value of 8 in the given number is: 8

Explanation:
We can find the position of any given number by using the place- value method.
From this, we can conclude that the place-value of 8 is: 8

Question 23.
379,021

Answer: The place-value of 7 in the given number is: 70,000

Explanation:
We can find the position of any given number by using the place- value method.
From this, we can conclude that the place-value of 7 in the given number is: 70,000

Explanation:

Question 24.
203,557

Answer: The place-value of 2 in the given number is: 200,000

Explanation:
We can find the position of any given number by using the place- value method.
From this, we can conclude that the place-value of 2 in the given number is: 200,000

Question 25.
497,384

Answer: The place-value of 3 in the given number is: 300

Explanation:
We can find the position of any given number by using the place- value method.
From this, we can conclude that the place-value of 3 in the given number is: 300

Lesson 4.2 Estimate Products

Explore and Grow
Choose an expression to estimate each product. Write the expression. You may use an expression more than once.
Big Ideas Math Answers 4th Grade Chapter 4 Multiply by Two-Digit Numbers 4.2 1
Big Ideas Math Answers 4th Grade Chapter 4 Multiply by Two-Digit Numbers 4.2 2
Compare your answers with a partner. Did you choose the same expressions?

Answer:
Let your Expression be 20 ×25.
The given Partner Expressions are:
A) 21 × 24  B) 26 × 38  C) 23 × 17  D) 42 × 23
By Comparing your Expression and your Partner Expression,
A) 500 is less than 504.

Explanation:
Let your Expression be 20 ×25.
By using the Associative Property of Multiplication,
20 × 25 = 25 × ( 2 × 10)
= (25 × 2) × 10
= (5 × 5 × 2) × 10
= 50 × 10
= 500
The given Partner Expressions are:
A) 21 × 24  B) 26 × 38  C) 23 × 17  D) 42 × 23
We can calculate the partner Expressions by Simplifying the given Expressions.
A)
21 × 24 = 3 × 7 × 3 × 8
= 9 × 56
= 504
By comparing your Expression with your Partner Expression, 500 is less than 504.
B)
26 × 38 = 13  × 2 × 19 × 2
= 4× 247
= 988
By comparing your Expression with your Partner Expression, 500 is less than 988.
C)
23 × 17
=391
By comparing your Expression with your Partner Expression, 500 is greater than 391.
D)
42 × 23 = 7  ×  2 ×3 ×  23
= 966
By comparing your Expression with your Partner Expression, 500 is less than 966.
Construct Arguments
Which estimated product do you think will be closer to the product of 29 and 37? Explain your reasoning.
Big Ideas Math Answers 4th Grade Chapter 4 Multiply by Two-Digit Numbers 4.2 3

Answer: 1,000 will be closer to 1,073

Explanation:
Given Product is
29 × 37 = 1,073
Given Expressions are:
25 × 40 = 1,000
30 × 40 = 1,200
Compare the given Product and the Expressions.
By comparison, we can conclude that 1,073 is close to 1,000.
Think and Grow: Estimate Products
You can estimate products using rounding or compatible numbers. Compatible numbers are numbers that are easy to multiply and are close to the actual numbers.
Example
Use rounding to estimate 57 × 38.
Step 1: Round each factor to the nearest ten.
Big Ideas Math Answers 4th Grade Chapter 4 Multiply by Two-Digit Numbers 4.2 4
Step 2: Multiply.
60 × 40 = 60 × 4 tens
= 240 tens
= 2400
So, 57 × 38 is about 2400.
Example
Use compatible numbers to estimate 24 × 31.
Step 1: Choose compatible numbers.
Big Ideas Math Answers 4th Grade Chapter 4 Multiply by Two-Digit Numbers 4.2 5
Step 2: Multiply.
25 × 30 = 25 × 3 tens
= 75 tens
= 750
So, 24 × 31 is about 750.
Show and Grow
Use rounding to estimate the product.

Question 1.
27 × 50

Answer: 1500

Explanation:
Let 27 be Rounded to 30.
Now, we have to find the result of 30 × 50.
We can find the product of multiples of ten by using two methods. They are:
A) The place-value method  B) Associative Property of Multiplication
A) By using the Associative Property of Multiplication,
30 × 50 = 30 × (5 × 10)
= (30 × 5) × 10
= ( 3 × 10 × 5) × 10
= 150 × 10
= 1,500
B) By using the place-value method,
30 × 50 = 30 × 5 tens
= 3 tens × 5 tens
= 15 × 1 ten × 1ten
= 15 × 10 × 10
= 1500
So,
27 × 50 can be rounded to 1,500

Question 2.
42 × 14

Answer: 600

Explanation:
Let 42 be Rounded to 40
Let 14 be Rounded to 15
Now, we have to find the result of 40 × 15.
We can find the product of multiples of ten by using two methods. They are:
A) The place-value method  B) Associative Property of Multiplication
A) By using the Associative Property of Multiplication,
40 × 15 = 15 × (4 × 10)
= (15 × 4) × 10
= ( 3 × 4 × 5) × 10
= 60 × 10
= 600
B) By using the place-value method,
40 × 15 = 15 × 4 tens
= 60 tens
= 60 × 10
= 600
So,
42 × 16 can be rounded to 600

Question 3.
61 × 73

Answer: 4,200

Explanation:
Let 61 be Rounded to 60
Let 73 be Rounded to 70
Now, we have to find the result of 60 × 70.
We can find the product of multiples of ten by using two methods. They are:
A) The place-value method  B) Associative Property of Multiplication
A) By using the Associative Property of Multiplication,
60 × 70 = 60 × (7 × 10)
= (60 × 7) × 10
= ( 6 × 10 × 7) × 10
= 420 × 10
= 4,200
B) By using the place-value method,
60 × 70 = 60 × 7 tens
= 6 tens ×7 tens
= 42 × 1 ten × 1ten
= 42 × 10 × 10
= 4,200
So,
61 ×73 can be rounded to 4,200
Use compatible numbers to estimate the product.

Question 4.
19 × 26

Answer: 500

Explanation:
Let 19 be Rounded to 20
Let 26 be Rounded to 25
Now, we have to find the result of 20 × 25.
We can find the product of multiples of ten by using two methods. They are:
A) The place-value method  B) Associative Property of Multiplication
A) By using the Associative Property of Multiplication,
20 × 25 = 25 × (2 × 10)
= (25 × 2) × 10
= ( 5 × 5 ×2) × 10
= 50 × 10
= 500
B) By using the place-value method,
20 × 25 = 25 × 2 tens
= 50 tens
= 50 × 1 ten
= 50 × 10
= 500
So,
19 ×26 can be rounded to 500

Question 5.
23 × 78

Answer: 2,000

Explanation:
Let 23 be Rounded to 25
Let 78 be Rounded to 80
Now, we have to find the result of 25 × 80.
We can find the product of multiples of ten by using two methods. They are:
A) The place-value method  B) Associative Property of Multiplication
A) By using the Associative Property of Multiplication,
25 × 80 = 25 × (8 × 10)
= (25 × 8) × 10
= ( 5× 5 × 8) × 10
= 200 × 10
= 2,000
B) By using the place-value method,
25 × 80 = 25 × 8 tens
= 200 tens
= 200 × 10
= 2,000
So,
23 ×78 can be rounded to 2,000

Question 6.
74 × 20

Answer: 1,500

Explanation:
Let 74 be Rounded to 75
Now, we have to find the result of 75 × 20.
We can find the product of multiples of ten by using two methods. They are:
A) The place-value method  B) Associative Property of Multiplication
A) By using the Associative Property of Multiplication,
75 × 20 = 75 × (2 × 10)
= (75 × 2) × 10
= ( 5 × 5 × 3 × 2) × 10
= 150 × 10
= 1,500
B) By using the place-value method,
75× 20 = 75 ×2 tens
= 150 tens
= 150× 10
= 1,500
So,
74 ×20 can be rounded to 1,500
Apply and Grow: Practice
Estimate the product.

Question 7.
41 × 73

Answer:  2,800

Explanation:
Let 41 be Rounded to 40
Let 73 be Rounded to 70
Now, we have to find the result of 40 × 70.
We can find the product of multiples of ten by using two methods. They are:
A) The place-value method  B) Associative Property of Multiplication
A) By using the Associative Property of Multiplication,
40 × 70 = 40 × (7 × 10)
= (40 × 7) × 10
= ( 4 × 10 × 7) × 10
= 280 × 10
= 2,800
B) By using the place-value method,
40 × 70 = 40 × 7 tens
= 4 tens ×7 tens
= 28 × 1 ten × 1ten
= 28 × 10 × 10
= 2,800
So,
41 ×73 can be rounded to 2,800

Question 8.
52 × 84

Answer: 4,250

Explanation:
Let 52 be Rounded to 50
Let 84 be Rounded to 85
Now, we have to find the result of 50 × 85.
We can find the product of multiples of ten by using two methods. They are:
A) The place-value method  B) Associative Property of Multiplication
A) By using the Associative Property of Multiplication,
50 × 85 = 85 × (5× 10)
= (85 × 5) × 10
= ( 17× 5 × 5) × 10
= 425 × 10
= 4,250
B) By using the place-value method,
50 × 85 = 85 × 5 tens
= 425 tens
= 425 × 10
= 4,250
So,
52 ×84 can be rounded to 4,250

Question 9.
26 × 68

Answer: 1,750

Explanation:
Let 26 be Rounded to 25
Let 68 be Rounded to 70
Now, we have to find the result of 25 × 70.
We can find the product of multiples of ten by using two methods. They are:
A) The place-value method  B) Associative Property of Multiplication
A) By using the Associative Property of Multiplication,
25 × 70 = 25 × (7 × 10)
= (25 × 7) × 10
= ( 5 × 5 × 7) × 10
= 175 × 10
= 1,750
B) By using the place-value method,
25 × 70 = 25 × 7 tens
= 175  tens
= 175 × 10
= 1,750
So,
26 ×68 can be rounded to 1,750

Question 10.
38 × 17

Answer: 600

Explanation:
Let 38 be Rounded to 40
Let 17 be Rounded to 15
Now, we have to find the result of 40 × 15.
We can find the product of multiples of ten by using two methods. They are:
A) The place-value method  B) Associative Property of Multiplication
A) By using the Associative Property of Multiplication,
15 × 40 = 15 × (4× 10)
= (15 × 4) × 10
= ( 5 × 3 × 4) × 10
= 60 × 10
= 600
B) By using the place-value method,
15 × 40 = 15 × 4 tens
= 60  tens
= 60 × 10
= 600
So,
38 ×17 can be rounded to 600

Question 11.
75 × 24

Answer: 1,875

Explanation:
Let 24 be Rounded to 25
Now, we have to find the result of 25 × 75.
We can find the Product by using the simplification method.
25 × 75
= 5 × 5 × 25 × 3
= 5 × 5× 5 × 5 × 3
= 25 × 25 × 3
= 625 × 3
= 1,875
So,
38 × 17 can be Rounded to 1,875

Question 12.
93 × 53

Answer: 4,500

Explanation:
Let 93 be Rounded to 90
Let 53 be Rounded to 50
Now, we have to find the result of 90 × 50.
We can find the product of multiples of ten by using two methods. They are:
A) The place-value method  B) Associative Property of Multiplication
A) By using the Associative Property of Multiplication,
90 × 50 = 90 × (5 × 10)
= (90 × 5) × 10
= ( 5 × 2 × 9 × 5) × 10
= 450 × 10
= 4,500
B) By using the place-value method,
90 × 50 = 90 × 5 tens
= 9 tens × 5 tens
= 45 × 1 ten × 1 ten
= 45 × 10 × 10
= 4,500
So,
93 ×53 can be rounded to 4,500

Question 13.
44 × 78

Answer: 3,600

Explanation:
Let 44 be Rounded to 45
Let 78 be Rounded to 80
Now, we have to find the result of 45 × 80.
We can find the product of multiples of ten by using two methods. They are:
A) The place-value method  B) Associative Property of Multiplication
A) By using the Associative Property of Multiplication,
45 × 80 = 45 × (8 × 10)
= (45 × 8) × 10
= ( 5 × 2 × 9 × 4) × 10
= 360 × 10
= 3,600
B) By using the place-value method,
45 × 80 = 45 × 8 tens
= 5 × 9 × 8 tens
= 45 × 8 × 10
= 360 × 10
= 3,600
So,
44 ×78 can be rounded to 4,500

Question 14.
21 × 33

Answer: 600

Explanation:
Let 21 be Rounded to 20
Let 33 be Rounded to 30
Now, we have to find the result of 20 × 30.
We can find the product of multiples of ten by using two methods. They are:
A) The place-value method  B) Associative Property of Multiplication
A) By using the Associative Property of Multiplication,
20 × 30 = 20 × (3× 10)
= (20 × 3) × 10
= ( 5× 4 × 3) × 10
= 60 × 10
= 600
B) By using the place-value method,
20 × 30 = 30 × 2 tens
= 3 tens × 2 tens
= 6 × 10 × 10
= 600
So,
21 ×33 can be rounded to 600

Question 15.
45 × 45

Answer: 2,500

Explanation:
Let 45 be Rounded to 50
Now, we have to find the result of 50 × 50.
We can find the product of multiples of ten by using two methods. They are:
A) The place-value method  B) Associative Property of Multiplication
A) By using the Associative Property of Multiplication,
50 × 50 = 50 × (5× 10)
= (50 × 5) × 10
= ( 10× 5 × 5) × 10
= 250 × 10
= 2,500
B) By using the place-value method,
50 × 50 = 50 × 5 tens
= 250 tens
= 25 × 10 × 10
= 2,500
So,
45 ×45 can be rounded to 2,500
Open-Ended
Write two possible factors that can be estimated as shown.

Question 16.
2,400
Big Ideas Math Answers 4th Grade Chapter 4 Multiply by Two-Digit Numbers 4.2 6

Answer:

Explanation:
The given number is: 2,400
The Products of 24 are:
4 × 6 = 24
6 × 4 =24
From the above two products, we can conclude that the two possible numbers that can give the product 2,400 are: 40, 60

Question 17.
1,200
Big Ideas Math Answers 4th Grade Chapter 4 Multiply by Two-Digit Numbers 4.2 7

Answer:              

Explanation:

Explanation:
The given number is: 1,200
The Products of 12 are:
3 × 4 =12
4 × 3 =12
6 ×2 =12
2 ×6 =12
From the above two products, we can conclude that the two possible numbers that can give the product 2,400 are: 40, 30, and 20, 60

Question 18.
DIG DEEPER!
You use 50 × 30 to estimate 46 × 29. Will your estimate be greater than or less than the actual product? Explain.

Answer: We will Estimate the Product greater than the actual Product

Explanation:
Given Product is: 46 × 29

Explanation:
Let 46 be Rounded to 50
Let 29 be Rounded to 30
Now, we have to find the result of 50 × 30.
We can find the product of multiples of ten by using two methods. They are:
A) The place-value method  B) Associative Property of Multiplication
A) By using the Associative Property of Multiplication,
50 × 30 = 50 × (3 × 10)
= (50 × 3) × 10
= ( 5 × 10 × 3) × 10
= 150 × 10
= 1,500
B) By using the place-value method,
50 × 30 = 30 × 5 tens
= 150  tens
= 150× 10
= 1,500
So,
46 ×29 can be rounded to 1,500

Question 19.
YOU BE THE TEACHER
Your friend uses rounding to estimate 15 × 72. She gets a product of 700. Is your friend’s estimate correct?
Explain.

Answer: No

Explanation:
Your friend is going to estimate 15 × 72 and she gets a product 700.
So, According to her Product, the Possible rounded off numbers to get the product 700 are 10 and 70
So, your Friend’s estimate is not correct.
Now,
Let 72 be Rounded to 70
Now, we have to find the result of 15 × 70.
We can find the product of multiples of ten by using two methods. They are:
A) The place-value method  B) Associative Property of Multiplication
A) By using the Associative Property of Multiplication,
15 × 70 = 15 × (7 × 10)
= (15 × 7) × 10
= (3 × 5 × 7) × 10
= 105 × 10
= 1,050
B) By using the place-value method,
15 × 70 = 15 × 7 tens
= 105  tens
= 105 × 10
= 1,050
So,
15 ×72 can be rounded to 1,050
Think and Grow: Modeling Real Life
Example
About how much does 1 year of phone service cost?
Big Ideas Math Answers 4th Grade Chapter 4 Multiply by Two-Digit Numbers 4.2 8
Think: What do you know? What do you need to find? How will you solve it?
There are 12 months in 1 year, so multiply the price per month by 12.
12 × 24 = 288
Estimate the product.
Big Ideas Math Answers 4th Grade Chapter 4 Multiply by Two-Digit Numbers 4.2 9

Answer: 
Show and Grow
Question 20.
Use the table above. About how much does 1 year of Internet service cost? About how much does 1 year of cable television service cost?

Answer:
Internet Service cost: $540
Cable television service cost: $1068

Explanation:
In the given table,
The Internet cost and cable television service costs are given per month.
For 1 year, there are 12 months.
So,
The cost of Internet service for a year is: 12 × $45 = $ 540
The cost of cable television service for a year is: 12 × $89 = 1068

Question 21.
A giant panda eats 28 pounds of food each day. An orca eats 17 times as much food as the panda eats each day. About how much food does the orca eat each day?
Big Ideas Math Answers 4th Grade Chapter 4 Multiply by Two-Digit Numbers 4.2 10

Answer: 476 Pounds

Explanation:
Given that a giant panda eats 28 pounds of food each day and an Orca eats 17 times as much food as the panda eats each day.
So,
The amount of food eaten by an Orca = 17 × The amount of food eaten by panda
= 17 × 28 = 476 pounds

Estimate Products Homework & Practice 4.2

Use rounding to estimate the product.

Question 1.
42 × 13

Answer:  600

Explanation:
Let 42 be Rounded to 40
Let 13 be Rounded to 15
Now, we have to find the result of 40 × 15.
We can find the product of multiples of ten by using two methods. They are:
A) The place-value method  B) Associative Property of Multiplication
A) By using the Associative Property of Multiplication,
40 × 15 = 15 × (4 × 10)
= (15 × 4) × 10
= ( 3 × 4 × 5) × 10
= 60 × 10
= 600
B) By using the place-value method,
40 × 15 = 15 × 4 tens
= 60 tens
= 60 × 10
= 600
So,
42 × 13 can be rounded to 600

Question 2.
56 × 59

Answer: 3,300

Explanation:
Let 56 be Rounded to 55
Let 59 be Rounded to 60
Now, we have to find the result of 55 × 60.
We can find the product of multiples of ten by using two methods. They are:
A) The place-value method  B) Associative Property of Multiplication
A) By using the Associative Property of Multiplication,
55 × 60 = 55 × (6 × 10)
= (55 × 6) × 10
= ( 5 × 11 × 6) × 10
= 330 × 10
= 3,300
B) By using the place-value method,
55 × 60 = 55 × 6 tens
= 330 tens
= 330 × 10
= 3,300
So,
42 × 16 can be rounded to 600

Question 3.
19 × 91

Answer: 1,800

Explanation:
Let 19 be Rounded to 20
Let91 be Rounded to 90
Now, we have to find the result of 20 × 90.
We can find the product of multiples of ten by using two methods. They are:
A) The place-value method  B) Associative Property of Multiplication
A) By using the Associative Property of Multiplication,
20 × 90 = 20 × (9× 10)
= (20 × 9) × 10
= ( 5 × 4 × 9) × 10
= 180 × 10
= 1,800
B) By using the place-value method,
20 × 90 = 20 × 9 tens
= 2 tens × 9 tens
= 18 × 1 ten × 1 ten
=18 × 10 × 10
= 18 × 100
= 1,800
So,
19 × 91 can be rounded to 1,800
Use compatible numbers to estimate the product.

Question 4.
23 × 78

Answer: 2,000

Explanation:
Let 23 be Rounded to 25
Let 78 be Rounded to 80
Now, we have to find the result of 25 × 80.
We can find the product of multiples of ten by using two methods. They are:
A) The place-value method  B) Associative Property of Multiplication
A) By using the Associative Property of Multiplication,
25 × 80 = 25 × (8 × 10)
= (25 × 8) × 10
= ( 5 × 5 × 8) × 10
= 200 × 10
= 2,000
B) By using the place-value method,
25 × 80 = 25 × 8 tens
= 200 tens
= 200 × 10
= 2,000
So,
23 × 78 can be rounded to 2,000

Question 5.
67 × 45

Answer:3,150

Explanation:
Let 67 be Rounded to 70
Now, we have to find the result of 70 × 45.
We can find the product of multiples of ten by using two methods. They are:
A) The place-value method  B) Associative Property of Multiplication
A) By using the Associative Property of Multiplication,
45 × 70 = 45 × (7 × 10)
= (45 × 7) × 10
= ( 5 × 9 × 7) × 10
= 315 × 10
= 3,150
B) By using the place-value method,
45 × 70 = 45 × 7 tens
= 315 tens
= 315 × 10
= 3,150
So,
67 × 45 can be rounded to 3,150

Question 6.
19 × 24

Answer: 500
Explanation;
Let 19 be Rounded to 20
Let 24 be Rounded to 25
Now, we have to find the result of 25 × 20.
We can find the product of multiples of ten by using two methods. They are:
A) The place-value method  B) Associative Property of Multiplication
A) By using the Associative Property of Multiplication,
20 × 25 =25 × (2 × 10)
= (25 × 2) × 10
= ( 5 × 5 × 2) × 10
= 50 × 10
= 500
B) By using the place-value method,
25 × 20 = 25 × 2 tens
= 50 tens
= 50 × 10
= 500
So,
19 × 24 can be rounded to 500
Estimate the product.

Question 7.
84 × 78

Answer: 6,800

Explanation:
Let 84 be Rounded to 85
Let 78 be Rounded to 80
Now, we have to find the result of 85 × 80.
We can find the product of multiples of ten by using two methods. They are:
A) The place-value method  B) Associative Property of Multiplication
A) By using the Associative Property of Multiplication,
85 × 80 = 85 × (8 × 10)
= (85 × 8) × 10
= ( 5 × 17 × 8) × 10
= 680 × 10
= 6,800
B) By using the place-value method,
85 × 80 = 85 × 8 tens
= 680 tens
= 680 × 10
= 6,800
So,
84 × 78 can be rounded to 600

Question 8.
92 × 34

Answer: 3,150

Explanation:
Let 92 be Rounded to 90
Let 34 be Rounded to 35
Now, we have to find the result of 90 × 35.
We can find the product of multiples of ten by using two methods. They are:
A) The place-value method  B) Associative Property of Multiplication
A) By using the Associative Property of Multiplication,
90 × 35 = 35 × (9 × 10)
= (35 × 9) × 10
= ( 5 × 7 × 9) × 10
= 315 × 10
= 3,150
B) By using the place-value method,
90 × 35 = 35 × 9 tens
= 315 tens
= 315 × 10
= 3,150
So,
92 × 34 can be rounded to 3,150

Question 9.
57 × 81

Answer: 4,800

Explanation:
Let 57 be Rounded to 60
Let 81 be Rounded to 80
Now, we have to find the result of 80 × 60.
We can find the product of multiples of ten by using two methods. They are:
A) The place-value method  B) Associative Property of Multiplication
A) By using the Associative Property of Multiplication,
80 × 60 = 80 × (6 × 10)
= (80 × 6) × 10
= ( 8 × 10 × 6) × 10
= 480 × 10
= 4,800
B) By using the place-value method,
80 × 60 = 80 × 6 tens
= 480 tens
= 480 × 10
= 4,800
So,
57 × 81 can be rounded to 600
Open-Ended
Write two possible factors that could be estimated as shown.

Question 10.
6,400
Big Ideas Math Answers 4th Grade Chapter 4 Multiply by Two-Digit Numbers 4.2 11

Answer:    

Explanation:
The Products of 64 are:
8 × 8 = 64
16 × 4 =64
From the above two products, we can conclude that the two possible numbers that can give the product 6,400 are: 80,80 and. 160,40

Question 11.
1,600
Big Ideas Math Answers 4th Grade Chapter 4 Multiply by Two-Digit Numbers 4.2 12

Answer:

Explanation:
The Products of 16 are:
4 × 4 = 16
8 × 2 =16
From the above two products, we can conclude that the two possible numbers that can give the product 1,600 are: 40, 40 and, 80,20

Question 12.
Reasoning
Are both Newton’s and Descartes’s estimates reasonable? Explain.
Big Ideas Math Answers 4th Grade Chapter 4 Multiply by Two-Digit Numbers 4.2 13

Answer: Both Newton’s and Descartes’s estimates are reasonable

Explanation:
According to Newton,
The estimated values of 27 and 68 are 30 and 70
According to Descartes,
The estimated values of 27 and 68 are 25 and 70
According to Newton:
27 × 68 = 2,100
We can find the product of multiples of ten by using two methods. They are:
A) The place-value method  B) Associative Property of Multiplication
A) By using the Associative Property of Multiplication,
30 × 70 = 30 × (7 × 10)
= (30 × 7) × 10
= ( 3 × 10 × 7) × 10
= 210 × 10
= 2,100
B) By using the place-value method,
30× 70 = 30 ×7 tens
= 210 tens
= 210 × 10
= 2,100
So,
27 × 68 can be rounded to 2,100 ( According to Newton)
According to Descartes:
27 × 68 = 1,750
We can find the product of multiples of ten by using two methods. They are:
A) The place-value method  B) Associative Property of Multiplication
A) By using the Associative Property of Multiplication,
25 × 70 = 25 × (7 × 10)
= (25 × 7) × 10
= ( 5 × 5 × 7) × 10
= 175 × 10
= 1,750
B) By using the place-value method,
25× 70 = 25 ×7 tens
= 175 tens
= 175 × 10
= 1,750
So,
27 × 68 can be rounded to 1,750 ( According to Descartes)

Question 13.
DIG DEEPER!
You use 90 × 30 to estimate 92 × 34. Will your estimate be greater than or less than the actual product? Explain.

Answer: No
Your friend is going to estimate 92 × 34 and she gets a product  2,700.
So, your Friend’s estimate is not correct.
Now,
Let 34 be Rounded to 35
Let 92 be Rounded to 90
Now, we have to find the result of 35 × 90.
We can find the product of multiples of ten by using two methods. They are:
A) The place-value method  B) Associative Property of Multiplication
A) By using the Associative Property of Multiplication,
35 × 90 = 35 × (9 × 10)
= (35 × 9) × 10
= (9× 5 × 7) × 10
= 315 × 10
= 3,150
B) By using the place-value method,
35 × 90 = 35 × 9 tens
= 315  tens
= 315 × 10
= 3,150
So,
92 ×34 can be rounded to 3,150

Question 14.
Modeling Real Life
About how many hours of darkness does Barrow, Alaska have in December?
Big Ideas Math Answers 4th Grade Chapter 4 Multiply by Two-Digit Numbers 4.2 14

Answer: 744 hours

Explanation:
From the above table,
The days of darkness in Barrow, Alaska are 31 days.
We know that there are 24 hours in a day.
So,
The number of hours of darkness does Barrow, Alaska have in December = 31 × 24 = 744 hours
Review & Refresh

Question 15.
Round 253,490 to the nearest ten thousand.

Answer: 250,000
Explanation;
The position of a given number is dependent on the place-value of that number.
So,
When 253,490 rounded off to the nearest ten thousand, the result ts 250,000

Question 16.
Round 628,496 to the nearest hundred thousand.

Answer: 630,000
Explanation;
The position of a given number is dependent on the place-value of that number.
So,
When 628,496 rounded off to the nearest ten thousand, the result ts 630,000

Lesson 4.3 Use Area Models to Multiply Two-Digit Numbers

Explore and Grow
Draw an area model that represents 15 × 18. Then break apart your model into smaller rectangles.
Big Ideas Math Answers Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.3 1
What is the total area of your model? Explain how you found your answer.

Answer: The Total Area of your Model = 400

Explanation:

Count the number of boxes in both the vertical and horizontal directions.
Since all the sides of the figure look the same, take any 1 row in the vertical direction and horizontal direction and count the number of boxes in it.
After counting, you get
The number of boxes in a row present in the vertical direction is: 20
The number of boxes in a row present in the horizontal direction is: 20
So, we can find the total area of your model by multiplying the number of boxes in both the vertical and horizontal directions. ( Remember, there is no need to count all the boxes in both directions since the figure has an equal number of boxes on all sides)
So,
The total area of your model = 20 × 20 = 400
Reasoning
Compare with a partner. Do you get the same answer? Explain.

Answer: Yes

Explanation:
My Partner counted the number of boxes in the middle of the model in both vertical and horizontal directions.
He got,
The number of boxes in a row present in the vertical direction is: 20
The number of boxes in a row present in the horizontal direction is: 20
So,
The total area of my partner’s model = 20 × 20 = 400
So,
My partner and I got the same answer.
Think and Grow: Use Area Models to Multiply
Example
Use an area model and partial products to find 12 × 14.
Model the expression. Break apart 12 as 10 + 2 and 14 as 10 + 4.
Big Ideas Math Answers Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.3 2
Big Ideas Math Answers Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.3 3

Answer: 168

Explanation:
So, 12 × 14 = 168
Show and Grow
Use the area model to find the product.

Question 1.
17 × 15 = _____
Big Ideas Math Answers Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.3 4

Answer: 255

Explanation:
So, 17 × 15 = 255

Question 2.
34 × 22 = _____

Big Ideas Math Answers Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.3 5

Answer: 748

Explanation:
  So, 34 × 22 = 748
Apply and Grow: Practice
Use the area model to find the product.

Question 3.
13 × 19 = _____
Big Ideas Math Answers Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.3 6

Answer: 247

Explanation:
So, 13 × 19 = 247
Now,
100 + 90 + 30 + 27 = 247
So,
                 13 × 19 = 247

Question 4.
25 × 39 = ____
Big Ideas Math Answers Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.3 7

Answer: 975

Explanation:

Now,
600 + 150 + 180 + 45 = 975
So,
25 × 39 = 975
Draw an area model to find the product.

Question 5.
11 × 13 = ______
Big Ideas Math Answers Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.3 8

Answer:  143

Explanation:
By using the Partial Products method,
10 × 6 + 3 × 6 + 10 × 5 + 3 × 5
= 60 + 18 + 50 + 15 = 143
So,
11 × 13 = 143

Question 6.
23 × 26 = ______

Answer: 598

Explanation:

By using the Partial products method,
20 × 20 + 3 × 20 + 20 × 6 + 3 × 6
= 400 + 60 + 120 + 18 = 598
So,  23 × 26 = 598

Question 7.
27 × 45 = ______

Answer: 1,215

Explanation:

By using the partial products method,
20 × 20 + 7 × 20 + 20 × 25 + 7 ×25
= 400 + 140 +500 + 175 = 1,215
So, 27 × 45 = 1,215

Question 8. Perseid meteors travel 59 kilometers each second. How far does a perseid meteor travel in 15 seconds?
Big Ideas Math Answers Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.3 9

Answer: 885 kilometers

Explanation:
Given that the Perseid meteors travel 59 kilometers each second.
So,
The distance traveled by the Perseid meteors in 15 seconds = 59 × 15 = 885 kilometers
From the above,
We can conclude that the Perseid meteors travel 885 kilometers in 15 seconds.

Question 9.
DIG DEEPER!
Write the multiplication equation represented by the area model.
Big Ideas Math Answers Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.3 10

Answer:

Explanation:
Using the Partial Products Model,
40 × 30 + 40 × 2 + 3 × 30 + 3 × 2
= 1,200 + 80 + 90 + 6 = 1,376
So, the multiplication equation represented by area model = 43 × 32
Think and Grow: Modeling Real Life
Example
A wind farm has 8 rows of new wind turbines and 3 rows of old wind turbines. Each row has 16 turbines. How many turbines does the wind farm have?
Big Ideas Math Answers Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.3 11

Answer: Add the number of rows of new turbines to the number of rows of old turbines
8 + 3 =11
So,
There are11 rows of turbines.
Multiply the number of rows by the number in each row.

So, the wind farm has 176  turbines.
Show and Grow

Question 10.
You can type 19 words per minute. Your cousin can type 33 words per minute. How many more words can your cousin type in 15 minutes than you?

Answer: 210 words

Explanation:
Given that you can type 19 words per minute and your cousin can type 33 words per minute.
So,
The number of words you can type in 15 minutes = 19 × 15 = 285
The number of words your cousin can type in 15 minutes = 495
Hence,
The number of words you have to type more than your cousin in 15 minutes = 495 – 285 =  310 words

Question 11.
A store owner buys 24 packs of solar eclipse glasses. Each pack has 12 glasses. The store did not sell 18 of the glasses. How many of the glasses did the store sell?
Big Ideas Math Answers Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.3 13

Answer: 270 glasses

Explanation:
Given that a store Owner buys 24 packs of solar eclipse glasses and each pack has 12 glasses.
So,
the total number of glasses that a store owner buy = 24 × 12 = 288 glasses
But,
It is also given that the store owner did not sell 18 glasses.
Hence, the total number of glasses the store owner sold = 288 – 18 = 270 glasses

Use Area Models to Multiply Two-Digit Numbers Homework & Practice 4.3

Use the area model to find the product.

Question 1.
12 × 13 = ______
Big Ideas Math Answers Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.3 14

Answer: 156

Explanation:
By using the partial products method,
10 × 10 + 2 ×10 + 10 × 3 + 2 ×3
= 100 + 20 +30 + 6 = 156
So, 12 × 13 = 156

Question 2.
38 × 24 = _____
Big Ideas Math Answers Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.3 15

Answer: 912

Explanation:
By using the Partial products method,
30 × 20 + 8 × 20 + 30 × 4 + 8 × 4
= 600 + 160 + 120 + 32 = 912
So,  38 × 24 = 912
Use the area model to find the product.

Question 3.
19 × 18 = ____
Big Ideas Math Answers Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.3 16

Answer: 342

Explanation:

By using the partial products method,
10 × 10 +8 × 10 + 10 × 9 +9 ×8
= 100 + 80 +90 + 72 = 342
So, 19 × 18 = 342

Question 4.
23 × 25 = _____
Big Ideas Math Answers Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.3 17

Answer: 575

Explanation:
By using the partial products method,
20 × 20 +5 × 20 + 3 × 20 +3 ×5
= 400 + 100 +60 + 15 = 575
So, 23 × 25 = 575
Draw an area model to find the product.

Question 5.
26 × 31 = _____

Answer: 806

Explanation:

By using the partial products method,
20 × 30 +6× 30 + 1 × 20 +6 ×1
= 600 + 180 +20 + 6 = 806
So, 26 × 31 = 806

Question 6.
22 × 47 = ______

Answer: 1,034

Explanation:

By using the partial products method,
20 × 40 +2×40 + 7 × 20 +2 ×7
= 800 + 80 +140 + 14 = 1,034
So, 22 × 47 = 1,034

Question 7.
YOU BE THE TEACHER
Your friend finds 12 × 42. Is your friend correct? Explain.
Big Ideas Math Answers Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.3 18

Answer: Yes, your friend is correct.

Explanation:

By using the partial products method,
10 × 40 +2×10 + 2 × 40 +2 ×2
= 400 + 20 +80 + 4 = 504
So, 12 × 42 = 504

Question 8.
Writing
Explain how to use an area model and partial products to multiply two-digit numbers.

Answer:
Let the Partial Products are: a, b, c, d
By using the Partial Products method,
a × c + b × c + a × d + b × d
= ac + bc + ad + bd
So, by using the above method, we can find the product of 2- digit numbers.

Question 9.
Modeling Real Life
A mega-arcade has 9 rows of single-player games and 5 rows of multi-player games. Each row has 24 games. How many games does the arcade have?
Big Ideas Math Answers Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.3 19

Answer:  The arcade has 336 games

Explanation:
Given that a mega arcade has 9 rows of single-player games and 5 rows of multi-player games.
So,
Total number of rows present in the arcade = 9 + 5 = 14
It is also given that each row has 24 games.
So,
The total number of games present in the arcade = 14 × 24 = 336

By using the partial products method,
10 × 20 +4 × 20 + 10 × 4 + 4 ×4
= 200 + 80 +40 + 16 = 336
So, 14 × 24 = 336
Review & Refresh
Find the sum. Check whether your answer is reasonable.

Question 10.
Big Ideas Math Answers Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.3 20

Answer: 84,016

Explanation:
To find the sum, add the digits starting from the Right-most position. If there is “Carry”, then add that carry to the result of the next Position value.

Question 11.
Big Ideas Math Answers Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.3 21

Answer: 71,585

Explanation:
To find the sum, add the digits starting from the Right-most position. If there is “Carry”, then add that carry to the result of the next Position value. (As shown in the above figure)

Question 12.
Big Ideas Math Answers Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.3 22

Answer: 569,821

Explanation:

To find the sum, add the digits starting from the Right-most position. If there is “Carry”, then add that carry to the result of the next Position value. (As shown in the above figure)

Lesson 4.4 Use the Distributive Property to Multiply Two-Digit Numbers

Explore and Grow
Use as few base ten blocks as possible to create an area model for 13 × 24. Draw to show your model.
Big Ideas Math Solutions Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.4 1
Color your model to show four smaller rectangles. Label the partial products.

Answer:
By using the partial products method,
10 × 20 +3 × 20 + 4 × 3 + 10 ×4
= 200 + 60 +12 + 40 = 312
So, 13 × 24 = 312
Reasoning
How do you think the Distributive Property relates to your area model? Explain.

Answer:
Distributive Property of Multiplication:
Let there are 3 numbers a, b, c.
The Distributive property is given as:
a × (b  + c) = ( a ×  b) + ( a ×  c)
SO, by using the above property, we can conclude that the Distributive Property relates to your area model.
Think and Grow: Use the Distributive Property to Multiply
Example
Find 17 × 25.
One Way: Use an area model and partial products.


Show and Grow

Question 1.
Use the area model and the Distributive Property to find 32 × 19.
Big Ideas Math Solutions Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.4 3

Answer: 608

Explanation:
Using the Distributive Property, we can find the product of 32 × 19
32 × 19 = 32 × ( 10 + 9)
= ( 32 × 10 ) + ( 32 × 9)
= ( 30 + 2 ) × 10 + ( 30 + 2 ) × 9
= ( 30 × 10 ) + ( 2 × 10 ) + ( 30 × 9 ) + ( 2 × 9)
= 300 + 20 + 270 + 18
= 608
So, 32 × 19 = 608
Apply and Grow: Practice

Question 2.
Use the area model and the Distributive Property to find 34 × 26.
Big Ideas Math Solutions Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.4 4

Answer: 884

Explanation:
Using the Distributive Property, we can find the product of 34 × 26
34 × 26 = 34 × ( 20 + 6)
= ( 34 × 20 ) + ( 34 × 6)
= ( 30 + 4 ) × 20 + ( 30 + 4 ) × 6
= ( 30 × 20 ) + ( 4 × 20 ) + ( 30 × 6 ) + ( 4 × 6)
= 600 + 80 + 180 + 24
= 884
So, 34 × 26 = 884
Use the Distributive Property to find the product.

Question 3.
28 × 47 = 28 × (40 + 7)
= (28 × 40) + (28 × 7)
= (20 + 8) × 40  + (20 + 8) × 7
= (20 × 40) + (8 × 40) + (20 × 7) + (8 × 7)
= 800 + 320 + 140 + 56
=1,316
So, 28 × 47 = 1,316

Answer: 28 × 47 = 1,316

Question 4.
39 × 41 = _____

Answer:
39 × 41 = 39 × (40 + 1)
= (39 × 40) + (39 × 1)
= (30 + 9) × 40  + (30 + 9) × 1
= (30 × 40) + (9 × 40) + (30 × 1) + (9 ×1)
=1,200 + 360 + 30 + 9
=1,599
So, 39 × 41 = 1,599

Question 5.
74 × 12 = ______

Answer:
74 × 12 = 74 × (10 + 2)
= (74 × 10) + (74 × 2)
= (70 + 4) × 10  + (70 + 4) × 2
= (70 × 10) + (4 × 10) + (70 × 2) + (4 ×2)
=700 + 40 + 140 + 8
=888
So, 74 × 12 = 888

Question 6.
83 × 65 = _____

Answer:
83 × 65 = 83 × (60 + 5)
= (83 × 60) + (83 × 5)
= (80 + 3) × 60  + (80 + 3) ×5
= (80 × 60) + (3 × 60) + (80 × 5) + (3 ×5)
=4,800 + 180 + 400 + 15
=5,395
So, 83 × 65 = 5,395

Question 7.
Which One Doesn’t Belong?
Which expression does not belong with the other three?
Big Ideas Math Solutions Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.4 5

Answer:
Let the given Expressions be ordered as A), B), C) and D)
From the Order, we can say that Expression C) does not belong to the other three.

Explanation:
The given Expressions are:
A) ( 40 + 7) × 52
B) ( 40 + 7) × (50 + 2)
C) ( 40 × 7) × ( 50 × 2)
D) 47 × ( 50 + 2)
The given Expressions are written using the Distributive Property of Multiplication.
The Distributive property is given as:
a × (b  + c) = ( a ×  b) + ( a ×  c)
So, from the above Property, we can conclude that Expression C) does not belong to the other three.
Think and Grow: Modeling Real Life
Example
The dunk tank at a school fair needs 350 gallons of water. There are 27 students in a class. Each student pours13 gallons of water into the tank. Is there enough water in the dunk tank?
Big Ideas Math Solutions Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.4 6
Find how many gallons of water the students put in the dunk tank.

Given that there are 350 gallons at a school fair.
But, we got 351 gallons of water.
So, there is 1 gallon enough water in the dunk tank.
Show and Grow

Question 8.
An event coordinator orders 35 boxes of T-shirts to give away at a baseball game. There are 48 T-shirts in each box. If 2,134 fans attend the game, will each fan get a T-shirt?
Big Ideas Math Solutions Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.4 8

Answer: No, each fan will not get a T-shirt.

Explanation:
Given that an event coordinator orders 35 boxes of T-shirts to give away at a baseball game and there are 48 T-shirts in each box.
So, we will get the total number of T-shirts due to the Product of 35 × 48.
We will get the product by using the Distributive Property of Multiplication.
35 × 48 = 35 × (40 + 8)
= (35 × 40) + (35 × 8)
= (30 + 5) × 40  + (30 + 5) × 8
= (30 × 40) + (5 × 40) + (30 × 8) + (8 ×5)
=1,200 + 20 + 240 + 40
=1,680
So, 35× 48 = 1,680
Note:
The Distributive property is given as:
a × (b  + c) = ( a ×  b) + ( a ×  c)

Question 9.
A horse owner must provide 4,046 square meters of pasture for each horse. Is the pasture large enough for 2 horses? Explain.
Big Ideas Math Solutions Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.4 9

Answer: Yes, the pasture is large enough for 2 horses.

Explanation:
Given that a horse owner must provide 4,046 square meters for each horse.
But, it is also given that the area of pasture is 86 × 96 square meters.
We have to find the Product of 86 × 96 by using the Distributive Property of Multiplication..
86 × 96 = 86 × (90 + 6)
= (86 × 90) + (86 × 6)
= (80 + 6) × 90  + (80 + 6) × 6
= (80 × 90) + (6 × 90) + (80 × 6) + (6 ×6)
=7,200 + 540 + 480 + 36
=8,256
So, 86× 96 = 8,256
By comparing the area of the pasture of each horse and the Product, we can conclude that the pasture is large enough for the 2 horses.
Note:
The Distributive property is given as:
a × (b  + c) = ( a ×  b) + ( a ×  c)

Use the Distributive Property to Multiply Two-Digit Numbers Homework & Practice 4.4

 

Question 1.
Use the area model and the Distributive Property to find 45 × 21.
Big Ideas Math Solutions Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.4 10

Answer: 945

Explanation: By using the Distributive Property of Multiplication,
45 × 21 = 45 × (20 + 1)
= (45 × 20) + (45 × 1)
= (40 + 5) × 20  + (40 + 5) ×1
= (20 × 40) + (5 × 20) + (40 × 1) + (1 ×5)
=800 + 100 + 40 + 5
=945
So, 45× 21 = 945
Note:
The Distributive property is given as:
a × (b  + c) = ( a ×  b) + ( a ×  c)

Question 2.
Use the Distributive Property to find the product.
34 × 49 = 34 × (40 + 9)
= (34 × 40) + (34 × 9)
= (30 + 4) × 40 + (30 + 4) × 9
= (30 × 40) + (4 × 40) + (30 × 9) + (4 × 9)
= 1,200 +160 + 270 + 36
= 1,666
So, 34 × 49 =1,666

Question 3.
14 × 27 = ______

Answer: 378

Explanation: Using the Distributive Property of Multiplication,
14 × 27 = 14 × (20 + 7)
= (14 × 20) + (14 × 7)
= (10 + 4) × 20  + (10 + 4) × 7
= (10 × 20) + (4 × 20) + (10 × 7) + (4 ×7)
=200 + 80 + 70 + 28
=378
So, 14× 27 = 378
Note:
The Distributive property is given as:
a × (b  + c) = ( a ×  b) + ( a ×  c)

Question 4.
38 × 31 = ______

Answer: 1,178

Explanation: Using the Distributive Property of Multiplication,
38 × 31 = 38 × (30 + 1)
= (38 × 30) + (38 × 1)
= (30 + 8) × 30  + (30 + 8) ×1
= (30 × 30) + (8× 30) + (30 × 1) + (8 ×1)
=900 + 240 + 30 + 8
=1,178
So, 38× 31 = 1,178
Note:
The Distributive property is given as:
a × (b  + c) = ( a ×  b) + ( a ×  c)

Question 5.
58 × 26 = ______

Answer: 1,508

Explanation: Using the Distributive Property of Multiplication,
58 × 26 = 58 × (20 + 6)
= (58 × 20) + (58 × 6)
= (50 + 8) × 20  + (50 + 8) × 6
= (50 × 20) + (8 × 20) + (50 × 6) + (8 ×6)
=1,000 + 160 + 300 + 48
=1,508
So, 58× 26 = 1,508
Note:
The Distributive property is given as:
a × (b  + c) = ( a ×  b) + ( a ×  c)

Question 6.
56 × 32 = ______

Answer: 1,792

Explanation: Using the Distributive Property of Multiplication,
56 × 32 = 56 × (30 + 2)
= (56 × 30) + (56× 2)
= (50 + 6) × 30  + (50 + 6) × 2
= (30 × 50) + (6 × 30) + (50 × 2) + (6 ×2)
=1,500 + 180 + 100 + 12
=1,792
So, 56× 32 = 1,792
Note:
The Distributive property is given as:
a × (b  + c) = ( a ×  b) + ( a ×  c)

Question 7.
87 × 23 = ______

Answer: 2,001

Explanation: Using the Distributive Property of Multiplication,
87 × 23 = 87 × (20 + 3)
= (87 × 20) + (87 × 3)
= (80 + 7) × 20  + (80 + 7) × 3
= (80 × 20) + (7 × 20) + (3 × 80) + (7 ×3)
=1,600 + 140 + 240 + 21
=2,001
So, 87× 23 = 2,001
Note:
The Distributive property is given as:
a × (b  + c) = ( a ×  b) + ( a ×  c)

Question 8.
95 × 81 = ______

Answer: 7,695

Explanation: Using the Distributive Property of Multiplication,
95 × 81 = 95 × (80 + 1)
= (95 × 80) + (95 × 1)
= (90 + 5) × 80  + (90 + 5) × 1
= (90 × 80) + (5 × 80) + (90 × 1) + (1 ×5)
=7,200 + 400 + 90 + 5
=7,695
So, 95× 81 = 7,695
Note:
The Distributive property is given as:
a × (b  + c) = ( a ×  b) + ( a ×  c)

Question 9.
DIG DEEPER!
Find 42 × 78 by breaking apart 42 first.

Answer: 3,276

Explanation: Using the Distributive Property of Multiplication,
42 × 78 = 78 × (40 + 2)
= (78 × 40) + (78 × 2)
= (70 + 8) × 40  + (70 + 8) × 2
= (70 × 40) + (8 × 40) + (70 × 8) + (8 ×2)
=2,800 + 320 + 560 + 16
=3,276
So, 42× 78 = 3,276
Note:
The Distributive property is given as:
a × (b  + c) = ( a ×  b) + ( a ×  c)

Question 10.
Modeling Real Life
The Elephant Building is 335 feet high. A real Asian elephant is 12 feet tall. If 29 real elephants could stand on top of each other, would they reach the top of the building?
Big Ideas Math Solutions Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.4 11

Answer: The 29 real elephants would reach the top of the building even when they stand on top of each other.

Explanation:
Give that the Elephant Building is 335 feet high. It is also given that a real Asian Elephant is 12 feet tall and there are 29 real elephants.
So, we have to find the height of 29 real elephants. We can find it using the product of 29 × 12.
We find the product by using the Distributive Property of Multiplication.
29 × 12 = 12 × (20 + 9)
= (12 × 20) + (12 × 9)
= (10 + 2) × 20  + (10 + 2) × 9
= (10 × 20) + (2 × 20) + (10 × 9) + (2 ×9)
=200 + 40 + 90 + 18
=348
So, 29× 12 = 348
We get the result of the Product as 348 feet but given that the Elephant Building is 335 feet high.
From this, we can conclude that if 29 real elephants could stand on top of each other,  they would reach the top of the building
Note:
The Distributive property is given as:
a × (b  + c) = ( a ×  b) + ( a ×  c)
Review & Refresh
Find the difference. Then check your answer.

Question 11.
Big Ideas Math Solutions Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.4 12

Answer: 25,259

Explanation:

The difference between the multi-digit numbers can be found by the difference taken from the left-most digit. If the number we want to subtract is less than the number to be subtracted, then we will take the carry from the Previous digit and the Previous digit contains 1 less number.

Question 12.
Big Ideas Math Solutions Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.4 13

Answer: 53,162

Explanation:
The difference between the multi-digit numbers can be found by the difference taken from the left-most digit. If the number we want to subtract is less than the number to be subtracted, then we will take the carry from the Previous digit and the Previous digit contains 1 less number.

Question 13.
Big Ideas Math Solutions Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.4 14

Answer: 140,938

Explanation:
The difference between the multi-digit numbers can be found by the difference taken from the left-most digit. If the number we want to subtract is less than the number to be subtracted, then we will take the carry from the Previous digit and the Previous digit contains 1 less number.

Lesson 4.5 Use Partial Products to Multiply Two-Digit Numbers

Explore and Grow

How can you use the rectangles to find 24 × 53? Complete the equation.
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.5 1
24 × 53 = _____

Answer: 1,272

Explanation:
By using the partial products method,
24 × 53 = 50 × 20 +3 × 20 + 4 × 3 + 50 ×4
= 1,000 + 60 +12 + 200 = 1,272
So, 53 × 24 = 1,272
Reasoning
What does the area of each rectangle represent.

Answer:
Think and Grow: Use Partial Products to Multiply Two-DigitNumbers
Example
Use place value and partial products to find 27 × 48.
Estimate: 30 × 50 = 1,500

So, 27 × 48 =1296
Check: Because 1,296 is close to the estimate, 1,500, the answer is reasonable.
Show and Grow
Find the product. Check whether your answer is reasonable.

Question 1.
Estimate: ______
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.5 3

Answer:  585

Explanation:
Using the Partial Products method,
39 × 15 = ( 30 + 9) × ( 10 + 5)
= 30 × 10 + 9 × 10 + 30 × 5 + 9 × 5
= 300 + 90 + 150 + 45
= 585
Estimate:
Let 39 be Rounded to 40.
So, 40 × 15 = 600
As the Estimate and the actual answer are near, the answer is reasonable.

Question 2.
Estimate: ______
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.5 4

Answer: 5,166

Explanation:
Using the Partial Products method,
82 × 63 = ( 80 + 2) × ( 60 + 3)
= 80 × 60 + 2 × 60 + 80 × 3 + 2 × 3
= 4,800 + 120 + 240 + 6
= 5,166
Estimate:
Let 82 be Rounded to 80.
Let 63 be Rounded to 65.
So, 80 × 65 = 5,200
As the Estimate and the actual answer are near, the answer is reasonable.

Question 3.
Estimate: _______
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.5 5

Answer: 3,976

Explanation:
Using the Partial Products method,
56 × 71 = ( 50 + 6) × ( 70 + 1)
= 50 × 70 + 1 × 50 + 70 × 6 + 6 × 1
= 3,500 + 50 + 420 + 6
= 3,976
Estimate:
Let 56 be Rounded to 55.
Let 71 be Rounded to 70.
So, 55 × 70 = 3,850
As the Estimate and the actual answer are not  near, the answer is not  reasonable.
Apply and Grow: Practice
Find the product. Check whether your answer is reasonable.

Question 4.
Estimate: _____
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.5 6

Answer: 364

Explanation:
Using the Partial Products method,
14 × 26 = ( 10 + 4) × ( 20 + 6)
= 10 × 20 + 10 × 6 + 20 × 4 + 4 × 6
= 200 + 60 + 80 + 24
= 364
Estimate:
Let 26 be Rounded to 25.
Let 14 be Rounded to 15.
So, 25 × 15 = 375
As the Estimate and the actual answer are near, the answer is reasonable.

Question 5.
Estimate: _____
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.5 7

Answer: 1,767

Explanation:
Using the Partial Products method,
57 × 31 = ( 50 + 7) × ( 30 + 1)
= 50 × 30 + 1 × 50 + 30 × 7 + 7 ×1
= 1,500 + 50 + 210 + 7
= 1,767
Estimate:
Let 57 be Rounded to 55. (or) Let 57 be Rounded to 60.
Let 31 be Rounded to 30.
So, 30 × 55 = 1,650 (or) 30 × 60 = 1,800
As the Estimate and the actual answer are near, the answer is reasonable. ( Depending on the Estimate value).

Question 6.
Estimate: _______
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.5 8

Answer: 2,116

Explanation:
Using the Partial Products method,
23 × 92 = ( 20 +3) × ( 90 + 2)
= 20 × 90 + 2 × 20 + 90 × 3 + 3 ×2
= 1,800 + 40 + 270 + 6
= 2,116
Estimate:
Let 23 be Rounded to 25.
Let 92 be Rounded to 90.
So, 90 × 25 = 2,250 
As the Estimate and the actual answer are near, the answer is reasonable.

Question 7.
Estimate: ______
13 × 98 = ______

Answer: 1,274

Explanation:
Using the Partial Products method,
13 × 98 = ( 10 + 3) × ( 90 +8)
= 10 × 90 + 3 × 90 + 10 × 8 + 3 ×8
= 900 + 270 + 80 + 24
= 1,274
Estimate:
Let 13 be Rounded to 15.
Let 98 be Rounded to 100.
So, 15 × 100 = 1,500 
As the Estimate and the actual answer are not near, the answer is not  reasonable.

Question 8.
Estimate: ______
65 × 22 = ______

Answer: 1,430

Explanation:
Using the Partial Products method,
65 × 22 = ( 60 + 5) × ( 20 + 2)
= 60 × 20 + 60 × 2 + 20 × 5 + 5 ×2
= 1,200 + 120 + 100 + 10
= 1,430
Estimate:
Let 22 be Rounded to 20.
So, 20 × 65 = 1,300
As the Estimate and the actual answer are not  near, the answer is not reasonable.

Question 9.
Estimate: ______
72 × 81 = _____

Answer: 5,832

Explanation:
Using the Partial Products method,
72 × 81 = ( 70 + 2) × (80 + 1)
= 70 × 80 + 1 × 70 + 80 × 2 + 2 ×1
= 5,600 + 70 + 160 + 2
= 5,832
Estimate:
Let 72 be Rounded to 70.
Let 81 be Rounded to 80.
So, 70 × 80 = 5,600
As the Estimate and the actual answer are not near, the answer is not  reasonable.

Question 10.
A farmer has 58 cows. Each cow produces 29 liters of milk. How many liters of milk do the cows produce in all?
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.5 9

Answer: 1,682 liters

Explanation:
Given that a farmer has 58 cows and each cow gives 29 liters of milk.
So, to find the total quantity of milk, we have to find the Product of 58 × 29 by using the Partial Products method.
Now,
Using the Partial Products method,
58 × 29 = ( 50 + 8) × ( 20 + 9)
= 50 × 20 + 9 × 50 + 20 ×8 + 8 ×9
= 1,000 + 450 + 160 + 72
= 1,682
From the above value, we can conclude that the total quantity of milk produces is 1,682 liters.

Question 11.
Number Sense
How much greater is the product of 12 and 82 than the product of 11 and 82? Explain how you know without multiplying.

Answer: The product of 12 and 82 is greater than 82 than the product of 11 and 82.

Explanation:
The given Products are
A) product of 12 and 82  B) product of 11 and 82
From the above, we can see that both the products have “82” as a Common number. So, find the difference between the remaining 2 numbers and we find that difference as ‘1’.
From this, we can conclude that the product of 12 and 82 is greater than 82 than the product of 11 and 82.
To verify this, we can find the products using the Partial Products method.
Now,
12 × 82 = ( 10 + 2) × ( 80 +2)
= 10 × 80 + 10 × 2 + 80 × 2 + 2 ×2
= 800 + 20 + 160 + 4
= 984
11 × 82 = ( 10 + 1) × ( 80 + 2)
= 10 × 80 + 10 × 2 + 80 ×1 + 2 × 1
= 800 + 20 + 80 + 2
= 902
Now, 984 – 902 = 82
By multiplication also, we can conclude that the product of 12 and 82 is greater than 82 than the product of 11 and 82.

Question 12.
DIG DEEPER!
Write the multiplication equation shown by the partial products.
200 + 60 + 50 + 15

Answer:  13 × 25

Explanation:
By using the Partial Products method,
200 + 60 + 50 + 15
= 20 × 10 + 20 × 3 + 5 ×10 + 5 × 3
= ( 10 + 3) × 20 +( 10 + 3) × 5
= ( 10 + 3) × ( 20 + 5)
= 13 × 25
Think and Grow: Modeling Real Life
Example
How many hours does a koala sleep in 2 weeks?
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.5 10
Find how many hours a koala sleeps each day.

We know that,
1 week = 7 days
So,
2 weeks = 2 × 7 days = 14 days
Hence,
A koala sleeps 308 hours in 2 weeks.
Show and Grow

Question 13.
Use the table above. How many hours does a python sleep in 3 weeks?
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.5 12

Answer: 378 hours

Explanation:
From the above table, we can conclude that the python sleeps for 18 hours a day.
We know that,
1 week = 7 days
So,
3 weeks = 3 × 7 = 21 days
Hence, to find how many hours a python sleep in 3 weeks, we have to find the product of 18 × 21.
Using the Partial Products method,
18 × 21 = ( 10 + 8) × ( 20 +1)
= 10 × 20 + 10 × 1 + 20 × 8 + 8 ×1
= 200 + 10 + 160 + 8
= 378
Hence, from the above,
we can conclude that the python sleeps for 378 hours in 3 weeks.

Question 14.
You have 12 packets of pea seeds and23 packets of cucumber seeds. How many fewer pea seeds do you have than cucumber seeds?
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.5 13

Answer:  102 pea seeds are less than cucumber seeds

Explanation:
Given that there are 12 packets of pea seeds and 23 packets of cucumber seeds.
From the table,
The number of seeds in each packet of pea seeds = 12 × 6 = 72
The number of seeds in each packet of cucumber = 12 × 3 = 36 + 6 = 42
So,
The total number of seeds in pea = 72  × 12
The total number of seeds in cucumber = 42 × 23
By using the Partial Products method,
72 × 12 = ( 10 + 2) × ( 70 +2)
= 10 × 70 + 10 × 2 + 2 × 70 + 2 ×2
= 700 + 20 + 140 + 4
= 864
42 × 23 = ( 40 + 2) × ( 20 +3)
= 40 × 20 + 40 × 3 + 20 × 2 + 2 ×3
= 800 + 120 + 40 + 6
= 966
So,
The number of pea seeds less than the cucumber seeds = 966 – 864 = 102 seeds

Use Partial Products to Multiply Two-Digit Numbers Homework & Practice 4.5

Find the product. Check whether your answer is reasonable.

Question 1.
Estimate: _______
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.5 14

Answer: 442

Explanation:
Using the Partial Products method,
17 × 26 = ( 10 + 7) × ( 20 + 6)
= 10 × 20 + 10 × 6 + 20 × 7 + 7 × 6
= 200 + 140 + 60 + 24
= 424
Estimate:
Let 17 be Rounded to 15.
Let 26 be Rounded to 25.
So, 25 × 15 = 375
As the Estimate and the actual answer are not near, the answer is not reasonable.

Question 2.
Estimate: _____
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.5 15

Answer: 2,356

Explanation:
Using the Partial Products method,
38 × 62 = ( 30 + 8) × ( 60 + 2)
= 30 × 60 + 30 × 2 + 60 × 8 + 8 × 2
= 1,800 + 60 + 480 + 16
= 2,356
Estimate:
Let 38 be Rounded to 40.
Let 62 be Rounded to 60.
So, 40 × 60 = 2,400
As the Estimate and the actual answer are near, the answer is reasonable.

Question 3.
Estimate: ______
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.5 16

Answer: 3,913

Explanation:
Using the Partial Products method,
91 × 43 = ( 90 + 1) × ( 40 + 3)
= 90 × 40 + 90 × 3 + 40 × 1 + 3 × 1
= 3,600 + 270 + 40 + 3
= 3,913
Estimate:
Let 91 be Rounded to 90.
Let 43 be Rounded to 45.
So, 90 × 45 = 4,050
As the Estimate and the actual answer are not near, the answer is not reasonable.
Find the product. Check whether your answer is reasonable.

Question 4.
Estimate: _____
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.5 17

Answer: 3,774

Explanation:
Using the Partial Products method,
51 × 74 = ( 50 + 1) × ( 70 + 4)
= 50 × 70 + 50 × 4 + 70 × 1 + 4 × 1
= 3,500 + 200 + 70 + 4
= 3,774
Estimate:
Let 51 be Rounded to 50.
Let 74 be Rounded to 75.
So, 50 × 75 = 3,750
As the Estimate and the actual answer are near, the answer is reasonable.

Question 5.
Estimate: ______
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.5 18

Answer: 532

Explanation:
Using the Partial Products method,
28 × 19 = ( 20 + 8) × ( 10 + 9)
= 10 × 20 + 20 × 9 + 10 × 8 + 9 × 8
= 200 + 180 + 80 + 72
= 532
Estimate:
Let 28 be Rounded to 30.
Let 19 be Rounded to 20.
So, 30 × 15 = 600
As the Estimate and the actual answer are near, the answer is reasonable.

Question 6.
Estimate: ______
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.5 19

Answer: 1,575

Explanation:
Using the Partial Products method,
35 × 45 = ( 30 + 5) × ( 40 + 5)
= 30 × 40 + 30 × 5 + 40 × 5 + 5 × 5
= 1,200 + 150 + 200 + 25
= 1,575
There is no need for Estimate because they are already rounded numbers. So, the answer is reasonable.

Question 7.

Estimate: ______
82 × 63 = ______

Answer: 5,166

Explanation:
Using the Partial Products method,
82 × 63 = ( 80 + 2) × ( 60 + 3)
= 80 × 60 + 80 × 3 + 60 × 2 + 2 × 3
= 4,800 + 240 + 120 + 6
= 5,166
Estimate:
Let 82 be Rounded to 80.
Let 63 be Rounded to 65.
So, 80 × 65 = 5,200
As the Estimate and the actual answer are near, the answer is reasonable.

Question 8.
Estimate: ______
36 × 93 = ______

Answer: 3,348

Explanation:
Using the Partial Products method,
36 × 93 = ( 30 + 6) × ( 90 + 3)
= 30 × 90 + 30 × 3 + 90 × 6 + 3 × 6
= 2,700 + 90 + 540 + 18
= 3,348
Estimate:
Let 36 be Rounded to 35.
Let 93 be Rounded to 95.
So, 35 × 95 = 3,325
As the Estimate and the actual answer are near, the answer is reasonable.

Question 9.
Estimate: _______
57 × 22 = ______

Answer: 1,254

Explanation:
Using the Partial Products method,
57 × 22 = ( 50 + 7) × ( 20 + 2)
= 50 × 20 + 50 × 2 + 20 × 7 + 7 × 2
= 1,000 + 100 + 140 + 14
= 1,254
Estimate:
Let 57 be Rounded to 55.
Let 22 be Rounded to 20.
So, 55 × 20 = 1,100
As the Estimate and the actual answer are not near, the answer is not reasonable.

Question 10.
DIG DEEPER!
Find the missing digits. Then find the product.
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.5 20

Answer: The missing digits are 1, 1

Explanation:
Using the Partial Products method,
100 + 50 + 60 + 30
= 10 × 10 + 10 × 5 + 10 × 6 + 5 × 6
= 10( 10 + 5) + 6( 10 + 5)
= ( 10 + 5) × ( 10 + 6)
= 15 × 16
So,
From the above, we can conclude that the missing digits are: 1, 1

Question 11.
Modeling Real Life
If Newton meets his goal each month, how many liters of water will he drink in 1 year?
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.5 21

Answer: 396 liters of water
Explanation;
Given,
Each ♦ = 6 liters
From this,
Half ♦ = 3 liters
From the above table,
Monthly water Intake by Newton = 6 × 5 + 3 = 33 liters
So, The yearly intake by Newton can find out by the product of 33 × 12. ( Since a year has 12 months)
Using the Partial Products method,
33 × 12 = ( 30 + 3) × ( 10 + 2)
= 30 × 10 + 30 × 2 + 10 × 3 + 3 × 2
= 300 + 60 + 30 + 6
= 396 liters
From the above,
we can conclude that the yearly intake by Newton is: 396 liters

Question 12.
Modeling Real Life
Use the table in Exercise 11. If you and Descartes each meet your goal each month, how many more liters will you drink in 1 year than Descartes?
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.5 22

Answer: 468 liters
Explanation;
From the above table,
Given that,
The monthly intake of water by you = 6 × 7 + 3 = 45 liters
The monthly intake of water by Descartes = 6 liters
We know that 1 year consists of 12 months.
So,
The yearly intake of water by Descartes = 12 × 6 = 72 liters
The yearly intake of water by you = 45 × 12
We have to find 45 × 12 using the Partial Products method.
45 × 12 = ( 40 + 5) × ( 10 +2)
= 10 × 40 + 40 × 2 + 5 × 10 + 5 ×2
= 400 + 80 + 50 + 10
= 540 liters
Hence,
The amount of water you drink more than Descartes = 540 – 72 = 468 liters
Review & Refresh
Add or subtract. Then check your answer.

Question 13.
512,006 + 318,071 = ______

Answer: 830,077

Explanation:
To find the sum, add the digits starting from the Right-most position. If there is “Carry”, then add that carry to the result of the next Position value.

Question 14.
746,620 – 529,706 = ______

Answer: 216,914

Explanation:
The difference between the multi-digit numbers can be found by the difference taken from the left-most digit. If the number we want to subtract is less than the number to be subtracted, then we will take the carry from the Previous digit and the Previous digit contains 1 less number.

Lesson 4.6 Multiply Two-Digit Numbers

Explore and Grow

Use base ten blocks to model each product. Draw each model.
Big Ideas Math Answers 4th Grade Chapter 4 Multiply by Two-Digit Numbers 4.6 1

Answer:
Reasoning
How are the models related to the product 17 × 13?

Answer:
The above model shows the pattern of the Partial Products method.
According to the Partial Products method,
(10 +7) × (10 + 3 ) ( As in the above model)
= 17 × 13
Think and Grow: Multiply Two-Digit Numbers

Show and Grow
Find the product. Check whether your answer is reasonable.

Question 1.
Estimate: ______
Big Ideas Math Answers 4th Grade Chapter 4 Multiply by Two-Digit Numbers 4.6 3

Answer: 1,312

Explanation: Using the Partial Products method,
41 × 32 = ( 40 + 1) × ( 30 +2)
= 30 × 40 + 40 × 2 + 1 × 30 + 1 ×2
= 1,200 + 80 + 30 + 2
= 1,312
Estimate:
Let 41 be Rounded to 40.
Let 32 be Rounded to 30.
So, 40 × 30 = 1,200
As the Estimate and the actual answer are not near, the answer is not reasonable.

Question 2.
Estimate: _____
Big Ideas Math Answers 4th Grade Chapter 4 Multiply by Two-Digit Numbers 4.6 4

Answer: 2,392

Explanation: Using the Partial Products method,
52 × 46 = ( 50 + 2) × ( 40 +6)
= 50 × 40 + 50 × 6 +2 × 40 + 6 ×2
= 2,000 + 300 + 80 + 12
= 2,392
Estimate:
Let 52 be Rounded to 50.
Let 46 be Rounded to 45.
So, 45 × 50 = 2,250
As the Estimate and the actual answer are not near, the answer is not reasonable.

Question 3.
Estimate: ______
Big Ideas Math Answers 4th Grade Chapter 4 Multiply by Two-Digit Numbers 4.6 5

Answer: 2,730

Explanation: Using the Partial Products method,
78 × 35 = ( 70 + 8) × ( 30 +5)
= 30 × 70 + 70 × 5 + 8 × 30 + 8 ×5
= 2,100 + 350 + 240 + 40
= 2,730
Estimate:
Let 78 be Rounded to 80.
So, 35 × 80 = 2,800
As the Estimate and the actual answer are  near, the answer is reasonable.
Apply and Grow: Practice
Find the product. Check whether your answer is reasonable.

Question 4.
Estimate: _____
Big Ideas Math Answers 4th Grade Chapter 4 Multiply by Two-Digit Numbers 4.6 6

Answer: 516

Explanation: Using the Partial Products method,
12 × 43 = ( 10 + 2) × ( 40 +3)
= 10 × 40 + 10 × 3 + 2 × 40 + 3 ×2
= 400 + 30 + 80 + 6
= 516
Estimate:
Let 12 be Rounded to 10.
Let 43 be Rounded to 45.
So, 45 × 10 = 450
As the Estimate and the actual answer are near, the answer is  reasonable.

Question 5.
Estimate: ______
Big Ideas Math Answers 4th Grade Chapter 4 Multiply by Two-Digit Numbers 4.6 7

Answer: 1,992

Explanation: Using the Partial Products method,
83 × 24 = ( 80 + 3) × ( 20 +4)
= 80 × 20 + 80 × 4 + 3 × 20 + 3 ×4
= 1,600 + 320 + 60 + 12
= 1,992
Estimate:
Let 83 be Rounded to 85.
Let 24 be Rounded to 25.
So, 85 × 25 = 2,152
As the Estimate and the actual answer are not near, the answer is not reasonable.

Question 6.
Estimate: _____
Big Ideas Math Answers 4th Grade Chapter 4 Multiply by Two-Digit Numbers 4.6 8

Answer: 4,484

Explanation: Using the Partial Products method,
59 × 76 = ( 50 + 9) × ( 70 +6)
= 50 × 70 + 50 × 6 + 9 × 70 + 9 ×6
= 3,500 + 300 + 630 + 54
= 4,484
EStimate:
Let 59 be Rounded to 60.
Let 76 be Rounded to 75.
So, 60 × 75 = 4,500
As the Estimate and the actual answer are near, the answer is  reasonable.

Question 7.
Estimate: ______
22 × 41 = ______

Answer: 902

Explanation: Using the Partial Products method,
41 × 22 = ( 40 + 1) × ( 20 +2)
= 20 × 40 + 40 × 2 + 1 × 20 + 1 ×2
= 800 + 80 + 20 + 2
= 902
Estimate:
Let 41 be Rounded to 40.
Let 22 be Rounded to 20.
So, 40 × 20 = 800
As the Estimate and the actual answer are not near, the answer is not reasonable.

Question 8.
Estimate: _____
94 × 32 = ______

Answer: 3,008

Explanation: Using the Partial Products method,
94 × 32 = ( 90 + 4) × ( 30 +2)
= 30 × 90 + 90 × 2 + 4 × 30 + 4 ×2
= 2,700 + 180 + 120 + 8
= 3,008
Estimate:
Let 94 be Rounded to 95.
Let 32 be Rounded to 30.
So, 95 × 30 = 2,850
As the Estimate and the actual answer are not near, the answer is not reasonable.

Question 9.
Estimate: _____
63 × 54 = _____

Answer: 3,402

Explanation: Using the Partial Products method,
63 × 54 = ( 60 + 3) × ( 50 +4)
= 60 × 50 + 60 × 4 + 3 × 50 + 3 ×4
= 3,000 + 240 + 150 + 12
= 3,402
Estimate:
Let 63 be Rounded to 65.
Let 54 be Rounded to 55.
So, 55 × 65 = 3,575
As the Estimate and the actual answer are not near, the answer is not reasonable.

Question 10.
Newton eats 14 treats each week. Each treat has 33 calories. How many treat calories does Newton eat each week?
Big Ideas Math Answers 4th Grade Chapter 4 Multiply by Two-Digit Numbers 4.6 9

Answer: 462 treat calories

Explanation:
Given that Newton eats 14 treats each week and each treat has 33 calories.
So, to find the total number of treat calories, we have to find the product of 14 × 33 by using the Partial Products method.
14 × 33 = ( 10 + 4) × ( 30 +3)
= 10 × 30 + 10 × 3 + 30 × 4 + 3 ×4
= 300 + 30 + 120 + 12
= 462
From the above,
we can conclude that Newton eats 462 treat calories each week.

Question 11.
YOU BE THE TEACHER
Your friend finds 43 × 26. Is he correct? Explain.
Big Ideas Math Answers 4th Grade Chapter 4 Multiply by Two-Digit Numbers 4.6 10

Answer: Your friend is not correct.

Explanation:
First, we will find 43 × 26 by using the Partial Products method.
43 × 26 = ( 40 + 3) × ( 20 +6)
= 40 × 20 + 40 × 6 + 20 × 3 + 3 ×6
= 800 + 240 + 60 + 18
= 860 + 258
= 1,118
Fro the above, we can conclude that 258 must be placed instead of 248

Question 12.
Open-Ended
Use 4 cards to write 2 two-digit numbers that have a product that is close to, but not greater than, 1,200.
Big Ideas Math Answers 4th Grade Chapter 4 Multiply by Two-Digit Numbers 4.6 11
_______ × ______

Answer:
Think and Grow: Modeling Real Life
Example
There are 16 hours in 1 day on Neptune. There are 88 times as many hours in 1 day on Mercury as 1 day on Neptune. There are 5,832 hours in 1 day on Venus. Are there more hours in 1 day on Mercury or 1 day on Venus?
Big Ideas Math Answers 4th Grade Chapter 4 Multiply by Two-Digit Numbers 4.6 12
Multiply to find how many hours Compare if there are in 1 day on Mercury.
Using the Multiplication method,

So, there are 1,408 hours on Mercury in 1 day
It is also given that there are 5,832 hours on venus in 1 day.
So, to compare how many hours are more on Mercury when compared to venus = 5,832 – 1,408 = 4,424 hours.
So, there are more hours in 1 day on Venus.
Show and Grow

Question 13.
A ninja lantern-shark is 18 inches longer than a whale. A whale-shark is 16 times as long as the ninja lantern-shark. A hammerhead shark is 228 inches long. Is the whale shark ? or the hammerhead shark longer?
Big Ideas Math Answers 4th Grade Chapter 4 Multiply by Two-Digit Numbers 4.6 14

Answer: The whale-shark is longer when compared to the hammerhead shark.

Explanation:
Given that a ninja lantern-shark is 18 inches long and a whale shark is 16 times longer than the ninja lantern-shark
So, to find the total length of a whale shark, we have to find the product of 18 × 16 using the Partial Products method.
18 × 16 = ( 10 + 8) × ( 10 +6)
= 10 × 10 + 10 × 6 + 10 × 8 + 8 ×6
= 100 + 60 + 80 + 48
= 288 inches
It is also given that the hammerhead shark is 228 inches long.
From the above, we can conclude that the whale shark is longer than the hammerhead shark.

Question 14.
There are 24 science classrooms in a school district. Each classroom receives 3 hot plates. Each hot plate costs $56. How much do all of the hot plates cost?
Big Ideas Math Answers 4th Grade Chapter 4 Multiply by Two-Digit Numbers 4.6 15

Answer: $4,032

Explanation:
Given that there are 24 science classrooms in a school district and each classroom receives 3 hot plates.
So, the total number of hot plates received in a school district = 24 × 3 = 72 hot plates
It is also given that each hot plate costs $56.
So, to find the total cost of the hot plates we have to find the product of 56 × 72 using the Partial Products method.
56 × 72 = (50 + 6) × ( 70 +2)
= 50 × 70 + 50 × 2 + 70 × 6 + 2 ×6
= 3,500 + 100 + 420 + 12
= $4,032
From the above, we can conclude that the cost of all the hotplates in a school district is: $4,032

Question 15.
Fourteen adults and 68 students visit the art museum. What is the total cost of admission?
Big Ideas Math Answers 4th Grade Chapter 4 Multiply by Two-Digit Numbers 4.6 16

Answer: $1,656

Explanation:
From the above table,
we can see that
the admission price of an adult is: $26
the admission price of a student is: $19
It is also given that there are 14 adults and 68 students.
So,
the total admission cost of the adults is: 14 × 26
the total admission cost of the children is: 68 × 19
Using  the Partial Products method,
14 × 26 = (20 + 6) × ( 10 +4)
= 20 × 10 + 20 × 4 + 10 × 6 + 4 ×6
= 200 + 80 + 60 + 24
= $364
Using  the Partial Products method,
68 × 19 = (60 + 8) × ( 10 +9)
= 60 × 10 + 60 × 9 + 10 × 8 + 8 ×9
= 600 + 540 + 80 + 72
= $1,292
Hence,
The total cost of admission (Both adults and children) = 364 + 1,292 = $1,656

Multiply Two-Digit Numbers Homework & Practice 4.6

Find the product. Check whether your answer is reasonable.

Question 1.
Estimate: ______
Big Ideas Math Answers 4th Grade Chapter 4 Multiply by Two-Digit Numbers 4.6 17

Answer: 1,196

Explanation: Using the Partial Products method,
31 × 92 = ( 30 + 1) × ( 90 +2)
= 30 × 90 + 30 × 2 + 1 × 90 + 1 ×2
= 2,700 + 60 + 90 + 2
= 2,852
Estimate:
Let 31 be Rounded to 30.
Let 92 be Rounded to 90.
So, 30 × 90 = 2,700
As the Estimate and the actual answer are not near, the answer is not reasonable.

Question 2.
Estimate: ______
Big Ideas Math Answers 4th Grade Chapter 4 Multiply by Two-Digit Numbers 4.6 18

Answer: 3,456

Explanation: Using the Partial Products method,
48 × 72 = ( 40 + 8) × ( 70 +2)
= 70 × 40 + 40 × 2 + 8 × 70 + 8 ×2
= 2,800 + 80 + 560 + 16
= 3,456
Estimate:
Let 48 be Rounded to 50.
Let 72 be Rounded to 70.
So, 50 × 70 = 3,500
As the Estimate and the actual answer are near, the answer is  reasonable.

Question 3.
Estimate: _____
Big Ideas Math Answers 4th Grade Chapter 4 Multiply by Two-Digit Numbers 4.6 19

Answer: 1,290

Explanation: Using the Partial Products method,
15 × 86 = ( 10 + 5) × ( 80 +6)
= 10 × 80 + 10 × 6 + 5 × 80 + 5 × 6
= 800 + 60 + 400 + 30
= 1,290
Estimate:
Let 86 be Rounded to 85.
So, 85 × 15 = 1,275
As the Estimate and the actual answer are near, the answer is reasonable.

Question 4.
Estimate: ______
Big Ideas Math Answers 4th Grade Chapter 4 Multiply by Two-Digit Numbers 4.6 20

Answer: 4,374

Explanation: Using the Partial Products method,
81 × 54 = ( 80 + 1) × ( 50 +4)
= 80 × 50 + 80 × 4 + 1 × 50 + 1 × 4
= 4,000 + 320 + 50 + 4
= 4,374
Estimate:
Let 81 be Rounded to 80.
Let 54 be Rounded to 55.
So, 55 × 80 = 4,400
As the Estimate and the actual answer are near, the answer is reasonable.

Question 5.
Estimate: ______
Big Ideas Math Answers 4th Grade Chapter 4 Multiply by Two-Digit Numbers 4.6 21

Answer: 1,426

Explanation: Using the Partial Products method,
23 × 62 = ( 20 + 3) × ( 60 +2)
= 20 × 60 + 20 × 2 + 3 × 60 + 3 ×2
= 1,200 + 40 + 180 + 6
= 1,426
Estimate:
Let 23 be Rounded to 25.
Let 62 be Rounded to 60.
So, 25 × 60 = 1,500
As the Estimate and the actual answer are near, the answer is  reasonable.

Question 6.
Estimate: _____
Big Ideas Math Answers 4th Grade Chapter 4 Multiply by Two-Digit Numbers 4.6 22

Answer: 5,335

Explanation: Using the Partial Products method,
97 × 55 = ( 90 + 7) × ( 50 +5)
= 90 × 50 + 90 × 5 + 7 × 50 + 7 × 5
= 4,500 + 450 + 350 + 35
= 5,335
Estimate:
Let 97 be Rounded to 95.
So, 55 × 95 = 5,225
As the Estimate and the actual answer are near, the answer is  reasonable.
Find the product. Check whether your answer is reasonable.

Question 7.
Estimate: ______
51 × 62 = ______

Answer: 3,162

Explanation: Using the Partial Products method,
51 × 62 = ( 50 +1) × ( 60 +2)
= 60 × 50 + 50 × 2 + 1 × 60 + 1 × 2
= 3,000 + 100 + 60 + 2
= 3,162
Estimate:
Let 51 be Rounded to 50.
Let 62 be Rounded to 60.
So, 50 × 60 = 3,000
As the Estimate and the actual answer are near, the answer is  reasonable.

Question 8.
Estimate: ______
37 × 13 = ______

Answer: 481

Explanation: Using the Partial Products method,
37 × 13 = ( 10 +3) × ( 30 +7)
= 10 × 30 + 10 × 7 + 3 × 30 + 3 × 7
= 300 + 70 + 90 + 21
= 481
Estimate:
Let 37 be Rounded to 40.
Let 13 be Rounded to 15.
So, 40 × 15= 600
As the Estimate and the actual answer are not near, the answer is not reasonable.

Question 9.
Estimate: _______
49 × 78 = ______

Answer: 3,822

Explanation: Using the Partial Products method,
49 × 78 = ( 40 +9) × ( 70 +8)
= 40 × 70 + 40 × 8 + 9 × 70 + 9 × 8
= 2,800 + 320 + 630 + 72
= 3,822
Estimate:
Let 49 be Rounded to 50.
Let 78 be Rounded to 80.
So, 50 × 80 = 4,000
As the Estimate and the actual answer are near, the answer is  reasonable.

Question 10.
Newton plays 21 basketball games. He scores 12 points each game. How many points does he score in all?
Big Ideas Math Answers 4th Grade Chapter 4 Multiply by Two-Digit Numbers 4.6 23

Answer: He scored 252 points.

Explanation:
Given that Newton plays 21 basketball games and he scores 12 points each game.
So, to find the total number of points in all games, we have to find the product of 21 × 12
By using the Partial Products method,
21 × 12 = ( 20 +1) × ( 10 + 2)
= 20 × 10 + 20 × 2 + 1 × 10 + 1 × 2
= 200 + 40 + 10 + 2
= 252 points
Hence,
From the above, we can conclude that Newton had scored 252 points in 21 basketball games.

Question 11.
DIG DEEPER!
When you use regrouping to multiply two-digit numbers, why does the second partial product always end in 0?

Answer: Because we divide the partial Products in terms of 10 only.
Ex:
13 × 15 = ( 10 + 3) × (10 + 5)

Question 12.
Number Sense
Find the missing digits.
Big Ideas Math Answers 4th Grade Chapter 4 Multiply by Two-Digit Numbers 4.6 24

Answer: The missing numbers are: 6, 7, 0 and 1

Explanation:

By using the Partial Products method,
34 × 65 = ( 30 +4) × ( 60 + 5)
= 30 × 60 + 30 × 5 + 4 × 60 + 4 × 5
= 1,800 + 150 + 240 + 20
= 2,040 + 170
= 2,210
So,
From the above, we can conclude that the missing numbers are: 6, 7, 0 and 1

Question 13.
Modeling Real Life
A tiger dives 12 feet underwater. An otter dives 25 times deeper than the tiger. A walrus dives 262 feet underwater. Does the otter or walrus dive deeper?
Big Ideas Math Answers 4th Grade Chapter 4 Multiply by Two-Digit Numbers 4.6 25

Answer:  Otter drives deeper than the Walrus.

Explanation:
Given that a tiger dives 12 feet underwater and an otter dives 25 times deeper than the tiger.
So,
The total distance dived by otter = 25 × 12 feet
By using the Partial Products method,
25 × 12 = ( 20 +5) × ( 10 + 2)
= 20 × 10 + 20 × 2 + 5 × 10 + 2 × 5
= 200 + 40 + 50 + 10
= 300
It is also given that a walrus dives 262 feet underwater.
From the above,
we can conclude that the otter dives deeper than the walrus.
Review & Refresh

Question 14.
Complete the table.
Big Ideas Math Answers 4th Grade Chapter 4 Multiply by Two-Digit Numbers 4.6 26

Answer:
A) 6,835
Word Form: Six thousand, eight hundred thirty-five
Expanded Form: 6,000 + 800 + 30 + 5

Explanation:
Any number can be written in 3 forms. They are:
A) Standard Form B) Word Form C) Expanded Form
So, we can write the given form in the remaining 2 forms.
B) 70,000 + 4,000 + 100 + 2
Standard Form: 74,102
Word Form: Seventy-four thousand, One hundred two

Explanation:
Any number can be written in 3 forms. They are:
A) Standard Form B) Word Form C) Expanded Form
So, we can write the given form in the remaining 2 forms.
C) Five hundred one thousand, three hundred twenty-nine
Standard Form: 501,329
Expanded Form: 500,000 +0 + 1,000 + 100 + 0 + 2

Explanation:
Any number can be written in 3 forms. They are:
A) Standard Form B) Word Form C) Expanded Form
So, we can write the given form in the remaining 2 forms.

Lesson 4.7 Practice Multiplication Strategies

Explore and Grow
Choose any strategy to find 60 × 80.
Big Ideas Math Answers Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.7 1

Answer: The strategy to find 60 × 80 is: Associative Property of Multiplication

Explanation: By using the Associative Property of Multiplication,
60 ×80 =60 × (8 × 10)
= (60 × 8) × 10
= (6 × 10 × 8) × 10
= 480 × 10
= 4,800
Choose any strategy to find 72 × 13.

Answer: The strategy to find 72 × 13 is: Distributive Property

Explanation:
72 × 13 = 72 × ( 10 + 3)
= ( 72 × 10 ) + ( 72 ×3)
= ( 70 + 2 ) × 10 + ( 70 + 2 ) × 3
= ( 70 × 10 ) + ( 2 × 10 ) + ( 70 × 3 ) + ( 2 × 3)
= 700 + 20 + 210 + 6
= 936
So, 72 × 13 = 936
Reasoning
Explain why you chose your strategies. Compare your strategies to your partner’s strategies. How are they the same or different?

Answer:
Think and Grow: Practice Multiplication Strategies
Example
Find 62 × 40.
One Way: Use place value.
62 × 40 = 62 × 4 tens
=248 tens
= 248 × 10
= 2,480
So, 62 × 40 = 2,480.
Another Way: Use an area model and partial products.

So, 62 × 40 =2,480 .
Example
Find 56 × 83.
One Way: Use place value and partial products.

So, 56 × 83 = 4,648.
Another Way: Use regrouping
Multiply 56 by 3 ones. Then multiply 56 by 8 tens. Regroup if necessary.

So, 56 × 83 = 4,648.
Show and Grow
Find the product.

Question 1.
90 × 37 = _____

Answer: 3,330

Explanation: Using the Associative Property of Multiplication,
37 × 90 = 37 × (9 × 10)
= (37 × 9) × 10
= (3 × 3 × 37) × 10
= 333 × 10
= 3,330

Question 2.
78 × 21 = ______

Answer: 1,638

Explanation: Using the Distributive Property,
78 × 21 = 78 × ( 20 + 1)
= ( 78 × 20 ) + ( 78 × 1)
= ( 70 + 8 ) × 20 + ( 70 + 8 ) × 1
= ( 70 × 20 ) + ( 8 × 20 ) + ( 70 × 1 ) + ( 8 × 1)
= 1,400 + 160 + 70 + 8
= 1,638
So, 78 × 21 = 1,638

Question 3.
14 × 49 = _____

Answer: 686

Explanation: Using the Associative Property of Multiplication,
14 × 49 = 14 × ( 40 + 9)
= ( 14 × 40 ) + ( 14 × 9)
= ( 10 + 4 ) × 40 + ( 10 + 4 ) × 9
= ( 40 × 10 ) + ( 4 × 40 ) + ( 10 × 9 ) + ( 4 × 9)
= 400 + 160 + 90 + 36
= 686
So, 14 × 49 = 686
Apply and Grow: Practice
Find the product.

Question 4.
74 × 30 = _____

Answer: 2,220

Explanation: Using the Associative Property of Multiplication,
74 × 30 = 74 ( 3 × 10)
= ( 74 × 3) × 10
= 222 × 10
= 2,220
So, 74 × 30 = 2,220

Question 5.
51 × 86 = _____

Answer: 4,386

Explanation: Using the Distributive Property ,
51 × 86 = 51 × ( 80 + 6)
= ( 51 × 80 ) + ( 51 × 6)
= ( 50 + 1 ) × 80 + ( 50 + 1 ) × 6
= ( 50 × 80 ) + ( 1 × 80 ) + ( 50 × 6 ) + ( 1 × 6)
= 4,000 + 80 + 300 + 6
= 4,386
So, 51 × 86 = 4,386

Question 6.
40 × 29 = ______

Answer: 1,160

Explanation: Using the Associative Property of Multiplication,
40 × 29 = 29 × (4 × 10)
= (29 × 4) × 10
= (29× 2 × 2) × 10
= 116 × 10
= 1,160

Question 7.
92 × 80 = _____

Answer: 7,360

Explanation: Using the Associative Property of Multiplication,
92 × 80 = 92 × (8 × 10)
= (92 × 8) × 10
= (92 × 2 × 4) × 10
= 736 × 10
= 7,360

Question 8.
41 × 17 = ______

Answer: 697

Explanation: Using the Distributive Property,
41 × 17 = 41 × ( 10 + 7)
= ( 41 × 10 ) + ( 41 × 7)
= ( 40 + 1 ) × 10 + ( 40 + 1 ) × 7
= ( 40 × 10 ) + ( 1 × 10 ) + ( 40 × 7 ) + ( 1 × 7)
= 400 + 10 + 280 + 7
= 697
So, 41 × 17 = 697

Question 9.
60 × 53 = _____

Answer: 3,180

Explanation: By using the Associative Property of Multiplication,
60 × 53 = 53 × (6 × 10)
= (53 × 6) × 10
= (53× 3 × 2) × 10
= 318 × 10
= 3,180
Logic
Find the missing factor.

Question 10.
Big Ideas Math Answers Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.7 5

Answer: The missing number is: 41

Explanation:

In the given figure, the partial fractions are given.
The 1st partial fraction represents the multiplication with the unit digit and the 2nd partial fraction represents the multiplication with the tens digit.
The first partial fraction will come when 72 is multiplied with 1 and the 2nd partial fraction will come when 72 is multiplied with 4.
So, the missing number is: 41

Question 11.
Big Ideas Math Answers Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.7 6

Answer: The missing number is: 34

Explanation:

In the given figure, the partial fractions are given.
The 1st partial fraction represents the multiplication with the unit digit and the 2nd partial fraction represents the multiplication with the tens digit.
The first partial fraction will come when 65 is multiplied with 4 and the 2nd partial fraction will come when 65 is multiplied with 3.
So, the missing number is: 34

Question 12.
Big Ideas Math Answers Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.7 7

Answer: The missing number is: 87

Explanation:

In the given figure, the partial fractions are given.
The 1st partial fraction represents the multiplication with the unit digit and the 2nd partial fraction represents the multiplication with the tens digit.
The first partial fraction will come when 93 is multiplied with 7 and the 2nd partial fraction will come when 93 is multiplied with 8.
So, the missing number is: 87

Question 13.
Writing
Explain why you start multiplying with the one’s place when using regrouping to multiply.

Answer: When using ” Regrouping” to multiply, we start multiplying from the rightmost position i.e.., one’s position because the place value of that position is 1.

Question 14.
DIG DEEPER!
Find the missing digit so that both products are the same.
Big Ideas Math Answers Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.7 8

Answer: The missing digit so that both products are the same is: 3

Explanation:

The product of 26 × 15 is:
26 × 15 = 26 × ( 10 + 5)
= ( 26 × 10 ) + ( 26 × 5)
= ( 20 + 6 ) × 10 + ( 20 + 6 ) × 5
= ( 20 × 10 ) + ( 6 × 10 ) + ( 20 × 5 ) + ( 6 × 5)
= 200 + 60 + 100 + 30
= 390
So, 26 × 15 = 390
To get the same result in the next product, we have to find the missing number.
So, 390 / 30 = 13
Hence, the missing digit so that both products are the same is: 3
Think and Grow: Modeling Real Life
Example
A swinging ship ride runs 50 times each afternoon. The ship has 10 rows of benches with 4 seats in each bench. If the ship is full each time it runs, how many people will ride the swinging ship in 1 afternoon?
Big Ideas Math Answers Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.7 9
Multiply to find how many people will ride the swinging ship ride each time.
10 × 4 = 40
So,
40 people will ride the swinging ship ride each time.
Multiply to find how many people will ride the swinging ship in 1 afternoon.
40 × 50 = 40 × (5 × 10)  (Rewrite 50 as 5 × 10)
= (40 × 5) × 10 ( By using Associative Property of Multiplication)
= 200 × 10
= 2,000
Hence,
2,000 people will ride the swinging ship in 1 afternoon.
Show and Grow

Question 15.
A teacher orders 25 rock classification kits. Each kit has 4 rows with 9 rocks in each row. How many rocks are there in all?

Answer: 900 rocks

Explanation:
Given that a teacher orders 25 rock classification kits and each kit has 4 rows with 9 rocks in each row.
So,
The total number of rows present in 25 rock classification kits = 25 × 4 = 100 rows
We have to find the number of rocks in 100 rows by multiplying 100 × 9.
Now, 100 × 9 = 900 rocks.
From the above,
We can conclude that there are 900 total rocks.

Question 16.
A hotel has 12 floors with 34 rooms on each floor. 239 rooms are in use. How many rooms are not in use?

Answer: The total number of rooms that are not in use: 169

Explanation:
Given that a hotel has 12 floors with 34 rooms on each floor.
So, the total number of rooms in a hotel = 34 × 12
By using the Distributive method,
34 × 12 = 34 ( 10 + 2 )
= ( 34 × 10 ) + ( 34 × 2 )
= ( 30 + 4 ) × 10 + ( 30 + 4 ) × 2
= ( 30 × 10 ) + ( 4 × 10 ) + ( 30 × 2 ) + ( 4 × 2)
= 300 + 40 + 60 + 8
= 408
But, it is also given that 239 rooms are in use.
So, the number of rooms that are not in use = 408 – 239 = 169 rooms

Question 17.
A child ticket for a natural history museum costs $13. An adult ticket costs twice as much as a child ticket. How much does it cost for 21 children and 14 adults to go to the museum?
Big Ideas Math Answers Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.7 10

Answer: $637

Explanation:
Given that a child ticket for a natural history museum costs $13 and an adult ticket costs twice as much as a child ticket.
So, the cost of an adult ticket = 2 × $13 = $26
So, to find the total cost for 21 children and 14 adults in a museum, we have to find
21 × $13  and 14 × $26
Now,
21 × 13 = 21 ( 10 + 3 )
= ( 21 × 10 ) + ( 21 × 3 )
= ( 20 + 1 ) × 10 + ( 20 +1 ) × 3
= ( 20 × 10 ) + ( 1 × 10 ) + ( 20 × 3 ) + ( 1 × 3)
= 200 + 10 + 60 + 3
= 273
14 × 26 = 14 ( 20 + 6 )
= ( 14 × 20 ) + ( 14 × 6 )
= ( 10 + 4 ) × 20 + ( 10 +4 ) × 6
= ( 10 × 20 ) + ( 4 × 20 ) + ( 10 × 6 ) + ( 4 × 6)
= 200 + 80 + 60 + 24
= 364
Hence,
The total cost for children and adults in a museum = 273 + 364 = $637

Practice Multiplication Strategies Homework & Practice 4.7

Find the product.

Question 1.
16 × 13 = _____

Answer: 208

Explanation: By using the Distributive method,
16 × 13 = 16 ( 10 + 3 )
= ( 16 × 10 ) + ( 16 × 3 )
= ( 10 + 6 ) × 10 + ( 10 + 6 ) × 3
= ( 10 × 10 ) + ( 6 × 10 ) + ( 10 × 3 ) + ( 6 × 3)
= 100 + 30 + 60 + 18
= 208

Question 2.
29 × 50 = _____

Answer: 1,450

Explanation: By using the Associative Property of Multiplication,
29 × 50 = 29 × ( 5 × 10 )
= ( 29 × 5 ) × 10
= 145 × 10
= 1,450

Question 3.
78 × 45 = _____

Answer: 3,510

Explanation: By using the Distributive method,
78 × 45 = 78 ( 40 + 5 )
= ( 78 × 40 ) + ( 78 × 5 )
= ( 70 + 8 ) × 40 + ( 70 + 8 ) × 5
= ( 70 × 40 ) + ( 8 × 40 ) + ( 70 × 5 ) + ( 8 × 5)
= 2,800 + 320 + 350 + 40
= 3,510

Question 4.
30 × 71 = ______

Answer: 2,130

Explanation: By using the Associative Property of Multiplication,
30 × 71 = 71 × ( 3 × 10 )
= ( 71 × 3 ) × 10
= 213 × 10
= 2,130

Question 5.
62 × 14 = _____

Answer: 868

Explanation: By using the Distributive method,
62 × 14 = 62 ( 10 + 4 )
= ( 62 × 10 ) + ( 62 × 4 )
= ( 60 + 2 ) × 10 + ( 60 + 2 ) × 4
= ( 60 × 10 ) + ( 2 × 10 ) + ( 60 × 4 ) + ( 2 × 4)
= 600 + 20 + 240 + 8
= 868

Question 6.
80 × 90 = _____

Answer: 7,200

Explanation: By using the Associative Property of Multiplication,
80 × 90 = 80 × ( 9 × 10 )
= ( 80 × 9 ) × 10
= 720 × 10
= 7,200
Find the product.

Question 7.
70 × 18 = _____

Answer: 1,260

Explanation: By using the Associative Property of Multiplication,
70 × 18 = 18 × ( 7 × 10 )
= ( 18 × 7 ) × 10
= 126 × 10
= 1,260

Question 8.
32 × 59 = _____

Answer: 1,888

Explanation: By using the Distributive method,
32 × 59 = 59 ( 30 + 2 )
= ( 59 × 30 ) + ( 59 × 2 )
= (50 + 9 ) × 30 + ( 50 + 9 ) × 2
= ( 50 × 30 ) + ( 9 × 30 ) + ( 50 × 2 ) + ( 2 × 9)
= 1,500 + 270 + 100 + 18
= 1,888

Question 9.
67 × 20 = ____

Answer: 1,340

Explanation: By using the Associative Property of Multiplication,
67 × 20 = 67 × ( 2 × 10 )
= ( 67 × 2 ) × 10
= 134 × 10
= 1,340

Question 10.
51 × 84 = _____

Answer: 4,284

Explanation: By using the Distributive method,
51 × 84 = 51 ( 80 + 4 )
= ( 51 × 80 ) + ( 51 × 4 )
= (50 + 1 ) × 80 + ( 50 + 1 ) × 4
= ( 50 × 80 ) + ( 1 × 80 ) + ( 50 × 4 ) + ( 1 × 4)
= 4,000 + 80 + 200 + 4
= 4,284

Question 11.
40 × 40 = _____

Answer: 1,600

Explanation: By using the Associative Property of Multiplication,
40 × 40 = 40 × ( 4 × 10 )
= ( 40 × 4 ) × 10
= 160 × 10
= 1,600

Question 12.
23 × 97 = ______

Answer: 2,231

Explanation: By using the Distributive method,
23 × 97 = 97 ( 20 + 3 )
= ( 97 × 20 ) + ( 97 × 3 )
= (90 + 7 ) × 20 + ( 90 + 7 ) × 3
= ( 90 × 20 ) + ( 7 × 20 ) + ( 90 × 3 ) + ( 7 × 3)
= 1,800 + 140 + 270 + 21
= 2,231

Question 13.
Writing
Which strategy do you prefer to use when multiplying two-digit numbers? Explain.

Answer: The strategy you have to prepare when multiplying 2-digit numbers must depend on the numbers.
some strategies to multiply 2-digit numbers are:
A) The place-value method  B) The Associative Property of Multiplication  C) Distributive Property D) Partial Products method E) Regrouping method

Question 14.
Patterns
What number can you multiply the number of tires by to find the total weight? Use this pattern to complete the table.
Big Ideas Math Answers Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.7 11

Answer: The number you can multiply the number of tires by to find the total weight is: 20

Explanation:

From the above table,
to find the multiple of total weight, divide the number of tires by the total weight. ( Take any 2 values)
So, 80 /4 = 20 ( Multiple of total weight)
From the above,
We can conclude that the number you can multiply the number of tires by to find the total weight is: 20

Question 15.
Modeling Real Life
Each bag of popcorn makes 13 cups. A school has a movie day, and the principal brings 15 boxes of popcorn. Each box has 3 bags of popcorn. How many cups of popcorn does the principal bring?
Big Ideas Math Answers Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.7 12

Answer: 585 cups of popcorn

Explanation:
Given that each bag of popcorn makes 13 cups and the principal brings 15 boxes of popcorn. It is also given that each box has 3 bags of popcorn.
So, to find the total number of cups of popcorn the principal bring = 13 × 15 × 3
Now,
By using the Distributive method,
13 × 45 = 45 ( 10 + 3 )
= (45 × 10 ) + ( 45 × 3 )
= (40 +5 ) × 10 + ( 40 + 5 ) × 3
= ( 40 × 10 ) + ( 5 × 10 ) + ( 40 × 3 ) + ( 5 × 3)
= 400 + 50 + 120 + 15
= 585
From the above,
we can conclude that the number of cups the principal bring is: 585 cups of popcorn
Review & Refresh
Find the product.

Question 16.
8 × 200 = _____

Answer: 1,600

Explanation: By using the Associative Property of Multiplication,
8 × 200 = 8 × (20 × 10)
= (8 × 20) × 10
= (8× 5 × 4) × 10
= 160 × 10
= 1,600

Question 17.
7 × 300 = _____

Answer: 2,100

Explanation: By using the Associative Property of Multiplication,
7 × 300 = 7 × (30 × 10)
= (7 × 30) × 10
= (7× 5 × 6) × 10
= 210 × 10
= 2,100

Question 18.
6,000 × 5 = _____

Answer: 30,000

Explanation: By using the Associative Property of Multiplication,
5 × 6,000 = 5 × (600 × 10)
= (5 × 600) × 10
= (6× 5 × 100) × 10
= 300 × 10
= 3,000

Question 19.
9 × 90 = ____

Answer: 810

Explanation: By using the Associative Property of Multiplication,
9 × 90 = 9 × (9 × 10)
= (9 × 9) × 10
= 81 × 10
= 810

Question 20.
3,000 × 6 = _____

Answer: 18,000

Explanation: By using the Associative Property of Multiplication,
6 × 3,000 = 6 × (300 × 10)
= (6 × 300) × 10
= (6× 3 × 100) × 10
= 1800 × 10
= 18,000

Question 21.
5 × 500 = _____

Answer: 2,500

Explanation: By using the Associative Property of Multiplication,
5 × 500 = 5 × (50 × 10)
= (5 × 50) × 10
= (5× 5 × 10) × 10
= 250 × 10
= 2,500

Lesson 4.8 Problem Solving: Multiplication with Two-Digit Numbers

Explore & Grow
Explain, in your own words, what the problem below is asking. Then explain how you can use multiplication to solve the problem.
Big Ideas Math Solutions Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.8 1
A ferry can transport 64 cars each time it leaves a port. The ferry leaves a port 22 times in 1 day. How many cars can the ferry transport in 1 day?

Answer: 1,408 cars can ferry transport in 1 day

Explanation:
Given that a ferry can transport 64 cars each time it leaves a port and the ferry leaves a port 22 times in 1 day.
From this, the total number of cars that can ferry transport in 1 day is: 64 × 22
By using the Distributive method,
64 × 22 = 64 ( 20 + 2 )
= (64 × 20 ) + ( 64 × 2 )
= (60 + 4 ) × 20 + ( 60 + 4 ) × 2
= ( 60 × 20 ) + ( 4 × 20 ) + ( 60 × 2 ) + ( 4 × 2)
= 1,200 + 80 + 120 + 8
= 1,408
From the above,
we can conclude that the total number of cars that ferry transport in 1 day is 1,408 cars.
Construct Arguments
Make a plan to find how many cars the ferry can transport in 1 week.

Answer: 9,856 cars

Explanation:
From the above problem, we know that the total number of cars a ferry can transport in 1 day is: 1,408 cars
We know that,
1 week = 7 days
First, we have to find the total number of cars the ferry can transport in 1 day using the Product 64 × 22.
By using the Distributive method,
64 × 22 = 64 ( 20 + 2 )
= (64 × 20 ) + ( 64 × 2 )
= (60 + 4 ) × 20 + ( 60 + 4 ) × 2
= ( 60 × 20 ) + ( 4 × 20 ) + ( 60 × 2 ) + ( 4 × 2)
= 1,200 + 80 + 120 + 8
= 1,408
Now,
the total number of cars a ferry can transport in 1 week = 1,408 × 7 = 9,856 cars
Think and Grow: Problem Solving: Multiplication with Two-Digit Numbers
Example
A pet store receives a shipment of 8 boxes of dog treats. Each box is 2 feet high and has 18 bags of dog treats. How many ounces of dog treats does the pet store receive in the shipment?
Big Ideas Math Solutions Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.8 2
Understand the Problem
What do you know?
• The store receives 8 boxes.
• Each box is 2 feet high.
• Each box has 18 bags of dog treats.
• Each bag weighs 32 ounces.
What do you need to find?
• You need to find how many ounces of dog treats the pet store receives in the shipment.
Make a Plan
How will you solve it?
• Multiply 32 by 18 to find how many ounces of dog treats are in each box.
• Then multiply the product by 8 to find how many ounces of dog treats the pet store receives in the shipment.
• The height of each box is unnecessary information.
Solve
Big Ideas Math Solutions Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.8 3

Answer: 4,608 ounces

Explanation:
Given that the store receives 8 boxes and Each box has 18 bags of dog treats. It is also given that each bag weighs 32 ounces.
The number of ounces of dog treats the pet store receives in the shipment = 18 × 8 × 32
It is also given that each box is 2 feet high. But it is not necessary for the calculation of a total number of ounces.
First, we will find the weight of 1 box = 18 × 32
By using the Distributive method,
18 × 32 = 18 ( 30 + 2 )
= (18 × 30 ) + ( 18 × 2 )
= (10 + 8 ) × 30 + ( 10 + 8 ) × 2
= ( 10 ×30 ) + ( 8 × 30 ) + ( 10 × 2 ) + ( 8 × 2)
= 300 + 240 + 20 + 16
= 576
So,
The total weight of dog treats the pet store receives in the shipment = 576 × 8 = 4,608 ounces
Hence,
The pet store receives 4,608 ounces of dog treats.
Show and Grow

Question 1.
Show how to solve the problem above using one equation.

Answer: 18 × 32 × 8
Apply and Grow: Practice
Understand the problem. What do you know? What do you need to find? Explain.

Question 2.
Thirteen students create a petition for longer recess. They need 5,000 signatures in all. So far, each student has 99 signatures. How many more signatures do they need?

Answer: 3,713 signatures

Explanation:
Given that there are 13 students that create a petition for longer recess and they need 5,000 signatures in all.
It is also given that each student has 99 signatures.
So, the total number of signatures that 13 students have are: 13 × 99
By using the Distributive method,
13 × 99 = 13 ( 90 + 9 )
= (13 × 90 ) + ( 13 × 9 )
= (10 +3 ) × 90 + ( 10 + 3 ) × 9
= ( 90 × 10 ) + ( 3 × 90 ) + ( 10 × 9 ) + ( 9 × 3)
= 900 + 270 + 90 + 27
= 1,287
But, the students required 5000 signatures in total.
So,
The remainin number of signatures required by students = 5,000 – 1,287 = 3,713 signatures

Question 3.
An activity book has 35 pages and costs $7. Each page has 4 puzzles. You have completed all of the puzzles on 16 of the pages. How many puzzles do you have left to complete?

Answer: 76 puzzles

Explanation:
Given that an activity book has 35 pages and each page has 4 puzzles.
So,
The total number of puzzles = 35 × 4 = 140 puzzles
But, it is also given that the puzzles completed in all of the 16 pages and we know that each page has 4 puzzles.
So,
The number of puzzles in 16 pages = 16 × 4 = 64 puzzles
Now,
The remaining number of puzzles in the remaining pages = 140 – 64 = 76 puzzles
Understand the problem. Then make a plan. How will you solve it? Explain.

Question 4.
Twelve classes provide items for a time capsule. There are 23 students in each class. Each student puts 2 small items in the time capsule. How many items are in the time capsule?
Big Ideas Math Solutions Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.8 4

Answer: 552 items

Explanation:
Given that there are 12 classes that provide items for a time capsule and there are 23 students in each class and each student puts 2 small items in the time capsule.
First, we have to find the total number of students by the product 23 × 12
By using the Distributive method,
12 × 23 = 23( 10 + 2 )
= (23 × 10 ) + ( 23 × 2 )
= (20 +3 ) × 10 + ( 20 + 3 ) × 3
= ( 20 × 10 ) + ( 3 × 10 ) + ( 20 × 3 ) + ( 3 × 3)
= 200 + 30 + 60 + 9
= 296 students
Now,
The total number of small items put by the total number of students = 296 × 2 = 552 small items

Question 5.
A craftsman cuts letters and numbers from license plates to make signs. He has 15 Florida plates and 25 Georgia plates. Each plate has a total of 7 letters and numbers. How many letters and numbers does the craftsman cut in all?

Answer: 280 letters and numbers
Explanation;
Given that a craftsman has 15 Florida plates and 25 Georgia plates and each plate has a total of 7 letters and numbers.
Hence,
The total number of plates that a craftsman have = 25 + 15 = 40 plates
The total number of letters and numbers each plate has = 40 × 7 = 280 letters and numbers

Question 6.
DIG DEEPER!
In 1 month, the solar panel and the wind turbine can produce the kilowatt-hours of electricity shown. How much electricity can 28 solar panels and1 small wind turbine produce each month?
Big Ideas Math Solutions Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.8 5

Answer: 1,673 kilowatt-hours of electricity
Explanation;
Given that each solar panel produces 30 kilowatt-hours and a wind turbine produces 833 kilowatt-hours.
Now, the electricity produced by 28 solar panels = 28 × 30
By using the Associative Property of Multiplication,
28 × 30 = 28 × (3 × 10)
= (28 × 3) × 10
= (7× 4 × 3) × 10
= 84 × 10
= 840 kilowatt-hours
Hence,
The total electricity produced by 28 solar panels and 1 wind turbine = 840 + 833 = 1,673 kilowatt-hours

Think and Grow: Modeling Real Life
Example
An adult ticket for a zip line course costs $48.A child ticket costs $19 less than an adult ticket. In 1 day, 86 adults and 42 children ride the zipline. How much more money was earned from adult tickets than from child tickets?
Big Ideas Math Solutions Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.8 6
Think: What do you know? What do you need to find? How will you solve?
Step 1: How much money was earned from adult tickets?
$48 × 86 = ‘a’
‘a’ is an unknown product.

So,
a = 4,128
Step 2: How much money was earned from child tickets?
A child ticket costs $48 – $19 = $29.
$29 × 42 = c
c is the unknown product.

So,
c = 1,218
Step 3: Use m to represent how much more money was earned from adult tickets.

So, $ 2910 more was earned from adult tickets.
Show and Grow

Question 7.
A theater has 45 rows of 72 seats on the floor level and 22 rows of 36 seats in the balcony. How many seats are there in all?
How many more seats are on the floor level than on the balcony?

Answer: 4,032 seats
Explanation;
Given that a theater has 45 rows of 72 seats on the floor level and 22 rows of 36 seats in the balcony.
So,
The total number of seats on the floor level = 45 × 72
The total number of seats on the balcony = 22 × 36
By using the Distributive method,
72 × 45 = 45 ( 70 + 2 )
= (45 × 70 ) + ( 45 × 2 )
= (40 +5 ) × 70 + ( 40 + 5 ) × 2
= ( 40 × 70 ) + ( 5 × 70 ) + ( 40 × 2 ) + ( 5 × 2)
= 2,800 + 350 + 80 + 6
= 3,240
By using the Distributive method,
22 × 36 = 36 ( 20 + 2 )
= (36 × 20 ) + ( 36× 2 )
= (30 +6 ) × 20 + ( 30 + 6) × 2
= ( 30 × 20 ) + ( 6 × 20 ) + ( 30 × 2 ) + ( 6 × 2)
= 600 + 120 + 60 + 12
= 792
So,
The total number of seats ( Floor level and balcony) = 3,240 + 792 = 4,032 seats

Problem Solving: Multiplication with Two-Digit Numbers Homework & Practice 4.8

Understand the problem. Then make a plan. How will you solve it? Explain.

Question 1.
Seventy-two mushers compete in a sled-dog race. Each musher has 16 dogs. How many dogs compete in the race than mushers?
Big Ideas Math Solutions Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.8 10

Answer: 1,152 dogs

Explanation:
Given that there are 72 mushers compete in a sled-dog race and each musher has 16 dogs.
So,
The total number of dogs that compete in the race = 16 × 72
By using the Distributive method,
16 × 72 = 72 ( 10 + 6 )
= (72 × 10 ) + ( 72× 6 )
= (70 +2 ) × 10 + ( 70 + 2) × 6
= ( 70 × 10 ) + ( 2 × 10 ) + ( 70 × 6 ) + ( 6 × 2)
= 700 + 20 + 420 + 12
= 1,152
Hence,
The total number of dogs that compete in the race are: 1,152 dogs

Question 2.
A photographer buys 3 USB drives that each cost $5. She puts 16 folders on each drive. Each folder has 75 photographs. How many photographs does the photographer put on the USB drives in all?

Answer: 3,600 photographs
Explanation;
Given that a photographer has 3 USB drives and there are 16 folders and 75 photographers on each drive.
So,
The total number of photographs on all the USB drives = 16 × 3 × 75 = 48 × 75
By using the Distributive method,
48 × 75 = 75 (40 + 8 )
= (75 × 40 ) + ( 75 × 8 )
= (70 +5 ) × 40 + ( 70 + 5) × 8
= ( 70 × 40 ) + ( 5 × 40 ) + ( 70 × 8 ) + ( 5 × 8)
= 2,800 + 200 + 560 + 40
= 3,600 photographs
Hence,
The total number of photographs on all the USB drives are: 3,600 photographs

Question 3.
A teacher has 68 students take 25
Question test. The teacher checks the answers for 9 of the tests. How many answers does the teacher have left to check?

Answer: 1,475 answers

Explanation:
Given that a teacher has 68 students take 25
Question test.
So,
The total number of answers the teacher have to check = 68 × 25
By using the Distributive method,
68 × 25 = 68 ( 20 + 5 )
= (68 × 20 ) + ( 68× 5 )
= (60 +8 ) × 20 + ( 60 + 8) × 5
= ( 60 × 20 ) + ( 8 × 20 ) + ( 60 × 5 ) + ( 8 × 5)
= 1,200 + 160 + 300 + 40
= 1,700
It is also given that the teacher checked 9 papers only.
So.
The number of answers checked = 9 × 25 = 225
Hence,
The number of answers remained unchecked = 1,700 – 225 = 1,475 answers

Question 4.
Each day, a cyclist bikes uphill for 17 miles and downhill for 18 miles. She drinks 32 fluid ounces of water after each bike ride. How many miles does the cyclist bike in 2 weeks?

Answer: 490 miles
Explanation;
Given that a cyclist bikes uphill for 17 miles and downhill for 18 miles.
SO,
The total distance that cyclist bikes each day = 17 + 18 = 35 miles
We know that,
1 week = 7 days
S0, 2 weeks = 2 × 7  = 14 days.
So, the distance traveled by a cyclist in 2 weeks = 35 × 14
By using the Distributive method,
35 × 14 = 35 ( 10 + 4 )
= (35 × 10 ) + ( 35× 4 )
= (30 +5 ) × 10 + ( 30 + 5) × 4
= ( 30 × 10 ) + ( 5 × 10 ) + ( 30 × 4 ) + ( 5 × 4)
= 300 + 50 + 120 + 20
= 490 miles
Hence,
The distance traveled by a cyclist in 2 weeks is: 490 miles

Question 5.
Precision
Which expressions can be used to solve the problem?
Twelve friends play a game that has 308 cards. Each player receives 16 cards. How many cards are left?
(308 – 12) × 16
308 – (16 × 12)
308 – (12 × 16)
(308 – 16) – 12

Answer: Let the Expressions be named as A),  B), C) and D)
Either B) or C) can be used to solve the problem.

Explanation:
The given Expressions are;
A) (308 – 12) × 16
B) 308 – (16 × 12)
C) 308 – (12 × 16)
D) (308 – 16) – 12
It is given that there are 12 friends that have 16 cards each and there are 308 cards in total.
So,
The number of cards that have left = 308 – ( 16 × 12 ) (or) 308 – ( 12 × 16 )
Hence,
From the above,
we can conclude that either B) or C) can be used to solve the problem.

Question 6.
Modeling Real Life
A child ticket costs $14 less than an adult ticket. What is the total ticket cost for 18 adults and 37 children?
Big Ideas Math Solutions Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.8 11

Answer: The total ticket cost is: $2,617

Explanation:
The given cost of an Adult ticket = $57
It is also given that a child ticket costs $14 less than an adult ticket.
So,
The given cost of a child ticket = 57 – 14 = $43
Now,
The cost for 18 adults = 18 × 57
The cost for 37 children = 37 × 43
By using the Distributive method,
18 × 57 = 57 ( 10 + 8 )
= (57 × 10 ) + ( 57× 8 )
= (50 + 7 ) × 10 + ( 50 + 7) × 8
= ( 50 × 10 ) + ( 7 × 10 ) + ( 50 × 8 ) + ( 7 × 8)
= 500 + 70 + 400 + 56
= 1,026
By using the Distributive method,
37 × 43 = 37 ( 40 + 3 )
= (37 × 40 ) + ( 37× 3 )
= (30 +7 ) × 40 + ( 30 + 7) × 3
= ( 30 × 40 ) + ( 7 × 40 ) + ( 30 × 3 ) + ( 7 × 3)
= 1,200 + 280 + 90 + 21
= 1,591
Hence,
The total cost of tickets = 1,591 + 1,026 = $2,617

Question 7.
Modeling Real Life
An artist creates a pattern by alternating square and rectangular tiles. The pattern has 14 square tiles and 13 rectangular tiles. How long is the pattern?
Big Ideas Math Solutions Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.8 12

Answer: 1,680 cm

Explanation:
Given that an artist creates a pattern by alternating square and rectangular tiles and the pattern 14 square tiles and 13 rectangular tiles.
From the given fig.,
Area of Square = 8 × 8 = 64 square cm
Area of Rectangle = 8 × 14 = 112 square cm
So,
Area of 14 squares = 14 × 64
Area of 13 rectangles = 112 × 13
By using the Distributive method,
14 × 64 = 64 ( 10 + 4 )
= (64 × 10 ) + ( 64× 4 )
= (60 +4 ) × 10 + ( 60 + 4) × 4
= ( 60 × 10 ) + ( 4 × 10 ) + ( 60 × 4 ) + ( 4 × 4)
= 600 + 40 + 240 + 16
= 896
By using the Distributive method,
112 × 13 = 13 ( 100 + 12 )
= (13 × 100 ) + ( 13× 12 )
= (10 +3 ) × 100 + ( 10 + 3) × 12
= ( 10 × 100 ) + ( 3× 100 ) + ( 10 × 12 ) + ( 12 × 3)
= 1,000 + 300 + 120 + 36
= 1,456
Hence,
The total length of the Pattern = 896 + 1,456 = 2,352 square cm

Question 8.
Modeling Real Life
A cargo ship has go ship has 34 rows of crates. Each row has 16 stacks of crates. There are 5 crates in each stack. The ship workers unload 862 crates. How many crates are still on the ship?
Big Ideas Math Solutions Grade 4 Chapter 4 Multiply by Two-Digit Numbers 4.8 13

Answer: 1,858 crates are still on the ship.

Explanation:
Given that a cargo ship has a go ship has 34 rows of crates and each row has 16 stacks of crates. It is also given that there are 5 crates in each stack.
So,
The total number of crates = 34 × 16 × 5 = 34 × 80
By using the Associative Property of Multiplication,
34 × 80 = 34 × (8 × 10)
= (34 × 8) × 10
= (17× 2 × 8) × 10
= 272 × 10
= 2,720
It is also given that the ship workers unload 862 crates.
Hence,
The number of crates that the ship had = 2,720 – 862 = 1,858 crates
Review & Refresh
Estimate the product.

Question 1.
4 × 85

Answer: 340

Explanation:
By using the partial products method,
4 × 85 = ( 80 + 5 ) × ( 2 + 2 )
= ( 80 × 2 ) + (80 × 2 ) + ( 5 × 2 ) + ( 5 × 2 )
= 160 + 160 + 10 + 10
= 340

Question 2.
6 × 705

Answer: 4,230

Explanation:
By using the partial products method,
6 × 705 = ( 700 + 5 ) × ( 2 + 4 )
= ( 700 × 2 ) + (700 × 4 ) + ( 5 × 2 ) + ( 5 × 4 )
= 1,400 + 2,800 + 10 + 20
= 4,230

Question 3.
8 × 7,923

Answer: 63,384

Multiply by Two-Digit Numbers Performance Task

Wind turbines convert wind to energy. Most wind or turbines have 3 blades. The blades rotate slower or faster depending on the speed of the wind. More energy is generated when the blades spin faster.
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers 1

Question 1.
A wind turbine rotates between 15 and 40 times in 1 minute.
a.What is the least number of times the turbine rotates in 1 hour?
b.What is the greatest number of times the turbine rotates in 1 hour?

Answer:
a) 900 times
b) 2,400 times

Explanation:
Given that a wind turbine rotates between 15 and 40 times in 1 minute.
We know that,
1 hour = 60 minutes
The wind turbine rotates minimum 15 times and maximum 40 times in a minute.
So,
The minimum number of times a wind turbine rotate in 1 hour = 15 × 60
The maximum number of times a wind turbine rotate in 1 hour = 40 × 60
Using the Associative Property of Multiplication,
15 × 60 = 15 ( 6 × 10)
= ( 15 × 6) × 10
= 90 × 10
= 900
Using the Associative Property of Multiplication,
40 × 60 = 40 ( 6 × 10)
= ( 40 × 6) × 10
= 240 × 10
= 2,400
Hence,
The minimum number of times a wind turbine rotate in 1 hour = 900 times
The maximum number of times a wind turbine rotate in 1 hour = 2,400 times

Question 2.
The tips of the turbine blades spin 5 times faster than the speed of the wind. The speed of the wind is 22 miles per hour. How fast do the blade tips spin?

Answer: 110 miles per hour
Explanation;
Given that the tips of the turbine blades spin 5 times faster than the speed of the wind. It is also given that the speed of the wind is 22 miles per hour.
So,
The speed of the blade tips = 22 × 5 = 110 miles per hour

Question 3.
A turbine farm has 7 large wind turbines. Each wind turbine can generate enough energy to power 1,485 houses. How many houses can the turbine farm power in all?

Answer: 10,395 houses

Explanation:
Given that a turbine farm has 7 large wind turbines and each wind turbine can generate enough energy to power 1,485 houses.
So,
The total number of houses that a turbine farm generates enough energy = 1,485 × 7 = 10,395 houses

Question 4.
Use the chart.
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers 2
a. How many times more houses are powered when the length of each blade is doubled?
b. A wind turbine has blades that are each 60 meters long. How many houses can the wind turbine power?

Answer:
a) 2 times
b) 33,000 houses
Explanation;
a) The given original length of each blade is 15 meters and 30 meters. It is also given that some houses are powered.
Now,
It is also given that the length of each blade is doubled.
So, 15 meters becomes 30 meters and 30 meters becomes 60 meters.
From this, we have to observe that the change in length also changes the energy powered in a directly proportional way.
From this, we can conclude that
The number of times more houses are powered when the length of each blade is doubled is: 2 times
b) Given that the length f each wind turbine is 60 meters long.
So, the number of houses that can be powered up = 60 × 550
Using the Associative Property of Multiplication,
550 × 60 = 550 × ( 10 × 6)
= ( 550 × 6 ) × 10
= 3300 × 10
= 33,000
Hence,
The number of houses can the wind turbine power is: 33,000 houses

Multiply by Two-Digit Numbers Activity

Multiplication Boss
Directions:
1. Each player flips 4 Number Cards and uses them in any order to create a multiplication problem with two-digit factors.
2. Each player finds the product of the two factors.
3. Players compare products. The player with the greater product takes all 8 cards.
4. If the products are equal, each player flips 4 more cards and plays again. The player with the greater product takes all 16 cards.
5. The player with the most cards at the end of the round wins
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers 3

Multiply by Two-Digit Numbers Chapter Practice

4.1 Multiply by Tens

Find the product.

Question 1.
50 × 20 = _____

Answer: 1,000

Explanation:
By using the Place-value method,
50 × 20 = 50 × 2 tens
= 5 tens × 2 tens
= 10 × 10 × 10
= 1,000
So, 50 × 20 = 1,000

Question 2.
30 × 60 = _____

Answer: 1,800

Explanation:
By using the Place-value method,
30 × 60 = 30 × 6 tens
= 3 tens × 6 tens
= 18 × 10 × 10
= 1,800
So, 30 × 60 = 1,800

Question 3.
80 × 10 = _____

Answer: 800

Explanation:
By using the Place-value method,
80 × 10 = 10 × 8 tens
= 1 ten × 8 tens
= 8 × 10 × 10
= 800
So, 80 × 10 = 800

Question 4.
40 × 70 = _____

Answer: 2,800

Explanation:
By using the Place-value method,
70 × 40 = 70 × 4 tens
= 7 tens × 4 tens
= 28 × 10 × 10
= 2,800
So, 70 × 40 = 2,800

Question 5.
60 × 50 = _____

Answer: 3,000

Explanation:
By using the Place-value method,
50 × 60 = 50 × 6 tens
= 5 tens × 6 tens
= 30 × 10 × 10
= 3,000
So, 50 × 60 = 3,000

Question 6.
90 × 90 = _____

Answer: 8,100

Explanation:
By using the Place-value method,
90 × 90 = 90 × 9 tens
= 9 tens × 9 tens
= 81 × 10 × 10
= 8,100
So, 90 × 90 = 8,100

Question 7.
70 × 11 = _____

Answer: 770

Explanation:
By using the Place-value method,
70 × 11 = 11 × 7 tens
= 77 tens
= 77 × 10
= 770
So, 70 × 11 = 770

Question 8.
18 × 30 = _____

Answer: 540

Explanation:
By using the Place-value method,
30 × 18 = 18 × 3 tens
= 54 tens
= 54 × 10
= 540
So, 30 × 18 = 540

Question 9.
20 × 75 = _____

Answer: 1,500

Explanation:
By using the Place-value method,
20 × 75 = 75 × 2 tens
= 150 tens
= 150 × 10
= 1,500
So, 20 × 75 = 1,500
Find the missing factor.

Question 10.
40 × ____ = 3,200

Answer: The missing number is: 80
Explanation;
Let the missing number be X
So, 40 × X = 3,200
X = 3,200 /40 = 80
Hence, the value of X is: 80

Question 11.
_____ × 20 = 1,200

Answer:
The missing number is: 60
Explanation;
Let the missing number be X
So, 20 × X = 1,200
X = 1,200 /20 = 60
Hence, the value of X is: 60

Question 12.
30 × ____ = 2,100

Answer:
The missing number is: 70
Explanation;
Let the missing number be X
So, 30 × X = 2,100
X = 2,100 /30 = 70
Hence, the value of X is: 70

4.2 Estimate Products

Estimate the product.

Question 13.
25 × 74

Answer: 1,850

Explanation:
Let 74 be rounded to 75
By using the Partial Fraction method,
25 × 75 = ( 20 + 5) × ( 70 + 5)
= ( 20 × 70) + ( 20 × 5) + ( 5 × 70) + ( 5 × 5)
= 1,400 + 100 + 350 + 25
= 1,875
So,
25 × 74 = 1,875

Question 14.
16 × 28

Answer: 448

Explanation:
Let 16 be rounded to 15
Let 28 be rounded to 30
By using the Partial Fraction method,
15 × 30 = ( 10 + 5) × ( 25 + 5)
= ( 10 × 25) + ( 10 × 5) + ( 5 × 25) + ( 5 × 5)
= 200 + 50 + 125 + 25
= 400
So,
16 × 28 = 400

Question 15.
42 × 81

Answer: 3,402

Explanation:
Let 42 be rounded to 40
Let 81 be rounded to 80
By using the tens method,
40 × 80 = 40 × 8 tens
= 4 tens × 8 tens
= 32 × 1 ten × 1 ten
= 32 × 10 × 10
= 3,200
So,
42 × 81 = 3,200
Open-Ended
Write two possible factors that can be estimated as shown.

Question 16.
8,100
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers chp 16

Answer:

Explanation:
The Products of 81 are:
9 × 9 = 81
From the above two products, we can conclude that the two possible numbers that can give the product 6,400 are: 90,90

Question 17.
400
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers chp 17

Answer:

Explanation:
The Products of 64 are:
2 × 2 = 4
4 × 1 =4
From the above two products, we can conclude that the two possible numbers that can give the product 400 are: 20,20 and. 10,40

4.3 Use Area Models to Multiply Two-Digit Numbers

Draw an area model to find the product.

Question 18.
13 × 19 = _____

Answer: 247

Explanation:
By using the Partial Products method,
13 × 19 = ( 10 + 3) × ( 10 + 9)
10 × 10 +  10× 9 + 10 × 3 + 3 × 9
= 100 + 90 + 30 + 27
= 247
So,
19 × 13 = 247

Question 19.
21 × 36 = _____

Answer: 756

Explanation:
By using the Partial Products method,
21 × 36 = ( 20 + 1) × ( 30 + 6)
= 20 × 30 +  20× 6 + 30 × 1 + 1 × 6
= 600 + 120 + 30 + 6
= 756
So,
21 × 36 = 756

Question 20.
YOU BE THE TEACHER
Your friend finds 28 × 24. Is your friend correct? Explain.
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers chp 20

Answer: Your friend is not correct.
Explanation;

By using the partial Products method,
400 + 160 +80 + 32 = 672
From the above,
we can conclude that in place of 16, 160 must be placed.

4.4 Use Distributive Property to Multiply Two-Digit Numbers

Use Distributive Property to find a product.

Question 21.
27 × 34 = 27 × (30 + 4)
= (27 × 30) + (27 × 4)
= (20 + 7) × 30 + (20 + 7) × 4
= (20 ×30) + (7 × 30) + (20 ×4) + (7 × 4)
= 600 + 210 + 80 + 28
= 918
So, 27 × 34 = 918.

Question 22.
43 × 18 = _____

Answer: 774

Explanation: Using the Partial Products method,
43 × 18 = 43× (10 + 8)
= (43 × 10) + (43 × 8)
= (40 + 3) × 10 + (40 + 3) × 8
= (40 ×10) + (3 × 10) + (40 ×8) + (3 × 8)
= 400 + 30 + 320 + 24
= 774
So, 43 × 18 = 774

Question 23.
35 × 57 = _____

Answer: 1,995

Explanation: Using the Partial products method,
35 × 57 = 35 × (50 + 7)
= (35 × 50) + (35 × 7)
= (30 + 5) × 50 + (30 + 5) × 7
= (50 ×30) + (5 × 50) + (30 ×7) + (7 × 5)
= 1,500 + 250 + 210 + 35
= 1,995
So, 35 × 57 = 1,995

Question 24.
81 × 76 = _____

Answer: 6,156
81 × 76 = 81 × (70 + 6)
= (81 × 70) + (81 × 6)
= (80 + 1) × 70 + (80 + 1) × 6
= (80 ×70) + (70 × 1) + (80 ×6) + (1 × 6)
= 5,600 + 70 + 480 + 6
= 6,156
So, 81 × 76 = 6,156

4.5 Use Partial Products to Multiply Two-Digit Numbers

Find the product. Check whether your answer is reasonable.

Question 25.
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers chp 25

Answer: 396

Explanation: Using the Partial Products method,
18 × 22 = 22 × (10 + 8)
= (22 × 10) + (22 × 8)
= (20 + 2) × 10 + (20 + 2) × 8
= (20 ×10) + (2 × 10) + (20 ×8) + (2 × 8)
= 200 + 20 + 160 + 16
= 396
So, 18 × 22 = 396

Question 26.
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers chp 26

Answer: 3,358

Explanation: Using the Partial Products method,
73 × 46 = 73 × (40 + 6)
= (73 × 40) + (73 × 6)
= (70 + 3) × 40 + (70 + 3) × 6
= (70 ×40) + (3 × 40) + (70 ×6) + (3 × 6)
= 2,800 + 120 + 420 + 18
= 3,358
So, 73 × 46 = 3,358.

Question 27.
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers chp 27

Answer: 3,276

Explanation: By using the Partial Products method,
39 × 84 = 39 × (80 + 4)
= (39 × 80) + (39 × 4)
= (30 + 9) × 80 + (30 + 9) × 4
= (80 ×30) + (9 × 80) + (30 ×4) + (9 × 4)
= 2,400 + 720 + 120 + 36
= 3,276
So, 39 × 84 = 3,276

Question 28.
57 × 19 = _____

Answer: 1,083

Explanation: By using the Partial Products method,
57 × 19 = 57 × (10 + 9)
= (57 × 10) + (57 × 9)
= (50 + 7) × 10 + (50 + 7) × 9
= (50 ×10) + (7 × 10) + (50 ×9) + (7 × 9)
= 500 + 70 + 450 + 63
= 1,083
So, 57 × 19 = 1,083

Question 29.
38 × 65 = _____

Answer: 2,470

Explanation: By using the Partial Products method,
38 × 65 = 65 × (30 + 8)
= (65 × 30) + (65 × 8)
= (60 + 5) × 30 + (60 + 5) × 8
= (60 ×30) + (5 × 30) + (60 ×8) + (5 × 8)
= 1,800 + 150 + 480 + 40
= 2,470
So, 38 × 65 = 2,470

Question 30.
94 × 26 = _____

Answer: 2,444

Explanation: By using the Partial Products method,
94 ×26 = 94 × (20 + 6)
= (94 × 20) + (94 × 6)
= (90 + 4) × 20 + (90 + 4) × 6
= (90 ×20) + (4 × 20) + (90 ×6) + (4 × 6)
= 1,800 + 80 + 540 + 24
= 2,444
So, 94 × 26 = 2,444
Reasoning
Find the missing digits. Then find the product.

Question 31.
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers chp 31

Answer:

Question 32.
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers chp 32

Answer:

4.6 Multiply Two-Digit Numbers

Find the product. Check whether your answer is reasonable.

Question 33.
Estimate: _____
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers chp 33

Answer: 2,108

Explanation:
Using the Partial Products method,
34 × 62 = ( 60 + 2) × ( 30 + 4)
= 30 × 60 + 4 × 60 + 30 × 2 + 2 × 4
= 1,800 + 240 + 60 + 8
= 2,108
Estimate:
Let 62 be Rounded to 60.
Let 34 be Rounded to 35.
So, 60 × 35 = 2,100
As the Estimate and the actual answer are near, the answer is reasonable.

Question 34.
Estimate: ____
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers chp 34

Answer: 7,917

Explanation:
Using the Partial Products method,
87 × 91 = ( 80 + 7) × ( 90 + 1)
= 80 × 90 + 7 × 90 + 80 × 1 + 7 × 1
= 7,200 + 630 + 80 + 7
= 7,917
Estimate:
Let 87 be Rounded to 85.
Let 91 be Rounded to 90.
So, 90 × 85 = 7,650
As the Estimate and the actual answer are not near, the answer is  not reasonable.

Question 35.
Estimate: _____
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers chp 35

Answer: 3,285

Explanation:
Using the Partial Products method,
73 × 45 = ( 70 + 3) × ( 40 + 5)
= 70 × 40 + 3 × 40 + 70 × 5 + 5 × 3
= 2,800 + 120 + 350 + 15
= 3,285
Estimate:
Let 73 be Rounded to 75.
So, 45 × 75 = 3,375
As the Estimate and the actual answer are near, the answer is reasonable.

Question 36.
Estimate: ____
13 × 21 = ______

Answer: 273

Explanation:
Using the Partial Products method,
13 × 21 = ( 20 + 1) × ( 10 + 3)
= 20 × 10 + 20 × 3 + 10 × 1 + 1 × 3
= 200 + 60 + 10 + 3
= 273
Estimate:
Let 13 be Rounded to 15.
Let 21 be Rounded to 20.
So, 20 × 15 = 300
As the Estimate and the actual answer are near, the answer is reasonable.

Question 37.
Estimate: _____
42 × 53 = _____

Answer: 2,226

Explanation:
Using the Partial Products method,
42 × 53 = ( 40 + 2) × ( 50 + 3)
= 40 × 50 + 3 × 40 + 50 × 2 + 2 × 3
= 2,000 + 120 + 100 + 6
= 2,226
Estimate:
Let 42 be Rounded to 40.
Let 53 be Rounded to 55.
So, 40 × 55 = 2,200
As the Estimate and the actual answer are near, the answer is reasonable.

Question 38.
Estimate: _____
29 × 66 = _____

Answer: 1,914

Explanation:
Using the Partial Products method,
29 × 66 = ( 20 + 9) × ( 60 + 6)
= 20 × 60 + 9 × 60 + 20 × 6 + 9 × 6
= 1,200 + 540 + 120 + 54
= 1,914
Estimate:
Let 29 be Rounded to 30.
Let 66 be Rounded to 65.
So, 30 × 65 = 1,950
As the Estimate and the actual answer are near, the answer is reasonable.

4.7 Practice Multiplication Strategies

Find the product.

Question 39.
80 × 30 = _____

Answer: 2,400

Explanation: Using the Place-value method,
80 × 30 = 80 × 3 tens
= 240 tens
= 2,400
So, 80 × 30 = 2,400

Question 40.
26 × 51 = _____

Answer: 1,326

Explanation: Using the Partial Products method,
26 × 51 = ( 20 + 6) × ( 50 + 1)
= ( 20 × 50) + ( 20 × 1) + ( 6 × 50) + (6 × 1)
= 1,000 + 20 + 300 + 6
= 1,326

Question 41.
94 × 70 = _____

Answer: 6,580

Explanation: Using the Place-value method,
94 × 70 = 94 × 7 tens
= 658 tens
= 6,580
So, 94 × 70 = 6,580

Question 42.
15 × 67 = _____

Answer: 1,005

Explanation: Using the Partial Products method,
15 × 67 = ( 10 + 5) × ( 60 + 7)
= ( 10 × 60) + ( 10 × 7) + ( 5 × 60) + ( 5 × 7)
= 600 + 70 + 300 + 35
=1,005

Question 43.
40 × 38 = _____

Answer: 1,520

Explanation: Using the Place-value method,
38 × 40 = 38 × 4 tens
= 152 tens
= 1,520
So, 38 × 40 = 1,520

Question 44.
29 × 92 = _____

Answer: 2,668

Explanation: Using the partial products method,
29 × 92 = ( 20 + 9) × ( 90 + 2)
= ( 20 × 90 ) + ( 20 × 2) + ( 9 × 90) + ( 9 × 2)
= 1,800 + 40 + 810 + 18
= 2,668

Question 45.
Modeling Real Life
A Ferris wheel runs 40 times each day. It has 16 cars with 4 seats in each car. If the Ferris wheel is full each time it runs, how many people will ride it in 1 day?
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers chp 45

Answer: 2,560 people will ride the Ferris wheel in 1 day

Explanation:
Given that a  Ferris wheel runs 40 times each day and it has 16 cars with 4 seats in each car.
So, the total number of people that can ride a Ferris wheel = 40 × 16 × 4 = 40 × 64
By using the place-value method,
64  × 40  = 64 × 4 tens
= 256 tens
= 2,560
From the above,
We can conclude that there are 2,560 people who will ride the Ferris wheel in 1 day
4.8 Problem Solving: Multiplication with Two-Digit Numbers

Question 46.
A music fan memorizes 59 songs for a concert. Her goal is to memorize all of the songs from 13 albums. There are 15 songs on each album. How many more songs does the music fan still need to memorize?
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers chp 46

Answer: 136 songs

Explanation:
Given that a music fan has to memorize all of the songs from 13 albums and there are 15 songs on each album.
So, the total number of songs the music fan has to memorize = 13 × 15
By using the Partial Products method,
13 × 15 = ( 10 + 3) × ( 10 + 5)
= ( 10 × 10 ) + ( 10 × 5) + ( 3 × 10 ) + ( 3 × 5)
= 100 + 50 + 30 + 15
= 195
It is also given that the music fan memorized 59 songs for the Concert.
Hence,
The number of songs that has to memorize by the music fan = 195 – 59 = 136 songs

Question 47.
Find the area of the Jamaican flag.
Big Ideas Math Answer Key Grade 4 Chapter 4 Multiply by Two-Digit Numbers chp 47

Answer: 2,592 inches
Explanation;
The given Jamaican flag is in the shape of a rectangle.
We know that,
Area of the rectangle = length × breadth
So,
Area of the Jamaican flag = 36 × ( 36 + 36) = 36 × 72
Using the Partial Products method,
36 × 72 = ( 30 + 6) × ( 70 + 2)
= ( 30 × 70 ) + ( 30 × 2 ) + ( 6 × 70) + ( 6 × 2)
= 2,100 + 60 + 720 + 12
= 2,592 inches
From this,
We can conclude that the area of the Jamaican flag is: 2,592 inches

Final Words:

We believe that the details provided in Big Ideas Math Answers Grade 4 Chapter 4 Multiply by Two-Digit Numbers are beneficial for you to overcome the difficulties in this chapter. Share this pdf link with your friends to help them to score good marks in the exams. Stay with us to get the latest updates regarding the solutions with explanations of all the BIM Grade 3 chapters.

Big Ideas Math Answers Grade 5 Chapter 9 Multiply Fractions

Big Ideas Math Answers Grade 5 Chapter 9 Multiply Fractions

Are you in the search of Big Ideas Math Answers Grade 5 Chapter 9 Multiply Fractions? If yes, you are on the correct page. Check out the detailed explanation for all the problems available on Big Ideas Grade 5 Chapter 9 Multiply Fractions Math Answers. You can easily become a math expert by referring to the BIM Grade 5 Chapter 9 Math Answers. Download Big Ideas Math Book 5th Grade Answer Key Chapter 9 Multiply Fractions now and start your preparation for the exam. You must prepare all the concepts and verify answers to test your preparation level.

Big Ideas 5th Grade Chapter 9 Multiply Fractions Math Book Answer Key

Every student must prepare each and every topic of ch 9 to score good marks in the exam. Also, take the math assignments and homework to get the perfect preparation. Prepare well by referring to our Big Ideas Math Book 5th Grade Answer Key Chapter 9 Multiply Fractions. Download Big Ideas Math Answers Grade 5 Chapter 9 Multiply Fractions for free. The below links have every topic problem along with explanations. So, check out every link and practice more where you lag.

Lesson: 1 Multiply Whole Numbers by Fractions

Lesson: 2 Use Models to Multiply Fractions by Whole Numbers

Lesson: 3 Multiply Fractions and Whole Numbers

Lesson: 4 Use Models to Multiply Fractions

Lesson: 5 Multiply Fractions

Lesson: 6 Find Areas of Rectangles

Lesson: 7 Multiply Mixed Numbers

Lesson: 8 Compare Factors and Products

Chapter: 9 – Multiply Fractions

Lesson 9.1 Multiply Whole Numbers by Fractions

Explore and Grow

Write any proper fraction that is not a unit fraction. Draw a model to represent your fraction. Draw a model to find a multiple of your fraction?
Answer:
The proper fraction that is not a unit fraction is:  \(\frac{5}{8}\)
Let the proper fraction be multiplied by 5
So,
We have to find the value of 5 × \(\frac{5}{8}\)
Now,
We know that,
a = \(\frac{a}{1}\)
a × \(\frac{a}{b}\) = \(\frac{a × a}{b × 1}\)
So,
5 × \(\frac{5}{8}\)
= \(\frac{5}{1}\) × \(\frac{5}{8}\)
= \(\frac{5 × 5}{1 × 8}\)
= \(\frac{25}{8}\)
Hence, from the above,
We can conclude that the multiple of your proper fraction is: \(\frac{25}{8}\)

Reasoning
How can you use a model to multiply a whole number by a fraction? Explain.
Answer:
We can multiply a whole number by a fraction using the properties of multiplication. They are:
A) a = \(\frac{a}{1}\)
B) a × \(\frac{a}{b}\) = \(\frac{a × a}{b × 1}\)
Hence, by using the above properties, we can multiply the whole number by a fraction.

Think and Grow: Multiply Whole Numbers by Fractions

Example
Find 3 × \(\frac{2}{5}\)

Show and Grow

Multiply.
Question 1.
2 × \(\frac{3}{4}\) = ______
Answer:
2 × \(\frac{3}{4}\) = \(\frac{6}{4}\)

Explanation:
The given numbers are: 2 and \(\frac{3}{4}\)
We know that,
a = \(\frac{a}{1}\)
a × \(\frac{a}{b}\) = \(\frac{a × a}{b × 1}\)
So,
2 × \(\frac{3}{4}\)
= \(\frac{2}{1}\) × \(\frac{3}{4}\)
= \(\frac{2 × 3}{1 × 4}\)
= \(\frac{6}{4}\)
Hence,
2 × \(\frac{3}{4}\) = \(\frac{6}{4}\)

Question 2.
4 × \(\frac{5}{8}\) = ____
Answer:
4 × \(\frac{5}{8}\) = \(\frac{20}{8}\)

Explanation:
The given numbers are: 4 and \(\frac{5}{8}\)
We know that,
a = \(\frac{a}{1}\)
a × \(\frac{a}{b}\) = \(\frac{a × a}{b × 1}\)
So,
4 × \(\frac{5}{8}\)
= \(\frac{4}{1}\) × \(\frac{5}{8}\)
= \(\frac{4 × 5}{1 × 8}\)
= \(\frac{20}{8}\)
Hence,
4 × \(\frac{5}{8}\) = \(\frac{20}{8}\)

Apply and Grow: Practice

Multiply.
Question 3.
5 × \(\frac{7}{10}\) = ______
Answer:
5 × \(\frac{7}{10}\) = \(\frac{35}{10}\)

Explanation:
The given numbers are: 5 and \(\frac{7}{10}\)
We know that,
a = \(\frac{a}{1}\)
a × \(\frac{a}{b}\) = \(\frac{a × a}{b × 1}\)
So,
5 × \(\frac{7}{10}\)
= \(\frac{5}{1}\) × \(\frac{7}{10}\)
= \(\frac{5 × 7}{1 × 10}\)
= \(\frac{35}{10}\)
Hence,
5 × \(\frac{7}{10}\) = \(\frac{35}{10}\)

Question 4.
8 × \(\frac{2}{3}\) = ______
Answer:
8 × \(\frac{2}{3}\) = \(\frac{16}{3}\)

Explanation:
The given numbers are: 8 and \(\frac{2}{3}\)
We know that,
a = \(\frac{a}{1}\)
a × \(\frac{a}{b}\) = \(\frac{a × a}{b × 1}\)
So,
8 × \(\frac{2}{3}\)
= \(\frac{8}{1}\) × \(\frac{2}{3}\)
= \(\frac{8 × 2}{1 × 3}\)
= \(\frac{16}{3}\)
Hence,
8 × \(\frac{2}{3}\) = \(\frac{16}{3}\)

Question 5.
7 × \(\frac{5}{6}\) = ______
Answer:
7 × \(\frac{5}{6}\) = \(\frac{35}{6}\)

Explanation:
The given numbers are: 7 and \(\frac{5}{6}\)
We know that,
a = \(\frac{a}{1}\)
a × \(\frac{a}{b}\) = \(\frac{a × a}{b × 1}\)
So,
7 × \(\frac{5}{6}\)
= \(\frac{7}{1}\) × \(\frac{5}{6}\)
= \(\frac{7 × 5}{1 × 6}\)
= \(\frac{35}{6}\)
Hence,
7 × \(\frac{5}{6}\) = \(\frac{35}{6}\)

Question 6.
9 × \(\frac{1}{2}\) = ______
Answer:
9 × \(\frac{1}{2}\) = \(\frac{9}{2}\)

Explanation:
The given numbers are: 9 and \(\frac{1}{2}\)
We know that,
a = \(\frac{a}{1}\)
a × \(\frac{a}{b}\) = \(\frac{a × a}{b × 1}\)
So,
9 × \(\frac{1}{2}\)
= \(\frac{9}{1}\) × \(\frac{1}{2}\)
= \(\frac{9 × 1}{1 × 2}\)
= \(\frac{9}{2}\)
Hence,
9 × \(\frac{1}{2}\) = \(\frac{9}{2}\)

Question 7.
6 × \(\frac{3}{100}\) = ______
Answer:
6 × \(\frac{3}{100}\) = \(\frac{18}{100}\)

Explanation:
The given numbers are: 6 and \(\frac{3}{100}\)
We know that,
a = \(\frac{a}{1}\)
a × \(\frac{a}{b}\) = \(\frac{a × a}{b × 1}\)
So,
6 × \(\frac{3}{100}\)
= \(\frac{6}{1}\) × \(\frac{3}{100}\)
= \(\frac{6 × 3}{1 × 100}\)
= \(\frac{18}{100}\)
Hence,
6 × \(\frac{3}{100}\) = \(\frac{18}{100}\)

Question 8.
15 × \(\frac{4}{7}\) = ______
Answer:
15 × \(\frac{4}{7}\) = \(\frac{60}{7}\)

Explanation:
The given numbers are: 15 and \(\frac{4}{7}\)
We know that,
a = \(\frac{a}{1}\)
a × \(\frac{a}{b}\) = \(\frac{a × a}{b × 1}\)
So,
15 × \(\frac{4}{7}\)
= \(\frac{15}{1}\) × \(\frac{4}{7}\)
= \(\frac{15 × 4}{1 × 7}\)
= \(\frac{60}{7}\)
Hence,
15 × \(\frac{4}{7}\) = \(\frac{60}{7}\)

Question 9.
10 × \(\frac{5}{3}\) = ______
Answer:
10 × \(\frac{5}{3}\) = \(\frac{50}{3}\)

Explanation:
The given numbers are: 10 and \(\frac{5}{3}\)
We know that,
a = \(\frac{a}{1}\)
a × \(\frac{a}{b}\) = \(\frac{a × a}{b × 1}\)
So,
10 × \(\frac{5}{3}\)
= \(\frac{10}{1}\) × \(\frac{5}{3}\)
= \(\frac{10 × 5}{1 × 3}\)
= \(\frac{50}{3}\)
Hence,
10 × \(\frac{5}{3}\) = \(\frac{50}{3}\)

Question 10.
4 × \(\frac{5}{2}\) = ______
Answer:
4 × \(\frac{5}{2}\) = 10

Explanation:
The given numbers are: 4 and \(\frac{5}{2}\)
We know that,
a = \(\frac{a}{1}\)
a × \(\frac{a}{b}\) = \(\frac{a × a}{b × 1}\)
So,
4 × \(\frac{5}{2}\)
= \(\frac{4}{1}\) × \(\frac{5}{2}\)
= \(\frac{4 × 5}{1 × 2}\)
= \(\frac{20}{2}\)
= 10
Hence,
4 × \(\frac{5}{2}\) = 10

Question 11.
3 × \(\frac{11}{8}\) = ______
Answer:
3 × \(\frac{11}{8}\) = \(\frac{33}{8}\)

Explanation:
The given numbers are: 3 and \(\frac{11}{8}\)
We know that,
a = \(\frac{a}{1}\)
a × \(\frac{a}{b}\) = \(\frac{a × a}{b × 1}\)
So,
3 × \(\frac{11}{8}\)
= \(\frac{3}{1}\) × \(\frac{11}{8}\)
= \(\frac{11 × 3}{1 × 8}\)
= \(\frac{33}{8}\)
Hence,
3 × \(\frac{11}{8}\) = \(\frac{33}{8}\)

Find the unknown number.
Question 12.
Big Ideas Math Answer Key Grade 5 Chapter 9 Multiply Fractions 9.1 2
Answer:
The missing number is: 3

Explanation:
The given fractions are: \(\frac{3}{20}\) and \(\frac{9}{20}\)
Let the missing number be X
So,
X × \(\frac{3}{20}\) = \(\frac{9}{20}\)
When multiplication goes from left to right or from right to left, it will become the division
When division goes from left to right or from right to left, it will become multiplication.
So,
X = \(\frac{9}{20}\) ÷ \(\frac{3}{20}\)
We know that,
\(\frac{a}{b}\) ÷ \(\frac{x}{y}\) = \(\frac{a}{b}\) × \(\frac{y}{x}\)
So,
\(\frac{9}{20}\) ÷ \(\frac{3}{20}\)
= \(\frac{9}{20}\) × \(\frac{20}{3}\)
= \(\frac{9 × 20}{20 × 3}\)
= 3
Hence,
The missing number is: 3

Question 13.
Big Ideas Math Answer Key Grade 5 Chapter 9 Multiply Fractions 9.1 3
Answer:
The missing number is: 6

Explanation:
The given fractions are: \(\frac{4}{9}\) and \(\frac{24}{9}\)
Let the missing number be X
So,
X × \(\frac{4}{9}\) = \(\frac{24}{9}\)
When multiplication goes from left to right or from right to left, it will become the division
When division goes from left to right or from right to left, it will become multiplication.
So,
X = \(\frac{24}{9}\) ÷ \(\frac{4}{9}\)
We know that,
\(\frac{a}{b}\) ÷ \(\frac{x}{y}\) = \(\frac{a}{b}\) × \(\frac{y}{x}\)
So,
\(\frac{24}{9}\) ÷ \(\frac{4}{9}\)
= \(\frac{24}{9}\) × \(\frac{9}{4}\)
= \(\frac{9 × 24}{9 × 4}\)
= 6
Hence,
The missing number is: 6

Question 14.
Big Ideas Math Answer Key Grade 5 Chapter 9 Multiply Fractions 9.1 4
Answer:
The missing number is: 10

Explanation:
The given fractions are: \(\frac{5}{12}\) and \(\frac{50}{12}\)
Let the missing number be X
So,
X × \(\frac{5}{12}\) = \(\frac{50}{12}\)
When multiplication goes from left to right or from right to left, it will become the division
When division goes from left to right or from right to left, it will become multiplication.
So,
X = \(\frac{50}{12}\) ÷ \(\frac{5}{12}\)
We know that,
\(\frac{a}{b}\) ÷ \(\frac{x}{y}\) = \(\frac{a}{b}\) × \(\frac{y}{x}\)
So,
\(\frac{50}{12}\) ÷ \(\frac{5}{12}\)
= \(\frac{50}{12}\) × \(\frac{12}{5}\)
= \(\frac{12 × 50}{12 × 5}\)
= 10
Hence,
The missing number is: 10

Question 15.
A recipe calls for \(\frac{3}{4}\) cup of dried rice noodles. You make 44 batches of the recipe. How many cups of dried rice noodles do you use?
Big Ideas Math Answer Key Grade 5 Chapter 9 Multiply Fractions 9.1 5
Answer:
The number of cups of dried rice noodles you used is: 33

Explanation:
It is given that a recipe calls for \(\frac{3}{4}\) cup of dried rice noodles.
It is also given that you make 44 batches of the recipe
So,
The number of cups of dried rice noodles you used = ( The number of batches of the recipe ) × ( The number of dried rice noodles for each batch )
= 44 × \(\frac{3}{4}\)
Now,
We know that,
a = \(\frac{a}{1}\)
a × \(\frac{a}{b}\) = \(\frac{a × a}{b × 1}\)
So,
44 × \(\frac{3}{4}\)
= \(\frac{44}{1}\) × \(\frac{3}{4}\)
= \(\frac{44 × 3}{1 × 4}\)
= \(\frac{33}{1}\)
= 33
Hence, from the above,
We can conclude that the number of cups of dried noodles you used is: 33 cups

Question 16.
YOU BE THE TEACHER
Your 5 × \(\frac{3}{5}\) friend says that 5 is equal to \(\frac{3}{5}\) + \(\frac{3}{5}\) + \(\frac{3}{5}\) + \(\frac{3}{5}\) + \(\frac{3}{5}\). Is your friend correct? Explain.
Answer:
Yes, your friend is correct.

Explanation:
It is given that your 5 × \(\frac{3}{5}\) friend says that 5 is equal to \(\frac{3}{5}\) + \(\frac{3}{5}\) + \(\frac{3}{5}\) + \(\frac{3}{5}\) + \(\frac{3}{5}\).
Now,
We know that,
n × a = a + a + a + a + …………….. + n times
Where,
‘n’ is the multiple of a
Here,
‘a’ will be the whole number or the fraction
So,
According to the above property,
5 × \(\frac{3}{5}\) = \(\frac{3}{5}\) + \(\frac{3}{5}\) + \(\frac{3}{5}\) +\(\frac{3}{5}\) + \(\frac{3}{5}\)
Hence, from the above,
We can conclude that your friend is correct.

Question 17.
Patterns
Describe and complete the pattern.
Big Ideas Math Answer Key Grade 5 Chapter 9 Multiply Fractions 9.1 6
Answer:
The completed pattern is:

From the above pattern,
We can observe that
The whole number is constant i.e., 9
The fraction part is changing i.e., the numerator part is increasing by 1 till 4 and the denominator is constant
So,
Now,
9 × \(\frac{1}{4}\)
= \(\frac{9}{1}\) × \(\frac{1}{4}\)
= \(\frac{9 × 1}{1 × 4}\)
= \(\frac{9}{4}\)
9 × \(\frac{2}{4}\)
= \(\frac{9}{1}\) × \(\frac{2}{4}\)
= \(\frac{9 × 2}{1 × 4}\)
= \(\frac{18}{4}\)
So,
The remaining two multiplication equations will also be solved in the same way.

Think and Grow: Modeling Real Life

Example
Your goal is to make a waterslide that is at least 10 meters long. You make the waterslide using 10 plastic mats that are each \(\frac{3}{2}\) meters long. Do you reach your goal?
Big Ideas Math Answer Key Grade 5 Chapter 9 Multiply Fractions 9.1 7
Find the length of the waterslide by multiplying the number of mats by the length of each mat.

So,
You reached your goal.

Show and Grow

Question 18.
An excavator is moving 4 piles of dirt that are the same size. Each pile requires \(\frac{3}{4}\) hour to move. Can the excavator move all of the piles in 2 hours?
Answer:
No, the excavator can’t move all of the piles in 2 hours

Explanation:
It is given that an excavator is moving 4 piles of dirt that are of the same size and each pile requires \(\frac{3}{4}\) hour to move.
So,
We know that,
1 hour = 60 minutes
So,
\(\frac{3}{4}\) of 1 hour = \(\frac{3}{4}\) × 60
= \(\frac{3}{4}\) × \(\frac{60}{1}\)
= \(\frac{3 × 60}{4 × 1}\)
= 45 minutes
So,
The time is taken to move 4 piles of dirt = ( The time is taken to move each pile )  × ( The total number of piles )
= 45 × 4
=180 minutes
It is given that you have to move the piles of dirt in 2 hours but the time taken is 4 hours.
Hence, from the above,
We can conclude that the excavator can’t move all the piles in 2 hours.

Question 19.
You walk dogs \(\frac{5}{4}\) miles two times each day. How far do you walk the dogs in 1 week?
Big Ideas Math Answer Key Grade 5 Chapter 9 Multiply Fractions 9.1 9
Answer:
The number of miles you walk the dogs in 1 week is: \(\frac{35}{2}\)

Explanation:
It is given that you walk dogs \(\frac{5}{4}\) miles two times each day
So,
The total number of miles you walk dogs in 1 day = 2 × \(\frac{5}{4}\)
= \(\frac{5}{4}\) × \(\frac{2}{1}\)
= \(\frac{5 × 2}{4 × 1}\)
= \(\frac{5}{2}\) miles
We know that,
1 week = 7 days
So,
The total number of miles you walk dogs in 1 week = ( The number of miles you walk dogs in 1 day ) × 7
= \(\frac{5}{2}\) × 7
= \(\frac{5}{2}\) × \(\frac{7}{1}\)
= \(\frac{5 × 7}{2 × 1}\)
= \(\frac{35}{2}\)
Hence, from the above,
We can conclude that the number of miles you walk dogs in 1 week is: \(\frac{35}{2}\) miles

Question 20.
DIG DEEPER
You have10 feet of string. You need \(\frac{5}{3}\) feet of string to make 1 necklace. You make 5 necklaces. Do you have enough string to make another necklace? Explain.
Answer:
Yes, we have enough string to make another necklace

Explanation:
It is given that you have 10 feet of string and you need \(\frac{5}{3}\) feet of string to make 1 necklace. and you make 5 necklaces.
So,
The length of the string required for 5 necklaces = ( The length of string required for 1 necklace ) × ( The total number of necklaces )
= \(\frac{5}{3}\) × 5
= \(\frac{5}{3}\) × \(\frac{5}{1}\)
= \(\frac{5 × 5}{3 × 1}\)
= \(\frac{25}{3}\) feet
We can write 10 feet as \(\frac{30}{3}\) feet
So,
When we compare the total length of the string and the length of the string to make 5 necklaces,
We can observe that the length of the string to make 5 necklaces is less than the total length of the string.
Hence, from the above,
We can conclude that we have enough string to make another necklace.

Multiply Whole Numbers by Fractions Homework & Practice 9.1

Multiply
Question 1.
5 × \(\frac{2}{3}\) = ______
Answer:
5 × \(\frac{2}{3}\) = \(\frac{10}{3}\)

Explanation:
The given numbers are: 5 and \(\frac{2}{3}\)
We know that,
a = \(\frac{a}{1}\)
a × \(\frac{a}{b}\) = \(\frac{a × a}{b × 1}\)
So,
5 × \(\frac{2}{3}\)
= \(\frac{5}{1}\) × \(\frac{2}{3}\)
= \(\frac{2 × 5}{1 × 3}\)
= \(\frac{10}{3}\)
Hence,
5 × \(\frac{2}{3}\) = \(\frac{10}{3}\)

Question 2.
9 × \(\frac{7}{8}\) = ______
Answer:
9 × \(\frac{7}{8}\) = \(\frac{63}{8}\)

Explanation:
The given numbers are: 9 and \(\frac{7}{8}\)
We know that,
a = \(\frac{a}{1}\)
a × \(\frac{a}{b}\) = \(\frac{a × a}{b × 1}\)
So,
9 × \(\frac{7}{8}\)
= \(\frac{9}{1}\) × \(\frac{7}{8}\)
= \(\frac{9 × 7}{1 × 8}\)
= \(\frac{63}{8}\)
Hence,
9 × \(\frac{7}{8}\) = \(\frac{63}{8}\)

Question 3.
4 × \(\frac{11}{12}\) = ______
Answer:
4 × \(\frac{11}{12}\) = \(\frac{44}{12}\)

Explanation:
The given numbers are: 4 and \(\frac{11}{12}\)
We know that,
a = \(\frac{a}{1}\)
a × \(\frac{a}{b}\) = \(\frac{a × a}{b × 1}\)
So,
4 × \(\frac{11}{12}\)
= \(\frac{4}{1}\) × \(\frac{11}{12}\)
= \(\frac{4 × 11}{1 × 12}\)
= \(\frac{44}{12}\)
Hence,
4 × \(\frac{11}{12}\) = \(\frac{44}{12}\)

Question 4.
8 × \(\frac{35}{100}\) = ______
Answer:
8 × \(\frac{35}{100}\) = \(\frac{280}{100}\)

Explanation:
The given numbers are: 8 and \(\frac{35}{100}\)
We know that,
a = \(\frac{a}{1}\)
a × \(\frac{a}{b}\) = \(\frac{a × a}{b × 1}\)
So,
8 × \(\frac{35}{100}\)
= \(\frac{8}{1}\) × \(\frac{35}{100}\)
= \(\frac{8 × 35}{1 × 100}\)
= \(\frac{280}{100}\)
Hence,
8 × \(\frac{35}{100}\) = \(\frac{280}{100}\)

Question 5.
3 × \(\frac{1}{2}\) = ______
Answer:
3 × \(\frac{1}{2}\) = \(\frac{3}{2}\)

Explanation:
The given numbers are: 3 and \(\frac{1}{2}\)
We know that,
a = \(\frac{a}{1}\)
a × \(\frac{a}{b}\) = \(\frac{a × a}{b × 1}\)
So,
3 × \(\frac{1}{2}\)
= \(\frac{3}{1}\) × \(\frac{1}{2}\)
= \(\frac{1 × 3}{1 × 2}\)
= \(\frac{3}{2}\)
Hence,
3 × \(\frac{1}{2}\) = \(\frac{3}{2}\)

Question 6.
7 × \(\frac{2}{5}\) = ______
Answer:
7 × \(\frac{2}{5}\) = \(\frac{14}{5}\)

Explanation:
The given numbers are: 7 and \(\frac{2}{5}\)
We know that,
a = \(\frac{a}{1}\)
a × \(\frac{a}{b}\) = \(\frac{a × a}{b × 1}\)
So,
7 × \(\frac{2}{5}\)
= \(\frac{7}{1}\) × \(\frac{2}{5}\)
= \(\frac{2 × 7}{1 × 5}\)
= \(\frac{14}{5}\)
Hence,
7 × \(\frac{2}{5}\) = \(\frac{14}{5}\)

Question 7.
6 × \(\frac{7}{4}\) = ______
Answer:
6 × \(\frac{7}{4}\) = \(\frac{42}{4}\)

Explanation:
The given numbers are: 6 and \(\frac{7}{4}\)
We know that,
a = \(\frac{a}{1}\)
a × \(\frac{a}{b}\) = \(\frac{a × a}{b × 1}\)
So,
6 × \(\frac{7}{4}\)
= \(\frac{6}{1}\) × \(\frac{7}{4}\)
= \(\frac{6 × 7}{1 × 4}\)
= \(\frac{42}{4}\)
Hence,
6 × \(\frac{7}{4}\) = \(\frac{42}{4}\)

Question 8.
12 × \(\frac{8}{7}\) = ______
Answer:
12 × \(\frac{8}{7}\) = \(\frac{96}{7}\)

Explanation:
The given numbers are: 12 and \(\frac{8}{7}\)
We know that,
a = \(\frac{a}{1}\)
a × \(\frac{a}{b}\) = \(\frac{a × a}{b × 1}\)
So,
12 × \(\frac{8}{7}\)
= \(\frac{12}{1}\) × \(\frac{8}{7}\)
= \(\frac{12 × 8}{1 × 7}\)
= \(\frac{96}{7}\)
Hence,
12 × \(\frac{8}{7}\) = \(\frac{96}{7}\)

Question 9.
25 × \(\frac{10}{9}\) = ______
Answer:
25 × \(\frac{10}{9}\) = \(\frac{250}{9}\)

Explanation:
The given numbers are: 25 and \(\frac{10}{9}\)
We know that,
a = \(\frac{a}{1}\)
a × \(\frac{a}{b}\) = \(\frac{a × a}{b × 1}\)
So,
25 × \(\frac{10}{9}\)
= \(\frac{25}{1}\) × \(\frac{10}{9}\)
= \(\frac{25 × 10}{1 × 9}\)
= \(\frac{250}{9}\)
Hence,
25 × \(\frac{10}{9}\) = \(\frac{250}{9}\)

Find the unknown number.
Question 10.
Big Ideas Math Answer Key Grade 5 Chapter 9 Multiply Fractions 9.1 10
Answer:
The missing number is: 5

Explanation:
The given fractions are: \(\frac{5}{7}\) and \(\frac{25}{7}\)
Let the missing number be X
So,
X × \(\frac{5}{7}\) = \(\frac{25}{7}\)
When multiplication goes from left to right or from right to left, it will become the division
When division goes from left to right or from right to left, it will become multiplication.
So,
X = \(\frac{25}{7}\) ÷ \(\frac{5}{7}\)
We know that,
\(\frac{a}{b}\) ÷ \(\frac{x}{y}\) = \(\frac{a}{b}\) × \(\frac{y}{x}\)
So,
\(\frac{25}{7}\) ÷ \(\frac{5}{7}\)
= \(\frac{25}{7}\) × \(\frac{7}{5}\)
= \(\frac{7 × 25}{7 × 5}\)
= 5
Hence,
The missing number is: 5

Question 11.
Big Ideas Math Answer Key Grade 5 Chapter 9 Multiply Fractions 9.1 11
Answer:
The missing number is: 7

Explanation:
The given fractions are: \(\frac{9}{10}\) and \(\frac{63}{10}\)
Let the missing number be X
So,
X × \(\frac{9}{10}\) = \(\frac{63}{10}\)
When multiplication goes from left to right or from right to left, it will become the division
When division goes from left to right or from right to left, it will become multiplication.
So,
X = \(\frac{63}{10}\) ÷ \(\frac{9}{10}\)
We know that,
\(\frac{a}{b}\) ÷ \(\frac{x}{y}\) = \(\frac{a}{b}\) × \(\frac{y}{x}\)
So,
\(\frac{63}{10}\) ÷ \(\frac{9}{10}\)
= \(\frac{63}{10}\) × \(\frac{10}{9}\)
= \(\frac{63 × 10}{10 × 9}\)
= 7
Hence,
The missing number is: 7

Question 12.
Big Ideas Math Answer Key Grade 5 Chapter 9 Multiply Fractions 9.1 12
Answer:
The missing number is: 9

Explanation:
The given fractions are: \(\frac{3}{5}\) and \(\frac{27}{5}\)
Let the missing number be X
So,
X × \(\frac{3}{5}\) = \(\frac{27}{5}\)
When multiplication goes from left to right or from right to left, it will become the division
When division goes from left to right or from right to left, it will become multiplication.
So,
X = \(\frac{27}{5}\) ÷ \(\frac{3}{5}\)
We know that,
\(\frac{a}{b}\) ÷ \(\frac{x}{y}\) = \(\frac{a}{b}\) × \(\frac{y}{x}\)
So,
\(\frac{27}{5}\) ÷ \(\frac{3}{5}\)
= \(\frac{27}{5}\) × \(\frac{5}{3}\)
= \(\frac{5 × 27}{5 × 3}\)
= 9
Hence,
The missing number is: 9

Question 13.
You make 5 servings of pancakes. You top each serving with \(\frac{1}{4}\) cup of strawberries. How many cups of strawberries do you use?
Answer:
The number of cups of strawberries you used is: 20 cups

Explanation:
It is given that you make 5 servings of pancakes and you top each serving with \(\frac{1}{4}\) cup of strawberries.
So,
The number of cups of strawberries = \(\frac{The number of servings of pancakes}{Each serving in pancake}\)
= 5 ÷ \(\frac{1}{4}\)
= \(\frac{5}{1}\) ÷ \(\frac{1}{4}\)
= \(\frac{5}{1}\) × \(\frac{4}{1}\)
= \(\frac{5 × 4}{1 × 1}\)
= 20 cups
hence, from the above,
We can conclude that the number of cups of strawberries is: 20 cups

Question 14.
Which One Doesn’tBelong?
Which one does not belong with the other three?
4 × \(\frac{3}{8}\)
3 × \(\frac{1}{8}\)
\(\frac{3}{8}\) + \(\frac{3}{8}\) + \(\frac{3}{8}\) + \(\frac{3}{8}\)
1 + \(\frac{1}{2}\)
Answer:
Let the expresseons be named A., B., C., D.
So,
A) 4 × \(\frac{3}{8}\)
B) 3 × \(\frac{1}{8}\)
C) \(\frac{3}{8}\) + \(\frac{3}{8}\) + \(\frac{3}{8}\) + \(\frac{3}{8}\)
D) 1 + \(\frac{1}{2}\)
So, from the above four expressions,
Expressions A), B), C) are in the normal form and C) is the expanded form of expression A)
Hence, from the above,
We can conclude that expression D) does not belong to the other three.

Question 15.
Modeling Real Life
You complete \(\frac{5}{2}\) inches of weaving each day for 5 days. The weaving needs to be at least 11 inches long. Is your weaving complete?
Big Ideas Math Answer Key Grade 5 Chapter 9 Multiply Fractions 9.1 13
Answer:
Yes, your weaving is complete

Explanation:
It is given that you complete \(\frac{5}{2}\) inches of weaving each day for 5 days.
So,
The total number of inches you completed in 5 days = ( The number of inches you weaved each day ) × 5
= \(\frac{5}{2}\) × 5
= \(\frac{5}{2}\) × \(\frac{5}{1}\)
= \(\frac{5 × 5}{2 × 1}\)
= \(\frac{25}{2}\)
It is also given that the weaving needs to be at least 11 inches long.
So,
11 inches can also be written as \(\frac{22}{2}\)
So,
When we compare the total number of inches needed for weaving and the total number of inches you weaved in 5 days, we can observe that
You weaved more than the amount of the weaving.
Hence, from the above,
We can conclude that your weaving is completed.

Question 16.
DIG DEEPER!
You spend \(\frac{7}{2}\) hours playing drums each day for 2 days. Your friend spends \(\frac{5}{4}\) hours playing drums each day for 6 days. Who spends more time playing drums? How much more?
Big Ideas Math Answer Key Grade 5 Chapter 9 Multiply Fractions 9.1 14
Answer:
Your friend spends more time playing drums than you
The amount you played drums more than your friend is: \(\frac{1}{2}\) hours

Explanation:
It is given that you spend \(\frac{7}{2}\) hours each day playing drums for 2 days
So,
The time you played drums for 2 days = The time you played drums for each day × 2
= \(\frac{7}{2}\) × 2
= \(\frac{7}{2}\) × \(\frac{2}{1}\)
= \(\frac{7 × 2}{2 × 1}\)
= \(\frac{14}{2}\)
= \(\frac{28}{4}\)
= 7 hours
It is also given that Your friend spends \(\frac{5}{4}\) hours playing drums each day for 6 days.
So,
The time your friend played drums for 6 days = The time your friend played drums for each day × 6
= \(\frac{5}{4}\) × 6
= \(\frac{5}{4}\) × \(\frac{6}{1}\)
= \(\frac{5 × 6}{4 × 1}\)
= \(\frac{30}{4}\) hours
Now,
When we compare the time played drums by you and your friends, your friend played more time than you
So,
The amount of time more your friend played than you = \(\frac{30}{4}\) – \(\frac{28}{4}\)
= \(\frac{30 – 28}{4}\)
= \(\frac{2}{4}\)
= \(\frac{1}{2}\) hour
Hence, from the above,
We can conclude that
Your friend spends more time playing drums than you
The amount you played drums more than your friend is: \(\frac{1}{2}\) hours

Review & Refresh

Find the product.
Question 17.
Big Ideas Math Answer Key Grade 5 Chapter 9 Multiply Fractions 9.1 15
Answer:
0.6 × 0.4 = 0.24

Explanation:
The given decimal numbers are: 0.6 and 0.4
the representation of the decimal numbers in the fraction form is: \(\frac{6}{10}\) and \(\frac{4}{10}\)
Now,
\(\frac{6}{10}\) × \(\frac{4}{10}\)
= \(\frac{6 × 4}{10 × 10}\)
= \(\frac{24}{100}\)
So,
The representation of \(\frac{24}{100}\) in the decimal form is: 0.24
Hence, 0.6 × 0.4 = 0.24

Question 18.
Big Ideas Math Answer Key Grade 5 Chapter 9 Multiply Fractions 9.1 16
Answer:
2.37 × 1.9 = 4.503

Explanation:
The given decimal numbers are: 2.37 and 1.9
the representation of the decimal numbers in the fraction form is: \(\frac{237}{100}\) and \(\frac{19}{10}\)
Now,
\(\frac{237}{100}\) × \(\frac{19}{10}\)
= \(\frac{237 × 19}{100 × 10}\)
= \(\frac{4,503}{1000}\)
So,
The representation of \(\frac{4503}{1000}\) in the decimal form is: 4.503
Hence, 2.37 × 1.9 = 4.503

Question 19.
Big Ideas Math Answer Key Grade 5 Chapter 9 Multiply Fractions 9.1 17
Answer:
52.8 × 0.75 = 39.6

Explanation:
The given decimal numbers are: 52.8 and 0.75
the representation of the decimal numbers in the fraction form is: \(\frac{6}{10}\) and \(\frac{4}{10}\)
Now,
\(\frac{528}{10}\) × \(\frac{75}{100}\)
= \(\frac{528 × 75}{10 × 100}\)
= \(\frac{3960}{1000}\)
So,
The representation of \(\frac{3960}{1000}\) in the decimal form is: 39.6
Hence, 52.8 × 0.75 = 39.6

Lesson 9.2 Use Models to Multiply Fractions by Whole Numbers

You need to give water to \(\frac{2}{3}\) of the dogs at a shelter. There are 12 dogs at the shelter. How many dogs need water? Draw a model to support your answer.
Answer:
The number of dogs that need water is: 8 dogs

Explanation:
It is given that you need to give water to \(\frac{2}{3}\) of the dogs at a shelter and there are 12 dogs at the shelter
So,
The number of dogs that need water = ( The fraction of dogs that need water ) × ( The total number of dogs )
= \(\frac{2}{3}\) × 12
We know that,
\(\frac{a}{b}\) × x = \(\frac{a}{b}\) × \(\frac{x}{1}\)
= \(\frac{a × x}{b × 1}\)
So,
\(\frac{2}{3}\) × 12
= \(\frac{2}{3}\) × \(\frac{12}{1}\)
= \(\frac{2 × 12}{3 × 1}\)
= \(\frac{8}{1}\)
= 8
Hence, from the above,
We can conclude that there are 8 dogs that need water

Reasoning
How can you use a model to multiply a fraction by a whole number? Explain.
Answer:
We can multiply a fraction by a whole number by using the following multiplication properties. They are:
A) \(\frac{a}{b}\) × x = \(\frac{a}{b}\) × \(\frac{x}{1}\)
B) x = \(\frac{x}{1}\)

Think and Grow: Multiply Fractions by Whole Numbers

You can use models to multiply a fraction by a whole number.
Example
Find \(\frac{3}{4}\) of 8.

Example
Find \(\frac{5}{8}\) × 4.

Show and Grow

Question 1.
Find \(\frac{2}{5}\) of 10.
Answer:
\(\frac{2}{5}\) × 10 = 4

Explanation:
The given numbers are: \(\frac{2}{5}\) and 10
We know that,
\(\frac{a}{b}\) × x = \(\frac{a}{b}\) × \(\frac{x}{1}\)
= \(\frac{a × x}{b × 1}\)
So,
\(\frac{2}{5}\) × 10 = \(\frac{2}{5}\) × \(\frac{10}{1}\)
= \(\frac{2 × 10}{5 × 1}\)
= \(\frac{4}{1}\)
= 4
Hence,
\(\frac{2}{5}\) × 10 = 4

Question 2.
Find \(\frac{7}{12}\) × 6.
Answer:
\(\frac{7}{12}\) × 6 = \(\frac{7}{2}\)

Explanation:
The given numbers are: \(\frac{7}{12}\) and 6
We know that,
\(\frac{a}{b}\) × x = \(\frac{a}{b}\) × \(\frac{x}{1}\)
= \(\frac{a × x}{b × 1}\)
So,
\(\frac{7}{12}\) × 6 = \(\frac{7}{12}\) × \(\frac{6}{1}\)
= \(\frac{7 × 6}{12 × 1}\)
= \(\frac{7}{2}\)
Hence,
\(\frac{7}{12}\) × 6 = \(\frac{7}{2}\)

Apply and Grow: Practice

Multiply. Use a model to help.
Question 3.
\(\frac{5}{6}\) of 12
Answer:
\(\frac{5}{6}\) × 12 = 10

Explanation:
The given numbers are: \(\frac{5}{6}\) and 12
We know that,
\(\frac{a}{b}\) × x = \(\frac{a}{b}\) × \(\frac{x}{1}\)
= \(\frac{a × x}{b × 1}\)
So,
\(\frac{5}{6}\) × 12 = \(\frac{5}{6}\) × \(\frac{12}{1}\)
= \(\frac{5 × 12}{6 × 1}\)
= \(\frac{10}{1}\)
= 10
Hence,
\(\frac{5}{6}\) × 12 = 10

Question 4.
\(\frac{2}{3}\) × 9
Answer:
\(\frac{2}{3}\) × 9 = 6

Explanation:
The given numbers are: \(\frac{2}{3}\) and 9
We know that,
\(\frac{a}{b}\) × x = \(\frac{a}{b}\) × \(\frac{x}{1}\)
= \(\frac{a × x}{b × 1}\)
So,
\(\frac{2}{3}\) × 9 = \(\frac{2}{3}\) × \(\frac{9}{1}\)
= \(\frac{2 × 9}{3 × 1}\)
= \(\frac{6}{1}\)
= 6
Hence,
\(\frac{2}{3}\) × 9 = 6

Question 5.
\(\frac{1}{5}\) × 10
Answer:
\(\frac{1}{5}\) × 10 = 2

Explanation:
The given numbers are: \(\frac{1}{5}\) and 10
We know that,
\(\frac{a}{b}\) × x = \(\frac{a}{b}\) × \(\frac{x}{1}\)
= \(\frac{a × x}{b × 1}\)
So,
\(\frac{1}{5}\) × 10 = \(\frac{1}{5}\) × \(\frac{10}{1}\)
= \(\frac{1 × 10}{5 × 1}\)
= \(\frac{2}{1}\)
= 2
Hence,
\(\frac{1}{5}\) × 10 = 2

Question 6.
\(\frac{3}{5}\) of 5
Answer:
\(\frac{3}{5}\) × 5 = 3

Explanation:
The given numbers are: \(\frac{3}{5}\) and 5
We know that,
\(\frac{a}{b}\) × x = \(\frac{a}{b}\) × \(\frac{x}{1}\)
= \(\frac{a × x}{b × 1}\)
So,
\(\frac{3}{5}\) × 5 = \(\frac{3}{5}\) × \(\frac{5}{1}\)
= \(\frac{3 × 5}{5 × 1}\)
= \(\frac{3}{1}\)
= 3
Hence,
\(\frac{3}{5}\) × 5 = 3

Question 7.
\(\frac{1}{6}\) of 3
Answer:
\(\frac{1}{6}\) × 3 = \(\frac{1}{2}\)

Explanation:
The given numbers are: \(\frac{1}{6}\) and 3
We know that,
\(\frac{a}{b}\) × x = \(\frac{a}{b}\) × \(\frac{x}{1}\)
= \(\frac{a × x}{b × 1}\)
So,
\(\frac{1}{6}\) × 3 = \(\frac{1}{6}\) × \(\frac{3}{1}\)
= \(\frac{1 × 3}{6 × 1}\)
= \(\frac{1}{2}\)
Hence,
\(\frac{1}{6}\) × 3 = \(\frac{1}{2}\)

Question 8.
\(\frac{3}{8}\) × 4
Answer:
\(\frac{3}{8}\) × 4 = \(\frac{3}{2}\)

Explanation:
The given numbers are: \(\frac{3}{8}\) and 4
We know that,
\(\frac{a}{b}\) × x = \(\frac{a}{b}\) × \(\frac{x}{1}\)
= \(\frac{a × x}{b × 1}\)
So,
\(\frac{3}{8}\) × 4 = \(\frac{3}{8}\) × \(\frac{4}{1}\)
= \(\frac{3 × 4}{8 × 1}\)
= \(\frac{3}{2}\)
Hence,
\(\frac{3}{8}\) × 4 = \(\frac{3}{2}\)

Question 9.
You have 25 beads. You use \(\frac{2}{5}\) of the beads to make a bracelet. How many beads do you use?
Answer:
The number of beads you used is: 10 beads

Explanation:
It is given that you have 25 beads and you use \(\frac{2}{5}\) of the beads to make a bracelet.
So,
The number of beads you used = ( The fraction of beads used to make 1 bracelet ) × ( The total number of beads )
= \(\frac{2}{5}\) × 25
We know that,
\(\frac{a}{b}\) × x = \(\frac{a}{b}\) × \(\frac{x}{1}\)
= \(\frac{a × x}{b × 1}\)
So,
\(\frac{2}{5}\) × 25 = \(\frac{2}{5}\) × \(\frac{25}{1}\)
= \(\frac{2 × 25}{5 × 1}\)
= \(\frac{10}{1}\)
= 10 beads
Hence, from the above,
We can conclude that the number of beads you used to make bracelets are: 10 beads

Question 10.
Writing
Write and solve a real-life problem for the expression.
\(\frac{3}{4}\) × 20
Answer:
Suppose there are 20 passengers in a bus and they occupied the place \(\frac{3}{4}\) of the bus.
Hence,
The total number of seats in the bus = ( The fraction of the place occupied by the passengers ) × ( The total number of passengers )
= \(\frac{3}{4}\) × 20
Now,
We know that,
\(\frac{a}{b}\) × x = \(\frac{a}{b}\) × \(\frac{x}{1}\)
= \(\frac{a × x}{b × 1}\)
So,
\(\frac{3}{4}\) × 20 = \(\frac{3}{4}\) × \(\frac{20}{1}\)
= \(\frac{3 × 20}{4 × 1}\)
= \(\frac{15}{1}\)
= 15 seats
Hence, from the above,
We can conclude that there are 15 seats in the bus.

Question 11.
YOU BE THE TEACHER
Descartes finds \(\frac{2}{3}\) × 6. Is he correct? Explain.
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.2 3
Answer:
No, Descartes is not correct

Explanation:
According to Descartes,
The given numbers are: \(\frac{2}{3}\) and 6
So,
According to Descartes, 6 is divided into 3 equal parts
So,
6 ÷ 3 = 2 equal parts
So,
According to Descartes, these 2 equal parts also divided into 2 parts.
So,
The total number of parts = 2 equal parts + 2 equal parts = 4 parts
So,
The product of \(\frac{2}{3}\) and 6 is 4
Now,
We know that,
\(\frac{a}{b}\) × x = \(\frac{a}{b}\) × \(\frac{x}{1}\)
= \(\frac{a × x}{b × 1}\)
So,
\(\frac{2}{3}\) × 6 = \(\frac{2}{3}\) × \(\frac{6}{1}\)
= \(\frac{6 × 2}{3 × 1}\)
= \(\frac{4}{1}\)
= 4
Hence, from the above,
We can conclude that Descartes is not correct.

Think and Grow: Modeling Real Life

Example
A recipe calls for 2 cups of rice. You only have \(\frac{3}{4}\) of that amount. How much more rice do you need?
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.2 4
Find the number of cups of rice that you have by finding \(\frac{3}{4}\) of 2.
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.2 5
So,
You have \(\frac{3}{2}\) cups of rice.
Subtract the amount of rice you have from the amount of rice you need.
2 – \(\frac{3}{2}\) = \(\frac{2}{1}\) – \(\frac{3}{2}\)
= \(\frac{(2 × 2) – 3}{2}\)
= \(\frac{1}{2}\)
So,
you need \(\frac{1}{2}\) cup more of rice.

Show and Grow

Question 12.
You have 12 tokens. You use \(\frac{3}{4}\) of them to play a pinball game. How many tokens do you have left?
Answer:
The number of tokens you have left is: 3 tokens

Explanation:
It is given that you have 12 tokens and you use \(\frac{3}{4}\) of them to play a pinball game.
So,
The number of tokens used to play a pinball game = ( The fraction of tokens used to play a pinball game ) × ( The total number of tokens )
= \(\frac{3}{4}\) × 12
= \(\frac{3}{4}\) × \(\frac{12}{1}\)
= \(\frac{3 × 12}{4 × 1}\)
= \(\frac{9}{1}\)
= 9 tokens
So,
The number of tokens left = ( The total number of tokens ) – ( The number of tokens used to play the pinball )
= 12 – 9
= 3 tokens
Hence, from the above,
We can conclude that the number of tokens left are: 3 tokens

Question 13.
A male lion sleeps \(\frac{5}{6}\) × 6 hours of each day. How many hours does the lion sleep in 1 week?
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.2 6
Answer:
The number of hours the lion sleeps in 1 week is: 35 hours

Explanation:
It is given that a male lion sleeps \(\frac{5}{6}\) × 6 hours of each day
So,
The number of hours the lion sleeps in 1 day = \(\frac{5}{6}\) × 6 hours
= \(\frac{5}{6}\) × \(\frac{6}{1}\)
= \(\frac{5 × 6}{6 × 1}\)
= \(\frac{5}{1}\)
= 5 hours
We know that,
1 week = 7 days
So,
The total number of hours the lion sleeps in 1 week = ( The number of hours the lion sleeps in 1 day ) × ( The number od fays in 1 week )
= 5 × 7
= 35 hours
Hence, from the above,
We can conclude that the lion sleeps for 35 hours in 1 week

Question 14.
DIG DEEPER!
In a class of 20 students, \(\frac{1}{10}\) of the students are 10 years old, \(\frac{4}{5}\)of the students are 11 years old, and the rest are 12 years old. How many more 11-year-olds than 12-year-olds are in the class?
Answer:
The number of 11-year-olds more than 12-year-olds is: 14

Explanation:
It is given that in a class of 20 students, \(\frac{1}{10}\) of the students are 10 years old, \(\frac{4}{5}\)of the students are 11 years old, and the rest are 12 years old.
So,
The number of 10-year-olds = ( The fraction of 10-year-olds ) × ( The total number of students )
= \(\frac{1}{10}\) × 20
= \(\frac{1 × 20}{10 × 1}\)
= 2 10-year-olds
The number of 11-year-olds = ( The fraction of 11-year-olds ) × ( The total number of students )
= \(\frac{4}{5}\) × 20
= \(\frac{4}{5}\) × \(\frac{20}{1}\)
= \(\frac{4 × 20}{5 × 1}\)
= \(\frac{16}{1}\)
= 16 11-year-olds
So,
The number of 12-year-olds = ( The total number of students ) – ( The number of 10-year-olds + The number of 11-year-olds )
= 20 – ( 16 + 2 )
= 2 12-year-olds
So,
The number of 11-year-olds more than 12-year-olds = ( The number of 11-year-olds ) – ( The numebr of 12-year-olds )
= 16 – 2
= 14 students
Hence, from the above,
We can conclude that there are 14 students who are 11-years-old more than 12-years-old.

Use Models to Multiply Fractions by Whole Numbers Homework & Practice 9.2

Multiply. Use a model to help.
Question 1.
\(\frac{2}{3}\) × 6
Answer:
\(\frac{2}{3}\) × 6 = 4

Explanation:
The given numbers are: \(\frac{2}{3}\) and 6
We know that,
\(\frac{a}{b}\) × x = \(\frac{a}{b}\) × \(\frac{x}{1}\)
= \(\frac{a × x}{b × 1}\)
So,
\(\frac{2}{3}\) × 6 = \(\frac{2}{3}\) × \(\frac{6}{1}\)
= \(\frac{2 × 6}{3 × 1}\)
= \(\frac{4}{1}\)
= 4
Hence,
\(\frac{2}{3}\) × 6 = 4

Question 2.
\(\frac{3}{5}\) of 10
Answer:
\(\frac{3}{5}\) × 10 = 6

Explanation:
The given numbers are: \(\frac{3}{5}\) and 10
We know that,
\(\frac{a}{b}\) × x = \(\frac{a}{b}\) × \(\frac{x}{1}\)
= \(\frac{a × x}{b × 1}\)
So,
\(\frac{3}{5}\) × 10 = \(\frac{3}{5}\) × \(\frac{10}{1}\)
= \(\frac{3 × 10}{5 × 1}\)
= \(\frac{6}{1}\)
= 6
Hence,
\(\frac{3}{5}\) × 10 = 6

Question 3.
\(\frac{1}{2}\) of 4
Answer:
\(\frac{1}{2}\) × 4 = 2

Explanation:
The given numbers are: \(\frac{1}{2}\) and 4
We know that,
\(\frac{a}{b}\) × x = \(\frac{a}{b}\) × \(\frac{x}{1}\)
= \(\frac{a × x}{b × 1}\)
So,
\(\frac{1}{2}\) × 4 = \(\frac{1}{2}\) × \(\frac{4}{1}\)
= \(\frac{1 × 4}{2 × 1}\)
= \(\frac{2}{1}\)
= 2
Hence,
\(\frac{1}{2}\) × 4 = 2

Question 4.
\(\frac{1}{4}\) × 12
Answer:
\(\frac{1}{4}\) × 12 = 3

Explanation:
The given numbers are: \(\frac{1}{4}\) and 12
We know that,
\(\frac{a}{b}\) × x = \(\frac{a}{b}\) × \(\frac{x}{1}\)
= \(\frac{a × x}{b × 1}\)
So,
\(\frac{1}{4}\) × 12 = \(\frac{1}{4}\) × \(\frac{12}{1}\)
= \(\frac{1 × 12}{4 × 1}\)
= \(\frac{3}{1}\)
= 3
Hence,
\(\frac{1}{12}\) × 4 = 3

Question 5.
\(\frac{5}{6}\) × 3
Answer:
\(\frac{5}{6}\) × 3 = \(\frac{5}{2}\)

Explanation:
The given numbers are: \(\frac{5}{6}\) and 3
We know that,
\(\frac{a}{b}\) × x = \(\frac{a}{b}\) × \(\frac{x}{1}\)
= \(\frac{a × x}{b × 1}\)
So,
\(\frac{5}{6}\) × 3 = \(\frac{5}{6}\) × \(\frac{3}{1}\)
= \(\frac{5 × 3}{6 × 1}\)
= \(\frac{5}{2}\)
Hence,
\(\frac{5}{6}\) × 3 = \(\frac{5}{2}\)

Question 6.
\(\frac{3}{4}\) of 2
Answer:
\(\frac{3}{4}\) × 2 = \(\frac{3}{2}\)

Explanation:
The given numbers are: \(\frac{3}{4}\) and 2
We know that,
\(\frac{a}{b}\) × x = \(\frac{a}{b}\) × \(\frac{x}{1}\)
= \(\frac{a × x}{b × 1}\)
So,
\(\frac{3}{4}\) × 2 = \(\frac{3}{4}\) × \(\frac{2}{1}\)
= \(\frac{3 × 2}{4 × 1}\)
= \(\frac{3}{2}\)
Hence,
\(\frac{3}{4}\) × 2 = \(\frac{3}{2}\)

Question 7.
You have 27 foam balls. You use \(\frac{1}{3}\) of the balls for a model. How many balls do you use?
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.2 7
Answer:
The number of balls you used is: 9 balls

Explanation:
It is given that there are 27 foam balls and you use \(\frac{1}{3}\) of the balls for a model.
So,
The number of balls you used = ( The fraction of balls you used ) × ( The total number of balls )
= \(\frac{1}{3}\) × 27
Now,
We know that,
\(\frac{a}{b}\) × x = \(\frac{a}{b}\) × \(\frac{x}{1}\)
= \(\frac{a × x}{b × 1}\)
So,
\(\frac{1}{3}\) × 27 = \(\frac{1}{3}\) × \(\frac{27}{1}\)
= \(\frac{1 × 27}{3 × 1}\)
= \(\frac{9}{1}\)
= 9 balls
Hence, from the above,
We can conclude that the number of foam balls you used are: 27

Question 8.
An object that weighs 1 pound on Earth weighs about \(\frac{1}{15}\) pound on Pluto. A man weighs 240 pounds on Earth. How many pounds does he weigh on Pluto?
Answer:
The number of pounds the man weigh in Pluto is: 16 pounds

Explanation:
It is given that an object that weighs 1 pound on Earth weighs about \(\frac{1}{15}\) pound on Pluto.
It is also given that a man weighs 240 pounds on Earth.
So,
The weight of a man on pluto = ( The fraction of weight of a man on pluto when compared to earth ) × ( The weight of the man on earth )
= \(\frac{1}{15}\) × 240
Now,
We know that,
\(\frac{a}{b}\) × x = \(\frac{a}{b}\) × \(\frac{x}{1}\)
= \(\frac{a × x}{b × 1}\)
So,
\(\frac{1}{15}\) × 240 = \(\frac{1}{15}\) × \(\frac{240}{1}\)
= \(\frac{1 × 240}{15 × 1}\)
= \(\frac{16}{1}\)
= 16 pounds
Hence, from the above,
We can conclude that the weight of the man on pluto is: 16 pounds

Question 9.
Structure
Write a multiplication equation represented by the model.
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.2 8
Answer:

Question 10.
DIG DEEPER!
Find each missing number.
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.2 9
Answer:
Let the expression be named A), B), C), and D)
So,
The missing number of A is: 12
The missing number of B is: 6
The missing number of C is: 2
The missing number of D is: 5

Explanation:
Let the expressions be named as A), B), C), and D)
So,
A) The given fractions are: \(\frac{1}{3}\) and 4
Let the missing number be X
So,
X × \(\frac{1}{3}\) = 4
When multiplication goes from left to right or from right to left, it will become the division
When division goes from left to right or from right to left, it will become multiplication.
So,
X = 4 ÷ \(\frac{1}{3}\)
We know that,
\(\frac{a}{b}\) ÷ \(\frac{x}{y}\) = \(\frac{a}{b}\) × \(\frac{y}{x}\)
So,
\(\frac{4}{1}\) ÷ \(\frac{1}{3}\)
= \(\frac{4}{1}\) × \(\frac{3}{1}\)
= \(\frac{4 × 3}{1 × 1}\)
= 12
Hence, the missing number is 12
B) The given numbers are: \(\frac{2}{5}\) and 15
We know that,
\(\frac{a}{b}\) × x = \(\frac{a}{b}\) × \(\frac{x}{1}\)
= \(\frac{a × x}{b × 1}\)
So,
\(\frac{2}{5}\) × 15 = \(\frac{2}{5}\) × \(\frac{15}{1}\)
= \(\frac{2 × 15}{5 × 1}\)
= \(\frac{6}{1}\)
= 6
Hence,
\(\frac{2}{5}\) × 15 = 6
Like the above two expressions, the remaining two expressions can also be solved.
hence, from the above,
We can conclude that
The missing number of A is: 12
The missing number of B is: 6
The missing number of C is: 2
The missing number of D is: 5

Question 11.
Modeling Real Life
You have 28 craft sticks. You use \(\frac{4}{7}\) of them for a project. How many craft sticks do you have left?
Answer:
The number of craft sticks you have left is: 12 craft sticks

Explanation:
It is given that you have 28 craft sticks and you used \(\frac{4}{7}\) of them for a project.
So,
The number of craft sticks you used = ( The fraction of craft sticks you used ) × ( The total number of craft sticks )
= \(\frac{4}{7}\) × 28
= \(\frac{4}{7}\) × \(\frac{28}{1}\)
= \(\frac{4 × 28}{7 × 1}\)
= \(\frac{16}{1}\)
= 16
So,
The number of craft sticks you left = ( The total number of craft sticks ) – ( The number of craft sticks you used )
= 28 – 16
= 12 craft sticks
Hence, from the above,
We can conclude that the number of craft sticks you left is: 12 craft sticks

Question 12.
Modeling Real Life
A mother otter spends \(\frac{1}{3}\) of each day feeding her baby. How many hours does the mother otter spend feeding her baby in 1 week?
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.2 10
Answer:
The number of hours the mother otter spend feeding her baby in 1 week is: 56 hours

Explanation:
It is given that a mother otter spends \(\frac{1}{3}\) of each day feeding her baby.
We know that,
1 day = 24 hours
So,
The number of hours the mother otter spends feeding her baby in 1 day = ( The fraction of time the mother otter spends in feeding milk ) × 24
= \(\frac{1}{3}\) × 24
= \(\frac{1}{3}\) × \(\frac{24}{1}\)
= \(\frac{1 × 24}{3 × 1}\)
= \(\frac{8}{1}\)
= 8 hours
Now,
We know that,
1 week = 7 days
So,
The number of hours the mother otter spends feeding her milk in 1 week = ( The number of hours the mother otter feeding her milk in 1 day ) × 7
= 8 × 7
= 56 hours
Hence, from the above,
We can conclude that the mother otter spends feeding her milk for 56 hours in 1 week

Review & Refresh

Estimate the quotient.
Question 13.
5,692 ÷ 5
Answer:
5,692 ÷ 5 =1,138 R 2

Explanation:
By using the partial quotients method,
5,692 ÷ 5 = ( 5,000 + 600 + 90 ) ÷ 5
= ( 5,000 ÷ 5 ) + ( 600 ÷ 5 ) + ( 90 ÷ 5 )
= 1,000 + 120 + 18
= 1,138 R 2
Hence, 5,692 ÷ 5 = 1,138 R 2

Question 14.
309 ÷ 12
Answer:
309 ÷ 12 = 25 R 9

Explanation:
By using the partial quotients method,
309 ÷ 12 = ( 240 + 60 ) ÷ 12
= ( 240 ÷ 12 ) + ( 60 ÷ 12 )
= 20 + 5
= 25 R 9
Hence, 309 ÷ 12 = 25 R 2

Question 15.
2,987 ÷ 53
Answer:
2,987 ÷ 53 = 2,987 ÷ ( 50 + 3 )
= ( 2,985 ÷ 50 ) + ( 2,985 ÷ 3 )
= 995 R 2 + 59.7
= 1,054.7 R 2

Lesson 9.3 Multiply Fractions and Whole Numbers

Explore and Grow

Use models to help you complete the table. What do you notice about each expression and its product?
Big Ideas Math Answers Grade 5 Chapter 9 Multiply Fractions 9.3 1
Answer:
The completed table is:

From the completed table,
We can observe the multiplication of the whole number and the fraction.
The product of a whole number and a fraction may be a whole number or a fraction.

Construct Arguments
Explain how to multiply fractions and whole numbers without using models.
Answer:
We can multiply the whole numbers and fractions by using the properties of the multiplication. They are:
A) \(\frac{a}{b}\) × p = \(\frac{a}{b}\) × \(\frac{p}{1}\)
= \(\frac{a × p}{ b × 1}\)
B) p = \(\frac{p}{1}\)

Think and Grow: Multiply Fractions and Whole Numbers

Key Idea
You can find the product of a fraction and a whole number by multiplying the numerator and the whole number. Then write the result over the denominator.
Example
Find 2 × \(\frac{5}{6}\).
Multiply the numerator and the whole number.

Example
Find \(\frac{5}{6}\) × 2
Multiply the numerator and the whole number.

Show and Grow

Multiply.
Question 1.
3 × \(\frac{5}{8}\) = _______
Answer:
3 × \(\frac{5}{8}\) = \(\frac{15}{8}\)

Explanation:
The given numbers are: 3 and \(\frac{5}{8}\)
We know that,
\(\frac{a}{b}\) × p = \(\frac{a}{b}\) × \(\frac{p}{1}\)
= \(\frac{a × p}{ b × 1}\)
p = \(\frac{p}{1}\)
So,
3 × \(\frac{5}{8}\) = \(\frac{3}{1}\) × \(\frac{5}{8}\)
= \(\frac{5 × 3}{8 × 1}\)
= \(\frac{15}{8}\)
Hence,
3 × \(\frac{5}{8}\) = \(\frac{15}{8}\)

Question 2.
6 × \(\frac{4}{9}\) = _______
Answer:
6 × \(\frac{4}{9}\) = \(\frac{8}{3}\)

Explanation:
The given numbers are: 6 and \(\frac{4}{9}\)
We know that,
\(\frac{a}{b}\) × p = \(\frac{a}{b}\) × \(\frac{p}{1}\)
= \(\frac{a × p}{ b × 1}\)
p = \(\frac{p}{1}\)
So,
6 × \(\frac{4}{9}\) = \(\frac{6}{1}\) × \(\frac{4}{9}\)
= \(\frac{6 × 4}{9 × 1}\)
= \(\frac{24}{9}\)
= \(\frac{8}{3}\)
Hence,
6 × \(\frac{4}{9}\) = \(\frac{8}{3}\)

Question 3.
\(\frac{2}{5}\) × 15 = _______
Answer:
15 × \(\frac{2}{5}\) = 6

Explanation:
The given numbers are: 15 and \(\frac{2}{5}\)
We know that,
\(\frac{a}{b}\) × p = \(\frac{a}{b}\) × \(\frac{p}{1}\)
= \(\frac{a × p}{ b × 1}\)
p = \(\frac{p}{1}\)
So,
15× \(\frac{2}{5}\) = \(\frac{15}{1}\) × \(\frac{2}{5}\)
= \(\frac{15 × 2}{5 × 1}\)
= \(\frac{6}{1}\)
= 6
Hence,
15 × \(\frac{2}{5}\) = 6

Apply and Grow: Practice

Multiply.
Question 4.
\(\frac{3}{5}\) × 2 = _______
Answer:
2 × \(\frac{3}{5}\) = \(\frac{6}{5}\)

Explanation:
The given numbers are: 2 and \(\frac{3}{5}\)
We know that,
\(\frac{a}{b}\) × p = \(\frac{a}{b}\) × \(\frac{p}{1}\)
= \(\frac{a × p}{ b × 1}\)
p = \(\frac{p}{1}\)
So,
2 × \(\frac{3}{5}\) = \(\frac{2}{1}\) × \(\frac{3}{5}\)
= \(\frac{2 × 3}{5 × 1}\)
= \(\frac{6}{5}\)
Hence,
2 × \(\frac{3}{5}\) = \(\frac{6}{5}\)

Question 5.
5 × \(\frac{2}{9}\) = _______
Answer:
5 × \(\frac{2}{9}\) = \(\frac{10}{9}\)

Explanation:
The given numbers are: 5 and \(\frac{2}{9}\)
We know that,
\(\frac{a}{b}\) × p = \(\frac{a}{b}\) × \(\frac{p}{1}\)
= \(\frac{a × p}{ b × 1}\)
p = \(\frac{p}{1}\)
So,
5 × \(\frac{2}{9}\) = \(\frac{5}{1}\) × \(\frac{2}{9}\)
= \(\frac{5 × 2}{9 × 1}\)
= \(\frac{10}{9}\)
Hence,
5 × \(\frac{2}{9}\) = \(\frac{10}{9}\)

Question 6.
\(\frac{5}{6}\) × 4 = _______
Answer:
4 × \(\frac{5}{6}\) = \(\frac{10}{3}\)

Explanation:
The given numbers are: 4 and \(\frac{5}{6}\)
We know that,
\(\frac{a}{b}\) × p = \(\frac{a}{b}\) × \(\frac{p}{1}\)
= \(\frac{a × p}{ b × 1}\)
p = \(\frac{p}{1}\)
So,
4 × \(\frac{5}{6}\) = \(\frac{4}{1}\) × \(\frac{5}{6}\)
= \(\frac{5 × 4}{6 × 1}\)
= \(\frac{20}{6}\)
= \(\frac{10}{3}\)
Hence,
4 × \(\frac{5}{6}\) = \(\frac{10}{3}\)

Question 7.
8 × \(\frac{3}{10}\) = _______
Answer:
8 × \(\frac{3}{10}\) = \(\frac{12}{5}\)

Explanation:
The given numbers are: 8 and \(\frac{3}{10}\)
We know that,
\(\frac{a}{b}\) × p = \(\frac{a}{b}\) × \(\frac{p}{1}\)
= \(\frac{a × p}{ b × 1}\)
p = \(\frac{p}{1}\)
So,
8 × \(\frac{3}{10}\) = \(\frac{8}{1}\) × \(\frac{3}{10}\)
= \(\frac{8 × 3}{10 × 1}\)
= \(\frac{24}{10}\)
= \(\frac{12}{5}\)
Hence,
8 × \(\frac{3}{10}\) = \(\frac{12}{5}\)

Question 8.
\(\frac{1}{5}\) × 7 = _______
Answer:
7 × \(\frac{1}{5}\) = \(\frac{7}{5}\)

Explanation:
The given numbers are: 7 and \(\frac{1}{5}\)
We know that,
\(\frac{a}{b}\) × p = \(\frac{a}{b}\) × \(\frac{p}{1}\)
= \(\frac{a × p}{ b × 1}\)
p = \(\frac{p}{1}\)
So,
7 × \(\frac{1}{5}\) = \(\frac{7}{1}\) × \(\frac{1}{5}\)
= \(\frac{7 × 1}{5 × 1}\)
= \(\frac{7}{5}\)
Hence,
7 × \(\frac{1}{5}\) = \(\frac{7}{5}\)

Question 9.
9 × \(\frac{5}{12}\) = _______
Answer:
9 × \(\frac{5}{12}\) = \(\frac{15}{4}\)

Explanation:
The given numbers are: 9 and \(\frac{5}{12}\)
We know that,
\(\frac{a}{b}\) × p = \(\frac{a}{b}\) × \(\frac{p}{1}\)
= \(\frac{a × p}{ b × 1}\)
p = \(\frac{p}{1}\)
So,
9 × \(\frac{5}{12}\) = \(\frac{9}{1}\) × \(\frac{5}{12}\)
= \(\frac{9 × 5}{12 × 1}\)
= \(\frac{45}{12}\)
= \(\frac{15}{4}\)
Hence,
9 × \(\frac{5}{12}\) = \(\frac{15}{4}\)

Question 10.
15 × \(\frac{5}{8}\) = _______
Answer:
15 × \(\frac{5}{8}\) = \(\frac{75}{8}\)

Explanation:
The given numbers are: 15 and \(\frac{5}{8}\)
We know that,
\(\frac{a}{b}\) × p = \(\frac{a}{b}\) × \(\frac{p}{1}\)
= \(\frac{a × p}{ b × 1}\)
p = \(\frac{p}{1}\)
So,
15 × \(\frac{5}{8}\) = \(\frac{15}{1}\) × \(\frac{5}{8}\)
= \(\frac{15 × 5}{8 × 1}\)
= \(\frac{75}{8}\)
Hence,
15 × \(\frac{5}{8}\) = \(\frac{75}{8}\)

Question 11.
\(\frac{3}{4}\) × 20 = _______
Answer:
20 × \(\frac{3}{4}\) = 15

Explanation:
The given numbers are: 20 and \(\frac{3}{4}\)
We know that,
\(\frac{a}{b}\) × p = \(\frac{a}{b}\) × \(\frac{p}{1}\)
= \(\frac{a × p}{ b × 1}\)
p = \(\frac{p}{1}\)
So,
20 × \(\frac{3}{4}\) = \(\frac{20}{1}\) × \(\frac{3}{4}\)
= \(\frac{20 × 3}{4 × 1}\)
= \(\frac{60}{4}\)
= \(\frac{15}{1}\)
= 15
Hence,
20 × \(\frac{3}{4}\) = 15

Question 12.
\(\frac{7}{9}\) × 5 = _______
Answer:
5 × \(\frac{7}{9}\) = \(\frac{35}{9}\)

Explanation:
The given numbers are: 5 and \(\frac{7}{9}\)
We know that,
\(\frac{a}{b}\) × p = \(\frac{a}{b}\) × \(\frac{p}{1}\)
= \(\frac{a × p}{ b × 1}\)
p = \(\frac{p}{1}\)
So,
5 × \(\frac{7}{9}\) = \(\frac{5}{1}\) × \(\frac{7}{9}\)
= \(\frac{5 × 7}{9 × 1}\)
= \(\frac{35}{9}\)
Hence,
5 × \(\frac{7}{9}\) = \(\frac{35}{9}\)

Question 13.
One-tenth of the 50 states in the United States of America have a mockingbird as their state bird. How many states have a mockingbird as their state bird?
Big Ideas Math Answers Grade 5 Chapter 9 Multiply Fractions 9.3 4
Answer:
The number of states that have mockingbird as their state bird is: 5 states

Explanation:
It is given that one-tenth of the 50 states in the United States of America have a mockingbird as their state bird
So,
The number of states that have a mockingbird as their state bird = ( The fraction of the states that have a mockingbird as their state bird ) × ( The total number of states in the United States of America )
= \(\frac{1}{10}\) × 50
Now,
We know that,
\(\frac{a}{b}\) × p = \(\frac{a}{b}\) × \(\frac{p}{1}\)
= \(\frac{a × p}{ b × 1}\)
p = \(\frac{p}{1}\)
So,
50 × \(\frac{1}{10}\) = \(\frac{50}{1}\) × \(\frac{1}{10}\)
= \(\frac{50 × 1}{10 × 1}\)
= \(\frac{50}{10}\)
= \(\frac{5}{1}\)
= 5 states
Hence, from the above,
We can conclude that there are 5 states that have mockingbird as their state bird

Question 14.
Writing
Explain why 9 × \(\frac{2}{3}\) is equivalent to \(\frac{2}{3}\) × 9.
Answer:
By using the Commutative property of multiplication,
a × b = b × a
So,
By using the above property,
In 9 × \(\frac{2}{3}\),
‘a’ is: 9
‘b’ is: \(\frac{2}{3}\)
Hence, from the above,
We can conclude that  9 × “\(\frac{2}{3}\)” is equivalent to “\(\frac{2}{3}\) × 9″ by using the Commutative property of multiplication.

Question 15.
Reasoning
Without calculating, determine which product is greater. Explain.
\(\frac{1}{8}\) × 24
\(\frac{7}{8}\) × 24
Answer:
The product of “\(\frac{7}{8}\) × 24″ is greater than “\(\frac{1}{8}\) × 24″

Explanation:
The given products are: \(\frac{1}{8}\) × 24 and \(\frac{7}{8}\) × 24
We know that,
\(\frac{a}{b}\) × p = \(\frac{a}{b}\) × \(\frac{p}{1}\)
= \(\frac{a × p}{ b × 1}\)
p = \(\frac{p}{1}\)
So, from the above fractions,
We can observe that except numerator, the denominator and the whole number is all the same.
So,
To find which product is greater, we just have to compare the numerators of the two fractions without calculating the value of the product.
In the comparison of the numerators,
We can observe that,
1 < 7
So,
First fraction numerator < Second fraction numerator
Hence, from the above,
We can conclude that the product of “\(\frac{7}{8}\) × 24″ is greater than “\(\frac{1}{8}\) × 24″

Think and Grow: Modeling Real Life

Example
Newton buys 27 songs. Two-thirds of them are classical songs. Descartes buys 16 songs. Seven-eighths of them are classical songs. Who buys more classical songs? How many more?
Big Ideas Math Answers Grade 5 Chapter 9 Multiply Fractions 9.3 5
Multiply \(\frac{2}{3}\) by 27 to find the number of classical songs Newton buys. Multiply \(\frac{7}{8}\) by 16 to find the number Descartes buys.

So, Newton buys more classical songs.
Subtract the products to find how many more.
18 – 14 = 4
Hence,
Newton buys 4 more classical songs than Descartes.

Show and Grow

Question 16.
You take 48 pictures on a walking tour. Five-twelfths of them is of buildings. Your friend takes 45 pictures. Six-fifteenths of them are of buildings. Who takes more pictures of buildings? How many more?
Answer:
You take more pictures
You take 2 pictures more than your friend

Explanation:
It is given that you take 48 pictures on a walking tour and five-twelfths if them is of buildings.
So,
The number of buildings taken by you = ( The total number of pictures taken by you ) × ( The fraction of pictures that is of buildings )
= 48 × \(\frac{5}{12}\)
= \(\frac{48}{1}\) × \(\frac{5}{12}\)
= \(\frac{48 × 5}{1 × 12}\)
= 20 pictures
It is also given that your friend taken 45 pictures and six-fifteenth of them is buildings
So,
The number of buildings taken by your friend = ( The total number of pictures taken by your friend ) × ( The fraction of pictures that is of buildings )
= 45 × \(\frac{6}{15}\)
= \(\frac{45}{1}\) × \(\frac{6}{15}\)
= \(\frac{45 × 6}{1 × 15}\)
= 18 pictures
So,
In the comparison of the pictures of buildings,
You take more pictures than your friend.
Now,
The number of pictures more taken by you than your friend = ( The number of pictures taken by you ) – ( The number of pictures taken by your friend )
= 20 – 18
= 2 pictures
Hence, from the above,
We can conclude that
You take more pictures
You take 2 pictures more than your friend

Question 17.
You have 72 rocks in your rock collection. Five-eighths of them are sedimentary, one-sixth of them are igneous, and the rest are metamorphic. How many of your rocks are metamorphic?
Answer:
The number of rocks that are metamorphic is: 15 metamorphous rocks

Explanation:
It is given that you have 72 rocks in your rock collection and five-eighths of them are sedimentary, one-sixth of them are igneous, and the rest are metamorphic.
So,
The number of rocks that are sedimentary = ( The total number of rocks ) × ( The fraction of rocks that are sedimentary )
= 72 × \(\frac{5}{8}\)
= \(\frac{72}{1}\) × \(\frac{5}{8}\)
= \(\frac{72 × 5}{1 × 8}\)
= 45 sedimentary rocks
The number of rocks that are igneous = ( The total number of rocks ) × ( The fraction of rocks that are igneous )
= 72 × \(\frac{1}{6}\)
= \(\frac{72}{1}\) × \(\frac{1}{6}\)
= \(\frac{72 × 1}{1 × 6}\)
= 12 igneous rocks
So,
The number of metamorphous rocks = ( The total number of rocks ) – ( The number of sedimentary rocks + The number of igneous rocks)
= 72 – ( 45 + 12 )
= 15 metamorphous rocks
Hence, from the above,
We can conclude that there are 15 metamorphous rocks

Question 18.
DIG DEEPER!
Each day, you spend \(\frac{3}{4}\) hour reading and \(\frac{1}{2}\) hour writing in a journal. How many total hours do you spend reading and writing in 1 week? Describe two ways to solve the problem.
Big Ideas Math Answers Grade 5 Chapter 9 Multiply Fractions 9.3 7
Answer:
The number of hours you spend on reading and writing in 1 week is:

Explanation:
It is given that, each day you spent \(\frac{3}{4}\) hour reading and \(\frac{1}{2}\) hour writing in a journal.
So,
The number of hours you spent on reading and writing a journal in 1 day = ( The number of hours spent on reading ) + ( The number of hours spent on writing )
= \(\frac{3}{4}\) + \(\frac{1}{2}\)
Multiply \(\frac{1}{2}\) with \(\frac{2}{2}\)
So,
\(\frac{3}{4}\) + \(\frac{2}{4}\)
= \(\frac{3 + 2}{4}\)
= \(\frac{5}{4}\) hours
We know that,
1 day = 24 hours
1 week = 7 days
So,
1 week = 7 × 24 hours
So,
The number of hours you spent on reading and writing in 1 week = ( The number of hours you spent on reading and writing in 1 day ) × ( The number of hours in 1 week )
= \(\frac{5}{4}\) × 7 × 24
= \(\frac{5}{4}\) × \(\frac{7}{1}\) × \(\frac{24}{1}\)
= \(\frac{5 × 7 × 24}{4 × 1}\)
= 5 × 7 × 6
= 210 hours
Hence, from the above,
We can conclude that the number of hours you spent on reading and writing a journal in 1 week is: 210 hours

Multiply Fractions and Whole Numbers Homework & Practice 9.3

Multiply.
Question 1.
\(\frac{5}{6}\) × 3 = _______
Answer:
3 × \(\frac{5}{6}\) = \(\frac{5}{2}\)

Explanation:
The given numbers are: 3 and \(\frac{5}{6}\)
We know that,
\(\frac{a}{b}\) × p = \(\frac{a}{b}\) × \(\frac{p}{1}\)
= \(\frac{a × p}{ b × 1}\)
p = \(\frac{p}{1}\)
So,
3 × \(\frac{5}{6}\) = \(\frac{3}{1}\) × \(\frac{5}{6}\)
= \(\frac{3 × 5}{6 × 1}\)
= \(\frac{5}{6}\)
= \(\frac{5}{2}\)
Hence,
3 × \(\frac{5}{6}\) = \(\frac{5}{2}\)

Question 2.
\(\frac{2}{3}\) × 6 = _______
Answer:
6 × \(\frac{2}{3}\) = 4

Explanation:
The given numbers are: 6 and \(\frac{2}{3}\)
We know that,
\(\frac{a}{b}\) × p = \(\frac{a}{b}\) × \(\frac{p}{1}\)
= \(\frac{a × p}{ b × 1}\)
p = \(\frac{p}{1}\)
So,
6 × \(\frac{2}{3}\) = \(\frac{6}{1}\) × \(\frac{2}{3}\)
= \(\frac{6 × 2}{3 × 1}\)
= \(\frac{12}{3}\)
= \(\frac{4}{1}\)
= 4
Hence,
6 × \(\frac{2}{3}\) = 4

Question 3.
7 × \(\frac{1}{8}\) = _______
Answer:
7 × \(\frac{1}{8}\) = \(\frac{7}{8}\)

Explanation:
The given numbers are: 7 and \(\frac{1}{8}\)
We know that,
\(\frac{a}{b}\) × p = \(\frac{a}{b}\) × \(\frac{p}{1}\)
= \(\frac{a × p}{ b × 1}\)
p = \(\frac{p}{1}\)
So,
7 × \(\frac{1}{8}\) = \(\frac{7}{1}\) × \(\frac{1}{8}\)
= \(\frac{7 × 1}{8 × 1}\)
= \(\frac{7}{8}\)
Hence,
7 × \(\frac{1}{8}\) = \(\frac{7}{8}\)

Question 4.
2 × \(\frac{1}{2}\) = _______
Answer:
2 × \(\frac{1}{2}\) = 1

Explanation:
The given numbers are: 2 and \(\frac{1}{2}\)
We know that,
\(\frac{a}{b}\) × p = \(\frac{a}{b}\) × \(\frac{p}{1}\)
= \(\frac{a × p}{ b × 1}\)
p = \(\frac{p}{1}\)
So,
2 × \(\frac{1}{2}\) = \(\frac{2}{1}\) × \(\frac{1}{2}\)
= \(\frac{2 × 1}{2 × 1}\)
= \(\frac{2}{2}\)
= 1
Hence,
2 × \(\frac{1}{2}\) = 1

Question 5.
\(\frac{4}{5}\) × 9 = _______
Answer:
9 × \(\frac{4}{5}\) = \(\frac{36}{5}\)

Explanation:
The given numbers are: 9 and \(\frac{4}{5}\)
We know that,
\(\frac{a}{b}\) × p = \(\frac{a}{b}\) × \(\frac{p}{1}\)
= \(\frac{a × p}{ b × 1}\)
p = \(\frac{p}{1}\)
So,
9 × \(\frac{4}{5}\) = \(\frac{9}{1}\) × \(\frac{4}{5}\)
= \(\frac{9 × 4}{5 × 1}\)
= \(\frac{36}{5}\)
Hence,
9 × \(\frac{4}{5}\) = \(\frac{36}{5}\)

Question 6.
4 × \(\frac{5}{12}\) = _______
Answer:
4 × \(\frac{5}{12}\) = \(\frac{5}{3}\)

Explanation:
The given numbers are: 4 and \(\frac{5}{12}\)
We know that,
\(\frac{a}{b}\) × p = \(\frac{a}{b}\) × \(\frac{p}{1}\)
= \(\frac{a × p}{ b × 1}\)
p = \(\frac{p}{1}\)
So,
4 × \(\frac{5}{12}\) = \(\frac{4}{1}\) × \(\frac{5}{12}\)
= \(\frac{4 × 5}{12 × 1}\)
= \(\frac{20}{12}\)
= \(\frac{5}{3}\)
Hence,
4 × \(\frac{5}{12}\) = \(\frac{5}{3}\)

Question 7.
\(\frac{1}{4}\) × 24 = _______
Answer:
24 × \(\frac{1}{24}\) = 6

Explanation:
The given numbers are: 24 and \(\frac{1}{4}\)
We know that,
\(\frac{a}{b}\) × p = \(\frac{a}{b}\) × \(\frac{p}{1}\)
= \(\frac{a × p}{ b × 1}\)
p = \(\frac{p}{1}\)
So,
24 × \(\frac{1}{4}\) = \(\frac{24}{1}\) × \(\frac{1}{4}\)
= \(\frac{24 × 1}{4 × 1}\)
= \(\frac{24}{4}\)
= 6
Hence,
24 × \(\frac{1}{4}\) = 6

Question 8.
16 × \(\frac{3}{8}\) = _______
Answer:
16 × \(\frac{3}{8}\) = 6

Explanation:
The given numbers are: 16 and \(\frac{3}{8}\)
We know that,
\(\frac{a}{b}\) × p = \(\frac{a}{b}\) × \(\frac{p}{1}\)
= \(\frac{a × p}{ b × 1}\)
p = \(\frac{p}{1}\)
So,
16 × \(\frac{3}{8}\) = \(\frac{16}{1}\) × \(\frac{3}{8}\)
= \(\frac{16 × 3}{8 × 1}\)
= \(\frac{48}{8}\)
= 6
Hence,
16 × \(\frac{3}{8}\) = 6

Question 9.
\(\frac{7}{10}\) × 25 = _______
Answer:
25 × \(\frac{7}{10}\) = \(\frac{35}{2}\)

Explanation:
The given numbers are: 25 and \(\frac{7}{10}\)
We know that,
\(\frac{a}{b}\) × p = \(\frac{a}{b}\) × \(\frac{p}{1}\)
= \(\frac{a × p}{ b × 1}\)
p = \(\frac{p}{1}\)
So,
25 × \(\frac{7}{10}\) = \(\frac{25}{1}\) × \(\frac{7}{10}\)
= \(\frac{7 × 25}{10 × 1}\)
= \(\frac{175}{10}\)
= \(\frac{35}{2}\)
Hence,
25 × \(\frac{7}{10}\) = \(\frac{35}{2}\)

Question 10.
You spend \(\frac{3}{4}\) hour jumping rope every week for 8 weeks. How many hours do you jump rope altogether?
Big Ideas Math Answers Grade 5 Chapter 9 Multiply Fractions 9.3 8
Answer:
The number of hours you jump rope altogether in 8 weeks is: 6 hours

Explanation:
It is given that you spent \(\frac{3}{4}\) hour jumping rope every week for 8 weeks.
So,
The number of hours you spent jumping rope in 8 weeks = ( The number of hours you jump rope in 1 week ) × ( The total number of weeks )
= \(\frac{3}{4}\) × 8
= \(\frac{3}{4}\) × \(\frac{8}{1}\)
= \(\frac{3 × 8}{4 × 1}\)
= 6 hours
hence, from the above,
We can conclude that you spent 6 hours on jumping rope in 8 weeks.

Question 11.
Logic
Your friend finds 25 items that are either insects or flowers. She says that \(\frac{1}{6}\) of the items are insects. Can this be true? Explain.
Answer:
No, this is not true

Explanation:
It is given that your friend finds 25 items that are either insects or flowers and she says that \(\frac{1}{6}\) of the items are insects.
But,
The number of either insects or flowers will always be a whole number.
But, from the given information,
25 will not be divided by \(\frac{1}{6}\)
So,
The statement “\(\frac{1}{6}\) of the items are insects” is false.

Question 12.
Open-Ended
Write two different pairs of fractions that could represent the insects and flowers your friend finds in Exercise 11.
Answer:
The different pairs of fractions that could represent the insects and flowers are:
\(\frac{1}{5}\) and \(\frac{1}{25}\)

Explanation:
These different pairs of fractions have to divide 25
So,
We have to take the fractions the multiples of 5 i.e., 5 and 25
Hence, from the above,
We can conclude that
The different pairs of fractions that could represent the insects and flowers are:
\(\frac{1}{5}\) and \(\frac{1}{25}\)

Question 13.
Modeling Real Life
Newton bakes 56 treats. Five-eighths of them contains peanut butter. Descartes bakes 120 treats. Five-sixths of them contain peanut butter. Who bakes more peanut butter treats? How many more?
Big Ideas Math Answers Grade 5 Chapter 9 Multiply Fractions 9.3 9
Answer:
Descartes bakes more peanut butter treats
The number of peanut butter treats Descartes made more than Descartes is: 65

Explanation:
It is given that Newton bakes 56 treats and five-eights of them contains peanut butter
So,
The number of peanut butter treats made by Newton = \(\frac{5}{8}\) × 56
= \(\frac{5}{8}\) × \(\frac{56}{1}\)
= \(\frac{5 × 56}{8 × 1}\)
= 35 peanut butter treats
It is also given that Descartes bakes 120 treats and five-sixths of them contain peanut butter
So,
the number of peanut butter treats made by Descartes = \(\frac{5}{6}\) × 120
= \(\frac{5}{6}\) × \(\frac{120}{1}\)
= \(\frac{5 × 120}{6 × 1}\)
= 100 peanut butter treats
So,
In the comparison of peanut butter treats,
Descartes bakes more
Now,
The number of peanut butter treats baked by Descartes more than Newton = 100 – 35
= 65 peanut butter treats
Hence, from the above,
We can conclude that
Descartes bakes more peanut butter treats
The number of peanut butter treats Descartes made more than Descartes is: 65

Question 14.
Modeling Real Life
Your class conducts an egg-dropping experiment with 60 eggs. Three-fifths of the eggs break open, one-sixth of the eggs crack, and the rest do not break at all. How many of the eggs do not crack or break open?
Answer:
The number of eggs that do not break is:

Explanation:
It is given that your class conducts an egg-dropping experiment with 60 eggs and three-fifths of the eggs break open, one-sixth of the eggs crack and the rest do not break.
So,
The number of eggs that break open = 60 × \(\frac{3}{5}\)
= \(\frac{60}{1}\) × \(\frac{3}{5}\)
= 36
The number of eggs that cracked = 60 × \(\frac{1}{6}\)
= \(\frac{60}{1}\) × \(\frac{1}{6}\)
= 10
So,
The number of eggs that do not breaked = ( The total number of eggs ) – ( The number of eggs that break open + The number of eggs that cracked )
= 60 – ( 36 + 10 )
= 16
Hence, from the above,
We can conclude that the number of eggs that do not break is: 16 eggs

Review & Refresh

Add.
Question 15.
5\(\frac{5}{8}\) + 6\(\frac{3}{4}\) = _______
Answer:
5\(\frac{5}{8}\) + 6\(\frac{3}{4}\) = \(\frac{99}{8}\)

Explanation:
The give mixed fractions are: 6\(\frac{3}{4}\) and 5\(\frac{5}{8}\)
The representation of mixed fractions in the improper form is: \(\frac{45}{8}\) and \(\frac{27}{4}\)
In addition,
We have to make the denominators equal.
So,
Multiply \(\frac{27}{4}\) with \(\frac{2}{2}\)
So,
\(\frac{45}{8}\) + \(\frac{54}{8}\) = \(\frac{45 + 54}{8}\)
= \(\frac{99}{8}\)
Hence,
5\(\frac{5}{8}\) + 6\(\frac{3}{4}\) = \(\frac{99}{8}\)

Question 16.
1\(\frac{5}{6}\) + 8\(\frac{1}{12}\) = ________
Answer:
1\(\frac{5}{6}\) + 8\(\frac{1}{12}\) = \(\frac{119}{12}\)

Explanation:
The give mixed fractions are: 1\(\frac{5}{6}\) and 8\(\frac{1}{12}\)
The representation of mixed fractions in the improper form is: \(\frac{11}{6}\) and \(\frac{97}{12}\)
In addition,
We have to make the denominators equal.
So,
Multiply \(\frac{11}{6}\) with \(\frac{2}{2}\)
So,
\(\frac{97}{12}\) + \(\frac{22}{12}\) = \(\frac{97 + 22}{12}\)
= \(\frac{119}{12}\)
Hence,
1\(\frac{5}{6}\) + 8\(\frac{1}{12}\) = \(\frac{119}{12}\)

Question 17.
3\(\frac{1}{2}\) + \(\frac{3}{5}\) + 2\(\frac{7}{10}\) = _______
Answer:
3\(\frac{1}{2}\) + 2\(\frac{7}{10}\) + \(\frac{3}{5}\)  = \(\frac{68}{10}\)

Explanation:
The give mixed fractions are: 3\(\frac{1}{2}\), \(\frac{3}{5}\) and 2\(\frac{7}{10}\)
The representation of mixed fractions in the improper form is: \(\frac{7}{2}\) and \(\frac{27}{10}\)
In addition,
We have to make the denominators equal.
So,
Multiply \(\frac{7}{2}\) with \(\frac{5}{5}\)
Multiply \(\frac{3}{5}\) with \(\frac{2}{2}\)
So,
\(\frac{27}{10}\) + \(\frac{35}{10}\) + \(\frac{6}{10}\)  = \(\frac{27 + 35 + 6}{10}\)
= \(\frac{68}{10}\)
Hence,
3\(\frac{1}{2}\) + 2\(\frac{7}{10}\) + \(\frac{3}{5}\)  = \(\frac{68}{10}\)

Lesson 9.4 Use Models to Multiply Fractions

Explore and Grow

Fold a sheet of paper in half. Shade \(\frac{1}{4}\) of either half. What fraction of the entire sheet of paper did you shade? Draw a model to support your answer.
Answer:
The fraction of the entire sheet of paper you shaded is: \(\frac{1}{8}\)

Explanation:
Take a full sheet of paper and fold in half
So,
The number of parts of full sheet of paper = \(\frac{1}{2}\)
Now,
Shade \(\frac{1}{4}\) of the half of the paper
So,,
The fraction of the paper you shaded = \(\frac{1}{4}\) × \(\frac{1}{2}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\) = \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
\(\frac{1}{4}\) × \(\frac{1}{2}\)
= \(\frac{1 × 1}{4 × 2}\)
= \(\frac{1}{8}\)
Hence, from the above,
We can conclude that the \(\frac{1}{8}\) part of the full sheet is shaded

Reasoning
What multiplication expression does your model represent? Explain your reasoning.
Answer:
The multiplication expression your model represents is: \(\frac{1}{8}\)
We can obtain the multiplication expression by using the following multiplication properties. They are:
A) \(\frac{a}{b}\) × \(\frac{p}{q}\) = \(\frac{a × p}{b × q}\)
B) a = \(\frac{a}{1}\)

Think and Grow: Use Models to Multiply Fractions

You can use models to multiply a fraction by a fraction.
Example
Find \(\frac{1}{2}\) × \(\frac{1}{3}\).

Show and Grow

Multiply. Use a model to help.
Question 1.
\(\frac{1}{3}\) × \(\frac{1}{4}\) = _______
Answer:
\(\frac{1}{3}\) × \(\frac{1}{4}\) = \(\frac{1}{12}\)

Explanation:
The given fractions are: \(\frac{1}{4}\) and \(\frac{1}{3}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\) = \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
\(\frac{1}{4}\) × \(\frac{1}{3}\)
= \(\frac{1 × 1}{4 × 3}\)
= \(\frac{1}{12}\)
Hence,
\(\frac{1}{4}\) × \(\frac{1}{3}\) = \(\frac{1}{12}\)

Question 2.
\(\frac{2}{3}\) × \(\frac{1}{2}\) = _______
Answer:
\(\frac{2}{3}\) × \(\frac{1}{2}\) = \(\frac{1}{3}\)

Explanation:
The given fractions are: \(\frac{2}{3}\) and \(\frac{1}{2}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\) = \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
\(\frac{1}{2}\) × \(\frac{2}{3}\)
= \(\frac{1 × 2}{2 × 3}\)
= \(\frac{1}{3}\)
Hence,
\(\frac{1}{2}\) × \(\frac{2}{3}\) = \(\frac{1}{3}\)

Apply and Grow: Practice

Multiply. Use a model to help.
Question 3.
\(\frac{1}{2}\) × \(\frac{1}{6}\) = _______
Answer:
\(\frac{1}{2}\) × \(\frac{1}{6}\) = \(\frac{1}{12}\)

Explanation:
The given fractions are: \(\frac{1}{2}\) and \(\frac{1}{6}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\) = \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
\(\frac{1}{2}\) × \(\frac{1}{6}\)
= \(\frac{1 × 1}{2 × 6}\)
= \(\frac{1}{12}\)
Hence,
\(\frac{1}{2}\) × \(\frac{1}{6}\) = \(\frac{1}{12}\)

Question 4.
\(\frac{1}{5}\) × \(\frac{1}{8}\) = _______
Answer:
\(\frac{1}{5}\) × \(\frac{1}{8}\) = \(\frac{1}{40}\)

Explanation:
The given fractions are: \(\frac{1}{5}\) and \(\frac{1}{8}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\) = \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
\(\frac{1}{5}\) × \(\frac{1}{8}\)
= \(\frac{1 × 1}{5 × 8}\)
= \(\frac{1}{40}\)
Hence,
\(\frac{1}{5}\) × \(\frac{1}{8}\) = \(\frac{1}{40}\)

Question 5.
\(\frac{1}{4}\) × \(\frac{1}{6}\) = _______
Answer:
\(\frac{1}{4}\) × \(\frac{1}{6}\) = \(\frac{1}{24}\)

Explanation:
The given fractions are: \(\frac{1}{4}\) and \(\frac{1}{6}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\) = \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
\(\frac{1}{4}\) × \(\frac{1}{6}\)
= \(\frac{1 × 1}{4 × 6}\)
= \(\frac{1}{24}\)
Hence,
\(\frac{1}{4}\) × \(\frac{1}{6}\) = \(\frac{1}{24}\)

Question 6.
\(\frac{2}{3}\) × \(\frac{1}{3}\) = _______
Answer:
\(\frac{2}{3}\) × \(\frac{1}{3}\) = \(\frac{2}{9}\)

Explanation:
The given fractions are: \(\frac{2}{3}\) and \(\frac{1}{3}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\) = \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
\(\frac{2}{3}\) × \(\frac{1}{3}\)
= \(\frac{2 × 1}{3 × 3}\)
= \(\frac{2}{9}\)
Hence,
\(\frac{2}{3}\) × \(\frac{1}{3}\) = \(\frac{2}{9}\)

Write a multiplication equation represented by the model.
Question 7.
Big Ideas Math Solutions Grade 5 Chapter 9 Multiply Fractions 9.4 2
Answer:
The multiplication equation represented by the model is:
\(\frac{1}{2}\) × \(\frac{1}{8}\)

Explanation:
The given model is:
Big Ideas Math Solutions Grade 5 Chapter 9 Multiply Fractions 9.4 2
From the above model,
The number of rows is: 2
The number of columns is: 8
So,
The value of 1 row = \(\frac{1}{2}\)
The value of 1 column = \(\frac{1}{8}\)
So,
Rows × Columns = \(\frac{1}{8}\) × \(\frac{1}{2}\)
Hence, from the above,
We can conclude that the multiplication equation represented by the above model is:
\(\frac{1}{8}\) × \(\frac{1}{2}\)

Question 8.
Big Ideas Math Solutions Grade 5 Chapter 9 Multiply Fractions 9.4 3
Answer:
The multiplication equation represented by the model is:
\(\frac{1}{4}\) × \(\frac{1}{5}\)

Explanation:
The given model is:
Big Ideas Math Solutions Grade 5 Chapter 9 Multiply Fractions 9.4 3
From the above model,
The number of rows is: 4
The number of columns is: 5
So,
The value of 1 row = \(\frac{1}{4}\)
The value of 1 column = \(\frac{1}{5}\)
So,
Rows × Columns = \(\frac{1}{4}\) × \(\frac{1}{5}\)
Hence, from the above,
We can conclude that the multiplication equation represented by the above model is:
\(\frac{1}{4}\) × \(\frac{1}{5}\)

Question 9.
One-fifth of the students in your school have tried skating. Of those students, \(\frac{1}{7}\) have tried ice skating. What fraction of students in your school have tried ice skating?
Answer:
The fraction of students that have tried ice skating is: \(\frac{1}{35}\)

Explanation:
It is given that one-fifth of the students in your school have tried skating and of these students, \(\frac{1}{7}\) have tried ice skating
So,
The fraction of students that have tried ice skating = ( The fraction of students that have tried skating ) × ( The fraction of students that tried ice skating out of the total students )
= \(\frac{1}{5}\) × \(\frac{1}{7}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\) = \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
\(\frac{1}{5}\) × \(\frac{1}{7}\)
= \(\frac{1 × 1}{5 × 7}\)
= \(\frac{1}{35}\)
Hence, from the above,
We can conclude that \(\frac{1}{35}\) of the total students tried ice skating.

Question 10.
DIG DEEPER!
Are both Newton and Descartes correct? Explain.
Big Ideas Math Solutions Grade 5 Chapter 9 Multiply Fractions 9.4 4
Answer: Yes, both Newton and Descartes are correct.

Explanation:
The given models of Newton and Descartes are:
Big Ideas Math Solutions Grade 5 Chapter 9 Multiply Fractions 9.4 4
From the model of Newton,
The number of rows is: 3
The number of columns is: 5
From the model of Descartes,
The number of rows is: 5
The number of columns is: 3
From the above models,
We can observe that the Descartes model is obtained by reversing the rows and columns of Newton i.e., the rows of Newton’s model become the columns of Descartes’s model and the columns of Newton’s model becomes the rows of Descartes’ model.
So,
The multiplication equation represented by Newton is:
\(\frac{1}{3}\) × \(\frac{1}{5}\)
The multiplication equation represented by Descartes is:
\(\frac{1}{5}\) × \(\frac{1}{3}\)
Hence, from the above,
We can conclude that Newton and Descartes are correct

Think and Grow: Modeling Real Life

Example
A recipe calls for \(\frac{3}{4}\) teaspoon of cinnamon. You4want to halve the recipe. What fraction of a teaspoon of cinnamon do you need?
Big Ideas Math Solutions Grade 5 Chapter 9 Multiply Fractions 9.4 5
Because you want to halve the recipe, multiply \(\frac{1}{2}\) by \(\frac{3}{4}\) to find how many teaspoons of cinnamon you need.

So,
You need \(\frac{3}{8}\) teaspoon of cinnamon.

Show and Grow

Question 11.
The mass of mango is \(\frac{2}{5}\) kilogram. The mass of guava is \(\frac{1}{4}\) as much as the mango. What is the mass of the guava?
Big Ideas Math Solutions Grade 5 Chapter 9 Multiply Fractions 9.4 7
Answer:
The mass of guava is: \(\frac{1}{10}\) kilograms

Explanation:
It is given that the mass of mango is \(\frac{2}{5}\) kilogram and the mass of guava is \(\frac{1}{4}\) as much as the mango.
So,
The mass of guava = ( The mass of mango ) × ( The fraction of mass of guava in the mass of mango )
= \(\frac{2}{5}\) × \(\frac{1}{4}\)
= \(\frac{1 × 2}{4 × 5}\)
= \(\frac{2}{20}\)
= \(\frac{1}{10}\) kilograms
Hnce, from the above,
We can conclude that the mass of guava is: \(\frac{1}{10}\) kilogram

Question 12.
A giant panda spends \(\frac{2}{3}\) of 1-day eating and foraging. It spends \(\frac{3}{4}\) of that time for eating bamboo. What fraction of 1 day does the panda spend eating bamboo?
Answer:
The fraction of 1 day the panda spent on eating bamboo is: \(\frac{1}{2}\)

Explanation:
It is given that a giant panda spends \(\frac{2}{3}\) of 1-day eating and foraging and it spends \(\frac{3}{4}\) of that time for eating bamboo.
So,
The fraction of 1 day the panda spent on eating bamboo = ( The fraction of time the panda spent on eating ) × ( The fraction of time the panda spent time on eating bamboo out of the total time )
= \(\frac{2}{3}\) × \(\frac{3}{4}\)
= \(\frac{3 × 2}{4 × 3}\)
= \(\frac{6}{12}\)
= \(\frac{1}{2}\) hour
Hence, from the above,
We can conclude that the time spent by the panda on eating bamboo is: \(\frac{1}{2}\) hour

Question 13.
DIG DEEPER!
You have a half-gallon carton of milk that you use only for cereal. You use the same amount each day for 5 days. There is \(\frac{3}{8}\) of the carton left. How many cups of milk do you use each day? Explain.
Answer:

Use Models to Multiply Fractions Homework & Practice 9.4

Multiply. Use a model to help.
Question 1.
\(\frac{1}{3}\) × \(\frac{1}{7}\) = _______
Answer:
\(\frac{1}{3}\) × \(\frac{1}{7}\) = \(\frac{1}{21}\)

Explanation:
The given fractions are: \(\frac{1}{7}\) and \(\frac{1}{3}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\) = \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
\(\frac{1}{7}\) × \(\frac{1}{3}\)
= \(\frac{1 × 1}{7 × 3}\)
= \(\frac{1}{21}\)
Hence,
\(\frac{1}{7}\) × \(\frac{1}{3}\) = \(\frac{1}{21}\)

Question 2.
\(\frac{1}{2}\) × \(\frac{1}{9}\) = _______
Answer:
\(\frac{1}{2}\) × \(\frac{1}{9}\) = \(\frac{1}{18}\)

Explanation:
The given fractions are: \(\frac{1}{2}\) and \(\frac{1}{9}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\) = \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
\(\frac{1}{2}\) × \(\frac{1}{9}\)
= \(\frac{1 × 1}{2 × 9}\)
= \(\frac{1}{18}\)
Hence,
\(\frac{1}{2}\) × \(\frac{1}{9}\) = \(\frac{1}{18}\)

Question 3.
\(\frac{2}{5}\) × \(\frac{1}{6}\) = _______
Answer:
\(\frac{2}{5}\) × \(\frac{1}{6}\) = \(\frac{1}{15}\)

Explanation:
The given fractions are: \(\frac{2}{5}\) and \(\frac{1}{6}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\) = \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
\(\frac{2}{5}\) × \(\frac{1}{6}\)
= \(\frac{2 × 1}{5 × 6}\)
= \(\frac{1}{15}\)
Hence,
\(\frac{2}{5}\) × \(\frac{1}{6}\) = \(\frac{1}{15}\)

Question 4.
\(\frac{3}{4}\) × \(\frac{2}{7}\) = _______
Answer:
\(\frac{3}{4}\) × \(\frac{2}{7}\) = \(\frac{3}{14}\)

Explanation:
The given fractions are: \(\frac{3}{4}\) and \(\frac{2}{7}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\) = \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
\(\frac{3}{4}\) × \(\frac{2}{7}\)
= \(\frac{3 × 2}{4 × 7}\)
= \(\frac{3}{14}\)
Hence,
\(\frac{3}{4}\) × \(\frac{2}{7}\) = \(\frac{3}{14}\)

Write a multiplication equation represented by the model.
Question 5.
Big Ideas Math Solutions Grade 5 Chapter 9 Multiply Fractions 9.4 8
Answer:
The multiplication equation represented by the model is:
\(\frac{1}{2}\) × \(\frac{1}{5}\)

Explanation:
The given model is:
Big Ideas Math Solutions Grade 5 Chapter 9 Multiply Fractions 9.4 8
From the above model,
The number of rows is: 2
The number of columns is: 5
So,
The value of 1 row = \(\frac{1}{2}\)
The value of 1 column = \(\frac{1}{5}\)
So,
Rows × Columns = \(\frac{1}{2}\) × \(\frac{1}{5}\)
Hence, from the above,
We can conclude that the multiplication equation represented by the above model is:
\(\frac{1}{2}\) × \(\frac{1}{5}\)

Question 6.
Big Ideas Math Solutions Grade 5 Chapter 9 Multiply Fractions 9.4 9
Answer:
The multiplication equation represented by the model is:
\(\frac{1}{3}\) × \(\frac{1}{6}\)

Explanation:
The given model is:
Big Ideas Math Solutions Grade 5 Chapter 9 Multiply Fractions 9.4 9
From the above model,
The number of rows is: 3
The number of columns is: 6
So,
The value of 1 row = \(\frac{1}{3}\)
The value of 1 column = \(\frac{1}{6}\)
So,
Rows × Columns = \(\frac{1}{3}\) × \(\frac{1}{6}\)
Hence, from the above,
We can conclude that the multiplication equation represented by the above model is:
\(\frac{1}{3}\) × \(\frac{1}{6}\)

Question 7.
One-sixth of the animals at a zoo are birds. Of the birds, \(\frac{1}{3}\) are female. What fraction of the animals at the zoo are female birds?
Answer:
The fraction of the animals at the zoo that is female birds is: \(\frac{1}{18}\)

Explanation:
It is given that one-sixth of the animals at the zoo are birds and of the birds, \(\frac{1}{3}\) is female.
So,
The fraction of the animals at the zoo are female birds = ( The fraction of animals in the zoo that is birds ) × ( The fraction of the animals in the zoo that is female birds )
= \(\frac{1}{3}\) × \(\frac{1}{6}\)
\(\frac{1 × 1}{3 × 6}\)
= \(\frac{1}{18}\)
Hence, from the above,
We can conclude that the number of animals that are female birds at the zoo is: \(\frac{1}{18}\)

Question 8.
Writing
Write and solve a real-life problem for the expression.
\(\frac{2}{3}\) × \(\frac{3}{7}\)
Answer:
Suppose in a school, there are boys ad girls.
In school, \(\frac{2}{3}\) of the boys and \(\frac{3}{7}\) of the girls passed the PET test
So,
The fraction of the students that passed the PET test out of the total number of students = ( The fraction of the boys that passed the PET test ) × ( The fraction of the girls that passed the PET test )
= \(\frac{2}{3}\) × \(\frac{3}{7}\)
= \(\frac{3 × 2}{7 × 3}\)
= \(\frac{6}{21}\)
= \(\frac{2}{7}\)
Hence, from the above,
We can conclude that the number of students who passed the PET test is: \(\frac{2}{7}\)

Question 9.
Modeling Real Life
A Gouldian finch is \(\frac{1}{2}\) the length of the sun conure. How long is the Gouldian finch?
Big Ideas Math Solutions Grade 5 Chapter 9 Multiply Fractions 9.4 10
Answer:
The length of Gouldian finch is: \(\frac{11}{24}\) feet

Explanation:
It is given that a Gouldian finch is \(\frac{1}{2}\) the length of the sun conure.
From the given figure,
The length of the sun conure = \(\frac{11}{12}\) feet
So,
The length of Gouldian finch = ( The length of Sun Conure ) ÷ 2
= \(\frac{11}{12}\) ÷ 2
= \(\frac{11}{12}\) ÷ \(\frac{2}{1}\)
= \(\frac{11}{12}\) × \(\frac{1}{2}\)
= \(\frac{11 × 1}{12 × 2}\)
= \(\frac{11}{24}\) feet
Hence, from the above,
We can conclude that the length of the Gouldian finch is: \(\frac{11}{24}\) feet

Question 10.
DIG DEEPER!
A recipe calls for \(\frac{2}{3}\) cup of chopped walnuts. You chop 4 walnuts and get \(\frac{1}{4}\) of the amount you need. How much more of a cup of chopped walnuts do you need? All of your walnuts are the same size. How many more walnuts should you chop? Explain.
Answer:
The cup of chopped walnuts more you need is: \(\frac{1}{3}\)
The number of walnuts you should chop is: 6

Explanation:
It is given that a recipe calls for \(\frac{2}{3}\) cup of chopped walnuts.
It is also given that you chop 4 walnuts and get \(\frac{1}{4}\) of the amount you need.
So,
The cup more of chopped walnuts you need = 1- ( The cup of chopped walnuts you need for recipe )
= 1 – \(\frac{2}{3}\)
= \(\frac{3}{3}\) – \(\frac{2}{3}\)
= \(\frac{3 – 2}{3}\)
= \(\frac{1}{3}\) more cup
Now,
The number of walnuts you need more = ( The number of walnuts you needed to get \(\frac{1}{4}\) of cup of chopped walnuts ) ÷ ( The amount of chopped walnuts you need for recipe )
= 4 ÷ \(\frac{2}{3}\)
= \(\frac{4}{1}\) ÷ \(\frac{2}{3}\)
= \(\frac{4}{1}\) × \(\frac{3}{2}\)
= \(\frac{4 × 3}{1 × 2}\)
= \(\frac{12}{2}\)
= 6
Hence, from the above,
We can conclude that
The cup of chopped walnuts more you need is: \(\frac{1}{3}\)
The number of walnuts you should chop is: 6

Review & Refresh

Subtract.
Question 11.
5 – 1\(\frac{3}{4}\) = ______
Answer:
5 – 1\(\frac{3}{4}\) = \(\frac{13}{4}\)

Explanation:
The given numbers are: 5 and 1\(\frac{3}{4}\)
The representation of 1\(\frac{3}{4}\) in the improper fraction form is: \(\frac{7}{4}\)
So,
5 – \(\frac{7}{4}\) = \(\frac{20}{4}\) – \(\frac{7}{4}\)
= \(\frac{20 – 7}{4}\)
= \(\frac{13}{4}\)
Hence,
5 – 1\(\frac{3}{4}\) = \(\frac{13}{4}\)

Question 12.
13\(\frac{1}{4}\) – 7\(\frac{5}{8}\) = ________
Answer:
13\(\frac{1}{4}\) – 7\(\frac{5}{8}\) = \(\frac{45}{8}\)

Explanation:
The given mixed fractions are: 13\(\frac{1}{4}\) and 7\(\frac{5}{8}\)
The representation of 13\(\frac{1}{4}\) in the improper fraction form is: \(\frac{53}{4}\)
The representation of 13\(\frac{1}{4}\) in the improper fraction form is: \(\frac{61}{8}\)
To subtract the fractions, we have to make the denominators equal.
So,
Multiply \(\frac{53}{4}\) by \(\frac{2}{2}\)
So,
\(\frac{53}{4}\) = \(\frac{106}{8}\)
So,
\(\frac{106}{8}\) – \(\frac{61}{8}\)
= \(\frac{106 – 61}{8}\)
= \(\frac{45}{8}\)
Hence,
13\(\frac{1}{4}\) – 13\(\frac{1}{4}\) = \(\frac{45}{8}\)

Question 13.
12\(\frac{7}{10}\) – 5\(\frac{3}{10}\) – 1\(\frac{1}{5}\) = ________
Answer:
12\(\frac{7}{10}\) – 5\(\frac{3}{10}\) – 1\(\frac{1}{5}\) = \(\frac{62}{10}\)

Explanation:
The given mixed fractions are: 12\(\frac{7}{10}\), 5\(\frac{3}{10}\) and 1\(\frac{1}{5}\)
The representation of 12\(\frac{7}{10}\) in the improper fraction form is: \(\frac{127}{10}\)
The representation of 5\(\frac{3}{10}\) in the improper fraction form is: \(\frac{53}{10}\)
The representation of 1\(\frac{1}{5}\) in the improper fraction form is: \(\frac{6}{5}\)
To subtract the fractions, we have to make the denominators equal.
So,
Multiply \(\frac{6}{5}\) by \(\frac{2}{2}\)
So,
\(\frac{6}{5}\) = \(\frac{12}{10}\)
So,
\(\frac{127}{10}\) – \(\frac{53}{10}\) – \(\frac{12}{10}\)
= \(\frac{127 – 53 – 12}{8}\)
= \(\frac{62}{10}\)
Hence,
12\(\frac{7}{10}\) – 5\(\frac{3}{10}\) – 1\(\frac{1}{5}\) = \(\frac{62}{10}\)

Lesson 9.5 Multiply Fractions

Explore and Grow

Use models to help you complete the table. What do you notice about each expression and its product?
Big Ideas Math Answer Key Grade 5 Chapter 9 Multiply Fractions 9.5 1
Answer:
The completed table is:

From the given table,
We can observe the product of fractions is also a product.

Construct Arguments
Explain how to multiply two fractions without using a model.
Answer:
We can multiply the fractions by using the properties of multiplication. They are:
A) \(\frac{a}{b}\) × \(\frac{p}{q}\)
= \(\frac{a × p}{b × q}\)
B) a = \(\frac{a}{1}\)

Think and Grow: Multiply Fractions

Key Idea
You can find the product of a fraction and a fraction by multiplying the numerators and multiplying the denominators.
Example
Find \(\frac{1}{2}\) × \(\frac{3}{2}\).
Multiply the numerators and multiply the denominators.

Example
Find \(\frac{5}{6}\) × \(\frac{3}{5}\).
Multiply the numerators and multiply the denominators.

Show and Grow

Multiply.
Question 1.
\(\frac{1}{2}\) × \(\frac{4}{3}\) = _______
Answer:
\(\frac{1}{2}\) × \(\frac{4}{3}\) = \(\frac{2}{3}\)

Explanation:
The given fractions are: \(\frac{1}{2}\) and \(\frac{4}{3}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\)
= \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
\(\frac{1}{2}\) × \(\frac{4}{3}\)
= \(\frac{1 × 4}{2 × 3}\)
= \(\frac{4}{6}\)
= \(\frac{2}{3}\)
Hence,
\(\frac{1}{2}\) × \(\frac{4}{3}\) = \(\frac{2}{3}\)

Question 2.
\(\frac{2}{5}\) × \(\frac{2}{3}\) = _______
Answer:
\(\frac{2}{5}\) × \(\frac{2}{3}\) = \(\frac{4}{15}\)

Explanation:
The given fractions are: \(\frac{2}{5}\) and \(\frac{2}{3}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\)
= \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
\(\frac{2}{5}\) × \(\frac{2}{3}\)
= \(\frac{2 × 2}{5 × 3}\)
= \(\frac{4}{15}\)
Hence,
\(\frac{2}{5}\) × \(\frac{2}{3}\) = \(\frac{4}{15}\)

Question 3.
\(\frac{3}{4}\) × \(\frac{5}{8}\) = _______
Answer:
\(\frac{3}{4}\) × \(\frac{5}{8}\) = \(\frac{15}{32}\)

Explanation:
The given fractions are: \(\frac{3}{4}\) and \(\frac{5}{8}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\)
= \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
\(\frac{3}{4}\) × \(\frac{5}{8}\)
= \(\frac{3 × 5}{4 × 8}\)
= \(\frac{15}{32}\)
Hence,
\(\frac{3}{4}\) × \(\frac{5}{8}\) = \(\frac{15}{32}\)

Apply and Grow: Practice

Multiply.
Question 4.
\(\frac{1}{4}\) × \(\frac{1}{4}\) = _______
Answer:
\(\frac{1}{4}\) × \(\frac{1}{4}\) = \(\frac{1}{16}\)

Explanation:
The given fractions are: \(\frac{1}{4}\) and \(\frac{1}{4}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\)
= \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
\(\frac{1}{4}\) × \(\frac{1}{4}\)
= \(\frac{1 × 1}{4 × 4}\)
= \(\frac{1}{16}\)
Hence,
\(\frac{1}{4}\) × \(\frac{1}{4}\) = \(\frac{1}{16}\)

Question 5.
\(\frac{5}{6}\) × \(\frac{7}{10}\) = _______
Answer:
\(\frac{5}{6}\) × \(\frac{7}{10}\) = \(\frac{7}{12}\)

Explanation:
The given fractions are: \(\frac{5}{6}\) and \(\frac{7}{10}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\)
= \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
\(\frac{5}{6}\) × \(\frac{7}{10}\)
= \(\frac{5 × 7}{6 × 10}\)
= \(\frac{35}{60}\)
= \(\frac{7}{12}\)
Hence,
\(\frac{5}{6}\) × \(\frac{7}{10}\) = \(\frac{7}{12}\)

Question 6.
\(\frac{6}{9}\) × \(\frac{8}{2}\) = _______
Answer:
\(\frac{6}{9}\) × \(\frac{8}{2}\) = \(\frac{8}{3}\)

Explanation:
The given fractions are: \(\frac{6}{9}\) and \(\frac{8}{2}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\)
= \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
\(\frac{6}{9}\) × \(\frac{8}{2}\)
= \(\frac{6 × 8}{2 × 9}\)
= \(\frac{48}{18}\)
= \(\frac{8}{3}\)
Hence,
\(\frac{6}{9}\) × \(\frac{8}{2}\) = \(\frac{8}{3}\)

Question 7.
\(\frac{21}{100}\) × \(\frac{3}{5}\) = _______
Answer:
\(\frac{21}{100}\) × \(\frac{3}{5}\) = \(\frac{63}{500}\)

Explanation:
The given fractions are: \(\frac{21}{100}\) and \(\frac{3}{5}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\)
= \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
\(\frac{21}{100}\) × \(\frac{3}{5}\)
= \(\frac{21 × 3}{100 × 5}\)
= \(\frac{63}{500}\)
Hence,
\(\frac{21}{100}\) × \(\frac{3}{5}\) = \(\frac{63}{500}\)

Question 8.
\(\frac{1}{12}\) × \(\frac{9}{4}\) = _______
Answer:
\(\frac{1}{12}\) × \(\frac{9}{4}\) = \(\frac{3}{16}\)

Explanation:
The given fractions are: \(\frac{1}{12}\) and \(\frac{9}{4}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\)
= \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
\(\frac{1}{12}\) × \(\frac{9}{4}\)
= \(\frac{1 × 9}{12 × 4}\)
= \(\frac{9}{48}\)
= \(\frac{3}{16}\)
Hence,
\(\frac{1}{12}\) × \(\frac{9}{4}\) = \(\frac{3}{16}\)

Question 9.
\(\frac{4}{7}\) × \(\frac{8}{8}\) = _______
Answer:
\(\frac{4}{7}\) × \(\frac{8}{8}\) = \(\frac{4}{7}\)

Explanation:
The given fractions are: \(\frac{4}{7}\) and \(\frac{8}{8}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\)
= \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
\(\frac{4}{7}\) × \(\frac{8}{8}\)
= \(\frac{8 × 4}{7 × 8}\)
= \(\frac{4}{7}\)
Hence,
\(\frac{4}{7}\) × \(\frac{8}{8}\) = \(\frac{4}{7}\)

Evaluate
Question 10.
\(\left(\frac{1}{2} \times \frac{7}{8}\right)\) × 2 = _______
Answer:
\(\left(\frac{1}{2} \times \frac{7}{8}\right)\) × 2 = \(\frac{7}{8}\)

Explanation:
The given fractions are: \(\frac{7}{8}\) and \(\frac{1}{2}\)

We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\)
= \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
( \(\frac{1}{2}\) × \(\frac{7}{8}\)  ) × 2
= ( \(\frac{1 × 7}{2 × 8}\) ) × \(\frac{2}{1}\)
= \(\frac{7}{16}\) × \(\frac{2}{1}\)
= \(\frac{7 × 2}{16 × 1}\)
= \(\frac{7}{8}\)
Hence,
\(\frac{4}{7}\) × \(\frac{8}{8}\) = \(\frac{4}{7}\)

Question 11.
\(\left(\frac{7}{6}-\frac{5}{6}\right)\) × \(\frac{2}{3}\) = _______
Answer:
\(\left(\frac{7}{6}-\frac{5}{6}\right)\) × \(\frac{2}{3}\) = \(\frac{2}{9}\)

Explanation:
The given fractions are: \(\frac{7}{6}\) , \(\frac{5}{6}\), and \(\frac{2}{3}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\)
= \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
( \(\frac{7}{6}\) – \(\frac{5}{6}\)  ) × \(\frac{2}{3}\)
= ( \(\frac{7 – 5}{6}\) ) × \(\frac{2}{3}\)
= \(\frac{2}{6}\) × \(\frac{2}{3}\)
= \(\frac{2 × 2}{6 × 3}\)
= \(\frac{4}{18}\)
= \(\frac{2}{9}\)
Hence,
\(\left(\frac{7}{6}-\frac{5}{6}\right)\) × \(\frac{2}{3}\) = \(\frac{2}{9}\)

Question 12.
\(\frac{9}{10}\) × \(\left(\frac{4}{9}+\frac{1}{3}\right)\) = _________
Answer:
\(\frac{9}{10}\) × \(\left(\frac{4}{9}+\frac{1}{3}\right)\) = \(\frac{7}{10}\)

Explanation:
The given fractions are: \(\frac{9}{10}\), \(\frac{4}{9}\) and \(\frac{1}{3}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\)
= \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
\(\frac{9}{10}\) × ( \(\frac{4}{9}\) + \(\frac{1}{3}\)  )
= ( \(\frac{4 + 3}{9}\) ) × \(\frac{9}{10}\)
= \(\frac{7}{9}\) × \(\frac{9}{10}\)
= \(\frac{7 × 9}{9 × 10}\)
= \(\frac{7}{10}\)
Hence,
\(\frac{9}{10}\) × \(\left(\frac{4}{9}+\frac{1}{3}\right)\) = \(\frac{7}{10}\)

Question 13.
At a school, \(\frac{3}{4}\) of the students play a sport. Of the students that play a sport, \(\frac{1}{5}\) play baseball. What fraction of the students at the school play baseball?
Big Ideas Math Answer Key Grade 5 Chapter 9 Multiply Fractions 9.5 4
Answer:
The fraction of the students at the school that play baseball is: \(\frac{3}{20}\)

Explanation:
It is given that at a school, \(\frac{3}{4}\) of the students play a sport and of the students that play a sport, \(\frac{1}{5}\) play baseball.
So,
The fraction of the students at the school that play baseall = ( The fraction of the students that play a sport ) × ( The fraction of the students that play baseball out of thetotal number of students )
= \(\frac{3}{4}\) × \(\frac{1}{5}\)
= \(\frac{3 × 1}{4 × 5}\)
= \(\frac{3}{20}\)
Hence, from the above,
We can conclude that the fraction of the students that play baseball is: \(\frac{3}{20}\)

Question 14.
Reasoning
Descartes says he can find the product of a whole number and a fraction the same way he finds the product of two fractions. Explain why his reasoning makes sense.
Answer:
The product of a whole number and a fraction follows the same procedure as that of the product of the two fractions because of the following properties of multiplication:
A) \(\frac{a}{b}\) × \(\frac{p}{q}\)
= \(\frac{a × p}{b × q}\)
B) a= \(\frac{a}{1}\)
Hence, from the above two properties of multiplication,
We can conclude that the reasoning of Descartes makes sense

Question 15.
Writing
Explain how multiplying fractions is different than adding and subtracting fractions.
Answer:
In the multiplication of the fractions, we multiply numerators and denominators.
Example:
\(\frac{a}{b}\) × \(\frac{p}{q}\)
= \(\frac{a × p}{b × q}\)
In addition of the fractions, we add only numerators making the denominators equal
Example:
\(\frac{a}{b}\) + \(\frac{p}{b}\)
= \(\frac{a + p}{b}\)
In subtraction of the fractions, we subtract only numerators making the denominators equal
Example:
\(\frac{a}{b}\) – \(\frac{p}{b}\)
= \(\frac{a – p}{b}\)

Think and Grow: Modeling Real Life

Example
A tourist is walking from the Empire State Building to Times Square. She is \(\frac{2}{3}\) of the way there. What fraction of a mile does she have left to walk?
Big Ideas Math Answer Key Grade 5 Chapter 9 Multiply Fractions 9.5 5
Find the distance she has walked. Because she has walked \(\frac{2}{3}\) of \(\frac{3}{4}\) mile, multiply \(\frac{2}{3}\) by \(\frac{3}{4}\).

Show and Grow

Question 16.
At a zoo, \(\frac{3}{5}\) of the animals are mammals. Of the mammals, \(\frac{5}{12}\) are primates.What fraction of the animals at the zoo are not primates?
Big Ideas Math Answer Key Grade 5 Chapter 9 Multiply Fractions 9.5 7
Answer:
The fractions of the animals at the zoo that are not primates is: \(\frac{3}{4}\)

Explanation:
It is given that at the zoo, \(\frac{3}{5}\) of the animals are mammals and of the animals, \(\frac{5}{12}\) are primates
So,
The fraction of the animals that are primates =( The fraction of the animals that are primates ) × ( The fraction of the animals that are primates )
= \(\frac{3}{5}\) × \(\frac{5}{12}\)
= \(\frac{3 × 5}{5 × 12}\)
= \(\frac{15}{60}\)
= \(\frac{1}{4}\)
Now,
Let the total number of animals that are mammals =1
So,
The fraction of the animals that are not primates = ( The total number of mammals ) – ( The fraction of the animals that are primates )
= 1 – \(\frac{1}{4}\)
= \(\frac{4}{4}\) – \(\frac{1}{4}\)
= \(\frac{4 – 1}{4}\)
= \(\frac{3}{4}\)
Hence, from the above,
We can conclude that the fraction of the animals that are not primates is: \(\frac{3}{4}\)

Question 17.
You have an album of 216 trading cards. One page contains \(\frac{1}{24}\) of the cards. On that page, \(\frac{2}{3}\) of the cards are epic. You only have one page with any epic cards. How many epic cards do you have?
Answer:
The number of epic cards is: 6 epic cards

Explanation:
It is given that you have an album of 216 trading cards
It is also given that one page contains \(\frac{1}{24}\) of the cards. On that page, \(\frac{2}{3}\) of the cards are epic.
So,
The fraction of the epic cards = ( The total number of cards on that page ) × ( The fraction of the epic cards that is on that page )
= \(\frac{1}{24}\) × \(\frac{2}{3}\)
= \(\frac{2 × 1}{3 × 24}\)
= \(\frac{1}{36}\)
So,
The number of epic cards = ( The total number of cards ) × ( The fraction of the epic cards )
= 216 × \(\frac{1}{36}\)
= \(\frac{216}{1}\) × \(\frac{}{36}\)
= \(\frac{216}{36}\)
= 6 epic cards
Hence, from the above,
We can conclude that there are 6 epic cards

Question 18.
DIG DEEPER!
In a class, \(\frac{2}{5}\) of the students play basketball and \(\frac{7}{10}\) play soccer.Of the students who play basketball, \(\frac{2}{3}\) also play soccer. There are 30 students in the class. How many students play soccer but do not play basketball?
Answer:
The number of students that play soccer but do not play basketball is: 4 students

Explanation:
It is given that in a class, \(\frac{2}{5}\) of the students play basketball and \(\frac{7}{10}\) play soccer.Of the students who play basketball, \(\frac{2}{3}\) also play soccer.
So,
The number of students who play basketball and soccer = ( The number of students who play basketball ) × ( The number of students playing soccer who are also playing basketball )
= \(\frac{2}{5}\) × \(\frac{2}{3}\)
= \(\frac{4}{15}\)
Now,
The number of students who only play soccer but not basketball = ( The number of students who play soccer only ) – ( The number of students who play both basketball and soccer )
= \(\frac{2}{5}\) – \(\frac{4}{15}\)
In subtraction, the denominators must be equal.
So,
Multiply \(\frac{2}{5}\) by \(\frac{3}{3}\)
So,
\(\frac{6}{15}\) – \(\frac{4}{15}\)
= \(\frac{4}{15}\)
It is also given that the total number of students are: 30
So,
The number of students who play soccer only = ( The fraction of the students who play soccer only ) × ( The total number of students )
= \(\frac{2}{15}\) × 30
= \(\frac{2}{15}\) × \(\frac{30}{1}\)
= 4 students
Hence, from the above,
We can conclude that there are 4 students who play soccer only.

Multiply Fractions Homework & Practice 9.5

Multiply.
\(\frac{1}{4}\) × \(\frac{1}{5}\) = _______
Answer:
\(\frac{1}{4}\) × \(\frac{1}{5}\) = \(\frac{1}{20}\)

Explanation:
The given fractions are: \(\frac{1}{4}\) and \(\frac{1}{5}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\)
= \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
\(\frac{1}{4}\) × \(\frac{1}{5}\)
= \(\frac{1 × 1}{4 × 5}\)
= \(\frac{1}{20}\)
Hence,
\(\frac{1}{4}\) × \(\frac{1}{5}\) = \(\frac{1}{20}\)

Question 2.
\(\frac{2}{7}\) × \(\frac{1}{2}\) = _______
Answer:
\(\frac{1}{2}\) × \(\frac{2}{7}\) = \(\frac{1}{7}\)

Explanation:
The given fractions are: \(\frac{1}{2}\) and \(\frac{2}{7}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\)
= \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
\(\frac{1}{2}\) × \(\frac{2}{7}\)
= \(\frac{1 × 2}{2 × 7}\)
= \(\frac{2}{14}\)
= \(\frac{1}{7}\)
Hence,
\(\frac{1}{2}\) × \(\frac{2}{7}\) = \(\frac{1}{7}\)

Question 3.
\(\frac{9}{10}\) × \(\frac{2}{3}\) = _______
Answer:
\(\frac{9}{10}\) × \(\frac{2}{3}\) = \(\frac{3}{5}\)

Explanation:
The given fractions are: \(\frac{9}{10}\) and \(\frac{2}{3}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\)
= \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
\(\frac{9}{10}\) × \(\frac{2}{3}\)
= \(\frac{9 × 2}{10 × 3}\)
= \(\frac{18}{30}\)
= \(\frac{3}{5}\)
Hence,
\(\frac{9}{10}\) × \(\frac{2}{3}\) = \(\frac{3}{5}\)

Question 4.
\(\frac{5}{8}\) × \(\frac{5}{6}\) = _______
Answer:
\(\frac{5}{8}\) × \(\frac{5}{6}\) = \(\frac{25}{48}\)

Explanation:
The given fractions are: \(\frac{5}{8}\) and \(\frac{5}{6}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\)
= \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
\(\frac{5}{8}\) × \(\frac{5}{6}\)
= \(\frac{5 × 5}{8 × 6}\)
= \(\frac{25}{48}\)
Hence,
\(\frac{5}{8}\) × \(\frac{5}{6}\) = \(\frac{25}{48}\)

Question 5.
\(\frac{9}{7}\) × \(\frac{3}{4}\) = _______
Answer:
\(\frac{9}{7}\) × \(\frac{3}{4}\) = \(\frac{27}{28}\)

Explanation:
The given fractions are: \(\frac{9}{7}\) and \(\frac{3}{4}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\)
= \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
\(\frac{9}{7}\) × \(\frac{3}{4}\)
= \(\frac{9 × 3}{7 × 4}\)
= \(\frac{27}{28}\)
Hence,
\(\frac{9}{7}\) × \(\frac{3}{4}\) = \(\frac{27}{28}\)

Question 6.
\(\frac{11}{100}\) × \(\frac{2}{5}\) = _______
Answer:
\(\frac{11}{100}\) × \(\frac{2}{5}\) = \(\frac{11}{250}\)

Explanation:
The given fractions are: \(\frac{11}{100}\) and \(\frac{2}{5}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\)
= \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
\(\frac{11}{100}\) × \(\frac{2}{5}\)
= \(\frac{11 × 2}{100 × 5}\)
= \(\frac{22}{500}\)
= \(\frac{11}{250}\)
Hence,
\(\frac{11}{100}\) × \(\frac{2}{5}\) = \(\frac{11}{250}\)

Question 7.
\(\frac{7}{20}\) × \(\frac{6}{2}\) = _______
Answer:
\(\frac{7}{20}\) × \(\frac{6}{2}\) = \(\frac{21}{20}\)

Explanation:
The given fractions are: \(\frac{7}{20}\) and \(\frac{6}{2}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\)
= \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
\(\frac{7}{20}\) × \(\frac{6}{2}\)
= \(\frac{7 × 6}{20 × 2}\)
= \(\frac{42}{40}\)
= \(\frac{21}{20}\)
Hence,
\(\frac{7}{20}\) × \(\frac{6}{2}\) = \(\frac{21}{20}\)

Question 8.
\(\frac{15}{16}\) × \(\frac{1}{3}\) = _______
Answer:
\(\frac{15}{16}\) × \(\frac{1}{3}\) = \(\frac{5}{16}\)

Explanation:
The given fractions are: \(\frac{15}{16}\) and \(\frac{1}{3}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\)
= \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
\(\frac{15}{16}\) × \(\frac{1}{3}\)
= \(\frac{15 × 1}{16 × 3}\)
= \(\frac{15}{48}\)
= \(\frac{5}{16}\)
Hence,
\(\frac{15}{16}\) × \(\frac{1}{3}\) = \(\frac{5}{16}\)

Question 9.
\(\frac{5}{12}\) × \(\frac{3}{10}\) = _______
Answer:
\(\frac{5}{12}\) × \(\frac{3}{10}\) = \(\frac{1}{8}\)

Explanation:
The given fractions are: \(\frac{5}{12}\) and \(\frac{3}{10}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\)
= \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
\(\frac{5}{12}\) × \(\frac{3}{10}\)
= \(\frac{5 × 3}{12 × 10}\)
= \(\frac{15}{120}\)
= \(\frac{1}{8}\)
Hence,
\(\frac{5}{12}\) × \(\frac{3}{10}\) = \(\frac{1}{8}\)

Evaluate.
Question 10.
3 × \(\frac{3}{10}\) = _______
Answer:
3 ×  \(\frac{3}{10}\) = \(\frac{9}{10}\)

Explanation:
The given numbers are: 3 and \(\frac{3}{10}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\)
= \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
3 × \(\frac{3}{10}\)
= \(\frac{3}{1}\) × \(\frac{3}{10}\)
= \(\frac{3 × 3}{10 × 1}\)
= \(\frac{9}{10}\)
Hence,
3 × \(\frac{3}{10}\) = \(\frac{9}{10}\)

Question 11.
\(\left(\frac{1}{3}+\frac{1}{3}\right)\) × \(\frac{4}{5}\) = _______
Answer:
\(\left(\frac{1}{3}+\frac{1}{3}\right)\) × \(\frac{4}{5}\) = \(\frac{8}{15}\)

Explanation:
The given fractions are: \(\frac{1}{3}\), \(\frac{1}{3}\) and \(\frac{4}{5}\)

We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\)
= \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
( \(\frac{1}{3}\) + \(\frac{1}{3}\)  ) × \(\frac{4}{5}\)
= ( \(\frac{1 + 1}{3}\) ) × \(\frac{4}{5}\)
= \(\frac{2}{3}\) × \(\frac{4}{5}\)
= \(\frac{4 × 2}{3 × 5}\)
= \(\frac{8}{15}\)
Hence,
\(\left(\frac{1}{3}+\frac{1}{3}\right)\) × \(\frac{4}{5}\) = \(\frac{8}{15}\)

Question 12.
\(\frac{6}{7}\) × \(\left(\frac{3}{4}-\frac{5}{12}\right)\) = _______
Answer:
\(\frac{6}{7}\) × \(\left(\frac{3}{4}-\frac{5}{12}\right)\) = \(\frac{2}{7}\)

Explanation:
The given fractions are: \(\frac{6}{7}\), \(\frac{3}{4}\) and \(\frac{5}{12}\)

We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\)
= \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
( \(\frac{3}{4}\) – \(\frac{5}{12}\)  ) × \(\frac{6}{7}\)
= ( \(\frac{9}{12}\) – \(\frac{5}{12}\)  ) × \(\frac{6}{7}\)
= \(\frac{4}{12}\) × \(\frac{6}{7}\)
= \(\frac{4 × 6}{12 × 7}\)
= \(\frac{2}{7}\)
Hence,
\(\frac{6}{7}\) × \(\left(\frac{3}{4}-\frac{5}{12}\right)\) = \(\frac{2}{7}\)

Question 13.
A pancake recipe calls for \(\frac{1}{3}\) cup of water. You want to halve the recipe. What fraction of a cup of water do you need?
Answer:
The fraction of a cup of water you need is: \(\frac{1}{6}\)

Explanation:
It is given that a pancake recipe calls for \(\frac{1}{3}\) cup of water and you want to halve the recipe.
So,
The fraction of a cup of water you need for half of the recipe = ( The cup of water you need for a recipe ) × \(\frac{1}{2}\)
= \(\frac{1}{3}\) × \(\frac{1}{2}\)
= \(\frac{1 × 1}{3 × 2}\)
= \(\frac{1}{6}\)
Hence, from the above,
We can conclude that the fraction of water you need for half of the recipe is: \(\frac{1}{6}\)

Question 14.
Number Sense
Which is greater, \(\frac{3}{4}\) × \(\frac{1}{5}\) or \(\frac{3}{4}\) × \(\frac{1}{8}\)? Explain.
Answer:
\(\frac{3}{4}\) × \(\frac{1}{5}\)” is greater than “\(\frac{3}{4}\) × \(\frac{1}{8}\)

Explanation:
The given fractions are: \(\frac{3}{4}\), \(\frac{1}{5}\) and \(\frac{1}{8}\)
Now,
\(\frac{3}{4}\) × \(\frac{1}{5}\)
= \(\frac{3 × 1}{4 × 5}\)
= \(\frac{3}{20}\)
Now,
\(\frac{3}{4}\) × \(\frac{1}{8}\)
= \(\frac{3 × 1}{4 × 8}\)
= \(\frac{3}{32}\)
So,
For the comparison of the products, equate the numerators.
Multiply \(\frac{3}{20}\) by \(\frac{32}{32}\)
Multiply \(\frac{3}{32}\) by \(\frac{20}{20}\)
So,
\(\frac{3}{20}\) = \(\frac{96}{640}\)
\(\frac{3}{32}\) = \(\frac{60}{640}\)
So,
By comparison of the products,
We can observe that
96 > 60
Hence, from the above,
We can conclude that “\(\frac{3}{4}\) × \(\frac{1}{5}\)” is greater than “\(\frac{3}{4}\) × \(\frac{1}{8}\)”

Question 15.
Reasoning
Is \(\frac{17}{24}\) × \(\frac{7}{8}\) equal to \(\frac{17}{8}\) × \(\frac{7}{24}\)? Explain.
Answer:
\(\frac{17}{24}\) × \(\frac{7}{8}\) is equal to \(\frac{17}{8}\) × \(\frac{7}{24}\)

Explanation:
By the commutative property of multiplication,
a × b = b × a
So,
By the above property of multiplication,
We can interchange the numbers
Hence, from the above,
We can conclude that \(\frac{17}{24}\) × \(\frac{7}{8}\) is equal to \(\frac{17}{8}\) × \(\frac{7}{24}\)

Question 16.
Number Sense
In which equations does k = \(\frac{5}{6}\)?
Big Ideas Math Answer Key Grade 5 Chapter 9 Multiply Fractions 9.5 8
Answer:
Let the multiplication equations be represented A), B), C) and D)
The given expressions are:
Big Ideas Math Answer Key Grade 5 Chapter 9 Multiply Fractions 9.5 8
So,
The product of the given expressions are:
A) \(\frac{1}{2}\) × \(\frac{5}{3}\) = \(\frac{5}{6}\)
B) \(\frac{1}{10}\) × k = \(\frac{1}{12}\)
C) \(\frac{1}{3}\) × k = \(\frac{2}{9}\)
D) \(\frac{5}{4}\) × \(\frac{2}{3}\) = k
Hence, from the above expressions,
We can conclude that equation A) fits the value of k perfectly.

Question 17.
Modeling Real Life
At a town hall meeting, \(\frac{37}{50}\) of the members are present. Of those who are present, \(\frac{1}{2}\) vote in favor of a new park. What fraction of the members do not vote in favor of the new park?
Answer:
The fraction of the members that do not vote in favor of the new park is: \(\frac{13}{50}\)

Explanation:
It is given that at a town hall meeting, \(\frac{37}{50}\) of the members are present. Of those who are present, \(\frac{1}{2}\) vote in favor of a new park.
So,
The fraction of the members that do not vote in favor of the new park = ( The total number of members at the town hall meeting ) – ( The fraction of the members that vote in the favor of a new park )
= 1 – \(\frac{37}{50}\)
= \(\frac{50}{50}\) – \(\frac{37}{50}\)
= \(\frac{50 – 37}{50}\)
= \(\frac{13}{50}\)
Hence, from the above,
We can conclude that the number of members that do not vote in the favor of a new park is: \(\frac{13}{50}\)

Question 18.
Modeling Real Life
There are 50 U.S. states. Seven twenty-fifths of the states share a land border with Canada or Mexico.Of those states, \(\frac{2}{7}\) share a border with Mexico. How many states share a border with Canada?
Answer:
The number of states that share a border with Canada is: 4 states

Explanation:
It is given that there are 50 U.S. states. Seven twenty-fifths of the states share a land border with Canada or Mexico and of those states, \(\frac{2}{7}\) share a border with Mexico.
So,
The number of states that share a border with Canada or Mexico = \(\frac{7}{25}\) × 50
= \(\frac{7}{25}\) × \(\frac{50}{1}\)
= 14 states
So,
The number of states that share a border with Canada = \(\frac{2}{7}\) × 14
= \(\frac{2}{7}\) × \(\frac{14}{1}\)
= 4 states
Hence, from the above,
We can conclude that there are 4 states that share a border with Canada

Review & Refresh

Evaluate. Check whether your answer is reasonable.
Question 19.
15.67 + 4 + 6.5 = _____
Answer:
15.67 + 4 + 6.5 = 26.17

Explanation:
The given numbers are: 15.67, 4, and 6.5
The representation of the numbers in the fraction form is: \(\frac{1567}{100}\), \(\frac{65}{10}\) and 4
As the highest number in the denominator is 100, make all the denominators hundred
So,
The representation of 4 with 100 in the denominator is: \(\frac{400}{100}\)
The representation of 6.5 with 100 in the denominator is: \(\frac{650}{100}\)
So,
\(\frac{1567}{100}\) + \(\frac{400}{100}\) + \(\frac{650}{100}\)
= \(\frac{  1567 + 400 + 650 }{100}\)
= \(\frac{2,617}{100}\)
= 26.17
Hence, 15.67 + 4 + 6.5 = 26.17

Question 20.
20.7 – 9.54 + 25.81 = _______
Answer:
20.7 – 9.54 + 25.81 = 36.97

Explanation:
The given numbers are: 20.7, 9.54, and 25.81
The representation of the numbers in the fraction form is: \(\frac{2581}{100}\), \(\frac{207}{10}\) and \(\frac{954}{100}\)
As the highest number in the denominator is 100, make all the denominators hundred
So,
The representation of 20.7 with 100 in the denominator is: \(\frac{2070}{100}\)
So,
\(\frac{2581}{100}\) + \(\frac{2070}{100}\) – \(\frac{954}{100}\)
= \(\frac{  2581 + 2070 – 954 }{100}\)
= \(\frac{3,697}{100}\)
= 36.97
Hence, 20.7 – 9.54 + 25.81 = 36.97

Lesson 9.6 Find Areas of Rectangles

Explore and Grow

Draw and cut out a rectangle that has any two of the side lengths below.
\(\frac{1}{2}\) ft
\(\frac{1}{3}\) ft
\(\frac{1}{4}\) ft
Use several copies of your rectangle to create a unit square. What is the area (in square feet) of each small rectangle? Explain your reasoning.
Answer:
Let there are two small rectangles with the following lengths:
\(\frac{1}{2}\) ft and \(\frac{1}{3}\) ft
\(\frac{1}{2}\) ft and \(\frac{1}{4}\) ft
So,
The area of the unit square with lengths \(\frac{1}{2}\) ft and \(\frac{1}{3}\) ft is:
\(\frac{1}{2}\) ft × \(\frac{1}{3}\) ft
= \(\frac{1 × 1}{2 × 3}\) ft
= \(\frac{1}{6}\) ft
The area of the unit square with lengths \(\frac{1}{2}\) ft and \(\frac{1}{4}\) ft is:
\(\frac{1}{2}\) ft × \(\frac{1}{4}\) ft
= \(\frac{1 × 1}{2 × 4}\) ft
= \(\frac{1}{8}\) ft
Hence, from the above,
We can conclude that the areas ofthe smaller rectngles with unit lengths is: \(\frac{1}{6}\) ft and \(\frac{1}{8}\) ft

Reasoning
How can you use a rectangle with unit fraction side lengths to find the area of the rectangle below? Explain your reasoning.
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.6 1
Answer:
The given rectangle is:
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.6 1
To find the area of the rectangle by the unit fractions, multiply the values of the length and the breadth
So,
\(\frac{3}{4}\) × \(\frac{3}{2}\)
= 3  (\(\frac{1}{4}\) ) × 3 ( \(\frac{1}{2}\) )
= 9 ( \(\frac{1}{4}\) × \(\frac{1}{2}\) )
= 9 × \(\frac{1 × 1}{4 × 2}\)
= \(\frac{9}{1}\) × \(\frac{1}{8}\)
= \(\frac{9}{8}\)
The unit fractions are the fractions that contain the value 1 in the numerator.
Here,
\(\frac{1}{4}\) and \(\frac{1}{2}\) are the unit fractions

Think and Grow: Find Areas of Rectangles

One way to find the area of a rectangle with fractional side lengths is to fill it with smaller rectangles.
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.6 2
Example
Find the area of the rectangle.

Show and Grow

Question 1.
Find the area of the shaded region.
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.6 4
Answer:
The area of the shaded region is: \(\frac{3}{12}\)

Explanation:
The given figure is:
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.6 4
From the given figure,
The area of the shaded region = \(\frac{3}{4}\) × \(\frac{1}{3}\)
= \(\frac{3 × 1}{4 × 3}\)
= \(\frac{3}{12}\)
Hence, from the above,
We can conclude that the area of the shaded region is: \(\frac{3}{12}\)

Apply and Grow: Practice

Find the area of the shaded region.
Question 2.
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.6 5
Answer:
The area of the shaded region is: \(\frac{2}{12}\)

Explanation:
The given figure is:
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.6 5
From the given figure,
The area of the shaded region = \(\frac{2}{4}\) × \(\frac{1}{3}\)
= \(\frac{2 × 1}{4 × 3}\)
= \(\frac{2}{12}\)
Hence, from the above,
We can conclude that the area of the shaded region is: \(\frac{2}{12}\)

Question 3.
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.6 6
Answer:
The area of the shaded region is: \(\frac{3}{16}\)

Explanation:
The given figure is:
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.6 6
From the given figure,
The area of the shaded region = \(\frac{3}{4}\) × \(\frac{1}{4}\)
= \(\frac{3 × 1}{4 × 4}\)
= \(\frac{3}{16}\)
Hence, from the above,
We can conclude that the area of the shaded region is: \(\frac{3}{16}\)

Use rectangles with unit fraction side lengths to find the area of the rectangle.
Question 4.
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.6 7
Answer:
The area of the rectangle is: \(\frac{6}{10}\)

Explanation:
The given figure is:
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.6 7
From the given figure,
The area of the rectangle = \(\frac{3}{2}\) × \(\frac{2}{5}\)
= \(\frac{3 × 2}{2 × 5}\)
= \(\frac{6}{10}\)
Hence, from the above,
We can conclude that the area of the shaded region is: \(\frac{6}{10}\)

Question 5.
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.6 8
Answer:
The area of the rectangle is: \(\frac{10}{18}\)

Explanation:
The given figure is:
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.6 8
From the given figure,
The area of the rectangle = \(\frac{5}{6}\) × \(\frac{2}{3}\)
= \(\frac{5 × 2}{6 × 3}\)
= \(\frac{10}{18}\)
Hence, from the above,
We can conclude that the area of the shaded region is: \(\frac{10}{18}\)

Question 6.
Find the area of a rectangle with side lengths of \(\frac{5}{8}\) and \(\frac{4}{3}\)?
Answer:
The area of a rectangle is: \(\frac{20}{24}\)

Explanation:
The given side lengths of a rectangle are: \(\frac{5}{8}\) and \(\frac{4}{3}\)
So,
The area of the rectangle = \(\frac{5}{8}\) × \(\frac{4}{3}\)
= \(\frac{5 × 4}{8 × 3}\)
= \(\frac{20}{24}\)
Hence, from the above,
We can conclude that the area of the rectangle is: \(\frac{20}{24}\)

Question 7.
Find the area of a rectangle with side lengths of \(\frac{7}{9}\) and \(\frac{1}{2}\)?
Answer:
The area of a rectangle is: \(\frac{7}{18}\)

Explanation:
The given side lengths of a rectangle are: \(\frac{7}{9}\) and \(\frac{1}{2}\)
So,
The area of the rectangle = \(\frac{7}{9}\) × \(\frac{1}{2}\)
= \(\frac{7 × 1}{9 × 2}\)
= \(\frac{7}{18}\)
Hence, from the above,
We can conclude that the area of the rectangle is: \(\frac{7}{18}\)

Question 8.
Reasoning
Can you find the area of a rectangle with fractional side lengths the same way you find the area of a rectangle with whole-number side lengths? Explain.
Answer:
Yes, we can find the area of a rectangle with fractional side lengths the same way you find the area of a rectangle with whole-number side lengths.
Example:
Let the fractional side lengths be: \(\frac{1}{5}\) and \(\frac{1}{9}\)
Let the whole number side lengths be: 3 and 4
So,
The area of the rectangle with the fractional side lengths = \(\frac{1}{5}\) × \(\frac{1}{9}\)
= \(\frac{1 × 1}{9 × 5}\)
= \(\frac{1}{45}\)
The area of the rectangle with the whole number side lengths = 3 × 4
= \(\frac{3}{1}\) × \(\frac{4}{1}\)
= \(\frac{3 × 4}{1 × 1}\)
= \(\frac{12}{1}\)
= 12
Hence, from the above,
We can conclude that we can find the area of a rectangle with fractional side lengths the same way you find the area of a rectangle with whole-number side lengths.

Question 9.
YOU BE THE TEACHER
Our friend says she can find the area of a square given only one fractional side length. Is your friend correct? Explain.
Answer:
Yes, she can find the area of a square by only one fractional side length

Explanation:
Let the unit fractional side length of the square be: \(\frac{1}{2}\)
We know that,
The length of all the sides in a square are equal.
So,
By using this property, we can find the area of the square by taking the fractional unit side length of 1 side as the same for all the sides
So,
The area of the square = \(\frac{1}{2}\) × \(\frac{1}{2}\)
= \(\frac{1 × 1}{2 × 2}\)
= \(\frac{1}{4}\)
Hence, from the above,
We can conclude that she can find the area of a square by only one fractional side length

Think and Grow: Modeling Real Life

Example
A zoo needs an outdoor enclosure with an area of at least \(\frac{3}{10}\) square kilometer10for a camel. Is the rectangular enclosure shown large enough for a camel?
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.6 9
Find the area of the enclosure by multiplying the length and the width.

So, the enclosure is large enough for a camel.

Show and Grow

Question 10.
The area of a square dog kennel is 4 square yards. Will the square mat fit in the kennel?
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.6 11
Answer:
Yes, the square mat will fit in the kennel

Explanation:
It is given that the area of a square dog kennel is 4 square yards and the side of the square mat is \(\frac{5}{3}\) yd
We know that,
The length of all the sides in a square are equal.
So,
By using this property,
The area of the square mat = \(\frac{5}{3}\) × \(\frac{5}{3}\)
= \(\frac{5 × 5}{3 × 3}\)
= \(\frac{25}{9}\) yd
Now,
The area of a square dog kennel can be written as: \(\frac{36}{9}\) yd
So,
When we compare the 2 areas,
We can get 25 < 36
Hence, from the above,
We can conclude that the square mat will fit in the kennel

Question 11.
DIG DEEPER!
The side lengths of each chalk art square are \(\frac{11}{4}\) meters. The side lengths of the square zone around each chalk square are an additional \(\frac{3}{4}\) meter. How many square meters of concrete is used to create the chalk walk shown?
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.6 12
Answer:
The amount of the concrete used to create the chalk walk is: \(\frac{196}{16}\) square meters

Explanation:
It is given that the side lengths of each chalk art square are \(\frac{11}{4}\) meters. The side lengths of the square zone around each chalk square are an additional \(\frac{3}{4}\) meter.
So,
The total side length of the chalk walk = \(\frac{11}{4}\) + \(\frac{3}{4}\)
= \(\frac{11 + 3}{4}\)
= \(\frac{14}{4}\) meters
We know that,
The length of all sides in the square are equal.
So,
By using this property,
The amount of concrete used for the chalk walk = \(\frac{14}{4}\) × \(\frac{14}{4}\)
= \(\frac{14 × 14}{4 × 4}\)
= \(\frac{196}{16}\) square meters
Hence, from the above,
We can conclude that the amount of the concrete used to create the chalk walk is: \(\frac{196}{16}\) square meters

Find Areas of Rectangles Homework & Practice 9.6

Find the area of the shaded region.
Question 1.
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.6 13
Answer:
The area of the shaded region is: \(\frac{1}{12}\)

Explanation:
The given figure is:
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.6 13
From the given figure,
The area of the shaded region = \(\frac{1}{2}\) × \(\frac{1}{6}\)
= \(\frac{1 × 1}{2 × 6}\)
= \(\frac{1}{12}\)
Hence, from the above,
We can conclude that the area of the shaded region is: \(\frac{1}{12}\)

Question 2.
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.6 14
Answer:
The area of the shaded region is: \(\frac{8}{25}\)

Explanation:
The given figure is:
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.6 14
From the given figure,
The area of the shaded region = \(\frac{2}{5}\) × \(\frac{4}{5}\)
= \(\frac{2 × 4}{5 × 5}\)
= \(\frac{8}{25}\)
Hence, from the above,
We can conclude that the area of the shaded region is: \(\frac{8}{25}\)

Use rectangles with unit fraction side lengths to find the area of the rectangle.
Question 3.
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.6 15
Answer:
The area of the rectangle is: \(\frac{2}{24}\)

Explanation:
The given figure is:
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.6 15
From the given figure,
The area of the rectangle = \(\frac{2}{3}\) × \(\frac{1}{8}\)
= \(\frac{1 × 2}{3 × 8}\)
= \(\frac{2}{24}\)
Hence, from the above,
We can conclude that the area of the shaded region is: \(\frac{2}{24}\)

Question 4.
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.6 16
Answer:
The area of the rectangle is: \(\frac{35}{90}\)

Explanation:
The given figure is:
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.6 16
From the given figure,
The area of the rectangle = \(\frac{7}{10}\) × \(\frac{5}{9}\)
= \(\frac{7 × 5}{10 × 9}\)
= \(\frac{35}{90}\)
Hence, from the above,
We can conclude that the area of the shaded region is: \(\frac{35}{90}\)

Question 5.
Find the area of a rectangle with a side length of \(\frac{3}{4}\) and \(\frac{5}{12}\).
Answer:
The area of a rectangle is: \(\frac{15}{48}\)

Explanation:
The given side lengths of a rectangle are: \(\frac{5}{12}\) and \(\frac{3}{4}\)
So,
The area of the rectangle = \(\frac{5}{12}\) × \(\frac{3}{4}\)
= \(\frac{5 × 3}{12 × 4}\)
= \(\frac{15}{48}\)
Hence, from the above,
We can conclude that the area of the rectangle is: \(\frac{15}{48}\)

Question 6.
Find the area of a square with side lengths of \(\frac{9}{16}\)
Answer:
The area of a square is: \(\frac{81}{256}\)

Explanation:
We know that,
The length of all the sides of the square is equal.
So,
The given side lengths of a square are: \(\frac{9}{16}\) and \(\frac{9}{16}\)
So,
The area of the square = \(\frac{9}{16}\) × \(\frac{9}{16}\)
= \(\frac{9 × 9}{16 × 16}\)
= \(\frac{81}{256}\)
Hence, from the above,
We can conclude that the area of the rectangle is: \(\frac{81}{256}\)

Question 7.
Open-Ended
The area of a rectangle is \(\frac{16}{24}\). What are the possible side lengths of the rectangle?
Answer:
The possible side lengths for the given area of a rectangle are:
A) \(\frac{4}{2}\) and \(\frac{4}{12}\)
B) \(\frac{1}{3}\) and \(\frac{4}{8}\)

Explanation:
The given area of a rectangle is: \(\frac{16}{24}\)
So,
To find the possible side lengths of a rectangle, find the factors for the numerator and denominator of \(\frac{16}{24}\)
So,
The factors of 16 are: 1, 2, 4, 8, 16
The factors of 24 are: 1, 2, 3, 4, 6, 8, 12, 24
So,
The possible side lengths of a rectangle are:
A) \(\frac{4}{2}\) and \(\frac{4}{12}\)
B) \(\frac{1}{3}\) and \(\frac{4}{8}\)
There are so many possible side lengths of a rectangle like above
Hence, from the above,
We can conclude that the possible side lengths of a rectangle are:
A) \(\frac{4}{2}\) and \(\frac{4}{12}\)
B) \(\frac{1}{3}\) and \(\frac{4}{8}\)

Question 8.
Structure
Write an expression that represents the area of the shaded rectangle.
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.6 17
Answer:
The expression that represents the area of the shaded region is: \(\frac{1}{3}\) × \(\frac{1}{3}\)

Explanation:
The given shaded rectangle is:
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.6 17
From the given shaded rectangle,
The number of shaded columns is: 3
The number of shaded rows is: 3
The total number of rows is: 1
The total number of columns are: 1
So,
The expression representing the shaded region = \(\frac{The total number of rows}{The number of shaded rows}\) × \(\frac{The total number of rows}{The number of shaded rows}\)
= \(\frac{1}{3}\) × \(\frac{1}{3}\)
Hence, from the above,
We can conclude that the expression representing the shaded region of a rectangle is: \(\frac{1}{3}\) × \(\frac{1}{3}\)

Question 9.
Modeling Real Life
The area of a square table is 9 square feet. Will the board game fit on the table?
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.6 18
Answer:
Yes, the board game fit on the table

Explanation:
The given area of a square table is 9 square feet
So,
The representation of the area of the square table in the fraction form is: \(\frac{36}{4}\)
Now,
The given board game is:
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.6 18
The area of the board game can find out by multiplying its side lengths
So,
The area of the board game = \(\frac{3}{2}\) × \(\frac{5}{2}\)
= \(\frac{3 × 5}{2 × 2}\)
= \(\frac{15}{4}\)
So,
Compare the area of the square table and the area of the board game,
We get
Area of the square table > Area of the board game
Hence, from the above,
We can conclude that the board game fit on the table

Review & Refresh

Divide. Then check your answer.
Question 10.
365 ÷ 14 = _______
Answer:
365 ÷ 14 = 26 R 1

Explanation:
By using the partial quotients method,
365 ÷ 14 = ( 280 ÷ 84 ) ÷ 14
= ( 280 ÷ 14 ) + ( 84 ÷ 14 )
= 20 + 6
= 26 R 1
Hence, 365 ÷ 14 = 26 R 1

Question 11.
282 ÷ 27 = ______
Answer:
282 ÷ 27 = 10 R 12

Explanation:
By using the partial quotients method,
282 ÷ 27 = 270 ÷ 27
= 10 R 12
Hence, 270 ÷ 27 = 10 R 12

Question 12.
601 ÷ 72 = _____
Answer:
601 ÷ 72 = 8 R 25

Explanation:
By using the partial quotients method,
601 ÷ 72 = 576 ÷ 72
= 8 R 25
Hence, 601 ÷ 72 = 8 R 25

Lesson 9.7 Multiply Mixed Numbers

Explore and Grow

Find the area of the rectangle. Explain how you found your answer.
Big Ideas Math Answers Grade 5 Chapter 9 Multiply Fractions 9.7 1
Answer:
The area of the rectangle is: \(\frac{35}{4}\)

Explanation:
The given rectangle is:
Big Ideas Math Answers Grade 5 Chapter 9 Multiply Fractions 9.7 1
From the above rectangle,
The side lengths are: 2\(\frac{1}{2}\) and 3\(\frac{1}{2}\)
The representation of the side lengths in the improper form is: \(\frac{5}{2}\) and \(\frac{7}{2}\)
So,
The area of rectangle = \(\frac{5}{2}\) × \(\frac{7}{2}\)
= \(\frac{5 × 7}{2 × 2}\)
= \(\frac{35}{4}\)
Hence,
2\(\frac{1}{2}\) × 3\(\frac{1}{2}\) = \(\frac{35}{4}\)

Structure
Find the area using a different method. Explain how you found your answer.
Answer:
The given mixed fractions of a rectangle are: 2\(\frac{1}{2}\) and 3\(\frac{1}{2}\)
To convert a mixed fraction into an improper fraction, we have to add the whole number part and the fractional part in the mixed fraction.
So,
2\(\frac{1}{2}\) = 2 + \(\frac{1}{2}\)
= \(\frac{4}{2}\) + \(\frac{1}{2}\)
= \(\frac{5}{2}\)
3\(\frac{1}{2}\) = 3 + \(\frac{1}{2}\)
= \(\frac{6}{2}\) + \(\frac{1}{2}\)
= \(\frac{7}{2}\)
So,
\(\frac{5}{2}\) × \(\frac{7}{2}\) = \(\frac{35}{4}\)

Think and Grow: Multiply Mixed Numbers

You can use a model to find the product of two mixed numbers. You can also write the mixed numbers as improper fractions and then multiply.
Example
Find 1\(\frac{1}{2}\) × 2\(\frac{3}{4}\).
One Way: Use an area model.
Step 1: Write each mixed number as a sum.
1\(\frac{1}{2}\) = 1 + \(\frac{1}{2}\) 2\(\frac{3}{4}\) = 2 + \(\frac{3}{4}\)
Step 2: Draw an area model that represents the product of the sums.
Step 3: Find the sum of the areas of the sections.

Another Way: Write each mixed number as an improper fraction, then multiply.

Show and Grow

Multiply.
Question 1.
2\(\frac{1}{2}\) × 1\(\frac{1}{2}\) = ________
Answer:
2\(\frac{1}{2}\) × 1\(\frac{1}{2}\) = \(\frac{15}{4}\)

Explanation:
The given fractions are: 2\(\frac{1}{2}\) and 1\(\frac{1}{2}\)
The representation of the fractions in the improper form is: \(\frac{5}{2}\) and \(\frac{3}{2}\)
So,
\(\frac{5}{2}\) × \(\frac{3}{2}\)
= \(\frac{5 × 3}{2 × 2}\)
= \(\frac{15}{4}\)
Hence,
2\(\frac{1}{2}\) × 1\(\frac{1}{2}\) = \(\frac{15}{4}\)

Question 2.
3\(\frac{1}{4}\) × 2\(\frac{2}{3}\) = ______
Answer:
3\(\frac{1}{4}\) × 2\(\frac{2}{3}\) = \(\frac{104}{12}\)

Explanation:
The given fractions are: 3\(\frac{1}{4}\) and 2\(\frac{2}{3}\)
The representation of the fractions in the improper form is: \(\frac{13}{4}\) and \(\frac{8}{3}\)
So,
\(\frac{13}{4}\) × \(\frac{8}{3}\)
= \(\frac{13 × 8}{4 × 3}\)
= \(\frac{104}{12}\)
Hence,
3\(\frac{1}{4}\) × 2\(\frac{2}{3}\) = \(\frac{104}{12}\)

Apply and Grow: Practice

Multiply.
Question 3.
1\(\frac{3}{4}\) × 2\(\frac{1}{6}\) = ______
Answer:
1\(\frac{3}{4}\) × 2\(\frac{1}{6}\) = \(\frac{91}{24}\)

Explanation:
The given fractions are: 1\(\frac{3}{4}\) and 2\(\frac{1}{6}\)
The representation of the fractions in the improper form is: \(\frac{7}{4}\) and \(\frac{13}{6}\)
So,
\(\frac{7}{4}\) × \(\frac{13}{6}\)
= \(\frac{13 × 7}{4 × 6}\)
= \(\frac{91}{24}\)
Hence,
1\(\frac{3}{4}\) × 2\(\frac{1}{6}\) = \(\frac{91}{24}\)

Question 4.
4\(\frac{1}{3}\) × 1\(\frac{5}{6}\) = ______
Answer:
4\(\frac{1}{3}\) × 1\(\frac{5}{6}\) = \(\frac{143}{18}\)

Explanation:
The given fractions are: 4\(\frac{1}{3}\) and 1\(\frac{5}{6}\)
The representation of the fractions in the improper form is: \(\frac{13}{3}\) and \(\frac{11}{6}\)
So,
\(\frac{13}{3}\) × \(\frac{11}{6}\)
= \(\frac{13 × 11}{6 × 3}\)
= \(\frac{143}{18}\)
Hence,
4\(\frac{1}{3}\) × 1\(\frac{5}{6}\) = \(\frac{143}{18}\)

Question 5.
3\(\frac{2}{5}\) × 1\(\frac{9}{10}\) = ______
Answer:
3\(\frac{2}{5}\) × 1\(\frac{9}{10}\) = \(\frac{323}{50}\)

Explanation:
The given fractions are: 3\(\frac{2}{5}\) and 1\(\frac{9}{10}\)
The representation of the fractions in the improper form is: \(\frac{17}{5}\) and \(\frac{19}{10}\)
So,
\(\frac{17}{5}\) × \(\frac{19}{10}\)
= \(\frac{17 × 19}{5 × 10}\)
= \(\frac{323}{50}\)
Hence,
3\(\frac{2}{5}\) × 1\(\frac{9}{10}\) = \(\frac{323}{50}\)

Question 6.
2\(\frac{3}{8}\) × 3\(\frac{1}{2}\) = ______
Answer:
2\(\frac{3}{8}\) × 3\(\frac{1}{2}\) = \(\frac{133}{16}\)

Explanation:
The given fractions are: 2\(\frac{3}{8}\) and 3\(\frac{1}{2}\)
The representation of the fractions in the improper form is: \(\frac{19}{8}\) and \(\frac{7}{2}\)
So,
\(\frac{19}{8}\) × \(\frac{7}{2}\)
= \(\frac{19 × 7}{8 × 2}\)
= \(\frac{133}{16}\)
Hence,
2\(\frac{3}{8}\) × 3\(\frac{1}{2}\) = \(\frac{133}{16}\)

Evaluate.
Question 7.
5\(\frac{1}{4}\) × \(\frac{2}{5}\) × 6\(\frac{1}{12}\) = _________
Answer:
5\(\frac{1}{4}\) × \(\frac{2}{5}\) × 6\(\frac{1}{12}\) = \(\frac{3,066}{240}\)

Explanation:
The given fractions are: 5\(\frac{1}{4}\), \(\frac{2}{5}\) and 6\(\frac{1}{12}\)
The representation of the fractions in the improper form is: \(\frac{21}{4}\) and \(\frac{73}{12}\)
So,
\(\frac{21}{4}\) × \(\frac{73}{12}\) × \(\frac{2}{5}\)
= \(\frac{21 × 73 × 2}{4 × 12 × 5}\)
= \(\frac{3,066}{240}\)
Hence,
5\(\frac{1}{4}\) × \(\frac{2}{5}\) × 6\(\frac{1}{12}\) = \(\frac{3,066}{240}\)

Question 8.
3\(\frac{2}{3}\) × (10\(\frac{7}{8}\) – 2\(\frac{1}{4}\)) = _________
Answer:
3\(\frac{2}{3}\) × (10\(\frac{7}{8}\) – 2\(\frac{1}{4}\)) = \(\frac{759}{24}\)

Explanation:
The given fractions are: 3\(\frac{2}{3}\), 10\(\frac{7}{8}\) and 2\(\frac{1}{4}\)
The representation of the fractions in the improper form is: \(\frac{11}{3}\), \(\frac{87}{8}\)  and \(\frac{9}{4}\)
So,
3\(\frac{2}{3}\) × (10\(\frac{7}{8}\) – 2\(\frac{1}{4}\))
= \(\frac{11}{3}\) × ( \(\frac{87}{8}\) – \(\frac{9}{4}\) )
= \(\frac{11}{3}\) × ( \(\frac{87 – 18}{8}\) )
= \(\frac{11}{3}\) × \(\frac{69}{8}\)
= \(\frac{11 × 69}{8 × 3}\)
= \(\frac{759}{24}\)
Hence,
3\(\frac{2}{3}\) × (10\(\frac{7}{8}\) – 2\(\frac{1}{4}\)) = \(\frac{759}{24}\)

Question 9.
YOU BE THE TEACHER
Your friend uses the model to find 4\(\frac{1}{2}\) × 3\(\frac{2}{3}\). Is your friend correct? Explain.
Big Ideas Math Answers Grade 5 Chapter 9 Multiply Fractions 9.7 4
Answer:
No, your friend is not correct

Explanation:
The given fractions are: 4\(\frac{1}{2}\) and 3\(\frac{2}{3}\)
The representation of the side lengths in the improper form is: \(\frac{9}{2}\) and \(\frac{11}{3}\)
So,
The area of rectangle = \(\frac{9}{2}\) × \(\frac{11}{3}\)
= \(\frac{9 × 11}{3 × 2}\)
= \(\frac{99}{6}\)
= \(\frac{33}{2}\)
= 15\(\frac{3}{2}\)
But according to your friend,
4\(\frac{1}{2}\) × 3\(\frac{2}{3}\) = 12\(\frac{1}{2}\)
Hence, from the above,
We can conclude that your friend is not correct.

Question 10.
Logic
Find the missing numbers.
Big Ideas Math Answers Grade 5 Chapter 9 Multiply Fractions 9.7 5
Answer:
The missing numbers are: 1 and \(\frac{8}{5}\)

Explanation:
According to the separation method,
To convert the mixed number into the improper fraction, we can add whole numbers and the fractions separately
Now,
Let the missing numbers be p and q
So,
p + 5 = 6
So,
p = 6 – 5 = 1
Now,
\(\frac{1}{4}\) + \(\frac{1}{q}\) = \(\frac{7}{8}\)
So,
\(\frac{1}{q}\) = \(\frac{7}{8}\) – \(\frac{1}{4}\)
= \(\frac{7}{8}\) – \(\frac{2}{8}\)
= \(\frac{7 – 2}{8}\)
= \(\frac{5}{8}\)
So,
q= \(\frac{8}{5}\)
Hence, from the above,
We can conclude that the missing numbers are: 1 and \(\frac{8}{5}\)

Think and Grow: Modeling Real Life

Example
A construction crew is paving 15 miles of a highway. The crew paves 4\(\frac{2}{10}\) miles each month. Does the crew finish paving the highway in 3\(\frac{1}{2}\) months?
Big Ideas Math Answers Grade 5 Chapter 9 Multiply Fractions 9.7 6
Find the length of the highway the crew paves by multiplying the number of months by the number of miles they pave each month. Write each mixed number as an improper fraction, then multiply.

Compare the length the crew paves to the amount that needs to be paved.
The Crew can finish paving the highway in 3\(\frac{1}{2}\) months.

Show and Grow

Question 11.
You have 3 cups of strawberries. You want to make 1\(\frac{1}{2}\) batches of the recipe. Do you have enough strawberries?
Big Ideas Math Answers Grade 5 Chapter 9 Multiply Fractions 9.7 8
Answer:
Yes, you have enough strawberries

Explanation:
The given recipe for a strawberry smoothie is:
Big Ideas Math Answers Grade 5 Chapter 9 Multiply Fractions 9.7 8
It is given that you want to make 1\(\frac{1}{2}\) batches of the recipe
From the table,
The number of cups of strawberry is: 1\(\frac{3}{4}\)
Now,
The representation of 1\(\frac{1}{2}\) in the iproper fraction is: \(\frac{3}{2}\)
The representation of 1\(\frac{3}{4}\) in the mixed form is: \(\frac{7}{4}\)
So,
The number of strawberries for the recipe = \(\frac{3}{2}\) × \(\frac{7}{4}\)
= \(\frac{3 × 7}{4 × 2}\)
= \(\frac{21}{8}\)
It is also given that you have 3 cups of strawberries
So,
We can write 3 in the fraction form and as the multiples of 8 is: \(\frac{24}{8}\)
So,
When we compare the number of cups we obtained and given,
We can say that we have enough strawberries to make the recipe
Hence, from the above,
We can conclude that we have enough strawberries

Question 12.
On Monday,you roller-skate 6\(\frac{1}{4}\) miles. On Tuesday, you skate 1\(\frac{2}{5}\) times as far as you did on Monday. How many total miles do you roller-skate on Monday and Tuesday combined?
Answer:
The number of miles you do roller-skate on Monday and Tuesday combined is: 15 miles

Explanation:
It is given that on Monday,you roller-skate 6\(\frac{1}{4}\) miles. On Tuesday, you skate 1\(\frac{2}{5}\) times as far as you did on Monday.
So,
The number of miles you do roller-skate on Tuesday = ( The number of miles you do roller-skate on Monday ) × 1\(\frac{2}{5}\)
= 6\(\frac{1}{4}\) × 1\(\frac{2}{5}\)
= \(\frac{25}{4}\) × \(\frac{7}{5}\)
= \(\frac{25 × 7}{5 × 4}\)
= \(\frac{175}{20}\)
Now,
The number of miles you do roller-skate on Monday and Tuesday combined = 6\(\frac{1}{4}\) + \(\frac{175}{20}\)
= \(\frac{25}{4}\) + \(\frac{175}{20}\)
= \(\frac{125}{20}\) + \(\frac{175}{20}\)
= \(\frac{175 + 125}{20}\)
= \(\frac{300}{20}\)
= 15 miles
Hence, from the above,
We can conclude that the number of miles you do roller- skate on Monday and Tuesday combined is: 15 miles

Question 13.
DIG DEEPER!
An artist paints a rectangular mural. The mural is 4\(\frac{1}{3}\) feet wide. The length is 2\(\frac{1}{4}\) times the width. What is the area of the mural?
Answer:
The area of the mural is: \(\frac{117}{12}\) square feet

Explanation:
It is given that an artist paints a rectangular mural. The mural is 4\(\frac{1}{3}\) feet wide. The length is 2\(\frac{1}{4}\) times the width.
So,
The area of the mural = ( The width of the mural ) × ( The length of the mural )
= 4\(\frac{1}{3}\) × 2\(\frac{1}{4}\)
= \(\frac{13}{3}\) × \(\frac{9}{4}\)
= \(\frac{13 × 9}{4 × 3}\)
= \(\frac{117}{12}\)
Hence, from the above,
We can conclude that the area of the mural is: \(\frac{117}{12}\) square feet

Multiply Mixed Numbers Homework & Practice 9.7

Multiply.
Question 1.
1\(\frac{1}{2}\) × 1\(\frac{1}{8}\) = ______
Answer:
1\(\frac{1}{2}\) × 1\(\frac{1}{8}\) = \(\frac{18}{16}\)

Explanation:
The given fractions are: 1\(\frac{1}{2}\) and 1\(\frac{1}{8}\)
The representation of the fractions in the improper form is: \(\frac{3}{2}\) and \(\frac{9}{8}\)
So,
\(\frac{9}{8}\) × \(\frac{3}{2}\)
= \(\frac{9 × 3}{8 × 2}\)
= \(\frac{18}{16}\)
Hence,
1\(\frac{1}{2}\) × 1\(\frac{1}{8}\) = \(\frac{18}{16}\)

Question 2.
1\(\frac{5}{6}\) × 2\(\frac{1}{4}\) = ______
Answer:
1\(\frac{5}{6}\) × 2\(\frac{1}{4}\) = \(\frac{99}{24}\)

Explanation:
The given fractions are: 1\(\frac{5}{6}\) and 2\(\frac{1}{4}\)
The representation of the fractions in the improper form is: \(\frac{11}{6}\) and \(\frac{9}{4}\)
So,
\(\frac{9}{4}\) × \(\frac{11}{6}\)
= \(\frac{9 × 11}{4 × 6}\)
= \(\frac{99}{24}\)
Hence,
1\(\frac{5}{6}\) × 2\(\frac{1}{4}\) = \(\frac{99}{24}\)

Evaluate.

Question 3.
2\(\frac{3}{8}\) × 2\(\frac{3}{4}\) = ______
Answer:
2\(\frac{3}{8}\) × 2\(\frac{3}{4}\) = \(\frac{209}{32}\)

Explanation:
The given fractions are: 2\(\frac{3}{8}\) and 2\(\frac{3}{4}\)
The representation of the fractions in the improper form is: \(\frac{19}{8}\) and \(\frac{11}{4}\)
So,
\(\frac{19}{8}\) × \(\frac{11}{4}\)
= \(\frac{19 × 11}{4 × 8}\)
= \(\frac{209}{32}\)
Hence,
2\(\frac{3}{8}\) × 2\(\frac{3}{4}\) = \(\frac{209}{32}\)

Question 4.
4\(\frac{1}{6}\) × 3\(\frac{2}{7}\) = ______
Answer:
4\(\frac{1}{6}\) × 3\(\frac{2}{7}\) = \(\frac{575}{42}\)

Explanation:
The given fractions are: 4\(\frac{1}{6}\) and 3\(\frac{2}{7}\)
The representation of the fractions in the improper form is: \(\frac{25}{6}\) and \(\frac{23}{7}\)
So,
\(\frac{23}{7}\) × \(\frac{25}{6}\)
= \(\frac{23 × 25}{7 × 6}\)
= \(\frac{575}{42}\)
Hence,
4\(\frac{1}{6}\) × 3\(\frac{2}{7}\) = \(\frac{575}{42}\)

Question 5.
2\(\frac{1}{3}\) × 3\(\frac{9}{10}\) × 5\(\frac{1}{5}\) = ______
Answer:
2\(\frac{1}{3}\)× 3\(\frac{9}{10}\) ×5\(\frac{1}{5}\) = \(\frac{7,098}{150}\)

Explanation:
The given fractions are: 2\(\frac{1}{3}\), 3\(\frac{9}{10}\) and 5\(\frac{1}{5}\)
The representation of the fractions in the improper form is: \(\frac{7}{3}\), \(\frac{39}{10}\)  and \(\frac{26}{5}\)
So,
\(\frac{7}{3}\) × \(\frac{39}{10}\) ×\(\frac{26}{5}\)
= \(\frac{7 × 39 × 26}{3 × 10 × 5}\)
= \(\frac{7,098}{150}\)
Hence,
2\(\frac{1}{3}\)× 3\(\frac{9}{10}\) ×5\(\frac{1}{5}\) = \(\frac{7,098}{150}\)

Question 6.
(1\(\frac{7}{8}\) + 4\(\frac{4}{5}\)) × 2\(\frac{1}{12}\) = _______
Answer:
(1\(\frac{7}{8}\) + 4\(\frac{4}{5}\)) × 2\(\frac{1}{12}\) = \(\frac{6,675}{480}\)

Explanation:
The given fractions are: 1\(\frac{7}{8}\), 4\(\frac{4}{5}\) and 2\(\frac{1}{12}\)
The representation of the fractions in the improper form is: \(\frac{15}{8}\), \(\frac{24}{5}\) and \(\frac{25}{12}\)
So,
(1\(\frac{7}{8}\) + 4\(\frac{4}{5}\)) × \(\frac{25}{12}\)
= ( \(\frac{15}{8}\) + \(\frac{24}{5}\) ) × \(\frac{25}{12}\)
= ( \(\frac{75}{40}\) + \(\frac{192}{40}\) ) × \(\frac{25}{12}\)
= \(\frac{75 + 192}{40}\) × \(\frac{25}{12}\)
= \(\frac{267}{40}\) × \(\frac{25}{12}\)
= \(\frac{267 × 25}{40× 12 }\)
= \(\frac{6,675}{480}\)
Hence,
(1\(\frac{7}{8}\) + 4\(\frac{4}{5}\)) × 2\(\frac{1}{12}\) = \(\frac{6,675}{480}\)

Question 7.
Structure
Find the missing numbers.
Big Ideas Math Answers Grade 5 Chapter 9 Multiply Fractions 9.7 9
Answer:
The missing numbers are: 5 and \(\frac{3}{10}\)

Explanation:
Let the missing numbers be p and q
By using the partial products method,
( p × 4 ) + ( p × \(\frac{1}{7}\) ) = 20 + \(\frac{5}{7}\)
p × ( 4 + \(\frac{1}{7}\) ) = 20 + \(\frac{5}{7}\)
p × \(\frac{29}{7}\) = \(\frac{145}{7}\)
So,
p = \(\frac{145}{7}\) ÷ \(\frac{29}{7}\)
= \(\frac{145}{7}\) × \(\frac{7}{29}\)
= \(\frac{145 × 7}{7 × 29}\)
= 5
Now,
(q × 4 ) + ( q × \(\frac{1}{7}\) ) = \(\frac{12}{10}\) + \(\frac{3}{70}\)
q × ( 4 + \(\frac{1}{7}\) ) = \(\frac{12}{10}\) + \(\frac{3}{70}\)
q × \(\frac{29}{7}\) = \(\frac{84 + 3}{70}\)
q × \(\frac{29}{7}\) = \(\frac{87}{70}\)
So,
q = \(\frac{87}{70}\) ÷ \(\frac{29}{7}\)
= \(\frac{87}{70}\) × \(\frac{7}{29}\)
= \(\frac{87 × 7}{70 × 29}\)
= \(\frac{3}{10}\)
Hence, from the above,
We can conclude that the missing numbers are: 5 and \(\frac{3}{10}\)

Question 8.
YOU BE THE TEACHER
Your friend finds 1\(\frac{11}{12}\) × 2\(\frac{3}{8}\) . Is your friend correct? Explain.
Big Ideas Math Answers Grade 5 Chapter 9 Multiply Fractions 9.7 10
Answer:
Yes your friend is correct

Explanation:
The given mixed numbers are: 1\(\frac{11}{12}\) and 2\(\frac{3}{8}\)
The representation of the mixed numbers in the fraction form is: \(\frac{23}{12}\) and \(\frac{19}{8}\)
So,
\(\frac{23}{12}\) × \(\frac{19}{8}\)
= \(\frac{23 × 19}{8 × 12}\)
= \(\frac{437}{96}\)
= 4\(\frac{53}{96}\)
Hence, from the above,
We can conclude that your friend is correct

Question 9.
Modeling Real Life
Your friend earns 7\(\frac{1}{2}\) dollars each hour. Will she earn enough money to buy a $35 toy after working 4\(\frac{3}{4}\) hours?
Answer:
she can’t earn enough money to buy $35 toy after working 4\(\frac{3}{4}\) hours

Explanation:
It is given that your friend earns 7\(\frac{1}{2}\) dollars each hour.
It is given that will she earn enough money to buy a $35 toy after working 4\(\frac{3}{4}\) hours
So,
The number of dollars she can earn = 7\(\frac{1}{2}\) × 4\(\frac{3}{4}\)
= \(\frac{15}{2}\) × \(\frac{19}{4}\)
= \(\frac{15 × 19}{4 × 2}\)
= \(\frac{285}{8}\)
Now,
The representation of $35 as the multiple of 8 is: \(\frac{280}{8}\)
So,
By comparing the given money with he money she earned,
We can say that she can’t buy the toy.
Hence, from the above,
We can conclude that she can’t earn enough money to buy a $35 toy after working 4\(\frac{3}{4}\) hours

Question 10.
Modeling Real Life
One class collects 8\(\frac{1}{4}\) pounds of recyclable materials. Another class collects 1\(\frac{1}{2}\) times as many pounds as the first class. How many pounds of recyclable materials do the two classes collect altogether?
Answer:
The number of pounds of recyclable materials the two classes collects together is: \(\frac{231}{8}\) pounds

Explanation:
It is given that one class collects 8\(\frac{1}{4}\) pounds of recyclable materials. Another class collects 1\(\frac{1}{2}\) times as many pounds as the first class.
So,
The number of pounds of recyclable materials collected by another class = 8\(\frac{1}{4}\) × 1\(\frac{1}{2}\)
= \(\frac{33}{4}\) × \(\frac{3}{2}\)
= \(\frac{99}{8}\) pounds
So,
The number of pouds of recyclable materials collected by two classes together = 8\(\frac{1}{4}\) + \(\frac{99}{8}\)
= \(\frac{33}{4}\) + \(\frac{99}{8}\)
= \(\frac{132 + 99}{8}\)
= \(\frac{231}{8}\) pounds
Hence, from the above,
We can conclude that the number of pounds of recyclable materials the two classes collects together is: \(\frac{231}{8}\) pounds

Review & Refresh

Find the product.
Question 11.
6 × 5.7 = ______
Answer:
6 × 5.7 = 34.2

Explanation:
By using the partial products method,
6 × 5.7 = 6 × ( 5 + 0.7 )
= ( 6 × 5 ) + ( 6 × 0.7 )
= 30 + 4.2
= 34.2
Hence, 6 × 5.7 = 34.2

Question 12.
0.84 × 9 = ______
Answer:
0.84 × 9 = 73.17

Explanation:
By using the partial quotients method,
0.84 × 9 = ( 0.81 + 0.03 ) × 9
= ( 0.81 × 9 ) + ( 0.03 × 9 )
= 72.9 + 0.27
= 73.17
Hence, 0.84 × 9 = 73.17

Lesson 9.8 Compare Factors and Products

Explore and Grow

Without calculating, order the rectangles by area from least to greatest. Explain your reasoning.
Big Ideas Math Solutions Grade 5 Chapter 9 Multiply Fractions 9.8 1
Answer:
Let the given rectangles be named A), B), C), and D)
So,
The side lengths of A are: 1 and 1\(\frac{1}{2}\)
The side lengths of B are: 1\(\frac{1}{3}\) and \(\frac{11}{12}\)
The side lengths of C are: \(\frac{3}{2}\) and \(\frac{11}{12}\)
The side lengths of D are: \(\frac{11}{10}\) and \(\frac{3}{2}\)
Now,
The area of A is: \(\frac{1}{2}\)
The area of B is: \(\frac{44}{36}\)
The area of C is: \(\frac{33}{24}\)
The area of D is: \(\frac{33}{20}\)
For comparison make the denominators of the areas of the four rectangles equal.
So,
The area of A is: \(\frac{18}{36}\)
The area of B is: \(\frac{44}{36}\)
So, by comparing these 2 areas,
We can say that B is greater
Now,
The area of C is: \(\frac{165}{120}\)
The area of D is: \(\frac{198}{120}\)
So, by comparing these 2 areas,
We can say that D is greater
Hence, from the above,
We can conclude that the order of rectangles by areas from the greatest to the least is: B > D> A > C

Construct Arguments
Explain your strategy to your partner. Compare your strategies.
Answer:
The strategy you followed is:
A) Write the side lengths of the rectangles
B) Find the areas of the four rectangles
C) Make the denominators of all the four areas of rectangles equal
D) Compare the numerators of all the four areas of the four rectangles

Think and Grow: Compare Factors and Products

Key Idea
When you multiply a number by a fraction greater than 1, the product is greater than the number. When you multiply a number by a fraction less than 1, the product is less than the number.
Example
Without calculating, tell whether the product 3\(\frac{1}{8}\) × \(\frac{5}{6}\) is less than, greater than, or equal to or each of its factors.

Show and Grow

Without calculating, tell whether the product is less than, greater than, or equal to each of its factors.
Question 1.
8 × \(\frac{3}{10}\)
Answer:
The value of 8 × \(\frac{3}{10}\) is less than 8

Explanation:
The given numbers are: 8 and \(\frac{3}{10}\)
We know that,
When you multiply a number by a fraction greater than 1, the product is greater than the number. When you multiply a number by a fraction less than 1, the product is less than the number.
So,
\(\frac{3}{10}\) is less than 1
So,
8 × \(\frac{3}{10}\) is less than 8
Hence, from the above,
We can conclude that the value of 8 × \(\frac{3}{10}\) is less than 8

Question 2.
\(\frac{4}{4}\) × 5\(\frac{2}{3}\)
Answer:
The value of \(\frac{4}{4}\) × 5\(\frac{2}{3}\) is greater than \(\frac{4}{4}\)

Explanation:
The given fractions are: \(\frac{4}{4}\) and 5\(\frac{2}{3}\)
We know that,
When you multiply a number by a fraction greater than 1, the product is greater than the number. When you multiply a number by a fraction less than 1, the product is less than the number.
We know that,
\(\frac{4}{4}\) is 1
Now,
The representation of 5\(\frac{2}{3}\) in the improper fraction form is: \(\frac{17}{3}\)
So,
\(\frac{17}{3}\) is greater than 1
So,
\(\frac{4}{4}\) × 5\(\frac{2}{3}\) is greater than 1
Hence, from the above,
We can conclude that the value of \(\frac{4}{4}\) × 5\(\frac{2}{3}\) is greater than \(\frac{4}{4}\)

Question 3.
\(\frac{4}{3}\) × \(\frac{1}{6}\)
Answer:
The value of \(\frac{4}{3}\) × \(\frac{1}{6}\) is greater than \(\frac{1}{6}\) or less than \(\frac{4}{3}\)

Explanation:
The given fractions are: \(\frac{4}{3}\) and \(\frac{1}{6}\)
We know that,
When you multiply a number by a fraction greater than 1, the product is greater than the number. When you multiply a number by a fraction less than 1, the product is less than the number.
Now,
\(\frac{4}{3}\) is greater than 1
\(\frac{1}{6}\) is less than 1
So,
\(\frac{4}{3}\) × \(\frac{1}{6}\) is less than \(\frac{4}{3}\)
Hence, from the above,
We can conclude that the value of \(\frac{4}{3}\) × \(\frac{1}{6}\) is greater than \(\frac{1}{6}\) or less than \(\frac{4}{3}\)

Apply and Grow

Without calculating, tell whether the product is less than, greater than, or equal to each of its factors.
Question 4.
\(\frac{1}{4}\) × \(\frac{1}{12}\)
Answer:
The value of \(\frac{1}{4}\) × \(\frac{1}{12}\) is less than 1

Explanation:
The given fractions are: latex]\frac{1}{4}[/latex] and \(\frac{1}{12}\)
We know that,
When you multiply a number by a fraction greater than 1, the product is greater than the number. When you multiply a number by a fraction less than 1, the product is less than the number.
So,
\(\frac{1}{4}\) is less than 1
\(\frac{1}{12}\) is less than 1
So,
\(\frac{1}{4}\) × \(\frac{1}{12}\) is less than 1
Hence, from the above,
We can conclude that the value of \(\frac{1}{4}\) × \(\frac{1}{12}\) is less than 1

Question 5.
3\(\frac{4}{5}\) × 6\(\frac{7}{8}\)
Answer:
The value of 3\(\frac{4}{5}\) × 6\(\frac{7}{8}\) is greater than 1

Explanation:
The given fractions are: 3\(\frac{4}{5}\) and 6\(\frac{7}{8}\)
The representation of 3\(\frac{4}{5}\) and 6\(\frac{7}{8}\) in the improper fractions form is: \(\frac{19}{5}\) and \(\frac{55}{8}\)
We know that,
When you multiply a number by a fraction greater than 1, the product is greater than the number. When you multiply a number by a fraction less than 1, the product is less than the number.
So,
\(\frac{19}{5}\) is greater than 1
\(\frac{55}{8}\) is greater than 1
So,
3\(\frac{4}{5}\) × 6\(\frac{7}{8}\) is greater than 1
Hence, from the above,
We can conclude that the value of 3\(\frac{4}{5}\) × 6\(\frac{7}{8}\) is greater than 1

Question 6.
\(\frac{1}{6}\) × \(\frac{10}{10}\)
Answer:
The value of \(\frac{1}{6}\) × \(\frac{10}{10}\) is less than 1

Explanation:
The given fractions are: \(\frac{1}{6}\) and \(\frac{10}{10}\)
We know that,
When you multiply a number by a fraction greater than 1, the product is greater than the number. When you multiply a number by a fraction less than 1, the product is less than the number.
We know that,
\(\frac{10}{10}\) is 1
So,
\(\frac{1}{6}\) is less than 1
So,
\(\frac{1}{6}\) × \(\frac{10}{10}\) is less than 1
Hence, from the above,
We can conclude that the value of \(\frac{1}{6}\) × \(\frac{10}{10}\) is less than 1

Question 7.
\(\frac{2}{3}\) × 5
Answer:
The value of \(\frac{2}{3}\) × 5 is less than 5

Explanation:
The given fractions are: \(\frac{2}{3}\) and 5
We know that,
When you multiply a number by a fraction greater than 1, the product is greater than the number. When you multiply a number by a fraction less than 1, the product is less than the number.
So,
\(\frac{2}{3}\) is less than 1
So,
\(\frac{2}{3}\) × 5 is less than 5
Hence, from the above,
We can conclude that the value of \(\frac{2}{3}\) × 5 is less than 5

Question 8.
\(\frac{7}{10}\) × 4\(\frac{8}{9}\)
Answer:
The value of \(\frac{7}{10}\) × 4\(\frac{8}{9}\) is greater than 1

Explanation:
The given fractions are: \(\frac{7}{10}\) and 4\(\frac{8}{9}\)
The representation of 4\(\frac{8}{9}\) in the improper fraction form is: \(\frac{44}{9}\)
We know that,
When you multiply a number by a fraction greater than 1, the product is greater than the number. When you multiply a number by a fraction less than 1, the product is less than the number.
So,
\(\frac{7}{10}\) is less than 1
\(\frac{44}{9}\) is greater than 1
So,
\(\frac{7}{10}\) × 4\(\frac{8}{9}\) is greater than 1
Hence, from the above,
We can conclude that the value of \(\frac{7}{10}\) × 4\(\frac{8}{9}\) is greater than 1

Question 9.
\(\frac{9}{2}\) × 1\(\frac{3}{4}\)
Answer:
The value of \(\frac{9}{2}\) × 1\(\frac{3}{4}\) is greater than 1

Explanation:
The given fractions are: \(\frac{9}{2}\) × 1\(\frac{3}{4}\)
The representation of 1\(\frac{3}{4}\) in the improper fraction form is: \(\frac{7}{4}\)
We know that,
When you multiply a number by a fraction greater than 1, the product is greater than the number. When you multiply a number by a fraction less than 1, the product is less than the number.
So,
\(\frac{9}{2}\) is greater than 1
\(\frac{7}{4}\) is greater than 1
So,
\(\frac{9}{2}\) × 1\(\frac{3}{4}\) is greater than 1
Hence, from the above,
We can conclude that the value of \(\frac{9}{2}\) × 1\(\frac{3}{4}\) is greater than 1

Without calculating, order the products from least to greatest.
Question 10.
Big Ideas Math Solutions Grade 5 Chapter 9 Multiply Fractions 9.8 3
Answer:
The order of products from the least to the greatest is:
\(\frac{5}{6}\) × \(\frac{1}{3}\) < \(\frac{5}{6}\) × \(\frac{7}{7}\) < \(\frac{5}{6}\) × 1\(\frac{8}{9}\)

Explanation:
The given products are:
A) \(\frac{5}{6}\) × \(\frac{1}{3}\)
B) \(\frac{5}{6}\) × \(\frac{7}{7}\)
C) \(\frac{5}{6}\) × 1\(\frac{8}{9}\)
So,
In A),
\(\frac{5}{6}\) and \(\frac{1}{3}\) are less than 1
So,
\(\frac{5}{6}\) × \(\frac{1}{3}\) is less than 1
In B),
\(\frac{5}{6}\) is less than 1
\(\frac{7}{7}\) is 1
So,
\(\frac{5}{6}\) × \(\frac{7}{7}\) is less than 1
In C),
\(\frac{5}{6}\) is less than1
The representation of  1\(\frac{8}{9}\) in the improper fraction form is: \(\frac{17}{9}\)
So,
\(\frac{17}{9}\) is greater than 1
So,
\(\frac{5}{6}\) × 1\(\frac{8}{9}\) is greater than 1
Hence, from the above,
We can conclude that the order of the products from the least to the greatest is:
\(\frac{5}{6}\) × \(\frac{1}{3}\) < \(\frac{5}{6}\) × \(\frac{7}{7}\) < \(\frac{5}{6}\) × 1\(\frac{8}{9}\)

Question 11.
Big Ideas Math Solutions Grade 5 Chapter 9 Multiply Fractions 9.8 4
Answer:
The order of products from the least to the greatest is:
\(\frac{1}{6}\) × \(\frac{1}{4}\) < \(\frac{1}{10}\) × \(\frac{1}{4}\) < 5\(\frac{7}{10}\) × \(\frac{1}{4}\)

Explanation:
The given products are:
A) \(\frac{1}{6}\) × \(\frac{1}{4}\)
B) \(\frac{1}{10}\) × \(\frac{1}{4}\)
C) \(\frac{1}{4}\) × 5\(\frac{7}{10}\)
So,
In A),
\(\frac{1}{6}\) and \(\frac{1}{4}\) are less than 1
So,
\(\frac{1}{6}\) × \(\frac{1}{4}\) is less than 1
In B),
\(\frac{1}{10}\) is less than 1
\(\frac{1}{4}\) is less than 1
So,
\(\frac{1}{10}\) × \(\frac{1}{4}\) is less than 1
In C),
\(\frac{1}{4}\) is less than1
The representation of  5\(\frac{7}{10}\) in the improper fraction form is: \(\frac{57}{10}\)
So,
\(\frac{57}{10}\) is greater than 1
So,
\(\frac{1}{4}\) × 5\(\frac{7}{10}\) is greater than 1
Hence, from the above,
We can conclude that the order of the products from the least to the greatest is:
\(\frac{1}{6}\) × \(\frac{1}{4}\) < \(\frac{1}{10}\) × \(\frac{1}{4}\) < 5\(\frac{7}{10}\) × \(\frac{1}{4}\)

Question 12.
YOU BE THE TEACHER
Your friend says that \(\frac{1}{2}\) × 8 is half as much as 8. Is your friend correct? Explain.
Answer:
Yes, your friend is correct

Explanation:
The given numbers are: \(\frac{1}{2}\) and 8
So,
\(\frac{1}{2}\) × 8 = \(\frac{1}{2}\) × \(\frac{8}{1}\)
= \(\frac{1 × 8}{2 × 1}\)
= 4
It is also given that according to your friend,
\(\frac{1}{2}\) × 8 is half as much as 8.
So,
8 ÷ 2 = 4
Hence, from the above,
We can conclude that your friend is correct.

Question 13.
DIG DEEPER!
Without calculating, tell whether the product is less than, greater than, or equal to 3\(\frac{3}{4}\). Explain.
\(\left(\frac{1}{2} \times 3 \frac{3}{4}\right)\) × \(\frac{2}{7}\)
Answer:
\(\left(\frac{1}{2} \times 3 \frac{3}{4}\right)\) × \(\frac{2}{7}\) is less than  3\(\frac{3}{4}\)

Explanation:
The representation of 3\(\frac{3}{4}\) in the improper fraction form is: \(\frac{15}{4}\)
Now,
\(\left(\frac{1}{2} \times 3 \frac{3}{4}\right)\) × \(\frac{2}{7}\)
= ( \(\frac{1}{2}\) × \(\frac{15}{4}\) ) × \(\frac{2}{7}\)
= \(\frac{15}{8}\) × \(\frac{2}{7}\)
= \(\frac{2 × 15}{7 × 8}\)
= \(\frac{30}{56}\)
= \(\frac{15}{28}\)
Now,
\(\frac{15}{4}\) is multiplied by \(\frac{7}{7}\)
So,
\(\frac{15}{4}\) = \(\frac{105}{28}\)
So,
When we compare \(\frac{105}{28}\) and \(\frac{15}{28}\)
We will get \(\frac{105}{28}\) is greater
Hence, from the above,
We can conclude that \(\left(\frac{1}{2} \times 3 \frac{3}{4}\right)\) × \(\frac{2}{7}\) is less than  3\(\frac{3}{4}\)

Think and Grow: Modeling Real Life

Example
Men’s shot put competitions use a shot with a mass of 7\(\frac{1}{4}\) kilograms. The mass of a bowling ball is \(\frac{7}{8}\) as much as the mass of the shot. Is the mass of the bowling ball less than, greater than, or equal to the mass of the shot?
Big Ideas Math Solutions Grade 5 Chapter 9 Multiply Fractions 9.8 5
Write an expression to represent the mass of the bowling ball.

So, the mass of the bowling ball is greater than the mass of the shot.

Show and Grow

Question 14.
You practice playing the keyboard for 3\(\frac{1}{2}\) hours. Your friend practices playing the keyboard for \(\frac{5}{4}\) as many hours as you. Does your friend practice for fewer hours, more hours, or the same number of hours as you?
Answer:
You practice more hours than your friend

Explanation:
It is given that you practice playing the keyboard for 3\(\frac{1}{2}\) hours. Your friend practices playing the keyboard for \(\frac{5}{4}\) as many hours as you.
So,
Now,
The representation of 3\(\frac{1}{2}\) in the improper fraction form is: \(\frac{7}{2}\)
So,
The number of hours you practice playing the keyboard is: \(\frac{7}{2}\) hours
The number of hours your friend practice playing the keyboard is: \(\frac{5}{4}\) hours
So,
For comparison, make the denominators equal.
So,
Multiply \(\frac{7}{2}\) with \(\frac{2}{2}\)
So,
\(\frac{7}{2}\) = \(\frac{14}{4}\)
By comparing the timings ,
We can say that
\(\frac{14}{4}\) > \(\frac{5}{4}\)
Hence, from the above,
We can conclude that you practice more hours than your friend

Question 15.
The original price of a telescope is $99. The sale price is \(\frac{4}{5}\) of the original price. An astrologist buys the telescope at its sale price and uses a half-off-coupon. What fraction of the original price does the astrologist pay for the telescope?
Answer:
The fraction of the original price the astrologist pays for the telescope is: \(\frac{4}{10}\)

Explanation:
It is given that the original price of a telescope is $99. The sale price is \(\frac{4}{5}\) of the original price. and an astrologist buys the telescope at its sale price and uses a half-off-coupon.
So,
The cost of the telescope that an astrologist bought = ( The \(\frac{4}{5}\)th of the original price ) × ( Half-off -coupon on the \(\frac{4}{5}\)th of the original price )
= \(\frac{4}{5}\) × \(\frac{1}{2}\)
= \(\frac{4 × 1}{5 × 2}\)
= \(\frac{4}{10}\)
Hence, from the above,
We can conclude that the fraction of the original price the astrologist pays for the telescope is: \(\frac{4}{10}\)

Question 16.
The Abraj Al-Bait Clock Tower is \(\frac{6}{10}\) kilometer tall. Zifeng Tower is \(\frac{3}{4}\) as tall as the clock tower. Is Zifeng Tower shorter than, taller than, or the same height as the Abraj Al-Bait Clock Tower? What is the height of each tower in meters?
Big Ideas Math Solutions Grade 5 Chapter 9 Multiply Fractions 9.8 7
Answer:
The Abraj al-Bait Clock tower is taller than the Zifeng tower
The height of Abraj al-Bait Clock tower is: \(\frac{24}{40}\)
The height of Zifeng tower is: \(\frac{18}{40}\)

Explanation:
It is given that the Abraj Al-Bait Clock Tower is \(\frac{6}{10}\) kilometer tall and Zifeng Tower is \(\frac{3}{4}\) as tall as the clock tower.
So,
The height of Abraj al-Bait Clock tower is: \(\frac{6}{10}\) kilometer
Now,
The height of Zifeng tower is = \(\frac{6}{10}\) × \(\frac{3}{4}\)
= \(\frac{6 × 3}{10 × 4}\)
= \(\frac{18}{40}\)
Now, for comparison, make the denomonators of both the towers equal.
So,
\(\frac{6}{10}\) is multiplied by \(\frac{4}{4}\)
So,
\(\frac{6}{10}\) = \(\frac{24}{40}\)
By comparing the heights of the two towers,
We can say that
\(\frac{24}{40}\) > \(\frac{18}{40}\)
Hence, from the above,
We can conclude that
The Abraj al-Bait Clock tower is taller than the Zifeng tower
The height of Abraj al-Bait Clock tower is: \(\frac{24}{40}\)
The height of Zifeng tower is: \(\frac{18}{40}\)

Compare Factors and Products Homework & Practice 9.8

Without calculating, tell whether the product is less than, greater than, or equal to each of its factors.
Question 1.
1\(\frac{3}{4}\) × 6
Answer:
The value of 1\(\frac{3}{4}\) × 6 is greater than 6

Explanation:
The given numbers are: 1\(\frac{3}{4}\) and 6
The representation of 1\(\frac{3}{4}\) in the improper fraction form is: \(\frac{7}{4}\)
We know that,
When you multiply a number by a fraction greater than 1, the product is greater than the number. When you multiply a number by a fraction less than 1, the product is less than the number.
So,
\(\frac{7}{4}\) is greater than 1
So,
1\(\frac{3}{4}\) × 6 is greater than 6
Hence, from the above,
We can conclude that the value of 1\(\frac{3}{4}\) × 6 is greater than 6

Question 2.
\(\frac{5}{12}\) × \(\frac{1}{6}\)
Answer:
The value of \(\frac{5}{12}\) × \(\frac{1}{6}\) is less than 1

Explanation;
The given fractions are: \(\frac{5}{12}\) and \(\frac{1}{6}\)
We know that,
When you multiply a number by a fraction greater than 1, the product is greater than the number. When you multiply a number by a fraction less than 1, the product is less than the number.
So,
\(\frac{5}{12}\) is less than 1
\(\frac{1}{6}\) is less than 1
So,
\(\frac{5}{12}\) × \(\frac{1}{6}\) is less than 1
Hence, from the above,
We can conclude that the value of \(\frac{5}{12}\) × \(\frac{1}{6}\) is less than 1

Question 3.
\(\frac{2}{7}\) × \(\frac{5}{5}\)
Answer:
The value of \(\frac{2}{7}\) × \(\frac{5}{5}\) is less than 1

Explanation:
The given fractions are: \(\frac{2}{7}\) and \(\frac{5}{5}\)
We know that,
When you multiply a number by a fraction greater than 1, the product is greater than the number. When you multiply a number by a fraction less than 1, the product is less than the number.
So,
\(\frac{2}{7}\) is less than 1
We know that
\(\frac{5}{5}\) is 1
So,
\(\frac{2}{7}\) × \(\frac{5}{5}\) is less than 1
Hence, from the above,
We can conclude that the value of \(\frac{2}{7}\) × \(\frac{5}{5}\) is less than 1

Question 4.
3\(\frac{4}{5}\) × 2\(\frac{9}{10}\)
Answer:
The value of 3\(\frac{4}{5}\) × 2\(\frac{9}{10}\) is greater than 1

Explanation:
the given mixed fractions are: 3\(\frac{4}{5}\) and 2\(\frac{9}{10}\)
We know that,
When you multiply a number by a fraction greater than 1, the product is greater than the number. When you multiply a number by a fraction less than 1, the product is less than the number.
So,
The representation of the mixed numbers in the fraction form is: \(\frac{19}{5}\) and \(\frac{29}{10}\)
So,
\(\frac{19}{5}\) is greater than 1
\(\frac{29}{10}\) is greater than
So,
3\(\frac{4}{5}\) × 2\(\frac{9}{10}\) is greater than 1
Hence, from the above,
We can conclude that the value of 3\(\frac{4}{5}\) × 2\(\frac{9}{10}\) is greater than 1

Question 5.
8 × \(\frac{2}{3}\)
Answer:
The value of  8 is greater than \(\frac{2}{3}\)

Explanation:
The given numbers are: 8 and \(\frac{2}{3}\)
We know that,
When you multiply a number by a fraction greater than 1, the product is greater than the number. When you multiply a number by a fraction less than 1, the product is less than the number.
So,
\(\frac{2}{3}\) is less than 1
So,
8 is greater than \(\frac{2}{3}\)
Hence, from the above,
We can conclude that the value of 8 is greater than \(\frac{2}{3}\)

Question 6.
1\(\frac{7}{8}\) × \(\frac{1}{4}\)
Answer:
The value of 1\(\frac{7}{8}\) × \(\frac{1}{4}\) is less than 1

Explanation:
The given fractions are: 1\(\frac{7}{8}\) and \(\frac{1}{4}\)
The representation of 1\(\frac{7}{8}\) in the fraction form is: \(\frac{15}{8}\)
We know that,
When you multiply a number by a fraction greater than 1, the product is greater than the number. When you multiply a number by a fraction less than 1, the product is less than the number.
So,
\(\frac{15}{8}\) is greater than 1
\(\frac{1}{4}\) is less than 1
So,
1\(\frac{7}{8}\) × \(\frac{1}{4}\) is less than 1
Hence, from the above,
We can conclude that the value of 1\(\frac{7}{8}\) × \(\frac{1}{4}\) is less than 1

Without calculating, order the products from least to greatest.
Question 7.
Big Ideas Math Solutions Grade 5 Chapter 9 Multiply Fractions 9.8 8
Answer:
The order of products from the least to the greatest is:
\(\frac{1}{3}\) × \(\frac{4}{5}\) < \(\frac{1}{3}\) × \(\frac{6}{6}\) < 8\(\frac{2}{9}\) × \(\frac{1}{3}\)

Explanation:
The given products are:
A) \(\frac{1}{3}\) × \(\frac{4}{5}\)
B) \(\frac{1}{3}\) × \(\frac{6}{6}\)
C) 8\(\frac{2}{9}\) × \(\frac{1}{3}\)
So,
In A),
\(\frac{1}{3}\) and \(\frac{4}{5}\) are less than 1
So,
\(\frac{1}{3}\) × \(\frac{4}{5}\) is less than 1
In B),
\(\frac{1}{3}\) is less than 1
\(\frac{6}{6}\) is 1
So,
\(\frac{1}{3}\) × \(\frac{6}{6}\) is less than 1
In C),
\(\frac{1}{3}\) is less than1
The representation of  8\(\frac{2}{9}\) in the improper fraction form is: \(\frac{74}{9}\)
So,
\(\frac{74}{9}\) is greater than 1
So,
\(\frac{1}{3}\) × 8\(\frac{2}{9}\) is greater than 1
Hence, from the above,
We can conclude that the order of the products from the least to the greatest is:
\(\frac{1}{3}\) × \(\frac{4}{5}\) < \(\frac{1}{3}\) × \(\frac{6}{6}\) < 8\(\frac{2}{9}\) × \(\frac{1}{3}\)

Question 8.
Big Ideas Math Solutions Grade 5 Chapter 9 Multiply Fractions 9.8 9
Answer:
The order of products from the least to the greatest is:
\(\frac{1}{12}\) × \(\frac{3}{5}\) < 4 × \(\frac{3}{5}\) < 2\(\frac{1}{2}\) × \(\frac{3}{5}\)

Explanation:
The given products are:
A) \(\frac{1}{12}\) × \(\frac{3}{5}\)
B) 4 × \(\frac{3}{5}\)
C) 2\(\frac{1}{2}\) × \(\frac{3}{5}\)
So,
In A),
\(\frac{1}{12}\) and \(\frac{3}{5}\) are less than 1
So,
\(\frac{1}{12}\) × \(\frac{3}{5}\) is less than 1
In B),
\(\frac{3}{5}\) is less than 1
So,
4 × \(\frac{3}{5}\) is less than 4
In C),
\(\frac{3}{5}\) is less than1
The representation of  2\(\frac{1}{2}\) in the improper fraction form is: \(\frac{5}{2}\)
So,
\(\frac{5}{2}\) is greater than 1
So,
\(\frac{3}{5}\) × 2\(\frac{1}{2}\) is greater than 1
Hence, from the above,
We can conclude that the order of the products from the least to the greatest is:
\(\frac{1}{12}\) × \(\frac{3}{5}\) < 4 × \(\frac{3}{5}\) < 2\(\frac{1}{2}\) × \(\frac{3}{5}\)

Question 9.
Logic
Without calculating, use >, <, or = to make the statement true. Explain.
Big Ideas Math Solutions Grade 5 Chapter 9 Multiply Fractions 9.8 10
Answer:
The value of 3\(\frac{1}{3}\) × \(\frac{1}{6}\) is less than 3\(\frac{1}{2}\)

Explanation:
The given fractions are: 3\(\frac{1}{3}\), \(\frac{1}{6}\) and 3\(\frac{1}{2}\)
The representations of the 3\(\frac{1}{3}\) and  3\(\frac{1}{2}\)  in the fraction form is: \(\frac{7}{2}\) and \(\frac{10}{3}\)
So,
3\(\frac{1}{3}\) × \(\frac{1}{6}\)
= \(\frac{10}{3}\)  × \(\frac{1}{6}\)
= \(\frac{10}{18}\)
To compare the fractions, equate the denominators
So,
Multiply \(\frac{7}{2}\) by \(\frac{9}{9}\)
= \(\frac{63}{18}\)
So,
By comparison,
We can say that
\(\frac{10}{18}\) < \(\frac{63}{18}\)
Hence, from the above,
We can conclude that the value of 3\(\frac{1}{3}\) × \(\frac{1}{6}\) is less than 3\(\frac{1}{2}\)

Question 10.
Logic
Without calculating, determine which number makes the statement true.
Big Ideas Math Solutions Grade 5 Chapter 9 Multiply Fractions 9.8 11
Answer:
Let the missing number be x.
The options for x are given as:
A) 1 B) \(\frac{1}{2}\) and C) 1\(\frac{4}{5}\)
So,
From the given three fractions,
The value of x is: 1\(\frac{4}{5}\)

Explanation:
The given multiplication equation is:
x × 1\(\frac{7}{8}\) is greater than 1\(\frac{7}{8}\)
So,
To find the value of x, there are 3 options. They are:
A) 1 B) \(\frac{1}{2}\) and C) 1\(\frac{4}{5}\)
Now,
Let x = 1\(\frac{4}{5}\)
So,
1\(\frac{4}{5}\) × 1\(\frac{7}{8}\)
= \(\frac{9}{5}\) × \(\frac{15}{8}\)
= \(\frac{27}{8}\)
So, by comparison,
We will get,
\(\frac{27}{8}\) is greater than \(\frac{15}{8}\)
Hence, from the above,
We can conclude that the missing number is: 1\(\frac{4}{5}\)

Question 11.
Reasoning
Why does multiplying by a fraction greater than one result in a product greater than the original number?
Answer:
In the given fraction, if the numerator is greater than the denominator, then the given fraction is greater than 1
So,
When the whole number is multiplied by the fraction which is greater than 1, the result in a product will be greater than the original number i.e, the whole number

Question 12.
Modeling Real Life
You snowboard 1\(\frac{7}{8}\) miles. Your friend snowboards 3\(\frac{2}{3}\) times as far as you. Does your friend snowboard fewer miles, more miles, or the same number of miles as you?
Answer:
The number of miles you snowboarded is less than your friend

Explanation:
It is given that you snowboard 1\(\frac{7}{8}\) miles. Your friend snowboards 3\(\frac{2}{3}\) times as far as you.
So,
The number of miles you snowboard is: 1\(\frac{7}{8}\) miles
So,
The number of miles your friend snowboard is = 1\(\frac{7}{8}\) × 3\(\frac{2}{3}\)
= \(\frac{15}{8}\) × \(\frac{11}{3}\)
= \(\frac{15 × 11}{3 × 8}\)
= \(\frac{165}{24}\) miles
So,
For comparison we have to make the denominators equal.
So,
Multiply \(\frac{15}{8}\) by \(\frac{3}{3}\)
So,
\(\frac{15}{8}\) = \(\frac{45}{24}\)
So,
By comparing, we will get
\(\frac{45}{24}\) is less than \(\frac{165}{24}\) miles
Hence, from the above,
We can conclude that the number of miles you snowboarded is less than your friend

Question 13.
Modeling Real Life
A pet owner has three dogs. The youngest dog weighs \(\frac{1}{4}\) as much as the second oldest dog. The oldest dog weighs 1\(\frac{1}{4}\) as much as the second oldest. The second oldest weighs 20 pounds. Which dog weighs the most? the least?
Big Ideas Math Solutions Grade 5 Chapter 9 Multiply Fractions 9.8 12

Answer:
The dog that weighs the most is: The oldest dog
The dog that weighs the least is: The youngest dog

Explanation:
It is given that a pet owner has three dogs. The youngest dog weighs \(\frac{1}{4}\) as much as the second oldest dog. The oldest dog weighs 1\(\frac{1}{4}\) as much as the second oldest. The second oldest weighs 20 pounds.
So,
The weight of the second oldest dog is: 20 pounds
Now,
The weight of the oldest dog = 1\(\frac{1}{4}\) × ( The weight of the oldest second dog )
= 1\(\frac{1}{4}\) × 20
= \(\frac{5}{4}\) × 20
= \(\frac{5 × 20}{4 × 1}\)
= 25 pounds
Now,
The weight of the youngest dog = \(\frac{1}{4}\) × ( The weight of the second oldest dog )
= \(\frac{1}{4}\) × 20
= \(\frac{1 × 20}{4}\)
= 5 pounds
Hence, from the above,
We can conclude that
The dog that weighs the most is: The oldest dog
The dog that weighs the least is: The youngest dog

Review & Refresh

Compare.
Question 14.
Big Ideas Math Solutions Grade 5 Chapter 9 Multiply Fractions 9.8 13
Answer:
40.5 is greater than 40.13

Explanation:
The given decimal numbers are: 40.5 and 40.13
For comparison, compare the place value of the given digits
If there are more than 2 digits after the decimal, first compare the tenths position and after that only compare the hundredths position
If we can get the result in the tenths position only, then there is no need for further comparison
Hence, from the above,
We can conclude that 40.5 is greater than 40.13

Question 15.
Big Ideas Math Solutions Grade 5 Chapter 9 Multiply Fractions 9.8 14
Answer:
13.90 is equal to 13.9

Explanation:
The given decimal numbers are: 13.90 and 13.9
For comparison, compare the place value of the given digits
If there are more than 2 digits after the decimal, first compare the tenths position and after that only compare the hundredths position
If we can get the result in the tenths position only, then there is no need for further comparison
Hence, from the above,
We can conclude that 13.90 is equal to 13.9

Question 16.
Big Ideas Math Solutions Grade 5 Chapter 9 Multiply Fractions 9.8 15
Answer:
32.006 is less than 32.06

Explanation:
The given decimal numbers are: 32.006 and 32.06
For comparison, compare the place value of the given digits
If there are more than 2 digits after the decimal, first compare the tenths position and after that only compare the hundredths position
If we can get the result in the tenths position only, then there is no need for further comparison
Hence, from the above,
We can conclude that 32.006 is less than 32.06

Multiply Fractions Performance Task

Question 1.
You see a rock formation at a national park. The formation has layers that formed millions of years ago when particles settled in the water and became rock. You make a model of the rock formation using \(\frac{3}{16}\) -inch foam sheets.
Big Ideas Math Answer Key Grade 5 Chapter 9 Multiply Fractions 1
a. The three types of sedimentary rocks are limestone, sandstone, and shale. Use the number of foam sheets to find the height of each sedimentary rock layer.
Answer:
The height of limestone is: \(\frac{9}{4}\) inches
The height of sandstone is: \(\frac{3}{2}\) inches
The height of shale is: \(\frac{15}{4}\) inches

Explanation:
It is given that the height of each foam sheet is: \(\frac{3}{16}\) -inches
Big Ideas Math Answer Key Grade 5 Chapter 9 Multiply Fractions 1
From the table,
The number of foam sheets for Limestone is: 12
The number of foam sheets for sandstone is: 8
The number of foam sheets for shale is: 20
So,
The height of limestone is: ( The height of each foam sheet ) × ( The total number of foam sheets )
= \(\frac{3}{16}\) × 12
= \(\frac{3}{16}\) × \(\frac{12}{1}\)
= \(\frac{9}{4}\) inches
The height of sandstone is: ( The height of each foam sheet ) × ( The total number of foam sheets )
= \(\frac{3}{16}\) × 8
= \(\frac{3}{16}\) × \(\frac{8}{1}\)
= \(\frac{3}{2}\) inches
The height of shale is: ( The height of each foam sheet ) × ( The total number of foam sheets )
= \(\frac{3}{16}\) × 20
= \(\frac{3}{16}\) × \(\frac{20}{1}\)
= \(\frac{15}{4}\) inches
Hence, from the above,
We can conclude that
The height of limestone is: \(\frac{9}{4}\) inches
The height of sandstone is: \(\frac{3}{2}\) inches
The height of shale is: \(\frac{15}{4}\) inches

b. What is the combined height of the sedimentary rock layers?
Answer:
The combined height of the sedimentary layers is: \(\frac{15}{2}\) inches

Explanation:
The combined height of the sedimentary rock layers = The height of limestone + The height of sandstone + The height of the shale
= \(\frac{9}{4}\) + \(\frac{3}{2}\) + \(\frac{15}{4}\)
= \(\frac{24}{4}\) + \(\frac{3}{2}\)
= \(\frac{12}{2}\) + \(\frac{3}{2}\)
= \(\frac{15}{2}\) inches
Hence, from the above,
We can conclude that the combined height of the sedimentary layers is: \(\frac{15}{2}\) inches

c. Will you use more foam sheets for the granite layers of the shale layers? Explain.
Answer:
We use  more foam sheets for the granite layers

Explanation:
From the above,
The height of the shale layers is: \(\frac{15}{4}\) inches
From the table,
The height of the granite layers is: 3 inches
When we compare the two values,
We can observe that the height of shale layers is greater than the granite layers
Hence, from the above,
We can conclude that we will use more foam sheets for granite layers

d. The height of the topsoil layer is 1\(\frac{1}{4}\) times the height of the sandstone layer. How many foam sheets do you use in the topsoil layer?
Answer:
The height of the topsoil layer is: \(\frac{15}{8}\) inches

Explanation:
From the above,
The height of the sandstone layer is: \(\frac{3}{2}\) inches
So,
The height of topsoil layer = 1\(\frac{1}{4}\) × \(\frac{3}{2}\)
= \(\frac{5}{4}\) × \(\frac{3}{2}\)
= \(\frac{15}{8}\) inches
Hence, from the above,
We can conclude that the height of the topsoil layer is: \(\frac{15}{8}\) inches

e. On your model, 1 inch represents 40 feet. What is the actual height of the rock formation?
Answer:
The actual height of the rock formation is: 495 feet

Explanation:
From the above,
The height of the sedimentary layers is: \(\frac{15}{2}\) inches
The height of the topsoil layer is: \(\frac{15}{8}\) inches
The height of the granite layer is: 3 inches
So,
The combined height of the rock formation = \(\frac{15}{2}\) + \(\frac{15}{8}\) + 3
= \(\frac{99}{8}\) inches
But, it is given that,
1 inch = 40 feet
So,
The combined height of the rock formation in feet = \(\frac{99}{8}\) inches × 40
= 495 feet
Hence, from the above,
We can conclude that the height of the rock formation in feet is: 495 feet

f. Why do you think the rock formation has layers?
Answer:
The rock formation has layers because of the tectonic plates.

Multiply Fractions Activity

Fraction Connection: Multiplication
Directions:
1. Players take turns rolling three dice.
2. On your turn, evaluate the expression indicated by your roll and cover the answer.
3. The first player to get four in a row, horizontally, vertically, or diagonally, wins?
Big Ideas Math Answer Key Grade 5 Chapter 9 Multiply Fractions 2
Big Ideas Math Answer Key Grade 5 Chapter 9 Multiply Fractions 3

Multiply Fractions Chapter Practice

9.1 Multiply Whole Numbers by Fractions

Multiply.
Question 1.
5 × \(\frac{1}{2}\) = _______
Answer:
5 × \(\frac{1}{2}\) = \(\frac{5}{2}\)

Explanation:
The given numbers are: 5 and \(\frac{1}{2}\)
So,
5 × \(\frac{1}{2}\)
= \(\frac{5}{1}\) × \(\frac{1}{2}\)
= \(\frac{5 × 1}{1 × 2}\)
= \(\frac{5}{2}\)
Hence,
5 × \(\frac{1}{2}\) = \(\frac{5}{2}\)

Question 2.
2 × \(\frac{7}{10}\) = _______
Answer:
2 × \(\frac{7}{10}\) = \(\frac{14}{10}\)

Explanation:
The given numbers are: 2 and \(\frac{7}{10}\)
So,
2 × \(\frac{7}{10}\)
= \(\frac{2}{1}\) × \(\frac{7}{10}\)
= \(\frac{2 × 7}{1 × 10}\)
= \(\frac{14}{10}\)
Hence,
2 × \(\frac{7}{10}\) = \(\frac{14}{10}\)

Question 3.
9 × \(\frac{5}{8}\) = _______
Answer:
9 × \(\frac{5}{8}\) = \(\frac{45}{8}\)

Explanation:
The given numbers are: 9 and \(\frac{5}{8}\)
So,
9 × \(\frac{5}{8}\)
= \(\frac{9}{1}\) × \(\frac{5}{8}\)
= \(\frac{5 × 9}{1 × 8}\)
= \(\frac{45}{8}\)
Hence,
9 × \(\frac{5}{8}\) = \(\frac{45}{8}\)

Question 4.
6 × \(\frac{71}{100}\) = _______
Answer:
6 × \(\frac{71}{100}\) = \(\frac{426}{100}\)

Explanation:
The given numbers are: 6 and \(\frac{71}{100}\)
So,
6 × \(\frac{71}{100}\)
= \(\frac{6}{1}\) × \(\frac{71}{100}\)
= \(\frac{6 × 71}{1 × 100}\)
= \(\frac{426}{100}\)
Hence,
6 × \(\frac{71}{100}\) = \(\frac{426}{100}\)

Question 5.
4 × \(\frac{8}{5}\) = _______
Answer:
4 × \(\frac{8}{5}\) = \(\frac{32}{5}\)

Explanation:
The given numbers are: 4 and \(\frac{8}{5}\)
So,
4 × \(\frac{8}{5}\)
= \(\frac{4}{1}\) × \(\frac{8}{5}\)
= \(\frac{4 × 8}{1 × 5}\)
= \(\frac{32}{5}\)
Hence,
4 × \(\frac{8}{5}\) = \(\frac{32}{5}\)

Question 6.
7 × \(\frac{5}{3}\) = _______
Answer:
7 × \(\frac{5}{3}\) = \(\frac{35}{3}\)

Explanation:
The given numbers are: 7 and \(\frac{5}{3}\)
So,
7 × \(\frac{5}{3}\)
= \(\frac{7}{1}\) × \(\frac{5}{3}\)
= \(\frac{5 × 7}{1 × 3}\)
= \(\frac{35}{3}\)
Hence,
7 × \(\frac{5}{3}\) = \(\frac{35}{3}\)

9.2 Use Models to Multiply Fractions by Whole Numbers

Multiply. Use a model to help.
Question 7.
\(\frac{2}{5}\) of 20
Answer:
20 × \(\frac{2}{5}\) = 8

Explanation:
The given numbers are: 20 and \(\frac{2}{5}\)
So,
20 × \(\frac{2}{5}\)
= \(\frac{20}{1}\) × \(\frac{2}{5}\)
= \(\frac{20 × 2}{1 × 5}\)
= \(\frac{8}{1}\)
= 8
Hence,
20 × \(\frac{2}{5}\) = 8

Question 8.
\(\frac{1}{6}\) × 12
Answer:
12 × \(\frac{1}{6}\) = 2

Explanation:
The given numbers are: 12 and \(\frac{1}{6}\)
So,
12 × \(\frac{1}{6}\)
= \(\frac{12}{1}\) × \(\frac{1}{6}\)
= \(\frac{12 × 1}{1 × 6}\)
= \(\frac{2}{1}\)
= 2
Hence,
12 × \(\frac{1}{6}\) = 2

Question 9.
\(\frac{1}{3}\) × 6
Answer:
6 × \(\frac{1}{3}\) = 2

Explanation:
The given numbers are: 6 and \(\frac{1}{3}\)
So,
6 × \(\frac{1}{3}\)
= \(\frac{6}{1}\) × \(\frac{1}{3}\)
= \(\frac{6 × 1}{1 × 3}\)
= \(\frac{2}{1}\)
= 2
Hence,
6 × \(\frac{1}{3}\) = 2

Question 10.
\(\frac{5}{16}\) of 8
Answer:
8 × \(\frac{5}{16}\) = \(\frac{5}{2}\)

Explanation:
The given numbers are: 8 and \(\frac{5}{16}\)
So,
8 × \(\frac{5}{16}\)
= \(\frac{8}{1}\) × \(\frac{5}{16}\)
= \(\frac{5 × 8}{1 × 16}\)
= \(\frac{5}{2}\)
Hence,
8 × \(\frac{5}{16}\) = \(\frac{5}{2}\)

Question 11.
Modeling Real Life
You have 24 apples. You use \(\frac{1}{4}\) of them to make a single serving of applesauce. How many apples do you not use?
Big Ideas Math Answer Key Grade 5 Chapter 9 Multiply Fractions chp 11
Answer:
The number of apples you do not use is: 18 apples

Explanation:
It is given that you have 24 apples and you use \(\frac{1}{4}\) of them to make a single serving of applesauce.
So,
The number of apples you used = ( The total number of apples ) × ( The fraction of apples you used to make a single serving of applesauce )
= 24 × \(\frac{1}{4}\)
= \(\frac{24}{1}\) × \(\frac{1}{4}\)
= \(\frac{6}{1}\)
= 6 apples
Now,
The number of apples that do not use = ( The total number of apples ) – ( The number of apples you used )
= 24 – 6
= 18 apples
Hence, from the above,
We can conclude that the number of apples that do not use is: 18 apples

9.3 Multiply Fractions and Whole Numbers

Multiply.
Question 12.
\(\frac{3}{5}\) × 15 = _______
Answer:
15 × \(\frac{3}{5}\) = 9

Explanation:
The given numbers are: 15 and \(\frac{3}{5}\)
So,
15 × \(\frac{3}{5}\)
= \(\frac{15}{1}\) × \(\frac{3}{5}\)
= \(\frac{15 × 3}{1 × 5}\)
= \(\frac{9}{1}\)
= 9
Hence,
15 × \(\frac{3}{5}\) = 9

Question 13.
\(\frac{9}{10}\) × 30 = _______
Answer:
30 × \(\frac{9}{10}\) = 27

Explanation:
The given numbers are: 30 and \(\frac{9}{10}\)
So,
30 × \(\frac{9}{10}\)
= \(\frac{30}{1}\) × \(\frac{9}{10}\)
= \(\frac{30 × 9}{1 × 10}\)
= \(\frac{27}{1}\)
= 27
Hence,
30 × \(\frac{9}{10}\) = 27

Question 14.
48 × \(\frac{3}{4}\) = _______
Answer:
48 × \(\frac{3}{4}\) = 36

Explanation:
The given numbers are: 48 and \(\frac{3}{4}\)
So,
48 × \(\frac{3}{4}\)
= \(\frac{48}{1}\) × \(\frac{3}{4}\)
= \(\frac{48 × 3}{1 × 4}\)
= \(\frac{36}{1}\)
= 36
Hence,
48 × \(\frac{3}{4}\) = 36

Question 15.
11 × \(\frac{5}{9}\) = _______
Answer:
11 × \(\frac{5}{9}\) = \(\frac{55}{9}\)

Explanation:
The given numbers are: 11 and \(\frac{5}{9}\)
So,
11 × \(\frac{5}{9}\)
= \(\frac{11}{1}\) × \(\frac{5}{9}\)
= \(\frac{5 × 11}{1 × 9}\)
= \(\frac{55}{9}\)
Hence,
11 × \(\frac{5}{9}\) = \(\frac{55}{9}\)

Question 16.
\(\frac{1}{6}\) × 19 = _______
Answer:
19 × \(\frac{1}{6}\) = \(\frac{19}{6}\)

Explanation:
The given numbers are: 19 and \(\frac{1}{6}\)
So,
19 × \(\frac{1}{6}\)
= \(\frac{19}{1}\) × \(\frac{1}{6}\)
= \(\frac{19 × 1}{1 × 6}\)
= \(\frac{19}{6}\)
Hence,
19 × \(\frac{1}{6}\) = \(\frac{19}{6}\)

Question 17.
7 × \(\frac{13}{50}\) = _______
Answer:
7 × \(\frac{13}{50}\) = \(\frac{91}{50}\)

Explanation:
The given numbers are: 7 and \(\frac{13}{50}\)
So,
7 × \(\frac{13}{50}\)
= \(\frac{7}{1}\) × \(\frac{13}{50}\)
= \(\frac{7 × 13}{1 × 50}\)
= \(\frac{91}{50}\)
Hence,
7 × \(\frac{13}{50}\) = \(\frac{91}{50}\)

9.4 Use Models to Multiply Fractions

Multiply. Use a model to help.
Question 18.
\(\frac{1}{2}\) × \(\frac{1}{10}\) = _______
Answer:
\(\frac{1}{2}\) × \(\frac{1}{10}\) = \(\frac{1}{20}\)

Explanation:
The given numbers are: \(\frac{1}{2}\) and \(\frac{1}{10}\)
So,
\(\frac{1}{2}\) × \(\frac{1}{10}\)
= \(\frac{1 × 1}{2 × 10}\)
= \(\frac{1}{20}\)
Hence,
\(\frac{1}{2}\) × \(\frac{1}{10}\) = \(\frac{1}{20}\)

Question 19.
\(\frac{1}{5}\) × \(\frac{1}{9}\) = _______
Answer:
\(\frac{1}{5}\) × \(\frac{1}{9}\) = \(\frac{1}{45}\)

Explanation:
The given numbers are: \(\frac{1}{5}\) and \(\frac{1}{9}\)
So,
\(\frac{1}{5}\) × \(\frac{1}{9}\)
= \(\frac{1 × 1}{5 × 9}\)
= \(\frac{1}{45}\)
Hence,
\(\frac{1}{5}\) × \(\frac{1}{9}\) = \(\frac{1}{45}\)

Question 20.
\(\frac{1}{6}\) × \(\frac{1}{7}\) = _______
Answer:
\(\frac{1}{6}\) × \(\frac{1}{7}\) = \(\frac{1}{42}\)

Explanation:
The given numbers are: \(\frac{1}{6}\) and \(\frac{1}{7}\)
So,
\(\frac{1}{6}\) × \(\frac{1}{7}\)
= \(\frac{1 × 1}{6 × 7}\)
= \(\frac{1}{42}\)
Hence,
\(\frac{1}{6}\) × \(\frac{1}{7}\) = \(\frac{1}{42}\)

Question 21.
\(\frac{1}{3}\) × \(\frac{1}{8}\) = _______
Answer:
\(\frac{1}{3}\) × \(\frac{1}{8}\) = \(\frac{1}{24}\)

Explanation:
The given numbers are: \(\frac{1}{3}\) and \(\frac{1}{8}\)
So,
\(\frac{1}{3}\) × \(\frac{1}{8}\)
= \(\frac{1 × 1}{3 × 8}\)
= \(\frac{1}{24}\)
Hence,
\(\frac{1}{3}\) × \(\frac{1}{8}\) = \(\frac{1}{24}\)

Question 22.
\(\frac{2}{5}\) × \(\frac{1}{3}\) = _______
Answer:
\(\frac{2}{5}\) × \(\frac{1}{3}\) = \(\frac{2}{15}\)

Explanation:
The given numbers are: \(\frac{2}{5}\) and \(\frac{1}{3}\)
So,
\(\frac{2}{5}\) × \(\frac{1}{3}\)
= \(\frac{2 × 1}{5 × 3}\)
= \(\frac{2}{15}\)
Hence,
\(\frac{2}{5}\) × \(\frac{1}{3}\) = \(\frac{2}{15}\)

Question 23.
\(\frac{2}{3}\) × \(\frac{3}{5}\) = _______
Answer:
\(\frac{2}{3}\) × \(\frac{3}{5}\) = \(\frac{2}{5}\)

Explanation:
The given numbers are: \(\frac{2}{3}\) and \(\frac{3}{5}\)
So,
\(\frac{2}{3}\) × \(\frac{3}{5}\)
= \(\frac{2 × 3}{3 × 5}\)
= \(\frac{2}{5}\)
Hence,
\(\frac{2}{3}\) × \(\frac{3}{5}\) = \(\frac{2}{5}\)

9.5 Multiply Fractions

Evaluate.
Question 24.
\(\frac{1}{3}\) × \(\frac{1}{8}\) = _______
Answer:
\(\frac{1}{3}\) × \(\frac{1}{8}\) = \(\frac{1}{24}\)

Explanation:
The given numbers are: \(\frac{1}{3}\) and \(\frac{1}{8}\)
So,
\(\frac{1}{3}\) × \(\frac{1}{8}\)
= \(\frac{1 × 1}{3 × 8}\)
= \(\frac{1}{24}\)
Hence,
\(\frac{1}{3}\) × \(\frac{1}{8}\) = \(\frac{1}{24}\)

Question 25.
\(\frac{5}{6}\) × \(\frac{1}{4}\) = _______
Answer:
\(\frac{5}{6}\) × \(\frac{1}{4}\) = \(\frac{5}{24}\)

Explanation:
The given numbers are: \(\frac{5}{6}\) and \(\frac{1}{4}\)
So,
\(\frac{5}{6}\) × \(\frac{1}{4}\)
= \(\frac{5 × 1}{6 × 4}\)
= \(\frac{5}{24}\)
Hence,
\(\frac{5}{6}\) × \(\frac{1}{4}\) = \(\frac{5}{24}\)

Question 26.
\(\frac{7}{2}\) × \(\frac{2}{5}\) = _______
Answer:
\(\frac{7}{2}\) × \(\frac{2}{5}\) = \(\frac{7}{5}\)

Explanation:
The given numbers are: \(\frac{7}{2}\) and \(\frac{2}{5}\)
So,
\(\frac{7}{2}\) × \(\frac{2}{5}\)
= \(\frac{7 × 2}{2 × 5}\)
= \(\frac{7}{5}\)
Hence,
\(\frac{7}{2}\) × \(\frac{2}{5}\) = \(\frac{7}{5}\)

Question 27.
\(\frac{9}{10}\) × \(\frac{3}{7}\) = _______
Answer:
\(\frac{9}{10}\) × \(\frac{3}{7}\) = \(\frac{27}{70}\)

Explanation:
The given numbers are: \(\frac{9}{10}\) and \(\frac{3}{7}\)
So,
\(\frac{9}{10}\) × \(\frac{3}{7}\)
= \(\frac{9 × 3}{7 × 10}\)
= \(\frac{27}{70}\)
Hence,
\(\frac{9}{10}\) × \(\frac{3}{7}\) = \(\frac{27}{70}\)

Question 28.
\(\frac{4}{5}\) × \(\frac{13}{100}\) = _______
Answer:
\(\frac{4}{5}\) × \(\frac{13}{100}\) = \(\frac{52}{500}\)

Explanation:
The given numbers are: \(\frac{4}{5}\) and \(\frac{13}{100}\)
So,
\(\frac{4}{5}\) × \(\frac{13}{100}\)
= \(\frac{4 × 13}{5 × 100}\)
= \(\frac{52}{500}\)
Hence,
\(\frac{4}{5}\) × \(\frac{13}{100}\) = \(\frac{52}{500}\)

Question 29.
\(\frac{11}{25}\) × \(\frac{3}{4}\) = _______
Answer:
\(\frac{11}{25}\) × \(\frac{3}{4}\) = \(\frac{33}{100}\)

Explanation:
The given numbers are: \(\frac{11}{25}\) and \(\frac{3}{4}\)
So,
\(\frac{11}{25}\) × \(\frac{3}{4}\)
= \(\frac{11 × 3}{25 × 4}\)
= \(\frac{33}{100}\)
Hence,
\(\frac{11}{25}\) × \(\frac{3}{4}\) = \(\frac{33}{100}\)

Question 30.
9 × \(\left(\frac{2}{9} \times \frac{1}{2}\right)\) = _______
Answer:
9 × ( \(\frac{1}{2}\) × \(\frac{2}{9}\) ) = 1

Explanation:
The given numbers are: 9, \(\frac{1}{2}\) and \(\frac{2}{9}\)
So,
9 × (\(\frac{1}{2}\) × \(\frac{2}{9}\) )
= 9 × (\(\frac{1 × 2}{2 × 9}\) )
=  9 × \(\frac{1}{9}\)
= 1
Hence,
9 × ( \(\frac{1}{2}\) × \(\frac{2}{9}\) ) = 1

Question 31.
\(\left(\frac{1}{10}+\frac{7}{10}\right)\) × \(\frac{3}{4}\) = _______
Answer:
\(\left(\frac{1}{10}+\frac{7}{10}\right)\) × \(\frac{3}{4}\) = \(\frac{3}{5}\)

Explanation:
The given fractions are: \(\frac{1}{10}\), \(\frac{7}{10}\) and \(\frac{3}{4}\)
So,
\(\frac{3}{4}\) × (\(\frac{1}{10}\) + \(\frac{7}{10}\) )
= \(\frac{3}{4}\) × (\(\frac{7 × 1}{10}\) )
=  \(\frac{3}{4}\) × \(\frac{8}{10}\)
= \(\frac{3 × 8}{4 × 10}\)
= \(\frac{24}{40}\)
= \(\frac{3}{5}\)
Hence,
\(\left(\frac{1}{10}+\frac{7}{10}\right)\) × \(\frac{3}{4}\) = \(\frac{3}{5}\)

Question 32.
\(\frac{4}{7}\) × \(\left(\frac{5}{8}-\frac{1}{4}\right)\) = _______
Answer:
\(\frac{4}{7}\) × \(\left(\frac{5}{8}-\frac{1}{4}\right)\) = \(\frac{3}{14}\)

Explanation:
The given fractions are: \(\frac{4}{7}\), \(\frac{5}{8}\) and \(\frac{1}{4}\)
So,
\(\frac{4}{7}\) × (\(\frac{5}{8}\) – \(\frac{1}{4}\) )
= \(\frac{4}{7}\) × (\(\frac{5 – 2}{8}\) )
=  \(\frac{4}{7}\) × \(\frac{3}{8}\)
= \(\frac{3 × 4}{7 × 8}\)
= \(\frac{12}{56}\)
= \(\frac{3}{14}\)
Hence,
\(\frac{4}{7}\) × \(\left(\frac{5}{8}-\frac{1}{4}\right)\) = \(\frac{3}{14}\)

9.6 Find Areas of Rectangles

Use rectangles with unit fraction side lengths to find the area of the rectangle.
Question 33.
Big Ideas Math Answer Key Grade 5 Chapter 9 Multiply Fractions chp 33
Answer:
The area of the rectangle is: \(\frac{1}{4}\)

Explanation:
The given figure is:
Big Ideas Math Answer Key Grade 5 Chapter 9 Multiply Fractions chp 33
From the given figure,
The side lengths of the rectangle are: \(\frac{3}{8}\) and \(\frac{2}{3}\)
So,
The area of the rectangle = \(\frac{3}{8}\) × \(\frac{2}{3}\)
= \(\frac{2 × 3}{3 × 8}\)
= \(\frac{1}{4}\)
Hence, from the above,
We can conclude that the area of the rectangle is: \(\frac{1}{4}\)

Question 34.
Big Ideas Math Answer Key Grade 5 Chapter 9 Multiply Fractions chp 34
Answer:
The area of the rectangle is: \(\frac{35}{48}\)

Explanation:
The given figure is:
Big Ideas Math Answer Key Grade 5 Chapter 9 Multiply Fractions chp 34
From the given figure,
The side lengths of the rectangle are: \(\frac{7}{12}\) and \(\frac{5}{4}\)
So,
The area of the rectangle = \(\frac{7}{12}\) × \(\frac{5}{4}\)
= \(\frac{7 × 5}{12 × 4}\)
= \(\frac{35}{48}\)
Hence, from the above,
We can conclude that the area of the rectangle is: \(\frac{35}{48}\)

Question 35.
Find the area of a rectangle with side lengths of \(\frac{2}{9}\) and \(\frac{1}{10}\).
Answer:
The area of the rectangle is: \(\frac{1}{45}\)

Explanation:
The given side lengths of a rectangle are: \(\frac{2}{9}\) and \(\frac{1}{10}\)
So,
The area of the rectangle = \(\frac{2}{9}\) × \(\frac{1}{10}\)
= \(\frac{2 × 1}{10 × 9}\)
= \(\frac{1}{45}\)
Hence, from the above,
We can conclude that the area of the rectangle is: \(\frac{1}{45}\)

Question 36.
Find the area of a with a side length of \(\frac{3}{4}\).
Answer:
The area of the square is: \(\frac{9}{16}\)

Explanation:
The given side length of a square is: \(\frac{3}{4}\)
We know that,
The length of all sides of a square are equal
So,
the area of the square = \(\frac{3}{4}\) × \(\frac{3}{4}\)
= \(\frac{3 × 3}{4 × 4}\)
= \(\frac{9}{16}\)
Hence, from the above,
We can conclude that the area of the square is: \(\frac{9}{16}\)

9.7 Multiply Mixed Numbers

Multiply.
Question 37.
1\(\frac{1}{4}\) × 2\(\frac{1}{4}\) = _______
Answer:
1\(\frac{1}{4}\) × 2\(\frac{1}{4}\) =\(\frac{45}{16}\)

Explanation:
The given mixed fractions are: 1\(\frac{1}{4}\) and 2\(\frac{1}{4}\)
The representation of 1\(\frac{1}{4}\) and 2\(\frac{1}{4}\) in the improper fraction form is: \(\frac{5}{4}\) and \(\frac{9}{4}\)
So,
\(\frac{5}{4}\) × \(\frac{9}{4}\)
= \(\frac{5 × 9}{4 × 4}\)
= \(\frac{45}{16}\)
Hence,
1\(\frac{1}{4}\) × 2\(\frac{1}{4}\) =\(\frac{45}{16}\)

Question 38.
3\(\frac{4}{5}\) × 1\(\frac{1}{2}\) = _______
Answer:
3\(\frac{4}{5}\) × 1\(\frac{1}{2}\) = \(\frac{57}{10}\)

Explanation:
The given mixed fractions are: 3\(\frac{4}{5}\) and 1\(\frac{1}{2}\)
The representation of 3\(\frac{4}{5}\) and 1\(\frac{1}{2}\) in the improper fraction form is: \(\frac{19}{5}\)  and \(\frac{3}{2}\)
So,
\(\frac{19}{5}\) × \(\frac{3}{2}\)
= \(\frac{19 × 3}{5 × 2}\)
= \(\frac{57}{10}\)
Hence,
3\(\frac{4}{5}\) × 1\(\frac{1}{2}\) = \(\frac{57}{10}\)

Question 39.
5\(\frac{1}{3}\) × 4\(\frac{7}{8}\) = _______
Answer:
5\(\frac{1}{3}\) × 4\(\frac{7}{8}\) = \(\frac{624}{24}\)

Explanation:
The given mixed fractions are: 5\(\frac{1}{3}\) and 4\(\frac{7}{8}\)
The representation of 5\(\frac{1}{3}\) and 4\(\frac{7}{8}\) in the improper fraction form is: \(\frac{16}{3}\)  and \(\frac{39}{8}\)
So,
\(\frac{16}{3}\) × \(\frac{39}{8}\)
= \(\frac{39 × 16}{3 × 8}\)
= \(\frac{624}{24}\)
Hence,
5\(\frac{1}{3}\) × 4\(\frac{7}{8}\) = \(\frac{624}{24}\)

Question 40.
4\(\frac{5}{6}\) × 2\(\frac{9}{10}\) × \(\frac{1}{8}\) = _______
Answer:
4\(\frac{5}{6}\) × 2\(\frac{9}{10}\) × \(\frac{1}{8}\) = \(\frac{841}{480}\)

Explanation:
The given mixed fractions are: 4\(\frac{5}{6}\), 2\(\frac{9}{10}\) and \(\frac{1}{8}\)
The representation of 4\(\frac{5}{6}\) and 2\(\frac{9}{10}\) in the improper fraction form is: \(\frac{29}{6}\) and \(\frac{29}{10}\)
So,
\(\frac{29}{6}\) × \(\frac{29}{10}\)  × \(\frac{1}{8}\)
= \(\frac{29 × 29 × 1}{8 × 10 × 6}\)
= \(\frac{841}{480}\)
Hence,
4\(\frac{5}{6}\) × 2\(\frac{9}{10}\) × \(\frac{1}{8}\) = \(\frac{841}{480}\)

Question 41.
Logic
Find the missing numbers.
Big Ideas Math Answer Key Grade 5 Chapter 9 Multiply Fractions chp 41
Answer:
The missing numbers are: 2 and 11

Explanation:
Let the missing numbers be: p and q
So,
4\(\frac{1}{p}\) × 2\(\frac{1}{5}\) = 9\(\frac{q}{10}\)
So,
To solve the mixed numbers, we can separate the whole numbers and the fraction numbers
So,
\(\frac{1}{p}\) × 2\(\frac{1}{5}\) = \(\frac{q}{10}\)
\(\frac{1}{p}\) × \(\frac{11}{5}\) = \(\frac{q}{10}\)
\(\frac{1 × 11}{p × 5}\) = \(\frac{q}{10}\)
\(\frac{11}{5p}\) = \(\frac{q}{10}\)
By comparing LHS and RHS
q = 11 and p = 2
Hence, from the above,
We can conclude that the missing numbers are: 2 and 11

9.8 Compare Factors and Products

Without calculating, tell whether the product is less than, greater than, or equal to each of its factors.
Question 42.
1\(\frac{1}{2}\) × 3
Answer:
The value of 1\(\frac{1}{2}\) × 3 is greater than 3

Explanation:
The given numbers are: 3 and 1\(\frac{1}{2}\)
The representation of 1\(\frac{1}{2}\) in the mixed form is: \(\frac{3}{2}\)
So,
\(\frac{3}{2}\) is greater than 1
So,
\(\frac{3}{2}\) × 3 is greater than 3
Hence, from the above,
We can conclude that the value of 1\(\frac{1}{2}\) × 3 is greater than 3

Question 43.
\(\frac{1}{8}\) × \(\frac{1}{6}\)
Answer:
The value of \(\frac{1}{8}\) × \(\frac{1}{6}\) is less than 1

Explanation:
The given numbers are: \(\frac{1}{8}\) and \(\frac{1}{6}\)
So,
\(\frac{1}{8}\) is less than 1
\(\frac{1}{6}\) is less than 1
So,
\(\frac{1}{8}\) × \(\frac{1}{6}\) is less than 1
Hence, from the above,
We can conclude that the value of \(\frac{1}{8}\) × \(\frac{1}{6}\) is less than 1

Question 44.
\(\frac{5}{5}\) × 2\(\frac{4}{7}\)
Answer:
The value of \(\frac{5}{5}\) × 2\(\frac{4}{7}\) is greater than 1

Explanation:
The given numbers are: \(\frac{5}{5}\) and 2\(\frac{4}{7}\)
The representation of 2\(\frac{4}{7}\) in the improper fraction form is: \(\frac{18}{7}\)
So,
\(\frac{18}{7}\) is greater than 1
\(\frac{5}{5}\) is 1
So,
\(\frac{5}{5}\) × 2\(\frac{4}{7}\) is greater than 1
Hence, from the above,
We can conclude that the value of \(\frac{5}{5}\) × 2\(\frac{4}{7}\) is greater than 1

Question 45.
\(\frac{3}{4}\) × 2\(\frac{11}{12}\)
Answer:
The value of \(\frac{3}{4}\) × 2\(\frac{11}{12}\) is greater than 1

Explanation:
The given numbers are: \(\frac{3}{4}\) and 2\(\frac{11}{12}\)
The representation of 2\(\frac{11}{12}\) in the improper fraction form is: \(\frac{35}{12}\)
So,
\(\frac{35}{12}\) is greater than 1
\(\frac{3}{4}\) is less than 1
So,
\(\frac{3}{4}\) × 2\(\frac{11}{12}\) is greater than 1
Hence, from the above,
We can conclude that the value of \(\frac{3}{4}\) × 2\(\frac{11}{12}\) is greater than 1

Question 46.
3\(\frac{1}{9}\) × 2\(\frac{7}{8}\)
Answer:
The value of 3\(\frac{1}{9}\) × 2\(\frac{7}{8}\) is greater than 1

Explanation:
The given mixed fractions are: 3\(\frac{1}{9}\) and 2\(\frac{7}{8}\)
The representation of 3\(\frac{1}{9}\) and 2\(\frac{7}{8}\) in the improper fractions form is: \(\frac{28}{9}\) and \(\frac{23}{8}\)
So,
\(\frac{28}{9}\) is greater than 1
\(\frac{23}{8}\) is greater than 1
So,
3\(\frac{1}{9}\) × 2\(\frac{7}{8}\) is greater than 1
Hence, from the above,
We can conclude that the value of 3\(\frac{1}{9}\) × 2\(\frac{7}{8}\) is greater than 1

Question 47.
\(\frac{9}{8}\) × \(\frac{5}{2}\)
Answer:
The value of \(\frac{9}{8}\) × \(\frac{5}{2}\) is greater than 1

Explanation:
The given fractions are: \(\frac{9}{8}\) and \(\frac{5}{2}\)
So,
\(\frac{9}{8}\) is greater than 1
So,
\(\frac{9}{8}\) × \(\frac{5}{2}\) is greater than 1
Hence, from the above,
We can conclude that the value of \(\frac{9}{8}\) × \(\frac{5}{2}\) is greater than 1

Conclusion:

Hope the information provided in the above answer key is beneficial for all the students of grade 5. Follow our Big Ideas Math Answers and make your learning fun and interesting.

Big Ideas Math Answers Grade 4 Chapter 10 Relate Fractions and Decimals

Big Ideas Math Answers Grade 4 Chapter 10

Do you want to become proficient in Math? Then, must follow our Big Ideas Math Answers Grade 4 Chapter 10 Relate Fractions and Decimals. Big Ideas Math Answers Grade 4 Chapter 10 Relate Fractions and Decimals is the one-stop destination for all your practice. You can get the topic-wise Big Ideas Math Answers Grade 4 Chapter 10 Relate Fractions and Decimals PDF for free of cost. Test your knowledge on the complete concepts by solving from the Big Ideas Math Answers Grade 4 Chapter 10 Relate Fractions and Decimals and identify the knowledge gap accordingly.

Big Ideas 4th Grade Chapter 10 Relate Fractions and Decimals Math Book Answer Key

Use our ultimate guide Big Ideas Math Book 4th Grade Answer Key Chapter 10 Relate Fractions and Decimals over here to clear all your doubts. How to grab the best source of BIM 4th grade Chapter 10 Answer Key? All you have to do is simply tap on the below quick links to get the respective topic from the 4th grade Chapter 10 Solution Key. Start your preparation right away by taking the help of the Relate Fractions and Decimals Big Ideas Math Grade 4 Solutions and clear the exam with flying colors.

Lesson: 1 Understand Tenths

Lesson: 2 Understand Hundredths

Lesson: 3 Fractions and Decimals

Lesson: 4 Compare Decimals

Lesson: 5 Add Decimal Fractions and Decimals

Lesson: 6 Fractions, Decimals and Money

Lesson: 7 Operations with Money

Performance Task

Lesson 10.1 Understand Tenths

Explore and Grow

How many dimes have a total value of one dollar? Draw a model.
Big Ideas Math Answers Grade 4 Chapter 10 Relate Fractions and Decimals 1

One dime is what fraction of one dollar? Write your answer in words and as a fraction.

Answer:
We know that,
1 dollar = 10 dimes
So,
The representation of one dime in the fraction of 1 dollar is: \(\frac{1}{10}\)
In terms of words, one Dime is equal to one-tenth of the Dollar

Explanation:
The given note is “Dollar note”
Big Ideas Math Answers Grade 4 Chapter 10 Relate Fractions and Decimals 1
We know that,
1 Dollar = 10 Dimes
So,
The representation of one Dime in the fraction of 1 dollar is: \(\frac{1}{10}\)

So,
The representation of a Dime in the model is:

From the above model,
We can say that one Dime is equal to one-tenth of one Dollar.

Structure
How is one whole related to one-tenth? How do you think you can write \(\frac{1}{10}\) in a place value chart?
Answer:
The representation of one-tenth in a place value chart is:

Explanation:
The one-tenth is the fraction value which is less than one whole. When we divide the one whole into ten parts,
then each divided part represents one-tenth of the 10 parts.
Hence,
The representation of \(\frac{1}{10}\) in the place-value chart is:

Think and Grow: Understand Tenths

A decimal is a number with one or more digits to the right of the decimal point. The first place to the right of the decimal point is the tenths place.
You can write tenths as fractions or decimals.
Big Ideas Math Answers Grade 4 Chapter 10 Relate Fractions and Decimals 2

Example
Write \(\frac{3}{10}\) as a decimal.
Shade the model. Use a place value chart.

Example
Write 2\(\frac{8}{10}\) as a decimal.
Shade the model. Use a place value chart.

Show and Grow

Write the fraction or mixed number as a decimal.

Question 1.
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 5
Answer: The representation of \(\frac{5}{10}\) in the place-value chart is:

Explanation:
The given fraction is: \(\frac{5}{10}\)
The given model fro \(\frac{5}{10}\) is:
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 5
The formula for converting fraction to a decimal is Decimal = Numerator ÷ Denominator
So,
The representation of \(\frac{5}{10}\) in the decimal form is: 0.5

Question 2.
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 6

Answer: The representation of 1\(\frac{7}{10}\) in the place-value chart is:

Explanation:
The given mixed fraction is: 1\(\frac{7}{10}\)
So,
The improper fraction of 1\(\frac{7}{10}\) is: \(\frac{17}{10}\)
The given model for 1\(\frac{7}{10}\) is:
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 6
Here,
In 1\(\frac{7}{10}\),
1 represents the one’s place and 7 represents the tenths place
The formula for converting fraction to a decimal is Decimal = Numerator ÷ Denominator
So,
The representation of 1\(\frac{7}{10}\) in the decimal form is: 1.7

Apply and Grow: Practice

Shade the model to represent the fraction or mixed number. Then write the fraction or mixed number as a decimal.

Question 3.
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 7
Answer: The representation of  \(\frac{2}{10}\) in the place- value chart is:

Explanation:
The given fraction is: \(\frac{2}{10}\)
The model representing the \(\frac{2}{10}\) is:

The formula for converting fraction to a decimal is Decimal = Numerator ÷ Denominator
Hence,
The representation of \(\frac{2}{10}\) in the decimal form is: 0.2

Question 4.
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 8
Answer: The representation of 1\(\frac{9}{10}\) in the place-value chart is:

Explanation:
The given mixed fraction is: 1\(\frac{9}{10}\)
So,
The improper fraction of 1\(\frac{9}{10}\) is: \(\frac{19}{10}\)
The model representiong 1\(\frac{9}{10}\) is:

The formula for converting fraction to a decimal is Decimal = Numerator ÷ Denominator
Hence,
The representation of 1\(\frac{9}{10}\) in the decimal form is: 1.9

Write the fraction or mixed number as a decimal.

Question 5.
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 9
Answer: The representation of \(\frac{7}{10}\) in the place-value chart is:

Explanation:
The given fraction is: \(\frac{7}{10}\)
The model representing the \(\frac{7}{10}\) is:

The formula for converting fraction to a decimal is Decimal = Numerator ÷ Denominator
Hence,
The representation of \(\frac{7}{10}\) in the decimal form is: 0.7

Question 6.
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 10
Answer: The representation of \(\frac{4}{10}\) in the place-value chart is:

Explanation:
The given fraction is: \(\frac{4}{10}\)
The model representing the \(\frac{4}{10}\) is:

The formula for converting fraction to a decimal is Decimal = Numerator ÷ Denominator
Hence,
The representation of \(\frac{4}{10}\) in the decimal form is: 0.4

Question 7.
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 11
Answer: The representation of 5\(\frac{1}{10}\) in the place-value chart is:

Explanation:
The given mixed fraction is:5\(\frac{1}{10}\)
So,
The improper fraction of 5\(\frac{1}{10}\) is: \(\frac{51}{10}\)
In 5\(\frac{1}{10}\),
5 represents the ones position and 1 represents the tenths position.
The formula for converting fraction to a decimal is Decimal = Numerator ÷ Denominator
Hence,
The representation of 5\(\frac{1}{10}\) in the decimal form is: 5.1

Question 8.
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 12
Answer: The representation of 24\(\frac{3}{10}\) in the place-value chart is:

Explanation:
The given mixed fraction is: 24\(\frac{3}{10}\)
So,
The improper fraction of 24\(\frac{3}{10}\) is: \(\frac{243}{10}\)
In 24\(\frac{3}{10}\),
2 represents the tens position
4 represents the one’s position
3 represents the tenths position
The formula for converting fraction to a decimal is Decimal = Numerator ÷ Denominator
Hence,
The representation of 24\(\frac{3}{10}\) in the decimal form is: 24.3

Write the number as a fraction or mixed number and as a decimal.

Question 9.
six tenths
Answer: The representation of six tenths in the fraction form is: \(\frac{6}{10}\)
The representation of \(\frac{6}{10}\) in the place-value chart is:

Explanation:
The given fraction is: \(\frac{6}{10}\)
In \(\frac{6}{10}\),
6 represents the tenth’s position
The formula for converting fraction to a decimal is Decimal = Numerator ÷ Denominator
Hence,
The representation of \(\frac{6}{10}\) in the Decimal form is: 0.6

Question 10.
eleven and five-tenths
Answer: The representation of eleven and five-tenths in the fraction form is: 11\(\frac{5}{10}\)
The representation of 11\(\frac{5}{10}\) in the place-value chart is:

Explanation:
The given mixed fraction is: 11\(\frac{5}{10}\)
So,
The improper fraction of 11\(\frac{5}{10}\) is: \(\frac{115}{10}\)
In 11\(\frac{5}{10}\),
1 represents Ten’s and 1 represents the one’s position
5 represents the tenth’s position
The formula for converting fraction to a decimal is Decimal = Numerator ÷ Denominator
Hence,
the representation of 11\(\frac{5}{10}\) in the decimal form is: 11.5

Question 11.
Newton passes 8 out of 10 obedience classes. What portion of the classes does Newton pass? Write your answer as a decimal.
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 13
Answer: The portion of the classes does Newton pass is: 0.8

Explanation:
It is given that Newton passes 8 out of 10 obedience classes.
The representation of the portion of the classes that Newton passed in the fraction form = \(\frac{The number of classes that Newton passed} {The total number of classes}\) = \(\frac{8}{10}\)
So,
The representation of \(\frac{8}{10}\) in the place-value chart is:

In \(\frac{8}{10}\),
8 represents the Tenth’s position.
So, the representation of \(\frac{8}{10}\) in the decimal form is: 0.8
The formula for converting fraction to a decimal is Decimal = Numerator ÷ Denominator
Hence,
We can conclude that the representation of the portion of the classes that Newton passed in the decimal form is: 0.8

Question 12.
You move a game piece around a game board 3\(\frac{2}{10}\) times before you lose a turn. Write this number as a decimal.
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 14.
Answer: The representation of 3\(\frac{2}{10}\) in the decimal form is: 3.2

Explanation:
It is given that you move a game piece around a game board 3\(\frac{2}{10}\) times before you lose a sum.
So,
The representation of 3\(\frac{2}{10}\) in the place-value chart is:

In 3\(\frac{2}{10}\),
3 represents one’s position
2 represents the tenth’s position
The formula for converting fraction to a decimal is Decimal = Numerator ÷ Denominator
Hence,
The representation of 3\(\frac{2}{10}\) in the decimal form is: 3.2

Question 13.
Writing
Do 0.5 and 5.0 have the same value? Explain.
Answer: 0.5 and 5.0 does not have the same value because the place-value of 5 is different in both 0.5 and 5.0

Explanation:
The given numbers are: 0.5 and 5.0
The position of 5 in 0.5 according to the place-value chart is:

The position of 5 in 5.0 according to the place-value chart is:

In 0.5, the position of 5 is: tenth’s position
In 5.0, the position of 5 is: One’s position
The formula for converting fraction to a decimal is Decimal = Numerator ÷ Denominator
Hence, from the above,
We can conclude that 0.5 and 5.0 do not have the same value.

Think and Grow: Modeling Real Life

Example
You have a collection of dinosaur figurines. What portion of the dinosaurs in your collection are carnivores? Write your answer as a decimal.
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 15
Draw a model to represent the collection. Shade the same number of parts as there are carnivore dinosaurs in the collection.
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 15.1
Write the decimal shown by the model.
Answer:
From the above table, 3 of the dinosaurs in the collection are carnivores.
The total number of dinosaurs in the collection is: 10
So,
The representation of carnivores in the collection of dinosaurs is: \(\frac{3}{10}\)

Explanation:
The given table is:
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 15
From the above table,
The total number of dinosaurs are: 10
The number of carnivores in the total number of dinosaurs is: 3
So,
The representation of carnivores in the total number of dinosaurs in the fraction form is: \(\frac{3}{10}\)
Now,
The representation of \(\frac{3}{10}\) in the place-value chart is:

In \(\frac{3}{10}\),
3 represents tenth’s position.
The formula for converting a fraction to a decimal is Decimal = Numerator ÷ Denominator
Hence,
The representation of carnivores from the total number of dinosaurs in the decimal form is: 0.3

Show and Grow

Question 14.
Use the table above. What portion of the dinosaurs in your collection are herbivores? Write your answer as a decimal.
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 16
Answer: The portion of dinosaurs in your collection that are herbivores is: 0.5

Explanation:
The given table is:
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 15
From the above table,
The total number of dinosaurs are: 10
The number of herbivores from the collection of dinosaurs is: 5
So,
The representation of the number of herbivores from the total dinosaurs in the fraction form is: \(\frac{5}{10}\)
Now,
The representation of \(\frac{3}{10}\) in the place-value chart is:

In \(\frac{3}{10}\),
3 represents the tenth’s position
The formula for converting fraction to a decimal is Decimal = Numerator ÷ Denominator
Hence,
The representation of the herbivores from the total number of dinosaurs in the decimal form is: 0.5

Question 15.
DIG DEEPER!
You have 10 apps on your tablet. Six of the apps are games. What portion of the apps on your tablet are not games? Write your answer as a decimal.
Answer: The portion of the apps on your tablet that is not games in the fraction form is: \(\frac{4}{10}\)

Explanation:
It is given that there are 10 apps on your tablet out of which 6 of the apps are games.
So,
The total number of apps on your tablet are: 10
The number of apps that are games on your tablet is: 6
So,
The number of apps that are not games on your tablet is: 4
So,
The representation of the apps that are not games out of the total apps in the fraction form is: \(\frac{4}{10}\)
So,
The representation of \(\frac{4}{10}\) in the place-value chart is:

In \(\frac{4}{10}\),
4 represents the tenth’s position
The formula for converting fraction to a decimal is Decimal = Numerator ÷ Denominator
Hence,
The portion of the apps that are not games out of the total number of apps in the decimal form is: 0.4

Question 16.
DIG DEEPER!
You make 3 pans of lasagna for a party. You cut each pan of lasagna into10 equal pieces. The guests eat 22 pieces. Write the fraction and decimal that represent how many pans of lasagna the guests eat.
Answer: The representation of the pans of lasagna the guests eat in the fraction form is: \(\frac{22}{30}\)

Explanation:
It is given that there are 3 pans of lasagna for a party and each pan of lasagna cut into 10 equal parts.
So,
The total number of pieces of Lasagna = The total number of pans × The number of parts that each Lasagna is cut
= 3 × 10 = 30 pieces
It is also given that the guests eat 22 pieces out of 30 pieces.
So,
The representation of the number of pieces that the guests eat is: \(\frac{22}{30}\)
The representation of \(\frac{22}{30}\) in the place-value chart is:

In \(\frac{22}{30}\),
22 represents the ten’s and one’s position
30 represents the tenth’s position
The formula for converting fraction to a decimal is Decimal = Numerator ÷ Denominator
Hence,
The representation of the number of pieces out of the total number of pieces in the decimal form is: 22.30

Understand Tenths Homework & Practice 10.1

Question 1
Write \(\frac{6}{10}\) as a decimal.
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 17
Answer: The representation of \(\frac{6}{10}\) in the decimal form is: 0.6

Explanation:
The given fraction is: \(\frac{6}{10}\)
The given model for \(\frac{6}{10}\) is:
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 17
The representation of \(\frac{6}{10}\) in the place-value chart is:

In \(\frac{6}{10}\),
6 represents the tenth’s position
The formula for converting fraction to a decimal is Decimal = Numerator ÷ Denominator
Hence,
The representation of \(\frac{6}{10}\) in the decimal form is: 0.6

Question 2.
Shade the model to represent 1\(\frac{8}{10}\). Then write the mixed number as a decimal.
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 18

Write the fraction or mixed number as a decimal.
Answer: The representation of 1\(\frac{8}{10}\) in the decimal form is: 1.8

Explanation:
The given mixed fraction is: 1\(\frac{8}{10}\)
So,
The improper fraction of 1\(\frac{8}{10}\) is: \(\frac{18}{10}\)
The model given for 1\(\frac{8}{10}\) is:

So,
The representation of 1\(\frac{8}{10}\) in the place-value chart is:

In 1\(\frac{8}{10}\),
1 represents the one’s position
8 represents the tenth’s position
The formula for converting fraction to a decimal is Decimal = Numerator ÷ Denominator
Hence,
Th representation of 1\(\frac{8}{10}\) in the decimal form is: 1.8

Question 3.
\(\frac{1}{10}\)
Answer: The representation of \(\frac{1}{10}\) in the decimal form is: 0.1

Explanation:
The given fraction is: \(\frac{1}{10}\)
Now,
The representation of \(\frac{1}{10}\) in the place-value chart is:

In \(\frac{1}{10}\),
1 represents the tenth’s position
The formula for converting fraction to a decimal is Decimal = Numerator ÷ Denominator
Hence,
The representation of \(\frac{1}{10}\) in the decimal form is: 0.1

Question 4.
\(\frac{5}{10}\)
Answer: The representation of \(\frac{5}{10}\) in the decimal form is: 0.5

Explanation:
The given fraction is: \(\frac{5}{10}\)
Now,
The representation of \(\frac{5}{10}\) in the place-value chart is:

In \(\frac{5}{10}\),
5 represents the tenth’s position
The formula for converting fraction to a decimal is Decimal = Numerator ÷ Denominator
Hence,
The representation of \(\frac{5}{10}\) in the decimal form is: 0.5

Question 5.
\(\frac{2}{10}\)
Answer: The representation of \(\frac{2}{10}\) in the decimal form is: 0.2

Explanation:
The given fraction is: \(\frac{2}{10}\)
Now,
The representation of \(\frac{2}{10}\) in the place-value chart is:

In \(\frac{2}{10}\),
2 represents the tenth’s position
The formula for converting fraction to a decimal is Decimal = Numerator ÷ Denominator
Hence,
The representation of \(\frac{2}{10}\) in the decimal form is: 0.2

Question 6.
\(\frac{8}{10}\)
Answer: The representation of \(\frac{8}{10}\) in the decimal form is: 0.8

Explanation:
The given fraction is: \(\frac{8}{10}\)
Now,
The representation of \(\frac{8}{10}\) in the place-value chart is:

In \(\frac{8}{10}\),
8 represents the tenth’s position
The formula for converting fraction to a decimal is Decimal = Numerator ÷ Denominator
Hence,
The representation of \(\frac{8}{10}\) in the decimal form is: 0.8

Question 7.
4\(\frac{3}{10}\)
Answer: The representation of 4\(\frac{3}{10}\) in the decimal form is: 4.3

Explanation:
The given mixed fraction is: 4\(\frac{3}{10}\)
So,
The improper fraction of 4\(\frac{3}{10}\) is: \(\frac{43}{10}\)
Now,
The representation of 4\(\frac{3}{10}\) in the place-value chart is:

In 4\(\frac{3}{10}\),
4 represents the one’s position
3 represents the tenth’s position
The formula for converting fraction to a decimal is Decimal = Numerator ÷ Denominator
Hence,
The representation of 4\(\frac{3}{10}\) in the decimal form is: 4.3

Question 8.
1\(\frac{4}{10}\)
Answer: The representation of 1\(\frac{4}{10}\) in the decimal form is: 1.4

Explanation:
The given mixed fraction is: 1\(\frac{4}{10}\)
So,
The improper fraction of 1\(\frac{4}{10}\) is: \(\frac{14}{10}\)
Now,
The representation of 1\(\frac{4}{10}\) in the place-value chart is:

In 1\(\frac{4}{10}\),
1 represents the one’s position
4 represents the tenth’s position
The formula for converting fraction to a decimal is Decimal = Numerator ÷ Denominator
Hence,
The representation of 1\(\frac{4}{10}\) in the decimal form is: 1.4

Question 9.
31\(\frac{7}{10}\)
Answer: The representation of 31\(\frac{7}{10}\) in the decimal form is: 31.7

Explanation:
The given mixed fraction is: 31\(\frac{7}{10}\)
So,
The improper fraction of 31\(\frac{7}{10}\) is: \(\frac{317}{10}\)
Now,
The representation of 31\(\frac{7}{10}\) in the place-value chart is:

In 31\(\frac{7}{10}\),
3 represents the ten’s position
1 represents the one’s position
7 represents the tenth’s position
The formula for converting fraction to a decimal is Decimal = Numerator ÷ Denominator
Hence,
The representation of 31\(\frac{7}{10}\) in the decimal form is: 31.7

Question 10.
40\(\frac{6}{10}\)
Answer: The representation of 40\(\frac{6}{10}\) in the decimal form is: 40.6

Explanation:
The given mixed fraction is: 40\(\frac{6}{10}\)
So,
The improper fraction of 40\(\frac{6}{10}\) is: \(\frac{406}{10}\)
Now,
The representation of 40\(\frac{6}{10}\) in the place-value chart is:

In 40\(\frac{6}{10}\),
4 represents the ten’s position
0 represents the one’s position
6 represents the tenth’s position
The formula for converting fraction to a decimal is Decimal = Numerator ÷ Denominator
Hence,
The representation of 40\(\frac{6}{10}\) in the decimal form is: 40.6

Write the number as a fraction or mixed number and as a decimal.

Question 11.
three tenths
Answer: The representation of the three-tenths in the decimal form is: 0.3

Explanation:
The given fraction in the word form is: Three-tenths
The representation of the three-tenths in the fraction form is: \(\frac{3}{10}\)
So,
The representation of \(\frac{3}{10}\) in the place-value chart is:

In \(\frac{3}{10}\),
3 represents the tenth’s place
Hence,
The representation of \(\frac{3}{10}\) in the decimal form is: 0.3

Question 12.
fourteen and nine-tenths
Answer: The representation of fourteen and nine-tenths in the decimal form is: 14.9

Explanation:
The given mixed fraction in Word form is: fourteen and nine-tenths
The representation of fourteen and nine-tenths in the fraction form is: 14\(\frac{9}{10}\)
So,
The improper fraction of 14\(\frac{9}{10}\) is: \(\frac{149}{10}\)
So,
The representation of 14\(\frac{9}{10}\) in the place-value chart is:

In 14\(\frac{9}{10}\),
1 represents the ten’s place
4 represents the one’s place
9 represents the tenth’s place
Hence,
The representation of 14\(\frac{9}{10}\) in the decimal form is: 14.9

Question 13.
You knock down 5 out of 10 bowling pins. What portion of the bowling pins do you knockdown? Write your answer as a decimal.
Answer: The portion the bowling pins did you knockdown is: \(\frac{5}{10}\) = 0.6

Explanation:
It is given that you knock down 5 bowling pins out of 10 bowling pins.
So,
The representation of the portion of bowling pins you knock down is: \(\frac{5}{10}\)
so,
The representation of \(\frac{5}{10}\) in the place-value chart is:

In \(\frac{5}{10}\),
5 represents the tenth’s position
Hence,
The representation of the portion of the bowling pins that you knockdown in the decimal form is: \(\frac{5}{10}\) = 0.5

Question 14.
You drive a go-kart around a track 8\(\frac{7}{10}\) times before you spin out. Write this number as a decimal.
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 19

Answer:  The representation of 8\(\frac{7}{10}\) in the decimal form is: 8.7

Explanation:
It is given that you drive a go-kart around a track 8\(\frac{7}{10}\) before you spin out.
So,
The given mixed fraction is: 8\(\frac{7}{10}\)
So,
The improper fraction of 8\(\frac{7}{10}\) is: \(\frac{87}{10}\)
So,
The representation of 8\(\frac{7}{10}\) in the place-value chart is:

In 8\(\frac{7}{10}\),
8 represents the one’s position
7 represents the tenth’s position
Hence,
The representation of 8\(\frac{7}{10}\) in the decimal form is: 8.7

Question 15.
DIG DEEPER!
Which number-cards are represented by the model?
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 20

Use the table.
Answer: The number- cards represented by the model are: 1\(\frac{3}{10}\) and 1.3

Explanation:
The given model is:
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 20
In the given model, the tables are represented by:
A) 1\(\frac{3}{10}\) B) 13 C) 0.13 D) 1.3
In the given model,
Consider the fully colored model as one whole unit.
So,
The total number of colored units is: 10
Now, in another model,
The number of colored parts is: 3
The number of total parts is: 10
So,
The portion of the colored part from the total number of parts is: \(\frac{3}{10}\)
So, by combining the 2 models,
The total number of colored parts = 1 + \(\frac{3}{10}\)
= 1 + 0.3
= 1.3
So,
The representation of the total number of colored parts in the mixed fraction form is: 1\(\frac{3}{10}\)
The representation of the total number of colored parts in the decimal form is: 1.3

Question 16.
Modeling Real Life
A photographer frames her photographs from a safari trip. What portion of the framed photographs are of mammals? Write your answer as a decimal.
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 21
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 21.1

Answer: The portion of the framed photographs that are of mammals in the decimal form is: 0.6

Explanation:
It is given that a photographer frames photos from a safari trip and the framed photos are all categorized according to the table given below:
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 21
So,
From the table,
The total number of framed photographs are: 10
The number of framed photographs that are mammals is: 6
So,
The portion of the mammals from the total number of framed photographs in the fraction form is: \(\frac{6}{10}\)
The representation of the \(\frac{6}{10}\) in the place-value chart is:

In \(\frac{6}{10}\),
6 represents the tenth’s position
Hence,
The portion of the mammals from the  framed photographs in the decimal form is:  0.6

Question 17.
DIG DEEPER!
What portion of the framed not photographs are of mammals? Write your answer as a decimal.
Answer: THe portion of the photographs that are not mammals in the decimal form is: 0.4

Explanation:
The table for all the framed photographs is given below:
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 21
From the table,
The total number of framed photographs are: 10
The number of framed photographs that are not mammals is: 4
So,
The portion of the photographs that are not mammals in the fraction form is: \(\frac{4}{10}\)
The representation of \(\frac{4}{10}\) in the place-value chart is:

In \(\frac{4}{10}\),
4 represents the tenth’s position
Hence,
The number of framed photographs that are not mammals in the decimal form is: 0.4

Review & Refresh

Find the product.

Question 18.
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 22
Answer: 47 × 6 = 282

Explanation:
By using the partial products method,
47 × 6 = ( 40 + 7 ) × 6
= ( 40 × 6 ) + ( 7 × 6 )
= 240 + 42
= 282
Hence, 47 × 6 = 282

Question 19.
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 23
Answer: 961 × 3 = 2,883

Explanation:
By using the partial products method,
961 × 3 = ( 900 + 60 + 1 ) × 3
= ( 900 × 3 ) + ( 60 × 3 ) + ( 1 × 3 )
= 2,700 + 180 + 3
= 2,883
Hence, 961 × 3 = 2,883

Question 20.
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 24
Answer: 2,405 × 8 = 19,240

Explanation:
By using the partial products method,
2,405 × 8 = ( 2,400 + 5 ) × 8
= ( 2,400 × 8 ) + ( 5 × 8 )
= 19,200 + 40
= 19,240
Hence, 2,405 × 8 = 19,240

Lesson 10.2 Understand Hundredths

Explore and Grow

How many pennies have a total value of one dollar? Draw a model.
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 25

One penny is what fraction of one dollar? Write your answer in words and as a fraction.

Answer: The representation of one penny into the dollar in the fraction form is: 0.01

Explanation:
The given note is:
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 25
We know that,
1 Dollar = 100 pennies
So,
The representation of  1 penny into the dollar is: \(\frac{1}{100}\) dollars

Hence, one dollar is equal to 100 pennies and 1 penny is equal to 0.01 dollars

Structure
How is one-tenth related to one hundredth? How do you think you can write \(\frac{1}{100}\) in a place value chart?

Answer: The one-tenth is related to the one-hundredth as: \(\frac{1}{10}\)

Explanation:
We know that,
one-hundredth = \(\frac{1}{10}\) × \(\frac{1}{10}\)
Now,
The representation of \(\frac{1}{100}\) in the place-value chart is:

In \(\frac{1}{100}\),
1 represents the hundredth’s position

Think and Grow: Understand Hundredths

In decimal, the second place to the right of the decimal point is the hundredths place. You can write hundredths as fractions or decimals. A fraction with a denominator of 10 or 100 is called a decimal fraction.
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 26

Show and Grow

Write the fraction or mixed number as a decimal.

Question 1.
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 29
Answer: The representation of \(\frac{45}{100}\) in the decimal form is: 0.45

Explanation:
The given fraction is: \(\frac{45}{100}\)
The model given for \(\frac{45}{100}\) is:
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 29

So,
The representation of \(\frac{45}{100}\) in the place-value chart is:

In \(\frac{45}{100}\),
5 represents the hundredth’s position
4 represents the tenth’s position
The formula for converting fraction to a decimal is Decimal = Numerator ÷ Denominator
Hence,
The representation of \(\frac{45}{100}\) in the decimal form is: 0.45

Question 2.
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 30
Answer: The representation of 1\(\frac{90}{100}\) in the decimal form is: 1.90

Explanation:
The given fraction is: 1\(\frac{90}{100}\)
The model given for 1\(\frac{90}{100}\) is:
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 30

So,
The representation of 1\(\frac{90}{100}\) in the place-value chart is:

In 1\(\frac{90}{100}\),
1 represents the one’s position
0 represents the hundredth’s position
9 represents the tenth’s position
The formula for converting fraction to a decimal is Decimal = Numerator ÷ Denominator
Hence,
The representation of 1\(\frac{90}{100}\) in the decimal form is: 1.90

Apply and Grow: Practice

Shade the model to represent the fraction or mixed number. Then write the fraction or mixed number as a decimal.

Question 3.
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 31
Answer: The representation of \(\frac{98}{100}\) in the decimal form is: 0.98

Explanation:
The given fraction is: \(\frac{98}{100}\)
The model given for \(\frac{98}{100}\) is:
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 31

So,
The representation of \(\frac{98}{100}\) in the place-value chart is:

In \(\frac{98}{100}\),
8 represents the hundredth’s position
9 represents the tenth’s position
The formula for converting fraction to a decimal is Decimal = Numerator ÷ Denominator
Hence,
The representation of \(\frac{98}{100}\) in the decimal form is: 0.98

Question 4.
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 32
Answer:

The representation of 1\(\frac{34}{100}\) in the decimal form is: 1.34

Explanation:
The given mixed fraction is: 1\(\frac{34}{100}\)
So,
The improper fraction of 1\(\frac{34}{100}\) is: \(\frac{134}{100}\)
The model given for 1\(\frac{34}{100}\) is:
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 32

So,
The representation of 1\(\frac{34}{100}\) in the place-value chart is:

In 1\(\frac{34}{100}\),
1 represents the one’s position
4 represents the hundredth’s position
3 represents the tenth’s position
The formula for converting fraction to a decimal is Decimal = Numerator ÷ Denominator
Hence,
The representation of 1\(\frac{34}{100}\) in the decimal form is: 1.34

Write the fraction or mixed number as a decimal.

Question 5.
\(\frac{42}{100}\)
Answer: The representation of \(\frac{42}{100}\) in the form of decimal number is: 0.42

Explanation:
The given fraction is: \(\frac{42}{100}\)
Now,
The representation of \(\frac{42}{100}\) in the place-value chart is:

In \(\frac{42}{100}\),
2 represents the hundredth position
4 represents the tenth position
Hence,
The representation of \(\frac{42}{100}\) in the decimal form is: 0.42

Question 6.
\(\frac{7}{100}\)
Answer: The representation of \(\frac{7}{100}\) in the form of decimal number is: 0.07

Explanation:
The given fraction is: \(\frac{7}{100}\)
Now,
The representation of \(\frac{7}{100}\) in the place-value chart is:

In \(\frac{7}{100}\),
7 represents the hundredth position
0 represents the tenth position
Hence,
The representation of \(\frac{7}{100}\) in the decimal form is: 0.07

Question 7.
4\(\frac{56}{100}\)
Answer: The representation of 4\(\frac{56}{100}\) in the form of a decimal number is: 4.56

Explanation:
The given mixed fraction is: 4\(\frac{56}{100}\)
So,
The improper fraction of 4\(\frac{56}{100}\) is: \(\frac{456}{100}\)
Now,
The representation of 4\(\frac{56}{100}\) in the place-value chart is:

In 4\(\frac{56}{100}\),
4 represents the one’s position
6 represents the hundredth position
5 represents the tenth position
Hence,
The representation of 4\(\frac{56}{100}\) in the decimal form is: 4.56

Question 8.
23\(\frac{9}{100}\)
Answer: The representation of 23\(\frac{9}{100}\) in the form of a decimal number is: 23.09

Explanation:
The given mixed fraction is: 23\(\frac{9}{100}\)
So,
The improper fraction of 23\(\frac{9}{100}\) is: \(\frac{2,309}{100}\)
Now,
The representation of 23\(\frac{9}{100}\) in the place-value chart is:

In 23\(\frac{9}{100}\),
2 represents the ten’s position
3 represents the one’s position
9 represents the hundredth position
0 represents the tenth position
Hence,
The representation of 23\(\frac{9}{100}\) in the decimal form is: 23.09

Write the fraction or mixed number as a decimal.

Question 9.
sixty-one hundredths
Answer:
The representation of sixty-one hundredths in the fraction form is: \(\frac{61}{100}\)
The representation of sixty-one hundredths in the decimal form is: 0.61

Explanation:
The given word form is: Sixty-one hundredths
So,
The representation of sixty-one hundredths in the fraction form is: \(\frac{61}{100}\)
Now,
The representation of \(\frac{61}{100}\) in the place-value chart is:

In \(\frac{61}{100}\),
1 represents the hundredth’s position
6 represents the tenth’s position
Hence,
The representation of \(\frac{61}{100}\) in the decimal form is: 0.61

Question 10.
twelve and eighty-three hundredths
Answer:
The representation of twelve and eighty-three hundredths in the fraction form is: 12\(\frac{83}{100}\)
The representation of twelve and eighty-three hundredths in the decimal form is: 12.83

Explanation:
The given word form is: twelve and eighty-three hundredths
So,
The representation of twelve and eighty-three hundredths in the fraction form is: 12\(\frac{83}{100}\)
Now,
The representation of 12\(\frac{83}{100}\) in the place-value chart is:

In 12\(\frac{83}{100}\),
1 represents the ten’s position
2 represents the one’s position
3 represents the hundredth’s position
8 represents the tenth’s position
Hence,
The representation of 12\(\frac{83}{100}\) in the decimal form is: 12.83

Question 11.
A shelter finds homes for 100 dogs. Five of the dogs are Doberman pinschers. What portion of the dogs are Doberman pinschers? Write your answer as a decimal.
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 33
Answer: The portion of the dogs that are Doberman pinschers in the fraction form is: 0.05

Explanation:
It is given that a shelter finds homes for 100 dogs and out of these 100 dogs, 5 dogs are Doberman pinschers
So,
The portion of the dogs that are Doberman pinschers = \(\frac{The number of dogs that are Doberman pinschers}{The total number of dogs}\)
= \(\frac{5}{100}\)
So,
The representation of \(\frac{5}{100}\) in the place-value chart is:

In \(\frac{5}{100}\),
0 represents the tenth’s position
5 represents the hundredth’s position
Hence,
The portion of the dogs that are Doberman pinschers in the decimal form is: 0.05

Question 12.
An athlete runs 3\(\frac{50}{100}\) lengths of a football field. Write this number as a decimal.
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 33.1
Answer: The representation of 3\(\frac{50}{100}\) in the decimal form is: 3.5

Explanation:
It is given that an athlete runs 3\(\frac{50}{100}\) lengths of a football field.
So,
The representation of 3\(\frac{50}{100}\) in the place-value chart is:

Now,
In 3\(\frac{50}{100}\),
3 represents the one’s place
0 represents the hundredth’s place
5 represents the tenth’s position
Now,
The given mixed fraction is: 3\(\frac{50}{100}\)
So,
The improper fraction of 3\(\frac{50}{100}\) is: \(\frac{350}{100}\)
So,
The representation of 3\(\frac{50}{100}\) in the decimal form is: 3.50

Question 13.
Number Sense
Which number of cards show three-hundredths?
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 34
Answer: From the given cards, \(\frac{3}{100}\) and 0.03 shows the form of three-hundredths

Explanation:
The given number of cards is:
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 34

The given fractions and decimal numbers in the given table are:
A) \(\frac{3}{100}\) B) \(\frac{3}{10}\) C) 300 D) 0.3 E) 0.03
We can observe that
From three-hundredths, we can say that,
The total number of cards are: 100
The portion of cards from 100 cards is: 3
So,
The representation of three-hundredths in the fraction form is: \(\frac{3}{100}\)
Now,
The representation of \(\frac{3}{100}\) in the place-value chart is:

In \(\frac{3}{100}\),
0 represents the tenth’s position
3 represents the hundredth’s position
So,
The representation of \(\frac{3}{100}\) in the decimal form is: 0.03
hence, from the given table,
We can conclude that the number of cards that represent the three-hundredths is: \(\frac{3}{100}\) and 0.03

Think and Grow: Modeling Real Life

Example
You use 51 toothpicks to make a bridge. What portion of the container of toothpicks do you use to make the bridge? Write your answer as a decimal.
Draw a model to represent the container of toothpicks. Shade the same number of parts as the number of toothpicks you use to make the bridge. Write the decimal shown by the model
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 35
Answer:
It is given that you are using 51 toothpicks to make a bridge.
So,
The total number of chopsticks are: 100
The number of chopsticks that are using to make a bridge is: 51
So,
The representation of the portion of the chopsticks that are used to make chopsticks = \(\frac{51}{100}\)
So, from \(\frac{3}{100}\),
We can say that you use 51 of the container of toothpicks to make the bridge.

Show and Grow

Question 14.
A book fair has 100 books. 60 of the books are chapter books. What portion of the books in the book fair are chapter books? Write your answer as a decimal.
Answer: The portion of the chapter books out of the total number of books in the decimal form is: 0.60

Explanation:
It is given that a book fair has 100 books and out of these 100 books, 60 books are chapter books.
So,
The portion of the chapter books out of the total number of books in the fraction form is: \(\frac{60}{100}\)
Now,
The representation of \(\frac{60}{100}\) in the place-value chart is:

From \(\frac{60}{100}\), we can say that
0 represents the hundredth’s position-
6 represents the tenth’s position
So,
The representation of \(\frac{60}{100}\) in the decimal form is: 0.60
Hence, from the above,
We can conclude that the portion of the books that are the chapter books in the decimal form is: 0.60

Question 15.
The model represents the members of a marching band. What portion of the marching band plays a brass instrument? percussion instrument? Write your answers as decimals.
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 36
Answer: The portion of the marching band that plays a brass instrument is: 0.47

Explanation:
It is given that the below model represents the members of a marching band.
The given model is:
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 36
In the given model, there are 3 categories of marching brand
In the given model, each box represents 1 unit.
So,
From the model,
The number of the marching band that plays a brass instrument is: 47
The number of the marching band that plays a woodwind instrument is: 36
The number of the marching band that plays a percussion instrument is: 17
The total number of instruments are: 100
So,
The portion of the marching band that plays a band instrument from the total number of instruments in the fraction form is: \(\frac{47}{100}\)
Hence, from the above,
The portion of the marching band that plays a band instrument from the total number of instruments i the decimal form is: 0.47

Question 16.
DIG DEEPER!
What portion of Earth’s surface is covered by water? Write your answer as a decimal.
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 37
Answer: The portion of Earth’s surface that is covered by water in the fraction form is: 0.71

Explanation:
It is given that about \(\frac{71}{100}\) of earth’s surface is covered by water.
So,
The representation of \(\frac{71}{100}\) in the place-value chart is:

From \(\frac{71}{100}\), we can say that
1 represents the hundredth’s position
7 represents the tenth’s position
Hence,
The portion of water on the Earth’s surface represented in decimal form is: 0.71

Understand Hundredths Homework & Practice 10.2

Write the fraction or mixed number as a decimal.

Question 1.
Write \(\frac{83}{100}\) as a decimal.
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 38
Answer: The representation of \(\frac{83}{100}\) in the decimal form is: 0.98

Explanation:
The given fraction is: \(\frac{83}{100}\)
The model given for \(\frac{83}{100}\) is:
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 38

So,
The representation of \(\frac{83}{100}\) in the place-value chart is:

In \(\frac{83}{100}\),
3 represents the hundredth’s position
8 represents the tenth’s position
The formula for converting fraction to a decimal is Decimal = Numerator ÷ Denominator
Hence,
The representation of \(\frac{83}{100}\) in the decimal form is: 0.83

Question 2.
Shade the model to represent 1\(\frac{65}{100}\). Then write the mixed number as a decimal.
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 39
Answer: The representation of \(\frac{65}{100}\) in the form of a decimal number is: 1.65

Explanation:
The given mixed fraction is: 1\(\frac{65}{100}\)
So,
The improper fraction of 1\(\frac{65}{100}\) is: \(\frac{165}{100}\)
Now,
The representation of 1\(\frac{65}{100}\) in the place-value chart is:

In 1\(\frac{65}{100}\),
1 represents the one’s position
5 represents the hundredth position
6 represents the tenth position
Hence,
The representation of 1\(\frac{65}{100}\) in the decimal form is: 1.65

Question 3.
\(\frac{12}{100}\)
Answer: The representation of \(\frac{12}{100}\) in the decimal form is: 0.12

Explanation:
The given fraction is: \(\frac{12}{100}\)
So,
The representation of \(\frac{12}{100}\) in the place-value chart is:

In \(\frac{12}{100}\),
2 represents the hundredth’s position
1 represents the tenth’s position
The formula for converting fraction to a decimal is Decimal = Numerator ÷ Denominator
Hence,
The representation of \(\frac{12}{100}\) in the decimal form is: 0.12

Question 4.
\(\frac{24}{100}\)
Answer: The representation of \(\frac{24}{100}\) in the decimal form is: 0.24

Explanation:
The given fraction is: \(\frac{24}{100}\)
So,
The representation of \(\frac{24}{100}\) in the place-value chart is:

In \(\frac{24}{100}\),
4 represents the hundredth’s position
2 represents the tenth’s position
The formula for converting fraction to a decimal is Decimal = Numerator ÷ Denominator
Hence,
The representation of \(\frac{24}{100}\) in the decimal form is: 0.24

Question 5.
\(\frac{2}{100}\)
Answer: The representation of \(\frac{2}{100}\) in the decimal form is: 0.02

Explanation:
The given fraction is: \(\frac{2}{100}\)
So,
The representation of \(\frac{2}{100}\) in the place-value chart is:

In \(\frac{2}{100}\),
2 represents the hundredth’s position
0 represents the tenth’s position
The formula for converting fraction to a decimal is Decimal = Numerator ÷ Denominator
Hence,
The representation of \(\frac{2}{100}\) in the decimal form is: 0.02

Question 6.
\(\frac{8}{100}\)
Answer: The representation of \(\frac{8}{100}\) in the decimal form is: 0.08

Explanation:
The given fraction is: \(\frac{8}{100}\)
So,
The representation of \(\frac{8}{100}\) in the place-value chart is:

In \(\frac{8}{100}\),
8 represents the hundredth’s position
0 represents the tenth’s position
The formula for converting fraction to a decimal is Decimal = Numerator ÷ Denominator
Hence,
The representation of \(\frac{8}{100}\) in the decimal form is: 0.08

Question 7.
2\(\frac{59}{100}\)
Answer: The representation of 2\(\frac{59}{100}\) in the form of a decimal number is: 2.59

Explanation:
The given mixed fraction is: 2\(\frac{59}{100}\)
So,
The improper fraction of 2\(\frac{59}{100}\) is: \(\frac{259}{100}\)
Now,
The representation of 2\(\frac{59}{100}\) in the place-value chart is:

In 2\(\frac{59}{100}\),
2 represents the one’s position
9 represents the hundredth position
5 represents the tenth position
Hence,
The representation of 2\(\frac{59}{100}\) in the decimal form is: 2.59

Question 8.
48\(\frac{31}{100}\)
Answer: The representation of 48\(\frac{31}{100}\) in the form of a decimal number is: 48.31

Explanation:
The given mixed fraction is: 48\(\frac{31}{100}\)
So,
The improper fraction of 48\(\frac{31}{100}\) is: \(\frac{4,831}{100}\)
Now,
The representation of 48\(\frac{31}{100}\) in the place-value chart is:

In 48\(\frac{31}{100}\),
4 represents ten’s position
8 represents the one’s position
1 represents the hundredth position
3 represents the tenth position
Hence,
The representation of 48\(\frac{31}{100}\) in the decimal form is: 48.31

Question 9.
6\(\frac{7}{100}\)
Answer: The representation of 6\(\frac{7}{100}\) in the form of a decimal number is: 6.07

Explanation:
The given mixed fraction is: 6\(\frac{7}{100}\)
So,
The improper fraction of 6\(\frac{7}{100}\) is: \(\frac{607}{100}\)
Now,
The representation of 6\(\frac{7}{100}\) in the place-value chart is:

In 6\(\frac{7}{100}\),
6 represents the one’s position
7 represents the hundredth position
0 represents the tenth position
Hence,
The representation of 6\(\frac{7}{100}\) in the decimal form is: 6.07

Question 10.
31\(\frac{6}{100}\)
Answer: The representation of 31\(\frac{6}{100}\) in the form of a decimal number is: 31.06

Explanation:
The given mixed fraction is: 31\(\frac{6}{100}\)
So,
The improper fraction of 31\(\frac{6}{100}\) is: \(\frac{3,106}{100}\)
Now,
The representation of 31\(\frac{6}{100}\) in the place-value chart is:

In 31\(\frac{6}{100}\),
3 represents the ten’s position
1 represents the one’s position
6 represents the hundredth position
0 represents the tenth position
Hence,
The representation of 31\(\frac{6}{100}\) in the decimal form is: 31.06

Write the number as a fraction or mixed number and as a decimal.

Question 11.
seventy-four hundredths
Answer:
The representation of seventy-four hundredths in the fraction form is: \(\frac{74}{100}\)
The representation of seventy-four hundredths in the decimal form is: 0.74

Explanation:
The given word form is: Seventy-four hundredths
So,
The representation of seventy-four hundredths in the fraction form is: \(\frac{74}{100}\)
Now,
The representation of \(\frac{74}{100}\) in the place-value chart is:

In \(\frac{74}{100}\),
4 represents the hundredth’s position
7 represents the tenth’s position
Hence,
The representation of \(\frac{74}{100}\) in the decimal form is: 0.74

Question 12.
sixteen and thirty-one hundredths
Answer:
The representation of sixteen and thirty-one hundredths in the fraction form is: 16\(\frac{31}{100}\)
The representation of sixteen and thirty-one hundredths in the decimal form is: 16.31

Explanation:
The given word form is: sixteen and thirty-one hundredths
So,
The representation of sixteen and thirty-one hundredths in the fraction form is: 16\(\frac{31}{100}\)
Now,
The representation of 16\(\frac{31}{100}\) in the place-value chart is:

In 16\(\frac{31}{100}\),
1 represents the ten’s position
6 represents the one’s position
1 represents the hundredth’s position
3 represents the tenth’s position
Hence,
The representation of 16\(\frac{31}{100}\) in the decimal form is: 16.31

Write the value of the underlined digit.

Question 13.
5.84
Answer: The value of 8 in 5.84 is: 0.8

Explanation:
The given number is: 5.84
The representation of 5.84 in the place-value chart is:

In 5.83, from the place-value chart,
We can say that the value of 8 in 5.83 is: 0.8

Question 14.
21.03
Answer: The value of 2 in 21.03 is: 20

Explanation:
The given number is: 21.03
The representation of 21.03 in the place-value chart is:

From the above place-value chart,
We can say that the value of 2 in 21.03 is: 20

Question 15.
67.32
Answer: The value of 2 in 67.32 is: 0.02

Explanation:
The given number is: 67.32
The representation of 67.32 in the place-value chart is:

From the above place-value chart,
We can say that the value of 2 in 67.32 is: 0.02

Question 16.
506.19
Answer: The value of 5 in 506.19 is: 500

Explanation:
The given number is: 506.19
The representation of 506.19 in the place-value chart is:

From the above place-value chart,
We can say that the value of 5 in 506.19 is: 500

Question 17.
A clown has 100 balloons. She uses 56 of the balloons to make animals. What portion of the balloons does she use? Write your answer as a decimal.
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 39.1
Answer: The portion of the balloons she used to make animals in the decimal form is: 0.56

Explanation:
It is given that a clown has 100 balloons and she uses 56 of the balloons to make animals.
So,
The total number of balloons are: 100
The number of ballons that are used to make animals is: 56
So,
The portion of the balloons that are used to make animals is:\(\frac{56}{100}\)
So,
The representation of \(\frac{56}{100}\) in the place-value chart is:

Hence,
We can conclude that the portion of the balloons that are used to make animals in the decimal form is: 0.56

Question 18.
You fill a beaker 4\(\frac{35}{100}\) times for an experiment. Write this number as a decimal.
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 40
Answer: The representation of 4\(\frac{35}{100}\) in the form of a decimal number is: 4.35

Explanation:
The given mixed fraction is: 4\(\frac{35}{100}\)
So,
The improper fraction of 4\(\frac{35}{100}\) is: \(\frac{435}{100}\)
Now,
The representation of 4\(\frac{35}{100}\) in the place-value chart is:

In 4\(\frac{35}{100}\),
4 represents the one’s position
5 represents the hundredth position
3 represents the tenth position
Hence,
The representation of 4\(\frac{35}{100}\) in the decimal form is: 4.35

Question 19.
YOU BE THE TEACHER
Descartes writes 2\(\frac{40}{100}\) as 2.04. Is he correct? Explain.
Answer:
The representation of 2\(\frac{40}{100}\) in the form of a decimal number is: 2.40
So, Descartes is wrong.

Explanation:
The given mixed fraction is: 2\(\frac{40}{100}\)
So,
The improper fraction of 2\(\frac{40}{100}\) is: \(\frac{240}{100}\)
Now,
The representation of 2\(\frac{40}{100}\) in the place-value chart is:

In 2\(\frac{40}{100}\),
2 represents the one’s position
0 represents the hundredth position
4 represents the tenth position
So,
The representation of 2\(\frac{40}{100}\) in the decimal form is: 2.40
But, according to Descartes,
The representation of 2\(\frac{40}{100}\) in the decimal form is: 2.04
Hence, from the above,
We can conclude that Descartes is not correct.

Question 20.
DIG DEEPER!
Shade each model to show 0.6 and 0.60. What do you notice?
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 41
Answer: From the values of 0.6 and 0.60, we can notice that the value of 6 in both numbers are the same.

Explanation:
The given numbers are: 0.6 and 0.60
The given models of 0.6 and 0.60 are:
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 41
Now,
The shaded models of 0.6 and 0.60 will be:

Now,
The representation of 0.6 in the place-value chart is:

The representation of 0.60 in the place-value chart is:

Hence,
From the above 2 place-value charts,
We can conclude that 0.6 and 0.60 are the same as the value of 6 in 0.6 and 0.60 is the same.

Question 21.
Modeling Real Life
You work on the puzzle shown. You connect 78 of the puzzle pieces. What portion of the puzzle have you completed? Write your answer as a decimal.
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 42
Answer: The portion of the puzzles you have completed in the decimal form is: 0.78

Explanation:
It is given that you are working on a 100-box puzzle and you connected 78 of the puzzle pieces.
So,
The total number of puzzles are: 100
The number of puzzles that are connected is: 78
So,
The portion of the puzzles that are connected in the fraction form is: \(\frac{78}{100}\)
Now,
The representation of \(\frac{78}{100}\) in the place-value chart is:

Hence, from the above,
We can conclude that the portion of the puzzles that are connected in the decimal form is: 0.78

Review & Refresh

Divide. Then check your answer.

Question 22.
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 43
Answer: 1,308 ÷ 5 = 261 R 3

Explanation:
By using the partial quotients method,
1,308 ÷ 5 = ( 1,000 + 300 + 5 ) ÷ 5
= ( 1,000 ÷ 5 ) + ( 300 ÷ 5 ) + ( 5 ÷ 5 )
= 200 + 60 + 1
= 261 R 3
Hence, 1,308 ÷ 5 = 261 R 3

Question 23.
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 44
Answer: 67 ÷ 4 = 16 R 3

Explanation:
By using the partial quotients method,
67 ÷ 4 = ( 56 + 8 ) ÷ 4
= ( 56 ÷ 4 ) + ( 8 ÷ 4 )
= 14 + 2
= 16 R 3
Henec, 67 ÷ 4 = 1 R 3

Question 24.
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 45
Answer: 725 ÷ 2 = 362 R 1

Explanation:
By using the partial quotients method,
725 ÷ 2 = ( 720 + 4 ) ÷ 2
= ( 720 ÷ 2 ) + ( 4 ÷ 2 )
= 360 + 2
= 362 R 1
Hence,
75 ÷ 2 = 362 R 1

Lesson 10.3 Fractions and Decimals

Explore and Grow

Plot each fraction or decimal on a number line.

Reasoning
What do you notice about the locations of the points? What can you conclude about the numbers?

Answer: From the above plots, we can see that the number of lines between any two numbers is the same whether the given number is in fraction form or in the decimal form
So, from the above plots, we can conclude that the location of the numbers and the gap between the two numbers is the same.

Think and Grow: Fractions and Decimals

Example
Write \(\frac{6}{10}\) as hundredths in fraction form and decimal form.
Fraction form: Shade the model to help write \(\frac{6}{10}\) as an equivalent fraction with a denominator of 100.

Two or more decimals that have the same value are equivalent decimals.
Example
Write 0.40 as tenths in decimal form and fraction form.

Show and Grow

Question 1.
Write \(\frac{9}{10}\) as hundredths in fraction form and decimal form.
Answer:
The representation of \(\frac{9}{10}\) as hundredths in the fraction form is: \(\frac{90}{100}\)
The representation of \(\frac{90}{100}\) in the decimal form is: 0.90

Explanation:
The given fraction is: \(\frac{9}{10}\)
So, to write \(\frac{9}{10}\) as hundredths, multiply the fraction and numerator of \(\frac{9}{10}\) with 10.
So,
Firstly the numerators 9 and 10 are multiplied and then the denominators 10 and 10 are multiplied
We know that,
Decimal = Numerator ÷ Denominator
So,
The representation of \(\frac{90}{100}\) in the place-value chart is:

Hence,
The representation of \(\frac{9}{10}\) as hundredths in the fraction form is: \(\frac{90}{100}\)
The representation of \(\frac{90}{100}\) in the decimal form is: 0.90

Question 2.
Write 0.20 as tenths in decimal form and fraction form.
Answer:
The representation of 0.20 as tenths in the fraction form is: \(\frac{2}{10}\)
The representation of \(\frac{2}{10}\) in the decimal form is: 0.2

Explanation:
The given fraction is: \(\frac{20}{100}\)
So, to write \(\frac{20}{100}\) as tenths, divide the fraction and numerator of \(\frac{20}{100}\) with 10.
So,
Firstly the numerators 20 and 10 are divided and then the denominators 100 and 10 are divided
So,
\(\frac{20}{100}=\frac{20 \div 10}{100 \div 10}=\frac{2}{10}\)
We know that,
Decimal = Numerator ÷ Denominator
So,
The representation of \(\frac{2}{10}\) in the place-value chart is:

Hence,
The representation of \(\frac{20}{100}\) as tenths in the fraction form is: \(\frac{2}{10}\)
The representation of \(\frac{2}{10}\) in the decimal form is: 0.2

Apply and Grow: Practice

Write the number as tenths in fraction form and decimal form.

Question 3.
\(\frac{80}{100}\)
Answer:
The representation of \(\frac{80}{100}\) as tenths in the fraction form is: \(\frac{8}{10}\)
The representation of \(\frac{8}{10}\) in the decimal form is: 0.8

Explanation:
The given fraction is: \(\frac{80}{100}\)
So, to write \(\frac{80}{100}\) as tenths, divide the fraction and numerator of \(\frac{80}{100}\) with 10.
So,
Firstly the numerators 80 and 10 are divided and then the denominators 100 and 10 are divided
So,
\(\frac{80}{100}=\frac{80 \div 10}{100 \div 10}=\frac{8}{10}\)
We know that,
Decimal = Numerator ÷ Denominator
So,
The representation of \(\frac{80}{100}\) in the place-value chart is:

Hence,
The representation of \(\frac{80}{100}\) as tenths in the fraction form is: \(\frac{8}{10}\)
The representation of \(\frac{8}{10}\) in the decimal form is: 0.8

Question 4.
\(\frac{50}{100}\)
Answer:
The representation of \(\frac{50}{100}\) as tenths in the fraction form is: \(\frac{5}{10}\)
The representation of \(\frac{5}{10}\) in the decimal form is: 0.5

Explanation:
The given fraction is: \(\frac{50}{100}\)
So, to write \(\frac{50}{100}\) as tenths, divide the fraction and numerator of \(\frac{50}{100}\) with 10.
So,
Firstly the numerators 50 and 10 are divided and then the denominators 100 and 10 are divided
So,
\(\frac{50}{100}=\frac{50 \div 10}{100 \div 10}=\frac{5}{10}\)
We know that,
Decimal = Numerator ÷ Denominator
So,
The representation of \(\frac{50}{100}\) in the place-value chart is:

Hence,
The representation of \(\frac{50}{100}\) as tenths in the fraction form is: \(\frac{5}{10}\)
The representation of \(\frac{5}{10}\) in the decimal form is: 0.5

Question 5.
0.30
Answer:
The representation of 0.30 as tenths in the fraction form is: \(\frac{3}{10}\)
The representation of \(\frac{3}{10}\) in the decimal form is: 0.3

Explanation:
The given decimal number is: 0.30
So,
The representation of 0.30 in the fraction form is: \(\frac{30}{100}\)
So, to write \(\frac{30}{100}\) as tenths, divide the fraction and numerator of \(\frac{30}{100}\) with 10.
So,
Firstly the numerators 30 and 10 are divided and then the denominators 100 and 10 are divided
So,
\(\frac{30}{100}=\frac{30 \div 10}{100 \div 10}=\frac{3}{10}\)
We know that,
Decimal = Numerator ÷ Denominator
So,
The representation of \(\frac{3}{10}\) in the place-value chart is:

Hence,
The representation of \(\frac{30}{100}\) as tenths in the fraction form is: \(\frac{3}{10}\)
The representation of \(\frac{3}{10}\) in the decimal form is: 0.3

Write the number as hundredths in fraction form and decimal form.

Question 6.
\(\frac{2}{10}\)
Answer:
The representation of \(\frac{2}{10}\) as hundredths in the fraction form is: \(\frac{20}{100}\)
The representation of \(\frac{20}{100}\) in the decimal form is: 0.20

Explanation:
The given fraction is: \(\frac{2}{10}\)
So, to write \(\frac{2}{10}\) as hundredths, multiply the fraction and numerator of \(\frac{2}{10}\) with 10.
So,
Firstly the numerators 2 and 10 are multiplied and then the denominators 10 and 10 are multiplied
We know that,
Decimal = Numerator ÷ Denominator
So,
The representation of \(\frac{20}{100}\) in the place-value chart is:

Hence,
The representation of \(\frac{2}{10}\) as hundredths in the fraction form is: \(\frac{20}{100}\)
The representation of \(\frac{20}{100}\) in the decimal form is: 0.20

Question 7.
0.7
Answer:
The representation of 0.7 as hundredths in the fraction form is: \(\frac{70}{100}\)
The representation of \(\frac{70}{100}\) in the decimal form is: 0.70

Explanation:
The given decimal number is: 0.7
So,
The representation of 0.7 in the fraction form is: \(\frac{7}{10}\)
So, to write \(\frac{7}{10}\) as hundredths, multiply the fraction and numerator of \(\frac{7}{10}\) with 10.
So,
Firstly the numerators 7 and 10 are multiplied and then the denominators 10 and 10 are multiplied
We know that,
Decimal = Numerator ÷ Denominator
So,
The representation of \(\frac{70}{100}\) in the place-value chart is:

Hence,
The representation of \(\frac{7}{10}\) as hundredths in the fraction form is: \(\frac{70}{100}\)
The representation of \(\frac{70}{100}\) in the decimal form is: 0.70

Question 8.
2\(\frac{1}{10}\)
Answer:
The representation of 2\(\frac{1}{10}\) as hundredths in the fraction form is: 2\(\frac{10}{100}\)
The representation of 2\(\frac{10}{100}\) in the decimal form is: 2.10

Explanation:
The given mixed fraction is: 2\(\frac{1}{10}\)
So,
To write 2\(\frac{1}{10}\) as hundredths, multiply the fraction and numerator of 2\(\frac{1}{10}\) with 10.
So,
Firstly the numerators 1 and 10 are multiplied and then the denominators 10 and 10 are multiplied
We know that,
Decimal = Numerator ÷ Denominator
So,
The representation of 2\(\frac{10}{100}\) in the place-value chart is:

Hence,
The representation of 2\(\frac{1}{10}\) as hundredths in the fraction form is: 2\(\frac{10}{100}\)
The representation of 2\(\frac{10}{100}\) in the decimal form is: 2.10

Write the number represented by the point as hundredths in fraction form and decimal form.

Answer:
From the above number line,
‘A’ is: 0.4
‘B’ is: 0.9
‘C’ is: 1.5

Question 9.
A
Answer:
From the above number line, ‘A’ is: 0.4
The representation of 0.4 as hundredths in the fraction form is: \(\frac{40}{100}\)
The representation of \(\frac{40}{100}\) in the decimal form is: 0.40

Explanation:
The given decimal number is: 0.4
So,
The representation of 0.4 in the fraction form is: \(\frac{4}{10}\)
So,
To write \(\frac{4}{10}\) as hundredths, multiply the fraction and numerator of \(\frac{4}{10}\) with 10.
So,
Firstly the numerators 4 and 10 are multiplied and then the denominators 10 and 10 are multiplied
We know that,
Decimal = Numerator ÷ Denominator
So,
The representation of \(\frac{40}{100}\) in the place-value chart is:

Hence,
The representation of \(\frac{4}{10}\) as hundredths in the fraction form is: \(\frac{40}{100}\)
The representation of \(\frac{40}{100}\) in the decimal form is: 0.40

Question 10.
B
Answer:
From the above number line, ‘B’ is: 0.9
The representation of 0.9 as hundredths in the fraction form is: \(\frac{90}{100}\)
The representation of \(\frac{90}{100}\) in the decimal form is: 0.90

Explanation:
The given decimal number is: 0.9
So,
The representation of 0.9 in the fraction form is: \(\frac{9}{10}\)
So,
To write \(\frac{9}{10}\) as hundredths, multiply the fraction and numerator of \(\frac{9}{10}\) with 10.
So,
Firstly the numerators 9 and 10 are multiplied and then the denominators 10 and 10 are multiplied
We know that,
Decimal = Numerator ÷ Denominator
So,
The representation of \(\frac{90}{100}\) in the place-value chart is:

Hence,
The representation of \(\frac{9}{10}\) as hundredths in the fraction form is: \(\frac{90}{100}\)
The representation of \(\frac{90}{100}\) in the decimal form is: 0.90

Question 11.
C
Answer:
From the above number line, ‘C’ is: 1.5
The representation of 1.5 as hundredths in the fraction form is: \(\frac{150}{100}\)
The representation of \(\frac{150}{100}\) in the decimal form is: 1.50

Explanation:
The given decimal number is: 1.5
So,
The representation of 1.5 in the fraction form is: \(\frac{15}{10}\)
So,
To write \(\frac{15}{10}\) as hundredths, multiply the fraction and numerator of \(\frac{15}{10}\) with 10.
So,
Firstly the numerators 15 and 10 are multiplied and then the denominators 10 and 10 are multiplied
We know that,
Decimal = Numerator ÷ Denominator
So,
The representation of \(\frac{150}{100}\) in the place-value chart is:

Hence,
The representation of \(\frac{15}{10}\) as hundredths in the fraction form is: \(\frac{150}{100}\)
The representation of \(\frac{150}{100}\) in the decimal form is: 1.50

Question 12.
DIG DEEPER!
Complete the table. Think: Can all of the numbers in the table be written as hundredths?

Answer: Yes, all the numbers in the table can be written as hundredths.

Explanation:
A) The given mixed fraction is: 23\(\frac{7}{10}\)
So,
The representation of 23\(\frac{7}{10}\) in the fraction form is: \(\frac{237}{10}\)
So,
To write 23\(\frac{7}{10}\) as hundredths, multiply the fraction and numerator of 23\(\frac{7}{10}\) with 10.
So,
Firstly the numerators 7 and 10 are multiplied and then the denominators 10 and 10 are multiplied
We know that,
Decimal = Numerator ÷ Denominator
So,
The representation of 23\(\frac{70}{100}\) in the place-value chart is:

Hence,
The representation of 23\(\frac{7}{10}\) as hundredths in the fraction form is: 23\(\frac{70}{100}\)
The representation of 23\(\frac{70}{100}\) in the decimal form is: 23.70

B) The given mixed fraction is: 18\(\frac{2}{10}\)
So,
The representation of 18\(\frac{2}{10}\) in the fraction form is: \(\frac{182}{10}\)
So,
To write 18\(\frac{2}{10}\) as hundredths, multiply the fraction and numerator of 18\(\frac{2}{10}\) with 10.
So,
Firstly the numerators 2 and 10 are multiplied and then the denominators 10 and 10 are multiplied
We know that,
Decimal = Numerator ÷ Denominator
So,
The representation of 18\(\frac{20}{100}\) in the place-value chart is:

Hence,
The representation of 18\(\frac{2}{10}\) as hundredths in the fraction form is: 18\(\frac{20}{100}\)
The representation of 18\(\frac{20}{100}\) in the decimal form is: 18.20
Hence, in the same way, C) and D) Expressions can also be written as hundredths and in the decimal form

Think and Grow: Modeling Real Life

Example
You use 100 tiles to make a mosaic. 80 of them are square tiles. Your friend uses 10 tiles to make a mosaic. Six of them are square tiles. Do the mosaics have the same fraction of square tiles?
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 53

Determine whether the fractions are equivalent.
Write your friend’s fraction as hundredths in fraction form. Then compare.

Hence,
The mosaics don’t have the same fraction of square tiles.

Show and Grow

Question 13.
You use 10 beads to make a bracelet. Seven of them are purple. Your friend uses 100 beads to make a bracelet. 70 of them are purple. Do the bracelets have the same fraction of purple beads?
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 56
Answer: Yes, the bracelets have the same fraction of purple beads.

Explanation:
It is given that you use 10 beads to make a bracelet and out of 10 beads, seven of them are purple.
So,
The portion of the bracelet that is purple is: \(\frac{7}{10}\)
So,
The given fraction is: \(\frac{7}{10}\)
So,
To write \(\frac{7}{10}\) as hundredths, multiply the fraction and numerator of \(\frac{7}{10}\) with 10.
So,
Firstly the numerators 7 and 10 are multiplied and then the denominators 10 and 10 are multiplied
We know that,
Decimal = Numerator ÷ Denominator
So,
The representation of \(\frac{70}{100}\) in the place-value chart is:

Hence,
The representation of \(\frac{7}{10}\) as hundredths in the fraction form is: \(\frac{70}{100}\)
It is also given that your friend uses 100 beads to make a bracelet and out of 100, 70 beads are purple.
So,
The portion of purple beads from the total number of beads is: \(\frac{70}{100}\)
Hence, from the above,
We can conclude that the bracelet has the same fraction of purple beads
Question 14.
DIG DEEPER!
The model represents the types of trees on a tree farm. What portion of the tree farm is blue spruce? Fraser fir? white pine? Write your answers as decimals in tenths.
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 57
Answer:
The portion of the tree farm that is blue spruce is: 0.4
The portion of the tree farm that is Fraser fir is: 0.4
The portion of the tree farm that is White pine is: 0.2

Explanation:
It is given that the below model represents the types of trees on a tree farm.
The given model is:
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 57
In the given model, there are 3 categories of trees
In the given model, each box represents 1 unit.
So,
From the model,
The number of trees of blue spruce is: 40
The number of trees of Frazer fir is: 40
The number of trees of White pine is: 20
The total number of trees in a tree farm is: 100
So,
The portion of the tree farm that is blue spruce from the total number of instruments in the fraction form is: \(\frac{40}{100}\)
The portion of the tree farm that is Frazer fir from the total number of instruments in the fraction form is: \(\frac{40}{100}\)
The portion of the tree farm that is White pine from the total number of instruments in the fraction form is: \(\frac{20}{100}\)
Now, for the Blue spruce,
So,
To write \(\frac{40}{100}\) as tenths, divide the fraction and numerator of \(\frac{40}{100}\) with 10.
So,
Firstly the numerators 40 and 10 are divided and then the denominators 100 and 10 are divided
So,
\(\frac{40}{100}=\frac{40 \div 10}{100 \div 10}=\frac{4}{10}\)
We know that,
Decimal = Numerator ÷ Denominator
So,
The representation of \(\frac{4}{10}\) in the place-value chart is:

Hence,
The representation of \(\frac{4}{10}\) as tenths in the decimal form is: 0.4
Now, for the Frazer fir,
So,
To write \(\frac{40}{100}\) as tenths, divide the fraction and numerator of \(\frac{40}{100}\) with 10.
So,
Firstly the numerators 40 and 10 are divided and then the denominators 100 and 10 are divided
So,
\(\frac{40}{100}=\frac{40 \div 10}{100 \div 10}=\frac{4}{10}\)
We know that,
Decimal = Numerator ÷ Denominator
So,
The representation of \(\frac{4}{10}\) in the place-value chart is:

Hence,
The representation of \(\frac{4}{10}\) as tenths in the decimal form is: 0.4
Now, for the White pine,
So,
To write \(\frac{20}{100}\) as tenths, divide the fraction and numerator of \(\frac{20}{100}\) with 10.
So,
Firstly the numerators 20 and 10 are divided and then the denominators 100 and 10 are divided
So,
\(\frac{20}{100}=\frac{20 \div 10}{100 \div 10}=\frac{4}{10}\)
We know that,
Decimal = Numerator ÷ Denominator
So,
The representation of \(\frac{2}{10}\) in the place-value chart is:

Hence,
The representation of \(\frac{2}{10}\) as tenths in the decimal form is: 0.2
Hence, from the above,
We can conclude that
The portion of the tree farm that is blue spruce is: 0.4
The portion of the tree farm that is Fraser fir is: 0.4
The portion of the tree farm that is White pine is: 0.2

Fractions and Decimals Homework & Practice 10.3

Write the number as tenths in fraction form and decimal form.

Question 1.
\(\frac{40}{100}\)
Answer:
The representation of \(\frac{40}{100}\) as tenths in the fraction form is: \(\frac{4}{10}\)
The representation of \(\frac{4}{10}\) in the decimal form is: 0.4

Explanation:
The given fraction is: \(\frac{40}{100}\)
So, to write \(\frac{40}{100}\) as tenths, divide the fraction and numerator of \(\frac{40}{100}\) with 10.
So,
Firstly the numerators 40 and 10 are divided and then the denominators 100 and 10 are divided
So,
\(\frac{40}{100}=\frac{40 \div 10}{100 \div 10}=\frac{5}{10}\)
We know that,
Decimal = Numerator ÷ Denominator
So,
The representation of \(\frac{4}{10}\) in the place-value chart is:

Hence,
The representation of \(\frac{40}{100}\) as tenths in the fraction form is: \(\frac{4}{10}\)
The representation of \(\frac{4}{10}\) in the decimal form is: 0.4

Question 2.
\(\frac{70}{100}\)
Answer:
The representation of \(\frac{70}{100}\) as tenths in the fraction form is: \(\frac{7}{10}\)
The representation of \(\frac{7}{10}\) in the decimal form is: 0.7

Explanation:
The given fraction is: \(\frac{70}{100}\)
So, to write \(\frac{70}{100}\) as tenths, divide the fraction and numerator of \(\frac{70}{100}\) with 10.
So,
Firstly the numerators 70 and 10 are divided and then the denominators 100 and 10 are divided
So,
\(\frac{70}{100}=\frac{70 \div 10}{100 \div 10}=\frac{7}{10}\)
We know that,
Decimal = Numerator ÷ Denominator
So,
The representation of \(\frac{7}{10}\) in the place-value chart is:

Hence,
The representation of \(\frac{70}{100}\) as tenths in the fraction form is: \(\frac{7}{10}\)
The representation of \(\frac{7}{10}\) in the decimal form is: 0.7

Question 3.
0.20
Answer:
The representation of 2.20 as tenths in the fraction form is: \(\frac{2}{10}\)
The representation of \(\frac{2}{10}\) in the decimal form is: 0.2

Explanation:
The given decimal number is: 0.20
So,
the representation of 0.20 in the fraction orm is: \(\frac{20}{100}\)
So, to write \(\frac{20}{100}\) as tenths, divide the fraction and numerator of \(\frac{20}{100}\) with 10.
So,
Firstly the numerators 20 and 10 are divided and then the denominators 100 and 10 are divided
So,
\(\frac{20}{100}=\frac{20 \div 10}{100 \div 10}=\frac{2}{10}\)
We know that,
Decimal = Numerator ÷ Denominator
So,
The representation of \(\frac{2}{10}\) in the place-value chart is:

Hence,
The representation of \(\frac{20}{100}\) as tenths in the fraction form is: \(\frac{2}{10}\)
The representation of \(\frac{2}{10}\) in the decimal form is: 0.2

Write the number as hundredths in fraction form and decimal form

Question 4.
\(\frac{8}{10}\)
Answer:
The representation of \(\frac{8}{10}\) as hundredths in the fraction form is: \(\frac{80}{100}\)
The representation of \(\frac{80}{100}\) in the decimal form is: 0.80

Explanation:
The given fraction is: \(\frac{8}{10}\)
So, to write \(\frac{8}{10}\) as hundredths, multiply the fraction and numerator of \(\frac{8}{10}\) with 10.
So,
Firstly the numerators 8 and 10 are multiplied and then the denominators 10 and 10 are multiplied
We know that,
Decimal = Numerator ÷ Denominator
So,
The representation of \(\frac{80}{100}\) in the place-value chart is:

Hence,
The representation of \(\frac{8}{10}\) as hundredths in the fraction form is: \(\frac{80}{100}\)
The representation of \(\frac{80}{100}\) in the decimal form is: 0.80

Question 5.
0.5
Answer:
The representation of 0.5 as hundredths in the fraction form is: \(\frac{50}{100}\)
The representation of \(\frac{50}{100}\) in the decimal form is: 0.50

Explanation:
The given decimal number is: 0.5
So,
The representation of 0.5 in the fraction form is: \(\frac{5}{10}\)
So, to write \(\frac{5}{10}\) as hundredths, multiply the fraction and numerator of \(\frac{5}{10}\) with 10.
So,
Firstly the numerators 5 and 10 are multiplied and then the denominators 10 and 10 are multiplied
We know that,
Decimal = Numerator ÷ Denominator
So,
The representation of \(\frac{50}{100}\) in the place-value chart is:

Hence,
The representation of \(\frac{5}{10}\) as hundredths in the fraction form is: \(\frac{50}{100}\)
The representation of \(\frac{50}{100}\) in the decimal form is: 0.50

Question 6.
9\(\frac{6}{10}\)
Answer:
The representation of 9\(\frac{6}{10}\) as hundredths in the fraction form is: 9\(\frac{60}{100}\)
The representation of 9\(\frac{60}{100}\) in the decimal form is: 9.60

Explanation:
The given mixed fraction is: 9\(\frac{6}{10}\)
So,
To write 9\(\frac{6}{10}\) as hundredths, multiply the fraction and numerator of 9\(\frac{6}{10}\) with 10.
So,
Firstly the numerators 6 and 10 are multiplied and then the denominators 10 and 10 are multiplied
We know that,
Decimal = Numerator ÷ Denominator
So,
The representation of 9\(\frac{60}{100}\) in the place-value chart is:

Hence,
The representation of 9\(\frac{6}{10}\) as hundredths in the fraction form is: 9\(\frac{60}{100}\)
The representation of 9\(\frac{60}{100}\) in the decimal form is: 9.60

Write the number represented by the point as hundredths in fraction form and decimal form.

Answer:
From the above number line,
‘A’ value is: 0.3
‘B’ value is: 0.6
‘c’ value is: 1.8

Question 7.
A
Answer:
From the above number line, ‘A’ is: 0.3
The representation of 0.3 as hundredths in the fraction form is: \(\frac{30}{100}\)
The representation of \(\frac{30}{100}\) in the decimal form is: 0.30

Explanation:
The given decimal number is: 0.3
So,
The representation of 0.3 in the fraction form is: \(\frac{3}{10}\)
So,
To write \(\frac{3}{10}\) as hundredths, multiply the fraction and numerator of \(\frac{3}{10}\) with 10.
So,
Firstly the numerators 3 and 10 are multiplied and then the denominators 10 and 10 are multiplied
We know that,
Decimal = Numerator ÷ Denominator
So,
The representation of \(\frac{30}{100}\) in the place-value chart is:

Hence,
The representation of \(\frac{3}{10}\) as hundredths in the fraction form is: \(\frac{30}{100}\)
The representation of \(\frac{30}{100}\) in the decimal form is: 0.30

Question 8.
B
Answer:
From the above number line, ‘A’ is: 0.6
The representation of 0.6 as hundredths in the fraction form is: \(\frac{60}{100}\)
The representation of \(\frac{60}{100}\) in the decimal form is: 0.60

Explanation:
The given decimal number is: 0.6
So,
The representation of 0.6 in the fraction form is: \(\frac{6}{10}\)
So,
To write \(\frac{6}{10}\) as hundredths, multiply the fraction and numerator of \(\frac{6}{10}\) with 10.
So,
Firstly the numerators 6 and 10 are multiplied and then the denominators 10 and 10 are multiplied
We know that,
Decimal = Numerator ÷ Denominator
So,
The representation of \(\frac{60}{100}\) in the place-value chart is:

Hence,
The representation of \(\frac{6}{10}\) as hundredths in the fraction form is: \(\frac{60}{100}\)
The representation of \(\frac{60}{100}\) in the decimal form is: 0.60

Question 9.
C
Answer:
From the above number line, ‘A’ is: 1.8
The representation of 1.8 as hundredths in the fraction form is: 1\(\frac{80}{100}\)
The representation of 1\(\frac{80}{100}\) in the decimal form is: 1.80

Explanation:
The given decimal number is: 1.8
So,
The representation of 1.8 in the fraction form is: 1\(\frac{8}{10}\)
So,
To write 1\(\frac{8}{10}\) as hundredths, multiply the fraction and numerator of 1\(\frac{8}{10}\) with 10.
So,
Firstly the numerators 8 and 10 are multiplied and then the denominators 10 and 10 are multiplied
We know that,
Decimal = Numerator ÷ Denominator
So,
The representation of 1\(\frac{80}{100}\) in the place-value chart is:

Hence,
The representation of 1\(\frac{8}{10}\) as hundredths in the fraction form is: 1\(\frac{80}{100}\)
The representation of 1\(\frac{80}{100}\) in the decimal form is: 1.80

Question 10.
Precision
Which of the following show forty-one and nine-tenths?
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 59
Answer:
Let the given Expressions be named as:
A) 41.9 B) 41.09  C) 40 + 1 + \(\frac{9}{10}\)  D) 40 + 1 + 9  E) 41\(\frac{90}{10}\) F) 40 + 1 + 0.9
Hence,
The representation of forty-one and nine-tenths is shown by: B), C) and F)

Explanation:
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 59
Let the given Expressions be named as:
A) 41.9 B) 41.09  C) 40 + 1 + \(\frac{9}{10}\)  D) 40 + 1 + 9  E) 41\(\frac{90}{10}\) F) 40 + 1 + 0.9
Now,
The given word form is: Forty-one and nine-tenths
The representation of forty-one and nine-tenths in the fraction form is: 41\(\frac{9}{10}\)
So,
The expanded form of 41\(\frac{9}{10}\) is: 40 + 1 + \(\frac{9}{10}\)
The expanded form of 41\(\frac{9}{10}\) in the decimal form is: 40 + 1 + 0.9
The standard form of 41\(\frac{9}{10}\) is: 41.09
Hence, from the above,
We can conclude that Expressions B), C), and F) representing 41\(\frac{9}{10}\)

Question 11.
Which One Doesn’t Belong? Which one does not belong with the other three?
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 60
Answer:
Let the Expressions be named as:
A) 0.70  B) \(\frac{7}{10}\)  C) 0.07  D) \(\frac{70}{100}\)
Hence, from above,
Expression C) does not belong to the other three.

Explanation:
Let the Expressions be named as:
A) 0.70  B) \(\frac{7}{10}\)  C) 0.07  D) \(\frac{70}{100}\)
Now,
The representation of \(\frac{7}{10}\) as hundredths in the fraction form is: \(\frac{70}{100}\)
The representation of \(\frac{70}{100}\) in the decimal form is: 0.70
So,
To write \(\frac{7}{10}\) as hundredths, multiply the fraction and numerator of \(\frac{7}{10}\) with 10.
So,
Firstly the numerators 7 and 10 are multiplied and then the denominators 10 and 10 are multiplied
We know that,
Decimal = Numerator ÷ Denominator
So,
The representation of \(\frac{70}{100}\) in the place-value chart is:

Hence, from the above,
We can conclude that Expression C) does not belong to the other three.

Question 12.
YOU BE THE TEACHER
Is Newton correct? Explain.
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 61
Answer: Yes, Newton is correct

Explanation:
The given numbers are: 80 and 0.8
When we compare 80 and 0.8,
We can say that 0.8 is the number that we can obtain when we divide 80 with 100
So,
We can observe that 80 is the whole number and 0.8 is the decimal number
We know that,
The whole number is always greater than the decimal number.
Hence, from the above,
We can conclude that 80 is greater than 0.8 as 80 is greater than 8 according to Newton

Question 13.
Number Sense
Write two equivalent fractions and two equivalent decimals represented by the model.
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 62
Answer:
The equivalent fractions of \(\frac{6}{10}\) are: \(\frac{60}{100}\) and \(\frac{6}{10}\)
The equivalent decimals represented by \(\frac{6}{10}\) are: 0.6 and 0.60

Explanation:
The given model is:
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 62
From the above model,
The total number of parts are: 10
The number of colored parts is: 6
Hence,
The portion of the colored part out of the total number of parts is: \(\frac{6}{10}\)
Now,
The given fraction is: \(\frac{6}{10}\)
So, to write \(\frac{6}{10}\) as hundredths, multiply the fraction and numerator of \(\frac{6}{10}\) with 10.
So,
Firstly the numerators 6 and 10 are multiplied and then the denominators 10 and 10 are multiplied
We know that,
Decimal = Numerator ÷ Denominator
So,
The representation of \(\frac{60}{100}\) in the place-value chart is:

The representation of \(\frac{6}{10}\) in the place-value chart is:

Hence,
The equivalent fractions of the given model is: \(\frac{60}{100}\) and \(\frac{6}{10}\)
The equivalent decimal numbers of \(\frac{6}{10}\) are: 0.6 and 0.60

Question 14.

Modeling Real Life
Does each player get a base hit on the same fraction of pitches? Explain.
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 63
Answer: Yes, each player gets a base hit on the same fraction of pitches.

Explanation:
It is given that there are 2 types of players. They are: Major League player and Youth League player
It is also given that the Major League player gets a base hit on 30 out of 100 pitches and youth League player gets a base hit on 3 out of 10 pitches
So, for comparison, convert the youth League player’s base hit into hundredths.
Now,
The representation of the youth League plyer’s base hit is: \(\frac{3}{10}\)
So,
To write \(\frac{3}{10}\) as hundredths, multiply the fraction and numerator of \(\frac{3}{10}\) with 10.
So,
Firstly the numerators 3 and 10 are multiplied and then the denominators 10 and 10 are multiplied
We know that,
Decimal = Numerator ÷ Denominator
So,
The representation of \(\frac{30}{100}\) in the place-value chart is:

Hence, from the above,
we can conclude that each player gets a base hit on the same fraction of pitches.

Review & Refresh

Find the equivalent fraction

Question 15.
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 64
Answer: The equivalent fraction of \(\frac{4}{6}\) is: \(\frac{2}{3}\)

Explanation:
The given fraction is \(\frac{4}{6}\)
From the above fraction, the numerator and denominator are: 4 and 6
4 and 6 are the multiples of 2.  ( Since the  the numerator and the denominator are the even numbers )
So,
We have to divide the \(\frac{4}{6}\) with 2
So,
\(\frac{4}{6}=\frac{4 \div 2}{6 \div 2}=\frac{2}{3}\)
Hence,
The equivalent fraction of \(\frac{4}{6}\) is: \(\frac{2}{3}\)

Question 16.
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 65
Answer: The equivalent fraction of \(\frac{25}{100}\) is: \(\frac{5}{20}\)

Explanation:
The given fraction is \(\frac{25}{100}\)
From the above fraction, the numerator and denominator are: 25 and 100
25 and 100 are the multiples of 5.  ( Since the  the numerator and the denominator are the multiples of 5 )
So,
We have to divide the \(\frac{25}{100}\) with 5
So,
\(\frac{25}{100}=\frac{25 \div 5}{100 \div 5}=\frac{5}{20}\)
Hence,
The equivalent fraction of \(\frac{25}{100}\) is: \(\frac{5}{20}\)

Question 17.
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 66
Answer: The equivalent fraction of \(\frac{14}{8}\) is: \(\frac{7}{4}\)

Explanation:
The given fraction is \(\frac{14}{8}\)
From the above fraction, the numerator and denominator are: 14 and 8
14 and 8 are the multiples of 2.  ( Since the  the numerator and the denominator are the even numbers )
So,
We have to divide the \(\frac{14}{8}\) with 2
So,
\(\frac{14}{8}=\frac{14 \div 2}{8 \div 2}=\frac{7}{4}\)
Hence,
The equivalent fraction of \(\frac{14}{8}\) is: \(\frac{7}{4}\)

Lesson 10.4 Compare Decimals

Use models to compare the decimals.


Answer:

By using the models, we can compare the number of boxes for the comparison of the decimal numbers.
Each box represents 1 unit out of 100 total units.
So,
The denominator will be the same i.e., the total number of boxes will be the same.
So, we have to compare only numerators i.e., the number of colored boxes so that we can compare the decimal numbers

Reasoning
How did you use your models to determine which decimal is greater?

Answer: For the comparison of the decimals, equate either the numerator or the denominator so that we can compare the numerators.
If we compare the numerators, then we have to make the denominators equal
If we compare the denominators, then we have to make the numerators equal
Hence, in this way, we can compare the decimal numbers.

Think and Grow: Compare Decimals

Example
Compare 0.7 and 0.07.
Answer: 0.7 is greater than 0.07

Explanation:
The given decimal numbers are: 0.7 and 0.07
The representation of 0.7 in the fraction form is: \(\frac{7}{10}\)
The representation of 0.07 in the fraction form is: \(\frac{7}{100}\)
So,
for comparison, we have to make the denominators equal.
So,
Multiply \(\frac{7}{10}\) with 10
So for \(\frac{7}{10}\),
The numerator is 7 and denominator is 10
So,
The numerator 7 is multiplied by 10 and the denominator is also multiplied by 10 to make the denominators equal.
So,
The representation of \(\frac{7}{10}\) in the hundredth’s form is: \(\frac{70}{100}\)
Hence, from the above,
We can conclude that 0.7 is greater than 0.07 by comparing their fraction forms.

Example
Use a place value chart. Start at the left. Compare the digits in each place until the digits differ.
The digits in the one’s place are the same. Compare the tenths.

So, 0.7 > 0.07

Example
Compare 0.25 and 0.3.

Use a number line. 0.25 is 25 hundredths. 0.3 and 0.30 are equivalent decimals. So, 0.3 is equivalent to 30 hundredths.

Show and Grow

Compare.

Question 1.
Big Ideas Math Answers 4th Grade Chapter 10 Relate Fractions and Decimals 73
Answer: 0.46 is less than 0.44

Explanation:
The given decimal numbers are: 0.46 and 0.44
The representation of 0.46 and 0.44 in the place-value chart is:

So,
From the above place-value chart,
we can observe that one’s and tenth’s positions are the same.
So,
Compare the hundredth’s position 6 and 4
So, 6 hundredths > 4 hundredths
Hence, from the above,
We can conclude that 0.46 is greater than 0.44

Question 2.
Big Ideas Math Answers 4th Grade Chapter 10 Relate Fractions and Decimals 74
Answer: 0.05 is less than 0.2

Explanation:
The given decimal numbers are 0.05 and 0.2
The representation of 0.05 in the fraction form is: \(\frac{5}{100}\)
The representation of 0.2 in the fraction form is: \(\frac{2}{10}\)
So, for comparison, change \(\frac{2}{10}\) in to hundredths
So, for the change of \(\frac{2}{10}\) in to hundredths, multiply \(\frac{2}{10}\) with 10
So,
The numerator 2 is multiplied by 10 and the denominator 10 is multiplied with 10
Hence,
The representation of 0.2 in hundredth’s form is: \(\frac{20}{100}\)
So, for comparison,
The given number line is:

So, from the above number line,
0.05 is to the left of 0.20
Hence, from the above,
We can conclude that 0.05 is less than 0.20

Apply and Grow: Practice

Use the number line to compare.

Question 3.
Big Ideas Math Answers 4th Grade Chapter 10 Relate Fractions and Decimals 76
Answer: 0.85 is less than 0.96

Explanation:
The given decimal numbers are: 0.85 and 0.96
The given number line is:

From the above number line, 0.85 is to the left of 0.90
Hence, from the above,
We can conclude that 0.85 is less than 0.90

Question 4.
Big Ideas Math Answers 4th Grade Chapter 10 Relate Fractions and Decimals 77
Answer: 0.25 is less than 0.52

Explanation:
The given decimal numbers are: 0.25 and 0.52
The given number line is:

From the above number line,
0.25 is to the left of 0.52
Hence, from the above,
We can conclude that 0.25 is less than 0.52

Explanation:

Question 5.
Big Ideas Math Answers 4th Grade Chapter 10 Relate Fractions and Decimals 78
Answer: 0.11 is greater than 0.09

Explanation:
The given decimal numbers are: 0.11 and 0.09
The given number line is:

From the above line, we can observe that,
0.11 is to the right of 0.09
Hence, from the above,
we can conclude that 0.11 is greater than 0.09

Question 6.
Big Ideas Math Answers 4th Grade Chapter 10 Relate Fractions and Decimals 79
Answer: 0.72 is greater than 0.59

Explanation:
The given decimal numbers are: 0.72 and 0.59
The given number line is:

From the above number line, we can observe that,
0.72 is to the right of 0.59
Hence, from the above,
We can conclude that 0.72 is greater than 0.59

Question 7.
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 80
Answer: 0.04 is less than 0.40

Explanation:
The given decimal numbers are: 0.04 and 0.40
The given number line is:

From the above number line, we can observe that
0.04 is to the left of 0.40
Hence, from the above,
we can conclude that 0.04 is less than 0.40

Question 8.
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 81
Answer: 0.90 is equal to 0.9

Explanation:
The given decimal numbers are: 0.90 and 0.9
Change the 0.9 into hundredths by multiplying 0.9 with 10.
Now,
The representation of 0.9 in the fraction form is: \(\frac{9}{10}\)
So,
When we multiply \(\frac{9}{10}\) with 10, we can get
The representation of \(\frac{9}{10}\) in hundredth’s in the fraction form is: \(\frac{90}{100}\)
Now,
the given number line is:

We get the numer \(\frac{90}{100}\) by convrting the \(\frac{9}{10}\) and the given number is also \(\frac{90}{100}\)
Hence, from the above,
we can conclude that 0.90 is equal to 0.9

Question 9.
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 82
Answer: 0.3 is greater than 0.03

Explanation:
The given decimal numbers are: 0.3 and 0.03
The representation of 0.3 in the fraction form is: \(\frac{3}{10}\)
The representation of 0.03 in the fraction form is: \(\frac{3}{100}\)
So, for the conversion of \(\frac{3}{10}\) into hundredth’s, we have to multiply \(\frac{3}{10}\) with 10
So,
The representation of \(\frac{3}{10}\) as hundredth’s in the fraction form is: \(\frac{30}{100}\)
The given number line is:

From the above number line, we can observe that
0.03 is to the left of 0.30
Hence, from the above,
We can observe that 0.03 is less than 0.30

Compare

Question 10.
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 83
Answer: 5.29 is greater than 5.24

Explanation:
The given decimal numbers are: 5.29 and 5.24
The representation of 5.29 and 5.24 in the place-value chart is:

From the above place-value chart, we can observe that one’s and the tenth’s positions are the same.
So,
We have to compare the hundredth’s position of the two decimal numbers.
So, 9 hundredths > 4 hundredths
Hence, from the above,
we can conclude that 5.29 is greater than 5.24

Question 11.
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 84
Answer: 25.94 is greater than 25.9

Explanation:
The given decimal numbers are: 25.94 and 25.9
The representation of 25.94 and 25.9 in the place-value chart is:

From the above place-value chart, we can observe that ten’s, one’s, and tenth’s positions are the same.
So, we have to compare the hundredth’s position.
So,
4 hundredths > 0 hundredths
Hence, from the above,
we can conclude that 25.94 is greater than 25.9

Open-Ended
Complete the statement to make it true.

Question 12.
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 85
Answer:
Let the missing numbers be 3 and 2
So,
0.31 > 0.21

Explanation:
Let the decimal numbers be: 0.31 and 0.21
The representation of 0.31 and 0.21 in the place-value chart is:

From the above place-value chart,
We can observe that one’s and hundredth’s positions are the same.
So, we have to compare the tenth’s position of the two decimal numbers
So,
3 tenths > 2 tenths
Hence, from the above,
We can conclude that 0.31 is greater than 0.21

Question 13.
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 86
Answer:
Le the missing number be 10
So,
12.10 = 12.10

Explanation:
Let the missing decimal numbers be: 12.10 and 12.10
The representation of 12.10 and 12.10 in the place-value chart is:

From the above place-value chart, we can observe that all the positions are the same.
Hence, from the above,
We can conclude that 12.10 is equal to 12.10

Question 14.
9.43 < ____
Answer:
Let the missing number be: 9.53
So,
9.43 < 9.53

Explanation:
Let the 2 decimal numbers be: 9.43 and 9.53
The representation of 9.43 and 9.53 in the place-value chart is:

From the above place-value chart, we can observe that one’s position and hundredths position is the same.
So, we can either compare the tenths position.
So,
When we compare the tenths position, 4 tenths < 5 tenths
Hence, from the above
We can conclude that 9.43 is less than 9.53

Question 15.
Precision
Write the number that is halfway between 3.6 and 3.7. Explain how you found your answer.
Answer: The number that is halfway between 3.6 and 3.7 is 3.65

Explanation:
The given decimal numbers are: 3.6 and 3.7
We know that,
The middle number between two numbers = (The given first number + The given second number) ÷ 2
So,
The number that is halfway between 3.6 and 3.7 = (3.6 + 3.7) ÷ 2
= 7.3 ÷ 2
= 3.65
Hence, from the above,
we can conclude that the number that is halfway between 3.6 and 3.7 is: 3.65

DIG DEEPER!
Write whether the statement is true or false. If false, explain why

Question 16.
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 87
Answer: \(\frac{1}{10}\) is greater than 0.07

Explanation:
The given numbers are: \(\frac{1}{10}\) and 0.07
The representation of 0.07 in the fraction form is: \(\frac{7}{100}\)
So, for comparison, we have to convert \(\frac{1}{10}\) into hundredths form by multiplying \(\frac{1}{10}\) with 10.
So,
The representation of \(\frac{1}{10}\) as hundredths in the fraction form is: \(\frac{10}{100}\)
The representation of 0.07 in the fraction form is: \(\frac{7}{100}\)
Hence, from the above,
we can conclude that \(\frac{1}{10}\) is greater than 0.07

Question 17.
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 88
Answer: 0.6 is greater than 0.36

Explanation:
The given numbers are: 0.6 and \(\frac{36}{100}\)
The representation of 0.6 in the fraction form is: \(\frac{6}{10}\)
So, for comparison, we have to convert \(\frac{6}{10}\) into hundredths form by multiplying \(\frac{6}{10}\) with 10.
So,
The representation of \(\frac{6}{10}\) as hundredths in the fraction form is: \(\frac{60}{100}\)
The representation of 0.36 in the fraction form is: \(\frac{36}{100}\)
Hence, from the above,
we can conclude that 0.6 is greater than \(\frac{36}{100}\)

Think and Grow: Modeling Real Life

Example
Newton and Descartes make paper airplanes. Newton’s paper Whose airplane flies 3.01 meters. Descartes’s paper airplane flies 3.10 meters. Whose paper airplane flies farther?
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 89
Use a place value chart. Compare the digits in each place until the digits differ.

Compare Newton’s distance to Descartes’s distance.
So,
Descartes’s paper airplane flies farther.

Show and Grow

Question 18.
Compare the thickness of a nickel and a quarter. Which coin is thinner?
Big Ideas Math Answers Grade 4 Chapter 10 Relate Fractions and Decimals 91
Answer: When we compare the thickness of nickel and quarter by observing the table, we can say that the Quarter is thinner.

Explanation:
The given table is:
Big Ideas Math Answers Grade 4 Chapter 10 Relate Fractions and Decimals 91
From the given table,
The thickness of the Nickel is: 1.95mm
The thickness of the Quarter is: 1.75mm
Now,
The representation of the thicknesses of the Nickel and the Quarter in the place-value chart is:

From the above place-value chart, we can observe that
7 tenths < 9 tenths
Hence, from the above,
We can conclude that the Quarter is thinner.

Question 19.
You, your cousin, and your friend run a 100-meter race. Who finishes first? second? third?
Big Ideas Math Answers Grade 4 Chapter 10 Relate Fractions and Decimals 92
Answer:
The one who finishes first is: You
The one who finishes second is: Friend
The one who finishes third is: Cousin

Explanation:
The given timings are:
The timing of yours is: 16.40 seconds
The timing of your friend is: 16.48 seconds
The timing of your cousin is: 16.58 seconds
Now, the representation of all the timings in the place-value chart is:


Hence, from the above,
We can conclude that:
The one who finishes first is: You
The one who finishes second is: Friend
The one who finishes third is: Cousin

Question 20.
DIG DEEPER!
Your water bottle is 0.25 full. Your friend’s water bottle is 0.5 full. You have more water than your friend. Explain how this is possible.
Answer: This is not possible.

Explanation:
It is given that your water bottle is 0.25 full and your friend’s water bottle is 0.5 full.
So, we have to compare 0.25 and 0.5
The representation of 0.5 in the fraction form is: \(\frac{5}{10}\)
The representation of 0.25 in the fraction form is: \(\frac{25}{100}\)
So, for comparison, we have to convert \(\frac{5}{10}\) as hundredths by multiplying \(\frac{5}{10} \) with 10.
So,
The representation of \(\frac{5}{10}\) as hundredths in the fraction form is: \(\frac{50}{100}\)
Hence, from the above,
we can conclude that your friend’s water bottle has more water than you.

Compare Decimals Homework & Practice 10.4

Compare

Question 1.
Big Ideas Math Answers Grade 4 Chapter 10 Relate Fractions and Decimals 93
Answer: 0.58 is greater than 0.52

Explanation:
The given decimal numbers are: 0.58 and 0.52
The representation of 0.58 and 0.52 in the place-value chart is:

From the above table, we can observe that one’s and tenth’s positions are the same.
So, we have to compare the hundredth’s position
So,
8 hundredths > 2 hundredths
Hence, from the above,
we can conclude that 0.58 is greater than 0.52

Question 2.
Big Ideas Math Answers Grade 4 Chapter 10 Relate Fractions and Decimals 94
Answer: 0.25 is greater than 0.05

Explanation:
The given decimal numbers are: 0.25 and 0.05
The given number line is:

From the above number line,
We can observe that 0.05 is to the left of 0.25
hence, from the above,
We can conclude that 0.25 is greater than 0.05

Use the number line to compare.

Question 3.
Big Ideas Math Answers Grade 4 Chapter 10 Relate Fractions and Decimals 96
Answer: 0.76 is greater than 0.59

Explanation:
The given decimal numbers are: 0.76 and 0.59
The given number line is:

From the above number line,
We can observe that 0.76 is to the right of 0.59
Hence, from the above,
we can conclude that 0.76 is greater than 0.59

Question 4.
Big Ideas Math Answers Grade 4 Chapter 10 Relate Fractions and Decimals 97
Answer: 0.21 is less than 0.23

Explanation:
the given decimal numbers are: 0.21 and 0.23
The given number line is:

From the above number line,
We can observe that 0.21 is to the left of 0.23
Hence, from the above,
we can conclude that 0.21 is less than 0.23

Question 5.
Big Ideas Math Answers Grade 4 Chapter 10 Relate Fractions and Decimals 98
Answer: 0.7 is greater than 0.07

Explanation:
The given decimal numbers are: 0.7 and 0.07
The representation of 0.7 in the fracton form is: \(\frac{7}{10}\)
So,
The \(\frac{7}{10}\) should be converted into hundredths by multiplying \(\frac{7}{10}\) with 10.
So,
The representation of \(\frac{7}{10}\) as hundredths in the fraction form is: \(\frac{70}{100}\)
Now,
The given number line is:

From the above number line,
We can observe that 0.7 is to the left of 0.07
Hence, from the above,
We can conclude that 0.7 is greater than 0.07

Question 6.
Big Ideas Math Answers Grade 4 Chapter 10 Relate Fractions and Decimals 99
Answer: 0.05 is less than 0.08

Explanation:
The given decimal numbers are: 0.05 and 0.08
The given number line is:

From the above number line,
We can observe that 0.05 is to the left of 0.08
Hence, from the above,
We can conclude that 0.05 is less than 0.08

Question 7.
Big Ideas Math Answers Grade 4 Chapter 10 Relate Fractions and Decimals 100
Answer: 0.10 is equal to 0.1

Explanation:
The given decimal numbers are: 0.10 and 0.1
Now,
The representation of 0.1 in the fraction form is: \(\frac{1}{10}\)
The representation of 0.10 in the fraction form is: \(\frac{1}{100}\)
Now, to make the denominators equal, multiply \(\frac{1}{10}\) by 2.
So,
The representation of \(\frac{1}{10}\) as hundredths, in the fraction form is: \(\frac{1}{100}\)
Now,
The given number line is:

Hence, from the above,
We can conclude that 0.10 is equal to 0.1

Question 8.
Big Ideas Math Answers Grade 4 Chapter 10 Relate Fractions and Decimals 101
Answer: 0.05 is less than 0.50

Explanation:
The given decimal numbers are: 0.05 and 0.50
The given number line is:

From the above number line,
We can observe that 0.05 is to the left of 0.50.
Hence, from the above,
we can conclude that 0.05 is less than 0.50

Question 9.
Big Ideas Math Answers Grade 4 Chapter 10 Relate Fractions and Decimals 102
Answer: 0.13 is less than 0.19

Explanation:
The given decimal numbers are: 0.13 and 0.19
The given number line is:


From the above number line,
We can observe that 0.13 is to the left of 0.10
Hence, from the above,
We can conclude that 0.13 is less than 0.19

Compare

Question 10.
Big Ideas Math Answers Grade 4 Chapter 10 Relate Fractions and Decimals 103
Answer: 2.2 is greater than 2.02

Explanation:
The given decimal numbers are: 2.2 and 2.02
The representation of 2.2 and 2.02 in the place-value chart is:

From the above place-value chart,
We can observe that only one’s position is the same
So,
2 tenths > 0 tenths
Hence, from the above,
We can conclude that 2.2 is greater than 2.02

Question 11.
Big Ideas Math Answers Grade 4 Chapter 10 Relate Fractions and Decimals 104
Answer: 4.70 is equal to 4.7

Explanation:
The given decimal numbers are: 4.70 and 4.7
The representation of 4.70 in the fraction form is: 4\(\frac{70}{100}\)
The representation of 4.7 in the fraction form is: 4\(\frac{7}{10}\)
So, to compare, we have to make the denominators equal.
So, multiply 4\(\frac{7}{10}\) with 10
So,
The representation of 4\(\frac{7}{10}\) as hundredth’s in the fraction form is: 4\(\frac{70}{100}\)
Hence, from the above,
We can conclude that 4.70 is equal to 4.7

Question 12.
Big Ideas Math Answers Grade 4 Chapter 10 Relate Fractions and Decimals 105
Answer: 8.35 is less than 8.53

Explanation:
The given decimal numbers are: 8.35 and 8.53
The representation of 8.35 and 8.53 in the place-value chart is:

From the above place-value chart,
We can observe that only the tenth’s position can be compared.
So,
3 tenths < 5 tenths
Hence, from the above,
We can conclude that 8.35 is less than 8.53

Question 13.
Big Ideas Math Answers Grade 4 Chapter 10 Relate Fractions and Decimals 106
Answer: 35.01 is greater than 32.98

Explanation:
The given decimal numbers are: 35.01 and 32.98
The representation of 35.01 and 32.98 in the place-value chart is:

From the above place-value chart,
We can compare one’s position since it is the position with the highest value after the ten’s position.
So,
5 ones > 2 ones
Hence, from the above,
we can conclude that 35.01 is greater than 32.98

Question 14.
Big Ideas Math Answers Grade 4 Chapter 10 Relate Fractions and Decimals 107
Answer: 14.9 is less than 14.92

Explanation:
The given decimal numbers are: 14.9 and 14.92
The representation of 14.9 and 14.92 in the place-value chart is:

From the above place-value chart,
We can compare only the hundredth’s positions.
So,
0 hundredths< 2 hundredths
Hence, from the above,
We can conclude that 14.9 is less than 14.92

Question 15.
Precision
Explain how to compare 0.46 and 0.48.
Answer: 0.46 is less than 0.48

Explanation:
The given decimal numbers are: 0.46 and 0.48
The representation of 0.46 and 0.48 in the place-value chart is:

From the above place-value chart,
We can compare the hundredth’s position
So,
6 hundredths < 8 hundredths
Hence, from the above
We can conclude that 0.46 is less than 0.48

Question 16.
Open-Ended
What might Descartes’s number be?
Big Ideas Math Answers Grade 4 Chapter 10 Relate Fractions and Decimals 108
Answer: Descartes’s number might be: 0.61, 0.62, 0.63, 0.64, 0.65

Explanation:
Given that my number is greater than 0.6 and less than 0.7.
So,
The numbers might be: 0.61, 0.62, 0.63, 0.64, 0.65, 0.66, 0.67, 0.68, 0.69
It is also given that the greatest digit in the number is in the tenths place.
So,
The numbers might be: 0.61, 0.62, 0.63, 0.64, 0.65
Hence, from the above,
We can conclude that Descartes’s number might be: 0.61, 0.62, 0.63, 0.64, 0.65

Question 17.
Modeling Real Life
A traffic light is red for 23.4 seconds and green for 23.6 seconds. Does the traffic light stay red or green longer?
Answer: The traffic light stay green for a long time

Explanation:
It is given that a traffic light is red for 23.4 seconds and green for 23.6 seconds.
So, we have to compare 23.4 and 23.6 to see which light stay for a long time
Now,
The representation of 23.4 and 23.6 in the place-value chart is:

From the above place-value chart, we can observe that we can compare only the tenths position.
So,
4 tenths < 6 tenths
So,
23.4 is less than 23.6
Hence, from the above,
We can conclude that the green light stays for a long time

Question 18.
Modeling Real Life
Order the caterpillars from longest to shortest.
Big Ideas Math Answers Grade 4 Chapter 10 Relate Fractions and Decimals 109
Answer: Caterpillar B > Caterpillar A > Caterpillar C

Explanation:
The given table is:
Big Ideas Math Answers Grade 4 Chapter 10 Relate Fractions and Decimals 109
From the table,
The length of caterpillar A is: 3.5 cm
The length of caterpillar B is: 3.65 cm
The length of caterpillar C is: 3.45 cm
So,
When we compare the tenth’s position in all the lengths of the caterpillars,
We can observe that B > A > C
Hence, from the above,
We can conclude that caterpillar B > caterpillar A > caterpillar C

Review & Refresh

Round the number to the nearest hundred thousand

Question 19.
695,023
Answer: The nearest hundred thousand of 695,023 is: 700,000

Explanation:
The given number is: 695,023
We know that,
The value of a digit depends on the position of the digit
So,
The value of the nearest hundred thousand in 695,023 is: 700,000

Question 20.
246,947
Answer: The value of the nearest hundred thousand in 246,947 is: 200,000

Explanation:
The given number is: 246,947
We know that,
The value of a digit depends on the position of the digit
So,
The value of the nearest hundred thousand in 246,947 is: 200,000

Lesson 10.5 Add Decimal Fractions and Decimals

Explore and Grow

How can you use a number line to find the sum?
Big Ideas Math Answers Grade 4 Chapter 10 Relate Fractions and Decimals 110
0.25 + 0.7
Big Ideas Math Answers Grade 4 Chapter 10 Relate Fractions and Decimals 111
How can you use models to check your answers?

Answer:
The representation of the sum in the fraction form is:
\(\frac{7}{10}\) + \(\frac{25}{100}\) = \(\frac{95}{100}\)
The representation of the sum in the decimal form is:
0.25 + 0.7 = 0.95

Explanation:
The given fractions are: \(\frac{25}{100}\) and \(\frac{7}{10}\)
For addition, we have to make either denominators or the numerators equal.
so,
For making \(\frac{7}{10}\) as the hundredths, we have to multiply \(\frac{7}{10}\) by 10
So,
The representation of \(\frac{7}{10}\) as hundredths in the fraction form is: \(\frac{70}{100}\)
So,
\(\frac{70}{100}\) + \(\frac{25}{100}\)
= \(\frac{70 + 25}{100}\)
= \(\frac{95}{100}\)
The representation of \(\frac{95}{100}\) in the decimal form is: 0.95
Now,

Hence, from the above,
We can conclude that
\(\frac{70}{100}\) + \(\frac{25}{100}\) = \(\frac{95}{100}\)
The representation of \(\frac{95}{100}\) in the decimal form is: 0.95

Reasoning
How can you add two decimal fractions with a denominator of 10? How can you add two decimal fractions with denominators of 10 and 100?

Answer: We add the two fractions only if their numerator or denominators are equal
If there are fractions with different denominators, then we will make the denominators equal either by multiplying or by dividing the fractions and add the different fractions.

Think and Grow: Add Decimal Fractions and Decimals

You have learned how to add fractions with the same denominator. You can use equivalent fractions to add fractions that do not have the same denominator.
Big Ideas Math Answers Grade 4 Chapter 10 Relate Fractions and Decimals 112
Step 1: Use equivalent fractions to write the fractions with the same denominator
Think: Rewrite \(\frac{3}{10}\) with a denominator of 100.


Example
Find 0.75 + 0.2.
Step 1: Write 0.75 and 0.2 as fractions.Think: 0.75 is 75 hundredths. 0.2 is 2 tenths.

Step 2: Use equivalent fractions to write the fractions with the same denominator.

Step 3: Add the numerators.

Step 4: Write the sum as a decimal.

So, 0.75 + 0.2 = 0.95

Show and Grow

Find the sum.

Question 1.
Big Ideas Math Answers Grade 4 Chapter 10 Relate Fractions and Decimals 119
Answer:
The representation of the sum in the fraction form is:
\(\frac{1}{10}\) + \(\frac{36}{100}\) = \(\frac{46}{100}\)
The representation of \(\frac{46}{100}\) in the decimal for is: 0.46

Explanation:
The given fractions are: \(\frac{36}{100}\) and \(\frac{1}{10}\)
For addition, we have to make either denominators or the numerators equal.
so,
For making \(\frac{1}{10}\) as the hundredths, we have to multiply \(\frac{1}{10}\) by 10
So,
The representation of \(\frac{1}{10}\) as hundredths in the fraction form is: \(\frac{10}{100}\)
So,
\(\frac{10}{100}\) + \(\frac{36}{100}\)
= \(\frac{10 + 36}{100}\)
= \(\frac{46}{100}\)
The representation of \(\frac{46}{100}\) in the decimal form is: 0.46
Hence from the above,
We can conclude that
\(\frac{1}{10}\) + \(\frac{36}{100}\) = \(\frac{46}{100}\)
The representation of \(\frac{46}{100}\) in the decimal for is: 0.46

Question 2.
0.5 + 0.25 = ____
Answer:
The representation of the sum in the fraction form is:
\(\frac{5}{10}\) + \(\frac{25}{100}\) = \(\frac{75}{100}\)
The representation of 0.25 + 0.5 in the decimal form is: 0.75

Explanation:
The given decimal numbers are: 0.5 and 0.25
So, convert the two decimal numbers in to respective fractions.
So,
The given fractions are: \(\frac{25}{100}\) and \(\frac{5}{10}\)
For addition, we have to make either denominators or the numerators equal.
so,
For making \(\frac{5}{10}\) as the hundredths, we have to multiply \(\frac{5}{10}\) by 10
So,
The representation of \(\frac{5}{10}\) as hundredths in the fraction form is: \(\frac{50}{100}\)
So,
\(\frac{50}{100}\) + \(\frac{25}{100}\)
= \(\frac{50 + 25}{100}\)
= \(\frac{75}{100}\)
The representation of \(\frac{75}{100}\) in the decimal form is: 0.75
Hence from the above,
We can conclude that
\(\frac{5}{10}\) + \(\frac{25}{100}\) = \(\frac{75}{100}\)
The representation of \(\frac{75}{100}\) in the decimal for is: 0.75

Apply and Grow: Practice

Find the sum.

Question 3.
Big Ideas Math Answers Grade 4 Chapter 10 Relate Fractions and Decimals 120
Answer:
The representation of the sum in the fraction form is:
\(\frac{4}{10}\) + \(\frac{37}{100}\) = \(\frac{77}{100}\)
The representation of \(\frac{77}{100}\) in the decimal for is: 0.77

Explanation:
The given fractions are: \(\frac{37}{100}\) and \(\frac{4}{10}\)
For addition, we have to make either denominators or the numerators equal.
so,
For making \(\frac{4}{10}\) as the hundredths, we have to multiply \(\frac{4}{10}\) by 10
So,
The representation of \(\frac{4}{10}\) as hundredths in the fraction form is: \(\frac{40}{100}\)
So,
\(\frac{40}{100}\) + \(\frac{37}{100}\)
= \(\frac{40 + 37}{100}\)
= \(\frac{77}{100}\)
The representation of \(\frac{77}{100}\) in the decimal form is: 0.77
Hence from the above,
We can conclude that
\(\frac{4}{10}\) + \(\frac{37}{100}\) = \(\frac{77}{100}\)
The representation of \(\frac{77}{100}\) in the decimal for is: 0.77

Question 4.
Big Ideas Math Answers Grade 4 Chapter 10 Relate Fractions and Decimals 121
Answer:
The representation of the sum in the fraction form is:
\(\frac{2}{10}\) + \(\frac{23}{100}\) = \(\frac{43}{100}\)
The representation of \(\frac{43}{100}\) in the decimal for is: 0.43

Explanation:
The given fractions are: \(\frac{23}{100}\) and \(\frac{2}{10}\)
For addition, we have to make either denominators or the numerators equal.
so,
For making \(\frac{2}{10}\) as the hundredths, we have to multiply \(\frac{2}{10}\) by 10
So,
The representation of \(\frac{2}{10}\) as hundredths in the fraction form is: \(\frac{20}{100}\)
So,
\(\frac{20}{100}\) + \(\frac{23}{100}\)
= \(\frac{20 + 23}{100}\)
= \(\frac{43}{100}\)
The representation of \(\frac{43}{100}\) in the decimal form is: 0.43
Hence from the above,
We can conclude that
\(\frac{2}{10}\) + \(\frac{23}{100}\) = \(\frac{43}{100}\)
The representation of \(\frac{43}{100}\) in the decimal for is: 0.43

Question 5.
Big Ideas Math Answers Grade 4 Chapter 10 Relate Fractions and Decimals 122
Answer:
The representation of the sum in the fraction form is:
\(\frac{7}{10}\) + \(\frac{19}{100}\) = \(\frac{89}{100}\)
The representation of \(\frac{89}{100}\) in the decimal for is: 0.89

Explanation:
The given fractions are: \(\frac{19}{100}\) and \(\frac{7}{10}\)
For addition, we have to make either denominators or the numerators equal.
so,
For making \(\frac{7}{10}\) as the hundredths, we have to multiply \(\frac{7}{10}\) by 10
So,
The representation of \(\frac{7}{10}\) as hundredths in the fraction form is: \(\frac{70}{100}\)
So,
\(\frac{70}{100}\) + \(\frac{19}{100}\)
= \(\frac{70 + 19}{100}\)
= \(\frac{89}{100}\)
The representation of \(\frac{89}{100}\) in the decimal form is: 0.89
Hence from the above,
We can conclude that
\(\frac{7}{10}\) + \(\frac{19}{100}\) = \(\frac{89}{100}\)
The representation of \(\frac{89}{100}\) in the decimal for is: 0.89

Question 6.
0.35 + 0.1 = ____
Answer:
The representation of the sum in the fraction form is:
\(\frac{1}{10}\) + \(\frac{35}{100}\) = \(\frac{45}{100}\)
The representation of 0.35 + 0.1 in the decimal for is: 0.45

Explanation:
The given decimal numbers are: 0.1 and 0.35
So, convert the two decimal numbers in to respective fractions.
So,
The given fractions are: \(\frac{35}{100}\) and \(\frac{1}{10}\)
For addition, we have to make either denominators or the numerators equal.
so,
For making \(\frac{1}{10}\) as the hundredths, we have to multiply \(\frac{1}{10}\) by 10
So,
The representation of \(\frac{1}{10}\) as hundredths in the fraction form is: \(\frac{10}{100}\)
So,
\(\frac{10}{100}\) + \(\frac{35}{100}\)
= \(\frac{10 + 35}{100}\)
= \(\frac{45}{100}\)
The representation of \(\frac{45}{100}\) in the decimal form is: 0.45
Hence from the above,
We can conclude that
\(\frac{1}{10}\) + \(\frac{35}{100}\) = \(\frac{45}{100}\)
The representation of \(\frac{45}{100}\) in the decimal for is: 0.45

Question 7.
0.8 + 0.15 = ____
Answer:
The representation of the sum in the fraction form is:
\(\frac{8}{10}\) + \(\frac{15}{100}\) = \(\frac{95}{100}\)
The representation of 0.15 + 0.8 in the decimal for is: 0.95

Explanation:
The given decimal numbers are: 0.8 and 0.15
So, convert the two decimal numbers in to respective fractions.
So,
The given fractions are: \(\frac{15}{100}\) and \(\frac{8}{10}\)
For addition, we have to make either denominators or the numerators equal.
so,
For making \(\frac{8}{10}\) as the hundredths, we have to multiply \(\frac{8}{10}\) by 10
So,
The representation of \(\frac{8}{10}\) as hundredths in the fraction form is: \(\frac{80}{100}\)
So,
\(\frac{80}{100}\) + \(\frac{15}{100}\)
= \(\frac{80 + 15}{100}\)
= \(\frac{95}{100}\)
The representation of \(\frac{95}{100}\) in the decimal form is: 0.95
Hence from the above,
We can conclude that
\(\frac{8}{10}\) + \(\frac{15}{100}\) = \(\frac{95}{100}\)
The representation of \(\frac{95}{100}\) in the decimal for is: 0.95

Question 8.
0.50 + 0.4 = __
Answer:
The representation of the sum in the fraction form is:
\(\frac{4}{10}\) + \(\frac{50}{100}\) = \(\frac{90}{100}\)
The representation of 0.50 + 0.4 in the decimal for is: 0.90

Explanation:
The given decimal numbers are: 0.4 and 0.50
So, convert the two decimal numbers in to respective fractions.
So,
The given fractions are: \(\frac{50}{100}\) and \(\frac{4}{10}\)
For addition, we have to make either denominators or the numerators equal.
so,
For making \(\frac{4}{10}\) as the hundredths, we have to multiply \(\frac{4}{10}\) by 10
So,
The representation of \(\frac{4}{10}\) as hundredths in the fraction form is: \(\frac{40}{100}\)
So,
\(\frac{40}{100}\) + \(\frac{50}{100}\)
= \(\frac{50 + 40}{100}\)
= \(\frac{90}{100}\)
The representation of \(\frac{90}{100}\) in the decimal form is: 0.90
Hence from the above,
We can conclude that
\(\frac{4}{10}\) + \(\frac{50}{100}\) = \(\frac{90}{100}\)
The representation of \(\frac{90}{100}\) in the decimal for is: 0.90

Question 9.
Big Ideas Math Answers Grade 4 Chapter 10 Relate Fractions and Decimals 123
Answer:
The representation of the sum in the fraction form is:
\(\frac{48}{100}\) + \(\frac{16}{100}\) + \(\frac{2}{10}\) = \(\frac{84}{100}\)
The representation of \(\frac{84}{100}\) in the decimal for is: 0.84

Explanation:
The given fractions are: \(\frac{48}{100}\), \(\frac{16}{100}\) and \(\frac{2}{10}\)
For addition, we have to make either denominators or the numerators equal.
so,
For making \(\frac{2}{10}\) as the hundredths, we have to multiply \(\frac{2}{10}\) by 10
So,
The representation of \(\frac{2}{10}\) as hundredths in the fraction form is: \(\frac{20}{100}\)
So,
\(\frac{48}{100}\) + \(\frac{16}{100}\) + \(\frac{20}{100}\)
= \(\frac{48 + 16 + 20}{100}\)
= \(\frac{84}{100}\)
The representation of \(\frac{84}{100}\) in the decimal form is: 0.84
Hence from the above,
We can conclude that
\(\frac{2}{10}\) + \(\frac{48}{100}\) + \(\frac{16}{100}\) = \(\frac{84}{100}\)
The representation of \(\frac{84}{100}\) in the decimal for is: 0.84

Question 10.
0.3 + 0.25 + 0.1 = ___
Answer:
The representation of the sum in the fraction form is:
\(\frac{3}{10}\) + \(\frac{1}{10}\) + \(\frac{25}{100}\) = \(\frac{65}{100}\)
The representation of 0.25 + 0.1 + 0.3 in the decimal for is: 0.65

Explanation:
The given decimal numbers are: 0.3, 0.1 and 0.25
So, convert the three decimal numbers in to respective fractions.
So,
The given fractions are: \(\frac{25}{100}\) , \(\frac{3}{10}\) and \(\frac{1}{10}\)
For addition, we have to make either denominators or the numerators equal.
so,
For making \(\frac{1}{10}\) and \(\frac{3}{10}\) as the hundredths, we have to multiply \(\frac{5}{10}\) and \(\frac{3}{10}\) by 10
So,
The representation of \(\frac{1}{10}\) as hundredths in the fraction form is: \(\frac{10}{100}\)
The representation of \(\frac{3}{10}\) as hundredths in the fraction form is: \(\frac{30}{100}\)
So,
\(\frac{10}{100}\) + \(\frac{25}{100}\) + \(\frac{30}{100}\)
= \(\frac{10 + 25 + 30}{100}\)
= \(\frac{65}{100}\)
The representation of \(\frac{65}{100}\) in the decimal form is: 0.65
Hence from the above,
We can conclude that
\(\frac{1}{10}\) + \(\frac{25}{100}\) + \(\frac{3}{10}\) = \(\frac{65}{100}\)
The representation of \(\frac{65}{100}\) in the decimal for is: 0.65

Number Sense
Find the sum.

Question 11.
Big Ideas Math Answers Grade 4 Chapter 10 Relate Fractions and Decimals 124
Answer:
The representation of the sum in the fraction form is:
\(\frac{5}{10}\) + \(\frac{29}{100}\) = \(\frac{79}{100}\)
The representation of 0.5 + \(\frac{29}{100}\) in the decimal for is: 0.79

Explanation:
The given numbers are: 0.5 and \(\frac{29}{100}\)
So, convert the decimal number in to respective fractions.
So,
The given fractions are: \(\frac{5}{10}\) and \(\frac{29}{100}\)
For addition, we have to make either denominators or the numerators equal.
so,
For making \(\frac{5}{10}\) as the hundredths, we have to multiply \(\frac{5}{10}\) by 10
So,
The representation of \(\frac{5}{10}\) as hundredths in the fraction form is: \(\frac{50}{100}\)
So,
\(\frac{50}{100}\) + \(\frac{29}{100}\)
= \(\frac{50 + 29}{100}\)
= \(\frac{79}{100}\)
The representation of \(\frac{79}{100}\) in the decimal form is: 0.79
Hence from the above,
We can conclude that
\(\frac{5}{10}\) + \(\frac{29}{100}\) = \(\frac{79}{100}\)
The representation of \(\frac{79}{100}\) in the decimal for is: 0.79

Question 12.
Big Ideas Math Answers Grade 4 Chapter 10 Relate Fractions and Decimals 125
Answer:
The representation of the sum in the fraction form is:
\(\frac{8}{10}\) + \(\frac{75}{100}\) = \(\frac{155}{100}\)
The representation of 0.75 + \(\frac{8}{10}\) in the decimal for is: 1.55

Explanation:
The given numbers are: 0.75 and \(\frac{8}{10}\)
So, convert the decimal number in to respective fractions.
So,
The given fractions are: \(\frac{8}{10}\) and \(\frac{75}{100}\)
For addition, we have to make either denominators or the numerators equal.
so,
For making \(\frac{8}{10}\) as the hundredths, we have to multiply \(\frac{8}{10}\) by 10
So,
The representation of \(\frac{8}{10}\) as hundredths in the fraction form is: \(\frac{80}{100}\)
So,
\(\frac{80}{100}\) + \(\frac{75}{100}\)
= \(\frac{80 + 75}{100}\)
= \(\frac{155}{100}\)
The representation of \(\frac{155}{100}\) in the decimal form is: 1.55
Hence from the above,
We can conclude that
\(\frac{8}{10}\) + \(\frac{75}{100}\) = \(\frac{155}{100}\)
The representation of \(\frac{155}{100}\) in the decimal for is: 1.55

Question 13.
YOU BE THE TEACHER
Your friend says Newton and Descartes are both correct. Is your friend correct? Explain.
Big Ideas Math Answers Grade 4 Chapter 10 Relate Fractions and Decimals 126
Answer: Yes, your friend is correct.

Explanation:
The representation of the sum in the fraction form is:
\(\frac{5}{10}\) + \(\frac{5}{100}\) + \(\frac{55}{100}\) = \(\frac{110}{100}\)
The representation of 0.55 + 0.5 + 0.05 in the decimal for is: 1.10

Explanation:
The given decimal numbers are: 0.05, 0.5 and 0.55
So, convert the three decimal numbers in to respective fractions.
So,
The given fractions are: \(\frac{55}{100}\) , \(\frac{5}{10}\) and \(\frac{5}{100}\)
For addition, we have to make either denominators or the numerators equal.
so,
For making \(\frac{5}{10}\)  as the hundredths, we have to multiply \(\frac{5}{10}\) by 10
So,
The representation of \(\frac{5}{10}\) as hundredths in the fraction form is: \(\frac{50}{100}\)
So,
\(\frac{50}{100}\) + \(\frac{55}{100}\) + \(\frac{5}{100}\)
= \(\frac{50 + 55 + 5}{100}\)
= \(\frac{110}{100}\)
The representation of \(\frac{110}{100}\) in the decimal form is: 1.10
Hence from the above,
We can conclude that
\(\frac{5}{10}\) + \(\frac{55}{100}\) + \(\frac{5}{100}\) = \(\frac{110}{100}\)
The representation of \(\frac{110}{100}\) in the decimal for is: 1.10 or 1.1

Question 14.
DIG DEEPER!
Write and solve a decimal addition problem represented by the model. Write your answer as a decimal and as a mixed number.
Big Ideas Math Answers Grade 4 Chapter 10 Relate Fractions and Decimals 127
Answer:
From the above model,
The representation of the sum in the fraction form is:
\(\frac{6}{10}\) + \(\frac{47}{100}\) = \(\frac{107}{100}\) = 1\(\frac{7}{100}\)
The representation of \(\frac{107}{100}\) in the decimal for is: 1.07

Explanation:
The given model is:
Big Ideas Math Answers Grade 4 Chapter 10 Relate Fractions and Decimals 127
From the model,
The given fractions are: \(\frac{47}{100}\) and \(\frac{6}{10}\)
For addition, we have to make either denominators or the numerators equal.
so,
For making \(\frac{6}{10}\) as the hundredths, we have to multiply \(\frac{6}{10}\) by 10
So,
The representation of \(\frac{6}{10}\) as hundredths in the fraction form is: \(\frac{60}{100}\)
So,
\(\frac{60}{100}\) + \(\frac{47}{100}\)
= \(\frac{60 + 47}{100}\)
= \(\frac{107}{100}\)
The representation of \(\frac{107}{100}\) in the decimal form is: 1.07
Hence from the above,
We can conclude that
\(\frac{6}{10}\) + \(\frac{47}{100}\) = \(\frac{107}{100}\) = 1\(\frac{7}{100}\)
The representation of \(\frac{107}{100}\) in the decimal for is: 1.07

Think and Grow: Modeling Real Life

Example
You use \(\frac{8}{10}\) pound of clay to make a cup. You make a handle for the cup with \(\frac{15}{100}\) pound of clay and attach the handle to the cup. What fraction of a pound does your cup weigh in all?
Add the fractions.
Big Ideas Math Answers Grade 4 Chapter 10 Relate Fractions and Decimals 136

Use equivalent fractions to write the fractions with the same denominator.
Rewrite \(\frac{8}{10}\) with a denominator of 100.

Show and Grow

Question 15.
Each morning, you walk \(\frac{25}{100}\) mile to your friend’s house and then \(\frac{5}{10}\) mile to school. What fraction of a mile do you walk each morning?
Answer:
The fraction of a mile you walk each morning is:
\(\frac{25}{100}\) + \(\frac{5}{10}\) = \(\frac{75}{100}\)

Explanation:
It is given that each morning, you walk \(\frac{25}{100}\) mile to your friend’s house and then \(\frac{5}{10}\) mile to school.
So,
To find the portion of a mile you walk in the morning, you have to add the fractions.
So,
Now, first, we have to convert \(\frac{5}{10}\) as hundredths so that we can add both the fractions because denominators have to be equal for the addition.
So,
Multiply \(\frac{5}{10}\) with 10 to convert it as hundredths
So,
The representation of \(\frac{5}{10}\) as hundredths in the fraction form is: \(\frac{50}{100}\)
So,
\(\frac{25}{100}\) + \(\frac{50}{100}\)
= \(\frac{75}{100}\)
Hence, from the above,
We can conclude that the portion of a mile you walk in the morning is: \(\frac{75}{100}\)

Question 16.
You ride a zip line that is mile-long \(\frac{15}{100}\) mile long. You ride another zip line that is \(\frac{3}{10}\) mile long. Your friend rides a total of \(\frac{40}{100}\) mile on zip lines. Who rides farther on zip lines?
Big Ideas Math Answers Grade 4 Chapter 10 Relate Fractions and Decimals 138
Answer: Your friend rides farther on zip lines.

Explanation:
It is given that you ride a zip line that is mile-long \(\frac{15}{100}\) mile long. You ride another zip line that is \(\frac{3}{10}\) mile long and your friend rides a total of \(\frac{40}{100}\) mile on zip
lines.
So, the given three fractions are:
\(\frac{15}{100}\), \(\frac{3}{10}\) and \(\frac{40}{100}\)
So, for the comparison of the three fractions, we have to make the denominators equal.
So, we have to convert \(\frac{3}{10}\) as hundredths.
So,
We have to multiply \(\frac{3}{10}\) with 10
So,
The representation of \(\frac{3}{10}\) as hundredths in the fraction form is: \(\frac{30}{100}\)
Now,
As the denominators are equal, compare the numerators.
So, we have to compare 15, 30, and 40
By comparing, we can observe that 40 > 30 > 15
Hence, from the above,
We can conclude that your friend rides farther on zip lines than you.

Question 17.
DIG DEEPER!
You ship a package that weighs 0.8 pounds. Your package is 0.75 pounds lighter than your friend’s package. How much does your friend’s package weigh? Write your answer in decimal form.
Answer: The weight of your friend’s package is: 1.55 pounds

Explanation:
The two given decimal numbers are: 0.8 and 0.75
It is given that your package is 0.75 pounds lighter than your friend’s package.
So,
The weight of your friend’s package = 0.8 + 0.75
Now, first, convert the given decimal numbers into fractions.
So, The representation of 0.8 and 0.75 in the fraction forms is: \(\frac{8}{10}\) and \(\frac{75}{100}\)
Now, for addition, we have to make the denominators of the two fractions equal.
So,
We have to multiply \(\frac{3}{10}\) with 10, to make the denominator equal to 100
So,
The representation of \(\frac{3}{10}\) as hundredths in the fraction form is: \(\frac{30}{100}\)
So,
\(\frac{30}{100}\) + \(\frac{75}{100}\)
= \(\frac{105}{100}\)
The representation of \(\frac{105}{100}\) in the decimal form is: 1.05
Hence, from the above,
We can conclude that the weight of your friend’s package in the decimal form is: 1.05 pounds

Add Decimal Fractions and Decimals Homework & Practice 10.5

Find the sum.

Question 1.
Big Ideas Math Answers Grade 4 Chapter 10 Relate Fractions and Decimals 139
Answer:
The representation of the sum in the fraction form is:
\(\frac{4}{10}\) + \(\frac{32}{100}\) = \(\frac{72}{100}\)
The representation of \(\frac{72}{100}\) in the decimal for is: 0.72

Explanation:
The given fractions are: \(\frac{32}{100}\) and \(\frac{4}{10}\)
For addition, we have to make either denominators or the numerators equal.
so,
For making \(\frac{4}{10}\) as the hundredths, we have to multiply \(\frac{4}{10}\) by 10
So,
The representation of \(\frac{4}{10}\) as hundredths in the fraction form is: \(\frac{40}{100}\)
So,
\(\frac{40}{100}\) + \(\frac{32}{100}\)
= \(\frac{40 + 32}{100}\)
= \(\frac{72}{100}\)
The representation of \(\frac{72}{100}\) in the decimal form is: 0.72
Hence from the above,
We can conclude that
\(\frac{4}{10}\) + \(\frac{32}{100}\) = \(\frac{72}{100}\)
The representation of \(\frac{72}{100}\) in the decimal for is: 0.72

Question 2.
Big Ideas Math Answers Grade 4 Chapter 10 Relate Fractions and Decimals 140
Answer:
The representation of the sum in the fraction form is:
\(\frac{8}{10}\) + \(\frac{3}{100}\) = \(\frac{83}{100}\)
The representation of \(\frac{83}{100}\) in the decimal for is: 0.83

Explanation:
The given fractions are: \(\frac{3}{100}\) and \(\frac{8}{10}\)
For addition, we have to make either denominators or the numerators equal.
so,
For making \(\frac{8}{10}\) as the hundredths, we have to multiply \(\frac{8}{10}\) by 10
So,
The representation of \(\frac{8}{10}\) as hundredths in the fraction form is: \(\frac{80}{100}\)
So,
\(\frac{80}{100}\) + \(\frac{3}{100}\)
= \(\frac{80 + 3}{100}\)
= \(\frac{83}{100}\)
The representation of \(\frac{83}{100}\) in the decimal form is: 0.83
Hence from the above,
We can conclude that
\(\frac{8}{10}\) + \(\frac{3}{100}\) = \(\frac{83}{100}\)
The representation of \(\frac{83}{100}\) in the decimal for is: 0.83

Question 3.
Big Ideas Math Answers Grade 4 Chapter 10 Relate Fractions and Decimals 141
Answer:
The representation of the sum in the fraction form is:
\(\frac{2}{10}\) + \(\frac{15}{100}\) = \(\frac{35}{100}\)
The representation of \(\frac{35}{100}\) in the decimal for is: 0.35

Explanation:
The given fractions are: \(\frac{15}{100}\) and \(\frac{2}{10}\)
For addition, we have to make either denominators or the numerators equal.
so,
For making \(\frac{2}{10}\) as the hundredths, we have to multiply \(\frac{2}{10}\) by 10
So,
The representation of \(\frac{2}{10}\) as hundredths in the fraction form is: \(\frac{20}{100}\)
So,
\(\frac{20}{100}\) + \(\frac{15}{100}\)
= \(\frac{20 + 15}{100}\)
= \(\frac{35}{100}\)
The representation of \(\frac{35}{100}\) in the decimal form is: 0.35
Hence from the above,
We can conclude that
\(\frac{2}{10}\) + \(\frac{15}{100}\) = \(\frac{35}{100}\)
The representation of \(\frac{35}{100}\) in the decimal for is: 0.35

Question 4.
Big Ideas Math Answers Grade 4 Chapter 10 Relate Fractions and Decimals 142
Answer:
The representation of the sum in the fraction form is:
\(\frac{1}{10}\) + \(\frac{45}{100}\) = \(\frac{55}{100}\)
The representation of \(\frac{55}{100}\) in the decimal for is: 0.55

Explanation:
The given fractions are: \(\frac{45}{100}\) and \(\frac{1}{10}\)
For addition, we have to make either denominators or the numerators equal.
so,
For making \(\frac{1}{10}\) as the hundredths, we have to multiply \(\frac{1}{10}\) by 10
So,
The representation of \(\frac{1}{10}\) as hundredths in the fraction form is: \(\frac{10}{100}\)
So,
\(\frac{10}{100}\) + \(\frac{45}{100}\)
= \(\frac{10 + 45}{100}\)
= \(\frac{55}{100}\)
The representation of \(\frac{55}{100}\) in the decimal form is: 0.55
Hence from the above,
We can conclude that
\(\frac{1}{10}\) + \(\frac{45}{100}\) = \(\frac{55}{100}\)
The representation of \(\frac{55}{100}\) in the decimal for is: 0.55

Question 5.
Big Ideas Math Answers Grade 4 Chapter 10 Relate Fractions and Decimals 143
Answer:
The representation of the sum in the fraction form is:
\(\frac{7}{10}\) + \(\frac{22}{100}\) = \(\frac{92}{100}\)
The representation of \(\frac{92}{100}\) in the decimal for is: 0.92

Explanation:
The given fractions are: \(\frac{22}{100}\) and \(\frac{7}{10}\)
For addition, we have to make either denominators or the numerators equal.
so,
For making \(\frac{7}{10}\) as the hundredths, we have to multiply \(\frac{7}{10}\) by 10
So,
The representation of \(\frac{7}{10}\) as hundredths in the fraction form is: \(\frac{70}{100}\)
So,
\(\frac{70}{100}\) + \(\frac{22}{100}\)
= \(\frac{70 + 22}{100}\)
= \(\frac{92}{100}\)
The representation of \(\frac{92}{100}\) in the decimal form is: 0.92
Hence from the above,
We can conclude that
\(\frac{7}{10}\) + \(\frac{22}{100}\) = \(\frac{92}{100}\)
The representation of \(\frac{92}{100}\) in the decimal for is: 0.92

Question 6.
Big Ideas Math Answers Grade 4 Chapter 10 Relate Fractions and Decimals 144
Answer:
The representation of the sum in the fraction form is:
\(\frac{5}{10}\) + \(\frac{17}{100}\) = \(\frac{67}{100}\)
The representation of \(\frac{67}{100}\) in the decimal for is: 0.67

Explanation:
The given fractions are: \(\frac{17}{100}\) and \(\frac{5}{10}\)
For addition, we have to make either denominators or the numerators equal.
so,
For making \(\frac{5}{10}\) as the hundredths, we have to multiply \(\frac{5}{10}\) by 10
So,
The representation of \(\frac{5}{10}\) as hundredths in the fraction form is: \(\frac{50}{100}\)
So,
\(\frac{50}{100}\) + \(\frac{17}{100}\)
= \(\frac{50 + 17}{100}\)
= \(\frac{67}{100}\)
The representation of \(\frac{67}{100}\) in the decimal form is: 0.67
Hence from the above,
We can conclude that
\(\frac{5}{10}\) + \(\frac{17}{100}\) = \(\frac{67}{100}\)
The representation of \(\frac{67}{100}\) in the decimal for is: 0.67

Question 7.
0.6 + 0.25 = ___
Answer:
The representation of the sum in the fraction form is:
\(\frac{6}{10}\) + \(\frac{25}{100}\) = \(\frac{85}{100}\)
The representation of 0.25 + 0.6 in the decimal form is: 0.85

Explanation:
The given decimal numbers are: 0.6 and 0.25
So, convert the two decimal numbers in to respective fractions.
So,
The given fractions are: \(\frac{25}{100}\) and \(\frac{6}{10}\)
For addition, we have to make either denominators or the numerators equal.
so,
For making \(\frac{6}{10}\) as the hundredths, we have to multiply \(\frac{6}{10}\) by 10
So,
The representation of \(\frac{6}{10}\) as hundredths in the fraction form is: \(\frac{60}{100}\)
So,
\(\frac{60}{100}\) + \(\frac{25}{100}\)
= \(\frac{60 + 25}{100}\)
= \(\frac{85}{100}\)
The representation of \(\frac{85}{100}\) in the decimal form is: 0.85
Hence from the above,
We can conclude that
\(\frac{6}{10}\) + \(\frac{25}{100}\) = \(\frac{85}{100}\)
The representation of \(\frac{85}{100}\) in the decimal for is: 0.85

Question 8.
0.3 + 0.40 = ___
Answer:
The representation of the sum in the fraction form is:
\(\frac{3}{10}\) + \(\frac{40}{100}\) = \(\frac{70}{100}\)
The representation of 0.40 + 0.3 in the decimal form is: 0.70

Explanation:
The given decimal numbers are: 0.3 and 0.40
So, convert the two decimal numbers in to respective fractions.
So,
The given fractions are: \(\frac{40}{100}\) and \(\frac{3}{10}\)
For addition, we have to make either denominators or the numerators equal.
so,
For making \(\frac{3}{10}\) as the hundredths, we have to multiply \(\frac{3}{10}\) by 10
So,
The representation of \(\frac{3}{10}\) as hundredths in the fraction form is: \(\frac{30}{100}\)
So,
\(\frac{30}{100}\) + \(\frac{40}{100}\)
= \(\frac{30 + 40}{100}\)
= \(\frac{70}{100}\)
The representation of \(\frac{70}{100}\) in the decimal form is: 0.70
Hence from the above,
We can conclude that
\(\frac{3}{10}\) + \(\frac{40}{100}\) = \(\frac{70}{100}\)
The representation of \(\frac{70}{100}\) in the decimal for is: 0.70

Question 9.
0.05 + 0.9 = ___
Answer:
The representation of the sum in the fraction form is:
\(\frac{9}{10}\) + \(\frac{5}{100}\) = \(\frac{95}{100}\)
The representation of 0.05 + 0.9 in the decimal form is: 0.95

Explanation:
The given decimal numbers are: 0.05 and 0.9
So, convert the two decimal numbers in to respective fractions.
So,
The given fractions are: \(\frac{5}{100}\) and \(\frac{9}{10}\)
For addition, we have to make either denominators or the numerators equal.
so,
For making \(\frac{9}{10}\) as the hundredths, we have to multiply \(\frac{9}{10}\) by 10
So,
The representation of \(\frac{9}{10}\) as hundredths in the fraction form is: \(\frac{90}{100}\)
So,
\(\frac{5}{100}\) + \(\frac{90}{100}\)
= \(\frac{5 + 90}{100}\)
= \(\frac{95}{100}\)
The representation of \(\frac{95}{100}\) in the decimal form is: 0.95
Hence from the above,
We can conclude that
\(\frac{9}{10}\) + \(\frac{5}{100}\) = \(\frac{95}{100}\)
The representation of \(\frac{95}{100}\) in the decimal for is: 0.95

Find the sum.

Question 10.
Big Ideas Math Answers Grade 4 Chapter 10 Relate Fractions and Decimals 145
Answer:
The representation of the sum in the fraction form is:
\(\frac{41}{100}\) + \(\frac{22}{100}\) + \(\frac{3}{10}\) = \(\frac{93}{100}\)
The representation of \(\frac{93}{100}\) in the decimal for is: 0.93

Explanation:
The given fractions are: \(\frac{41}{100}\), \(\frac{22}{100}\) and \(\frac{3}{10}\)
For addition, we have to make either denominators or the numerators equal.
so,
For making \(\frac{3}{10}\) as the hundredths, we have to multiply \(\frac{3}{10}\) by 10
So,
The representation of \(\frac{3}{10}\) as hundredths in the fraction form is: \(\frac{30}{100}\)
So,
\(\frac{41}{100}\) + \(\frac{22}{100}\) + \(\frac{30}{100}\)
= \(\frac{41 + 22 + 30}{100}\)
= \(\frac{93}{100}\)
The representation of \(\frac{93}{100}\) in the decimal form is: 0.93
Hence from the above,
We can conclude that
\(\frac{3}{10}\) + \(\frac{41}{100}\) + \(\frac{22}{100}\) = \(\frac{93}{100}\)
The representation of \(\frac{93}{100}\) in the decimal for is: 0.93

Question 11.
0.8 + 0.25 + 0.75 = ___
Answer:
The representation of the sum in the fraction form is:
\(\frac{8}{10}\) + \(\frac{75}{100}\) + \(\frac{25}{100}\) = \(\frac{180}{100}\)
The representation of 0.25 + 0.75 + 0.8 in the decimal for is: 1.80

Explanation:
The given decimal numbers are: 0.8, 0.75 and 0.25
So, convert the three decimal numbers in to respective fractions.
So,
The given fractions are: \(\frac{25}{100}\) , \(\frac{8}{10}\) and \(\frac{75}{100}\)
For addition, we have to make either denominators or the numerators equal.
so,
For making \(\frac{8}{10}\)  as the hundredths, we have to multiply \(\frac{}{10}\) by 10
So,
The representation of \(\frac{8}{10}\) as hundredths in the fraction form is: \(\frac{80}{100}\)
So,
\(\frac{80}{100}\) + \(\frac{25}{100}\) + \(\frac{75}{100}\)
= \(\frac{80 + 25 + 75}{100}\)
= \(\frac{180}{100}\)
The representation of \(\frac{180}{100}\) in the decimal form is: 1.80
Hence from the above,
We can conclude that
\(\frac{8}{10}\) + \(\frac{25}{100}\) + \(\frac{75}{100}\) = \(\frac{180}{100}\)
The representation of \(\frac{180}{100}\) in the decimal for is: 1.80

Question 12.
Patterns
Describe and complete the pattern.

Answer:
The representation of the sum in the fraction form is:
\(\frac{1}{10}\) + \(\frac{9}{100}\) = \(\frac{19}{100}\)
The representation of \(\frac{19}{100}\) in the decimal for is: 0.19

The representation of the sum in the fraction form is:
\(\frac{2}{10}\) + \(\frac{18}{100}\) = \(\frac{38}{100}\)
The representation of \(\frac{38}{100}\) in the decimal for is: 0.38

The representation of the sum in the fraction form is:
\(\frac{3}{10}\) + \(\frac{27}{100}\) = \(\frac{57}{100}\)
The representation of \(\frac{114}{100}\) in the decimal for is: 0.57
Now,
Whwn we add all the three results, we will get
The last result’s representation in the fraction form is:
\(\frac{19}{100}\) + \(\frac{38}{100}\) + \(\frac{57}{100}\) = \(\frac{114}{100}\)
The last result’s representation in the decimal form is:
0.19 + 0.38 + 0.57 = 1.14

Question 13.
Which One Doesn’t Belong? Which expression does not belong with the other three?
Big Ideas Math Solutions Grade 4 Chapter 10 Relate Fractions and Decimals 147
Answer:
Let the Expressions be named as A, B, C, and D
The given expressions are:
Big Ideas Math Solutions Grade 4 Chapter 10 Relate Fractions and Decimals 147
From the above expressions, Expression D does not belong with the other three

Explanation:
Let the expressions be named  A, B, C, and D
So,
The expressions are:
Big Ideas Math Solutions Grade 4 Chapter 10 Relate Fractions and Decimals 147
So, from the above expressions,
We can observe that the Expression D does not belong with the other three

Question 14.
Modeling Real Life
In an aquarium, \(\frac{5}{10}\) of the fish are red and \(\frac{3}{100}\) of the fish are yellow. What fraction of the fish is more?
Answer: The fraction of the fish that is red is more in the aquarium

Explanation:
It is given that in the aquarium,
There are \(\frac{5}{10}\) of the fish are red and \(\frac{3}{100}\) of the fish are yellow.
So, for comparison, we have to equal the denominators
So,
\(\frac{5}{10}\) has to be multiplied by 10 to make the denominator of \(\frac{5}{10}\) as hundredths
So,
The representation of \(\frac{5}{10}\) as hundredths in the fraction form is: \(\frac{50}{100}\)
So, when we compare the fishes that are red and yellow,
We will get that
Red fishes > Yellow fishes
hence, from the above,
We can conclude that the red fishes are more than yellow fishes in the aquarium

Question 15.
DIG DEEPER!
Which gecko is longer? Explain.
Big Ideas Math Solutions Grade 4 Chapter 10 Relate Fractions and Decimals 148
Answer: Leopard Gecko is longer

Explanation:
It is given that
The total length of Leopard Gecko is: 0.05 + 0.06 = 0.11 m = \(\frac{11}{100}\)
The length of Adult Electric Blue Gecko is: \(\frac{7}{100}\)
So, from the above,
We can say that the numerators of both the fractions are equal.
So, when we compare,
We can see that leopard gecko is longer than adult Electric Blue Gecko
Hence, from the above,
We can conclude that the Leopard Gecko is longer than the Adult Electric Blue Gecko

Review & Refresh

Question 16.
38 ÷ 4
Answer: 38 ÷ 4 = 9 R 2

Explanation:
By using the partial quotients method,
38 ÷ 4 = ( 32 + 4 ) ÷ 4
= ( 32 ÷ 4 ) + ( 4 ÷ 4 )
= 8 + 1
= 9 R 2
Hence, 38 ÷ 4 = 9 R 2

Question 17.
641 ÷ 9
Answer: 641 ÷ 9 = 71 R 2

Explanation:
By using the partial quotients method,
641 ÷ 9 = ( 630 + 9 ) ÷ 9
= ( 630 ÷ 9 ) + ( 9 ÷ 9 (
= 70 + 1
= 71 R 2
Hence, 641 ÷ 9 = 71 R 2

Question 18.
52 ÷ 7
Answer: 52 ÷ 7 = 7 R 3

Explanation:
By using the partial quotients method,
52 ÷ 7 = (42 + 7 ) ÷ 7
= ( 42 ÷ 7 ) + ( 7 ÷ 7 )
= 6 + 1
= 7 R 3
Hence, 52 ÷ 7 = 7 R 3

Lesson 10.6 Fractions, Decimals and Money

Explore and Grow

Shade the model to show each money moment.

Answer: Here, we have taken the reference of 1 dollar and find out the amount by them that is equal to 1 dollar

Explanation:
We know that,
1 Quarter = 0.25 dollar
1 dime = 0.1 dollar
1 nickel = 0.05 dollar
1 penny = 0.01 dollar
Hence, the above values are marked in the above-given model.

Reasoning

How can you write each money amount as a fraction and a decimal in terms of dollars?

Answer:
The representation of the Quarter, in the dollar in the fraction form is: \(\frac{25}{100}\) dollar
The representation of the Quarter, in the dollar in the decimal form is: 0.25

The representation of the dime, in the dollar in the fraction form is: \(\frac{10}{100}\) dollar
The representation of the dime, in the dollar in the decimal form is: 0.10

The representation of the nickel, in the dollar in the fraction form is: \(\frac{5}{100}\) dollar
The representation of the nickel, in the dollar in the decimal form is: 0.05

The representation of the penny, in the dollar in the fraction form is: \(\frac{1}{100}\) dollar
The representation of the penny, in the dollar in the decimal form is: 0.01

Explanation:
We know that,
1 Quarter = 0.25 dollar
1 dime = 0.1 dollar
1 nickel = 0.05 dollar
1 penny = 0.01 dollar
So, all the money moments are represented in the terms of dollars.
So, the representation of all the money moments in terms of dollars in the fraction and the decimal forms is:
The representation of the Quarter, in the dollar in the fraction form is: \(\frac{25}{100}\) dollar
The representation of the Quarter, in the dollar in the decimal form is: 0.25

The representation of the dime, in the dollar in the fraction form is: \(\frac{10}{100}\) dollar
The representation of the dime, in the dollar in the decimal form is: 0.10

The representation of the nickel, in the dollar in the fraction form is: \(\frac{5}{100}\) dollar
The representation of the nickel, in the dollar in the decimal form is: 0.05

The representation of the penny, in the dollar in the fraction form is: \(\frac{1}{100}\) dollar
The representation of the penny, in the dollar in the decimal form is: 0.01

Think and Grow: Fractions, Decimals, and Money

You can use a dollar sign and a decimal point to write a money amount. Just as a decimal point separates ones from tenths and hundredths, it also separates whole dollars from cents.
Big Ideas Math Solutions Grade 4 Chapter 10 Relate Fractions and Decimals 150

Use the total money amount to complete the table.

Show and Grow

Find the total money amount. Then write the amount as a fraction or mixed number and as a decimal.

Question 1.
Big Ideas Math Solutions Grade 4 Chapter 10 Relate Fractions and Decimals 152
Answer: The total amount of money is: $0.66

Explanation:
We know that,
1 Quarter = $0.25
1 nickel = $0.05
1 penny = $0.01
The given money is: 2 Quarters, 3 nickels, and 1 penny
So,
The total amount of money = ( 2 × 0.25 ) + ( 3 × 0.05 ) + ( 1 × 0.01 )
= 0.50 + 0.15 + 0.01
= 0.06
Hence,
The total amount of money is: $0.06
The representation of $0.06 in the fraction form is: \(\frac{6}{100}\) dollar
The representation of $0.06 in the decimal form is: $0.06

Question 2.
Big Ideas Math Solutions Grade 4 Chapter 10 Relate Fractions and Decimals 153
Answer: The total amount of money is: $2.30

Explanation:
We know that,
1 Quarter = $0.25
1 nickel = $0.05
The given money is: 2 dollars, 1 Quarter, and 1 nickel
So,
The total amount of money = ( 2 × 1 ) + ( 1 × 0.25 ) + ( 1 × 0.05 )
= 2 + 0.25 + 0.05
= 2.30
Hence,
The total amount of money is: $2.30
The representation of $2.30 in the fraction form is: 2\(\frac{30}{100}\) dollar
The representation of $2.30 in the decimal form is: $2.30

Apply and Grow: Practice

Find the total money amount. Then write the amount as a fraction or mixed number and as a decimal.

Question 3.
Big Ideas Math Solutions Grade 4 Chapter 10 Relate Fractions and Decimals 154
Answer: The total amount of money is: $0.46

Explanation:
We know that,
1 Quarter = $0.25
1 nickel = $0.05
1 dime = $0.10
1 penny = $0.01
The given money is: 6 pennies, 1 Quarter, 1 dime, and 1 nickel
So,
The total amount of money = ( 6 × 0.01 ) + ( 1 × 0.25 ) + ( 1 × 0.05 ) + ( 1 × 0.10 )
= 0.06 + 0.25 + 0.05 + 0.10
= 0.46
Hence,
The total amount of money is: $0.46
The representation of $0.46 in the fraction form is: \(\frac{46}{100}\) dollar
The representation of $0.46 in the decimal form is: $0.46

Question 4.
Big Ideas Math Solutions Grade 4 Chapter 10 Relate Fractions and Decimals 155
Answer: The total amount of money is: $0.46

Explanation:
We know that,
1 Quarter = $0.25
1 nickel = $0.05
4 Quarters = 1 dollar
The given money is: 4 Quarters, 1 dollar, and 1 nickel
So,
The total amount of money = ( 1 × 1 ) + ( 4 × 0.25 ) + ( 1 × 0.05 )
= 1 + 1 + 0.05
= 2.05
Hence,
The total amount of money is: $2.05
The representation of $2.05 in the fraction form is: 2\(\frac{5}{100}\) dollar
The representation of $2.05 in the decimal form is: $2.05

Write the fraction or mixed number as a money amount and as a decimal.

Question 5.
\(\frac{53}{100}\)
Answer: The representation of \(\frac{53}{100}\) as the total amount of money is: $0.53

Explanation:
The given fraction is: \(\frac{53}{100}\)
The given fraction will be given as an amount in dollars.
So,
The total amount of money in the decimal form is: $0.53

Question 6.
\(\frac{4}{100}\)
Answer: The representation of \(\frac{4}{100}\) as the total amount of money is: $0.04

Explanation:
The given fraction is: \(\frac{4}{100}\)
The given fraction will be given as an amount in dollars.
So,
The total amount of money in the decimal form is: $0.04

Question 7.
\(\frac{100}{100}\)
Answer: The representation of \(\frac{100}{100}\) as the total amount of money is: $1

Explanation:
The given fraction is: \(\frac{100}{100}\)
The given fraction will be given as an amount in dollars.
So,
The total amount of money in the decimal form is: $1

Question 8.
1\(\frac{22}{100}\)
Answer: The representation of 1\(\frac{22}{100}\) as the total amount of money is: $1.22

Explanation:
The given fraction is: 1\(\frac{22}{100}\)
The given fraction will be given as an amount in dollars.
So,
The total amount of money in the decimal form is: $1.22

Question 9.
1\(\frac{18}{100}\)
Answer: The representation of 1\(\frac{18}{100}\) as the total amount of money is: $1.18

Explanation:
The given fraction is: 1\(\frac{18}{100}\)
The given fraction will be given as an amount in dollars.
So,
The total amount of money in the decimal form is: $1.18

Question 10.
1\(\frac{70}{100}\)
Answer: The representation of 1\(\frac{70}{100}\) as the total amount of money is: $1.70

Explanation:
The given fraction is: 1\(\frac{70}{100}\)
The given fraction will be given as an amount in dollars.
So,
The total amount of money in the decimal form is: $1.70

Question 11.
You find 1 dime, 3 nickels, and 2 pennies on the ground. How much money do you find? Write your answer in three different ways.
Answer: The total amount of money you find on the ground is: $0.27

Explanation:
The given amount that you find on the ground is: 1 dime, 3 nickels, and 2 pennies
We know that,
1 dime = 0.10 dollars
1 nickel = $0.05 dollars
1 penny = $0.01 dollars
So,
The total amount of money you find on the ground = ( 1 × 0.10 ) + ( 3 × 0.05 ) + (2 × 0.01 )
= 0.10 + 0.15 + 0.02
= 0.27
Hence from the above,
We can conclude that the three ways of representing the total amount of money is:
The total amount of money you find on the ground is: $0.27
The representation of the total amount of money in the fraction form is: \(\frac{27}{100}\) dollar
The representation of the total amount of money in the decimal form is: $0.27

Question 12.
YOU BE THE TEACHER
Your friend has three $1 bills and 2 pennies. Your friend writes, “I have $ 3.2.” Is your friend correct? Explain.
Answer: No, your friend is wrong

Explanation:
It is given that your friend has three $1 bills and 2 pennies.
So,
The given amount your friend has: 3 $1 bills and 2 pennies
We know that,
1 penny = 0.01 dollars
So,
The total amount of money your friend has = ( 3 × 1 ) + ( 2 × 0.01 )
= 3 + 0.02
= 3.02
Hence,
The total amount of money your friend has: $3.02 dollars or 3 dollars and 2 pennies
But, your friend has written $3.02 as $3.20 which means 3 dollars and 2 dimes
Hence, from the above,
We can conclude that your friend is wrong

Question 13.
DIG DEEPER!
You have \(\frac{1}{4}\) dollar in coins. Draw two possible groups of coins that you could have.
Answer: The possible group of coins that you could have for \(\frac{1}{4}\) is: Quarter

Explanation:
It is given that you have \(\frac{1}{4}\) dollar in coins.
But, we have al, the money moments in terms of 100.
So, to make the denominator of \(\frac{1}{4}\) 100, multiply \(\frac{1}{4}\) by 25
So,
The representation of \(\frac{1}{4}\) as hundredths in the fraction form is: \(\frac{25}{100}\)
We know that,
1 Quarter = $0.25
Hence, from the above,
We can conclude that the possible group of coins that you could have for \(\frac{1}{4}\) is: Quarter

Think and Grow: Modeling Real Life

Example
Newton has \(\frac{85}{100}\) dollar. Can he buy the spinning toy? Explain.
Big Ideas Math Solutions Grade 4 Chapter 10 Relate Fractions and Decimals 156
Write the fraction as a money amount.
\(\frac{85}{100}\) as a money amount is $0.85.
Compare the amount of money Newton has to the price of the toy.
Newton can’t buy the spinning toy.

Explanation:
It is given that Newton has \(\frac{85}{100}\) dollar
So,
The amount of money that Newton has in the decimal form is: $0.85
But, it is also given that
The cost of spinning toy is: \(\frac{99}{100}\) dollar
So, for comparison, we have to make either the numerators or the denominators equal.
Here, both the denominators are equal.
So, we can compare both the fractions directly.
So, by comparing, we get,
0.85 < 0.99
Hence,
We can conclude that Newton can’t buy the spinning toy.

Show and Grow

Question 14.
Descartes has \(\frac{76}{100}\) dollar. Can he buy the bouncy ball? Explain.
Big Ideas Math Solutions Grade 4 Chapter 10 Relate Fractions and Decimals 156.1
Answer: Yes, he can buy the bouncy ball

Explanation:
It is given that Descartes has \(\frac{76}{100}\) dollar
So,
The amount of money that Descartes has in the decimal form is: $0.76
But, it is also given that
The cost of bouncy ball is: \(\frac{50}{100}\) dollar
So, for comparison, we have to make either the numerators or the denominators equal.
Here, both the denominators are equal.
So, we can compare both the fractions directly.
So, by comparing, we get,
0.76 > 0.50
Hence,
We can conclude that Descartes can buy the bouncy ball.

Question 15.
You throw 3 dimes, 3 nickels, and 8 pennies into a fountain. Your friend throws 1 quarter, 4 nickels, and 5 pennies. Who throws a greater amount of money into the fountain?
Answer: You throw a greater amount of money into the fountain when compared to your friend.

Explanation:
It is given that you throw 3 dimes, 3 nickels, and 8 pennies and your friend throws 1 quarter, 4 nickels, and 5 pennies
So,
The amount that you have is: 3 dimes, 3 nickels, and 8 pennies
The amount that your friend has is: 1 quarter, 4 nickels, and 5 pennies
We know that,
1 quarter = $0.25
1 dime = $0.10
1 nickel = $0.05
1 penny = $0.01
So,
The total amount of money you have = ( 3 × 0.10 ) + ( 3 × 0.05 ) + ( 8 × 0.01 )
= 0.30 + 0.15 + 0.08
= $0.53
The total amount of money your friend has = ( 1 × 0.25 ) + ( 4 × 0.05 ) + ( 5 × 0.01 )
= 0.25 + 0.20 + 0.05
= $0.50
Now,
The representation of the amount of money you have in the fraction form is: \(\frac{53}{100}\) dollars
The representation of the amount of money your friend has in the fraction form is: \(\frac{50}{100}\) dollars
So, by comparing these two amounts,
We can conclude that you have a greater amount of money than your friend

Question 16.
DIG DEEPER!
Complete the table. Which piggy bank has the greatest amount of money? the least amount of money?

Answer:
The piggy bank which has the greatest amount of money is: B
the piggy bank which has the least amount of money is: C

Explanation:
The given table is:

From the given table,
The amount of money present in all the three piggy banks is: Quarters, Dimes, Nickels,  and pennies
We know that,
1 Quarter = $0.25
1 dime = $0.10
1 nickel = $0.05
1 penny = $0.01
So,
The total amount of money that all the three piggy banks have:
A: ( 3 × 0.25 ) + ( 1 × 0.10 ) + ( 4 × 0.05 ) + ( 2 × 0.01 ) = $1.07
B: ( 1 × 0.25 ) + ( 7 × 0.10 ) + ( 3 × 0.05 ) + ( 0 × 0.01 ) = $1.10
C: ( 2 × 0.25 ) + ( 0 × 0.10 ) + ( 8 × 0.05 ) + ( 11 × 0.01 ) = $1.01
Hence, from the above,
we can conclude that
The piggy bank that has the greatest amount of money is: B
The piggy bank that has the least amount of money is: C

Fractions, Decimals and Money Homework & Practice 10.6

Write the money amount as a fraction or mixed number and as a decimal.

Question 1.
$ 0.53
Answer: The representation of $0.53 as the total amount of money in the fraction form is: \(\frac{53}{100}\) dollar

Explanation:
The given amount of money in the decimal form is: $0.53
The given decimal form will be given as an amount in dollars.
So,
The total amount of money in the fraction form is: \(\frac{53}{100}\) dollars
The total amount of money in the decimal form is: $0.53

Question 2.
$ 0.40
Answer: The representation of $0.40 as the total amount of money in the fraction form is: \(\frac{40}{100}\) dollar

Explanation:
The given amount of money in the decimal form is: $0.40
The given decimal form will be given as an amount in dollars.
So,
The total amount of money in the fraction form is: \(\frac{40}{100}\) dollars
The total amount of money in the decimal form is: $0.40

Question 3.
$1.01
Answer: The representation of $1.01 as the total amount of money in the fraction form is: 1\(\frac{1}{100}\) dollar

Explanation:
The given amount of money in the decimal form is: $1.01
The given decimal form will be given as an amount in dollars.
So,
The total amount of money in the fraction form is: 1\(\frac{1}{100}\) dollars
The total amount of money in the decimal form is: $1.01

Find the total money amount. Then write the amount as a fraction or mixed number and as a decimal.

Question 4.
Big Ideas Math Solutions Grade 4 Chapter 10 Relate Fractions and Decimals 158
Answer: The total amount of money is: $0.65

Explanation:
We know that,
1 Quarter = $0.25
1 nickel = $0.05
1 dime = $0.10
The given money is: 1 Quarter, 2 dimes, and 4 nickel
So,
The total amount of money = ( 1 × 0.25 ) + ( 4 × 0.05 ) + ( 2 × 0.10 )
=  0.25 + 0.20 + 0.20
= 0.65
Hence,
The total amount of money is: $0.65
The representation of $0.65 in the fraction form is: \(\frac{65}{100}\) dollar
The representation of $0.65 in the decimal form is: $0.65

Question 5.
Big Ideas Math Solutions Grade 4 Chapter 10 Relate Fractions and Decimals 159
Answer: The total amount of money is: $2.13

Explanation:
We know that,
1 dime = $0.10
1 penny = $0.01
The given money is: 3 pennies, 1 dime, and 2 dollars
So,
The total amount of money = ( 3 × 0.01 ) + ( 2 × 1 ) + ( 1 × 0.10 )
= 0.03 + 2 + 0.10
= 2.13
Hence,
The total amount of money is: $2.13
The representation of $2.13 in the fraction form is: 2\(\frac{13}{100}\) dollar
The representation of $2.13 in the decimal form is: $2.13

Question 6.
Big Ideas Math Solutions Grade 4 Chapter 10 Relate Fractions and Decimals 160
Answer: The total amount of money is: $0.92

Explanation:
We know that,
1 Quarter = $0.25
1 dime = $0.10
1 penny = $0.01
The given money is: 2 pennies, 2 Quarters, 4 dimes
So,
The total amount of money = ( 2 × 0.01 ) + ( 2 × 0.25 ) + ( 4 × 0.10 )
= 0.02 +0.50 + 0.40
= 0.92
Hence,
The total amount of money is: $0.92
The representation of $0.92 in the fraction form is: \(\frac{92}{100}\) dollar
The representation of $0.92 in the decimal form is: $0.92

Question 7.
Big Ideas Math Solutions Grade 4 Chapter 10 Relate Fractions and Decimals 161
Answer: The total amount of money is: $3.15

Explanation:
We know that,
1 Quarter = $0.25
1 nickel = $0.05
1 dime = $0.10
The given money is: 4 Quarters, 2 dollars, 1 dime, and 1 nickel
So,
The total amount of money = ( 2 × 1 ) + ( 4 × 0.25 ) + ( 1 × 0.05 ) + ( 1 × 0.10 )
= 2 + 1 + 0.05 + 0.10
= 3.15
Hence,
The total amount of money is: $3.15
The representation of $3.15 in the fraction form is: 3\(\frac{15}{100}\) dollar
The representation of $3.15 in the decimal form is: $3.15

Write the fraction or mixed number as a money amount and as a decimal.

Question 8.
\(\frac{87}{100}\)
Answer: The representation of \(\frac{87}{100}\) as the total amount of money is: $0.87

Explanation:
The given fraction is: \(\frac{87}{100}\)
The given fraction will be given as an amount in dollars.
So,
The total amount of money in the decimal form is: $0.87

Question 9.
\(\frac{12}{100}\)
Answer: The representation of \(\frac{12}{100}\) as the total amount of money is: $0.12

Explanation:
The given fraction is: \(\frac{12}{100}\)
The given fraction will be given as an amount in dollars.
So,
The total amount of money in the decimal form is: $0.12

Question 10.
1\(\frac{9}{100}\)
Answer: The representation of 1\(\frac{9}{100}\) as the total amount of money is: $1.09

Explanation:
The given fraction is: 1\(\frac{9}{100}\)
The given fraction will be given as an amount in dollars.
So,
The total amount of money in the decimal form is: $1.09

Question 11.
You find 3 quarters, 2 nickels, and 1 penny in your backpack. How much money do you find? Write your answer in three different ways.
Answer: The amount of money you find is: $0.86

Explanation:
It is given that you find 3 quarters, 2 nickels, and 1 penny in your backpack
So,
The amount of money you have is: 3 quarters, 2 nickels, and 1 penny
We know that,
1 quarter = $0.25
1 nickel = $0.05
1 penny = $0.01
So,
The total amount of money you have = ( 3 × 0.25 ) + ( 2 × 0.05 ) + ( 1 × 0.01 )
= 0.75 + 0.10 + 0.01
= 0.86
Hence,
The total amount of money you have is: $0.86
The total amount of money you have in the fraction form is: \(\frac{86}{100}\)
The total amount of money you have in the decimal form is: $0.86

Question 12.
Which One Doesn’t Belong? Which one does not belong with the other three?
Big Ideas Math Solutions Grade 4 Chapter 10 Relate Fractions and Decimals 162
Answer:
Let the amounts be named A, B, C, and D
So, from the amounts,
we can say that A) does not belong to the other three.

Explanation:
Le the amounts named as A, B, C, and D
Now,
A) 3 pennies  B) \(\frac{3}{10}\)  C) 0.30 dollars  D) \(\frac{30}{100}\)
We know that,
1 penny = $0.01
So,
3 pennies = $0.03
Hence, from the above,
we can conclude that A) does not belong to the other three.

Question 13.
Reasoning
Would you rather have \(\frac{2}{10}\) of a dollar or 6 nickels? Explain.
Answer: You would rather have 6 nickels

Explanation:
The given fraction is: \(\frac{2}{10}\) of a dollar
So,
The representation of \(\frac{2}{10}\) as hundredths in the fraction form is: \(\frac{20}{100}\)
The representation of \(\frac{20}{100}\) in the decimal form is: 0.20
We know that,
1 nickel = $0.05
So,
6 nickels = 6 × 0.05 = $0.30
Now,
When we compare the given fraction and 6 nickels, w can observe that
6 nickels > \(\frac{2}{10}\) of a dollar
Hence, from the above,
we can conclude that you would have to rather have 6 nickels than \(\frac{2}{10}\) of a dollar

Question 14.
Modeling Real Life
Newton has \(\frac{46}{100}\) dollar. Can he buys the key chain ? Explain.
Big Ideas Math Solutions Grade 4 Chapter 10 Relate Fractions and Decimals 163
Answer: No, Newton can’t buy the key chain

Explanation:
It is given that Newton has \(\frac{46}{10}\) dollar
So,
The representation of \(\frac{46}{10}\) in the decimal form is: $0.46
It is also given that,
The cost of the key chain is: $0.49
Now,
When we compare the fractions, we have to equate either the numerators or the denominators.
Here, both the denominators are equal.
So,
When we compare, we will observe
0.46 < 0.49
Hence, from the above,
We can conclude that Newton can’t buy the key chain

Question 15.
DIG DEEPER!
Descartes has $1. Can he buy 2 key chains? Explain how you know without calculating.
Answer: Yes, Descartes can buy 2 key chains

Explanation:
From the above problem,
The cost of 1 key chain is: $0.49
It is given that Descartes has $1 and he wants to buy the 2 key chains
So,
The cost of 2 key chains = 0.49 + 0.49 = $0.98
But, Descartes has $1
So,
The money that Descartes left = 1 – 0.98 = 0.02
Hence, from the above,
We can conclude that Descartes can buy the 2 key chains

Review & Refresh

Add.

Question 16.
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 164
Answer: \(\frac{24}{100}\) + \(\frac{35}{100}\) = \(\frac{59}{100}\)

Explanation:
The given fractions are: \(\frac{24}{100}\) and \(\frac{35}{100}\)
So, for the addition of the fractions, we have to equate either both the numerators or both the denominators.
Here, both the denominators are equal.
Hence,
\(\frac{24}{100}\) + \(\frac{35}{100}\) = \(\frac{59}{100}\)

Question 17.
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 165
Answer: \(\frac{10}{8}\) + \(\frac{3}{8}\) = \(\frac{13}{8}\)

Explanation:
The given fractions are: \(\frac{10}{8}\) and \(\frac{3}{8}\)
So, for the addition of the fractions, we have to see whether the denominators are equal or the numerators are equal.
Here, both the denominators are equal.
Hence,
\(\frac{10}{8}\) + \(\frac{3}{8}\) = \(\frac{13}{8}\)

Question 18.
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 166
Answer: \(\frac{1}{10}\) + \(\frac{3}{10}\) + \(\frac{6}{10}\) = \(\frac{10}{10}\)

Explanation:
The given fractions are: \(\frac{1}{10}\), \(\frac{3}{10}\) and \(\frac{6}{10}\)
So, in addition,
We have to equate either both the numerators or both the denominators.
Here,
The denominators of all the fractions are equal.
Hence,
\(\frac{1}{10}\) + \(\frac{3}{10}\) + \(\frac{6}{10}\) = \(\frac{10}{10}\)

Lesson 10.7 Operations with Money

Explore and Grow

Draw bills and coins to solve each problem. How much do all of the toys cost?
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 167

  1. You pay for one of the toys with a $10 bill. What is your change?

Answer:
Let the toy be an Action figure.
Now,
It is given that the cost of an Action figure is: $5.50
It is also given that you pay for one of the toys with a $10 bill
So,
Your change = 10 – 5.50 = $4.50
Hence, from the above,
We can conclude that the change is: $4.50

2. You buy three of the same toys. How much do the toys cost in all?

Answer:
The given table is:
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 167
From the given table,
The cost of an Action figure is: $5.50
The cost of a whistle is: $1.25
The cost of a Board game is: $3.75
So,
The cost of 3 Action figures = 3 × 5.50 = $16.50
The cost of 3 whistles = 3 × 1.25 = $3.75
The cost of 3 board games = 3 × 3.75 = $11.15

3. You and your friend put your money together to buy some of the toys. The cashier gives you a $4.50 change. You want to share the change equally. How much money does each of you get?

Answer: It is given that you and your friend put your money together to buy some of the toys. It is also given that the cashier gives you a $4.50 change.
So,
The amount that should be shared equally between you and your friend = 4.50 ÷ 2
= (4 ÷ 2 ) + ( 0.50 ÷ 2 )
= 2 + 0.25
= 2.25
Hence, from the above,
we can conclude that the amount that should be shared equally between you and your friend is: $2.25

Precision
Compare your work to your partner’s.

Think and Grow:
Newton has $1.35. Descartes has $1.25. How much money do they have altogether?
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 168
Answer:
It is given that Newton has $1.35 and Descartes has $1.25.
So,
The total amount of money they have altogether = $1.35 + $1.25
= $2.60
Hence, from the above,
We can conclude that they have $2.60 altogether.

Example
Newton has $2.45. He spends $1.10. How much money does he have left?

He has $1.35 left.
Example
Three friends each have$0.60. How much money do they have in all?
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 170
Answer: It is given that the three friends each have $0.60
So,
the total amount of money that the three friends have = 0.60 × 3 = $1.80
Hence, from the above,
We can conclude that they have $1.80 in all.
Example
You and a friend have a total of $ 1.48. You want to share the money equally. How much money should each of you get?
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 171
Answer: It is given that you and a friend have a total of $1.48
It is also given that you want to share the money equally
So,
The amount of money that you and your friend share the money equally = 1.48 ÷ 2
= ( 1 ÷ 2 ) + ( 0.48 ÷ 2 )
= 0.5 + 0.24
= 0.74
Hence, from the above,
We can conclude that you each should get $0.74.

Show and Grow

Question 1.
You pay a total of $2.25 for 3 granola bars. How much money does each bar cost? Draw bills and coins to solve.
Answer: The amount of money does each bar cost is: $0.75

Explanation:
It is given that you pay a total of $2.25 for 3 granola bars
So,
The amount of money that each bar cost = The total cost of 3 granola bars ÷ 3
= 2.25 ÷ 3
= 0.75
Hence, from the above,
We can conclude that the amount of money does each ba cost is: $0.75

Apply and Grow: Practice

Question 2.
You buy 2 stamps. Each stamp costs $0.49. How much money do you spend in all?
Answer: The total amount of money you spend is: $0.98

Explanation:
It is given that you buy 2 stamps and each stamp costs $0.49
So,
The total amount of money you spent on buying stamps = The cost of each stamp × 2
= 0.49 × 2
= $0.98
Hence, from the above,
we can conclude that the amount of money you spent on buying the 2 stamps is: $0.98

Question 3.
Newton has $2.50. He spends $1.07 on a flying disk. How much money does Newton have left?
Answer: The amount of money does Newton have left is: $1.43

Explanation:
It is given that Newton has $2.50 and he spends $1.07 on a flying disk
So,
The amount of money does Newton have left = The amount of money that Newton has – The amount of money that Newton spent
= 2.50 – 107
= $1.43
Hence, from the above,
We can conclude that the amount of money does Newton have left is: $1.43

Question 4.
A tube of toothpaste costs $2.71 and a toothbrush costs $1.62. How much more money does the toothpaste cost more than the toothbrush?
Answer: The amount of money that the toothpaste cost more than the toothbrush is: $1.09

Explanation:
It is given that a tube of toothpaste costs $2.71 and a toothbrush costs $1.62.
So,
The amount of money that the toothpaste cost more than the toothbrush = The cost of toothpaste -The cost of the toothbrush
= 2.71 – 1.62
= $1.07
Hence, from the above,
We can conclude that the more money does the toothpaste cost more than the toothbrush is: $1.07

Question 5.
Two fingerboards cost a total of $7.20. Each fingerboard costs the same amount. How much does each fingerboard cost?
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 172
Answer: Each fingerboard costs: 0.36

Explanation:
It is given that the two fingerboards cost a total of $7.20
It is also given that each fingerboard costs the same amount.
So,
The cost of each fingerboard = The cost of 2 fingerboards ÷ 2
= 7.20 ÷ 2
= ( 7 ÷ 2 ) + ( 0.2 ÷ 2 )
= 3.5 + 0.1
= 3.6
Hence, from the above,
We can conclude that the cost of each fingerboard is: $0.36

Question 6.
In Exercise 2, you pay for the stamps using a $1 bill. What is your change?
Answer:
From Exercise 2,
The total cost of buying 2 stamps is: $0.98
In this exercise, it is given that you pay for the stamps using a $1 bill.
So,
The change = 1 – 0.98
= 0.02
Hence, from the above,
We can conclude that The change after paying the $1 bill for paying the stamps is: $0.02

Question 7.
You have four $1 bills and 3 dimes. Do you have enough money to buy the tube of toothpaste and the toothbrush in Exercise 4? Explain.
Answer: No, you don’t have enough money to buy the tube of toothbrush and toothpaste which is explained in Exercise 4

Explanation:
From Exercise 4,
The total cost of money to buy the tube of toothpaste and the toothbrush = 2.71 + 1.62 = $4.33
It is given that you have 4 $1 bills and 3 dimes
We know that,
1 dime = $0.10
So,
The total amount of money you have = (4 × 1 ) + ( 3 × 0.10 )
= 4 + 0.30
= 4.30
Now,
We have to compare 4.33 and 4.30
So, by comparing the 2 values, we will observe
4.33 > 4.30
Hence, from the above,
We can conclude that you don’t have enough money to buy the toothpaste and the toothbrush

Question 8.
DIG DEEPER!
You have $1.10 less than Descartes. How much money do you, Newton, and Descartes have altogether?
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 173
Answer:
It is given that you have $1.10 less than Descartes
From the above figure,
Newton has $1.50 and Descartes has $1.25 more than Newton
So,
The amount of money that Descartes has = The amount of money Newton has + 1.25
= 1.50 + 1.25
= $2.75
The amount of money you have = The amount of money Descartes has – 1.10
= 2.75 – 1.10
= $1.65
Hence, from the above,
We can conclude that
The amount of money you have is: $1.65
The amount of money Newton has is: $1.50
The amount of money Descartes has is: $2.75

Think and Grow: Modeling Real Life

Example
You buy a joke book that costs $3.50 and a book about science experiments that costs $4.25. You give the cashier $8. What is your change?
Think: What do you know? What do you need to find? How will you solve?
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 174
Step 1: Draw bills and coins to show the money you give the cashier, $8.
Step 2: Subtract the cost of the joke book. Subtract $3.50 by taking away
3 $1 bills and 2 quarters.
The given model is:
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 175
Step 3: Subtract the cost of the science experiment book.
Subtract $4.25 by taking away
4 $1 bills and 1 quarter.
Step 4: The remaining coin represents your change.
Your change is $1.75

Show and Grow

Question 9.
You buy a gel pen that costs $1.10 and a school shirt that costs $5.85. You give the cashier $7. What is your change?
Answer: The change is: $0.05

Explanation:
It is given that you buy a gel pen that costs $1.10 and a school shirt that costs $5.85.
So,
The total cost of the gel pen and the school skirt = 5.85 + 1.10 = $ 6.95
It is also given that you give the cashier $7
So,
the change = The money given to the cashier – The total cost of the gel pen and the school skirt
= 7 – 6.95
= 0.05
Hence, from the above,
We can conclude that the change is: $0.05

Question 10.
You have $2.50. Your friend has 2 times as much money as you. How much money do you and your friend have altogether?
Answer: The amount of money you and your friend have altogether is: $7.50

Explanation:
It is given that you have $2.50 and your friend has 2 times as much money as you.
So,
The amount of money your friend has = 2.50 × 2 = $5
So,
The amount of money you and your friend have altogether = The amount of money you have + The amount of money your friend has
= 5 + 2.5
= $7.5
Hence, from the above,
We can conclude that the amount of money you and your friend have altogether is: $7.50

Question 11.
DIG DEEPER!
You have $8.38. Your friend has $3.16. How much money can you give to your friend so that you each have the same amount?
Answer: The amount of money you can give to your friend so that you each have the same amount = $5.22

Explanation:
It is given that you have $8.38 and your friend has $3.16
So,
The amount of money you can give to your friend so that you each have the same amount = The amount of money you have – The amount of money your friend has
= 8.38 – 3.16
= $5.22
Hence, from the above,
We can conclude that the amount of money you can give to your friend so that you each have the same amount is: $5.22

Operations with Money Homework & Practice 10.7

Draw bills and coins to solve.

Question 1.
A sketch pad is $2.85 and a sketching pencil is $1.25. How much more money is the sketch pad than the sketching pencil?
Answer: The amount of money the sketch pad needs more than the sketching pencil is: $1.60

Explanation:
It is given that a sketch pad is $2.85 and a sketching pencil is $1.25.
So,
The cost of a sketch pad is: $2.85
The cost of a sketching pencil is: $1.25
So,
the amount of money more than the sketching pencil = The cost of sketch pad – The cost of the sketching pencil
= 2.85 – 1.25
= $1.60
Hence, from the above,
We can conclude that the amount of money needed more than the sketching pencil is: $.1.60

Question 2.
Descartes buys two toys for a total of $2.54. Each toy costs the same amount. How much does each toy cost?
Answer: The cost of each boy is: $1.27

Explanation:
It is given that Descartes buys 2 toys for a total of $2.54 and it is also given that each toy costs the same amount
So,
The cost of each toy = The total cost of the two toys ÷ 2
= 2.54 ÷ 2
= $1.27
Hence, from the above,
We can conclude that the cost of each toy is: $1.27

Question 3.
You buy 4 bags of the water balloons shown. How much money do you spend in all?
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 177
Answer: The total cost of the 4 bags of balloons are: $4.80

Explanation:
It is given that you have 4 bags of balloons and it is also given that the cost of each bag of balloon is $1.20
So,
The total cost of the four bags of balloons = 4 × 1.20
= $4.80
hence, from the above,
We can conclude that the cost of the four bags of balloons is: $4.80

Question 4.
You have four $1 bills and 2 nickels. Do you have enough money to buy the sketch pad and the sketching pencil in Exercise 1? Explain.
Answer: No, we have enough money to buy the sketch pad and the sketching pencil

Explanation:
From Exercise 1,
The total cost of the sketch pen and the sketching pencil = 2.85 + 1.25
= $4.10
In this exercise, it is given that you have 4 $1 bills and 2 nickels.
We know that,
1 nickel = $0.05
So,
The total amount of money = ( 4 × 1 ) + ( 2 × 0.05 )
=  + 0.1
= $4.1
So,
The total amount from exercise 1 and the money you have are equal
Hence, from the above,
We can conclude that we have enough money to buy the sketch pen and the sketching pencil

Question 5.
Reasoning
You have 3 jars, with $2.32 in each jar. Do you have enough money to buy the model car? If not, how much more money do you need?
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 178
Answer: You don’t have enough money to buy the model car
The amount of money that you needed more to buy the model car is: $0.53

Explanation:
It is given that you have 3 jars, with $2.32 in each jar
So,
The total amount of money n the three jars = 3 × 2.32 = $6.96
It is also given that
The cost of the model car is: $7.49
So,
The amount of money needed more to buy the model car = The cost of the model car – The total amount of money in the three jars
= 7.49 – 6.96
= $0.53
Hence, from the above,
We can conclude that we don’t have enough money to buy the model car
The amount of money needed more to buy the car is: $0.53

Question 6.
DIG DEEPER!
Descartes has 3 quarters, 1 dime, and 3 nickels. He wants to put the same amount of money into each of the two piggy banks. How can he do this with these coins?
Answer: The amount of money in each of the two piggy banks is: $0.5

Explanation:
It is given that Descartes has 3 quarters, 1 dime, and 3 nickels.
So,
The amount of money Descartes possess is: 3 quarters, 1 dime, and 3 nickels
We know that,
1 quarter = $0.25
1 dime = $0.10
1 nickel = $0.05
So,
The total amount of money Descartes possesses = ( 3 × 0.25 ) + ( 1 × 0.10 ) + ( 3 × 0.05 )
= 0.75 + 0.10 + 0.15
= $1.00
So,
The amount of money in each of the two piggy banks = 1 ÷ 2 = $0.5
Hence, from the above,
We can conclude that the amount of money in each of the two piggy banks is: $0.5

Question 7.
Modeling Real Life
You buy the key chains shown. You pay with a $5 bill. What is your change?
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 179
Answer: The change is: $1.10

Explanation:
It is given that you have bought some key chains as shown in the below figure.

Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 179
From the above figure,
The cost of a basketball key chain is: $2.53
The cost of a football chain is: $2.37
So,
The total cost of the two key chains = 2.53 + 2.37
= $4.90
It is also given that you pay with a $5 bill
So,
The change = 5 – The total cost of the key chains
= 5 – 4.90
= $1.10
Hence, from the above,
We can conclude that the change obtained after buying the key chains is: $1.10

Question 8.
DIG DEEPER!
Your class collects pennies and nickels in separate jars. Your class collects $5.87 in pennies and $2.65 in nickels. You divide the total amount of money collected between two charities. How many pennies do you put in the nickel jar so that both jars have the same amount of money?
Answer:
The total amount of money divided between the 2 charities is: $4.26
The number of pennies you need to put in the nickel jar is: $3.22

Explanation:
It is given that your class collects pennies and nickels in separate jars. It is also given that your class collects $5.87 in pennies and $2.65 in nickels.
It is given that you divide the total amount of money collected between two charities.
So,
The total amount of money collected by the class = 5.87 + 2.65 = $8.52
Now,
The total amount of money divided between the 2 charities = 8.52 ÷ 2
= ( 8 ÷ 2 ) + ( 0.52 ÷ 2 )
= 4 + 0.26
= $ 4.26
Now,
The amount of money needed more to add pennies into the nickel jars = 5.87 – 2.65 = $3.22
Hence, from the above,
We can conclude that
The total amount of money divided between the 2 charities is: $4.26
The number of pennies you need to put in the nickel jar is: $3.22

Review & Refresh

Write the product as a multiple of a unit fraction. Then find the product.

Question 9.
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 180
Answer: The product of 6 and \(\frac{7}{12}\) is: \(\frac{14}{4}\)

Explanation:
The given fractions are: \(\frac{6}{1}\) and \(\frac{7}{12}\)
For multiplication,
Multiply numerators and denominators separately.
So,
\(\frac{6}{1}\) × \(\frac{7}{12}\) = \(\frac{6 × 7 }{1 × 12}\)
= \(\frac{42}{12}\)
For the simplified form of \(\frac{42}{12}\), divide \(\frac{42}{12}\) by 3 as 42 and 12 are the multiples of 3
Hence,
\(\frac{7}{12}\) × \(\frac{6}{1}\) = \(\frac{42}{12}\) = \(\frac{14}{4}\)

Question 10.
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 181
Answer: The product of 2 and \(\frac{5}{6}\) is: \(\frac{5}{3}\)

Explanation:
The given fractions are: \(\frac{2}{1}\) and \(\frac{5}{6}\)
For multiplication,
Multiply numerators and denominators separately.
So,
\(\frac{2}{1}\) × \(\frac{5}{6}\) = \(\frac{2 × 5 }{1 × 6}\)
= \(\frac{10}{6}\)
For the simplified form of \(\frac{10}{6}\), divide \(\frac{10}{6}\) by 2 as 10 and 6 are the multiples of 2
Hence,
\(\frac{5}{6}\) × \(\frac{2}{1}\) = \(\frac{10}{6}\) = \(\frac{5}{3}\)

Question 11.
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 182
Answer: The product of 10 and \(\frac{3}{8}\) is: \(\frac{15}{4}\)

Explanation:
The given fractions are: \(\frac{10}{1}\) and \(\frac{3}{8}\)
For multiplication,
Multiply numerators and denominators separately.
So,
\(\frac{10}{1}\) × \(\frac{3}{8}\) = \(\frac{10 × 3 }{1 × 8}\)
= \(\frac{30}{8}\)
For the simplified form of \(\frac{30}{8}\), divide \(\frac{30}{8}\) by 2 as 30 and 8 are the multiples of 2
Hence,
\(\frac{5}{8}\) × \(\frac{10}{1}\) = \(\frac{30}{8}\) = \(\frac{15}{4}\)

Relate Fractions and Decimals Performance Task 10

You have a recipe to make one loaf of home made whole wheat bread. You want to make 8 loaves of bread.

1. You need between 6.5 cups and 7 cups of whole wheat flour for one loaf of bread.
a. So far, you measure 3\(\frac{1}{4}\) cups of flour for one loaf. What is the least amount of cups you need to add?
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 183

Answer: The least amount of cups you need to add is: 3.25

Explanation:
It is given that you need between 6.5 and 7 cups of whole wheat flour for one loaf of bread.
It is given that you measured so far 3\(\frac{1}{4}\) cups of flour for one loaf
So,
The least number of cups you need to add to make one loaf of bread = 6.5 – 3\(\frac{1}{4}\)
The representation of 3\(\frac{1}{4}\) in the decimal form is: 3.25
So,
The least number of cups you need = 6.5 – 3.25 = 3.25 cups
Hence, from the above,
We can conclude that you need a minimum of 3.25 cups of wheat flour to make a loaf of bread

b. There are about 4 cups of flour in 1 pound. How many 5-pound bags of whole wheat flour should you buy to make all of the bread?
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 184

Answer: You should buy about 20 cups of flour to make all the bread

Explanation:
It is given that we need about 4 cups of flour in 1 pound.
So,
The number of cups you need for five-pound bags = 4 × 5 = 20 cups
Hence, from the above,
We can conclude that we will need about 20 cups of flour to make all the bread

c. You use a $10 bill to buy enough bags of whole wheat flour for 8 loaves. What is your change?
Answer: Your change is: $4.8

Explanation:
It is given that you use a $10 bill to buy enough bags of whole wheat flour for 8 loaves.
So,
The total cost of wheat flour = 2.69 × 2 = $5.38
So,
Your change = 10 – 5.38 = $4.62
Hence, from the above,
We can conclude that the change is: $4.62

Question 2.
You need to add 2\(\frac{1}{4}\) cups of warm water for one loaf of bread. The temperature of the water should be about 110°F.
a. How many cups of water do you need for all of the bread?
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 185

Answer: The number of cups of warm water you need for all the bread is: 18 cups

Explanation:
It is given that you need to add 2\(\frac{1}{4}\) cups of warm water for one loaf of bread.
But, it is given for the whole read, there are 8 loaves of bread.
So,
The total number of cups you need to make all the bread = 2\(\frac{1}{4}\) × 8
= 2.25 × 8
= 18 cups
Hence, from the above,
We can conclude that we will need 18 cups of warm water to make all the whole bread

b. You find the temperatures of 3 different samples of water. Which sample of water should you use? Explain.
Answer: We will use a C sample of water

Explanation:
The given temperatures are:
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 185
From the above table,
The temperature of A is: 105.5
The temperature of B is: 114.4
The temperature of C is: 109.6
It is given that the temperature to make the bread is: 110
So,
We will choose sample C to make the bread.
Hence, from the above,
we can conclude that we will use C’s sample of water to make the bread

Relate Fractions and Decimals Activity

Decimal Boss

Directions:

  1. Divide the Decimal Boss Cards equally between both players.
  2. Each player flips a Decimal Boss Card.
  3. Players compare their numbers. The player with the greater number takes both cards.
  4. The player with the most cards at the end of the round wins!

Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 186

Relate Fractions and Decimals Chapter Practice 10

10.1 Understand Tenths

Write the fraction or mixed number as a decimal.

Question 1.
\(\frac{8}{10}\)
Answer: The representation of \(\frac{8}{10}\) in the decimal form is: 0.8

Explanation:
The given fraction is: \(\frac{8}{10}\)
Now,
The representation of \(\frac{8}{10}\) in the place-value chart is:

In \(\frac{8}{10}\),
8 represents the tenth’s position
The formula for converting fraction to a decimal is Decimal = Numerator ÷ Denominator
Hence,
The representation of \(\frac{8}{10}\) in the decimal form is: 0.8

Question 2.
\(\frac{3}{10}\)
Answer: The representation of \(\frac{3}{10}\) in the decimal form is: 0.3

Explanation:
The given fraction is: \(\frac{3}{10}\)
Now,
The representation of \(\frac{3}{10}\) in the place-value chart is:

In \(\frac{3}{10}\),
3 represents the tenth’s position
The formula for converting fraction to a decimal is Decimal = Numerator ÷ Denominator
Hence,
The representation of \(\frac{3}{10}\) in the decimal form is: 0.3

Question 3.
6\(\frac{7}{10}\)
Answer: The representation of 6\(\frac{7}{10}\) in the decimal form is: 6.7

Explanation:
The given fraction is: 6\(\frac{7}{10}\)
Now,
The representation of 6\(\frac{7}{10}\) in the place-value chart is:

In 6\(\frac{7}{10}\),
6 represents the one’s position
7 represents the tenth’s position
The formula for converting fraction to a decimal is Decimal = Numerator ÷ Denominator
Hence,
The representation of 6\(\frac{7}{10}\) in the decimal form is: 6.7

Question 4.
15\(\frac{4}{10}\)
Answer: The representation of 15\(\frac{4}{10}\) in the decimal form is: 15.4

Explanation:
The given fraction is: 15\(\frac{4}{10}\)
Now,
The representation of 15\(\frac{4}{10}\) in the place-value chart is:

In 15\(\frac{4}{10}\),
1 represents the ten’s position
5 represents the one’s position
4 represents the tenth’s position
The formula for converting fraction to a decimal is Decimal = Numerator ÷ Denominator
Hence,
The representation of 15\(\frac{4}{10}\) in the decimal form is: 15.4

Write the number as a fraction or mixed number and as a decimal.

Question 5.
two tenths
Answer:
The representation of two-tenths in the fraction form is: \(\frac{2}{10}\)
The representation of two-tenths in the decimal form is: 0.2

Explanation:
The given word form is: Two-tenths
So,
The representation of two-tenths in the fraction form is: \(\frac{2}{10}\)
Now,
The representation of \(\frac{2}{10}\) in the place-value chart is:

In \(\frac{2}{10}\),
2 represents the tenth’s position
Hence,
The representation of \(\frac{2}{10}\) in the decimal form is: 0.2

Question 6.
thirteen and six tenths
Answer:
The representation of thirteen and six-tenths in the fraction form is: 13\(\frac{6}{10}\)
The representation of thirteen and six-tenths in the decimal form is: 13.6

Explanation:
The given word form is: Thirteen and six-tenths
So,
The representation of thirteen and six-tenths in the fraction form is: 13\(\frac{6}{10}\)
Now,
The representation of 13\(\frac{6}{10}\) in the place-value chart is:

In 13\(\frac{6}{10}\),
1 represents the ten’s position
3 represents the one’s position
6 represents the tenth’s position
Hence,
The representation of 13\(\frac{6}{10}\) in the decimal form is: 13.6

Question 7.
Modeling Real Life
You bake 2 loaves of banana bread for a party. You cut each loaf into10 equal pieces. The guests eat 18 pieces. Write the fraction and decimal that represent how many loaves the guests eat in all.
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 187
Answer:
The representation of the loaves that the guests eat in the fraction form is: \(\frac{18}{20}\)
The representation of the loaves that the guests eat in the decimal form is: 0.9

Explanation:
It is given that you bake 2 loaves of banana bread and you cut each banana bread into 10 pieces
So,
The total number of pieces is: 20 pieces
It is also given that the guests eat 18 pieces.
So,
The number of pieces eaten by guests is: 18
So,
The representation of the loaves that the guests eat in the fraction form is: \(\frac{18}{20}\)
The representation of the loaves that the guests eat in the decimal form is: 0.9

10.2 Understand Hundredths

Write the fraction or mixed number as a decimal.

Question 8.
\(\frac{10}{100}\)
Answer: The representation of \(\frac{10}{100}\) in the form of decimal number is: 0.10

Explanation:
The given fraction is: \(\frac{10}{100}\)
Now,
The representation of \(\frac{10}{100}\) in the place-value chart is:

In \(\frac{10}{100}\),
0 represents the hundredth position
1 represents the tenth position
Hence,
The representation of \(\frac{10}{100}\) in the decimal form is: 0.10

Question 9.
\(\frac{6}{100}\)
Answer: The representation of \(\frac{6}{100}\) in the form of decimal number is: 0.06

Explanation:
The given fraction is: \(\frac{6}{100}\)
Now,
The representation of \(\frac{6}{100}\) in the place-value chart is:

In \(\frac{6}{100}\),
6 represents the hundredth position
0 represents the tenth position
Hence,
The representation of \(\frac{6}{100}\) in the decimal form is: 0.06

Question 10.
8\(\frac{75}{100}\)
Answer: The representation of 8\(\frac{75}{100}\) in the form of a decimal number is: 8.75

Explanation:
The given fraction is: 8\(\frac{75}{100}\)
Now,
The representation of 8\(\frac{75}{100}\) in the place-value chart is:

In 8\(\frac{75}{100}\),
8 represents the one’s position
5 represents the hundredth position
7 represents the tenth position
Hence,
The representation of 8\(\frac{75}{100}\) in the decimal form is: 8.75

Question 11.
34\(\frac{2}{100}\)
Answer: The representation of 34\(\frac{2}{100}\) in the form of a decimal number is: 34.02

Explanation:
The given fraction is: 34\(\frac{2}{100}\)
Now,
The representation of 34\(\frac{2}{100}\) in the place-value chart is:

In 8\(\frac{75}{100}\),
3 represents the ten’s position
4 represents the one’s position
2 represents the hundredth position
0 represents the tenth position
Hence,
The representation of 34\(\frac{2}{100}\) in the decimal form is: 34.02

Write the number as a fraction or mixed number and as a decimal.

Question 12.
thirty-seven hundredths
Answer:
The representation of thirty-seven hundredths in the fraction form is: \(\frac{37}{100}\)
The representation of thirty-seven hundredths in the decimal form is: 0.37

Explanation:
The given word form is: Thirty-seven hundredths
So,
The representation of thirty-seven hundredths in the fraction form is: \(\frac{37}{100}\)
Now,
The representation of \(\frac{37}{100}\) in the place-value chart is:

In \(\frac{37}{100}\),
7 represents the hundredth’s position
3 represents the tenth’s position
Hence,
The representation of \(\frac{37}{100}\) in the decimal form is: 0.37

Question 13.
nineteen and forty-one hundredths
Answer:
The representation of nineteen and forty-seven hundredths in the fraction form is: 19\(\frac{47}{100}\)
The representation of nineteen and forty-seven hundredths in the decimal form is: 19.47

Explanation:
The given word form is: Nineteen and forty-seven hundredths
So,
The representation of nineteen and forty-seven hundredths in the fraction form is: 19\(\frac{47}{100}\)
Now,
The representation of 19\(\frac{47}{100}\) in the place-value chart is:

In 19\(\frac{47}{100}\),
1 represents the ten’s position
9 represents the one’s position
7 represents the hundredth’s position
4 represents the tenth’s position
Hence,
The representation of 19\(\frac{47}{100}\) in the decimal form is: 19.47

10.3 Fractions and Decimals

Write the number as tenths in fraction form and decimal form.

Question 14.
\(\frac{30}{100}\)
Answer:
The representation of \(\frac{30}{100}\) as tenths in the fraction form is: \(\frac{3}{10}\)
The representation of \(\frac{3}{10}\) in the decimal form is: 0.3

Explanation:
The given fraction is: \(\frac{30}{100}\)
So, to write \(\frac{30}{100}\) as tenths, divide the fraction and numerator of \(\frac{30}{100}\) with 10.
So,
Firstly the numerators 30 and 10 are divided and then the denominators 100 and 10 are divided
So,
\(\frac{30}{100}=\frac{30 \div 10}{100 \div 10}=\frac{8}{10}\)
We know that,
Decimal = Numerator ÷ Denominator
So,
The representation of \(\frac{30}{100}\) in the place-value chart is:

Hence,
The representation of \(\frac{30}{100}\) as tenths in the fraction form is: \(\frac{3}{10}\)
The representation of \(\frac{3}{10}\) in the decimal form is: 0.3

Question 15.
\(\frac{90}{100}\)
Answer:
The representation of \(\frac{90}{100}\) as tenths in the fraction form is: \(\frac{39{10}\)
The representation of \(\frac{39}{10}\) in the decimal form is: 0.9

Explanation:
The given fraction is: \(\frac{90}{100}\)
So, to write \(\frac{390}{100}\) as tenths, divide the fraction and numerator of \(\frac{90}{100}\) with 10.
So,
Firstly the numerators 90 and 10 are divided and then the denominators 100 and 10 are divided
So,
\(\frac{90}{100}=\frac{90 \div 10}{100 \div 10}=\frac{8}{10}\)
We know that,
Decimal = Numerator ÷ Denominator
So,
The representation of \(\frac{90}{100}\) in the place-value chart is:

Hence,
The representation of \(\frac{90}{100}\) as tenths in the fraction form is: \(\frac{9}{10}\)
The representation of \(\frac{9}{10}\) in the decimal form is: 0.9

Question 16.
0.50
Answer:
The representation of 0.50 as tenths in the fraction form is: \(\frac{5}{10}\)
The representation of \(\frac{5}{10}\) in the decimal form is: 0.5

Explanation:
The given decimal number is: 0.50
So,
The representation of 0.50 in the fraction form is: \(\frac{50}{100}\)
So, to write \(\frac{50}{100}\) as tenths,divide the fraction and numerator of \(\frac{50}{100}\) with 10.
So,
Firstly the numerators 50 and 10 are divided and then the denominators 100 and 10 are divided
We know that,
Decimal = Numerator ÷ Denominator
So,
The representation of \(\frac{5}{10}\) in the place-value chart is:

Hence,
The representation of \(\frac{50}{100}\) as tenths in the fraction form is: \(\frac{5}{10}\)
The representation of \(\frac{5}{10}\) in the decimal form is: 0.5

Write the number as hundredths in fraction form and decimal form.

Question 17.
\(\frac{7}{10}\)
Answer:
The representation of \(\frac{7}{10}\) as hundredths in the fraction form is: \(\frac{70}{100}\)
The representation of \(\frac{70}{100}\) in the decimal form is: 0.70

Explanation:
The given fraction is: \(\frac{7}{10}\)
So, to write \(\frac{7}{10}\) as hundredths, multiply the fraction and numerator of \(\frac{7}{10}\) with 10.
So,
Firstly the numerators 7 and 10 are multiplied and then the denominators 10 and 10 are multiplied
We know that,
Decimal = Numerator ÷ Denominator
So,
The representation of \(\frac{70}{100}\) in the place-value chart is:

Hence,
The representation of \(\frac{7}{10}\) as hundredths in the fraction form is: \(\frac{70}{100}\)
The representation of \(\frac{70}{100}\) in the decimal form is: 0.70

Question 18.
\(\frac{4}{10}\)
Answer:
The representation of \(\frac{4}{10}\) as hundredths in the fraction form is: \(\frac{40}{100}\)
The representation of \(\frac{40}{100}\) in the decimal form is: 0.40

Explanation:
The given fraction is: \(\frac{4}{10}\)
So, to write \(\frac{4}{10}\) as hundredths, multiply the fraction and numerator of \(\frac{4}{10}\) with 10.
So,
Firstly the numerators 4 and 10 are multiplied and then the denominators 10 and 10 are multiplied
We know that,
Decimal = Numerator ÷ Denominator
So,
The representation of \(\frac{40}{100}\) in the place-value chart is:

Hence,
The representation of \(\frac{4}{10}\) as hundredths in the fraction form is: \(\frac{40}{100}\)
The representation of \(\frac{40}{100}\) in the decimal form is: 0.40

Question 19.
0.6
Answer:
The representation of 0.6 as hundredths in the fraction form is: \(\frac{60}{100}\)
The representation of \(\frac{60}{100}\) in the decimal form is: 0.60

Explanation:
The given decimal number is: 0.6
So,
The representation of 0.6 in the fraction form is: \(\frac{6}{10}\)
So, to write \(\frac{6}{10}\) as hundredths,multiply the fraction and numerator of \(\frac{6}{10}\) with 10.
So,
Firstly the numerators 6 and 10 are multiplied and then the denominators 100 and 10 are multiplied
We know that,
Decimal = Numerator ÷ Denominator
So,
The representation of \(\frac{60}{100}\) in the place-value chart is:

Hence,
The representation of \(\frac{6}{10}\) as hundredths in the fraction form is: \(\frac{60}{100}\)
The representation of \(\frac{60}{100}\) in the decimal form is: 0.60

10.4 Compare Decimals

Compare

Question 20.
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 188
Answer: 0.79 is greater than 0.72

Explanation:
The given decimal numbers are: 0.79 and 0.72
The representation of 0.79 and 0.72 in the place-value chart is:

So,
From the above place-value chart,
we can observe that one’s and tenth’s positions are the same.
So,
Compare the hundredth’s position 9 and 2
So, 9 hundredths > 2 hundredths
Hence, from the above,
We can conclude that 0.79 is greater than 0.72

Question 21.
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 189
Answer: 9.16 is less than 9.56

Explanation:
The given decimal numbers are: 9.16 and 9.56
The representation of 9.16 and 9.56 in the place-value chart is:

So,
From the above place-value chart,
we can observe that one’s and hundredth’s positions are the same.
So,
Compare the tenth’s position 1 and 5
So, 1 hundredths < 5 hundredths
Hence, from the above,
We can conclude that 9.16 is less than 9.56

Question 22.
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 190
Answer: 11.40 is equal to 11.4

Explanation:
The given decimal numbers are: 11.40 and 11.4
The representation of 11.40 and 11.4 in the place-value chart is:

So,
From the above place-value chart,
we can observe that one’s, tenth’s, and hundredth’s positions are the same.
So,
Compare the tenth’s position 4 and 4
So, 4 hundredths = 4 hundredths
Hence, from the above,
We can conclude that 11.40 is equal to 11.4

Open-Ended
Complete the statement to make it true.

Question 23.
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 191
Answer: 0.19 is greater than 0.15

Explanation:
Let the missing number be 1
So,
The given decimal numbers are: 0.19 and 0.15
The representation of 0.19 and 0.15 in the place-value chart is:

So,
From the above place-value chart,
we can observe that one’s and tenth’s positions are the same.
So,
Compare the hundredth’s position 9 and 5
So, 9 hundredths > 5 hundredths
Hence, from the above,
We can conclude that 0.19 is less than 0.15

Question 24.
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 192
Answer: 6.30 is equal to 6.3

Explanation:
Let the missing number be 0
So,
The given decimal numbers are: 6.30 and 6.3
The representation of 6.30 and 6.3 in the place-value chart is:

So,
From the above place-value chart,
we can observe that one’s, hundredth’s, and tenth’s positions are the same.
So,
Compare the hundredth’s position 0 and 0
So, 0 hundredths = 0 hundredths
Hence, from the above,
We can conclude that 6.30 is equal to 6.3

Question 25.
___ > 40.48
Answer: 40.58 is greater than 40.48

Explanation:
Let the missing number be 40.58
So,
The given decimal numbers are: 40.58 and 40.48
The representation of 40.58 and 40.48 in the place-value chart is:

So,
From the above place-value chart,
we can observe that one’s and hundredth’s positions are the same.
So,
Compare the tenth’s position 9 and 5
So, 5 tenths > 4 tenths
Hence, from the above,
We can conclude that 40.58 is greater than 40.48

Question 26.
Open-Ended
What might Newton’s number be?
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 193
Answer: Newton’s number might be: 0.23, 0.24, 0.25, 0.26, 0.27, 0.28, 0.29

Explanation:
It is given that Newton’s number will be between 0.2 and 0.3
So,
Newton’s number might be: 0.21, 0.22, 0.23, 0.24, 0.25, 0.26, 0.27, 0.28, 0.29
It is also given that the highest digit will be in the hundredth’s place
Hence,
Newton’s number might be: 0.23, 0.24, 0.25, 0.26, 0.27, 0.28, 0.29

10.5 Add Decimal Fractions and Decimals

Find the sum.

Question 27.
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 194
Answer:
The representation of the sum in the fraction form is:
\(\frac{6}{10}\) + \(\frac{14}{100}\) = \(\frac{74}{100}\)
The representation of \(\frac{74}{100}\) in the decimal for is: 0.74

Explanation:
The given fractions are: \(\frac{14}{100}\) and \(\frac{6}{10}\)
For addition, we have to make either denominators or the numerators equal.
so,
For making \(\frac{6}{10}\) as the hundredths, we have to multiply \(\frac{6}{10}\) by 10
So,
The representation of \(\frac{6}{10}\) as hundredths in the fraction form is: \(\frac{60}{100}\)
So,
\(\frac{60}{100}\) + \(\frac{14}{100}\)
= \(\frac{60 + 14}{100}\)
= \(\frac{74}{100}\)
The representation of \(\frac{74}{100}\) in the decimal form is: 0.74
Hence from the above,
We can conclude that
\(\frac{6}{10}\) + \(\frac{14}{100}\) = \(\frac{74}{100}\)
The representation of \(\frac{74}{100}\) in the decimal for is: 0.74

Question 28.
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 195
Answer:
The representation of the sum in the fraction form is:
\(\frac{3}{10}\) + \(\frac{52}{100}\) = \(\frac{82}{100}\)
The representation of \(\frac{82}{100}\) in the decimal for is: 0.82

Explanation:
The given fractions are: \(\frac{52}{100}\) and \(\frac{3}{10}\)
For addition, we have to make either denominators or the numerators equal.
so,
For making \(\frac{3}{10}\) as the hundredths, we have to multiply \(\frac{3}{10}\) by 10
So,
The representation of \(\frac{3}{10}\) as hundredths in the fraction form is: \(\frac{30}{100}\)
So,
\(\frac{30}{100}\) + \(\frac{52}{100}\)
= \(\frac{30 + 52}{100}\)
= \(\frac{82}{100}\)
The representation of \(\frac{82}{100}\) in the decimal form is: 0.82
Hence from the above,
We can conclude that
\(\frac{3}{10}\) + \(\frac{52}{100}\) = \(\frac{82}{100}\)
The representation of \(\frac{82}{100}\) in the decimal for is: 0.82

Question 29.
0.12 + 0.6 = ___
Answer:
The representation of the sum in the fraction form is:
\(\frac{6}{10}\) + \(\frac{12}{100}\) = \(\frac{72}{100}\)
The representation of 012 + 0.6 in the decimal form is: 0.72

Explanation:
The given decimal numbers are: 0.6 and 0.12
So, convert the two decimal numbers in to respective fractions.
So,
The given fractions are: \(\frac{12}{100}\) and \(\frac{6}{10}\)
For addition, we have to make either denominators or the numerators equal.
so,
For making \(\frac{6}{10}\) as the hundredths, we have to multiply \(\frac{6}{10}\) by 10
So,
The representation of \(\frac{6}{10}\) as hundredths in the fraction form is: \(\frac{60}{100}\)
So,
\(\frac{60}{100}\) + \(\frac{12}{100}\)
= \(\frac{60 + 12}{100}\)
= \(\frac{72}{100}\)
The representation of \(\frac{72}{100}\) in the decimal form is: 0.72
Hence from the above,
We can conclude that
\(\frac{6}{10}\) + \(\frac{12}{100}\) = \(\frac{72}{100}\)
The representation of \(\frac{72}{100}\) in the decimal for is: 0.72

Question 30.
0.4 + 0.72 = ___
Answer:
The representation of the sum in the fraction form is:
\(\frac{4}{10}\) + \(\frac{72}{100}\) = \(\frac{112}{100}\)
The representation of 0.4 + 0.72 in the decimal form is: 1.12

Explanation:
The given decimal numbers are: 0.4 and 0.72
So, convert the two decimal numbers in to respective fractions.
So,
The given fractions are: \(\frac{72}{100}\) and \(\frac{4}{10}\)
For addition, we have to make either denominators or the numerators equal.
so,
For making \(\frac{4}{10}\) as the hundredths, we have to multiply \(\frac{4}{10}\) by 10
So,
The representation of \(\frac{4}{10}\) as hundredths in the fraction form is: \(\frac{40}{100}\)
So,
\(\frac{40}{100}\) + \(\frac{72}{100}\)
= \(\frac{72 + 40}{100}\)
= \(\frac{112}{100}\)
The representation of \(\frac{112}{100}\) in the decimal form is: 1.12
Hence from the above,
We can conclude that
\(\frac{4}{10}\) + \(\frac{72}{100}\) = \(\frac{112}{100}\)
The representation of \(\frac{112}{100}\) in the decimal for is: 1.12

Question 31.
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 196
Answer:
The representation of the sum in the fraction form is:
\(\frac{23}{100}\) + \(\frac{36}{100}\) + \(\frac{2}{10}\) = \(\frac{79}{100}\)
The representation of \(\frac{79}{100}\) in the decimal for is: 0.79

Explanation:
The given fractions are: \(\frac{23}{100}\), \(\frac{36}{100}\) and \(\frac{2}{10}\)
For addition, we have to make either denominators or the numerators equal.
so,
For making \(\frac{2}{10}\) as the hundredths, we have to multiply \(\frac{2}{10}\) by 10
So,
The representation of \(\frac{2}{10}\) as hundredths in the fraction form is: \(\frac{20}{100}\)
So,
\(\frac{23}{100}\) + \(\frac{36}{100}\) + \(\frac{20}{100}\)
= \(\frac{23 + 36 + 20}{100}\)
= \(\frac{79}{100}\)
The representation of \(\frac{79}{100}\) in the decimal form is: 0.79
Hence from the above,
We can conclude that
\(\frac{2}{10}\) + \(\frac{23}{100}\) + \(\frac{36}{100}\) = \(\frac{79}{100}\)
The representation of \(\frac{79}{100}\) in the decimal for is: 0.79

Question 32.
0.18 + 0.2 + 0.07 = ___
Answer:
The representation of the sum in the fraction form is:
\(\frac{2}{10}\) + \(\frac{18}{100}\) + \(\frac{7}{100}\) = \(\frac{45}{100}\)
The representation of 0.18 + 0.2 + 0.07 in the decimal for is: 0.45

Explanation:
The given decimal numbers are: 0.18, 0.2 and 0.07
So, convert the three decimal numbers in to respective fractions.
So,
The given fractions are: \(\frac{18}{100}\) , \(\frac{2}{10}\) and \(\frac{7}{100}\)
For addition, we have to make either denominators or the numerators equal.
so,
For making \(\frac{2}{10}\)  as the hundredths, we have to multiply \(\frac{2}{10}\)  by 10
So,
The representation of \(\frac{2}{10}\) as hundredths in the fraction form is: \(\frac{20}{100}\)
So,
\(\frac{20}{100}\) + \(\frac{18}{100}\) + \(\frac{7}{100}\)
= \(\frac{20 + 18 + 7}{100}\)
= \(\frac{45}{100}\)
The representation of \(\frac{45}{100}\) in the decimal form is: 0.45
Hence from the above,
We can conclude that
\(\frac{2}{10}\) + \(\frac{18}{100}\) + \(\frac{7}{100}\) = \(\frac{45}{100}\)
The representation of \(\frac{45}{100}\) in the decimal for is: 0.45

Number Sense
Find the sum.

Question 33.
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 197
Answer:
The representation of the sum in the fraction form is:
\(\frac{5}{10}\) + \(\frac{48}{100}\) = \(\frac{98}{100}\)
The representation of 0.5 + \(\frac{48}{100}\) in the decimal for is: 0.98

Explanation:
The given numbers are: 0.5 and \(\frac{48}{100}\)
So, convert the decimal number in to respective fractions.
So,
The given fractions are: \(\frac{5}{10}\) and \(\frac{48}{100}\)
For addition, we have to make either denominators or the numerators equal.
so,
For making \(\frac{5}{10}\) as the hundredths, we have to multiply \(\frac{5}{10}\) by 10
So,
The representation of \(\frac{5}{10}\) as hundredths in the fraction form is: \(\frac{50}{100}\)
So,
\(\frac{50}{100}\) + \(\frac{48}{100}\)
= \(\frac{50 + 48}{100}\)
= \(\frac{98}{100}\)
The representation of \(\frac{98}{100}\) in the decimal form is: 0.98
Hence from the above,
We can conclude that
\(\frac{5}{10}\) + \(\frac{48}{100}\) = \(\frac{98}{100}\)
The representation of \(\frac{98}{100}\) in the decimal for is: 0.98

Question 34.
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 198
Answer:
The representation of the sum in the fraction form is:
\(\frac{9}{10}\) + \(\frac{25}{100}\) = \(\frac{115}{100}\)
The representation of 0.25 + \(\frac{9}{10}\) in the decimal for is: 1.15

Explanation:
The given numbers are: 0.25 and \(\frac{9}{10}\)
So, convert the decimal number in to respective fractions.
So,
The given fractions are: \(\frac{9}{10}\) and \(\frac{25}{100}\)
For addition, we have to make either denominators or the numerators equal.
so,
For making \(\frac{9}{10}\) as the hundredths, we have to multiply \(\frac{9}{10}\) by 10
So,
The representation of \(\frac{9}{10}\) as hundredths in the fraction form is: \(\frac{90}{100}\)
So,
\(\frac{90}{100}\) + \(\frac{25}{100}\)
= \(\frac{90 + 25}{100}\)
= \(\frac{115}{100}\)
The representation of \(\frac{115}{100}\) in the decimal form is: 1.15
Hence from the above,
We can conclude that
\(\frac{9}{10}\) + \(\frac{25}{100}\) = \(\frac{115}{100}\)
The representation of \(\frac{115}{100}\) in the decimal for is: 1.15

10.6 Fractions, Decimals, and Money

Find the total money amount. Then write the amount as a fraction or mixed number and as a decimal.

Question 35.
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 199
Answer: The total amount of money is: $1.13

Explanation:
We know that,
1 Quarter = $0.25
1 nickel = $0.05
1 penny = $0.01
The given money is: 3 pennies, 4 Quarters, and 2 nickels
So,
The total amount of money =  ( 4 × 0.25 ) + ( 2 × 0.05 ) + ( 3 × 0.01 )
= 1 + 0.10 + 0.03
= 1.13
Hence,
The total amount of money is: $1.13
The representation of $1.13 in the fraction form is: 1\(\frac{13}{100}\) dollar
The representation of $1.13 in the decimal form is: 1.13

Question 36.
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 200
Answer: The total amount of money is: $2.11

Explanation:
We know that,
1 Quarter = $0.25
1 nickel = $0.05
1 penny = $0.01
The given money is: 1 penny, 4 Quarters, 1 dollar, and 2 nickels
So,
The total amount of money = ( 1 × 1 ) + ( 4 × 0.25 ) + ( 2 × 0.05 ) + ( 1 × 0.01 )
= 1 + 1 + 0.1 + 0.01
= 2.11
Hence,
The total amount of money is: $2.11
The representation of $2.11 in the fraction form is: 2\(\frac{11}{100}\) dollar
The representation if $2.11 in the decimal form is: 2.11

Question 37.
Write \(\frac{18}{100}\) as a money amount and as a decimal.
Answer: The representation of \(\frac{18}{100}\) as the total amount of money is: $0.18

Explanation:
The given fraction is: \(\frac{18}{100}\)
The given fraction will be given as an amount in dollars.
So,
The total amount of money in the decimal form is: $0.18

Question 38.
Write $0.94 as a fraction and as a decimal.
Answer: The representation of 0.94 as the total amount of money is: $0.94

Explanation:
The given decimal number is: 0.94
So,
The representation of 0.94 in the fraction form is: \(\frac{94}{100}\)
The given fraction will be given as an amount in dollars.
So,
The total amount of money in the decimal form is: $0.94

10.7 Operations with Money

Draw bills and coins to solve.

Question 39.
Bananas cost $0.29 per pound. You buy 3 pounds of bananas. How much money do you spend in all?
Answer: The total amount of money you spent is: $0.87

Explanation:
It is given that bananas cost $0.29 per pound and you bought 3 pounds of bananas
So,
The cost of bananas per pound is: $0.29
So,
The cost of 3 pounds of bananas = 3 × 0.29 = $0.87
hence, from the above,
We can conclude that the total amount of money you spent is: $0.89

Question 40.
Descartes has $3.50. He spends $1.75 on a journal. How much money does Descartes have left?
Answer: The amount of money Descartes left is: $1.75

Explanation:
It is given that Descartes has $3.50 and he spent $1.75 on a journal
So,
The total amount of money Descartes has: $3.50
The amount of money Descartes spent is: $1.75
So,
The amount of money Descartes left = The total amount of money Descartes has – The amount of money Descartes spent
= 3.50 – 1.75
= $1.75
Hence, from the above,
We can conclude that the amount of money Descartes left is: $1.75

Question 41.
You buy the items shown at a book fair. How much money do you spend in all?
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 201
Answer: The total amount of money you spent is: $7.49

Explanation:
The given figure is:
Big Ideas Math Answer Key Grade 4 Chapter 10 Relate Fractions and Decimals 201
From the given figure,
The cost of a book is: $6.99
The cost of the pen is: $0.50
Hence,
The total amount of money you spent = The cost of a book + The cost of a pen
= 6.99 + 0.50
= $7.49
Hence, from the above,
We can conclude that the amount of money you spent is: $7.49

Conclusion:

All detailed and step by step explanations are covered in the Big Ideas Math Answers Grade 4 Chapter 10 Relate Fractions and Decimals. The solutions are prepared by the highly experienced subject experts after the ample research. This will help you to score the highest marks in the exams. Keep in touch with us to get the solution key of all Big Ideas Math Grade 4 Chapters.

Big Ideas Math Answers Grade 7 Chapter 7 Probability

Big Ideas Math Answers Grade 7 Chapter 7 Probability

To solve real-life mathematical problems, candidates have to understand the topics behind the problems. Before solving the problems, know what is probability and how it works in real-life situations. Follow the tips and tricks for the most interesting chapter of Grade 7. Download Big Ideas Math Answers Grade 7 Chapter 7 Probability problems pdf and start your exam preparation to score good marks.

Refer to all the important questions and solutions which help you in scoring the highest marks in the exam. With the help of the given syllabus, you can score marks and also can know the tricks and tips of solving the problems quickly. Follow the syllabus and attend mock tests to get perfection in all the subjects. BIM Grade 7 Chapter 7 pdf is accessible for free of cost, check all the problems, and get thorough with the syllabus.

Big Ideas Math Book 7th Grade Answer Key Chapter 7 Probability

Probability is one such concept where you feel it is difficult at the initial stages, but as you practice it you will love this concept. Big Ideas Math Answer Key for Grade 7 Chapter 7 Probability is mandatory to practice if you want to secure more marks in the exam. Tap on the links and download the pdf for free of cost. Refer to all the problems and know the importance of each concept in the exam.

With the given pdf and problems, you can manage the exam quickly and efficiently. Look at the guide and answer key which clears all your doubts regarding the exam. A serious practice must be needed to get perfection in the probability concept. Check the Big Ideas Math Book 7th Grade Answer Key Chapter 7 Probability before going to attend the exam.

Performance Task

Lesson: 1 Probability

Lesson: 2 Experimental and Theoretical Probability

Lesson: 3 Compound Events

Lesson: 4 Simulations

Chapter: 7 – Probability 

Probability STEAM Video/Performance Task

STEAM Video

Massively Multiplayer Rock Paper Scissors.
You can use experimental probability to describe the percent of times that you win, lose, or tie in Rock Paper Scissors. Describe a real-life situation where it is helpful to describe the percent of times that a particular outcome occurs.

Watch the STEAM Video “Massively Multiplayer Rock Paper Scissors.”Then answer the following questions.
Big Ideas Math Answer Key Grade 7 Chapter 7 Probability 1
1. The table shows the ways that you can win, lose, or tie in Rock Paper Scissors. You and your opponent throw the signs for rock, paper, or scissors at random. What percent of the time do you expect to win? lose? tie?
2. You play Rock Paper Scissors 15 times. About how many times do you expect to win? Explain your reasoning.

Performance Task

Fair and Unfair Carnival Games
After completing this chapter, you will be able to use the concepts you learned to answer the questions in the STEAM Video Performance Task.
Big Ideas Math Answer Key Grade 7 Chapter 7 Probability 2
You will be given information about a version of Rock Paper Scissors used at a carnival. Then you will be asked to design your own “unfair” carnival game using a spinner or a number cube, and test your game with a classmate.
In what ways can a game of chance be considered fair? unfair? Explain your reasoning.

Probability Getting Ready for Chapter 7

Chapter Exploration
Work with a partner.
1. Play Rock Paper Scissors 30 times. Tally your results in the table.
2. How many possible results are there?
3. Of the possible results, in how many ways can Player A win? In how many ways can Player B win? In how many ways can there be a tie?
4. Is one of the players more likely to win than the other player? Explain your reasoning.
Big Ideas Math Answer Key Grade 7 Chapter 7 Probability 3

Vocabulary
The following vocabulary terms are defined in this chapter. Think about what each term might mean and record your thoughts.
probability
theoretical probability
simulation
relative frequency
sample space
experimental probability
compound event

Lesson 7.1 Probability

EXPLORATION 1

Determining Likelihood
Work with a partner. Use the spinners shown.
Big Ideas Math Answer Key Grade 7 Chapter 7 Probability 7.1 1
a. For each spinner, determine which numbers you are more likely to spin and which numbers you are less likely to spin. Explain your reasoning.
b. Spin each spinner 20 times and record your results in two tables. Do the data support your answers in part(a)? Explain why or why not.
Big Ideas Math Answer Key Grade 7 Chapter 7 Probability 7.1 2
c. How can you use percents to describe the likelihood of spinning each number? Explain.
Answer:

Big Ideas Math Answer Key Grade 7 Chapter 7 Probability 7.1 3

Try It

Question 1.
You randomly choose one of the tiles shown from a hat.
Big Ideas Math Answer Key Grade 7 Chapter 7 Probability 7.1 4
a. How many possible outcomes are there?

Answer:
There are 5 number of unique possible outcomes
They are { A,B,C,D,E}
Therefore, there are 5 possible outcomes for the event.

b. What are the favorable outcomes of choosing a vowel ?

Answer:
There are 2 favorable outcomes of choosing an vowel
They are {A,E}
Therefore, there are 2 favorable outcomes of choosing an vowel.

C. In how many ways can choosing a consonant occur?
Answer: 3 ways

Explanation:
There are 3 unique consonant tiles. They are {B,C,D}

Describe the likelihood of the event given its probability.
Question 2.
The probability that you land a jump on a snowboard is \(\frac{1}{10}\).
Answer:
Given, Probability of landing a jump on snowboard = 1/10 = 0.1
So, the likelihood of the event of jumping on to the snowboard after every jump is 1 out of 10 times.

Question 3.
There is a 100% chance that the temperature will be less than 120°F tomorrow.
Answer:
Certain event .
Given that , there is 100% chance that the temperature will be less than 120F tomorrow.
The probability for a certain event is P(E)= 1.
So, the probability that the temperature being 120F tomorrow is 1.

Question 4.
You attempt three-point shots on a basketball court and record the number of made and missed shots. Describe the likelihood of each event.̇̇̇̇̇̇̇̇̇
Big Ideas Math Answer Key Grade 7 Chapter 7 Probability 7.1 5
̇̇a. You make your next shot.
b. You miss your next shot.
Answer:
a.
Explanation:
Total number of shots = 15
Frequency of shots made = 9
The likelihood of making next shot is
Shots made/ Total = 9/15=0.6
So, the next shot is likely to occur

b.
Explanation:
Total number of shots = 15
Frequency of missed shots = 6
The likelihood of the next shot will miss is shots missed/ Total =6/15 0.4
So, the next shot is not likely to occur.

Self-Assessment for Concepts & Skills
Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 5.
IDENTIFYING OUTCOMES
You roll a number cube. What are the possible outcomes?
Big Ideas Math Answer Key Grade 7 Chapter 7 Probability 7.1 6

Answer: 6
A cube has 6 faces .
They are {1,2,3,4,5,6} .
There are 6 possible outcomes.
Therefore , there are 6 possible outcomes for a cube.

Question 6.
USING RELATIVE FREQUENCIES
A bag contains only red marbles and blue marbles. You randomly draw a marble from the bag and replace it. The table shows the results of repeating this experiment. Find the likelihood of each event.̇̇̇̇̇̇̇̇̇̇
Big Ideas Math Answer Key Grade 7 Chapter 7 Probability 7.1 7
a. The next marble you choose is red. ̇̇
b. The next marble you choose is neither red nor blue.
Answer:

a.
Explanation:
Total number of balls = 42
Number of red balls = 21
The likelihood of drawing the red marble in the next turn=
Number of red balls/ Total number of balls
= 21/42 =0.5
There is only 50% chance of chance of drawing a red ball in the next turn
So, the likelihood of the event is may or ,ay not occur evenly.

b.The likelihood of drawing neither red nor blue ball is 0
0%= Impossible
Because, there are only red and blue marbles
So, it is impossible to draw another ball other than red and blue.
Therefore, the likelihood of the event is cannot occur

Self-Assessment for Problem Solving
Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 7.
The table shows the number of days you have a pop quiz and the number of days you do not have a pop quiz in three weeks of school. How many days can you expect to have a pop quiz during a 180-day school year? Explain.
Big Ideas Math Answer Key Grade 7 Chapter 7 Probability 7.1 8
Answer:
Total number of days for pop quiz for 3 weeks = 2
Total frequency = 15
Now, the number of times to have a pop quiz in 180 days be x
Total number of days =180
2/15 = x/180
15x = 360
x =  360/ 15
x = 24
Therefore, we can expect 24 days to have a pop quiz during a 180-day school year.

Question 8.
In a football game, the teams pass the ball on 40% of the plays. Of the passes thrown, greater than 75% are completed. You watch the film of a randomly chosen play. Describe the likelihood that the play results in a complete pass. Explain your reasoning.
Answer:
Given ,
The team pass the ball on 40% of the plays
And also, 75% are completed
75% = 0.75
Therefore 75% = 3/4
Therefore , the likelihood is likely to happen of the plays results in a complete pass = 0.34

Probability Homework & Practice 7.1

Review & Refresh

An account earns simple interest. Find the interest earned.
Question 1
$700 at 3% for 4 years
Answer: $84
The formula for simple interest= Principal x rate x time
I = P x r x t
= $700 x0.03 x 4 =$84
Therefore , the interest earned =  $84

Question 2.
$650 at 2% for 6 years
Answer: $78
The formula for simple interest= Principal x rate x time
I = P x r x t
= $650 x0.02 x 6 =$78
Therefore , the interest earned =  $78

Question 3.
$480 at 1.5% for 5 years
Answer: $36
The formula for simple interest= Principal x rate x time
I = P x r x t
= $480 x0.015 x 5 =$36
Therefore , the interest earned =  $36

Question 4.
$1200 at 2.8% for 30 months
Answer: $84
The formula for simple interest= Principal x rate x time
I = P x r x t
= $1200 x0.028 x 2.5 =$84
Therefore , the interest earned =  $84

Write the indicated ratio. Then find and interpret the value of the ratio.
Big Ideas Math Answer Key Grade 7 Chapter 7 Probability 7.1 9
Question 5.
rolled oats : chopped peanuts
Answer:
2 : 1/2
2/1 : 1/2
Multiplying with 2
2/1 x 2 : 1/2 x 2
4/2 : 2/4
2/1 : 1/2
2/1 x 2 : 1/2:2
4 : 1
Therefore, the value of ratio = 4: 1

Question 6.
sunflower seeds : pumpkin seeds
Answer:
1/3 :1/4
Multiplying with 12
1/3 x 12 : 1/4 x 12
12/36 : 12/ 48
1/3: 1 /4
1/3 x 12 : 1/4 x12
12/3 : 12/4
4 : 3
Therefore, the ratio = 4 : 3

Question 7.
pumpkin seeds : rolled oats
Answer:
1/4 :2
On multiplying with 4
1/4 x 4 : 2/1 x 4
4/16 : 8/4
1/4: 2
1/4 x 4 : 2 x 4
1 : 8
Therefore , ratio = 1:8

Solve the inequality. Graph the solution.
Question 8.
x + 5 < 9
Answer: x<4

Explanation:
x + 5 < 9
=x < 9-5
x<4
So, the inequality form = x<4
Graph:

Question 9.
b – 2 ≥ – 7
Answer: b≥-5

Explanation:
b – 2 ≥ – 7
= b ≥ -7+2
= b≥-5
Graph:

Question 10.
1 > – \(\frac{w}{3}\)
Answer:

Question 11.
6 ≤ – 2g
Answer: g≤ -3

Explanation:
= g≤ 6/-3
= g≤ -3
The inequality = g≤ -3
Graph:

Concepts, Skills, & Problem Solving

DETERMINING LIKELIHOOD Determine which numbers you are more likely to spin and which numbers you are less likely to spin. Explain your reasoning. (See Exploration 1, p. 283.)
Question 12.
Big Ideas Math Answer Key Grade 7 Chapter 7 Probability 7.1 10

Answer:
The spinner is divided into 6 parts (unequal)
They are numbered as {1,2,3,4,5,6}
The numbers {2,4,5} are more likely to happen
Because, they occupy more space than remaining.
And , the numbers {1,3,6] are less likely to happen
Because , they occupy less space .

Question 13.
Big Ideas Math Answer Key Grade 7 Chapter 7 Probability 7.1 11
Answer:
The spinner is divided into 4 equal parts
The 4 parts are numbered as {1,2,3,4}
So, each number have equal chance of spinning
Such events are also called as equally likely happen events.

IDENTIFYING OUTCOMES You spin the spinner shown.
Big Ideas Math Answer Key Grade 7 Chapter 7 Probability 7.1 12
Question 14.
How many possible outcomes are there?
Answer: 8

Explanation:
The spinner is divided into 8 equal parts
The number of possible outcomes for the spinner are 8
{ 1,2,3,4,5,6,7,8}

Question 15.
What are the favorable outcomes of spinning a number not greater than 3?
Answer: 3

Explanation:
There are 3 numbers those are not greater than 3, they are  {1,2,3}.
There are 3 favorable outcomes.

Question 16.
In how many ways can spinning an even number occur?
Answer:  4

Explanation:
There are 4 even numbers , they are {2,4,6,8}
So, there are 4 ways of getting an even number.

Question 17.
In how many ways can spinning a prime number occur?

Answer:4

Explanation:
There are 4 prime numbers , they are {2,3,5,7}
So, there are 4 ways of getting a prime number.

IDENTIFYING OUTCOMES You randomly choose one marble from the bag.(a) Find the number of ways the event can occur. (b) Find the favorable outcomes of the event.
Big Ideas Math Answer Key Grade 7 Chapter 7 Probability 7.1 13

There are 9 colored balls in the bag in which ,
The number of red balls = 3 .
The number of blue balls = 2
The number of yellow balls = 1
The number of purple balls = 2.
The number of green balls =1
The number of favorable outcomes = 9.

Question 18.
Choosing blue
Answer: 2/9

Explanation :
The total number of out comes = 9.
The number of blue balls in the bag = 2
Consider B is the event of getting Blue ball
So, the probability of getting blue ball, P(B)=  (Number of Favorable outcomes) / (Total number of outcomes)
P(B) = 2 / 9.
Therefore, the probability of getting blue ball is 2/9 = 0.22.

Question 19.
Choosing green
Answer: 1/9

Explanation :
The total number of out comes = 9.
The number of green balls in the bag = 1
Consider G is the event of getting Blue ball
So, the probability of getting green ball, P(G)=  (Number of Favorable outcomes) / (Total number of outcomes)
P(G) = 2 / 9.
Therefore, the probability of getting green ball is 1/9 = 0.11.

Question 20.

Choosing purple

Answer: 2/9

Explanation :
The total number of out comes = 9.
The number of purple balls in the bag = 2
Consider P is the event of getting purple ball
So, the probability of getting purple ball, P(P)=  (Number of Favorable outcomes) / (Total number of outcomes)
P(P) = 2 / 9.
Therefore, the probability of getting purple ball is 2/9 = 0.22.

Question 21.

Choosing yellow

Answer: 1/9

Explanation :
The total number of out comes = 9.
The number of yellow balls in the bag = 1
Consider Y is the event of getting yellow ball
So, the probability of getting yellow ball, P(Y)=  (Number of Favorable outcomes) / (Total number of outcomes)
P(Y) = 2 / 9.
Therefore, the probability of getting yellow ball is 1/9 = 0.11.

Question 22.

Choosing not red
Answer: 0.67
There are 3 red balls in the bag
The total number of colored balls in the bag are 9
The balls other than red balls are 9 – 3 = 6
Therefore, the probability of getting a ball other than red is
P(E) = Favorable outcomes / Total number of outcomes .
Where, E is the event of getting a ball other than red
P(E) = 6/9 = 2/3
Therefore , the probability of choosing not red ball is 2/3 = 0.67.

Question 23.
Choosing not blue

Answer: 7/9 = 0.78

Explanation:
There are 2 blue balls in the bag
The total number of colored balls in the bag are 9
The balls other than blue balls are 9 – 2 = 7
Therefore, the probability of getting a ball other than blue is
P(E) = Favorable outcomes / Total number of outcomes .
Where, E is the event of getting a ball other than blue
P(E) = 7/9 = 0.78
Therefore , the probability of choosing not blue ball is 7/9.

Question 24.
YOU BE THE TEACHER
Your friend finds the number of ways that choosing not purple can occur. Is your friend correct? Explain your reasoning.
Big Ideas Math Answer Key Grade 7 Chapter 7 Probability 7.1 14

Answer: Yes, he is correct
Given, the number of not purple colors=4
They are red, blue, green, yellow
Therefore the number of ways that choosing not purple can occur = 4 .

CRITICAL THINKING Tell whether the statement is true or false. If it is false, change the italicized word to make the statement true.
Big Ideas Math Answer Key Grade 7 Chapter 7 Probability 7.1 15
Question 25.
Spinning blue and spinning green have the same number of favorable outcomes on Spinner A.
Answer: False
Spinning blue and spinning Red have the same number of favorable outcomes on spinner A.

Question 26.
There are three possible outcomes of spinning Spinner A.
Answer: True

Question 27.
Spinning red can occur in four ways on Spinner B.
Answer: True

Question 28.
Spinning not green can occur in three ways on Spinner B.
Answer: False

Spinning not green can occur in four ways on Spinner B.

DESCRIBING LIKELIHOOD Describe the likelihood of the event given its probability.
Question 29.
Your soccer team wins \(\frac{3}{4}\) of the time.
Answer:50
The likelihood of the event is may or may not happen .

Question 30.
There is a 0% chance that you will grow12 feet.
Answer:
The probability =0
Thus, the likelihood of the event is Impossible
Therefore , It is impossible to to grow 12 feet.

Question 31.
The probability that the sun rises tomorrow is 1.
Answer:
The probability 1=100%
The likelihood of the event is  certain
So, it is sure that the sun rises tomorrow

Question 32.
It rains on \(\frac{1}{5}\) of the days in June.
Answer:
The likelihood of the event is not likely to happen .

Question 33.
MODELING REAL LIFE
You have a 50% chance of being chosen to explain a math problem in front of the class. Describe the likelihood that you are chosen.
Answer:
Given,  there is 50% chance of being chosen to explain a math problem in front of the class
50% = 0.5
So, the likelihood of the event is may or may not occur .
there is 50% chance

Question 34.
MODELING REAL LIFE
You roll a number cube and record the number of times you roll an even number and the number of times you roll an odd number. Describe the likelihood of each event.̇̇̇̇̇̇̇̇̇̇̇̇̇̇̇̇̇̇̇̇
Big Ideas Math Answer Key Grade 7 Chapter 7 Probability 7.1 16
a. You roll an even number on your next roll.
b. You roll an odd number on your next roll.
Answer:

a.Explanation:
Total number times cube rolled = 50
The frequency for even number =26
Consider E be the event of rolling an even number
P(E)=(Number of times an event occurs)/ (Total number of trails)
P(E)= 26/50 = 0.72
Therefore , the chances of rolling an even number = 0.72

b.Explanation:
Total number times cube rolled = 50
The frequency for odd number =24
Consider O be the event of rolling a odd number
P(O)=(Number of times an event occurs)/ (Total number of trails)
P(O)= 24/50 = 0.48
Therefore , the chances of rolling a odd number = 0.48

Question 35.
REASONING
You want to determine whether a coin is fair. You flip the coin and record the number of times you flip heads and the number of times you flip tails.
Big Ideas Math Answer Key Grade 7 Chapter 7 Probability 7.1 17
a Describe the likelihood that you flip heads on your next flip.
b. Describe the likelihood that you flip tails on your next flip.
c. Do you think the coin is a fair coin? Explain. ̇̇̇̇̇
Answer:

a.Total number of flips = 25
The total frequency for heads = 22
Now, the likelihood that you flip heads on your next flip
= Number of heads/ total
= 22 / 25 = 0.88
=88%
The likelihood of the event is more likely to occur

b.Total number of flips =25
The total frequency for tails = 3
Now, the likelihood that you flip tails on your next flip
= Number of tails/ Total
= 3/25 =0.12
Therefore, the likelihood of the event is not likely to occur

c.Every coin has both sides, { heads, tails}
But, the probability of tossing coin is not fair
They do not have 50/50 outcomes for both heads and tails .
So the coin is not fair in tossing.

Question 36.
LOGIC
At a carnival, each guest randomly chooses 1 of 50 rubber ducks and then replaces it. The table shows the numbers of each type of duck that have been drawn so far. Out of 150 draws, how many can you expect to not be a losing duck? Justify your answer.̇̇̇̇̇̇̇̇̇̇̇̇̇̇̇̇
Big Ideas Math Answer Key Grade 7 Chapter 7 Probability 7.1 18

Answer:
Total number of ducks = 25
Total number of win ducks = 6
Now,
The probability to expect to not be a losing duck after 105 draws =
won ducks/ total =x/150
6/25 =x/150
25x= 900
x= 900/25
x= 36 =0.36
The chance to expect to not be a losing duck after 105 draws is ‘not likely to occur

Question 37.
CRITICAL THINKING
A dodecahedron has twelve sides numbered 1 through 12. Describe the likelihood that each event will occur when you roll the dodecahedron. Explain your reasoning.
Big Ideas Math Answer Key Grade 7 Chapter 7 Probability 7.1 19
a. rolling a 1
b. rolling a multiple of 3
c. rolling a number greater than 6

a. rolling a 1

Answer: The probability of rolling a 1 is 1/12

Explanation :
There are 12 number of outcomes for a dodecahedron,
{1,2,3,4,5,6,7,8,9,10,11,12}
P{O}= favorable outcomes / Total number of outcomes.
Where O is the event of getting number 1
p{O}= 1 /12
Therefore , the probability of rolling 1 is 1/12.

b. Rolling a multiple of 3

Answer:  1/3

Explanation:
There are 4 multiples of 3 in a dodecahedron.
They are, {3,6,9,12}
There are 12 number of outcomes for a dodecahedron,
{1,2,3,4,5,6,7,8,9,10,11,12}
P{M}= Favorable outcomes/ Total number of outcomes
Where, M is the event of rolling a multiple of 3
P{M}= 4/12 = 1/3
Therefore, the probability of getting a multiple of 3 is 1/3

C. Rolling a number greater than 6.

Answer: 1/2

Explanation:
There are 6 numbers of which are greater than 6 . they are, {7,8,9,10,11,12}
There are 12 number of outcomes for a dodecahedron,
{1,2,3,4,5,6,7,8,9,10,11,12}
P{G}= Favorable outcome / Total number of outcomes
Where, G is the event of getting a number greater than 6
P{G} = 6/12=1/2
Therefore, the probability of getting a number greater than 6 is 1/2.

Question 38.
DIG DEEPER!
A bargain bin contains classical CDs and rock CDs. There are 60 CDs in the bin. Choosing a rock CD and not choosing a rock CD have the same number of favorable outcomes. How many rock CDs are in the bin?

Answer: 30 Rock CDs

Explanation:
Given that,
There are 60 CDs in the bin
So, There are 60 total number of outcomes
Also given that, choosing a rock CD and not choosing a rock CD have the same number of favorable outcomes
So, there is equal probability .
We can conclude that there are same number of classical CDs and rock CDs
Which means , 30 +30 = 60
Therefore, there are 30 rock CDs in the bin.

Question 39.
REASONING
You randomly choose one of the cards and set it aside. Then you randomly choose a second card. Describe how the number of possible outcomes changes after the first card is chosen.
Big Ideas Math Answer Key Grade 7 Chapter 7 Probability 7.1 20
Answer:
Initially,
Total number cards =5
So, the number of possible outcome =5
If you choose one card and set it aside , the number of possible outcomes decreases by 1
Then when you randomly choose second card , there are only four cards left
Therefore, the number of possible outcomes decreases by 1 after the first card is chosen.

STRUCTURE A Punnett square is a grid used to show possible gene combinations for the offspring of two parents. In the Punnett square shown, a boy is represented by XY. A girl is represented by XX.
Big Ideas Math Answer Key Grade 7 Chapter 7 Probability 7.1 21
Question 40.
Complete the Punnett square. Explain why the events “having a boy” and “having a girl” are equally likely.
Answer:

There is a chance of 50% female and 50% male
The probability = 50%
Therefore , the likelihood of the event is equally likely.

Question 41.
Two parents each have the gene combination Cs. The gene C is for curly hair. The gene s is for straight hair. Any gene combination that includes a C results in curly hair. When all outcomes are equally likely, what is the probability of a child having curly hair?
Answer: 4
C is considered as X which results for curly hair
since, there are 4 outcomes includes X
Thus, the probability of child having curly hair = 4/4 =1
Therefore , the probability = 1
So, the likelihood or the chances of a child having curly hair is certain.

Lesson 7.2 Experimental and Theoretical Probability

EXPLORATION 1

Conducting Experiments
Work with a partner. Conduct the following experiments and find the relative frequencies.
Experiment 1
• Flip a quarter 25 times and record whether each flip lands heads up or tails up.
Big Ideas Math Answers 7th Grade Chapter 7 Probability 7.2 1
Experiment 2
• Toss a thumbtack onto a table 25 times and record whether each toss lands point up or on its side.
Big Ideas Math Answers 7th Grade Chapter 7 Probability 7.2 2
a. Combine your results with those of your classmates. Do the relative frequencies change? What do you notice?
b. Everyone in your school conducts each experiment and you combine the results. How do you expect the relative frequencies to change?
c. How many times in 1000 flips do you expect a quarter to land heads up? How many times in 1000 tosses do you expect a thumbtack to land point up? Explain your reasoning.
d. In a uniform probability model, each outcome is equally likely to occur. Can you use a uniform probability model to describe either experiment? Explain.
Answer:

Big Ideas Math Answers 7th Grade Chapter 7 Probability 7.2 3

Try It

The table shows the results of rolling a number cube 50 times. Find the experimental probability of the event.
Big Ideas Math Answers 7th Grade Chapter 7 Probability 7.2 4
Question 1.
rolling a 3
Answer: 8/50 =  0.16

Explanation:
Total number of trails = 50
Frequency for  3 = 8
Consider T be the event of rolling a 3
Experimental probability = P(T)(Number of times an event occurs)/ (Total number of trails)
P(T)= 8/50= 0.16
Therefore , the probability of rolling a 3 is 0.16.

Question 2.
rolling an odd number.
Answer:
The probability of rolling an odd number = 0.58

Explanation:
Total number of odd numbers = 3
They are {1,3,5,}
The frequency for 1 = 10
The frequency for 3 =8
The frequency for 5= 11
Total frequency = 10 +8+11 = 29
Total number of trails = 50
Consider O be the event of rolling a Odd number
Experimental probability = P(O)=(Number of times an event occurs)/ (Total number of trails)
P(O)= 29/50= 0.58
Therefore , the probability of rolling an odd number = 0.58

Question 3.
What is the theoretical probability of randomly choosing an X?
Big Ideas Math Answers 7th Grade Chapter 7 Probability 7.2 5
Answer: 1/7 =0.142

Explanation:
Number of favorable outcomes = 7
Consider X be the event of choosing an X
Theoretical probability P(X)=(Number of favorable outcomes)/(Total number of outcomes)
P(X)= 1/7 = 0.142
Therefore, the probability of randomly choosing X =0.142

Question 4.
How does the experimental probability of rolling a number greater than 1 compare with the theoretical probability?
Answer:
Theoretical probability  is what we expect to happen, where experimental probability is what actually happens when we try it out.
The experimental probability of an event cannot be greater than 1 since the number of trials in which the event can happen cannot be greater than the total number of trials.

Question 5.
An inspector randomly selects 200 pairs of jeans and finds 5 defective pairs. About how many pairs of jeans do you expect to be defective in a shipment of 5000?
Big Ideas Math Answers 7th Grade Chapter 7 Probability 7.2 6
Answer: 125 defective jeans

Explanation:
Total number of pairs = 200
Number of defective pairs = 5
Let ,x be the number of defective pairs of jeans in a shipment of 5000
Now, the number of defective pair of jeans in a shipment of 5000 =
Defective / Total= x/5000
5/200 = X/5000
200 x x =  5000 x 5
200x = 25000
x = 25000/200
x = 125
Therefore there are 125 pairs of jeans are expected  to be defective in a shipment of 5000

Self-Assessment for Concepts & Skills
Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 6.
Explain what it means for an event to have a theoretical probability of 0.25 and an experimental probability of 0.3.
Answer:
If you have a fair coin, and flip it twice,
The  theoretical probability of getting 2 heads = 0.25
There’s 4 outcomes: HH, TT, TH, and HT.
1 out of 4 is 0.25 chance of 2 heads.
But if you run the experiment and do actual flips, maybe 100 times,
The average times you get 2 heads might be 0.3 if you got 2 heads 30 times out of a 100.
It could be just random chance or maybe the coin is slightly heavier on the head side, causing slightly more heads than the theoretical probability.

Question 7.
DIFFERENT WORDS, SAME QUESTION
You flip a coin and record the results in the table. Which is different? Find “both” answers.
Big Ideas Math Answers 7th Grade Chapter 7 Probability 7.2 7
Answer:

  1. Experimental probability of flipping heads = frequency of heads / Total number of trails = 32/60 =0.53

2.       The fraction of the flips you can expect a result of heads = 32/60 =0.53

3.        The percent of flips result in heads = 0.53 =53%

4.        The relative frequency of flipping heads =  32/60 = 0.53

Self-Assessment for Problem Solving
Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 8.
Contestants randomly draw a ticket from a hat and replace it. The table shows the results after 40 draws. There are 7 winning tickets in the hat. Predict the total number of tickets in the hat. Explain.
Big Ideas Math Answers 7th Grade Chapter 7 Probability 7.2 8
Answer:
Let, the total numbers of tickets in the hat = x
Given that , Number of winning tickets = 7
Number of winning won after 40 draws = 2
Total number of tickets =
2/40 = 7/x
2x = 280
x = 280/2
x= 140
Therefore, The total number of tickets in the hat = 140

Question 9.
DIG DEEPER!
You choose two different songs on a music play list at random. Out of 80 songs on the playlist, 36 are hip hop songs. The first song you choose is a hip hop song. What is the probability that the second song is also a hip hop song? Explain your reasoning.
Big Ideas Math Answers 7th Grade Chapter 7 Probability 7.2 9
Answer:
Total number of songs = 80
Number of hip pop songs = 36
Given that the first song is hip pop
The probability of playing second song also hip pop =
Total number of hip pop songs / total number of songs
Consider H be the event of playing hip pop
P(H)= 36/ 80 =0.45
If there is no repeat :
Total number of hip pop songs after playing one song = 35
P(H)= 0.43
Therefore ,the likelihood of the event is unlikely to happen.

Experimental and Theoretical Probability Homework & Practice 7.2

Review & Refresh

Describe the likelihood of the event given its probability.
Question 1.
You randomly guess the correct answer of a multiple choice question \(\frac{1}{4}\) of the time.
Answer:
The likelihood of the event is not likely to happen .

Question 2.
There is a 95% chance that school will not be cancelled tomorrow.
Answer:
Given , 95% chance that school will be cancelled tomorrow
95%= 0.95
Probability = 0.95
So the likelihood of the event is more likely to happen

Find the annual interest rate.
Question 3.
I = $16, P = $200, t = 2 years
Answer: 4%
The formula for simple interest= Principal x rate x time
I = P x r x t
$16 = $200 x r x 2
$16= 400r
r = 16/400
r= 0.04
0.04 x 100= 4
Therefore, the annual interest rate = 4%

Question 4.
I = $26.25, P = $500, t = 18 months
Answer:3.5
The formula for simple interest= Principal x rate x time
I = P x r x t
$26.25 = $500 x r x 1.5
$26.25= 750r
r = 26.25/750
r= 0.035
0.035 x 100= 3.5
Therefore, the annual interest rate = 3.5%

Tell whether x and y are proportional.
Question 5.
Big Ideas Math Answers 7th Grade Chapter 7 Probability 7.2 10
Answer:
The ratio of x and y for the given values is
1/8 = 0.12
3/24= 0.12
9/75 = 0.12
We get equal values for all the ratios
Therefore, the relationship given in the table is proportional.

Question 6.
Big Ideas Math Answers 7th Grade Chapter 7 Probability 7.2 11
Answer:
The ratio of x and y for the given values is
0.75/0.3 =2.5
1.5/0.6 = 2.5
2.25/0.9 = 2.5
We get equal values for all the ratios
Therefore, the relationship given in the table is proportional.

Concepts, Skills, & Problem Solving

CONDUCTING AN EXPERIMENT Use the bar graph below to find the relative frequency of the event. (See Exploration 1, p. 291.)
Question 7.
spinning a 6
Answer: 0.14

Explanation:
Total number of spins =50
The number of times 6 spun =7
Consider X be the event of spinning a 6
Experimental probability = P(T)=(Number of times an event occurs)/ (Total number of trails)
P(X)= 7/50= 0.14
Therefore, the probability spinning a 6= 0.14

Question 8.
spinning an even number
Answer:0.48

Explanation:
Total number of spins =50
Numbers of even numbers = 3
They are {2,4,6}
Number of times 2 spun = 6
Number of times 4 spun = 11
Number of times 6 spun =7
Total : 24
Consider E be the event of spinning a number less than 3
Experimental probability  P(E)=(Number of times an event occurs)/ (Total number of trails)
P(T)= 24/50=0.48
Therefore the probability of spinning a even number = 0.48

FINDING AN EXPERIMENTAL PROBABILITY Spinning a Spinner .Use the bar graph to find the experimental probability of the event.
Big Ideas Math Answers 7th Grade Chapter 7 Probability 7.2 12
Question 9.
spinning a number less than 3
Answer:  14/50=0.28

Explanation:
Total number of spins =50
Numbers less than 3 are {1,2}
Number of times 1 spun =8
Number of times 2 spun =6
Total : 8 +6 = 14
Consider T be the event of spinning a number less than 3
Experimental probability = P(T)=(Number of times an event occurs)/ (Total number of trails)
P(T)= 14/50=0.28
Therefore, the probability spinning a number less than 3 =0.28

Question 10.
not spinning a 1
Answer: 42/50= 0.84

Explanation:
Total number of spins =50
The numbers other than 1 are{2,3,4,5,6}
The number of times 2 spun = 6
The number of times 3 spun= 9
The number of times 4 spun= 11
The number of times 5 spun= 9
The number of times 6 spun= 7
Total : 6+9+11+9+7=42
Consider O be the event of spinning a number less than 3
Experimental probability = P(O)=(Number of times an event occurs)/ (Total number of trails)
P(O)= 42/ 50= 0.84
Therefore, the probability of spinning a number other than 1 is 0.84.

Question 11.
spinning a 1 or a 3
Answer:
Total number of spins =50
Number of times 1 spun = 8
The number of times 3 spun= 9
Total : 8+9 = 17
Consider S be the event spinning a 1 or 3
Experimental probability = P(S)=(Number of times an event occurs)/ (Total number of trails)
P(S)= 17/ 50= 0.34
Therefore, the probability of spinning a  1 OR 3 is 0.34.

Question 12.
spinning a 7
Answer: 0

Explanation:
The bar graph consists 6 numbers
They are, {1,2,3,4,5,6}
So, it is impossible to spin 7
Therefore the probability of spinning is 0.

Question 13.
YOU BE THE TEACHER
Your friend uses the bar graph above to find the experimental probability of spinning. Is your friend correct? Explain your reasoning.
Big Ideas Math Answers 7th Grade Chapter 7 Probability 7.2 13
Answer:
No, he is wrong

Explanation:
The formula for finding experimental probability is
Experimental probability P(O)=(Number of times an event occurs)/ (Total number of trails)
Consider O be the event of spinning 4
P(O)=11/50
Therefore, the experimental probability of spinning a 4 is 0.22
So, he is wrong

Question 14.
MODELING REAL LIFE
You check 20 laser pointers at random. Three of the laser pointers are defective. What is the experimental probability that a laser pointer is defective?
Big Ideas Math Answers 7th Grade Chapter 7 Probability 7.2 14
Answer: 0.3
Experimental probability = P(T)=(Number of times an event occurs)/ (Total number of trails)
Total number of lasers = 20
Number of defective lasers =3
P(T) = 3/20
= 0.3
Therefore, the experimental probability that a laser pointer is defective is 0.3

FINDING A THEORETICAL PROBABILITY Use the spinner to find the theoretical probability of the event.
Big Ideas Math Answers 7th Grade Chapter 7 Probability 7.2 15
Question 15.
spinning red
Answer: 0.33

Explanation:
The spinner is divided into 6 equal parts.
So, the total number of outcomes of a spinner = 6
In which 2 parts are red and remaining parts are blue, green , orange and purple
Consider R is the event of getting a red color.
So, the probability of spinning red color P(R) = ( number of favorable outcomes)/(Total number of outcomes)
P(R)= 2/6= 0.33

Question 16.
spinning a 1
Answer: 1/6 = 0.16

Explanation:
The spinner is divided into 6 equal parts.
So, the total number of outcomes of a spinner = 6
{1,2,3,4,5,6}
Consider ,O is the event of getting 1
So, the probability of getting 1 is P(O)=( number of favorable outcomes)/(Total number of outcomes)
P(O)= 1/6 = 0.16
Therefore the probability of spinning a 1= 0.16

Question 17.
spinning an odd number
Answer: 3/6= 0.5

Explanation:
The number of total outcomes of a spinner = 6
They are , {1,2,3,4,5,6}
There are 3 odd numbers, They are {1,3,5}
Consider N is the event of getting an odd number
So, the probability of spinning an odd number P(N) =( number of favorable outcomes)/(Total number of outcomes)
P(N)= 3/6= 0.5.
Therefore, the probability of spinning an odd number = 3/6 =0.5

Question 18.
spinning a multiple of 2
Answer: 3/6= 0.5

Explanation:
The total number of outcomes of a spinner = 6
They are , {1,2,3,4,5,6}
There are 3 multiples of 2 , They are {2,4,6}
Consider M is the event of getting an odd number
So, the probability of spinning a multiple of 2is  P(M) =( number of favorable outcomes)/(Total number of outcomes)
P(M)= 3/6= 0.5.
Therefore, the probability of spinning a multiple of 2 = 3/6 =0.5

Question 19.
spinning a number less than 7
Answer: 1

Explanation:
The spinner is divided into 6 equal parts.
So, the total number of outcomes of a spinner = 6
{1,2,3,4,5,6}
Consider, S is the event of spinning a number less than 7
So, the probability of spinning a number less than 7 P(S) =( number of favorable outcomes)/(Total number of outcomes)
P(S)= 6/6 =1
So, the probability of spinning a number less than 7 = 1

Question 20.
spinning a 9
Answer: 0
The spinner is divided into 6 equal parts numbering from 1 to 6
={1,2,3,4,5,6}
There is no 9 in the spinner
So, the probability of spinning 9 is 0
It is also called as impossible event .

Question 21.
REASONING
Each letter of the alphabet is printed on an index card. What is the theoretical probability of randomly choosing any letter except Z?
Answer: 25/26 =0.96

Explanation:
We know that, there are 26 alphabets.
Also, the number of  alphabets other than Z are: 25
Let us consider Z is the event of choosing any letter except Z .
P(Z)= (number of favorable outcomes) / (Total number of outcomes )
p(Z)= 25/ 26 =0.96
Therefore, the theoretical probability of choosing a letter except Z is 25/26 = 0.96

COMPARING PROBABILITIES The bar graph shows the results of spinning the spinner below 200 times. Compare the theoretical and experimental probabilities of the event.
Big Ideas Math Answers 7th Grade Chapter 7 Probability 7.2 16.
Question 22.
spinning a 4
Answer:
Theoretical probability:
Total number of outcomes = 5
Consider F be the event of spinning 4
Theoretical probability P(F)= (Number of favorable outcome )/(Total number of outcomes)
P(F)= 1/5= 0.2
Experimental probability :
Total number of spins = 200
Number of times 4 spun =37
Consider F be the event of spinning 4
Experimental probability = P(F)=(Number of times an event occurs)/ (Total number of trails)
P(F)= 37/200= 0.185
Therefore, the theoretical probability = 0.2
The experimental probability = 0.18

Question 23.
spinning a 3
Answer:
Theoretical probability:
Total number of outcomes = 5
Consider T be the event of spinning 3
Theoretical probability P(T)= (Number of favorable outcome )/(Total number of outcomes)
P(T)= 1/5= 0.2
Experimental probability :
Total number of spins = 200
Number of times 3 spun = 39
Consider T be the event of spinning 3
Experimental probability = P(T)=(Number of times an event occurs)/ (Total number of trails)
P(T)= 39/200= 0.195
Therefore, the theoretical probability = 0.16
The experimental probability = 0.195

Question 24.
spinning a number greater than 4
Answer:
Theoretical probability:
Total number of outcomes = 5
Consider N be the event of spinning a number greater than 4
Numbers greater than 4 = {5}
Theoretical probability P(T)= (Number of favorable outcome )/(Total number of outcomes)
P(T)= 1/5= 0.2
Experimental probability :
Total number of spins = 200
Consider N be the event of spinning a number greater than 4
Number of times 5 spun =40
Experimental probability = P(T)=(Number of times an event occurs)/ (Total number of trails)
P(T)= 40/200= 0.2
Therefore, the theoretical probability = 0.2
The experimental probability = 0.2

Question 25.
spinning an odd number
Answer:
Theoretical probability:
Total number of outcomes = 5
Consider S be the event of spinning an odd number
Total odd numbers = 3
They are {1,3,5}
Theoretical probability P(T)= (Number of favorable outcome )/(Total number of outcomes)
P(T)= 3/5= 0.6
Experimental probability :
Total number of spins = 200
Consider S be the event of spinning an odd number
The number of time 1 spun=41
The number of time 3 spun=39
The number of time 5 spun=40
Total : 41+39+40= 120
Experimental probability = P(T)=(Number of times an event occurs)/ (Total number of trails)
P(T)= 120/200= 0.6
Therefore, the theoretical probability = 0.6
The experimental probability = 0.6

Question 26.
REASONING
Should you use theoretical or experimental probability to predict the number of times you will spin a 3 in 10,000 spins? Explain.
Answer: theoretical probability

Explanation:
Theoretical probability of event =
Theoretical probability = P(T)=(Number of favorable outcomes)/ (Total number of outcomes)
Now,
Consider T be the event of spinning 3
P(T)= 1/6.

Question 27.
MODELING REAL LIFE
A board game uses a bag of 105 lettered tiles. You randomly choose a tile and then return it to the bag. The table shows the number of vowels and the number of consonants after 50 draws. Predict the number of vowels in the bag.
Big Ideas Math Answers 7th Grade Chapter 7 Probability 7.2 16
Answer:
let , X be the number of vowels in the bag.
Total number of tiles = 105
Number of vowels after 50 draws = 18
Now, the number of vowels in the bag =
X/150 = 18/50
50X=2700
X = 2700/50
X = 54
Therefore, 54  vowels are expected to be in the bag.

Question 28.
MODELING REAL LIFE
On a game show, a contestant randomly draws a chip from a bag and replaces it. Each chip says either win or lose. The theoretical probability of drawing a winning chip is \(\frac{3}{10}\). The bag 10contains 9 winning chips.
a. How many chips are in the bag?
b. Out of 20 contestants, how many do you expect to draw a winning chip?
Answer:
Given , the probability of drawing a winning chip = 3/10
Consider , x be the number of chips in the bag
The number of chips = 9
The ration of strikes in the bag = 9/x
3/10 = 9/x
3x = 90
x= 30
Therefore , there are 30 chips in the bag

Question 29.
PROBLEM SOLVING
There are 8 females and 10 males in a class.
a. What is the theoretical probability that a randomly chosen student is female?
b. One week later, there are 27 students in the class. The theoretical probability that a randomly chosen student is a female is the same as last week. How many males joined the class?
Answer:

a.Number of females=8
Total number of students = 8+10= 18
The theoretical probability that a randomly chosen student is female
Consider F be the event of chosen a female =
P(F)= Number of favorable outcomes / Total number of outcomes
P(F)= 8/18 =0.44

b.Initially,
Total number of students = 18
In which , number of females =8
Number of males =10
one week later , the total number of students = 27
Given, the theoretical probability for choosing girl is same
So, Number of males joined = 27 -18 = 9
Therefore, 9 males joined

Question 30.

NUMBER SENSE
The table at the right shows the results of flipping two coins 12 times each.
Big Ideas Math Answers 7th Grade Chapter 7 Probability 7.2 17
a. What is the experimental probability of flipping two tails? Using this probability, how many times can you expect to flip two tails in 600 trials?
b. The table at the left shows the results of flipping the same two coins 100 times each. What is the experimental probability of flipping two tails? Using this probability, how many times can you expect to flip two tails in 600 trials?
Big Ideas Math Answers 7th Grade Chapter 7 Probability 7.2 18
c. Why is it important to use a large number of trials when using experimental probability to predict results?
Answer:

a.Explanation:
According to table,
Number of turns = 12
Frequency of flipping 2 tails = 1
The probability of flipping two tails in 600 trails =
1/12 = x/600
12x =600
x = 600/ 12
x = 50
In 600 trails , flipping 2 tails is expected to be 50 times.

b.Explanation:
According to table,
Number of turns = 100
Frequency of flipping 2 tails = 22
The probability of flipping two tails in 600 trails =
22/100= x/600
100x =13,200
x = 13200/100
x = 132
In 600 trails , flipping 2 tails is expected to be 132 times.

c.In experimental probability, as the number of trials increases, the experimental probability gets closer to the theoretical probability.
So, it is important to use a large number of trails.

Question 31.
COMPARING PROBABILITIES
The table shows the possible outcomes of rolling a pair of number cubes. You roll a pair of number cubes 60 times and record your results in the bar graph shown.
Big Ideas Math Answers 7th Grade Chapter 7 Probability 7.2 19
a. Compare the theoretical and experimental probabilities of rolling each sum.
b. Which sum do you expect to be most likely after 500 trials? 1000 trials? Explain your reasoning.
c. Predict the experimental probability of rolling each sum after 10,000 trials. Explain your reasoning.
Answer:

a.

b. 6 sum is more likely to happen after after 500 trials, 1000 trials, and even after 10,000 trails because the the number of times 6 rolled is greater than other numbers sum.

Question 32.
PROJECT
When you toss a paper cup into the air, there are three ways for the cup to land: open-end up, open-end down, or on its side.
a. Toss a paper cup 100 times and record your results. Do the outcomes for tossing the cup appear to be equally likely? Explain.
b. Predict the number of times each outcome will occur in 1000 tosses. Explain your reasoning.
c. Suppose you tape a quarter to the bottom of the cup. Do you think the cup will be more likely less likely or to land open-end up? Justify your answer.
Big Ideas Math Answers 7th Grade Chapter 7 Probability 7.2 20
Answer:
The answer will vary

a. The results of tossing a cup 100 times are :
Open end up      = 18
Open end down = 31
On its side           = 51
The events of tossing the cup are not equally likely.

b.The probability of the events in 1000 tosses
open end up =18/ 100 x 1000 =180
Open end down = 31/100 x 1000 =310
On its side = 51 / 100 x1000 =510

c.If you tape quarter to the bottom, it will probably be more likely to land open end up since the weight of the bottom
will pull the bottom of the cup down and the top of the cap up.

Lesson 7.3 Compound Events

EXPLORATION 1

Comparing Combination Locks
Work with a partner. You are buying a combination lock. You have three choices.
a. One lock has 3 wheels. Each wheel is numbered from 0 to 9. How many possible outcomes are there for each wheel? How many possible combinations are there?
Big Ideas Math Answers Grade 7 Chapter 7 Probability 7.3 1
b. How can you use the number of possible outcomes on each wheel to determine the number of possible combinations?
c. Another lock has one wheel numbered from 0 to 39. Each combination uses a sequence of three numbers.How many possible combinations are there?
Big Ideas Math Answers Grade 7 Chapter 7 Probability 7.3 2
d. Another lock has 4 wheels as described. How many possible combinations are there?
Big Ideas Math Answers Grade 7 Chapter 7 Probability 7.3 3
e. For which lock are you least likely to guess the combination? Why?
Answer:

The set of all possible outcomes of one or more events is called the sample space. You can use tables and tree diagrams to find the sample space of two or more events.
Big Ideas Math Answers Grade 7 Chapter 7 Probability 7.3 4

Try It

Question 1.
WHAT IF?
The sandwich shop adds a multi-grain bread. Find the sample space. How many sandwiches are possible?
Answer: sample space = {Sandwich , multi grain bread }

Question 2.
Find the total number of possible outcomes of spinning the spinner and randomly choosing a number from 1 to 5.
Big Ideas Math Answers Grade 7 Chapter 7 Probability 7.3 5
Answer:
Number of colors in the spinner(a) =4
They are {red, blue, green, yellow}
Total numbers (b)= 5
{1,2,3,4,5}
Total number of possible outcomes =
According to fundamental principle ,(a x b)=
4 x 5 = 20
Therefore , total number of outcomes = 20

Question 3.
How many different outfits can you make from 4 T-shirts, 5 pairs of jeans, and 5 pairs of shoes?
Big Ideas Math Answers Grade 7 Chapter 7 Probability 7.3 6

Answer: 100

Explanation:
Number of T-shirts(a) = 4
Number of jeans pairs(b) = 5
Number of pair of shoes (C)= 5
According to fundamental principle,
Total number of outcomes = a x b x c
=4 x 5 x5
=100
Therefore , about 100 outfits can be made .

Question 4.
In Example 2, what is the probability of rolling at most 4 and flipping heads?
Answer:

Self-Assessment for Concepts & Skills
Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 5.
FINDING THE SAMPLE SPACE
You randomly choose a flower and ornament for a display case. Find the sample space. How many different displays are possible?
Big Ideas Math Answers Grade 7 Chapter 7 Probability 7.3 7

Answer:
Total number of flowers =3
They are {daffodil, hyacinth, tulip}
Total number of ornament =2
They are {figurine, trophy}
Total number of outcomes = 3 x 2 = 6
sample space =
{daffodil, figurine} {daffodil, trophy}
{hyacinth, figurine } {hyacinth, trophy }
{tulip, figurine}{tulip, trophy }.

Question 6.
FINDING THE TOTAL NUMBER OF POSSIBLE OUTCOMES
You randomly choose a number from 1 to 5 and a letter from A to D. Find the total number of possible outcomes.
Answer: 20
Total numbers (a) = 5
They are {1,2,3,4,5}
Total number of letters (b)=4
They are {A,B,C,D}
According to fundamental theory ,
Total number of possible outcomes = a x b
= 4 x 5
=20
Therefore , the total number of possible outcomes = 20

Question 7.
WHICH ONE DOESN’T BELONG?
You roll a number cube and flip a not coin. Which probability does belong with the other three? Explain your reasoning.
Big Ideas Math Answers Grade 7 Chapter 7 Probability 7.3 8
Answer: P(greater than 2 and tails ) probability does not belong with the other three.

Self-Assessment for Problem Solving
Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 8.
A tour guide organizes vacation packages at a beach side town. There are 7 hotels, 5 cabins, 4 meal plans, 3 escape rooms, and 2 amusement parks. The tour guide chooses either a hotel or a cabin and then selects one of each of the remaining options. Find the total number of possible vacation packages.
Answer:
Given, the tour guide chooses either a hotel or a cabin and then selects one of each of the remaining options.
Event 1 :
If the tour guide choose hotel , the total number of possible vacation packages.
According to fundamental counting principle ,
7 hotels x 4 meal plans x 3 escape rooms x 2 amusement parks =168
Therefore, The total number of possible vacation packages = 168
Event 2 :
If the tour guide choose cabin , the total number of possible vacation packages.
According to fundamental counting principle ,
5 cabins x4 meal plans x 3 escape rooms x 2 amusement parks =120
Therefore, The total number of possible vacation packages = 120

Question 9.
DIG DEEPER!
A fitness club with 100 members offers one free training session per member in either running, swimming, or weightlifting. Thirty of the fitness center members sign up for the free session. The running and swimming sessions are each twice as popular as the weightlifting session. What is the probability that a randomly chosen fitness club member signs up for a free running session?
Big Ideas Math Answers Grade 7 Chapter 7 Probability 7.3 9
Answer:
Total number of fitness club = 100
The number of members signed for free up session = 30
Given , the running and swimming sessions are each twice as popular as the weightlifting session
let , x be the event of weight lifting and 2x be the event of running and swimming
2x + 2x + x = 30
5x = 30
x = 6
Now,
the probability that a randomly chosen fitness club member signs up for a free running session=
P(R) =2x /30
= 2×6/ 30
= 12/30
P(R) = 0.4
Therefore , the probability of that a randomly chosen fitness club member signs up for a free running session= 0.4

Compound Events Homework & Practice 7.3

Review & Refresh

Use the bar graph to find the experimental probability of the event.
Big Ideas Math Answers Grade 7 Chapter 7 Probability 7.3 10
Question 1.
rolling a 5
Answer:

Explanation:
Total number of spins =100
Total number of times 5 rolled= 19
Consider T be the event of rolling a 5
Experimental probability = P(T)=(Number of times an event occurs)/ (Total number of trails)
P(T)= 19/100=0.19
Therefore, the probability of rolling 5 =19/100=0.19

Question 2.
rolling a 2 or 6
Answer:
Total number of spins =100
The number of times 2 rolled =16
The probability of rolling 2 = 16 /100 =0.16
The number of times 6 rolled = 20
The probability of rolling 6 = 20/100 =0.20

Question 3.
rolling at least a 3
Answer:

Explanation:
The number of spins = 100
The number of times 3 =15
Consider S be the event of spinning at least 3 =
P(S) = 15/100
P(S) = 0.15

Question 4.
rolling a number less than or equal to 4
Answer: 0.6

Explanation:
Total number of spins =100
Total number of times 1 rolled = 12
Total number of times 2 rolled= 16
Total number of times 3 rolled = 15
Total number of times 4 rolled= 17
Total : 12 + 16 + 15 +17 = 60
Consider L be the event of rolling a number less than or equal to 4
Experimental probability = P(T)=(Number of times an event occurs)/ (Total number of trails)
P(T)= 60 /100=0.6
Therefore, the probability of rolling a number less than or equal to 4 = 0.6

Find the product.
Question 5.
3 . 2
Answer: 6

Question 6.
5(- 3)
Answer:
= -12

Question 7.
– 6(- 2)
Answer:
= 12

Concepts, Skills, & Problem Solving

COMPARING PASSWORDS Determine which password is less likely to be guessed. (See Exploration 1, p. 299.)
Question 8.
a password with 3 numbers or a password with 3 capital letters
Answer: the likelihood of the event is certain to happen with the passwords

Question 9.
a password with 6 numbers or a password with 4 capital letters
Answer: The password with 4 capital letters is more likely to guess than the password with 6 numbers .

USING A TREE DIAGRAM Use a tree diagram to find the sample space and the total number of possible outcomes.
Question 10.
Big Ideas Math Answers Grade 7 Chapter 7 Probability 7.3 11

Answer:

Total number of possible outcomes = 3x 2 =6

Question 11.
Big Ideas Math Answers Grade 7 Chapter 7 Probability 7.3 12
Answer:

Question 12.
Big Ideas Math Answers Grade 7 Chapter 7 Probability 7.3 13
Answer:

Question 13.
Big Ideas Math Answers Grade 7 Chapter 7 Probability 7.3 14
Answer:

Question 14.
YOU BE THE TEACHER
Your friend finds the total number of ways that you can answer a quiz with five true-false questions. Is your friend correct? Explain your reasoning.
Big Ideas Math Answers Grade 7 Chapter 7 Probability 7.3 15
Answer: No,

Explanation ,
The number of outcomes for each question =
Possible answers for question number 1 = 2
Possible answers for question number 2 = 2
Possible answers for question number 3= 2
Possible answers for question number 4= 2
Possible answers for question number 5 = 2
According to fundamental counting principle ,
Total number of outcomes = 2 x 2 x 2 x 2 x 2 = 32
Therefore , You can answer the quiz in 32 different ways

USING THE FUNDAMENTAL COUNTING PRINCIPLE Use the Fundamental Counting Principle to find the total number of possible outcomes.
Question 15.
Big Ideas Math Answers Grade 7 Chapter 7 Probability 7.3 16
Answer: 12

Explanation:
Total number of sizes (a) = 3
Number of flavors (b) = 4
By using fundamental counting principle ,(a x b)
The total number of possible outcomes = 3 x 4 = 12
Therefore, total number of outcomes = 12

Question 16.
Big Ideas Math Answers Grade 7 Chapter 7 Probability 7.3 17
Answer:20

Explanation:
Number of batteries (a)= 4
Number of colors (b)= 5
By using fundamental counting principle ,( a x b)
The total number of possible outcomes = 4 x 5 = 20
Therefore, total number of outcomes = 20

Question 17.
Big Ideas Math Answers Grade 7 Chapter 7 Probability 7.3 18
Answer:24

Explanation:
Number of suits (a)= 3
Number of wigs (b)= 2
Number of talents (c)= 4
By using fundamental counting principle ,( a x b x c)
The total number of possible outcomes = 3 x 2 x 4=24
Therefore, total number of outcomes = 24

Question 18.
Big Ideas Math Answers Grade 7 Chapter 7 Probability 7.3 19
Answer:36

Explanation:
Number of appetizer (a)= 3
Number of Entre (b)= 4
Number of dessert (c)= 3
By using fundamental counting principle ,( a x b x c)
The total number of possible outcomes = 3 x 4 x 3 = 36
Therefore, total number of outcomes = 36

Question 19.
CHOOSE TOOLS
You randomly choose one of the marbles. Without replacing the first marble, you choose a second marble.
Big Ideas Math Answers Grade 7 Chapter 7 Probability 7.3 20
a. Name two ways you can find the total number of possible outcomes.
b. Find the total number of possible outcomes.
Answer:

a. Explanation:
The tree diagram and fundamental counting principle

b.There are 12 number of possible outcomes
Total number of balls = 4
Therefore, there are 4 possible outcomes
without replacing the first ball, the number of possible outcomes =3
According to fundamental principle ,( a x b )
= 4 x 3 = 12
Therefore there are 12 number of outcomes
Tree diagram:

Question 20.
FINDING A PROBABILITY
You roll two number cubes. What is the probability of rolling double threes?
Answer:
Number of cubes =2
{1,2,3,4,5,6} , {1,2,3,4,5,6}
Total number of  possible outcomes = 6 x 6 = 36
The probability of rolling double threes =
P(T) = 2/36 = 0.05
Therefore, the probability of rolling double threes = 0.05

FINDING THE PROBABILITY OF A COMPOUND EVENT You spin the spinner and flip a coin. Find the probability of the compound event.
Big Ideas Math Answers Grade 7 Chapter 7 Probability 7.3 21
Question 21.
spinning a 1 and flipping heads
Answer: 0.1

Explanation:
The spinner is divided into 5 equal parts
They are {12,3,4,5}
Total number of outcomes =5
Consider O be the event of spinning a 1
P(O)= (Number of favorable outcomes )/ (Total number of outcomes)
P(O)= 1/ 5
Total number of outcomes for a coin =2
They are {Heads, Tails}
The probability of flipping a heads
P(H)= 1/2
The probability of compound event =
1/5 x 1/2 = 1/10 =0.1
Therefore ,the probability of compound event = 1/10

Question 22.
spinning an even number and flipping heads
Answer: 0.2

Explanation:
Number of even numbers = 2
They are {2,4}
Total number of outcomes = 5
Consider E be the event of spinning an even number
P(E) = 2/5
Total number of outcomes for a coin =2
They are {Heads, Tails}
The probability of flipping a heads
P(H)= 1/2
The probability of compound event (a x b)=
2/5 x 1/2 = 2/10 =0.2
Therefore, the probability of compound event = 0.2

Question 23.
spinning a number less than 3 and flipping tails
Answer: 0.2

Explanation:
Total number of outcomes = 5
Number of numbers less than 3 =2
They are {1,2}
Consider, S be the event of spinning a number less than 3
P(S) = 2/5
Total number of outcomes for a coin =2
They are {Heads, Tails}
The probability of flipping a heads
P(H)= 1/2
The probability of compound event (a x b)=
2/5 x 1/2 =2/10 =0.2
Therefore the probability of compound event = 0.2

Question 24.
spinning a 6 and flipping tails
Answer: 0

Explanation:
The spinner is divided into 5 equal parts
They are {12,3,4,5}
Total number of outcomes =5
Consider O be the event of spinning a 6
P(O)= (Number of favorable outcomes )/ (Total number of outcomes)
P(O)= 0
Therefore it is impossible to spin 6
Total number of outcomes for a coin =2
They are {Heads, Tails}
The probability of flipping a tails
P(H)= 1/2
The probability of compound event =
0 x 1/2 =0
Therefore ,the probability of compound event =0

Question 25.
not spinning a 5 and flipping heads
Answer:0.4

Explanation:
The spinner is divided into 5 equal parts
They are {12,3,4,5}
Total number of outcomes =5
Consider N be the event of  not spinning a 5
Favorable outcomes = 4 .They are {1,2,3,4}
P(N)= (Number of favorable outcomes )/ (Total number of outcomes)
P(N)= 4/ 5
Total number of outcomes for a coin =2
They are {Heads, Tails}
The probability of flipping a heads
P(H)= 1/2
The probability of compound event =
4/5 x 1/2 = 4/10 =0.4
Therefore ,the probability of compound event = 0.4

Question 26.
spinning a prime number and not flipping heads
Answer:0.3

Explanation:
Total number of outcomes = 5
Number of prime numbers = 3
They are {2,3,5}
Consider P be the event of spinning a prime number
P(P) = (Number of favorable outcomes )/ (Total number of outcomes)
P(P) = 3/5
Total number of outcomes for a coin =2
They are {Heads, Tails}
The probability of not flipping a heads
P(H)= 1/2
The probability of compound event =
3/5 x 1/2 = 3/10 =0.3
Therefore ,the probability of compound event = 0.3

FINDING THE PROBABILITY OF A COMPOUND EVENT You spin the spinner, flip a coin, and then spin the spinner again. Find the probability of the compound event.
Big Ideas Math Answers Grade 7 Chapter 7 Probability 7.3 22
Question 27.
spinning blue, flipping heads, then spinning a 1
Answer:
The spinner is divided into 3 equal parts
Indicated numerally as {1,2,3} also colored in {red, blue , yellow}
Consider B be the event of spinning blue
P(B) = 1/3
Number of outcomes for a coin =2
They are {Heads , Tails }
Consider H be the event of flipping heads
P(H) = 1/2
Consider O be the event of spinning 1
P(O)= 1/3
The compound probability of compound event =
= 1/3 x 1/2 x1/3
= 1/18
Therefore , the compound probability of the event is 1/18.

Question 28.
spinning an odd number, flipping heads, then spinning yellow
Answer:
The spinner is divided into 3 equal parts
Indicated numerally as {1,2,3} also colored in {red, blue , yellow}
Consider Y be the event of spinning yellow
P(Y) = 1/3
Number of outcomes for a coin =2
They are {Heads , Tails }
Consider H be the event of flipping heads
P(H) = 1/2
Consider O be the event of spinning Odd  number
Number of odd numbers = 2
They are {1,3}
P(O)= 2/3
The compound probability of compound event =
= 1/3 x 1/2 x2/3
= 2/18 =0.11
Therefore , the compound probability of the event is 2/18.

Question 29.
spinning an even number, flipping tails, then spinning an odd number
Answer:
The spinner is divided into 3 equal parts
Indicated numerally as {1,2,3} also colored in {red, blue , yellow}
Consider E be the event of spinning an even number
Number of even numbers =1
They are {2}
P(B) = 1/3
Number of outcomes for a coin =2
They are {Heads , Tails }
Consider T be the event of flipping Tails
P(T) = 1/2
Consider O be the event of spinning an odd number
Number of odd numbers =2
They are {1,3}
P(O)= 2/3
The compound probability of compound event =
= 1/3 x 1/2 x2/3
= 2/18 =0.11
Therefore , the compound probability of the event is 2/18.

Question 30.
not spinning red, flipping tails, then not spinning an even number
Answer:
The spinner is divided into 3 equal parts
Indicated numerally as {1,2,3} also colored in {red, blue , yellow}
Consider R be the event of  not spinning red
The colors other than red are {blue , yellow}
P(R) = 2/3
Number of outcomes for a coin =2
They are {Heads , Tails }
Consider T be the event of flipping tails
P(T) = 1/2
Consider E be the event of not spinning an even number
number of not even numbers =2
They are {1,3}
P(E)= 2/3
The compound probability of compound event =
= 2/3 x 1/2 x2/3
= 4/18 =0.22
Therefore , the compound probability of the event is 0.22

Question 31.
REASONING
You randomly guess the answers to two questions on a multiple-choice test. Each question has three choices: A,B, and C.
Big Ideas Math Answers Grade 7 Chapter 7 Probability 7.3 23
a. What is the probability that you guess the correct answers to both questions?
b. Suppose you can eliminate one of the choices for each question. How does this change the probability that both of your guesses are correct?
Answer:
Probability of an event = Number of favorable outcomes / Total number of outcomes
There is one correct answer for each question.
So, the number of favorable outcomes = 2
Given that, there are 3 choices {a, b, c} for each question
So, total number of outcomes = 6
Consider, C be event that you guess the correct answers to both questions=
P(C)= 2/6 = 1/3
Therefore, the probability that you guess the correct answers to both questions =1/3

b.
Answer:
There is a correct answer for each question
So, the favorable outcomes =2
If you eliminate one of the choices for each question ,
Now, the choices for the questions are {a, b}
So, the total number of outcomes =4
Consider, S be the event of choosing correct answer
P(S) = 2/4
= 0.5
Therefore, the probability of choosing a correct answer is 0.5

Question 32.
REASONING
You forget the last two digits of your cell phone password.
Big Ideas Math Answers Grade 7 Chapter 7 Probability 7.3 24
a. What is the probability that you randomly choose the correct digits?
b. Suppose you remember that both digits are even. How does this change the probability that you choose the correct digits?
Answer:

a.Number of digits to be guessed = 2
There are 10 possible numbers from 0 to 9
By using fundamental counting principle,
Number of possible outcomes = 10 x 10 =100
Each digit has one correct answer
So, the number of favorable outcomes = 2
Probability = Number of favorable outcomes/ Total number of outcomes
Consider C be the event of choosing correct digits
P(C) = 2/100 =1/50
Therefore the probability of choosing correct digits is 1/50

b.Given that, the two digits to be guessed are even numbers
The number of even numbers from 0 to 9 = 5
They are {0,2,4,6,8}
According to fundamental counting principle ,
So, total number of possible outcomes = 5 x 5= 25
Each digit has only one correct answer,
So number of favorable outcomes = 2
Consider S be the event of choosing correct digits
P(S) = 2/25 =0.08
Therefore, the probability of choosing correct answer = 0.08

Question 33.
MODELING REAL LIFE
A combination lock has 3 wheels, each numbered from 0 to 9. You try to guess the combination by writing five different numbers from 0 to 999 on a piece of paper. Find the probability that the correct combination is written on the paper.
Answer:
The number of digits numbered from 0 to 9 = 10
You get to try 5 of the 1000 possible combinations
So, the probability of getting them all right is 5/1000 or 1/200
Therefore ,5 tries, with a 1/1000 chance each time = 5/1000 = 1/200

Question 34.

MODELING REAL LIFE
A train has one engine and six train cars. Find the total number of ways an engineer can arrange the train. (The engine must be first.)
Big Ideas Math Answers Grade 7 Chapter 7 Probability 7.3 25
Answer:
Total number of train cars = 6
The engine must be first So, the total outcomes for train cars
So , there are 6 possible places for the train cars
Number of possible outcomes for the first train car = 6
Number of possible outcomes for the second train car =5
Number of possible outcomes for the third train car = 4
Number of possible outcomes for the fourth train car = 3
Number of possible outcomes for the fifth train car = 2
Number of possible outcomes for the sixth train car = 1
According to fundamental principle ,
6 x 5 x 4 x 3 x 2 x 1 = 720
Therefore, the number of ways you can arrange the train = 720

Question 35.
REPEATED REASONING
You have been assigned a nine-digit identification number.
a. Should you use the Fundamental Counting Principle or a tree diagram to find the total number of possible identification numbers? Explain.
b. How many identification numbers are possible?
c. RESEARCH Use the Internet to find out why the possible number of Social Security numbers is not the same as your answer to part(b).
Answer:

a.It is difficult to use tree diagram to find the total number of possible identification numbers because, the total number of identification is very large . So, it is impossible to use tree diagram
Therefore , it is easy to easy to use the fundamental counting principle .

b.Total number of identification digits are from 0 to 9
They are {0,1,2,3,4,5,6,7,8,9}
The number of possible numbers from 0 to 9 for first digit = 10
The number of possible numbers from 0 to 9 for second digit = 10
The number of possible numbers from 0 to 9 for third  digit = 10
The number of possible numbers from 0 to 9 for fourth digit = 10
The number of possible numbers from 0 to 9 for fifth digit = 10
The number of possible numbers from 0 to 9 for sixth digit = 10
The number of possible numbers from 0 to 9 for seventh digit = 10
The number of possible numbers from 0 to 9 for eighth digit = 10
The number of possible numbers from 0 to 9 for ninth digit = 10
According to fundamental principle
10x10x10x10x10x10x10x10x10 = 1,000,000,000
Therefore there are 1,000,000,000 possible identification numbers

c.The possible number of Social Security numbers is not the same as your answer to part(b)
Because , some special numbers are never allocated.

Question 36
DIG DEEPER!
A social media account password includes a number from 0 to 9, an uppercase letter, a lowercase letter, and a special character, in that order.
a. There are 223,080 password combinations. How many special characters are there?
b. What is the probability of guessing the account password if you know the number and uppercase letter, but forget the rest?
Answer:

a.From 0 to 9, there are a total of 10 values.
For uppercase letters, there are a total of 26 values from A, B, C, D …Z
For lower case letters, there are also a total of 26 values from a, b, c, d …z.
So out of these three characters, we have a total of 10 × 26 × 26 = 6,760 different combinations.
If there are 223,080 password combinations
We need to divide this by 6,760 to calculate the possible values of the special characters.
6,760 × Number of possible special characters = 223,080,
Number of special characters = 223,080 / 6760= 33.
So there are 33 special characters.
If the number and uppercase values are known then the various lowercase letters and special characters are the unknown values.
The number of possible combinations = number of lowercase letters × number of special characters = 26 × 33 = 858.
So the probability of guessing the password is 1 out of 858 combinations.

Question 37.
PROBLEM SOLVING
From a group of 5 scientists, an environmental committee of 3 people is selected. How many different committees are possible?
Answer:
Let,
The 5 scientists be indicated as A,B,C,D,E
The environmental committee of three people =
ABC, ABD, ABE ,ACD,ACE , ADE
BCD,BCE, BDE,CDE .
Therefore, there are 10 different ways to form a committee.

Lesson 7.4 Simulations

EXPLORATION 1

Using a Simulation
Work with a partner. A basketball player makes 80% of her free throw attempts.
a. Is she likely to make at least two of her next three free throws? Explain your reasoning.
b. The table shows 30 randomly generated numbers from 0 to 999. Let each number represent three shots. How can you use the digits of these numbers to represent made shots and missed shots?
Big Ideas Math Solutions Grade 7 Chapter 7 Probability 7.4 1
c. Use the table to estimate the probability that of her next three free throws, she makes

  • exactly two free throws.
  • at most one free throw.
  • at least two free throws.
  • at least two free throws in a row.

d. The experiment used in parts (b) and (c) is called a simulation. Another player makes \(\frac{3}{5}\) of her free throws. Describe a simulation that can be 5used to estimate the probability that she makes three of her next four free throws.
Answer:
A simulation is an experiment that is designed to reproduce the conditions of a situation or process.Simulations allow you to study situations that are impractical to create in real life.

Big Ideas Math Solutions Grade 7 Chapter 7 Probability 7.4 2

Try It

Question 1.
You randomly guess the answers to four true-false questions.
a. Design a simulation that you can use to model the answers.
b. Use your simulation to find the experimental probability that you answer all four questions correctly.

Answer:a.  There are two answers in a true-false question.
There is an equal chance of being correct or incorrect.
Therefore, we can use a coin to simulate answers where heads is correct and tails is incorrect.
Flip 4 coins in each trial to simulate the four answers. Run the simulation 50 times.

b.The table below shows the  results of simulation 50 times
Out of 50 trails , there are only 4 times all the answers are correct
So, the probability that you answer all four questions correct =
Consider C be the event of answer all four questions correct
P(C) =4 /50 = 0.08

   

Question 2.
A baseball team wins 70% of the time. Design and use a simulation to estimate the probability that the team wins the next three games.
Answer:
Given that there is 70% chance of winning
let us use numbers to determine the win or lose of game
Assume 0-6 numbers to represent win and 7-9 numbers represent lose since, there is 70% chance of winning.
The table below shows the possible results after simulation 3 numbers 50 times on a random number on a graphing calculator.
Out of 50 trails , 19 of them had three wins
Therefore, the probability = 19/50 =0.38

Self-Assessment for Concepts & Skills
Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 3.
SIMULATING OUTCOMES
Four multiple-choice questions on a quiz each have five answer choices. You randomly guess the answer to each question. Design and use a simulation to find the experimental probability that you answer all of the questions correctly.
Answer:
Given , the number of questions = 4
The number of answer choices for each question = 5
The experimental probability that you answer all of the questions correctly = 0.5%

Question 4.
SIMULATING OUTCOMES
You select a marble from a bag and a chip from a box. You have a 20% chance of choosing a green marble and a 90% chance of choosing a red chip. Estimate the probability that you choose a green marble and a red chip.
Big Ideas Math Solutions Grade 7 Chapter 7 Probability 7.4 3
Answer:
Given ,
The chance of  choosing green marble = 20%
The chance of choosing a red chip =90%
The simulation about the probability that you choose green marble and red marble
88  5  86  13  31  49  33  21  99  97
30  62  18  4  63  3  32  94  8  77
24  87  74  56  19  42  61  75  81
45  84  51  17  15  46  66  69  34  28
36  9  64  53  59  10  58  57  39  43  93
The digits from 1 through 2 in tens place indicate green marble
The digits from 1 through 9 in tens place indicates red chip
The experimental probability of choosing green marble and red marble
P(B) = 8/50 = 0.16

Self-Assessment for Problem Solving
Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 5.
Each day there is a 50% chance that your tablet overheats. Estimate the probability that your tablet overheats on exactly 2 of the next 3 days.
Answer:
Given that ,
The percent of  tablet overheats each day = 50%
The digits from 1 through 5 in the numbers
460  677  330  467  654  942  875  217  734  692
646  421  222  804  399  380  103  977  991  996
367  337  760  495  280  359  726  349  605  649
573  836  413  196  475  470  315  911  122  970
369  666  908  431  396  230  885  863  333  762
The digits 1 through 5 at least two in a three digit number indicate that your tablet overheats on exactly 2 of the next 3 days.
The experimental probability that your tablet overheats on exactly 2 of the next 3 days
P(S) = 15/ 50 = 0.3

Question 6.
DIG DEEPER
The probability that a homeowner needs a plumber this year is 22%. The probability that the homeowner needs a septic tank specialist is 14%. Estimate the probability that the homeowner needs a plumber, but not a septic tank specialist.
Big Ideas Math Solutions Grade 7 Chapter 7 Probability 7.4 4
Answer:
Given that,
The probability home owner needs a plumber this year = 22%
The probability that home owner needs septic tank specialist = 14%
The probability that the home owner needs plumber but not septic tank specialist
80  52  96  94  77  49  74  16  79  68
45  98  34  47  48  57  12  62  21  71
60  82  78  11  64  28  43  22  75  19
76  65  17  81  87  14  38  27  6  42
37  40  51  15  2  13  92  90  73  7
The numbers 1 through 2 in tens place indicates owner needs plumber
The number 2 through 9 in once place  indicate owner does not needs tank specialist
P(S) = 11 / 50 = 0.22

Simulations Homework & Practice 7.4

Review & Refresh
You flip a coin and roll the 20-sided figure. Find the probability of the compound event.
Big Ideas Math Solutions Grade 7 Chapter 7 Probability 7.4 5
Question 1.
Flipping tails and rolling at least a 14
Answer:
The total numbers of outcomes for a coin = 2
They are {Tails, heads }
Consider T be the event of flipping tails
P(T) = 1/2
The number of outcomes for a 20 sided figure  = 20
The probability of rolling at least a 14
Consider F be the event of rolling at least a 14
P(F) = 1/ 20
Compound event ( a x b )
= 1/2 x 1/20 = 1/40

Question 2.
Flipping heads and rolling less than 3
Answer:
The total numbers of outcomes for a coin = 2
They are {Tails, heads }
Consider H be the event of flipping heads
P(H ) = 1/2
The number of outcomes for a 20 sided figure  = 20
Consider L be the event of rolling less than 3
Number of numbers less than 3 = 2
They are {1,2}
P(L) = 2/ 20 = 1/10
Compound event ( a x b )
= 1/2 x 1/10 = 1/20

Simplify the expression.
Question 3.
5(a – 2)
Answer:
5(a – 2) =0
5a -10 =0
5a =10
a=10/5
a =2

Question 4.
– 7(1 + 3x)
Answer:
– 7(1 + 3x) =0

-7 – 21x=0
-21x= 7
x= 7/-21
x = -0.33

Question 5.
– 1(3p – 8)
Answer:
– 1(3p – 8) =0
-3p +8 =0
p =-8/-3  =-2.6

Concepts, Skills, & Problem Solving

USING A SIMULATION A medicine is effective for 80% of patients. The table shows 30 randomly generated numbers from 0 to 999. Use the table to estimate the probability of the event. (See Exploration 1, p. 307.)
Big Ideas Math Solutions Grade 7 Chapter 7 Probability 7.4 6
Question 6.
The medicine is effective on each of three patients.
Answer:
Let the, digits 1 through 8 in the tens place represent the medicine is effective
P(E) = 24/30

Question 7.
The medicine is effective on fewer than two of the next three patients.
Answer:
P(N) = 6/30 = 0.2

SIMULATING OUTCOMES Design and use a simulation to find the experimental probability.
Question 8.
In your indoor garden, 50% of seeds sprout. What is the experimental probability that at least one of your next three seeds sprouts?
Big Ideas Math Solutions Grade 7 Chapter 7 Probability 7.4 7
Answer:
Use the random number generator on a graphing calculator.
Randomly generate 50 numbers from 0 to 99. The table below shows the results.
Let the digits 1 through 5 in the tens place represent the seeds are sprout

P(rain both days) =6/50
The experimental probability is 6/50 = 0.12, or 12%.

Question 9.
An archer hits a target 50% of the time. What is the experimental probability that the archer hits the target exactly four of the next five times?
Answer:
Given , the archer hits the target = 50%
25  82  53  49  24  95  31  66  40  90
42  30  9  78  4  80  16  99  23  85  39
44  6  61  46  5  87  64  36  21  57  58
98  81  13  97  29  18  92  22  77  35
55  14  60  28  38  89  56  69
Because, the makes 50 % of shots 1 to 50 represents successful shots .
the experimental probability that the archer hits the target exactly four of the next five times
P(X) = 25/50 =0.5 = 50%.

Question 10.
A bank randomly selects one of four free gifts to send to each new customer. Gifts include a calculator, a key chain, a notepad, and a pen. What is the experimental probability that the next two new customers both receive calculators? that neither receives a calculator?
Big Ideas Math Solutions Grade 7 Chapter 7 Probability 7.4 8
Answer:
Given that ,
Gifts include a calculator, a key chain, a notepad, and a pen.
The experimental probability that the next two new customers both receive calculators=
P(C) =1/4 x1/4 = 1/16
The experimental probability that the next two new customers both neither receive calculators=
P(N) = 3/4 x 3/4 = 9/16

Question 11.
Employees spin a reward wheel. The wheel is equally likely to stop on each of six rewards labeled A–F. What is the experimental probability that fewer than two of the next three spins land on reward A?
Answer:
Given , The six rewards are labelled as {A,B,C,D,E,F}
What is the experimental probability that fewer than two of the next three spins land on reward A
92  72  41  33  83  4  60  32  6  81
1  12  61  57  93  27  46  29  42  47
21  79  23  45  16  63  26  87  14  68
75  38  94  24  20  86  82  70  8  97
39  59  19  64  55  25  77  9  88  37
the experimental probability that fewer than two of the next three spins land on reward A = 4/50

USING NUMBER CUBES Design and use a simulation with number cubes to estimate the probability.
Question 12.
Your lawnmower does not start on the first try \(\frac{1}{6}\) of the time. Estimate the probability that your lawnmower will not start on the first try exactly one of the next two times you move the lawn.
Big Ideas Math Solutions Grade 7 Chapter 7 Probability 7.4 9
Answer:
Given the probability of the event is 1/6 = 16
The simulation of random generate calculator about 50 numbers
The digits from 1 through 16 indicates the event
50  99  88  6  98  76  63  21  65  42
96  84  92  37  27  33  11  57  69  68
48  40  10  30  24  26  67  47  60  34
59  87  18  78  7  79  12  95  9  8  1
86  38  45  93  44  4  15  73  39
The experimental probability of the event is 10/50
P(S) = 0.2

Question 13.
An application on your phone correctly identifies four out of every six songs. Estimate the probability that at least three of the next four songs are correctly identified.
Answer:

SIMULATING OUTCOMES Design and use a simulation to find the experimental probability.
Question 14.
Two beakers are used in a lab test. What is the experimental probability that there are reactions in both beakers during the lab test?
Big Ideas Math Solutions Grade 7 Chapter 7 Probability 7.4 10
Answer:
The simulation of beaker 1 and beaker 2 have reactions
The digits 1 through 8 in tens place indicate probability of beaker 1
The digits 1 through 5 in once place indicate probability of beaker 2
52 66 73 68 75 28 35 47 48 2
16 68 49 3 77 35 92 78 6 6
58 18 89 39 24 80 32 41 77 21
32 40 96 59 86 1 12 0 94 73
40 71 28 61 1 24 37 25 3 25
The experimental probability that there are reactions in both beakers during the lab test
P(R) =32/50 = 0.64

Question 15.
You use a stain remover on two separate stains on a shirt. What is the experimental probability that the stain remover removes both the mud stain and the food stain?
Big Ideas Math Solutions Grade 7 Chapter 7 Probability 7.4 11
Answer:
The simulation of random generator calculator is as follows:
The digits 1 through 9 in tens place indicate the event in mud
The digits 1 through 8 in once place indicate the event of stain removal in food
62  35  50  43  96  37  70  31  59  40
55  92  89  14  10  41  87  95  11  99
12  53  71  26  38  24  80  36  16  42
85  32  3  33  47  1  49  51  8  23  88
69  15  64  84  81  60  66  72  74
The experimental probability that the stain remover removes both the mud stain and the food stain
P(S) = 41/50 = 0.82

Question 16.
DIG DEEPER!
The probability that a computer crashes one or more times in a month is 10%. Estimate the probability that the computer crashes at least one or more times per month for two months in a row during the first half of the year.
Answer:

Question 17.
MODELING REAL LIFE
You visit an orchard. The probability that you randomly select a ripe apple is 92%. The probability that you randomly select a ripe cherry is 86%. Estimate the probability that you pick an apple that is ripe and a cherry that is not ripe.
Big Ideas Math Solutions Grade 7 Chapter 7 Probability 7.4 12
Answer:
Given , the probability that you randomly select a ripe apple = 92%
The probability  that you pick a ripe cherry = 86%
The digits from 1 through 8 in tens place indicates event of choosing a ripe apple
The digits 9  in once place indicates event of not choosing a ripe cherry
75  99  69  33  67  1  22  17  18  37
29  27  9  12  54  7  31  39  26  87
10  72  82  42  36  85  74  3  5  92
29  11  21  81  76  77  52  13  90
15  73  69  70  8  23  53  59  51  44  48
The probability that you pick an apple that is ripe and a cherry that is not ripe = 9/50

Question 18
CRITICAL THINKING
You use a simulation to find an experimental probability. How does the experimental probability compare to the theoretical probability as the number of trials increases?
Answer:
Experimental probability is the result of an experiment. Theoretical probability is what is expected to happen.
In experimental probability, as the number of trials increases, the experimental probability gets closer to the theoretical probability.

Question 19.
LOGIC
At a restaurant,30% of customers donate to charity in exchange for a coupon. Estimate the probability that it will take at least four customers to find one who donates.
Answer:

Probability Connecting Concepts

Using the Problem-Solving Plan
Question 1.
In an Internet contest, gift cards and bicycles are given as prizes in the ratio 9 : 1. Estimate the probability that at least two of three randomly selected winners receive bicycles.
Big Ideas Math Answer Key Grade 7 Chapter 7 Probability cc 1
Understand the problem.
You know the ratio of gift cards to bicycles awarded in the contest. You want to find the probability that atleast two of three randomly selected winners receive bicycles.
Make a plan.
Use the ratio to find the theoretical probability that a randomly selected winner receives a bicycle. Then use a simulation involving 50 randomly generated three-digit numbers to estimate the probability that at least two of three randomly selected winners receive bicycles.
Solve and check.
Use the plan to solve the problem. Then check your solution.

Answer:

Question 2.
A board game uses the spinner shown.
Big Ideas Math Answer Key Grade 7 Chapter 7 Probability cc 2
a. Use theoretical probability to predict the number of times you will spin a number greater than or equal to 8 in 30 spins.
b. You play the game and record the results of 30 spins. Find the percent error of your prediction in part(a).
Answer:
a. Theoretical probability = number of favorable/ outcomes by total number of outcomes
Total number of outcomes = 10
The probability of spinning a number greater than or equal to 8 P(S) = 3/8 = 0.375
b. The experimental probability =
P(S) = 12/30= 0.4

Question 3.
The tiles shown are placed in a bag. You randomly select one of the tiles, return it to the bag, and then randomly select another tile. What is the probability that the product of the numbers on the tiles selected is greater than zero? Justify your answer.
Big Ideas Math Answer Key Grade 7 Chapter 7 Probability cc 3
Answer:
Total number of tiles = 5
= 5×5 = 25 total number of outcomes
The probability that the product of the numbers on the tiles selected is greater than zero
P(Z) = 2/ 25
They are {1 x 2= 2 } and {2x 1 = 2 }
Therefore the probability of the product of the numbers on the tiles selected is greater than zero = 2/25 = 0.08

Performance Task

Fair and Unfair Carnival Games
At the beginning of this chapter, you watched a STEAM Video called “Massively Multi player Rock Paper Scissors.” You are now ready to complete the performance task related to this video, available at BigIdeasMath.com. Be sure to use the problem-solving plan as you work through the performance task.
Big Ideas Math Answer Key Grade 7 Chapter 7 Probability cc 4

Probability Chapter Review

Review Vocabulary

Write the definition and give an example of each vocabulary term.
Big Ideas Math Answer Key Grade 7 Chapter 7 Probability cr 1

Graphic Organizers

You can use a Four Square to organize information about a concept. Each of the four squares can be a category, such as definition, vocabulary, example, non-example, words, algebra, table, numbers, visual, graph,or equation. Here is an example of a Four Square for probability.
Big Ideas Math Answer Key Grade 7 Chapter 7 Probability cr 2
Choose and complete a graphic organizer to help you study the concept.
Big Ideas Math Answer Key Grade 7 Chapter 7 Probability cr 3
1. favorable outcomes
2. relative frequency
3. experimental probability
4. theoretical probability
5. Fundamental Counting Principle
6. compound event
7. simulation

Chapter Self-Assessment

As you complete the exercises, use the scale below to rate your understanding of the success criteria in your journal.
Big Ideas Math Answer Key Grade 7 Chapter 7 Probability cr 4

7.1 Probability (pp. 283–290)
Learning Target: Understand how the probability of an event indicates its likelihood.

You randomly choose one toy race car.
Big Ideas Math Answer Key Grade 7 Chapter 7 Probability cr 5
Question 1.
How many possible outcomes are there?
Answer:
There are 5 green cars , one blue car and one red car
So the possible outcomes are 7
They are {green , red, blue}

Question 2.
What are the favorable outcomes of choosing a car that is not green?
Answer: 2
Number of green cars = 5
Number of cars that is not green = 2
They are {blue , red}
Therefore the possible outcomes of choosing a car that is not green = 2

Question 3.
In how many ways can choosing a green car occur?
Answer: 5
Number of green cars = 5
So, choosing a green car can occur in 5 ways
You spin the spinner. (a) Find the number of ways the event can occur. (b) Find the favorable outcomes of the event.
Big Ideas Math Answer Key Grade 7 Chapter 7 Probability cr 6
Question 4.
spinning a 1
Answer:
The spinner is divided into 8 equal parts
numbered with two 1s , three 2s and three 3s
{1,1,2,2,2,3,3,3}
The number of favorable outcomes of spinning a 1 = 2
Total number of outcomes = 8
Consider O be the event of spinning a 1
Probability P(O) = Number of favorable outcomes/ Total number of outcomes
P(O) = 2/8 =1/4
So, the probability of spinning a 1 is 0.25

Question 5.
spinning a 3
Answer:
The spinner is divided into 8 equal parts
numbered with two 1s , three 2s and three 3s
{1,1,2,2,2,3,3,3}
The number of favorable outcomes of spinning a 3 = 3
Total number of outcomes = 8
Consider T be the event of spinning a 3
Probability P(T) = Number of favorable outcomes/ Total number of outcomes
P(T) = 3/8 = 0.37
Therefore, the probability of spinning a 3 = 0.37

Question 6.
spinning an odd number
Answer:
The spinner is divided into 8 equal parts
numbered with two 1s , three 2s and three 3s
{1,1,2,2,2,3,3,3}
The number of favorable outcomes of spinning an odd number  = 5
They are {1,1,3,3,3}
Total number of outcomes = 8
Consider O be the event of spinning an odd number
Probability P(O) = Number of favorable outcomes/ Total number of outcomes
P(O) = 5/8 =0.62
Therefore , the probability of spinning an odd number = 0.62

Question 7.
spinning an even number
Answer:
The spinner is divided into 8 equal parts
numbered with two 1s , three 2s and three 3s
{1,1,2,2,2,3,3,3}
The number of favorable outcomes of spinning an even number = 3
They are {2,2,2}
Total number of outcomes = 8
Consider E be the event of spinning an even number
Probability P(E) = Number of favorable outcomes/ Total number of outcomes
P(E) = 3/8 = 0.37

Question 8.
spinning a number greater than 0
Answer:
The spinner is divided into 8 equal parts
numbered with two 1s , three 2s and three 3s
{1,1,2,2,2,3,3,3}
The number of favorable outcomes of spinning a number greater than 0 =8
They are {1,1,2,2,2,3,3,3}
Total number of outcomes = 8
Consider S be the event of spinning a 1
Probability P(S) = Number of favorable outcomes/ Total number of outcomes
P(S) = 8/8 =1

Question 9.
spinning a number less than 3
Answer:
The spinner is divided into 8 equal parts
numbered with two 1s , three 2s and three 3s
{1,1,2,2,2,3,3,3}
The number of favorable outcomes of spinning a number less than a 3 = 5
They are {1,1,2,2,2}
Total number of outcomes = 8
Consider T be the event of spinning a number less than a 3
Probability P(T) = Number of favorable outcomes/ Total number of outcomes
P(T) = 5/8 = 0.62

Describe the likelihood of the event given its probability.
Question 10.
There is a 0% chance of snow in July for Florida.
Answer:
Given , the probability = 0
So the likelihood of the event is impossible

Question 11.
The probability that you are called on to answer a question in class is \(\frac{1}{25}\).
Answer:

Question 12.
There is an 85% chance the bus is on time.
Answer:
Given, 85% chance = 0.85
So, the probability is more likely to happen

Question 13.
The probability of flipping heads on a coin is 0.5.
Answer:
Given 0.5 = 50%
So, the probability of flipping heads is may or may not occur

Question 14.
During a basketball game, you record the number of rebounds from missed shots for each team. (a)Describe the likelihood that your team rebounds the next missed shot. (b) How many rebounds should your team expect to have in 15 missed shots?
Big Ideas Math Answer Key Grade 7 Chapter 7 Probability cr 7
Answer:
a.Total number of shots = 10
Consider M be the event of rebounds the next missed shot
The probability  = Number of missed shots / total number of trails
P(M) = 7/10 = 0.7
Therefore , the likelihood that your team rebounds the next missed shot is likely to happen

b.Total number of shots = 15
Consider S be the event of rebounds the your team misses the shot
probability P(S) = Number of rebounds / total number of trails
P(S) = 7/15 = 0.46
Therefore, the likelihood that the rebounds are expected to happen in 15 shots is not likely to occur

7.2 Experimental and Theoretical Probability (pp. 291–298)
Learning Target :Develop probability models using experimental and theoretical probability.

The bar graph shows the results of spinning a spinner 100 times. Use the bar graph to find the experimental probability of the event.
Big Ideas Math Answer Key Grade 7 Chapter 7 Probability cr 15
Question 15.
spinning a 2
Answer: 0.18

Explanation :
Total number of spins = 100
The number of times 2 spun = 18
Experimental probability = Number of times an event occur / Total number of spins
Consider T be the event of spinning a 2
P(T) = 18/100 =0.18
Therefore , experimental probability of spinning a 2 = 0.18

Question 16.
spinning an even number
Answer: 0.39

Explanation:
Total number of spins = 100
Number of even numbers = 2
They are {2,4}
The number of times 2 spun = 18
The number of times 4 spun = 21
Total : 18+21 = 39
Experimental probability = Number of times an event occur / Total number of spins
Consider T be the event of spinning an even number
P(T) = 38/100 =0.39
Therefore , experimental probability of spinning an even number = 0.39

Question 17.
not spinning a 5
Answer: 0.81
Total number of spins = 100
Number of numbers other than 5 = 4
They are {1,2,3,4}
The number of times 1 spun = 20
The number of times 2 spun = 18
The number of times 3 spun = 22
The number of times 4 spun = 21
Total : 81
Experimental probability = Number of times an event occur / Total number of spins
Consider F be the event of not spinning a 5
P(F) = 81/100 =0.81
Therefore , experimental probability of not spinning a 5 = 0.81

Question 18.
spinning a number less than 3
Answer:
Total number of spins = 100
The numbers less than 3 = 2
They are {1,2}
The number of times 1 spun = 20
The number of times 2 spun = 18
Total : 18 + 20 = 38
Experimental probability = Number of times an event occur / Total number of spins
Consider T be the event of spinning a number less than 3
P(T) = 38/100 =0.38
Therefore , experimental probability of spinning a number less than 3  = 0.38

Question 19.
In Exercise 16, how does the experimental probability of spinning an even number compare with the theoretical probability?
Answer:

Use the spinner to find the theoretical probability of the event.
Big Ideas Math Answer Key Grade 7 Chapter 7 Probability cr 20
Question 20.
spinning blue
Answer:
The spinner is divided into 8 equal parts
They are numbered with three 1 s ,one 2 ,one 4 ,one 6 , one 8 , and one 10
And colored in 2 red , 2 blue , 2 purple and 2 orange in color
Consider , B be the event of spinning a blue
P(B) = 2/8
P(B) = 0.25

Question 21.
spinning a 1
Answer:
The spinner is divided into 8 equal parts
They are numbered with three 1 s ,one 2 ,one 4 ,one 6 , one 8 , and one 10
Consider , O be the event of spinning a 1
P(O) = 3/8
P(O) = 0.37

Question 22.
spinning an even number
Answer:
The spinner is divided into 8 equal parts
They are numbered with three 1 s ,one 2 ,one 4 ,one 6 , one 8 , and one 10
Total number of even numbers = 5
They are {2,4,6,8,10}
Consider , E be the event of spinning an even number
P(E) = 5/8
P(E) = 0.62

Question 23.
spinning a 4
Answer:
The spinner is divided into 8 equal parts
They are numbered with three 1 s ,one 2 ,one 4 ,one 6 , one 8 , and one 10
Consider , F be the event of spinning a blue
P(F) = 1/8
P(F) = 0.125

Question 24.
The theoretical probability of choosing a red grape from a bag of grapes is \(\frac{2}{9}\). There are 8 red grapes in the bag. How many grapes are in the bag?
Answer:
Given theoretical probability = 2/9
Number of red grapes = 8
Total number of grapes = x
8/x = 2/9
72 = 2x
x = 72/2
x = 35

Question 25.
The theoretical probability of choosing Event A is \(\frac{2}{7}\). What is the theoretical probability of not choosing Event A? Explain your reasoning.
Answer:

7.3 Compound Events (pp. 299–306)
Learning Target: Find sample spaces and probabilities of compound events.

Question 26.
You have 6 bracelets and 15 necklaces. Find the number of ways you can wear one bracelet and one necklace.
Answer:
Number of bracelets = 6
Number of necklaces = 15
According to fundamental counting principle,
a x b = 6 x 15 = 90
Therefore, In 90 ways you can wear one bracelet and one necklace

Question 27.
Use a tree diagram to find how many different home theater systems you can make from 6 DVD players,8 TVs, and 3 brands of speakers.
Answer:
Number of DVDs = 6
Number of TVs = 8
Number of brands of speaker = 3
Total number of outcomes = 6 x 8 x 3 = 144

Question 28.
A red, green, and blue book are on a shelf. You randomly pick one of the books. Without replacing the first book, you choose another book. What is the probability that you picked the red and blue book?
Answer:
Number of books in shelf = 3

Question 29.
You flip two coins and roll a number cube. What is the probability of flipping two tails and rolling an even number?
Answer:
number of total outcomes for coin = 2
The probability of flipping tails P(T ) = 1/2
For two coins , probability of flipping tails = 1/2 x 2 = 1/4
The number of total outcomes for 6
even numbers = {2,4,6}
The probability for flipping an even number = 3/6 = 1/2
So, the probability of flipping two tails and rolling an even number
= 1/4 x 1/2 = 1/8

Question 30.
Describe a compound event that has a probability between 50% and 80%.
Answer:

Question 31.
Your science teacher sets up six flasks. Two of the flasks contain water and four of the flasks contain hydrogen peroxide. A reaction occurs when you add yeast to hydrogen peroxide. You add yeast to two of the flasks. What is the probability that at least one reaction will occur?
Big Ideas Math Answer Key Grade 7 Chapter 7 Probability cr 31
Answer:

7.4 Simulations (pp. 307–312)
Learning Target: Design and use simulations to find probabilities of compound events.

Question 32.
You select a marble from two different bags. You have a 30% chance of choosing a blue marble from the first bag and a 70% chance of choosing a blue marble from the second bag. Design and use a simulation to estimate the probability that you choose a blue marble from both bags.
Answer:
Given , the chance of choosing blue marble from first bag =30%
The chance of choosing blue marble from second bag = 70%
82  4  90  96  18  65  11  71  97  53
48  63 32  54  52  17  14  5  83  50
61  36  10  81 58  84  9  80  94  95
15  42  2  45  68  64  33 38  12  60
35  16  21  99  59  55  22  20  37  87
Let the digits 1 through 3 in the tens place represents choosing blue marble from first bag
the digits 1 and 7 in the ones place represent choosing blue marble from second bag
let P(S) be the event choosing both blue marbles from both bags = 18/50
Therefore, the experimental probability = 0.36 = 36%

Question 33.
A cereal company is including a prize in each box. There are 5 different prizes, all of which are equally likely.
a. Describe a simulation involving 50 trials that you can use to model the prizes in the next 3 boxes of cereal you buy.
b. Use your simulation to find the experimental probability that all three boxes contain a different prize.
Answer:
Given ,
There are 5 different prizes, all of which are equally likely.

Question 34.
In the past month, your cell phone has lost its entire charge on 40% of days. Design and use a simulation to estimate the experimental probability that your cell phone loses its entire charge on exactly 2 of the next 5 days.
Answer:
Given,
Cell phone has lost its entire charge on 40%
Let the numbers 1 through 40 represents possible
6  79  3  23  82
84  67  14  59 88
11  47  26 68  18
 69  91  51  56  42
 15  99  58  92  62
 13  41  55  17  35
 28  54  85  97  65
 98  57  71  73  38
 89  90  87  39  72
 19  8  21  45  31
There are 2 or more successes in trails {1,3,6,10 }
The experimental probability that your cell phone loses its entire charge on exactly 2 of the next 5 days = 4/50
P(S) = 0.08 = 8%

Question 35.
You and your friends form a team in gym class. You have an 80% chance of winning a game of basketball and a 10% chance of winning a game of soccer. Design and use a simulation involving 50 randomly generated numbers to estimate the probability of winning both games.
Big Ideas Math Answer Key Grade 7 Chapter 7 Probability cr 35
Answer:
Given , The chance of winning a game of basket ball = 80%
The probability of winning a soccer game = 10%
33  91  74  87  53  67  63  9  47  52  17  8
40  58  100  11  44  20  49  72  60  66  79  51
69  73  76  43  77  97  2  93  12  36  6  86  92 59
 84  29  15  85  88  96  32  38  64  71  98  34
Let the digits 1 through 8 in the tens place represents winning a basketball
Let digits 1 and 2 in the ones place represent winning a soccer game.
let P(S) be the event of winning both the games = 7/50
Therefore, the experimental probability = 0.14 = 14%

Probability Practice Test

You randomly choose one game piece. (a) Find the number of ways the event can occur. (b) Find the favorable outcomes of the event.
Big Ideas Math Answers 7th Grade Chapter 7 Probability pt 1
Question 1.
choosing green
Answer:
Number of game pieces = 10
In which , 3 are yellow , 2 blue , 2 red and 1 are green in color
Probability P(G)= Number of favorable outcomes / Total number of outcomes
Consider G be the event of choosing green
P(G) = 1 / 10= 0..1
Therefore, the probability of choosing green = 0.1

Question 2.
choosing not yellow
Answer:
Number of game pieces = 10
In which , 3 are yellow , 2 blue , 2 red and 1 are green in color
Number of not yellow colors = 5
They are { blue , green , red }
P(N) = Number of favorable outcomes / Total umber of outcomes
P(N) = 5/10 = 0.5
Therefore, the probability of not choosing yellow= 0.5

Find the sample space and the total number of possible outcomes.
Question 3.
Big Ideas Math Answers 7th Grade Chapter 7 Probability pt 3
Answer:
Number of SPF = 5
Number of types = 3
The total number of outcomes = 5 x 3 =15
Therefore number of possible outcomes = 15
{10,lotion} {15,lotion} {30, lotion} {45, lotion} {50, lotion}
{10, spray } {15, spray} {30, spray } {45, spray} {50, spray}
{10,gel} {15, gel} {30 , gel} {45 ,gel} {50 , gel}

Question 4.
Big Ideas Math Answers 7th Grade Chapter 7 Probability pt 4
Answer:
Number of types = 4
Number of colors = 3
Total number of possible outcomes = 4 x3 = 12
{Basic display, Black  } {Basic display, white } {Basic display , silver}
{Scientific, Black} {Scientific, white} {Scientific , silver}
{Graphing ,black } {Graphing , white} {Graphing , silver}

Use the bar graph to find the experimental probability of the event.
Big Ideas Math Answers 7th Grade Chapter 7 Probability pt 5
Question 5.
rolling a 1 or a 2
Answer:
Total Number of rolls = 90
The number of times 1 rolled = 12
The number of times 2 rolled = 18
Total = 30
The probability of rolling a 1 or a 2 =
P(R) = 30/90
P(R) = 0.33

Question 6.
rolling an odd number
Answer:
The number of odd numbers = 3
They are {1,3,5}
Number of times 1 rolled = 12
Number of times 3 rolled = 14
Number of times 5 rolled =  16
Total : 12 + 14 + 16 =42
Consider O be the event of spinning an odd number
P(O) = 42/90 = 0.46

Question 7.
not rolling a 5
Answer:
The total number of rolls = 90
The number of  numbers other than 5 are =5
They are 1,2,3,4,6
The number of times 1 rolled = 12
The number of times 2 rolled = 18
The number of times 3 rolled = 14
The number of times 4 rolled = 17
The number of times 6 rolled = 13
Total = 74
The probability of rolling not 5 = 74/ 90
P(G) = 0.82

Question 8.
rolling a number less than 7
Answer:
The total number of rolls = 90
The number of  numbers less than 7 are = 6
They are 1,2,3,4,5,6
The number of times 1 rolled = 12
The number of times 2 rolled = 18
The number of times 3 rolled = 14
The number of times 4 rolled = 17
The number of times 5 rolled = 16
The number of times 6 rolled = 13
Total = 90
The probability of rolling a number less than 7 = 90/ 90
P(L) =1

Use the spinner to find the theoretical probability 93of the event(s).
Big Ideas Math Answers 7th Grade Chapter 7 Probability pt 9
Question 9.
spinning an even number
Answer:
The spinner is divided into 9 equal parts
Numbered as {1,2,3,4,5,6,7,8,9}
So, total number of outcomes = 9
Number of even numbers = 4
They are {2,4,6,8}
Theoretical probability = Number of favorable outcomes / Total number of outcomes
Consider , E be the event of choosing an even number
P(E) = 4/ 9 = 0.44
Therefore , the probability for spinning an even number =0.44

Question 10.
spinning a 1 and then a 2
Answer:
The spinner is divided into 9 equal parts
Numbered as {1,2,3,4,5,6,7,8,9}
So, total number of outcomes = 9
The probability of spinning 1 = 1/9
Then the probability of spinning 2 = 1/9
Total probability of the event spinning a 1 and then 2 = 1/9 x 1/9
Total P(B) = 1/ 81
Therefore , the probability of spinning a 1 and then 2 = 0.012

Big Ideas Math Answers 7th Grade Chapter 7 Probability pt 11
Question 11.
You randomly choose one of the pens shown. What is the theoretical probability of choosing a black pen?
Answer:
Theoretical probability = Number of favorable outcomes / Total number of outcomes
Total number of pens = 5
Number of black pens = 2
Consider B be the event of choosing a black pen
P(B) = 2/5 = 0.4
Therefore, the theoretical probability of choosing a black pen = 0.4

Question 12.
You randomly choose one of the pens shown. Your friend randomly chooses one of the remaining pens. What is the probability that you and your friend both choose a blue pen?
Answer:
Total number of pens = 5
Number of blue pens = 2
The probability that you and your friend both choose a blue pen
P(B) = 2/5
P(B) = 0.4

Question 13.
There is an 80% chance of a thunderstorm on Saturday. Describe the likelihood that there is not a thunderstorm on Saturday.
Answer:
Given probability 80% = 0.8
Not thunder storm probability = 20 % = 0.2
The likelihood of the event is not likely to happen

Probability Cumulative Practice

Big Ideas Math Answers 7th Grade Chapter 7 Probability cp 1
Question 1.
A school athletic director asked each athletic team member to name his or her favorite professional sports team. The results are below:

  • D.C.United: 3
  • Florida Panthers: 8
  • Jacksonville Jaguars: 26
  • Jacksonville Sharks: 7
  • Miami Dolphins: 22
  • Miami Heat: 15
  • Miami Marlins: 20
  • Minnesota Lynx: 4
  • New YorkKnicks: 5
  • Orlando Magic: 18
  • Tampa Bay Buccaneers: 17
  • Tampa Bay Lightning: 12
  • Tampa Bay Rays: 28
  • Other: 6

One athletic team member is picked at random. What is the likelihood that this team member’s favorite professional sports team is not located in Florida?
A. certain
B. likely, but not certain
C. unlikely, but not impossible
D. impossible
Answer: B
13/14 = 0.92
The likelihood of the event is likely, but not certain

Question 2.
Each student in your class voted for his or her favorite day of the week. Their votes are shown in the circle graph:
Big Ideas Math Answers 7th Grade Chapter 7 Probability cp 2
A student from your class is picked at random. What is the probability that this student’s favorite day of the week is Sunday?
Answer:
Total number of students =30
Total number of Sunday = 6
The probability of that this student’s favorite day of the week is Sunday
P(S) = 6/30
= 0.2

Question 3.
What value makes the equation 11 – 3x = – 7 true?
F. – 6
G. \(-\frac{4}{3}\)
H. 6
I. 54
Answer:  F

Explanation:
11 – 3x = -7
Consider the value of x = -6
L.H.S  = 11-3X
11 – 3 (-6)
11 – 18
= -4
L.H.S = R.H.S
Therefore, the value of x =-6

Question 4.
Your friend solved the proportion in the box below.
Big Ideas Math Answers 7th Grade Chapter 7 Probability cp 4
What should your friend do to correct the error that he made?
A. Add 40 to 16 and 27 to p.
B. Subtract 16 from 40 and 27 from p.
C. Multiply 16 by 27 and p by 40.
D. Divide 16 by 27 and p by 40.
Answer: C
Multiply 16 by 27 and p by 40

Question 5.
Which value is a solution of the inequality?
3 – 2y < 7
F. – 6
G. – 3
H. – 2
I. – 1
Answer: H
3- 2y < 7 =0
3-2y >7
-2y > 7-3
-2y >4
y > -4/2
y > -2

Question 6.
A spinner is divided into eight equal sections, as shown. You spin the spinner twice. What is the probability that the arrow will stop in a yellow section both times?
Big Ideas Math Answers 7th Grade Chapter 7 Probability cp 6
Answer:
The spinner is divided 8 equal parts
In which , 3 are red , 2 are blue, 2 are yellow and 1 part is in pink color
so the probability of landing on yellow would be
P(Y) = 3/8 on first spun.
Therefore, for spinning the spinner twice then the probability of landing it on yellow would be
P(Y) = (3/8)/2
P(Y) =3/16

Question 7.
A pair of running shoes is on sale for 25% off the original price. Which price is closest to the sale price of the running shoes?
Big Ideas Math Answers 7th Grade Chapter 7 Probability cp 7
A. $93
B. $99
C. $124
D. $149
Answer: A

Explanation :
Original price = $122.76
Percent of discount = 25% = 0.25
The amount of discount :
Amount of discount = Discount x original price
= 0.25 x $125
= $30.39
Sale price :
Sale price = Original price – discount
= $122.76 – $30.39
= $92.07
It is very close to the sale price of the running shoes = $ 93

Question 8.
The value of a baseball card was $6 when it was sold. The value of this card is now $15. What is the percent increase in the value of the card?
F. 40%
G. 90%
H. 150%
I. 250%
Answer: H
Given,
Original price of the basket ball  = $6
New price = $15
The original price increases so, the percent of change is called percent of increase
Percent of increase = New price – original price/ Original price
On substituting ,
= 15 – 6/ 6
= 9 / 6
=3/2
On multiplying with 100
= 3/2 x 100
= 150%
Therefore , the percent increase in the value of the card = 150%

Question 9.
You roll a number cube twice. You want to roll two even numbers.
Big Ideas Math Answers 7th Grade Chapter 7 Probability cp 9
Part A Find the number of favorable outcomes and the number of possible outcomes of each roll.
Part B Find the probability of rolling two even numbers. Explain your reasoning.
Answer:
Part A :
The even numbers on a number cube are 2, 4, and 6
The favorable outcomes for 2 are =3
They are : 2 and 2 , 2 and 4 , 2 and 6
The favorable outcomes for 4 are =3
They are : 4 and 2 , 4 and 4 , 4 and 6
The favorable outcomes for 6 are =3
They are : 6 and 2 , 6 and 4 , 6 and 6
So,  total number favorable outcomes = 9
There are 6 outcomes for each cube so there are 6(6) = 36 total outcomes.

Part B :
There are 9 favorable outcomes of a total of 36 outcomes
So, the probability is 9/36 = 1/4

Question 10.
You put $600 into an account. The account earns 5% simple interest per year. What is the balance after 4 years?
A. $120
B. $720
C. $1800
D. $12,600
Answer: B
Simple interest = principle x annual rate of interest x Time
I = P x r x t
5% = 0.5
I = 600 x 0.05 x 4
I = $120
The interest earned = $120
The balance after 4 years = principle + Interest
= 600 + 120
=$720

Question 11.
You are comparing the prices of four boxes of cereal. Two of the boxes contain free extra cereal.

  • Box F costs $3.59 and contains 16 ounces.
  • Box G costs $3.79 and contains 16 ounces, plus an additional 10% for free.
  • Box H costs $4.00 and contains 500 grams.
  • Box  I costs $4.69 and contains 500 grams, plus an additional 20% for free.

Which box has the least unit cost?
F. Box F
G. Box G
H. Box H
I. Box I
Answer: Box G

Explanation :
Box F
The unit cost of box F = 3.59/16
The unit cost = $0.224 per ounces
Box G
Given 10% free
Which means, 10% = 16.01 = 0.6 ounces
The unit cost box G = $3.79 / 16 +0.6
=$0.215 ounces
Box H
Given, 500 grams
Which means , 500/28.35 = 17. 64 ounces

The unit cost for box H = $4.00/ 17.64
= 0.227 ounces
Box I
Given 500 grams and 20 % additional
Which means , 500/28.35 = 17. 64 ounces
20% = 17.64 .02 = 3.53 ounces
The unit cost = 4.69/ 17.64+3.54
= 0.222 ounces
Therefore, Box G has the least unit cost .
Chapter 7
Final Words:
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Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions

Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions

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Big Ideas Math Book Algebra 1 Answer Key Chapter 3 Graphing Linear Functions

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Graphing Linear Functions Maintaining Mathematical Proficiency

Plot the point in a coordinate plane. Describe the location of the point.

Question 1.
A(3, 2)
Answer:
The given point is: A (3,2)
Compare the given point with (x, y)
So,
x = 3 and y = 2
Hence,
The representation of the point in the coordinate plane is:

Question 2.
B(-5, 1)
Answer:
The given point is: B (-5,1)
Compare the given point with (x, y)
So,
x = -5 and y = 1
Hence,
The representation of the given point in the coordinate plane is:

Question 3.
C(0, 3)
Answer:
The given point is: c (0,3)
Compare the given point with (x,y)
So,
x = 0 and y = 3
Hence,
The representation of the given point in the coordinate plane is:

Question 4.
D(-1, -4)
Answer:
The given point is: D (-1,-4)
Compare the given point with (x,y)
So,
x = -1 and y = -4
Hence,
The representation of the given point in the coordinate plane is:

Question 5.
E(-3, 0)
Answer:
The given point is: E (-3,0)
Compare the given point with (x,y)
So,
x = -3 and y = 0
Hence,
The representation of the given point in the coordinate plane is:

Question 6.
F(2, -1)
Answer:
The given point is: F (2,-1)
Compare the given point with (x,y)
So,
x = 2 and y = -1
Hence,
The representation of the given point in the coordinate plane is:

Evaluate the expression for the given value of x.

Question 7.
3x – 4; x = 7
Answer:
The value of the expression for the given value of x is: 17

Explanation:
The given expression is:
3x – 4 with x = 7
Hence,
The value of the expression is:
3 (7) – 4 = 21 – 4 = 17
Hence, from the above,
We can conclude that the value of the expression for the given value of x is: 17

Question 8.
-5x + 8; x = 3
Answer:
The value of the expression for the given value of x is: -7

Explanation:
The given expression is:
-5x + 8 with x = 3
Hence,
The value of the expression is:
-5 (3) + 8 = -15 + 8 = -7
Hence, from the above,
We can conclude that the value of the expression for the given value of x is: -7

Question 9.
10x + 18; x = 5
Answer:
The value of the expression for the given value of x is: 68

Explanation:
The given expression is:
10x + 18 with x = 5
Hence,
The value of the expression is:
10 (5) + 18 = 50 + 18 = 68
Hence, from the above,
We can conclude that the value of the expression for the given value of x is: 68

Question 10.
-9x – 2; x = -4
Answer:
The value of the expression for the given value of x is: 34

Explanation:
The given expression is:
-9x – 2 with x = -4
Hence,
The value of the expression is:
-9 (-4) – 2 = 36 – 2 = 34
Hence, from the above,
We can conclude that the value of the expression for the given value of x is: 34

Question 11.
24 – 8x; x = -2
Answer:
The value of the expression for the given value of x is: 40

Explanation:
The given expression is:
24 – 8x with x = -2
Hence,
The value of the expression is:
24 – 8 (-2) = 24 + 16 = 40
Hence, from the above,
We can conclude that the value of the expression for the given value of x is: 40

Question 12.
15x + 9; x = -1
Answer:
The value of the expression for the given value of x is: -6

Explanation:
The given expression is:
15x + 9 with x = -1
Hence,
The value of the expression is:
15 (-1) + 9 = -15 + 9 = -6
Hence, from the above,
We can conclude that the value of the expression for the given value of x is: -6

Question 13.
ABSTRACT REASONING
Let a and b be positive real numbers. Describe how to plot (a, b), (-a, b), (a, -b), and (-a, -b).
Answer:
It is given that a and b are positive real numbers
The given points are (a, b), (-a, b), (a, -b) and (-a, -b)
Let the names of the points be:
A (a, b), B (-a, b), C (a, -b), and D (-a, -b)
We know that,
The coordinate plane is divided into 4 parts. These parts are called  “Quadrants”
So,
The representation of a and b in the 4 quadrants are:
1st Quadrant: (a, b)
2nd Quadrant: (-a, b)
3rd Quadrant: (-a, -b)
4th Quadrant: (a, -b)
Hence,
The representation of the given points in the coordinate plane is:

Graphing Linear Functions Mathematical Practices

Monitoring Progress

Determine whether the viewing window is square. Explain.

Question 1.
-8 ≤ x ≤ 7, -3 ≤ y ≤ 7

Question 2.
-6 ≤ x ≤ 6, -9 ≤ y ≤ 9

Question 3.
-18 ≤ x ≤ 18, -12 ≤ y ≤ 12

Use a graphing calculator to graph the equation. Use a square viewing window.

Question 4.
y = x + 3
Answer:
The given equation is:
y = x + 3
Now,
We can find the values of x and y by putting the values 0, 1, 2…..
Hence,
The representation of the given equation in the coordinate plane is:

Question 5.
y = -x – 2
Answer:
The given equation is:
y = -x – 2
Now,
We can find the values of x and y by putting the values 0, 1, 2…..
Hence,
The representation of the given equation in the coordinate plane is:

Question 6.
y = 2x – 1
Answer:
The given equation is:
y = 2x – 1
Now,
We can find the values of x and y by putting the values 0, 1, 2…..
Hence,
The representation of the given equation in the coordinate plane is:

Question 7.
y = -2x + 1
Answer:
The given equation is:
y = -2x + 1
Now,
We can find the values of x and y by putting the values 0, 1, 2…..
Hence,
The representation of the given equation in the coordinate plane is:

Question 8.
y = –\(\frac{1}{3}\)x – 4
Answer:
The given equation is:
y = –\(\frac{1}{3}\)x – 4
Now,
We can find the values of x and y by putting the values 0, 1, 2…..
Hence,
The representation of the given equation in the coordinate plane is:

Question 9.
y = \(\frac{1}{3}\)x + 2
Answer:
The given equation is:
y = \(\frac{1}{3}\)x + 2
Now,
We can find the values of x and y by putting the values 0, 1, 2…..
Hence,
The representation of the given equation in the coordinate plane is:

Question 10.
How does the appearance of the slope of a line change between a standard viewing window and a square viewing window?
Answer:
A typical graphing calculator screen has a height to width ratio of 2 to 3. This means that when you use the standard viewing window of -10 to 10 ( on each axis ), the graph will not be in its true perspective.
To see a graph in its true perspective, you need to use a square viewing window, in which the tick marks on the x-axis are spaced the same as the tick marks on the y-axis.

Lesson 3.1 Functions

Essential Question

A relation pairs inputs with outputs. When a relation is given as ordered pairs, the x-coordinates are inputs and the y-coordinates are outputs. A relation that pairs each input with exactly one output is a function.

EXPLORATION 1
Describing a Function
Work with a partner.
Functions can be described in many ways.

  • by an equation
  • by an input-output table
  • using words
  • by a graph
  • as a set of ordered pairs

a. Explain why the graph is shown represents a function.
Answer:
The vertical line test can be used to determine whether a graph represents a function. If we can draw any vertical line that intersects a graph more than once, then the graph does not define a function because that x value has more than one output. A function has only one output value for each input value.

b. Describe the function in two other ways.
Answer:
The function can be described in 4 ways. They are:
a. A function can be represented verbally.
Example:
The circumference of a square is four times one of its sides.
b. A function can be represented algebraically.
Example:
3 x + 6 .
c. A function can be represented numerically.
d. A function can be represented graphically.

EXPLORATION 2

Identifying Functions
Work with a partner. Determine whether each relation represents a function. Explain your reasoning.

Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 1
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 1.1
e. (-2, 5), (-1, 8), (0, 6), (1, 6), (2, 7)
f. (-2, 0), (-1, 0), (-1, 1), (0, 1), (1, 2), (2, 2)
g. Each radio frequency x in a listening area has exactly one radio station y.
h. The same television station x can be found on more than one channel y.
i. x = 2
j. y = 2x + 3
Answer:
We know that,
Functions can be described in different ways. They are:
a. By an equation
b. By an input-output table
c. Using words
d. By a graph
e. As a set of ordered pairs
We know that,
A function has only a single input for each or multiple outputs
Hence,
The given Exercises above i.e., a, e, g,  j are the relations that represent a function and the remaining relations are not functions

Communicate Your Answer

Question 3.
What is a function? Give examples of relations, other than those in Explorations 1 and 2, that (a) are functions and (b) are not functions.
Answer:
Definition of Function:
A relation from a set of inputs to a set of possible outputs where each input is related to exactly one output is known as ” Function”
Examples of relation that is a function:
a. y = x + 3
b. y = -x² + 1
Example of relation that is not a function:

We know that,
A relation is a set of inputs for specific outputs.
But from the above,
A single input has multiple outputs

3.1 Lesson

Monitoring Progress

Determine whether the relation is a function. Explain.

Question 1.
(-5, 0), (0, 0), (5, 0), (5, 10)
Answer:
The given relation is:
(-5, 0), (0, 0), (5, 0), (5, 10)
From the given relation,
We know that,
x represents the input
y represents the output
We know that,
For a function, each input should correspond with only one output
But,
When we observe the given relation, there are 2 outputs for a single input
Hence, from the above,
We can conclude that the given relation is not a function

Question 2.
(-4, 8), (-1, 2), (2, -4), (5, -10)
Answer:
The given relation is:
(-4, 8), (-1, 2), (2, -4), (5, -10)
From the given relation,
We know that,
x represents the input
y represents the output
We know that,
For a function, each input should correspond with only one output
Hence, from the above,
We can conclude that the given relation is a function

Question 3.
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 2
Answer:
The given table is:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 2
The representation of the table in the form of relation is:
(2, 2.6), (4, 5.2), (6, 7.8)
From the given relation,
We know that,
x represents the input
y represents the output
We know that,
For a function, each input should correspond with only one output
Hence, from the above,
We can conclude that the given relation is a function

Question 4.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 3
Answer:
The given figure is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 3
The representation of the given figure in the form of relation is:
(1, -2), (1, 0), (2, 4)
From the given relation,
We know that,
x represents the input
y represents the output
We know that,
For a function, each input should correspond with only one output
But,
When we observe the given relation, there are 2 outputs for a single input
Hence, from the above,
We can conclude that the given relation is not a function

Monitoring Progress

Determine whether the graph represents a function. Explain.

Question 5.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 4
Answer:
The given graph is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 4
The representation of the points from the graph is:
(0, 3), (1, 3), (2, 3), (3, 3), (4, 3), (5, 3)
From the given relation,
We know that,
x represents the input
y represents the output
We know that,
For a function, each input should correspond with only one output
We have to remember that the inputs must be different but outputs may be the same or different
Hence, from the above,
We can conclude that the given graph is a function

Question 6.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 5
Answer:
The given figure is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 5
From the graph,
The representation of the points are:
(2, 1), (1, 2), (1, 3), (1, 4), (0.5, 5), (0.5, 6), (3, 0), (4, 1), (4.2, 2), (4.8, 3), (5, 4), (5.1, 5), (5.1, 6)
From the points,
We can observe that the output is different for the same input
Hence, from the above,
We can conclude that the given graph is not a function

Question 7.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 6
Answer:
The given figure is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 6
From the given graph,
We can observe that for the same value of x, there are different values of y
Where,
x represents the input
y represents the output
We know that,
For a function, each input must be matched with a single output
Hence, from the above,
We can conclude that the given graph is not a function

Question 8.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 7
Answer:
The given figure is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 7
From the graph,
We can observe that there are multiple outputs for a single input
Hence, from the above,
We can conclude that the given graph is not a function

Monitoring Progress

Find the domain and range of the function represented by the graph.

Question 9.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 8

Answer:
The given graph is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 8
We know that,
“Domain” is defined as the set of all values present in the x-axis
“Range” is defined as the set of all values present in the y-axis
Hence,
The domain of the given graph is: -2, -1, 0, 1, 2
The range of the given graph is: 1, 2, 3,  4

Question 10.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 9
Answer:
The given graph is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 9
We know that,
“Domain” is defined as the set of all values present in the x-axis
“Range” is defined as the set of all values present in the y-axis
Hence,
The domain of the given graph is: 1, 2, 3, 4, 5
The range of the given graph is: 0, 1, 2, 3,  4

Monitoring Progress

Question 11.
The function a = -4b + 14 represents the number ‘a’ of avocados you have left after making b batches of guacamole.
a. Identify the independent and dependent variables.
Answer:
The “Independent variable” represents the input values of a function and can be any value in the domain.
The dependent variable represents the output values of the function and depends on the value of the independent variable
Hence, from the above,
The independent variable is: b (batches of guacamole)
The dependent variable is: a ( The number of avocados )

b. The domain is 0, 1, 2, and 3. What is the range?
Answer:
The given function is:
a = -4b + 14
In the given function,
Input: b
Output: a
We know that,
The “Domain” is defined as the set of all values present in the input or the x-axis
The given Domain is: 0, 1, 2, and 3
So,
To find the range, we have to find the values of b for each value present in the domain
So,
a = -4 (0) + 14 = 14
a = -4 (1) + 14 = 10
a = -4 (2) + 14 = 6
a = -4 (3) + 14 = 2
Hence, from the above,
We can conclude that the range for the given equation is: 2,  6, 10, and 14

Question 12.
The function t = 19m + 65 represents the temperature t (in degrees Fahrenheit) of an oven after preheating for m minutes.
a. Identify the independent and dependent variables.
Answer:
The “Independent variable” represents the input values of a function and can be any value in the domain.
The dependent variable represents the output values of the function and depends on the value of the independent variable
Hence, from the above,
The independent variable is: Minutes
The dependent variable is: Temperature

b. A recipe calls for an oven temperature of 350°F. Describe the domain and range of the function.
Answer:
The given function is:
t = 19m + 65
Compare the given function with
y = mx + c
It is given that a recipe calls for an oven temperature of 350°F
So,
350 = 19m + 65
19m = 350 – 65
19m = 285
m = 285 / 19
m = 15 minutes
So,
t = 19 (15) + 350
t = 285 + 65
t = 350°F
Hence, from the above,
We can conclude that
The domain of the given function is: 0 ≤ m ≤ 15 [ Since the minutes will not be -ve ]
The range of the given function is: 65 ≤ t ≤ 350 [ Since the  minimum temperature is the value of c ]

Functions 3.1 Exercises

Vocabulary and Core Concept Check

Question 1.
WRITING
How are independent variables and dependent variables different?
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.1 Question 1

Question 2.
DIFFERENT WORDS, SAME QUESTION
Which is different? Find “both” answers.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 9.1
Answer:
The given statements are:
a. Find the range of the function represented by the table?
b. Find the inputs of the function represented by the table?
c. Find the x values of the function represented by (-1, 7), (0, 5), and (1, -1)?
d. Find the domain of the function represented by (-1, 7), (0, 5), and (1, -1)?
Now,
From the given table,
The values of x are: -1, 0, 1
The values of y are: 7, 5, -1
We know that,
For a function,
The x values represent the input and the domain
The y values represent the output and the range
So,
a.
The range of the function represented by the table is: 7, 5, -1
b.
The inputs of the function represented by the table are: -1, 0, 1
c.
The x values of the function represented by the given points are: -1, 0, 1
d.
The domain of the function represented by the given points are: -1, 0, 1
Hence, from the above,
We can conclude that all the given four are the same

Monitoring Progress and Modeling with Mathematics

In Exercises 3–8, determine whether the relation is a function. Explain.

Question 3.
(1, -2), (2, 1), (3, 6), (4, 13), (5, 22)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.1 Question 3

Question 4.
(7, 4), (5, -1), (3, -8), (1, -5), (3, 6)
Answer:
The given points are:
(7, 4), (5, -1), (3, -8), (1, -5), (3, 6)
We know that,
For a function, each input has only a single output
But from the above,
We can observe that 3 has multiple outputs
Hence, from the above,
We can conclude that the given relation is not a function

Question 5.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 10
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.1 Question 5

Question 6.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 11
Answer:
The given relation is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 11
We know that,
For a function, a single input has a single output
Hence, from the above,
We can say that each input has only 1 output
Hence, from the above,
We can conclude that the given relation is a function

Question 7.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 12
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.1 Question 7

Question 8.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 13
Answer:
The given table is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 13
From the given table,
We can observe that each input has a single output
Hence, from the above,
We can conclude that the given relation is a function

In Exercises 9–12, determine whether the graph represents a function. Explain.

Question 9.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 14
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.1 Question 9

Question 10.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 15
Answer:
The given graph is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 15
From the given graph,
We can observe that the vertical line can be drawn through more than one point on the graph i.e., input 2 has repeated 2 times i.e., (2, 1) and (2, 5)
Hence, from the above,
We can conclude that the given graph is not a function

Question 11.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 16
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.1 Question 11

Question 12.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 17
Answer:
The given graph is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 17
From the given graph,
We can observe that each input corresponds to a single output
Hence, from the above,
We can conclude that the given graph is a function

In Exercises 13–16, find the domain and range of the function represented by the graph.

Question 13.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 18
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.1 Question 13

Question 14.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 19
Answer:
The given graph is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 19
From the given graph,
The ordered pairs are: (0, 4), (-2, 4), (2, 4), (4, 4)
Hence,
The domain of the given graph is: 0, -2, 2, 4
The range of the given graph is: 4

Question 15.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 20
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.1 Question 15

Question 16.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 21
Answer:
The given graph is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 21
From the given graph,
Identify the x and y values
So,
The values of x range from 2 to 7 excluding 2 and 7
The values of y range from 1 to 6 excluding 1 and 6
Hence, from the above,
We can conclude that
The domain of the given graph is: 2 < x < 7
The range of the given graph is: 1 < y < 6

Question 17.
MODELING WITH MATHEMATICS
The function y = 25x + 500 represents your monthly rent y (in dollars) when you pay x days late.
a. Identify the independent and dependent variables.
b. The domain is 0, 1, 2, 3, 4, and 5. What is the range?
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.1 Question 17

Question 18.
MODELING WITH MATHEMATICS
The function y = 3.5x + 2.8 represents the cost y (in dollars) of a taxi ride of x miles.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 22
a. Identify the independent and dependent variables.
Answer:
The given function is:
y = 3.5x + 2.8
From the above function,
The independent variable is: x which represents the number of miles
The dependent variable is: y which represents the cost in dollars

b. You have enough money to travel at most 20 miles in a taxi. Find the domain and range of the function.
Answer:
The given function is:
y = 3.5x + 2.8
It is given that you have enough money to travel at most 20 miles i.e., the value of x in a taxi i.e.,
x ≥ 0 and x ≤ 20
So,
The value of x ranges from 0 ≤ x ≤ 20
Now,
y = 3.5 (20) + 2.8
y = 7 + 2.8
y = $9.8
y = 3.5 (0) + 2.8
y = 0 + 2.8
y = $2.8
Hence, from the above,
We can conclude that
The domain of the given function is: 0 ≤ x ≤ 20 miles
The range of the given function is: $2.8 ≤ y ≤ $9.8

ERROR ANALYSIS
In Exercises 19 and 20, describe and correct the error in the statement about the relation shown in the table.

Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 23

Question 19.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 24
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.1 Question 19

Question 20.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 25
Answer:
It is given that the relation is a function and the range is 1, 2, 3, 4, and 5
We know that,
The relation is a function only when a single input pairs with an output
For the function, the domain, and the range exist
From the given table,
The domain is: 1, 2, 3, 4, 5
The range is: 6, 7, 8, 6, 9
Hence, from the above,
We can conclude that the given statement is not correct

ANALYZING RELATIONSHIPS
In Exercises 21 and 22, identify the independent and dependent variables.

Question 21.
The number of quarters you put into a parking meter affects the amount of time you have on the meter.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.1 Question 21

Question 22.
The battery power remaining on your MP3 player is based on the amount of time you listen to it.
Answer:
The given statement is:
The battery power remaining on your MP3 player is based on the amount of time you listen to it.
Hence, from the above,
We can conclude that
The Independent variable: Amount of time
The dependent variable: Battery power

Question 23.
MULTIPLE REPRESENTATIONS
The balance y (in dollars) of your savings account is a function of the month x.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 26
a. Describe this situation in words.
b. Write the function as a set of ordered pairs.
c. Plot the ordered pairs in a coordinate plane.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.1 Question 23

Question 24.
MULTIPLE REPRESENTATIONS
The function 1.5x + 0.5y = 12 represents the number of hardcover books x and softcover books y you can buy at a used book sale.
a. Solve the equation for y.
Answer:
The given function is:
1.5x + 0.5y = 12
So,
0.5y = 12 – 1.5x
y = \(\frac{12 – 1.5x}{0.5}\)
y = \(\frac{12}{0.5}\) – \(\frac{1.5x}{0.5}\)
y = 24 – 3x
Hence, from the above,
We can conclude that the equation for y is:
y = 24 – 3x

b. Make an input-output table to find ordered pairs for the function.
Answer:
From part (a),
The equation for y is:
y = 24 – 3x
Now,
Put the values of 0, 1, 2, 3 in x
So,
y = 24 – 3(0) = 24
y = 24 – 3(1) = 21
y = 24 – 3(2) = 18
y = 24 – 3 (3) = 15
Hence,
The input-output table for the given equation of y is:

c. Plot the ordered pairs in a coordinate plane.
Answer:
From part (b),
The table is:

We know that,
The ordered pair is in the form of (x, y)
From the table,
The ordered pairs are:
(0, 24), (1, 21), (2, 18), (3, 15)
Hence,
The representation of the ordered pairs in the coordinate plane is:

Question 25.
ATTENDING TO PRECISION
The graph represents a function. Find the input value corresponding to an output of 2.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 27
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.1 Question 25

Question 26.
OPEN-ENDED
Fill in the table so that when t is the independent variable, the relation is a function, and when t is the dependent variable, the relation is not a function.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 28
Answer:
Let the function in terms of t and v  such as t is the independent variable and v is the dependent variable is:
v = t + 4 ———–(1)
Let the function in terms of t and v  such as v is the independent variable and t is the dependent variable is:
t = 4v ———–(2)
Now,
place the values 0, 1, 2, 3 …….. in the independent variables of both functions
Now,
In equation (1),
v = 0 + 4 = 4
v = 1 + 4 = 5
v = 2 + 4 = 6
v = 3 + 4 = 7
In equation (2),
t = 4(0) = 0
t = 4(1) = 4
t = 4(2) = 8
t = 4(3) = 12
Hence,
     
Hence, from the above,
We can conclude that equation (1) is a function and equation (2) is not a function

Question 27.
ANALYZING RELATIONSHIPS
You select items in a vending machine by pressing one letter and then one number.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 29
a. Explain why the relation that pairs letter-number combinations with food or drink items is a function.
b. Identify the independent and dependent variables.
c. Find the domain and range of the function
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.1 Question 27

Question 28.
HOW DO YOU SEE IT?
The graph represents the height h of a projectile after t seconds.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 30
a. Explain why h is a function of t.
b. Approximate the height of the projectile after 0.5 seconds and after 1.25 seconds.
c. Approximate the domain of the function.
d. Is t a function of h? Explain.
Answer:
a.
The given graph is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 30
From the graph,
We can observe that when we draw the vertical lines, each vertical line corresponds to only 1 value.
Hence, from the above,
We can conclude that h is a function of t by using the vertical line method

b.
From the given graph,
We can observe that the height of the projectile after 0.5 seconds increases and after some time, the height of the projectile decreases
We can observe that the height of the projectile after 1.25 seconds decreases steadily
Hence, from the above,
We can conclude that
The approximate maximum height after 0.5 seconds is: 30 feet
The approximate maximum height after 1.25 seconds is: 25 feet

c.
From the given graph,
The values of t in the x-axis vary from 0 to 2.5
Hence,
The domain of the given graph is: 0 ≤ t ≤ 2.5 seconds

d.
t is not a function of h
Reason:
When we observe the graph,
We can see that for a single value of t, there are multiple values of h.
We know that a relation can be considered as a function only when a single input pairs with a single output
hence, from the above,
We can conclude that that t is not a function of h

Question 29.
MAKING AN ARGUMENT
Your friend says that a line always represents a function. Is your friend correct? Explain.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.1 Question 29

Question 30.
THOUGHT-PROVOKING
Write a function in which the inputs and/or the outputs are not numbers. Identify the independent and dependent variables. Then find the domain and range of the function.
Answer:
It is given that the inputs and/or the outputs will not be numbers.
So,
The given function in which the inputs and/or the outputs are not numbers is:
°C = 32 + °F
From the given function,
The independent variable is: °F
The dependent variable is: °C
Now,
From the given function,
We can observe that the values of °F vary from -∞ to ∞
Now,
Place the values of -∞ to ∞ in the place of °F
So,
°C = 32 + 0 = 32°F
°C = 32 + 1 = 33°F
°C = 32 – 1 = 31°F
°C = 32 – 40 = -8°F
Hence, from the above values,
We can observe that the values of °C vary from -∞ to ∞
Hence, from the above,
We can conclude that
The domain of the function is: -∞ to ∞
The range of the function is: -∞ to ∞

ATTENDING TO PRECISION In Exercises 31–34, determine whether the statement uses the word function in a way that is mathematically correct. Explain your reasoning.

Question 31.
The selling price of an item is a function of the cost of making the item.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.1 Question 31

Question 32.
The sales tax on a purchased item in a given state is a function of the selling price.
Answer:
The given statement is:
The sales tax on a purchased item in a given state is a function of the selling price
We know that,
The sales tax is a percentage applied to the selling price
Hence, from the above,
We can conclude that the given statement uses the word function in a way that is mathematically correct.

Question 33.
A function pairs each student in your school with a homeroom teacher.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.1 Question 33

Question 34.
A function pairs each chaperone on a school trip with 10 students.
Answer:
The given statement is:
A function pairs each chaperone on a school trip with 10 students.
We know that,
Each chaperone on a school trip pairs with more than 1 student i.e., the number may be 2 or ∞ but not exactly 10
Hence, from the above,
We can conclude that the given statement does not use the word function in a way that is mathematically correct.

REASONING
In Exercises 35–38, tell whether the statement is true or false. If it is false, explain why.

Question 35.
Every function is a relation.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.1 Question 35

Question 36.
Every relation is a function.
Answer:
The given statement is false

Explanation:
We know that,
Every function is a relation that has only 1 output for a single input
But we can not say that every relation has a single output for a single input
Hence, from the above,
We can conclude that the given statement is false

Question 37.
When you switch the inputs and outputs of any function, the resulting relation is a function.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.1 Question 37

Question 38.
When the domain of a function has an infinite number of values, the range always has an infinite number of values.
Answer:
The given statement is false

Explanation:
We know that,
The domain is defined as the set of all the values of x
The range is defined as the set of all the values of y
Now, consider an example
Let the input be x
Let the output be a constant
Now,
The domain of the input can vary from -∞ to ∞
But the range of the output is only a constant even though we put any value of x
Hence, from the above,
We can conclude that the given statement is false

Question 39.
MATHEMATICAL CONNECTIONS
Consider the triangle shown.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 31
a. Write a function that represents the perimeter of the triangle.
b. Identify the independent and dependent variables.
c. Describe the domain and range of the function. (Hint: The sum of the lengths of any two sides of a triangle is greater than the length of the remaining side.)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.1 Question 39

REASONING
In Exercises 40–43, find the domain and range of the function.

Question 40.
y = | x |
Answer:
The given function is:
y = | x |
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
We can put the values of x from -∞ to ∞
So,
The values of x vary from 0 to ∞ since x can’t be negative
Hence, from the above,
We can conclude that
The domain of the given function is: -∞ to ∞
The range of the given function is: 0 to ∞

Question 41.
y = – | x |
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.1 Question 41

Question 42.
y = | x | – 6
Answer:
The given function is:
y = | x | – 6
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
We can vary the values of x from -∞ to ∞
The values of y vary from -6 to ∞
Hence, from the above,
We can conclude that
The domain of the given function is: -∞ to ∞
The range of the given function is: y ≥ -6

Question 43.
y = 4 – | x |
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.1 Question 43

Maintaining Mathematical Proficiency

Write the sentence as an inequality. (Section 2.1)

Question 44.
A number y is less than 16.
Answer:
The given worded form is:
A number y is less than 16
Hence,
The representation of the given worded form in the form of inequality is:
y < 16

Question 45.
Three is no less than a number x.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.1 Question 45

Question 46.
Seven is at most the quotient of a number d and -5.
Answer:
The given worded form is:
Seven is at most the quotient of a number d and -5
Hence,
The representation of the given worded form in the form of inequality is:
7 ≤ d ÷ (-5)

Question 47.
The sum of a number w and 4 is more than -12.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.1 Question 47

Evaluate the expression.

Question 48.
112
Answer:
The product of 11² is:
11² = 11 × 11 = 121

Question 49.
(-3)4
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.1 Question 49

Question 50.
-52
Answer:
The product of -5² is:
-5² = -5 × -5 = 25 [ We know that – × – = + ]

Question 51.
25
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.1 Question 51

Lesson 3.2 Linear Functions

Essential Question

How can you determine whether a function is linear or non-linear?
Answer:
Simplify the equation as closely as possible to the form of y = mx + c.
Check to see if your equation has exponents.
If it has exponents, it is nonlinear. If your equation has no exponents, it is linear.

EXPLORATION 1
Finding Patterns for Similar Figures

Work with a partner. Copy and complete each table for the sequence of similar figures. (In parts (a) and (b), use the rectangle shown.) Graph the data in each table. Decide whether each pattern is linear or nonlinear. Justify your conclusion.
a. perimeters of similar rectangles
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 32
Answer:
We know that,
The perimeter of the rectangle (P) = 2 (Length + Width)
P = 2 (x + 2x)
P = 2 (3x)
P = 6x
Hence,
The completed table for the perimeters of similar rectangles is:

The representation of the perimeters of the similar rectangles in the coordinate plane is:

b. areas of similar rectangles
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 32
Answer:
We know that,
Area of the rectangle (A) = Length × Width
A = x × (2x)
A = 2x²
Hence,
The complete table for the area of the similar rectangles is:

The representation of the areas of the similar rectangles in the coordinate plane is:

c. circumferences of circles of radius r
Answer:
We know that,
The circumference of circle = 2πr
Take the value of π as 3
Hence,
The complete table for the circumferences of circles of radius r is:

The representation of the circumferences of the circles in the coordinate plane is:

d. areas of circles of radius r
Answer:
We know that,
The area of the circle = πr²
Take the value of π as 3
Hence,
The complete table for the areas of the similar circles is:


The representation of the areas of the similar circles is:

Communicate Your Answer

Question 2.
How do you know that the patterns you found in Exploration 1 represent functions?
Answer:
In Exploration 1,
From the graphs,
We can observe that from the vertical test, only one point passes through each vertical line i.e., each input hs only 1 output
Our observation coincides with the definition of function
Hence, from the above,
We can conclude that the given patterns in Exploration 1 represent functions

Question 3.
How can you determine whether a function is linear or nonlinear?
Answer:
Simplify the equation as closely as possible to the form of y = mx + c.
Check to see if your equation has exponents.
If it has exponents, it is nonlinear. If your equation has no exponents, it is linear.

Question 4.
Describe two real-life patterns: one that is linear and one that is nonlinear. Use patterns that are different from those described in Exploration 1.
Answer:
The real-life pattern that is linear is:
The distance you travel when you go for a jog, you can graph the function and make some assumptions with only two points. The slope of a function is the same as the rate of change for the dependent variable (y), For instance, if you’re graphing distance Vs
The real-life pattern that is non-linear is:
Triangulation of GPS signals
Example:
A device like your cellphone receives signals from GPS satellites, which have known orbital positions around the Earth.

3.2 Lesson

Monitoring Progress

Does the graph or table represent a linear or nonlinear function? Explain.

Question 1.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 36
Answer:
The given graph is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 36
The graph represents a straight line
We know that,
The straight line will be in the form of
y = mx + c or y = mx
Where,
m is the slope-intercept
c is the y-intercept that cuts through the y-axis
Hence, from the above,
We can conclude that the given graph is linear function

Question 2.

Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 37
Answer:
The given graph is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 37
From the given graph,
We can observe that the graph is not a straight line i.e., it is not linear
Hence, from the above,
We can conclude that the given graph is a non-linear function

Question 3.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 38
Answer:
The given table is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 38
From the given table,
We can observe that
There is a constant difference of 1 between the values of x and there is a constant difference of 2 between the values of y
The difference is constant for both the values of x and y
Hence, from the above,
We can conclude that the given table is a linear function

Question 4.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 39
Answer:
The given table is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 39
From the given table,
We can observe that
There is a constant difference of 1 between the values of x and there is a constant ratio of 2 between the values of y.
Since the operations are different between the values of x and y,
We can conclude that the given table is a non-linear function

Does the equation represent a linear or nonlinear function? Explain.

Question 5.
y = x + 9
Answer:
The given equation represents the linear function

Explanation:
The given function is:
y = x + 9
We know that,
The standard representation of the linear function is:
y = mx + c
When we compare the given equation with the standard representation of linear function,
We can conclude that the given equation is a linear function

Question 6.
y = \(\frac{3 x}{5}\)
Answer:
The given equation represents a linear function

Explanation:
The given function is:
y = \(\frac{3 x}{5}\)
5y = 3x
3x – 5y = 0
y = \(\frac{3}{5}\)x + 0
We know that,
The standard representation of a linear function is:
y = mx + c
When we compare the above function with the standard representation of a linear function,
We can conclude that the given function is a linear function

Question 7.
y = 5 – 2x²
Answer:
The given equation represents a non-linear function

Explanation:
The given function is:
y = 5 – 2x²
We know that,
The standard representation of a linear function is:
y = mx + c
When we compare the given function with the standard representation of a linear function,
We can conclude that the given function is a non-linear function

Question 8.
The linear function m = 50 – 9d represents the amount m (in dollars) of money you have after buying d DVDs.
(a) Find the domain of the function. Is the domain discrete or continuous? Explain.
Answer:
The given linear function is:
m = 50 – 9d
Where,
m is the amount in dollars you have after buying d DVDs
m is the dependent variable
d is the independent variable
We know that,
The domain is defined for the independent variables
So,
Let
d = 0, 1, 2, 3, 4, 5……
Now,
m = 50 – 9 (0) = 50
m = 50 – 9 (1) = 41
m = 50 – 9 (2) = 32
m = 50 – 9 (3) = 23
m = 50 – 9 (4) = 14
m = 50 – 9 (5) = 5
m = 50 – 9 (6) = -4
Hence, from the above,
We can conclude that
The domain of the given linear function is: 0, 1, 2, 3, 4, and 5 [ Since from 6, -ve values are coming and the money will not be -ve ]
The domain is discrete [ Since a discrete graph is a series of unconnected points ]

(b) Graph the function using its domain.
Answer:
From part (a),
The domain of the given function is: 0, 1, 2, 3, 4, and 5
Hence,
The representation of the domain in the coordinate plane is:

Question 9.
Is the domain discrete or continuous? Explain.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 40
Answer:
The given table is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 40
From the given table,
The ordered pairs are:
(1, 12), (2, 24), (3, 36)
Hence,
The representation of the ordered pairs in the coordinate plane is:

From the above graph,
We can say that the points are scattered or discrete and they are unconnected
Hence, from the above,
We can conclude that the domain for the given table is discrete

Question 10.
A 20-gallon bathtub is draining at a rate of 2.5 gallons per minute. The number g of gallons remaining is a function of the number m of minutes.
a. Does this situation represent a linear function? Explain.
Answer:
It is given that a 20-gallon bathtub is draining at a rate of 2.5 gallons per minute.
Where,
The number g of gallons remaining is a function of the number m of minutes.
So,
From the above,
We can say that the bathtub is draining at the constant rate
We know that,
From the property of linear function, the change will be constantly increasing or decreasing
So,
The representation of the linear function for this situation is:
g = 20 – 2.5x
Hence, from the above,
We can conclude that the given situation is a linear function

b. Find the domain of the function. Is the domain discrete or continuous? Explain.
Answer:
From part (a),
We can conclude that the given situation is a linear function
So,
The domain of the function = \(\frac{Total volume of the bathtub}{The rate of draining}\)
= \(\frac{20}{2.5}\)
= \(\frac{200}{25}\)
= 8
Hence,
The domain of the function is: 0 ≤ x ≤ 8 [ Since the draining rate will not be -ve ]
From the representation of the domain,
We can conclude that the domain is continuous

c. Graph the function using its domain.
Answer:
From part (a),
The linear function is:
g = 20 – 2.5x
We know that,
The domain is: 0 ≤ x ≤ 8
So,
g = 20 – 2.5 (0) = 20
g = 20 – 2.5 (1) = 17.5
g = 20 – 2.5 (2) = 15
g = 20 – 2.5 (3) = 12.5
g = 20 – 2.5 (4) = 10
g = 20 – 2.5 (5) = 7.5
g = 20 – 2.5 (6) = 5
g = 20 – 2.5 (7) = 2.5
g = 20 – 2.5 (8) = 0
Hence,
The representation of the linear function using the domain in the coordinate plane is:

Write a real-life problem to fit the data shown in the graph. Is the domain of the function discrete or continuous? Explain.

Question 11.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 41
Answer:
The given graph is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 41
The real-life problem that fits the given graph is:
An escalator moves upwards at a constant rate of 1step/minute.
So,
What is the rate escalator moves upwards after 8 minutes?
From the given graph,
We can say that the points are connected
Hence, from the above,
We can conclude that the domain of the function is continuous

Question 12.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 42
Answer:
The given graph is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 42
The real-life problem that fits the given graph is:
If one company offers to pay you $450 per week and the other offers $10 per hour, and both ask you to work 40 hours per week, which company is offering the better rate of pay?
From the graph,
We can observe that the points are connected
Hence, from the above,
We can conclude that the domain is continuous

Linear Functions 3.2 Exercises

Vocabulary and Core Concept Check

Question 1.
COMPLETE THE SENTENCE
A linear equation in two variables is an equation that can be written in the form ________, where m and b are constants.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.2 Question 1

Question 2.
VOCABULARY
Compare linear functions and nonlinear functions.
Answer:
Linear Function:
A Linear function is a relation between two variables that produces a straight line when graphed.
Example:
y = 2x + 3
Non-Linear Function:
A non-linear function is a function that does not form a line when graphed.
Example:
y = 6x³

Question 3.
VOCABULARY
Compare discrete domains and continuous domains.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.2 Question 3

Question 4.
WRITING
How can you tell whether a graph shows a discrete domain or a continuous domain?
Answer:
A “Discrete domain” is a set of input values that consists of only certain numbers in an interval.
A “continuous domain” is a set of input values that consists of all numbers in an interval.
Sometimes, the set of points that represent the solutions of an equation are distinct, and other times the points are connected.

Monitoring Progress and Modeling with Mathematics

In Exercises 5–10, determine whether the graph represents a linear or nonlinear function. Explain.

Question 5.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 43
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.2 Question 5

Question 6.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 44
Answer:
The given graph is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 44
From the above graph,
By using the vertical test, there is only one point going through each point and the given graph is a straight line
Hence, from the above,
We can conclude that the given graph is a linear function

Question 7.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 45
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.2 Question 7

Question 8.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 46
Answer:
The given graph is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 46
From the above graph,
By using the vertical test, each line passes through only one point and the given graph is a straight line
Hence, from the above,
We can conclude that the given graph is a linear function

Question 9.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 47
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.2 Question 9

Question 10.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 48
Answer:
The given graph is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 48
From the above graph,
By using the vertical test, each line passes through each point and the given graph is not a straight line
Hence, from the above,
We can conclude that the given graph is a non-linear function

In Exercises 11–14, determine whether the table represents a linear or nonlinear function. Explain.

Question 11.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 49
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.2 Question 11

Question 12.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 50
Answer:
The given table is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 50
From the above table,
The difference between all the values of x is 2 which is a constant through all the values of x
The difference between all the values of y is not constant
Since the difference between all the values of y is not constant
The given table is a non-linear function

Question 13.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 51
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.2 Question 13

Question 14.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 52
Answer:
The given table is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 52
From the above table,
The difference between all the values of x is +1 throughout all the values of x
The difference between all the values of y is -15 throughout all the values of y
Since the differences are constant for all the values of x and y,
The given table is a linear function

ERROR ANALYSIS
In Exercises 15 and 16, describe and correct the error in determining whether the table or graph represents a linear function.

Question 15.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 53
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.2 Question 15

Question 16.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 54
Answer:
By using the vertical test in the graph,
We can say that each line passes through one point and the given graph is a straight line
Hence, from the above,
We can conclude that the given graph is a linear function

In Exercises 17–24, determine whether the equation represents a linear or nonlinear function. Explain.

Question 17.
y = x2 + 13
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.2 Question 17

Question 18.
y = 7 – 3x
Answer:
The given equation is a linear function

Explanation:
The given equation is:
y = 7 – 3x
Compare the given equation with the standard representation of the given linear function
The standard representation of the linear function is:
y = mx + c
Hence, from the above,
We can conclude that the given equation is a linear function

Question 19.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 55
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.2 Question 19

Question 20.
y = 4x(8 – x)
Answer:
The given equation is not a linear function

Explanation:
The given equation is:
y = 4x (8 – x)
so,
y = 4x (8) – 4x (x)
y = 32x – 4x²
Compare the above equation with the standard representation of the linear function
The standard representation of the linear function is:
y = mx + c
Hence, from the above,
We can conclude that the given equation is not a linear function

Question 21.
2 + \(\frac{1}{6}\) y = 3x + 4
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.2 Question 21

Question 22.
y – x = 2x – \(\frac{2}{3}\)y
Answer:
The given equation is a linear fraction

Explanation:
The given equation is:
y – x = 2x – \(\frac{2}{3}\)y
So,
y + \(\frac{2}{3}\)y = 2x + x
\(\frac{3y}{3}\) + \(\frac{2y}{3}\) = 3x
\(\frac{5}{3}\)y = 3x
y = 3x × \(\frac{3}{5}\)
y = \(\frac{9}{5}\)x + 0
Compare the above equation with the standard representation of the linear function
The standard representation of the linear function is:
y = mx + c
Hence, from the above,
We can conclude that the given equation is a linear function

Question 23.
18x – 2y = 26
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.2 Question 23

Question 24.
2x + 3y = 9xy
Answer:
The given equation is not a linear fraction

Explanation:
The given equation is:
2x + 3y = 9xy
2x = 9xy – 3y
2x = y (9x – 3)
y = \(\frac{2}{9x – 3}\)x
Compare the above equation with the standard representation of the linear function
The standard representation of the linear function is:
y = mx + c
Hence, from the above,
We can conclude that the given equation is not a linear function

Question 25.
CLASSIFYING FUNCTIONS
Which of the following equations do not represent linear functions? Explain.
A. 12 = 2x2 + 4y2
B. y – x + 3 = x
C. x = 8
D. x = 9 – \(\frac{3}{4}\)y
E. y = \(\frac{5x}{11}\)
F = \(\sqrt{x}\) + 3
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.2 Question 25

Question 26.
USING STRUCTURE
Fill in the table so it represents a linear function.
Big Ideas Math Algebra 1 Solutions Chapter 3 Graphing Linear Functions 56
Answer:
The given table is:
Big Ideas Math Algebra 1 Solutions Chapter 3 Graphing Linear Functions 56
From the above table,
we can observe that the difference between all the values of x is 5 which is a constant
Now,
To find the difference between all the values of y which is a constant, we have to use the trial and error method.
Now,
If we add +1 to -1 and continue adding +1 to all the values of y, then
-1 + 1 = 0
0 + 1 = 1
1 + 1 = 2
2 + 1 = 3
But the last value is: 11
Now,
If we add +2 to -1 and continue adding +2 to all the values of y, then
-1 + 2 = 1
1 + 2 = 3
3 + 2 = 5
5 + 2 = 7
But the last value is: 11
Now,
If we add +3 to -1 and continue adding +3 to all the values of y, then
-1 + 3 = 2
2 + 3 = 5
5 + 3 = 8
8 + 3 = 11
The last value is also: 11
Hnece, from the above,
We can conclude that we have to add +3 to make all the values of y constant so that the given table represents a linear function
The completed table is:

In Exercises 27 and 28, find the domain of the function represented by the graph. Determine whether the domain is discrete or continuous. Explain.

Question 27.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 57
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.2 Question 27

Question 28.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 58
Answer:
The given graph is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 58
We know that,
The domain is defined as the range of all the values of x
So,
From the above graph,
The domain is: 0, 1, 2, 3, 4, 5, 6, and 7
Hence, from the above,
We can conclude that the domain of the given graph is: 0, 1, 2, 3, 4, 5, 6, and 7

In Exercises 29–32, determine whether the domain is discrete or continuous. Explain.

Question 29.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 59
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.2 Question 29

Question 30.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 60
Answer:
The given table is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 60
From the given table,
The difference between the values of x is 1 which is constant throughout all of the values of x
The difference between the values of y is 3 which is constant throughout all of the values of y
Hence,
Since the difference is constant for both the values of x and y,
The domain of the given table is continuous

Question 31.

Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 61
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.2 Question 31

Question 32.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 62
Answer:
The given table is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 62
From the above table,
The difference between all the values of x is 1 which is a constant
The difference between all the values of y is 4 which is a constant
Hence,
Since the difference between all the values of x and y is constant,
The given function is a linear function
The domain of the given function is continuous

ERROR ANALYSIS
In Exercises 33 and 34, describe and correct the error in the statement about the domain. 33. xy214324682.5 is in the domain.

Question 33.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 63
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.2 Question 33

Question 34.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 64
Answer:
We know that,
The domain is the range of all the values of x
Now,
From the given graph,
The domain of the given function is: 0, 1, 2, 3, 4, 5, and 6
From the given graph,
We can observe that the domain of the given graph is continuous because there are not any unconnected points in the graph

Question 35.
MODELING WITH MATHEMATICS
The linear function m = 55 – 8.5b represents the amount m (in dollars) of money that you have after buying b books.
a. Find the domain of the function. Is the domain discrete or continuous? Explain.
b. Graph the function using its domain.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 65
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.2 Question 35.1
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.2 Question 35.2

Question 36.
MODELING WITH MATHEMATICS
The number y of calories burned after x hours of rock climbing is represented by the linear function y = 650x.
a. Find the domain of the function. Is the domain discrete or continuous? Explain.
b. Graph the function using its domain.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 65.1
Answer:
a.
It is given that
The number y of calories burned after x hours of rock climbing is represented by the linear function
y = 650x.
It is given that x is the number of hours
Hence,
The domain of the given function is:
1 ≤ x ≤ 24
Since, the domain of the function is inequality,
The domain of the function is continuous

b.
The given function is:
y = 650x
From part (a),
The domain of the function is: 1 ≤ x ≤ 24
So,
y = 650 (1) = 650
y = 650 (2) = 1300
y = 650 (3) = 1950
.
.
.
.
y = 650 (12) = 7800
Hence,
The representation of the function with the domain in the coordinate plane is:

Question 37.
MODELING WITH MATHEMATICS
You are researching the speed of sound waves in dry air at 86°F. The table shows the distances d (in miles) sound waves travel in t seconds.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 66
a. Does this situation represent a linear function? Explain.
b. Find the domain of the function. Is the domain discrete or continuous? Explain.
c. Graph the function using its domain.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.2 Question 37

Question 38.
MODELING WITH MATHEMATICS
The function y = 30 + 5x represents the cost y (in dollars) of having your dog groomed and buying x extra services.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 67
a. Does this situation represent a linear function? Explain.
Answer:
The given function is:
y = 30 + 5x
We know that,
The standard representation of the linear function is:
y = mx + c
Compare the given function with the standard representation
Hence, from the above,
We can conclude that the given situation represents a linear function

b. Find the domain of the function. Is the domain discrete or continuous? Explain.
Answer:
The given function is:
y = 30 + 5x
Where,
y is the amount in dollars
x is the cost of extra grooming services
From the above,
The maximum number of given extra grooming services is: 5
So,
We can use extra grooming service or not
Hence, from the above,
We can conclude that
The domain of the given function is: 0 ≤ x ≤ 5
Hence,
The domain of the given function is continuous

c. Graph the function using its domain.
Answer:
The given function is:
y = 30 + 5x
We know that,
The domain of the function is: 0 ≤ x ≤ 5
So,
y = 30 + 5(0) = 30
y = 30 + 5 (1) = 35
y = 30 + 5(2) = 40
y = 30 + 5 (3) = 45
y = 30 + 5 (4) = 50
y = 30 + 5 (5) = 55
Hence,
The representation of the function using its domain in the coordinate plane is:

WRITING In Exercises 39–42, write a real-life problem to fit the data shown in the graph. Determine whether the domain of the function is discrete or continuous. Explain.

Question 39.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 68
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.2 Question 39

Question 40.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 69
Answer:
The given graph is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 69
From the above graph,
The real-life situation is:
The temperature of a country in the winter season falls by 2°C
The domain of the graph is continuous as there are not any unconnected points

Question 41.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 70
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.2 Question 41

Question 42.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 71
Answer:
The given graph is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 71
From the above graph,
The real-life situation is:
The number of ants in a colony increase by 2 times per day. So, the number of ants increases by how many times in 5 days?
The domain of the given graph is continuous since there is not any unconnected point in the graph

Question 43.
USING STRUCTURE
The table shows your earnings y (in dollars) for working x hours.
a. What is the missing y-value that makes the table represent a linear function?
b. What is your hourly pay rate?
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 72
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.2 Question 43

Question 44.
MAKING AN ARGUMENT
The linear function d = 50t represents the distance d (in miles) Car A is from a car rental store after t hours. The table shows the distances Car B is from the rental store.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 73
a. Does the table represent a linear or nonlinear function? Explain.
Answer:
The given table is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 73
From the above table,
The values of x increases at a constant rate of 2 but the value of y increases by 120 1st time and by 130 2nd time
Hence,
We can observe that for the constant difference of the values of x, there is no constant difference of the values of y
Hence, from the above,
We can conclude that the given table is a linear function

b. Your friend claims Car B is moving at a faster rate. Is your friend correct? Explain.
Answer:
It is given that
The function represented by car A is:
d = 50t
The function represented by car B is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 73
From the above functions,
We can say that the distance traveled by car A increases at a constant rate whereas the distance traveled by car B increases at a random rate
Hence, from the above
We can conclude that the car B is moving at a faster rate when we observe the above table

MATHEMATICAL CONNECTIONS
In Exercises 45–48, tell whether the volume of the solid is a linear or nonlinear function of the missing dimension(s). Explain.

Question 45.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 74
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.2 Question 45

Question 46.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 74.1
Answer:
The given figure is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 74.1
From the above figure,
We can observe that the given figure is a prism
We know that,
The volume of a prism = Area × height
= Length × Width × Height
So,
The volume of a prism (V) =3 × b × 4
V = 12b
Compare the above volume with y = mx + c
So,
V = 12b + 0
Hence, from the above,
We can conclude that the equation represents a Linear function

Question 47.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 74.2
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.2 Question 47

Question 48.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 74.3
Answer:
The given figure is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 74.3
From the above figure,
We can observe that the given figure is a cone
We know that,
The volume of a cone = \(\frac{1}{3}\) πr²h
So,
The volume of a cone (V) = \(\frac{1}{3}\) × \(\frac{22}{7}\) × r² × 15
V = 770r²
Compare the above equation with
y = mx + c
But, the given equation is not in the form of y = mx + c
Hence, from the above,
We can conclude that the equation represents a non-linear function

Question 49.
REASONING
A water company fills two different-sized jugs. The first jug can hold x gallons of water. The second jug can hold y gallons of water. The company fills A jugs of the first size and B jugs of the second size. What does each expression represent? Does each expression represent a set of discrete or continuous values?
a. x + y
b. A + B
c. Ax
d. Ax + By
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 75
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.2 Question 49

Question 50.
THOUGHT-PROVOKING
You go to a farmer’s market to buy tomatoes. Graph a function that represents the cost of buying tomatoes. Explain your reasoning.
Answer:
It is given that you go to a farmer’s market to buy tomatoes.
So,
To draw the graph that represents the cost of buying tomatoes,
The required relation is:
Cost of tomatoes ∝ Quantity or weight of tomatoes
We know that,
∝ represents the direct relation. In a graph, this relation can be represented in a straight line
Hence,
The representation of the relation of cost of buying tomatoes and weight of tomatoes is:

Question 51.
CLASSIFYING A FUNCTION
Is the function represented by the ordered pairs linear or nonlinear? Explain your reasoning.
(0, 2), (3, 14), (5, 22), (9, 38), (11, 46)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.2 Question 51

Question 52.
HOW DO YOU SEE IT?
You and your friend go running. The graph shows the distances you and your friend run.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 76
a. Describe your run and your friend’s run. Who runs at a constant rate? How do you know? Why might a person not run at a constant rate?
Answer:
The given graph is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 76
From the graph,
The running represented by you is a straight line
The running represented by your friend is not a straight line
We know that,
A straight line has a constant rate
Hence, from the above,
We can conclude that you run at a constant rate and your friend does not run at a constant rate

b. Find the domain of each function. Describe the domains using the context of the problem.
Answer:
From part (a),
The given graph is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 76
We know that,
The domain is defined as the range of the values of x
Hence, from the above,
We can conclude that
The domain of the function related to you is: 0 ≤ x ≤ 50
The domain of the function related to your friend is: 0 ≤ x ≤ 50

WRITING
In Exercises 53 and 54, describe a real-life situation for the constraints.

Question 53.
The function has at least one negative number in the domain. The domain is continuous.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.2 Question 53

Question 54.
The function gives at least one negative number as an output. The domain is discrete.
Answer:
When you go on a world tour and for some days, you stayed in Antarctica,
The temperatures in Antarctica is at a negative temperature around the year and only for some months in the year, the temperature will be positive

Maintaining Mathematical Proficiency

Tell whether x and y show direct variation. Explain your reasoning.

Question 55.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 77
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.2 Question 55

Question 56.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 78
Answer:
The given graph is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 78
From the graph,
We can observe that it is a straight line and passes through the origin
Hence, from the above,
We can conclude that x and y shows direct variation

Question 57.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 79
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.2 Question 57

Evaluate the expression when x = 2.

Question 58.
6x + 8
Answer:
The given expression is: 6x + 8
When x = 2,
6x + 8 = 6 (2) + 8
= 12 + 8 = 20
Hence,
The value of the expression when x = 2 is: 20

Question 59.
10 – 2x + 8
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.2 Question 59

Question 60.
4(x + 2 – 5x)
Answer:
The given expression is: 4 (x + 2 – 5x)
When x = 2,
4 (x + 2 – 5x) = 4 (2 + 2 – 5 (2) )
= 4 (4 – 10 )
= 4 (-6) = -24
Hence,
The value of the expression when x = 2 is: -24

Question 61.
\(\frac{x}{2}\) + 5x – 7
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.2 Question 61

Lesson 3.3 Function Notation

Essential Question

How can you use function notation to represent a function?
The notation f(x), called function notation, is another name for y. This notation is read as “the value of f at x” or “f of x.” The parentheses do not imply multiplication. You can use letters other than f to name a function. The letters g, h, j, and k are often used to name functions.

EXPLORATION 1
Matching Functions with Their Graphs
Work with a partner.
Match each function with its graph.
a. f (x) = 2x – 3
b. g(x) = -x + 2
c. h(x) = x2 – 1
d. j(x) = 2x2 – 3
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 80
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 81
Answer:
The given equations are:
a. f (x) = 2x – 3
b. g(x) = -x + 2
c. h(x) = x2 – 1
d. j(x) = 2x2 – 3
Now,
a.
The given equation is:
f(x) = 2x – 3
So,
The representation of the given equation in the coordinate plane is:

Hence, from the above,
We can conclude that graph B) matches this equation
b.
The given equation is:
f(x) = 2x – 3
So,
The representation of the given equation in the coordinate plane is:

Hence, from the above,
We can conclude that graph D) matches this equation
c.
The given equation is:
h(x) = x² – 1
So,
The representation of the given equation in the coordinate plane is:

Hence, from the above,
We can conclude that graph A) matches this equation
d.
The given equation is:
j(x) = 2x² – 3
So,
The representation of the given equation in the coordinate plane is:

Hence, from the above,
We can conclude that graph C) matches this equation

EXPLORATION 2
Evaluating a Function
Work with a partner.
Consider the function
f(x) = -x + 3.
Locate the points (x, f(x)) on the graph. Explain how you found each point.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 82
a. (-1, f(-1))
b. (0, f(0))
c. (1, f(1))
d. (2, f(2))
Answer:
The given function is:
f(x) = -x + 3
The graph for the given function is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 82
Now,
The simplified points are:
a.
(-1, f(-1)) = (-1, [-(-1) + 3]) = (-1, 4)
b.
(0, f(0)) = (0, [0 + 3]) = (0, 3)
c.
(1, f(1)) = (1, [-1 + 3]) = (1, 2)
d.
(2, f(2)) = (2, [-2 + 3]) = (2, 1)
So,
The simplified points are:
(-1, 4), (0, 3), (1, 2), (2, 1)
Hence,
The representation of the above points in the graph is:

Communicate Your Answer

Question 3.
How can you use function notation to represent a function? How are standard notation and function notation similar? How are they different?
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 82.1
Answer:
“Function notation” is a simpler method of describing a function without a lengthy written explanation. The most frequently used function notation is f(x) which is read as “f” of “x”.
The standard notation and function notation are similar in the way of simplification
Difference betwwen Function notataion and standard notation:
The representation of function notation largely depends on the number of variables present in the function
Ex:
f(x,y) = 2xy + 3
f(x) = x + 3
The representation of the standard notation does not depend on the number of variables present in the equation.

3.3 Lesson

Monitoring Progress

Evaluate the function when x = −4, 0, and 3.

Question 1.
f(x) = 2x – 5
Answer:
The given function is:
f(x) = 2x – 5
Now,
When x = -4,
f(-4) = 2 (-4) – 5 = -8 – 5 = -13
When x = 0,
f(0) = 2 (0) – 5 = 0 – 5 = -5
When x = 3,
f(3) = 2 (3) – 5 = 6 – 5 = 1
Hence, from the above,
We can conclude that the values of f(x) when x = -4, 0, 3 are: -13, -5, and 1

Question 2.
g(x) = -x – 1
Answer:
The given function is:
g(x) = -x – 1
Now,
When x = -4,
g(-4) = -[-4] – 1 = 4 – 1 = 3
When x = 0,
g(0) = 0 – 1 = -1
When x = 3,
g(3) = -3 – 1 = -4
Hence, from the above,
We can conclude that the values of g(x) when x = -4, 0, 3 are: 3, -1, and -4

Question 3.
WHAT IF? In Example 2, let f(t) be the outside temperature (°F) t hours after 9 A.M. Explain the meaning of each statement.
a. f(4) = 75
b. f(m) = 70
c. f(2) = f(9)
d. f(6) > f(0)

Monitoring Progress

Find the value of x so that the function has the given value.

Question 4.
f(x) = 6x + 9; f(x) = 21
Answer:
The value of x is: 2

Explanation:
The give function is:
f(x) = 6x + 9 with f(x) = 21
So,
21 = 6x + 9
6x = 21 – 9
6x = 12
x = 12 / 6
x = 2
Hence, from the above,
We can conclude that the value of x in the given function is: 2

Question 5.
g(x) = \(-\frac{1}{2}\)x + 3; g(x) = -1
Answer:
The value of x is: 8

Explanation:
The given function is:
g(x) = \(-\frac{1}{2}\)x + 3 with g(x) = -1
So,
-1 = \(-\frac{1}{2}\)x + 3
\(-\frac{1}{2}\)x = -1 – 3
\(-\frac{1}{2}\)x = -4
–\(\frac{1}{2}\)x = -4
\(\frac{1}{2}\)x = 4
x = 2(4)
x = 8
Hence, from the above,
We can conclude that the value of x in the given function is: 8

Graph the linear function.

Question 6.
f(x) = 3x – 2
Answer:
The given function is:
f(x) = 3x – 2
Now,
put the values -2, -1, 0, 1, 2 in the place of x and find the values of f(x) toplot a graph [ Remember you can take any value and any number of values]
So,
f(-2) = 3 (-2) – 2 = -6 – 2 = -8
f(-1) = 3(-1) – 2 = -3 – 2 = -5
So,
The completed table for the given function is:

Hence,
The representation of the given function in the coordinate plane is:

Question 7.
g(x) = -x + 4
Answer:
The given function is:
g(x) = -x + 4
Now,
put the values -2, -1, 0, 1, 2 in the place of x and find the values of f(x) toplot a graph [ Remember you can take any value and any number of values]
So,
g(-2) = -[-2] + 4 = 2 + 4 = 6
g(-1) = -[-1] + 4 = 4 + 1 = 5
So,
The completed table for the given function is:

Hence,
The representation of the given function in the coordinate plane is:

Question 8.
h(x) = \(-\frac{3}{4}\)x – 1
Answer:
The given function is:
h(x) = \(-\frac{3}{4}\)x – 1
Now,
put the values -2, -1, 0, 1, 2 in the place of x and find the values of f(x) toplot a graph [ Remember you can take any value and any number of values]
So,
h(-2) = \(-\frac{3}{4}\) (-2) – 1
= \(\frac{3}{2}\) – 1
= \(\frac{1}{2}\)
h(-1) = \(-\frac{3}{4}\) (-1) – 1
= \(\frac{3}{4}\) – 1
= \(-\frac{1}{4}\)
So,
The completed table for the given function is:

Hence,
The representation of the given function in the coordinate plane is:

Question 9.
WHAT IF?
Let f(x) = 250 – 75x represent the second flight, where f(x) is the number of miles the helicopter is from its destination after x hours. Which flight takes less time? Explain.

Function Notation 3.3 Exercises

Vocabulary and Core Concept Check

Question 1.
COMPLETE THE SENTENCE
When you write the function y = 2x + 10 as f(x) = 2x + 10, you are using ______________.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.3 Question 1

Question 2.
REASONING
Your height can be represented by a function h, where the input is your age. What does h(14) represent?
Answer:
It is given that your height can be represented by a function h, where the input is your age
So,
h (14) means you are 14 years old

Monitoring Progress and Modeling with Mathematics

In Exercises 3–10, evaluate the function when x = –2, 0, and 5.

Question 3.
f(x) = x + 6
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.3 Question 3

Question 4.
g(x) = 3x
Answer:
The given function is:
g (x) = 3x
When x = -2,
g (-2) = 3 (-2) = -6
When x = 0,
g (0) = 3 (0) = 0
When x = 5,
g (5) = 3 (5) = 15
Hence, from the above,
We can conclude that the values of g (x) when x = -2, 0, 5 is: -6, 0, and 15

Question 5.
h(x) = -2x + 9
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.3 Question 5

Question 6.
r(x) = -x – 7
Answer:
The given function is:
r (x) = -x – 7
When x = -2,
r (-2) = -[-2] – 7
= 2 – 7 = -5
When x = 0,
r (0) = 0 – 7 = -7
When x = 5,
r (5) = -5 – 7 = -12
Hence, from the above,
We can conclude taht the values of r (x) when x = 0, -2, 5 is: -5, -7, and -12

Question 7.
p(x) = -3 + 4x
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.3 Question 7

Question 8.
b(x) = 18 – 0.5x
Answer:
The given function is:
b (x) = 18 – 0.5x
When x = -2,
b (-2) = 18 – 0.5 (-2)
= 18 + 1 = 19
When x = 0,
b (0) = 18 – 0.5 (0) = 18
When x = 5,
b (5) = 18 – 0.5 (5)
18 – 2.5 = 15.5
Hence, from the above,
We can conclude that the values of b (x) when x = 0, -2, 5 is: 19, 18, and 15.5

Question 9.
v(x) = 12 – 2x – 5
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.3 Question 9

Question 10.
n(x) = -1 – x + 4
Answer:
The given function is:
n (x) = -1 – x + 4
When x = -2,
n (-2) = -1 – [-2] + 4
= -1 + 2 + 4
= 6 – 1
= 5
When x = 0,
n (0) = -1 – 0 + 4
= -1 + 4 = 3
When x = 2,
n (2) = -1 – 2 + 4
= 4 – 3
= 1
Hence, from the above,
We can conclude that the values of n (x) when x = -2, 0, 5 is: 5, 3, and 1

Question 11.
INTERPRETING FUNCTION NOTATION
Let c(t) be the number of customers in a restaurant t hours after 8 A.M. Explain the meaning of each statement.
a. c(0) = 0
b. c(3) = c(8)
c. c(n) = 29
d. c(13) < c(12)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.3 Question 11

Question 12.
INTERPRETING FUNCTION NOTATION
Let H(x) be the percent of U.S. households with Internet use x years after 1980. Explain the meaning of each statement.
a. H(23) = 55
b. H(4) = k
c. H(27) ≥ 61
d. H(17) + H(21) ≈ H(29)
Answer:
It is given that H (x) is the percent of U.S households with internet use x years after 1980
Now,
a.
The given function is:
H (23) = 55
The meaning of the above function is:
23 years after 1980, 55% of U.S households will be using the internet

b.
The given function is:
H(4) = k
The meaning of the above function is:
4 years after 1980, k% of U.S households will be using the internet

c.
The given function is:
H(27) ≥ 61
The meaning of the above function is:
27 years after 1980, more than 61% of U.S households will be using the internet

d.
The given function is:
H(17) + H(21) ≈ H(29)
The meaning of the above function is:
The percentage of U.S households using the internet after 29 years is equal to the sum of percentage of household using the internet after 17 years and 21 years

In Exercises 13–18, find the value of x so that the function has the given value.  

Question 13.
h(x) = -7x; h(x) = 63
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.3 Question 13

Question 14.
t(x) = 3x; t(x) = 24
Answer:
The given function is:
t (x) = 3x with t (x) = 24
So,
24 = 3x
x = 24 / 3
x = 8
Hence, from the above,
We can conclude that the value of the given function is: 8

Question 15.
m(x) = 4x + 15; m(x) = 7
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.3 Question 15

Question 16.
k(x) = 6x – 12; k(x) = 18
Answer:
The given function is:
k (x) = 6x – 12 with k (x) = 18
So,
6x – 12 = 18
6x = 12 + 18
6x = 30
x = 30 / 6
x = 5
Hence, from the above,
We can conclude that the value of the given function is: 5

Question 17.
q(x) = \(\frac{1}{2}\)x – 3; q(x) = -4
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.3 Question 17

Question 18.
j(x) = –\(\frac{4}{5}\)x + 7; j(x) = -5
Answer:
The given function is:
j (x) = —\(\frac{4}{5}\)x + 7 with j (x) = -5
So,
-5 = –\(\frac{4}{5}\)x + 7
-5 – 7 = –\(\frac{4}{5}\)x
–\(\frac{4}{5}\)x = -12
\(\frac{4}{5}\)x = 12
x = 12 × –\(\frac{5}{4}\)
x = 15
Hence, from the above,
We can conclude that the value of the given function is: 15

In Exercises 19 and 20, find the value of x so that f(x) = 7. 

Question 19.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 83
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.3 Question 19

Question 20.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 84
Answer:
The given function is:
f (x) = 7
So,
We know that,
y = f (x)
So,
From the above,
We can say that the value of y is: 7
Now,
To find the value of x, observe the location corresponding to the value of y
So,
The value of x is: -2
Hence,
The point of the graph where f (x) = 7 is: (-2, 7)

Question 21.
MODELING WITH MATHEMATICS
The function C(x) = 17.5x – 10 represents the cost (in dollars) of buying x tickets to the orchestra with a $10 coupon.
a. How much does it cost to buy five tickets?
b. How many tickets can you buy for $130?
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.3 Question 21

Question 22.
MODELING WITH MATHEMATICS
The function d(t) = 300,000t represents the distance (in kilometers) that light travels in t seconds.
a. How far does light travel in 15 seconds?
b. How long does it take light to travel 12 million kilometers?
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 85
Answer:
The given function is:
d (t) = 300,000t
represents the distance in kilometers that travels in t seconds
a.
The distance traveled by light in 15 seconds is:
d (15) = 300,000 (15)
= 4,500,000 kilometers
Hence, from the above,
We can conclude that the distance traveled by light in 15 seconds is: 4,500,000 kilometers

b.
It is given that the total distance traveled by light is 12 million kilometers
So,
d (t) = 12 million kilometers
So,
12,000,000 = 300,000 (t)
t = 12,000,000 / 300,000
t = 40 seconds
Hence, from the above,
We can conclude that the time taken by light to travel 12 million kilometers is: 40 seconds

In Exercises 23–28, graph the linear function. 

Question 23.
p(x) = 4x
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.3 Question 23

Question 24.
h(x) = -5
Answer:
The given function is:
h (x) = -5
Now,
put the values -2, -1, 0, 1, 2 in the place of x and find the values of f(x) to plot a graph [ Remember you can take any value and any number of values]
So,
The completed table for the given function is:

Hence,
The representation of the given function in the coordinate plane is:

Question 25.
d(x) = \(-\frac{1}{2} x\) – 3
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.3 Question 25

Question 26.
w(x) = \(\frac{3}{5} x\) + 2
Answer:
The given function is:
w (x) = \(\frac{3}{5} x\) + 2
So,
w (x) = \(\frac{3}{5}\) x + 2
Now,
put the values -2, -1, 0, 1, 2 in the place of x and find the values of f(x) to plot a graph [ Remember you can take any value and any number of values]
So,
The completed table for the given function is:

Hence,
The representation of the given function in the coordinate plane is:

Question 27.
g(x) = -4 + 7x
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.3 Question 27

Question 28.
f(x) = 3 – 6x
Answer:
The given function is:
f (x) = 3 – 6x
Now,
put the values -2, -1, 0, 1, 2 in the place of x and find the values of f(x) to plot a graph [ Remember you can take any value and any number of values]
So,
The completed table for the given function is:

Hence,
The representation of the given function in the coordinate plane is:

Question 29.
PROBLEM-SOLVING
The graph shows the percent p(in decimal form) of battery power remaining in a laptop computer after t hours of use. A tablet computer initially has 75% of its battery power remaining and loses 12.5% per hour. Which computer’s battery will last longer? Explain.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 86
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.3 Question 29.1
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.3 Question 29.2

Question 30.
PROBLEM-SOLVING
The function C(x) = 25x + 50 represents the labor cost (in dollars) for Certified Remodeling to build a deck, where x is the number of hours of labor. The table shows sample labor costs from its main competitor, Master Remodeling. The deck is estimated to take 8 hours of labor. Which company would you hire? Explain.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 87
Answer:
It is given that the given function
C(x) = 25x + 50
represents the labor cost (in dollars) for certified Remodeling to build a deck
Where,
x is the number of hours of labor.
The given table is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 87
The given table shows the sample labor costs from its main competitor, Master Remodeling.
It is also given that the deck is estimated to take 8 hours of labor.
Now,
From the given table,
We can observe that the competitor ‘Master Remodeling’ completes the deck in 6 hours whereas for certified Remodeling, it will take 8 hours of labor to complete a deck thereby increases the cost. i.e., the avlue of C(x) also increases when compared to the Master Remodeling
Hence, from the above,
We can conclude that we choose “Master Remodeling” company

Question 31.
MAKING AN ARGUMENT
Let P(x) be the number of people in the U.S. who own a cell phone x years after 1990. Your friend says that P(x + 1) > P(x) for any x because x + 1 is always greater than x. Is your friend correct? Explain.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.3 Question 31

Question 32.
THOUGHT-PROVOKING
Let B(t) be your bank account balance after t days. Describe a situation in which B(0) < B(4) < B(2).
Answer:
The given relation is:
B (0) < B (4) < B (2)
Let us consider
Sunday – Day 0
Monday -Day 1
Tuesday -Day 2
Wednesday – Day 3
Thursday – Day 4
Friday – Day 5
Saturday – Day 6
Now,
By using the above relation,
The situation we can assume is:
A man named A works at a company where he receives his salary every Tuesday (Day 2) of a normal week. He then spends some of his salary paying bills on Thursday (Day 4). On Sunday, he decides to spend all of his remaining salary on food, groceries, and transportation allowance for the following week

Question 33.
MATHEMATICAL CONNECTIONS
Rewrite each geometry formula using function notation. Evaluate each function when r = 5 feet. Then explain the meaning of the result.
a. Diameter, d = 2r
b. Area, A = πr2
c. Circumference, C = 2πr
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 88
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.3 Question 33

Question 34.
HOW DO YOU SEE IT?
The function y = A(x) represents the attendance at a high school x weeks after a flu outbreak. The graph of the function is shown.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 89
a. What happens to the school’s attendance after the flu outbreak?
Answer:
The given graph is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 89
From the above graph,
We can observe that after the flu outbreak, the school’s attendance first decreased and then again steadily increased

b. Estimate A(13) and explain its meaning.
Answer:
The given graph is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 89
Now,
We know that
The function notation and the standard notations are similar
So,
y = A (x)
So,
y = A (13)
So,
The value of x is: 13
To find the value of y or the given function, observe the graph for the value of y corresponding to the value of x i.e., 13
So,
From the graph,
y = 430 [Approximately]
Hence, from the above,
We can conclude that
The estimation of A(13) is:
A (13) = 430 {Approximately]

c. Use the graph to estimate the solution(s) of the equation A(x) = 400. Explain the meaning of the solution(s).
Answer:
The given graph is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 89
The given equation is:
A (x) = 400
From the above graph,
For the solution of the given equation i.e., the value of x,
Observe the location of 400 in the graph and its corresponding x-value
Hence,
From the graph,
We can observe that the value of x is: 1
Hence, from the above,
We can conclude that the solution of the given equation is: 1

d. What was the least attendance? When did that occur?
Answer:
The given graph is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 89
From the above graph,
We can observe that there is deep depreciation in the attendance i.e., least attendance of the students in a week
So,
By observing the graph,
We can say that
The least attendance of the students occurs in the 4th week
The least attendance of the students is: 350
Hence, from the above,
We can conclude that
The week that has the least attendance is: 4th week
The number of students of the least attendance is: 350students

e. How many students do you think are enrolled at this high school? Explain your reasoning.
Answer:
The given graph is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 89
From the given graph,
We can observe that the y-axis represents the number of students enrolled in the high school
So,
The highest number on the y-axis represents the total number of students enrolled in the high school
Hence, from the above,
We can conclude that the total number of students in the high school is: 450 students

Question 35.
INTERPRETING FUNCTION NOTATION
Let f be a function. Use each statement to find the coordinates of a point on the graph of f.
a. f(5) is equal to 9.
b. A solution of the equation f(n) = -3 is 5.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.3 Question 35

Question 36.
REASONING
Given a function f, tell whether the statement f(a + b) = f(a) + f(b) is true or false for all inputs a and b. If it is false, explain why.
Answer:
Let the given function is:
f or f(x) = mx + c
Now,
Let the values of a and b be integers
So,
Let,
a = 0 and b = 1
So,
f( a + b ) = f (1)
So,
f(1) = m + c
Now,
f(a) = f (0) = c
f (b) = f(1) = m + c
Hence, from the above,
We can conclude that
f (a + b) is not equal to f (a) + f(b)

Maintaining Mathematical Proficiency

Solve the inequality. Graph the solution. (Section 2.5)

Question 37.
-2 ≤ x – 11 ≤ 6
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.3 Question 37

Question 38.
5a < -35 or a – 14 > 1
Answer:
The given inequality is:
5a < -35 or a – 14 > 1
So,
a < -35 / 5 or a > 1 + 14
a < -7 or a > 15
Hence,
The solutions of the given inequality are:
a < -7 or a > 15
The representation of the solution of the given inequality in the graph is:

Question 39.
-16 < 6k + 2 < 0
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.3 Question 39

Question 40.
2d + 7 < -9 or 4d – 1 > -3
Answer:
The given inequality is:
2d + 7 < -9 or 4d – 1 > -3
2d < -9 – 7 or 4d > -3 + 1
2d < -16 or 4d > -2
d < -16 / 2 or d > -2 / 4
d < -8 or d > -1 /  2
Hence,
The solutions of the given inequality are:
d < -8 or d > -1 / 2
The representation of the solutions of the given inequality in the graph is:

Question 41.
5 ≤ 3y + 8 < 17
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.3 Question 41

Question 42.
4v + 9 ≤ 5 or -3v ≥ -6
Answer:
The given inequality is:
4v + 9 ≤5 or -3v ≥ -6
So,
4v ≤ 5 – 9 or 3v ≥ 6
4v ≤ -4 or v ≥ 6 / 3
v ≤ -4 / 4 or v ≥ 2
v ≤ -1 or v ≥ 2
Hence,
The solutions of the given inequality are:
v ≤ -1 or v ≥ 2
The representation of the solutions of the given inequality is:

Graphing Linear Functions Study Skills: Staying Focused During Class

Core Vocabulary
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 90

Core Concepts
Section 3.1
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 91

Section 3.2
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 92

Section 3.3
Using FunctionNotation, p. 122

Mathematical Practices

Question 1.
How can you use technology to confirm your answers in Exercises 40–43 on page 110?
Answer:
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
By using the above properties,
We can find the domain and the range of the given Exercises
Hence,
We can confirm the answers in Exercises 40 – 43 on page 110

Question 2.
How can you use patterns to solve Exercise 43 on page 119?
Answer:
In Exercise 43 on page 119,
We can observe from the table that the difference between the values of x and y is constant
So,
By using the above property, we can find the constant difference between the values of x and y to complete the pattern

Question 3.
How can you make sense of the quantities in the function in Exercise 21 on page 125?
Answer:
In Exercise 21 on page 125,
Compare the given function with the standard representation of the linear function y = mx + c
So,
In the above function,
We know that,
m is the cost of x tickets in dollars
c is the constant
f (x) or y is the function notation of Cost function

Study Skills

Staying Focused during Class

As soon as class starts, quickly review your notes from the previous class and start thinking about math.
Repeat what you are writing in your head.
When a particular topic is difficult, ask for another example.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 94

Graphing Linear Functions 3.1 – 3.3 Quiz

Determine whether the relation is a function. Explain. (Section 3.1)

Question 1.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 95
Answer:
The given table is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 95
We know that for a relationship to be a function,
Each input value has to match with each output value
Hence,
From the above table,
We can conclude that the given table is a function since each input value matches with each output value

Question 2.
(-10, 2), (-8, 3), (-6, 5), (-8, 8), (-10, 6)
Answer:
The given ordered pairs are:
(-10, 2), (-8, 3), (-6, 5), (-8, 8), (-10, 6)
We know that,
A relation is said to be a function if each input matches with only 1 output
So,
By observing the above-ordered pairs,
We can say that the input -8 comes 2 times
Hence, from the above,
We can conclude that the given ordered pairs do not represent a function

Find the domain and range of the function represented by the graph. (Section 3.1)

Question 3.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 96
Answer:
The given graph is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 96
We know that,
The domain is the set of all x-values
The range is the set of all y-values
So,
The domain of the given graph is: {0, 1, 2, 3, 4}
The range of the given graph is: {1, -1, -3}

Question 4.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 97
Answer:
The given graph is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 97

We know that,
The domain is the set of all x-values
The range is the set of all y-values
So,
The domain of the given graph is: {-1, -2, 1, 2}
The range of the given graph is: {-2, -1, 0, 1, 2}

Question 5.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 98
Answer:
The given graph is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 98
We know that,
The domain is the set of all x-values
The range is the set of all y-values
So,
The domain of the given graph is: {-3, -2, -1, 1, 2, 3}
The range of the given graph is: {-1, 0, 1, 2, 3}

Determine whether the graph, table, or equation represents a linear or nonlinear function. Explain. (Section 3.2)

Question 6.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 99
Answer:
The given graph is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 99
We know that,
A linear function must always represent a straight line irrespective of the straight line passes through the origin or passes through any other point
Hence, from the above,
We can conclude that the given graph is a linear function

Question 7.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 100
Answer:
The given table is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 100
The representation of the values of x and y in the table in the form of ordered pairs is:
(-5, 3), (0, 7), and (5, 10)
Represent the ordered pairs in the coordinate plane
The representation of the ordered pairs in the coordinate plane is:

From the above points, we can say it forms a straight line
A linear function must always represent a straight line irrespective of the straight line passes through the origin or passes through any other point
Hence, from the above,
We can conclude that the given table represents a linear function

Question 8.
y = x(2 – x)
Answer:
The given function is:
y = x (2 – x)
So,
y = 2 (x) – x (x)
y = 2x – x²
Compare the above function with the standard representation of the linear function
The standard representation of the linear function is:
y = mx + c
Hence, from the above comparison,
We can conclude that the given function is a non-linear function

Determine whether the domain is discrete or continuous. Explain. (Section 3.2)

Question 9.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 101
Answer:
The given table is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 101
From the above table,
The values of x have specific values of y
We know that,
The domain is defined as the set of all the values of x
Hence, from the above,
We can conclude that the domain of the given table is discrete

Question 10.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 102
Answer:
The given table is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 102
From the above table
We can observe that the values of x have specific values of y
We know that,
The domain is defined as the set of all the values of x
Hence, from the above,
We can conclude that the domain of the given table is discrete

Question 11.
For w(x) = -2x + 7, find the value of x for which w(x) = -3. (Section 3.3)
Answer:
The given function is:
w (x) = -2x + 7 with w (x)  = -3
So,
-3 = -2x + 7
-2x = -3 – 7
-2x = -10
2x = 10
x = 10 / 5
x = 2
Hence, from the above,
We can conclude that the value of the given function is: 2

Graph the linear function. ( Section 3.3)

Question 12.
g(x) = x + 3
Answer:
The given function is:
g (x) = x + 3
Now,
put the values -2, -1, 0, 1, 2 in the place of x and find the values of f(x) toplot a graph [ Remember you can take any value and any number of values]
So,
The completed table for the given function is:

Hence,
The representation of the given function in the coordinate plane is:

Question 13.
p(x) = -3x – 1
Answer:
The given function is:
p (x) = -3x – 1
Now,
put the values -2, -1, 0, 1, 2 in the place of x and find the values of f(x) toplot a graph [ Remember you can take any value and any number of values]
So,
The completed table for the given function is:

Hence,
The representation of the given function in the coordinate plane is:

Question 14.
m(x) = \(\frac{2}{3}\)x
Answer:
The given function is:
m (x) = \(\frac{2}{3}\)x
Now,
put the values -2, -1, 0, 1, 2 in the place of x and find the values of f(x) to plot a graph [ Remember you can take any value and any number of values]
So,
The completed table for the given function is:

Hence,
The representation of the given function in the coordinate plane is:

Question 15.
The function m = 30 – 3r represents the amount m (in dollars) of money you have after renting r video games. (Section 3.1 and Section 3.2)
a. Identify the independent and dependent variables.
Answer:
The given function is:
m = 30 – 3r
We know that,
The independent variables are the values of x
The dependent variables are the values of y
Now,
Compare the given function with
y = mx + c
Hence,
The independent variable of the given function is: r
The dependent variable of the given function is: m

b. Find the domain and range of the function. Is the domain discrete or continuous? Explain.
Answer:
The given function is:
m = 30 – 3r
Now,
put the values -2, -1, 0, 1, 2 in the place of x and find the values of f(x) to plot a graph [ Remember you can take any value and any number of values]
So,
The completed table of the given function is:

We know that,
The domain is defined as the set of all the values of x
the range is defined as the set of all the values of y
Hence,
The domain of the given function is: {-2, -1, 0, 1, 2}
The range of the given function is: {36, 33, 30, 27, 24, 21}

c. Graph the function using its domain.
Answer:
The completed table for the given function from part (b) is:

Hence,
The representation of the given function using its domain in the coordinate plane is:

Question 16.
The function d(x) = 1375 – 110x represents the distance (in miles) a high-speed train is from its destination after x hours. (Section 3.3)
a. How far is the train from its destination after 8 hours?
Answer:
The given function is:
d (x) = 1375 – 110x
Where,
d (x) represents the distance (in miles)
x represents the time
Now,
It is given that we have to find the distance traveled by train after 8 hours i.e., the value of x is given
So,
d (x) = 1375 – 110 (8)
= 1375 – 880
= 495 miles
Hence, from the above,
We can conclude that the distance traveled by train after 8 hours is: 495 miles

b. How long does the train travel before reaching its destination?
Answer:
The given function is:
d (x) = 1375 – 110x
From part (a),
We find the value of d (x) as: 495 miles
So,
495 = 1375 – 110x
110x = 1375 – 495
110x = 880
x = 880 /110
x = 8 hours
Hence, from the above,
We can conclude that train travel for 8 hours before reaching its destination

Lesson 3.4 Graphing Linear Equations in Standard Form

Essential Question
How can you describe the graph of the equation Ax + By = C?
Answer:
When A and B are not both zero,
The graph of Ax + By = C is always a line.
Now,
Divide both sides by B
Because the form Ax + By = C can describe any line,
It is called the standard form of an equation for a line.

EXPLORATION 1
Using a Table to Plot Points
Work with a partner.
You sold a total of $16 worth of tickets to a fundraiser. You lost track of how many of each type of ticket you sold. Adult tickets are $4 each. Child tickets are $2 each.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 102.1
a. Let x represent the number of adult tickets. Let y represent the number of child tickets. Use the verbal model to write an equation that relates to x and y.
Answer:
It is given that you sold a total of $16 worth of tickets to a fundraiser. You lost track of how many of each type of ticket you sold. Adult tickets are $4 each. Child tickets are $2 each.
It is also given that
Let x be the number of adult tickets
Let y be the number of child tickets
So,
The total cost of tickets = (The cost of the child tickets + The cost of the adult tickets)
The total cost of tickets = (The number of children) × ( The cost of each child ticket ) + ( The numebr of adults ) × ( The cost of each adult ticket )
So,
16 = 2x + 4y
2 (x + 2y) = 16
x + 2y = 16 / 2
So,
x + 2y = 8
Hence, from the above,
We can conclude that the equation that relates x and y is:
x + 2y = 8

b. Copy and complete the table to show the different combinations of tickets you might have sold.
Answer:
The equation that represents the number of different tickets sold is:
x + 2y = 8
2y = 8 – x
y = \(\frac{8 – x}{2}\)
So,
The completed table that shows the different combinations of tickets you might have sold is:

c. Plot the points from the table. Describe the pattern formed by the points.
Answer:
The completed table that shows the different combinations of tickets you might have sold is from part (b) is:

The representation of the ordered pairs from the above table is:
(1, 4), (2, 3), (3, 3), (4, 2), (5, 2)
Hence,
The representation of the points from the table in the coordinate plane is:

Hence, from the graph,
We can observe that the pattern drawn by the points is a straight line

d. If you remember how many adult tickets you sold, can you determine how many child tickets you sold? Explain your reasoning.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 104
Answer:
The given function from part (a) is in the form of
y = mx + c
Where,
y is the number of adult tickets
x is the number of child tickets
So,
If we know the number of adult tickets you sold, i.e., the value of y, then we can find the number of child tickets, i.e., the value of x by putting the value of y in the equation
y = mx + c
Hence, from the above,
We can conclude that if we know the number of adult tickets you sold, then we can find the number of child tickets you sold

EXPLORATION 2
Rewriting and Graphing an Equation
Work with a partner.
You sold a total of $48 worth of cheese. You forgot how many pounds of each type of cheese you sold. Swiss cheese costs $8 per pound. Cheddar cheese costs $6 per pound.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 105
a. Let x represent the number of pounds of Swiss cheese. Let y represent the number of pounds of cheddar cheese. Use the verbal model to write an equation that relates to x and y.
Answer:
it is given that you sold a total of $48 worth of cheese. You forgot how many pounds of each type of cheese you sold. Swiss cheese costs $8 per pound. Cheddar cheese costs $6 per pound.
It is also given that,
x represents the number of pounds of Swiss cheese
y represents the number of pounds of Cheddar cheese
Now,
The total cost = ( The cost of Swiss Cheese ) + ( The cost of Cheddar Cheese )
The total cost = ( The number of pounds of Swiss cheese ) × ( The cost of Swiss cheese ) + ( The number of pounds of Cheddar cheese ) × ( The cost of Cheddar cheese )
48 = 8x + 6y
8x + 6y = 48
2 (4x + 3y) = 48
4x + 3y = 48 / 4
4x + 3y = 12
Hence, from the above,
We can conclude that the equation that relates both x and y is:
4x + 3y = 12

b. Solve the equation for y. Then use a graphing calculator to graph the equation. Given the real-life context of the problem, find the domain and range of the function.
Answer:
The equation that relates to x and y from part (a) is:
4x + 3y = 12
3y = 12 – 4x
y = \(\frac{12 – 4x}{3}\)
So,
The representation of the equation in the coordinate plane is:

We know that,
The domain is the set of all the values of x in the graph
The range is the set of all the values of y in the graph
Now,
From the graph,
The domain of the equation is: { -4, -3, -2, -1, 0, 1, 2, 3 }
The range of the given equation is: {-1, -2, -3, -4, -5, -6, -7, -8, -9, 5, 6, 7, 8, 9, 10}

c. The x-intercept of a graph is the x-coordinate of a point where the graph crosses the x-axis. The y-intercept of a graph is the y-coordinate of a point where the graph crosses the y-axis. Use the graph to determine the x- and y-intercepts.
Answer:
The given equation is:
y = \(\frac{12 – 4x}{3}\)
It is given that the x-intercept of a graph is the x-coordinate of a point where the graph crosses the x-axis. The y-intercept of a graph is the y-coordinate of a point where the graph crosses the y-axis.
Now,
The graph from part (b) is:

Hence,
By observing the graph,
We can conclude that
The x-intercept is: 3
The y-intercept is: 4

d. How could you use the equation you found in part (a) to determine the x- and y-intercepts? Explain your reasoning.
Answer:
The given equation is:
y = \(\frac{12 – 4x}{3}\)
We know that,
We can obtain the x-intercept by making y term zero
We can obtain the y-intercept making x term zero
So,
The x-intercept is:
0 = \(\frac{12 – 4x}{3}\)
12 – 4x = 0
4x = 12
x = 12 / 4
x = 3
Hence,
The x-intercept is: 3
The y-intercept is:
y = \(\frac{12 – 4(0)}{3}\)
y = 12 / 3
y = 4
Hence,
The y-intercept is: 4

e. Explain the meaning of the intercepts in the context of the problem.
Answer:
An intercept of any function is a point where the graph of the function crosses, or intercepts, the x-axis or y-axis.  When the linear function is used to represent a real-world situation, the intercepts have significant meaning in the context of the problem.

Communicate Your Answer

Question 3.
How can you describe the graph of the equation Ax + By = C?
Answer:
When A and B are not both zero,
The graph of Ax + By = C is always a line.
Now,
Divide both sides by B
Because the form Ax + By = C can describe any line,
It is called the standard form of an equation for a line.

Question 4.
Write a real-life problem that is similar to those shown in Explorations 1 and 2.
Answer:
The real-life problem that is similar to those in Explorations  1 and 2 is:
The age of A is twice the age of B and the sum of the twice the age of A and the age of B is 24

3.4 Lesson

Monitoring Progress

Question 1.
y = -2.5
Answer:
The given equation is:
y = -2.5
Hence,
The representation of the given equation in the coordinate plane is:

Question 2.
x = 5
Answer:
The given equation is:
x = 5
Hence,
The representation of the given equation in the coordinate plane is:

Use intercepts to graph the linear equation. Label the points corresponding to the intercepts.

Question 3.
2x – y = 4
Answer:
The given equation is:
2x – y = 4
Rewrite the given equation in the standard form
We know that,
The standard form of the linear equation is:
y = mx + c
So,
y = 2x – 4
To find the value of x-intercept, put y = 0
2x – 4 = 0
2x = 4
x = 4 / 2
x = 2
To find the value of y-intercept, put x = 0
y = 2(0) – 4
y = -4
Hence,
The representation of the given equation corresponding to the intercepts in the coordinate plane is:

Question 4.
x + 3y = -9
Answer:
The given equation is:
x + 3y = -9
Rewrite the given equation in the standard form
We know that,
The standard form of the linear equation is:
y = mx + c
So,
3y = -9 – x
y = (-9 -x ) / 3
Now,
To find the value of x-intercept, put y = 0
-9 – x = 0
x = -9
To find the value of y-intercept, put x = 0
y = (-9 – 0) / 3
t = -9 / 3
y = -3
Hence,
The representation of the given equation corresponding to the intercepts in the coordinate plane is:

Question 5.
WHAT IF? You decide to rent tables from a different company. The situation can be modeled by the equation 4x + 6y = 180, where x is the number of small tables and y is the number of large tables. Graph the equation and interpret the intercepts.
Answer:
It is given that you decide to rent tables from a different company.
The given equation corresponding to the above situation is given as:
4x + 6y = 180
Where,
x is the number of small tables
y is the number of large tables
Now,
Rewrite the given equation in the standard form of the linear equation
We know that,
The standard form of the linear equation is:
y = mx + c
So,
6y = 180 – 4x
y = (180 – 4x) / 6
Now,
To find the value of the x-intercept, put y = 0
180 – 4x = 0
4x = 180
x = 180 / 4
x = 45
To find the value of the y-intercept, put x = 0
y = (180 – 4(0) ) / 6
y = 180 / 6
y = 30
Hence,
The representation of the given equation corresponding to the intercepts in the coordinate plane is:

Graphing Linear Equations in Standard Form 3.4 Exercises

Vocabulary and Core Concept Check

Question 1.
WRITING
How are x-intercepts and y-intercepts alike? How are they different?
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.4 Question 1

Question 2.
WHICH ONE did DOESN’T BELONG?
Which point does not belong with the other three? Explain your reasoning.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 105.1
Answer:
The given points are:
a. (0, -3)
b. (0, 0)
c. (4, -3)
d. (4, 0)
From the above points,
We can observe that
(0, -3) belongs to the y-axis
(0, 0) belongs to the origin
(4, -3) does not belong either to the x-axis or y-axis
(4, 0) belongs to the x-axis
Hence, from the above,
We can conclude that (4, -3) does not belong with the other three

Monitoring Progress and Modeling with Mathematics

In Exercises 3–6, graph the linear equation.

Question 3.
x = 4
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.4 Question 3

Question 4.
y = 2
Answer:
The given equation is:
y = 2
Hence,
The representation of the given equation in the coordinate plane is:

Question 5.
y = -3
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.4 Question 5

Question 6.
x = -1
Answer:
The given equation is:
x = -1
Hence,
The representation of the given equation in the coordinate plane is:

In Exercises 7–12, find the x- and y-intercepts of the graph of the linear equation.

Question 7.
2x + 3y = 12
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.4 Question 7

Question 8.
3x + 6y = 24
Answer:
The given equation is:
3x + 6y = 24
Now,
To find the value of the x-intercept, put y = 0
So,
3x + 0 = 24
3x = 24
x = 24 / 3
x = 8
To find the value of the y-intercept, put x = 0
So,
0 + 6y = 24
6y = 24
y = 24 / 6
y = 4
Hence, from the above,
We can conclude that
The value of the x-intercept is: 8
The value of the y-intercept is: 4

Question 9.
-4x + 8y = -16
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.4 Question 9

Question 10.
-6x + 9y = -18
Answer:
The given equation is:
-6x + 9y = -18
Now,
To find the x-intercept, put y = 0
So,
-6x + 0 = -18
-6x = -18
6x = 18
x = 18 / 6
x = 3
To find the y-intercept, put x = 0
So,
0 + 9y = -18
9y = -18
y = -18 / 9
y = -2
Hence, from the above,
We can conclude that
The value of the x-intercept is: 3
The value of the y-intercept is: -2

Question 11.
3x – 6y = 2
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.4 Question 11

Question 12.
-x + 8y = 4
Answer:
The given equation is:
-x + 8y = 4
Now,
To find the x-intercept, put y = 0
So,
-x + 0 = 4
-x = 4
x = -4
To find the y-intercept, put x = 0
So,
0 + 8y = 4
8y = 4
y = 4 / 8
y = 1 / 2
Hence, from the above,
We can conclude that
The value of the x-intercept is: -4
The value of the y-intercept is: 1/2

In Exercises 13–22, use intercepts to graph the linear equation. Label the points corresponding to the intercepts.

Question 13.
5x + 3y = 30
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.4 Question 13

Question 14.
4x + 6y = 12
Answer:
The given equation is:
4x + 6y = 12
Now,
To find the x-intercept, put y = 0
So,
4x + 0 = 12
4x = 12
x = 12 / 4
x = 3
To find the y-intercept, put x = 0
0 + 6y = 12
6y = 12
y = 12 / 6
y = 2
Hence,
The representation of the given equation along with the intercepts in the coordinate plane is:

Question 15.
-12x + 3y = 24
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.4 Question 15

Question 16.
-2x + 6y = 18
Answer:
The given equation is:
-2x + 6y = 18
Now,
To find the x-intercept, put y = 0
So,
-2x + 0 = 18
-2x = 18
x = -18 / 2
x = -9
To find the y-intercept, put x = 0
So,
0 + 6y = 18
6y = 18
y = 18 / 6
y = 3
Hence,
The representation of the given equation along with the intercepts in the coordinate plane is:

Question 17.
-4x + 3y = -30
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.4 Question 17

Question 18.
-2x + 7y = -21
Answer:
The given equation is:
-2x + 7y = -21
Now,
To find the value of the x-intercept, put y = 0
-2x + 0 = -21
-2x = -21
2x = 21
x = 21/2
To find the value of the y-intercept, put x = 0
So,
0 + 7y = -21
7y = -21
y = -21 / 7
y = -3
Hence,
The representation of the given equation along with the intercepts in the coordinate plane is:

Question 19.
-x + 2y = 7
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.4 Question 19

Question 20.
3x – y = -5
Answer:
The given equation is:
3x – y = -5
Now,
To find the x-intercept, put y = 0
So,
3x – 0 = -5
3x =  -5
x = -5/3
To find the y-intercept, put x = 0
So,
0 – y = -5
-y = -5
y = 5
Hence,
The representation of the given equation along with the intercepts in the coordinate plane is:

Question 21.
–\(\frac{5}{2}\)x + y = 10
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.4 Question 21.1
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.4 Question 21.2

Question 22.
–\(\frac{1}{2}\)x + y = -4
Answer:
The given equation is:
–\(\frac{1}{2}\)x + y = -4
Now,
To find the x-intercept, put y =0
So,
–\(\frac{1}{2}\)x + 0 = -4
–\(\frac{1}{2}\)x = -4
-x = -4(2)
-x = -8
x = 8
To find the y-intercept, put x = 0
So,
0 + y = -4
y = -4
Hence,
The representation of the given equation along with the intercepts in the coordinate plane is:

Question 23.
MODELING WITH MATHEMATICS
A football team has an away game, and the bus breaks down. The coaches decide to drive the players to the game in cars and vans. Four players can ride in each car. Six players can ride in each van. There are 48 players on the team. The equation 4x + 6y = 48 models this situation, where x is the number of cars and y is the number of vans.
a. Graph the equation. Interpret the intercepts.
b. Find four possible solutions in the context of the problem.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.4 Question 23.1
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.4 Question 23.2
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.4 Question 23.3

Question 24.
MODELING WITH MATHEMATICS
You are ordering shirts for the math club at your school. Short-sleeved shirts cost $10 each. Long-sleeved shirts cost $12 each. You have a budget of $300 for the shirts. The equation 10x + 12y = 300 models the total cost, where x is the number of short-sleeved shirts and y is the number of long-sleeved shirts.
a. Graph the equation. Interpret the intercepts.
Answer:
It is given that you are ordering shirts for the math club at your school. Short-sleeved shirts cost $10 each. Long-sleeved shirts cost $12 each. You have a budget of $300 for the shirts.
The above situation is represented by the equation
10x + 12y = 300
where,
x is the number of short-sleeved shirts
y is the number of long-sleeved shirts.
Now,
The given equation is:
10x + 12y = 300
To find the x-intercept, put y =0
So,
10x + 12 (0) = 300
10x = 300
x = 300 / 10
x = 30
To find the y-intercept, put x = 0
So,
10 (0) + 12y = 300
12y = 300
y = 300 / 12
y = 25
Hence,
The representation of the given equation along with the intercepts in the coordinate plane is:

b. Twelve students decide they want short-sleeved shirts. How many long-sleeved shirts can you order?
Answer:
The given equation is:
10x + 12y = 300
where,
x is the number of short-sleeved shirts
y is the number of long-sleeved shirts.
It is given that 12 students decided they want short-sleeved shirts
So,
The number of short-sleeved shirts = 12
So,
10 (12) + 12y = 300
120 + 12y = 300
12y = 300 – 120
12y = 180
y = 180 / 12
y = 15
Hence, from the above,
We can conclude that the number of long-sleeved shirts is: 15

ERROR ANALYSIS
In Exercises 25 and 26, describe and correct the error in finding the intercepts of the graph of the equation.

Question 25.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 106
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.4 Question 25

Question 26.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 107
Answer:
The given equation is:
4x + 10y = 20
Now,
To find the value of the x-intercept, put y = 0
So,
4x + 10 (0) = 20
4x = 20
x = 20 / 4
x = 5
To find the value of the y-intercept, put x = 0
So,
4 (0) + 10y = 20
10y = 20
y = 20 / 10
y = 2
Hence, from the above,
We can conclude that
The x-intercept is: (5, 0)
The y-intercept is: (0, 2)

Question 27.
MAKING AN ARGUMENT
You overhear your friend explaining how to find intercepts to a classmate. Your friend says, “When you want to find the x-intercept, just substitute 0 for x and continue to solve the equation.” Is your friend’s explanation correct? Explain.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.4 Question 27

Question 28.
ANALYZING RELATIONSHIPS
You lose track of how many 2-point baskets and 3-point baskets a team makes in a basketball game. The team misses all the 1-point baskets and still scores 54 points. The equation 2x + 3y = 54 models the total points scored, where x is the number of 2-point baskets made and y is the number of 3-point baskets made.
a. Find and interpret the intercepts.
Answer:
It is given that you lose track of how many 2-point baskets and 3-point baskets a team makes in a basketball game.
It is also given that the team misses all the 1-point baskets and still scores 54 points.
The above situation can be described by the equation:
2x + 3y = 54
where,
x is the number of 2-point baskets made
y is the number of 3-point baskets made.
So,
To find the value of the x-intercept, put y = 0
So,
2x = 54
x = 54 / 2
x = 27
To find the value of the y-intercept, put x = 0
3y = 54
y = 54 / 3
y = 18
Hence, from the above,
We can conclude that
The value of the x-intercept is: 27
The value of the y-intercept is: 18

b. Can the number of 3-point baskets made be odd? Explain your reasoning.
Answer:
From part (a),
The number of 3-point baskets can be represented as: y
So,
To make the 3-point baskets odd, the multiples of 3 must be odd
But, in this scenario, it is not possible
Hence, from the above,
We can conclude that the number of 3-point baskets made not be odd

c. Graph the equation. Find two more possible solutions in the context of the problem.
Answer:

The given equation is:
2x + 3y = 54
From the graph,
We can observe that there are many solutions like (3, 16), (6, 14), (10, 11) and so on
Now,
To satisfy the above equation, substitute the points we obtained from the graph
Hence, from the above,
We can conclude that the solutions to the given equation are: (3, 16) and (6, 14)

MULTIPLE REPRESENTATIONS
In Exercises 29–32, match the equation with its graph.

Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 108

Question 29.
5x + 3y = 30
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.4 Question 29

Question 30.
5x + 3y = -30
Answer:
The given equation is:
5x + 3y = -30
Now,
To find the value of the x-intercept, put y = 0
So,
5x = -30
x = -30 / 5
x = -6
To find the value of the y-intercept, put x = 0
So,
3y = -30
y = -30 / 3
y = -10
Hence, from the above,
We can conclude that the graph C) matches the given equation

Question 31.
5x – 3y = 30
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.4 Question 31

Question 32.
5x – 3y = -30
Answer:
The given equation is:
5x – 3y = -30
Now,
To find the value of the x-intercept, put y = 0
So,
5x = -30
x = -30 / 5
x = -6
To find the value of the y-intercept, put x = 0
So,
-3y = -30
3y = 30
y = 30 / 3
y = 10
Hence, from the above,
We can conclude that the graph B) matches the given equation

Question 33.
MATHEMATICAL CONNECTIONS
Graph the equations x = 5, x = 2, y = -2, and y = 1. What enclosed shape do the lines form? Explain your reasoning.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.4 Question 33

Question 34.
HOW DO YOU SEE IT? You are organizing a class trip to an amusement park. The cost to enter the park is $30. The cost to enter with a meal plan is $45. You have a budget of $2700 for the trip. The equation 30x + 45y = 2700 models the total cost for the class to go on the trip, where x is the number of students who do not choose the meal plan and y is the number of students who do choose the meal plan.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 109
a. Interpret the intercepts of the graph.
Answer:
The given graph is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 109
We know that,
The x-intercept is the point that cuts the x-axis and y must be 0
The y-intercept is the point that cuts the y-axis and x must be 0
Hence,
From the graph,
We can conclude that
The x-intercept is: (90, 0)
The y-intercept is: (0, 60)

b. Describe the domain and range in the context of the problem.
Answer:
The given graph is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 109
We know that,
The domain is defined as the set of all the values of x
The range is defined as the set of all the values of y
Hence,
The domain of the given function is: {0, 10, 20, 30, 40, 50, 60, 70, 80, 90}
The range of the given function is: {0, 10, 20, 30, 40, 50, 60}

Question 35.
REASONING
Use the values to fill in the equation Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 110 so that the x-intercept of the graph is -10 and the y-intercept of the graph is 5.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 110.1
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.4 Question 35

Question 36.
THOUGHT-PROVOKING
Write an equation in the standard form of a line whose intercepts are integers. Explain how you know the intercepts are integers.
Answer:
We know that,
The equation in the standard form is:
y = mx + c
Now,
To find the value of the x-intercept, put y = 0
So,
mx = -c
x = –\(\frac{c}{m}\)
To find the value of the y-intercept, put x = 0
So,
y = 0 + c
y = c
Now, from the above
We can observe that to make the x and y-intercepts integers, the values of the intercepts must be the multiples of the variable c

Question 37.
WRITING
Are the equations of horizontal and vertical lines written in standard form? Explain your reasoning.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.4 Question 37

Question 38.
ABSTRACT REASONING
The x- and y-intercepts of the graph of the equation 3x + 5y = k are integers. Describe the values of k. Explain your reasoning.
Answer:
The given equation is:
3x + 5y = k
To find the value of the x-intercept, put y = 0
So,
3x = k
So,
The values of k so that k become integer is 0, 1, 2, and so on
To find the value of the y-intercept, put x = 0
So,
5y = k
So,
The values of k so that k becomes integer is 0, 1, 2, and so on
Hence, from the above,
We can conclude that the values of k so that the x and y-intercepts become integers are: 0, 1, 2, and so on

Maintaining Mathematical Proficiency

Simplify the expression.

Question 39.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 111
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.4 Question 39

Question 40.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 112
Answer:
The given expression is:
\(\frac{14 -18}{0 – 2}\)
= \(\frac{-4}{-2}\)
= \(\frac{4}{2}\)
= 2
Hence, from the above,
We can conclude that the value of the given expression is: 2

Question 41.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 113
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.4 Question 41

Question 42.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 114
Answer:
The given expression is:
\(\frac{12 – 17}{-5 – (-2)}\)
= \(\frac{12 – 17}{-5 + 2}\)
= \(\frac{-5}{-3}\)
= \(\frac{5}{3}\)
Hence, from the above,
We can conclude that the value of the given expression is: \(\frac{5}{3}\)

Lesson 3.5 Graphing Linear Equations in Slope-Intercept Form

EXPLORATION 1
Finding Slopes and y-Intercepts
Work with a partner.
Find the slope and y-intercept of each line.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 115
Answer:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 115
We know that,
The standard form of a linear equation is:
y = mx + c
Where,
m is the slope
c is the y-intercept
Now,
a.
The given equation is:
y = \(\frac{2}{3}\)x + 2
Now,
Compare the given equation with the standard form of the linear equation
By comparison,
We get,
m = \(\frac{2}{3}\) and c = 2

b.
The given equation is:
y = -2x – 1
Now,
Compare the given equation with the standard form of the linear equation
By comparison,
We get,
m = -2 and c = -1

EXPLORATION 2
Writing a Conjecture
Work with a partner.
Graph each equation. Then copy and complete the table. Use the completed table to write a conjecture about the relationship between the graph of y = mx + b and the values of m and b.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 116
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 117
Answer:
The given table is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 117
Now,
The representation of the given equations in the coordinate plane is as follows:
Now,
a.
The given equation is:
y = –\(\frac{2}{3}\)x + 3
Hence,
The representation of the given equation in the coordinate plane is:

b.
The given equation is:
y = 2x – 2
Hence,
The representation of the given equation in the coordinate plane is:

c.
The given equation is:
y = -x + 1
Hence,
The representation of the given equation in the coordinate plane is:

d.
The given equation is:
y = x – 4
Hence,
The representation of the given equation in the coordinate plane is:

Now,
We know that,
The standard form of a linear equation is:
y = mx + b
Where,
m is the slope
b is the y-intercept
Hence,
The completed table for the above equations’ slopes and y-intercepts is:

Communicate Your Answer

Question 3.
How can you describe the graph of the equation y = mx + b?
a. How does the value of m affect the graph of the equation?
b. How does the value of b affect the graph of the equation?
c. Check your answers to parts (a) and (b) by choosing one equation from Exploration 2 and (1) varying only m and (2) varying only b.
Answer:
The graph of the equation
y = mx + b
is a straight line
Now,
a.
The value of m affects the steepness of the line. It also describes the direction (positive or negative). The value of m defines the constant rate of change of variables

b.
The value of b affects the line to where it should have the point of intersection with the y-axis

c.
The examples of how m and b varies are as follows:
y =-3x – 3
y = 2x + 8
y = -3x + 6
Hence,
The slopes and y-intercepts of the above equations are:
m = -3 and b = -3
m = 2 and b = 8
m = -3 and b = 6

3.5 Lesson

Monitoring Progress

Describe the slope of the line. Then find the slope.

Question 1.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 118
Answer:
The given graph is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 118
From the graph,
The given points are:
(-4, 3), (1, 1)
We know that,
The slope of the line when the two points are given is:
m = \(\frac{y2 – y1}{x2 – x1}\)
So,
The points are represented as (x, y)
So,
The first point is represented as (x1, y1)
The second point is represented as (x2, y2)
So,
(x1, y1) = (-4, 3) and (x2, y2) = (1, 1)
Hence,
m = \(\frac{1 – 3}{1 – [-4]}\)
m = \(\frac{-2}{1 + 4}\)
m = \(\frac{-2}{5}\)
Hence, from the above,
We can conclude that the slope of the given graph is: \(\frac{-2}{5}\)

Question 2.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 119
Answer:
The given graph is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 119
From the graph,
The given points are:
(3, 3), (-3, 1)
We know that,
The slope of the line when the two points are given is:
m = \(\frac{y2 – y1}{x2 – x1}\)
So,
The points are represented as (x, y)
So,
The first point is represented as (x1, y1)
The second point is represented as (x2, y2)
So,
(x1, y1) = (3, 3) and (x2, y2) = (-3, 1)
Hence,
m = \(\frac{1 – 3}{-3 – 3}\)
m = \(\frac{-2}{6}\)
m = \(\frac{-1}{3}\)
Hence, from the above,
We can conclude that the slope of the given graph is: \(\frac{-1}{3}\)

Question 3.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 120
Answer:
The given graph is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 120
From the graph,
The given points are:
(5, 4), (2, -3)
We know that,
The slope of the line when the two points are given is:
m = \(\frac{y2 – y1}{x2 – x1}\)
So,
The points are represented as (x, y)
So,
The first point is represented as (x1, y1)
The second point is represented as (x2, y2)
So,
(x1, y1) = (5, 4) and (x2, y2) = (2, -3)
Hence,
m = \(\frac{-3 – 4}{2 – 5}\)
m = \(\frac{-7}{-3}\)
m = \(\frac{7}{3}\)
Hence, from the above,
We can conclude that the slope of the given graph is: \(\frac{7}{3}\)

Monitoring Progress

The points represented by the table lie on a line. How can you find the slope of the line from the table? What is the slope of the line?

Question 4.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 121
Answer:
The given table is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 121
From the above table,
The representations of the values of x and y in the form of ordered pairs are:
(2, 10), (4, 15), (6, 20), (8, 25)
Now,
To find the slope of a line,
Take any 2 ordered pairs and find the slope
Now,
Let,
(x1, y1) = (2, 10) and (x2, y2) = (4, 15)
We know that,
The slope of the line when the two points are given is:
m = \(\frac{y2 – y1}{x2 – x1}\)
Hence,
m = \(\frac{15 – 10}{4 – 2}\)
m = \(\frac{5}{2}\)
Hence, from the above,
We can conclude that the slope of the given graph is: \(\frac{5}{2}\)

Question 5.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 122
Answer:
The given table is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 122
From the above table,
The representations of the values of x and y in the form of ordered pairs are:
(5, -12), (5, -9), (5, -6), (5, -3)
Now,
To find the slope of a line,
Take any 2 ordered pairs and find the slope
Now,
Let,
(x1, y1) = (5, -12) and (x2, y2) = (5, -9)
We know that,
The slope of the line when the two points are given is:
m = \(\frac{y2 – y1}{x2 – x1}\)
Hence,
m = \(\frac{-9 – [-12]}{5 – 5}\)
m = \(\frac{3}{0}\)
Hence, from the above,
We can conclude that the slope of the given graph is: undefined

Find the slope and the y-intercept of the graph of the linear equation.

Question 6.
y = -6x + 1
Answer:
The given equation is:
y = -6x + 1
We know that,
The standard representation of a linear equation is:
y  mx + c
Where,
m is the slope
c is the y-intercept
Now,
Compare the given equation with the standard representation of the linear equation
Hence,
The values of m and c are: -6 and 1

Question 7.
y = 8
Answer:
The given equation is:
y = 8
Rewrite the given equation as:
y = 0x + 8
We know that,
The standard representation of a linear equation is:
y  mx + c
Where,
m is the slope
c is the y-intercept
Now,
Compare the given equation with the standard representation of the linear equation
Hence,
The values of m and c are: 0 and 8

Question 8.
x + 4y = -10
Answer:
The given equation is:
x + 4y = -10
Rewrite the given equation as:
4y = -10 – x
y = \(\frac{-10 – x}{4}\)
y = \(\frac{-x}{4}\) – \(\frac{10}{4}\)
We know that,
The standard representation of a linear equation is:
y  mx + c
Where,
m is the slope
c is the y-intercept
Now,
Compare the given equation with the standard representation of the linear equation
Hence,
The values of m and ‘c’ are: –\(\frac{1}{4}\) and –\(\frac{10}{4}\)

Graph the linear equation. Identify the x-intercept.

Question 9.
y = 4x – 4
Answer:
The given equation is:
y = 4x – 4
Now,
To find the x-intercept, put y = 0
So,
0 = 4x – 4
4x = 4
x = 4 / 4
x = 1
Hence,
The representation of the given equation along with the x-intercept in the coordinate plane is:

Question 10.
3x + y = -3
Answer:
The given equation is:
3x + y = -3
Now,
To find the x-intercept, put y = 0
So,
3x + 0 = -3
3x = -3
x = -3 / 3
x = -1
Hence,
The representation of the given equation along with the x-intercept in the coordinate plane is:

Question 11.
x + 2y = 6
Answer:
The given equation is:
x + 2y = 6
Now,
To find the x-intercept, put y = 0
So,
x + 0 = 6
x = 6
Hence,
The representation of the given equation along with the x-intercept in the coordinate plane is:

Question 12.
A linear function h models a relationship in which the dependent variable decreases 2 units for every 5 units the independent variable increases. Graph h when h(0) = 4. Identify the slope, y-intercept, and x-intercept of the graph.
Answer:
It is given that a linear function h models a relationship in which the dependent variable decreases 2 units for every 5 units the independent variable increases.
We know that,
The independent variable is: x
the dependent variable is: y
So,
So,
x = -2 and y = +5
So,
We can say that the slope is represented as:
m = \(\frac{-2}{5}\)
It is also given that
h (0) = 2
So,
y = 2 at x = 0
Hence,
The y-intercept is: (0, 2)
Hence,
The representation of the equation in the standard form is:
y = mx + c
y = –\(\frac{2}{5}\)x + 2
Now,
The value of the x-intercept can be obtained by putting y = 0
So,
0 = –\(\frac{2}{5}\)x + 2
–\(\frac{2}{5}\)x = -2
\(\frac{2}{5}\)x = 2
x = \(\frac{5 × 2}{2}\)
x = 5
Hence, from the above,
We can conclude that
The slope is: –\(\frac{2}{5}\)
The x-intercept is: (5, 0)
The y-intercept is: (0, 2)

Question 13.
WHAT IF? The elevation of the submersible is modeled by h(t) = 500t – 10,000.
(a) Graph the function and identify its domain and range.
Answer:
The given function is:
h(t) = 500t – 10,000
So,
The representation of the given function in the coordinate plane is:

From the above graph,
We can say that the given equation is parallel to the y-axis
Hence,
There are no values for the domain since there are no values of x
The range of the given function is: -20,000 ≤ t ≤ 20,000

(b) Interpret the slope and the intercepts of the graph.
Answer:
The given function is:
h (t) = 500t – 10,000
Compare the given equation with the standard linear equation y = mx + c
So,
m = 500
c = -10,000
Now,
To find the x-intercept, put y = 0 or h (t) = 0
So,
500t – 10,000 = 0
500t = 10,000
t = 10,000 / 500
t = 20 or x = 20
To find the y-intercept, put x = 0 or t = 0
So,
h (t) = 500 (0) – 10,000
h (t) = -10,000
Hence, from the above,
We can conclude that
The slope is: 500
The x-intercept is: 20
The y-intercept is: -10,000

Graphing Linear Equations in Slope-Intercept Form 3.5 Exercises

Vocabulary and Core Concept

Question 1.
COMPLETE THE SENTENCE
The ________ of a nonvertical line passing through two points is the ratio of the rise to the run.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.5 Question 1

Question 2.
VOCABULARY
What is a constant function? What is the slope of a constant function?
Answer:
A “Constant function” is a function where the output (y) is the same for every input (x) value.
The slope for a constant function will be 0

Question 3.
WRITING
What is the slope-intercept form of a linear equation? Explain why this form is called the slope-intercept form.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.5 Question 3

Question 4.
WHICH ONE did DOESN’T BELONG? Which equation does not belong with the other three? Explain your reasoning.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 122.1
Answer:
The given equations are:
a. y = -5x – 1
b. 2x – y = 8
c. y = x + 4
d. y = -3x + 13
Now,
Rewrite all the above equations in the form of
y = mx + c
So,
Now,
a.
The given equation is:
y = -5x – 1
Compare the above equation with
y = mx + c
So,
We will get,
m = -5 and c = -1

b.
The given equation is:
2x – y = 8
y = 2x – 8
Compare the above equation with
y = mx + c
So,
We will get
m = 8 and c = -8

c.
The given equation is:
y = x + 4
Compare the above equation with
y = mx + c
So,
We will get
m = 1 and c = 4

d.
The given equation is:
y = -3x + 13
Compare the above equation with
y = mx + c
So,
We will get
m = -3 and c = 3
Hence, from the above,
We can conclude that equation (b) does not belong with the other three since the slope of equation (b) is even whereas all the other slops are odd

Monitoring Progress and Modeling with Mathematics

In Exercises 5–8, describe the slope of the line. Then find the slope.

Question 5.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 123
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.5 Question 5

Question 6.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 124
Answer:
The given graph is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 124
From the graph,
We can observe that the slope falls from right to left
So,
The slope is positive
Now,
We have to represent the first pair as (x1, y1) and the second pair as (x2, y2)
Now,
The representation of the values of x and y in the form of ordered pairs to find a slope is:
(4, 3), (1, -1)
We know that,
m = \(\frac{y2 – y1}{x2 – x1}\)
m = \(\frac{-1 – 3}{1 – 4}\)
m = \(\frac{-4}{-3}\)
m = \(\frac{4}{3}\)
Hence, from the above,
We can conclude that the slope of the line is: \(\frac{4}{3}\)

Question 7.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 125
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.5 Question 7

Question 8.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 126
Answer:
The given graph is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 126
From the graph,
We can observe that the slope falls from left to right
So,
The slope is negative
Now,
We have to represent the first pair as (x1, y1) and the second pair as (x2, y2)
Now,
The representation of the values of x and y in the form of ordered pairs to find a slope is:
(0, 3), (5, -1)
We know that,
m = \(\frac{y2 – y1}{x2 – x1}\)
m = \(\frac{-1 – 3}{5 – 0}\)
m = \(\frac{-4}{5}\)
m = –\(\frac{4}{5}\)
Hence, from the above,
We can conclude that the slope of the line is: –\(\frac{4}{5}\)

In Exercises 9–12, the points represented by the table lie on a line. Find the slope of the line.

Question 9.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 127
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.5 Question 9

Question 10.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 128
Answer:
The given table is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 128
We know that,
To find the slope from a given table, we can take any x and y pair from the table
We have to represent the first pair as (x1, y1) and the second pair as (x2, y2)
Now,
The representation of the values of x and y in the form of ordered pairs to find a slope is:
(-1, -6), (2, -6)
We know that,
m = \(\frac{y2 – y1}{x2 – x1}\)
m = \(\frac{-6 + 6}{2 + 1}\)
m = \(\frac{0}{3}\)
m = 0
Hence, from the above,
We can conclude that the slope of the line is: 0

Question 11.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 129
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.5 Question 11

Question 12.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 130
Answer:
The given table is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 130
We know that,
To find the slope from a given table, we can take any x and y pair from the table
We have to represent the first pair as (x1, y1) and the second pair as (x2, y2)
Now,
The representation of the values of x and y in the form of ordered pairs to find a slope is:
(-4, 2), (-3, -5)
We know that,
m = \(\frac{y2 – y1}{x2 – x1}\)
m = \(\frac{-5 – 2}{-3 – [-4]}\)
m = \(\frac{-7}{1}\)
m = -7
Hence, from the above,
We can conclude that the slope of the line is: -7

Question 13.
ANALYZING A GRAPH
The graph shows the distance y(in miles) that a bus travels in x hours. Find and interpret the slope of the line.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 131
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.5 Question 13

Question 14.
ANALYZING A TABLE
The table shows the amount x(in hours) of time you spend at a theme park and the admission fee y (in dollars) to the park. The points represented by the table lie on a line. Find and interpret the slope of the line.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 132
Answer:
The given table is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 132
It is given that the table shows the amount x(in hours) of time you spend at a theme park and the admission fee y (in dollars) to the park and the points represented by the table lie on a line.
Now,
The representations of the x and y values in the form of ordered pairs i.e., (x, y)
So,
Let (6, 54.99) and (7, 54.99) be the 2 ordered pairs
We know that,
m = \(\frac{y2 – y1}{x2 – x1}\)
m = \(\frac{54.99 – 54.99}{7 – 6}\)
m = \(\frac{0}{1}\)
m = 0
Hence, from the above,
We can conclude that the slope of the line is: 0

In Exercises 15–22, find the slope and the y-intercept of the graph of the linear equation.

Question 15.
y = -3x + 2
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.5 Question 15

Question 16.
y = 4x – 7
Answer:
The given equation is:
y = 4x – 7
Compare the given equation with the standard form of the linear equation
y = mx + c
By comparing, we will get
m = 4 and c = -7
Hence, from the above,
We can conclude that
The slope of the given equation (m) is: 4
The y-intercept of the given equation is: -7

Question 17.
y = 6x
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.5 Question 17

Question 18.
y = -1
Answer:
The given equation is:
y = -1
Rewrite the given equation in the form of
y = mx + c
So,
y = 0x – 1
Compare the given equation with the standard form of the linear equation
y = mx + c
By comparing, we will get
m = 0 and c = -1
Hence, from the above,
We can conclude that
The slope of the given equation (m) is: 0
The y-intercept of the given equation is: -1

Question 19.
-2x + y = 4
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.5 Question 19

Question 20.
x + y = -6
Answer:
The given equation is:
x + y = -6
Rewrite the given equation in the form of the linear equation
y = mx + c
So,
y = -x – 6
Compare the given equation with the standard form of the linear equation
y = mx + c
By comparing, we will get
m = -1 and c = -6
Hence, from the above,
We can conclude that
The slope of the given equation (m) is: -1
The y-intercept of the given equation is: -6

Question 21.
-5x = 8 – y
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.5 Question 21

Question 22.
0 = 1 – 2y + 14x
Answer:
The given equation is:
0 =1 – 2y + 14x
Rewrite the give equation in the form of
y = mx + c
So,
2y = 14x + 1
Divide by 2 into both sides
y = (14 / 2)x + (1 / 2)
y = 7x + \(\frac{1}{2}\)
Compare the given equation with the standard form of the linear equation
y = mx + c
By comparing, we will get
m = 7 and c = \(\frac{1}{2}\)
Hence, from the above,
We can conclude that
The slope of the given equation (m) is: 7
The y-intercept of the given equation is:\(\frac{1}{2}\)

ERROR ANALYSIS
In Exercises 23 and 24, describe and correct the error in finding the slope and the y-intercept of the graph of the equation.

Question 23.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 133
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.5 Question 23

Question 24.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 134
Answer:
The given equation is:
y = 3x – 6
Compare the above equation with the standard form of the linear equation
y = mx + c
So,
We will get
m = 3 and c = -6
Hence, from the above,
We can conclude that
The slope of the given equation (m) is: 3
The y-intercept of the given equation is: -6

In Exercises 25–32, graph the linear equation. Identify the x-intercept.

Question 25.
y = -x + 7
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.5 Question 25

Question 26.
y = \(\frac{1}{2}\)x + 3
Answer:
The given linear equation is:
y = \(\frac{1}{2}\)x + 3
To find the x-intercepy, put y = 0
0 = \(\frac{1}{2}\)x + 3
–\(\frac{1}{2}\)x = 3
x = 3 (-2)
x = -6
Hence,
The representation of the given linear equation in the coordinate plane is:

Question 27.
y = 2x
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.5 Question 27

Question 28.
y = -x
Answer:
The given linear equation is:
y = -x
Rewrite the given equation in the form of y = mx + c
So,
y = -x + 0
To find the x-intercept, put y = 0
So,
0 = -x + 0
x = 0
Hence,
The representation of the given linear equation along with the x-intercept in the coordinate plane is:

Question 29.
3x + y = -1
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.5 Question 29

Question 30.
x + 4y = 8
Answer:
The given linear equation is:
x + 4y = 8
4y = -x + 8
Divide by 4 into both sides
y = –\(\frac{1}{4}\)x + (8 / 4)
y = –\(\frac{1}{4}\)x + 2
To find the x-intercept, put y = 0
0 = –\(\frac{1}{4}\)x + 2
\(\frac{1}{4}\)x = 2
x = 2(4)
x = 8
Hence,
The representation of the given linear equation along with the x-intercept in the coordinate plane is:

Question 31.
-y + 5x = 0
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.5 Question 31

Question 32.
2x – y + 6 = 0
Answer:
The given linear equation is:
2x – y + 6 = 0
So,
y = 2x + 6
To find the x-intercept, put y = 0
So,
0 = 2x + 6
-2x = 6
x = -6 / 2
x = -3
Hence,
The representation of the given linear equation along with the x-intercept in the coordinate plane is:

In Exercises 33 and 34, graph the function with the given description. Identify the slope, y-intercept, and x-intercept of the graph.

Question 33.
A linear function f models a relationship in which the dependent variable decreases 4 units for every 2 units the independent variable increases. The value of the function at 0 is -2.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.5 Question 33

Question 34.
A linear function h models a relationship in which the dependent variable increases 1 unit for every 5 units the independent variable decreases. The value of the function at 0 is 3.
Answer:
It is given that a linear function h models a relationship in which the dependent variable increases 1 unit for every 5 units the independent variable decreases.
We know that,
The independent variable is: x
The dependent variable is: y
So,
Slope (m) = y / x
m = 1 / (-5)
m = –\(\frac{1}{5}\)
It is given that the value of the function at 0 is 3
So,
The value of y at x = 0 is: 3
Hence,
The y-intercept of the given function is: 3
Hence,
The representation of the slope and the y-intercept in the standard form of the linear function s:
y = mx + c
So,
y = –\(\frac{1}{5}\)x + 3
To find the x-intercept, put y = 0
So,
0 = –\(\frac{1}{5}\)x + 3
\(\frac{1}{5}\)x = 3
x = 3(5)
x = 15
To find the y-intercept, put x = 0
So,
y = 3
Hence, from the above,
We can conclude that
The slope of the given function is: –\(\frac{1}{5}\)
The x-intercept is: 15
The y-intercept is: 3
The representation of the given function in the coordinate plane is:

Question 35.
GRAPHING FROM A VERBAL DESCRIPTION
A linear function r models the growth of your right index fingernail. The length of the fingernail increases 0.7 millimeters every week. Graph r when r(0) = 12. Identify the slope and interpret the y-intercept of the graph.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.5 Question 35

Question 36.
GRAPHING FROM A VERBAL DESCRIPTION
A linear function m models the amount of milk sold by a farm per month. The amount decreases 500 gallons for every $1 increase in price. Graph m when m(0) = 3000. Identify the slope and interpret the x- and y-intercepts of the graph.
Answer:
It is given that the amount of milk decreases 500 gallons for every $1 increase
From the above,
We can say that,
The price is the independent variable
So,
x represents the price
The amount of milk is the dependent variable
So,
y represents the amount of milk
So,
Slope = y / x
Slope = -500 / 1
So,
Slope (m) = -500
It is given that,
m (0) = 3000
We know that,
m (0) is the functional representation of y
So,
y = 3000 when x = 0
Hence,
The y-intercept is: 3000
Now,
The representation of the slope and the y-intercept in the standard form of the linear equation is:
y = ,x + c
y = -500x + 3000
Now,
To find the x-intercept, put y = 0
So,
0 = -500x + 3000
500x = 3000
x = 3000 / 500
x = 6
To find the y-intercept, put x = 0
So,
y = -500 (0) + 3000
y = 3000
Hence, from the above,
We can conclude that
The slope of the given equation is: -500
The c-intercept or y-intercept of the given equation is: 3000
The x-intercept of the given equation is: 6

Question 37.
MODELING WITH MATHEMATICS
The function shown models the depth d (in inches) of snow on the ground during the first 9 hours of a snowstorm, where t is the time (in hours) after the snowstorm begins.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 135
a. Graph the function and identify its domain and range.
b. Interpret the slope and the d-intercept of the graph.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.5 Question 37

Question 38.
MODELING WITH MATHEMATICS
The function c(x) = 0.5x + 70 represents the cost c (in dollars) of renting a truck from a moving company, where x is the number of miles you drive the truck.
a. Graph the function and identify its domain and range.
Answer:
The given function is:
c (x) = 0.5x + 70
Where,
represents the cost c (in dollars) of renting a truck from a moving company,
x is the number of miles you drive the truck.
Hence,
The representation of the given function in the coordinate plane is:

We know that,
The domain is the set of all the values of x that holds the given equation true
The range is defined as the set of all the values of y that holds the given equation true
Hence,
The domain of the given function is: -100 ≤ x ≤ 60
The range of the given function is: 20 ≤y ≤ 70

b. Interpret the slope and the c-intercept of the graph.
Answer:
The given function is:
c (x) = 0.5x + 70
Compare the above equation with the standard form of the linear function
y = mx + c
SO,
By comparing, we get
m = 0.5 and c = 70
Hence, from the above,
We can conclude that
The slope of the given function is: 0.5
The c-intercept is: 70

Question 39.
COMPARING FUNCTIONS
A linear function models the cost of renting a truck from a moving company. The table shows the cost y (in dollars) when you drive the truck x miles. Graph the function and compare the slope and the y-intercept of the graph with the slope and the c-intercept of the graph in Exercise 38.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 136
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.5 Question 39

ERROR ANALYSIS
In Exercises 40 and 41, describe and correct the error in graphing the function.

Question 40.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 137
Answer:
The given equation is:
y + 1 = 3x
Rewrite the above equation in the form of
y = mx + c
So,
The given equation can be rewritten as:
y = 3x – 1
Now,
Compare the above equation with the standard form of the linear equation
y = mx + c
So,
By comparison,
We get
m = 3 and c = -1
Hence,
The representation of the given equation in the coordinate plane is:

Question 41.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 138
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.5 Question 41

Question 42.
MATHEMATICAL CONNECTIONS
Graph the four equations in the same coordinate plane.
3y = -x – 3
2y – 14 = 4x
4x – 3 – y = 0
x – 12 = -3y
a. What enclosed shape do you think the lines form? Explain.
Answer:
The given equations are:
a. 3y = -x – 3
b. 2y – 14 = 4x
c. 4x – 3 – y = 0
d. x – 12 = -3y
Now,
Rewrite the given equations in the form of
y = mx + c
So,
a.
The given equation is:
3y = -x – 3
Divide by 3 into both sides
y = –\(\frac{x}{3}\) – (3 / 3)
y = –\(\frac{1}{3}\)x – 1

b.
The given equation is:
2y – 14 = 4x
Divide by 2 into both sides
y – (14 / 2) = 2x
y – 7 = 2x
y = 2x + 7

c.
The given equation is:
4x – 3 – y = 0
y = 4x – 3

d.
The given equation is:
x – 12 = -3y
Divide by -3 into both sides
y = –\(\frac{x}[3]\) + (12 / 3)
y = –\(\frac{1}{3}\)x + 4
Hence,
The representations of the given four equations in the coordinate plane is:

Hence,
By observing the graph,
We can conclude that the enclosed lines in the graph form a rectangle

b. Write a conjecture about the equations of parallel lines.
Answer:
We can determine from their equations whether two lines are parallel by comparing their slopes.
If the slopes are the same and the y-intercepts are different, the lines are parallel.
If the slopes are different, the lines are not parallel.
Unlike parallel lines, perpendicular lines do intersect.

Question 43.
MATHEMATICAL CONNECTIONS The graph shows the relationship between the width y and the length x of a rectangle in inches. The perimeter of a second rectangle is 10 inches less than the perimeter of the first rectangle.
a. Graph the relationship between the width and length of the second rectangle.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 139
b. How does the graph in part (a) compare to the graph shown?
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.5 Question 43.1
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.5 Question 43.2

Question 44.
MATHEMATICAL CONNECTIONS
The graph shows the relationship between the base length x and the side length (of the two equal sides) y of an isosceles triangle in meters. The perimeter of a second isosceles triangle is 8 meters more than the perimeter of the first triangle.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 140
a. Graph the relationship between the base length and the side length of the second triangle.
The given graph is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 140
It is given that the graph shows the relationship between the base length x and the side length (of the two equal sides) y of an isosceles triangle in meters. The perimeter of a second isosceles triangle is 8 meters more than the perimeter of the first triangle.
Now,
From the graph,
The equation for the first Isosceles triangle is:
y = 6 – \(\frac{1}{2}\)x
Multiply by 2 into both sides
2y = 6 (2) – x
2y = 12 – x
2y + x = 12
The above equation represents the perimeter of the first Isosceles triangle
Where,
x is the base length
y is the side of the equal length like the third side
It is also given that the perimeter of a second isosceles triangle is 8 meters more than the perimeter of the first triangle.
So,
2y + x = 12 + 8
2y + x = 20
The above equation represents the perimeter of the second Isosceles triangle
Hence,
The representation of the perimeter of the second isosceles triangle in the coordinate plane is:

b. How does the graph in part (a) compare to the graph shown?
Answer:
The equation of the perimeter of the first isosceles triangle is:
y = 6 – \(\frac{1}{2}\)x
The equation of the perimeter of the second isosceles triangle is:
y = 10 – \(\frac{1}{2}\)x
Hence,
From comparing the above 2 equations with the standard form of the linear equation
y = mx + c
We can observe that the slopes are equal and only the y-intercepts are different
Hence, from the above,
We can conclude that the graphs of the perimeters of the first isosceles triangle and the second isosceles triangle are parallel

Question 45.
ANALYZING EQUATIONS
Determine which of the equations could be represented by each graph.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 141
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.5 Question 45

Question 46.
MAKING AN ARGUMENT
Your friend says that you can write the equation of any line in slope-intercept form. Is your friend correct? Explain your reasoning.
Answer:
No, your friend is not correct

Explanation:
We can write the equation in slope-intercept form only when hat equation is in the linear form i.e., only x and y terms.
If there are exponential terms in an equation, then we can not write that equation in slope-intercept form
Hence, from the above,
We can conclude that your friend is not correct

Question 47.
WRITING
Write the definition of the slope of a line in two different ways.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.5 Question 47

Question 48.
THOUGHT-PROVOKING
Your family goes on vacation to a beach 300 miles from your house. You reach your destination 6 hours after departing. Draw a graph that describes your trip. Explain what each part of your graph represents.
Answer:

Question 49.
ANALYZING A GRAPH
The graphs of the functions g(x) = 6x + a and h(x) = 2x + b, where a and b are constants, are shown. They intersect at the point (p, q).
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 142
a. Label the graphs of g and h.
b. What do a and b represent?
c. Starting at the point (p, q), trace the graph of g until you get to the point with the x-coordinate p + 2. Mark this point C. Do the same with the graph of h. Mark this point D. How much greater is the y-coordinate of point C than the y-coordinate of point D?
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.5 Question 49.1
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.5 Question 49.2

Question 50.
HOW DO YOU SEE IT? Your commute to school by walking and by riding a bus. The graph represents your commute.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 143
a. Describe your commute in words.
Answer:
The given graph is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 143
From the graph,
We can observe that your commute to school is different at different time intervals
Now,
From 0 to 10 seconds,
The distance you commuted gradually increases
From 10 to 14 seconds,
The distance you commuted is constant
From 14 to 18 seconds,
The distance you commuted abruptly increased

b. Calculate and interpret the slopes of the different parts of the graph.
Answer:
The given graph is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 143
From the graph,
We can observe that there are different commutes at different time intervals,
Now,
The x-axis represents the time
The y-axis represents the distance in miles
Now,
From 0 to 10 seconds,
The ordered pairs from the coordinate plane are:
(0, 0), (10, 0.5)
We know that,
The slope when 2 ordered pairs are given is represented as:
m = \(\frac{y2 – y1}{x2 – x1}\)
So,
From 0 to 10 seconds,
m = \(\frac{0.5 – 0}{10 – 0}\)
m = \(\frac{0.5}{10}\)
m = \(\frac{5}{100}\)
m = 0.05
From 10 to 14 seconds,
The ordered pairs from the coordinate plane are:
(10, 0.5), (14, 0.5)
We know that,
The slope when 2 ordered pairs are given is represented as:
m = \(\frac{y2 – y1}{x2 – x1}\)
So,
From 10 to 14 seconds,
m = \(\frac{0.5 – 0.5}{14 – 10}\)
m = \(\frac{0}{4}\)
m = 0
From 14 to 18 seconds,
The ordered pairs from the coordinate plane are:
(14, 0.5), (18, 2.5)
We know that,
The slope when 2 ordered pairs are given is represented as:
m = \(\frac{y2 – y1}{x2 – x1}\)
So,
From 14 to 18 seconds,
m = \(\frac{2.5 – 0.5}{18 – 14}\)
m = \(\frac{2}{4}\)
m = \(\frac{1}{2}\)
m = 0.5
Hence, fromthe above,
We can conclude that
The slope from 0 to 10 seconds is: 0.05
The slope from 10 to 14 seconds is: 0
The slope from 14 to 18 seconds is: 0.5

PROBLEM-SOLVING
In Exercises 51 and 52, find the value of k so that the graph of the equation has the given slope or y-intercept.

Question 51.
y = 4kx – 5; m = \(\frac{1}{2}\)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.5 Question 51

Question 52.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 144
Answer:
The given equation is:
y = –\(\frac{1}{3}\)x + \(\frac{5}{6}\)k
Compare the above equation with the standard representation of the linear equation.
We know that,
The standard representation of the linear equation is:
y = mx + b
On comparison,
We get,
m = –\(\frac{1}{3}\) and b = \(\frac{5}{6}\)k
It is given that
The value of b is: -10
So,
\(\frac{5}{6}\)k = -10
5k = -10 × 6
k = \(\frac{-10 × 6}{5}\)
k = -12
Hence, from the above,
We can conclude that the value of k is: -12

Question 53.
ABSTRACT REASONING
To show that the slope of a line is constant, let (x1, y1) and (x2, y2) be any two points on the line y = mx + b. Use the equation of the line to express y1 in terms of x1 and y2 in terms of x2. Then use the slope formula to show that the slope between the points is m.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.5 Question 53

Maintaining Mathematical Proficiency

Find the coordinates of the figure after the transformation.

Question 54.
Translate the rectangle with 4 units left.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 145
Answer:
The given graph is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 145
From the given graph,
The rectangle is covered up to 1 to 3 units at the x-axis and 0 to -4 units at the y-axis
So,
When we move the rectangle 4 units left, i.e., move the 1 and 3 units left for 4 units and move the 0 and -4 units left for 4 units
So,
The new rectangle in the graph is formed at:
At the x-axis:
1 – 4 = -3
3 – 4 = -1
At the y-axis:
0 – 4 = -4
-4 – 4 = -8
Hence,
The new rectangle is formed at (-1, -3) at the x-axis and at (-4, -8) at the y-axis
Hence,
The representation of the rectangle after the transformation in the coordinate plane is:

Question 55.
Dilate the triangle with respect to the origin using a scale factor of 2.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 146
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.5 Question 55

Question 56.
Reflect the trapezoid in the y-axis.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 147
Answer:
Determine whether the equation represents a linear or nonlinear function. Explain.

Question 57.
y – 9 = \(\frac{2}{x}\)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.5 Question 57

Question 58.
x = 3 + 15y
Answer:
The given function is:
x = 3 + 15y
Rewrite the above given function in the form of
y = mx + c
So,
15y = 3 – x
y = \(\frac{3 – x}{15}\)
y = \(\frac{3}{15}\) – \(\frac{1}{15}\)x
Hence,
The above function is in the form of
y = mx + c
Hence, from the above,
We can conclude that the given function is linear

Question 59.
\(\frac{x}{4}\) + \(\frac{y}{12}\) = 1
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.5 Question 59

Question 60.
y = 3x4 – 6
Answer:
The given function is:
y = 3x4 – 6
Rewrite the above given function in the form of
y = mx + c
But,
The above function is not in the form of
y = mx + c
Hence, from the above,
We can conclude that the given function is non-linear

Lesson 3.6 Transformations of Graphs of Linear Functions

Essential Question
How does the graph of the linear function f(x) = x compare to the graphs of g(x) = f(x) + c and h(x) = f(cx)? Comparing Graphs of Functions
Answer:
The given linear function is:
f (x) = x
The graph corresponding to the above function will be the vertical line
Now,
The given functions are:
g (x) = f (x) + c and h (x) = f (cx)
Now,
The graph of g (x) will be the vertical line translated i.e., addition or subtraction by c units
The graph of h (x) will be the graph of the function f(x) stretched or compressed by 1/c units

EXPLORATION 1
Comparing Graphs of Functions
Work with a partner.
The graph of f(x) = x is shown. Sketch the graph of each function, along with f, on the same set of coordinate axes. Use a graphing calculator to check your results. What can you conclude?
a. g(x) = x + 4
b. g(x) = x + 2
c. g(x) = x – 2
d. g(x) = x – 4
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 148
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 149
Answer:
The given functions are:
a. g(x) = x + 4
b. g(x) = x + 2
c. g(x) = x – 2
d. g(x) = x – 4
The other given function is:
f (x) = x
Now,
a.
The given function is:
g (x) = x + 4
Hence,
The representations of f(x) and g (x) in the same coordinate plane are:

Hence, from the above,
We can conclude that g (x) translated 4 units away from f(x) to the left side i.e., towards the positive y-axis

b.
The given function is:
g (x) = x + 2
Hence,
The representations of f(x) and g (x) in the same coordinate plane are:

Hence, from the above,
We can conclude that g (x) translated 2 units away from f(x) to the left side i.e., towards the positive y-axis

c.
The given function is:
g (x) = x – 2
Hence,
The representations of f(x) and g (x) in the same coordinate plane are:

Hence, from the above,
We can conclude that g (x) translated 2 units away from f(x) to the right side i.e., towards the negative y-axis

d,
The given function is:
g (x) = x – 4
Hence,
The representations of f(x) and g (x) in the same coordinate plane are:

Hence, from the above,
We can conclude that g (x) translated 4 units away from f(x) to the right side i.e., towards the negative y-axis

EXPLORATION 2
Comparing Graphs of Functions
Work with a partner.
Sketch the graph of each function, along with f(x) = x, on the same set of coordinate axes. Use a graphing calculator to check your results. What can you conclude?
a. h(x) = \(\frac{1}{2}\)x
b. h(x) = 2x
c. h(x) = –\(\frac{1}{2}\)x
d. h(x) = -2x
Answer:
The given functions are:
a. h(x) = \(\frac{1}{2}\)x
b. h(x) = 2x
c. h(x) = –\(\frac{1}{2}\)x
d. h(x) = -2x
The other given function is:
f (x) = x
Now,
a.
The given function is:
h(x) = \(\frac{1}{2}\)x
Hence,
The representations of f (x) and h (x) in the same coordinate plane are:

Hence, from the above,
We can conclude that h (x) and f (x) passes through the origin and h (x) is steeper than f (x)

b.
The given function is:
h(x) = 2x
Hence,
The representation of f (x) and h(x) in the same coordinate plane is:

Hence, from the above,
We can conclude that h(x) and f(x) passes through the origin and f (x) is steeper than h (x)

c.
The given function is:
h(x) = –\(\frac{1}{2}\)x
Hence,
The representation of f (x) and h (x) in the same coordinate plane is:

Hence, from the above,
We can conclude that h (x) and f (x) are on the opposite axes

d.
The given function is:
h(x) = -2x
Hence,
The representation of f (x) and h (x) in the same coordinate plane is:

Hence, from the above,
We can conclude that f (x) and h (x) are in the opposite axes

EXPLORATION 3
Matching Functions with Their Graphs
Work with a partner.
Match each function with its graph. Use a graphing calculator to check your results. Then use the results of Explorations 1 and 2 to compare the graph of k to the graph of f(x) = x.
a. k(x) = 2x – 4
b. k(x) = -2x + 2
c. k(x) = –\(\frac{1}{2}\)x + 4
d. k(x) = –\(\frac{1}{2}\)x — 2
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 150
Answer:
The given functions are:
a. k(x) = 2x – 4
b. k(x) = -2x + 2
c. k(x) = –\(\frac{1}{2}\)x + 4
d. k(x) = –\(\frac{1}{2}\)x — 2
The other given function is:
f (x) = x
Now,
a.
The given equation is:
k(x) = 2x – 4
Hence,
The representation of k (x) and f (x) in the coordinate plane is:

Hence, from the above,
We can conclude that graph C) matches k (x)
By comparing f (x) and k (x),
We can say that k (x) translates 2 units from f (x) towards the positive x-axis

b.
The given function is:
k(x) = -2x + 2
Hence,
The representation of k (x) and f (x) in the coordinate plane is:

Hence, from the above,
We can conclude that graph A) matches k (x)
By comparing f (x) and k (x),
We can say that k (x) and f (x) are on the opposite axes

c.
The given function is:
k(x) = –\(\frac{1}{2}\)x + 4
Hence,
The representation of k (x) and f (x) in the coordinate plane is:

Hence, from the above,
We can conclude that graph D) matches k (x)
By comparing f (x) and k (x),
We can say that k (x) and f (x) are on opposite axes and k (x) only translates through only the positive and negative x-axes

d.
The given function is:
k(x) = –\(\frac{1}{2}\)x — 2
Hence,
The representation of k (x) and f (x) in the coordinate plane is:

Hence, from the above,
We can conclude that graph B) matches k (x)
By comparing f (x) and k (x),
We can say that k (x) and f(x) are on the opposite axes and k (x) translates through only the negative x-axis and the negative y-axis

Communicate Your Answer

Question 4.
How does the graph of the linear function f(x) = x compare to the graphs of g(x) = f(x) + c and h(x) = f(cx)?
Answer:
The given linear function is:
f (x) = x
The graph corresponding to the above function will be the vertical line
Now,
The given functions are:
g (x) = f (x) + c and h (x) = f (cx)
Now,
The graph of g (x) will be the vertical line translated i.e., addition or subtraction by c units
The graph of h (x) will be the graph of the function f(x) stretched or compressed by 1/c units

3.6 Lesson

Monitoring Progress
Using f, graph (a) g and (b) h. Describe the transformations from the graph of f to the graphs of g and h.

Question 1.
f(x) = 3x + 1; g(x) = f(x) – 2; h(x) = f(x – 2)
Answer:
The given functions are:
f (x) = 3x + 1; g (x) = f (x) – 2 and h (x) = f (x – 2)
Now,
It is given that,
g (x) = f (x) – 2
So,
g (x) = 3x + 1 – 2
g (x) = 3x – 1
It is given that,
h (x) = f (x – 2)
h (x) = 3 (x – 2) + 1
h (x) = 3 (x) + 3 (2) + 1
h (x) = 3x + 6 + 1
h (x) = 3x + 7
Hence,
The representations of f (x), g (x), and h (x) in a coordinate plane is:

Hence, from the above,
We can conclude that
g (x) is translated 2 units away from f (x) toward the positive x-axis and h (x) is translated 5 units away from f (x) toward the positive y-axis

Question 2.
f(x) = -4x – 2; g(x) = -f(x); h(x) = f(-x)
Answer:
The given functions are:
f (x) = -4x – 2; g (x) = -f (x) and h (x) = f (-x)
Now,
It is given that,
g (x) = -f (x)
So,
g (x) = – (-4x – 2)
g (x) = 4x + 2
It is given that,
h (x) = f (-x)
h (x) = -4 (-x) – 2
h (x) = 4x – 2
Hence,
The representations of f (x), g (x), and h (x) in a coordinate plane is:

Hence, from the above,
We can conclude that
g (x) is translated 2.5 units away from f (x) toward the positive y-axis and h (x) is translated 1 unit away from f (x) toward the positive x-axis

Using f, graph (a) g and (b) h. Describe the transformations from the graph of f to the graphs of g and h.

Question 3.
f(x) = 4x – 2; g(x) = f (\(\frac{1}{2}\)x ); h(x) = 2f(x)
Answer:
The given functions are:
f (x) = 4x – 2; g (x) = f (\(\frac{1}{2}\)x ) and h (x) = 2f (x)
Now,
It is given that,
g (x) = f (\(\frac{1}{2}\)x )
So,
g (x) = 4 ( f (\(\frac{1}{2}\)x ) – 2
g (x) = 2x – 2
It is given that,
h (x) = 2f (x)
h (x) = 2 (4x – 2)
h (x) = 2 (4x) – 2(2)
h (x) = 8x – 4
Hence,
The representations of f (x), g (x), and h (x) in a coordinate plane is:

Hence, from the above,
We can conclude that
g (x) is translated 0.5 units away from f (x) toward the positive x-axis and h (x) is translated 0 units away from f (x) toward the origin

Question 4.
f(x) = -3x + 4; g(x) = f(2x); h(x) = \(\frac{1}{2}\)f(x)
Answer:
The given functions are:
f (x) = -3x + 4; g (x) = f (2x) and h (x) = \(\frac{1}{2}\)f(x)
Now,
It is given that,
g (x) = f (2x)
So,
g (x) = -3 (2x) + 4
g (x) = -6x + 4
It is given that,
h (x) = \(\frac{1}{2}\)f(x)
h (x) = \(\frac{1}{2}\) (3x + 4)
h (x) = \(\frac{3}{2}\)x + \(\frac{4}{2}\)
h (x) = \(\frac{3}{2}\)x + 2
Hence,
The representations of f (x), g (x), and h (x) in a coordinate plane is:

Hence, from the above,
We can conclude that
g (x) and f (x) are on the same axis and h (x) and f (x) are on the opposite axis

Question 5.
Graph f(x) = x and h(x) = \(\frac{1}{2}\)x – 2. Describe the transformations from the graph of f to the graph of h.
Answer:
The given functions are:
f (x) = x and h (x) = \(\frac{1}{2}\)x – 2
Hence,
The representation of f (x) and h (x) in the same coordinate plane is:

Hence, from the above,
We can conclude that h (x) translated 4 units away from f (x) towards the positive x-axis

Transformations of Graphs of Linear Functions 3.6 Exercises

Vocabulary and Core Concept Check

Question 1.
WRITING
Describe the relationship between f(x) = x and all other nonconstant linear functions.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.6 Question 1

Question 2.
VOCABULARY
Name four types of transformations. Give an example of each and describe how it affects the graph of a function.
Answer:
The four types of transformations that affect the graph of a function are:
a. Translation
Ex:
f (x) = x and g (x) = 2 f(x)
b. Rotation
Ex:
f (x) = 3x + 2 and g (x) = 3x – 2
c. Reflection
Ex:
f (x) =x and g (x) =-f (x)
d. Dilation
Ex:
f (x) = 3x + 6 and g (x) = f (2x)

Question 3.
WRITING
How does the value of a in the equation y = f(ax) affect the graph of y = f(x)? How does the value of a in the equation y = af(x) affect the graph of y = f(x)?
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 151
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.6 Question 3

Question 4.
REASONING
The functions f and g are linear functions. The graph of g is a vertical shrink of the graph of f. What can you say about the x-intercepts of the graphs of f and g? Is this always true? Explain.
Answer:
It is given that the functions f and g are linear functions
We know that,
f can be written as f (x)
g can be written as g (x)
It is also given that the graph of g is a vertical shrink of the graph of f.
So,
Since the graph of g shrinks, then the x-intercept of g will also shrink if we observe the functions of f and g
Hence, from the above,
We can conclude that the x-intercept of g will shrink and this is always true

In Exercises 5–10, use the graphs of f and g to describe the transformation from the graph of f to the graph of g.

Question 5.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 152
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.6 Question 5

Question 6.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 153
Answer:
The given graph is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 153
From the above graph,
The given functions are:
f (x) = x – 3
g (x) = f (x + 4)
So,
g (x) = (x + 4) – 3
g (x) = x + 1
Hence,
When we observe f (x) and g (x), we can say that
g (x) shrinks by 2 units of f (x)

Question 7.
f(x) = \(\frac{1}{3}\)x + 3; g(x) = f(x) = -3
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.6 Question 7

Question 8.
f(x) = -3x + 4; g(x) = f(x) + 1
Answer:
The given functions are:
f (x) = -3x + 4
g (x) = f (x) + 1
So,
g (x) = -3x + 4 + 1
g (x) = -3x + 5
Hence,
The representation of f and g in the same coordinate plane is:

Hence, from the above,
We can conclude that g (x) translates 1 unit away from f (x)

Question 9.
f(x) = -x – 2; g(x) = f(x + 5)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.6 Question 9

Question 10.
f(x) = \(\frac{1}{2}\)x – 5; g(x) = f(x – 3)
Answer:
The given functions are:
f (x) = \(\frac{1}{2}\)x – 5
g (x) = f (x – 3)
So,
g (x) = \(\frac{1}{2}\) ( x – 3 ) – 5
Hence,
The representation of f (x) and g (x) in the same coordinate plane is:

Hence, from the above,
We can conclude that g (x) is dilated from f (x)

Question 11.
MODELING WITH MATHEMATICS
You and a friend start biking from the same location. Your distance d (in miles) after t minutes is given by the function d(t) = \(\frac{1}{5}\)t. Your friend starts biking 5 minutes after you. Your friend’s distance f is given by the function f(t) = d(t – 5). Describe the transformation from the graph of d to the graph of f.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 154
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.6 Question 11

Question 12.
MODELING WITH MATHEMATICS
The total cost C (in dollars) to cater an event with p people is given by the function C(p) = 18p + 50. The set-up fee increases by $25. The new total cost T is given by the function T(p) = C(p) + 25. Describe the transformation from the graph of C to the graph of T.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 155
Answer:
It is given that the total cost C (in dollars) to cater an event with p people is given by the function
C(p) = 18p + 50
It is also given that the set-up fee increases by $25.
So,
The new total cost T is given by the function
T(p) = C(p) + 25
So,
T (p) = 18p + 50 + 25
T (p) = 18p + 75
Hence,
The representation of T (p) and C (p) in the same coordinate plane is:

Hence, from the above,
We can conclude that T (p) translated 25 units away from C (p)

In Exercises 13–16, use the graphs of f and h to describe the transformation from the graph of f to the graph of h.

Question 13.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 156
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.6 Question 13

Question 14.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 157
Answer:
The given graph is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 157
From the above graph,
The given functions are:
f (x) = -3x + 1
h (x) = f (-x)
So,
h (x) = -3 (-x) + 1
h (x) = 3x + 1
Hence,
The representations of f (x) and h (x) in the same coordinate plane is:

Hence, from the above,
We can conclude that f (x) and h (x) are perpendicular lines

Question 15.
f(x) = -5 – x; h(x) = f(-x)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.6 Question 15

Question 16.
f(x) = \(\frac{1}{4}\)x – 2; h(x) = -f(x)
Answer:
The given functions are:
f (x) = \(\frac{1}{4}\)x – 2
h (x) = – f(x)
So,
h (x) = – (\(\frac{1}{4}\)x – 2)
Hence,
The representation of f (x) and h (x) in the same coordinate plane is:

Hence, from the above,
We can conclude that h (x) is a reflection of f (x) on the y-axis

In Exercises 17–22, use the graphs of f and r to describe the transformation from the graph of f to the graph of r.

Question 17.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 158
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.6 Question 17

Question 18.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 159
Answer:
The given graph is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 159
From the above graph,
The given functions are:
f (x) = -x
r (x) = f (\(\frac{1}{4}\) x)
So,
r (x) = –\(\frac{1}{4}\)x
Hence,
The representation of f (x) and r (x) in the same coordinate plane is:

Hence, from the above,
We can conclude that r (x) shrinks by \(\frac{1}{4}\) of f (x)

Question 19.
f(x) = -2x – 4; r(x) = f(\(\frac{1}{2}\)x )
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.6 Question 19

Question 20.
f(x) = 3x + 5; r(x) = f (\(\frac{1}{3}\)x)
Answer:
The given functions are:
f (x) = 3x + 5
r (x) = f (\(\frac{1}{3}\)x)
So,
r (x) = 3 ( \(\frac{1}{3}\) x ) + 5
r (x) = x + 5
Hence,
The representation of f (x) and r (x) in the same coordinate plane is:

Hence, from the above,
We can conclude that r (x) shrinks by 1 unit of f (x)

Question 21.
f(x) = \(\frac{2}{3}\)x + 1; r(x) = 3f(x)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.6 Question 21

Question 22.
f(x) = –\(\frac{1}{4}\)x – 2; r(x) = 4f(x)
Answer:
The given functions are:
f (x) = –\(\frac{1}{4}\)x – 2
r (x) = 4 f (x)
So,
r (x) = 4 [-\(\frac{1}{4}\)x – 2]
r (x) = -x – 8
Hence,
The representation of f (x) and r (x) in the same coordinate plane is:

Hence, from the above,
We can conclude that r (x) translates 4 units away from  f (x)

In Exercises 23–28, use the graphs of f and h to describe the transformation from the graph of f to the graph of h.

Question 23.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 160
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.6 Question 23

Question 24.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 161
Answer:
The given graph is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 161
From the above grah,
The given functions are:
f (x) = -2x – 6
h (x) = \(\frac{1}{3}\) f(x)
So,
h (x) = –\(\frac{1}{3}\) (2x + 6)
Hence,
The representation of f (x) and h (x) in the same coordinate plane is:

Hence, from the above,
We can conclude that h (x) stretches by \(\frac{1}{3}\) of f (x)

Question 25.
f(x) = 3x – 12; h(x) = \(\frac{1}{6}\)f(x)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.6 Question 25

Question 26.
f(x) = -x + 1; h(x) = f(2x)
Answer:
The given functions are:
f (x) = -x + 1
h (x) = f (2x)
So,
h (x) = -(2x) + 1
h (x) = -2x + 1
Hence,
The representation of f (x) and h (x) in the same coordinate plane is:

Hence, from the above,
We can conclude that h (x) stretches by 1 unit from f (x)

Question 27.
f(x) = -2x – 2; h(x) = f(5x)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.6 Question 27

Question 28.
f(x) = 4x + 8; h(x) = \(\frac{3}{4}\)f(x)
Answer:
The given functions are:
f (x) = 4x + 8
h (x) = \(\frac{3}{4}\) f (x)
So,
h (x) = \(\frac{3}{4}\) (4x + 8)
h (x) = \(\frac{3}{4}\) (4x) + \(\frac{3}{4}\) (8)
h (x) = 3x + 6
Hence,
The representation of f (x) and h (x) in the same coordinate plane is:

Hence, from the above,
We can conclude that h (x) shrinks by 2 units from f (x)

In Exercises 29–34, use the graphs of f and g to describe the transformation from the graph of f to the graph of g.

Question 29.
f(x) = x – 2; g(x) = \(\frac{1}{4}\)f(x)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.6 Question 29

Question 30.
f(x) = -4x + 8; g(x) = -f(x)
Answer:
The given functions are:
f (x) = -4x + 8
g (x) = -f (x)
So,
g (x) = -4 (-x) + 8
g (x) = 4x + 8
Hence,
The representation of f (x) and g (x) in the same coordinate plane is:

Hence, from the above graph,
We can conclude that g (x) is a reflection of f (x)

Question 31.
f(x) = -2x – 7; g(x) = f(x – 2)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.6 Question 31

Question 32.
f(x) = 3x + 8; g(x) = f(\(\frac{2}{3}\)x)
Answer:
The given functions are:
f (x) = 3x + 8
g (x) = f (\(\frac{2}{3}\)x)
So,
g (x) = 3 ( \(\frac{2}{3}\) ) x + 8
g (x) = 2x + 8
Hence,
The representation of f (x) and g (x) in the same coordinate plane is:

Hence, from the above,
We can conclude that g (x) shrink by 1 unit of the graph of f (x)

Question 33.
f(x) = x – 6; g(x) = 6f(x)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.6 Question 33

Question 34.
f(x) = -x; g(x) = f(x) -3
Answer:
The given functions are:
f (x) = -x
g (x) = f (x) – 3
So,
g (x) = -x – 3
Hence,
The representation of f (x) and g (x) in the same coordinate plane is:

Hence, from the above,
We can conclude that g (x) is translated by 3 units away from f (x)

In Exercises 35–38, write a function g in terms of f so that the statement is true.

Question 35.
The graph of g is a horizontal translation 2 units right of the graph of f.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.6 Question 35

Question 36.
The graph of g is a reflection in the y-axis of the graph of f.
Answer:
We know that,
f can be writen as f (x)
g can be written as g (x)
Now,
The given statement is:
The graph of g is a reflection of f(x) in the y-axis of the graph of f (x)
So,
Reflection in the y-axis means if f (x) is +ve, then the reflection of f (x) will be -ve and vice-versa
Hence,
The representation of the given statement in terms of f(x) is:
g (x) = f (-x)

Question 37.
The graph of g is a vertical stretch by a factor of 4 of the graph of f.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.6 Question 37

Question 38.
The graph of g is a horizontal shrink by a factor of \(\frac{1}{5}\) of the graph of f.
Answer:
We know that,
f can be writen as f (x)
g can be written as g (x)
Now,
The given statement is:
The graph of g is a horizontal shrink by a factor of \(\frac{1}{5}\) of the graph
We know that,
“Shrink” is represented by ‘-‘
“Stretch” is represented by ‘+’
Hence,
The representation of the given statement in terms of f (x) is:
g (x) = – \(\frac{1}{5}\) f (x)

ERROR ANALYSIS In Exercises 39 and 40, describe and correct the error in graphing g.

Question 39.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 162
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.6 Question 39

Question 40.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 163
Answer:
The given functions are:
f (x) = -x + 3 and g (x) = f (-x)
So,
g (x) = -[-x] + 3
g (x) = x + 3
Hence,
The representations of f (x) and g (x) in the same coordinate plane is:

Hence, from the above,
We can conclude that f (x) and g (x) are perpendicular lines from the graph

In Exercises 41–46, graph f and h. Describe the transformations from the graph of f to the graph of h.

Question 41.
f(x) = x; h(x) = \(\frac{1}{3}\)x + 1
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.6 Question 41

Question 42.
f(x) = x; h(x) = 4x – 2
Answer:
The given functions are:
f (x) = x and h (x) = 4x – 2
Hence,
The representation of f (x) and h (x) in the same coordinate plane is:

Hence, from the above,
We can conclude that the transformations are a vertical stretch by a factor of 4 followed by a vertical translation of 2 units down

Question 43.
f(x) = x; h(x) = -3x – 4
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.6 Question 43

Question 44.
f(x) = x; h(x) = –\(\frac{1}{2}\)x + 3
Answer:
The given functions are:
f (x) = x and h (x) = –\(\frac{1}{2}\)x + 3
Hence,
The representation of f (x) and h (x) in the same coordinate plane is:

Hence, from the above,
We can conclude that the transformations are a vertical shrink by a factor of \(\frac{1}{2}\) followed by a vertical translation of 3 units up

Question 45.
f(x) = 2x; h(x) = 6x – 5
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.6 Question 45

Question 46.
f(x) = 3x; h(x) = -3x – 7
Answer:
The given functions are:
f (x) = 3x and h (x) = -3x – 7
Hence,
The representation of f (x) and h (x) in the same coordinate plane is:

Hence, from the above,
We can conclude that the transformations are a vertical shrink by a factor of 3 followed by a vertical translation of 7 units down

Question 47.
MODELING WITH MATHEMATICS
The function t(x) = -4x +72 represents the temperature from 5 P.M. to 11 P.M., where x is the number of hours after 5 P.M. The function d(x) = 4x + 72 represents the temperature from 10 A.M. to 4 P.M., where x is the number of hours after 10 A.M. Describe the transformation from the graph of t to the graph of d.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 164
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.6 Question 47

Question 48.
MODELING WITH MATHEMATICS
A school sells T-shirts to promote school spirit. The school’s profit is given by the function P(x) = 8x – 150, where x is the number of T-shirts sold. During the playoffs, the school increases the price of the T-shirts. The school’s profit during the play-offs is given by the function Q(x) = 16x – 200, where x is the number of T-shirts sold. Describe the transformations from the graph of P to the graph of Q.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 164.1
Answer:
It is given that a school sells T-shirts to promote school spirit. The school’s profit is given by the function
P(x) =8x- 150
where,
x is the number of T-shirts sold
It is also given that during the playoffs, the school increases the price of the T-shirts. The school’s profit during the play-offs is given by the function
Q(x) = 16x – 200
where,
x is the number of T-shirts sold.
Hence,
The representation of P (x) and Q(x) in the same coordinate plane is:

Hence, from the above graph,
We can conclude that
The x-axis shrink by a factor of 2
The y-axis shrink by 50

Question 49.
USING STRUCTURE
The graph of g(x) = a • f(x – b) + c is a transformation of the graph of the linear function f. Select the word or value that makes each statement true.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 165
a. The graph of g is a vertical ______ of the graph of f when a = 4, b = 0, and c = 0.
b. The graph of g is a horizontal translation ______ of the graph of f when a = 1, b = 2, and c = 0.
c. The graph of g is a vertical translation 1 unit up of the graph of f when a = 1, b = 0, and c = ____.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.6 Question 49

Question 50.
USING STRUCTURE
The graph of h(x) = a • f(bx – c) + d is a transformation of the graph of the linear function f. Select the word or value that makes each statement true.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 166
a. The graph of h is a ______ shrink of the graph of f when a = \(\frac{1}{3}\), b = 1, c = 0, and d = 0.
b. The graph of h is a reflection in the ______ of the graph of f when a = 1, b = -1, c = 0, and d = 0.
c. The graph of h is a horizontal stretch of the graph of f by a factor of 5 when a = 1, b = _____, c = 0, and d = 0.
Answer:
The given original function is:
f (x) = x
The given choice of words to make the given statements true is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 166
The transformation of the function of f (x) is given as:
h (x) = a f (bx – c) + d
Where,
a, b, c, and d are constants
Now,
a.
The given values of constants are:
a = \(\frac{1}{3}\),
b = 1
c = 0
d = 0
So,
h (x) = \(\frac{1}{3}\)f (1 (x) – 0) + 0
h (x) = \(\frac{1}{3}\) f (x)
So,
h (x) = \(\frac{1}{3}\)x
Hence,
The representations of f (x) and h (x) in the coordinate plane is:

Hence, from the above,
We can conclude that the graph of h is a vertical shrink of the graph of f when a = \(\frac{1}{3}\), b = 1, c = 0, and d = 0.

b.
The given values of constants are:
a = 1
b = -1
c = 0
d = 0
So,
h (x) = 1f (-1 (x) – 0) + 0
h (x) =  f (-x)
So,
h (x) = -x
Hence,
The representations of f (x) and h (x) in the coordinate plane is:

Hence, from the above,
We can conclude that the graph of h is a reflection in the y-axis of the graph of f when a = 1, b = -1, c = 0, and d = 0

c.
The given values of constants are:
a = 1
b = p
c = 0
d = 0
So,
h (x) = 1f (p (x) – 0) + 0
h (x) =  f (px)
So,
h (x) = px
Hence,
The representations of f (x) and h (x) in the coordinate plane is:

Hence, from the above,
We can conclude that the graph of h is a horizontal stretch of the graph of f by a factor of 5 when a = 1, b = \(\frac{1}{5}\), c = 0, and d = 0

Question 51.
ANALYZING GRAPHS
Which of the graphs are related by only a translation? Explain.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 167
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.6 Question 51

Question 52.
ANALYZING RELATIONSHIPS
A swimming pool is filled with water by a hose at a rate of 1020 gallons per hour. The amount v (in gallons) of water in the pool after t hours is given by the function v(t) = 1020t. How does the graph of v change in each situation?
a. A larger hose is found. Then the pool is filled at a rate of 1360 gallons per hour.
Answer:
It is given that a swimming pool is filled with water by a hose at a rate of 1020 gallons per hour. The amount v (in gallons) of water in the pool after t hours is given by the function
v(t) = 1020t
Now,
It is also given that a pool is filled at a rate of 1360 gallons per hour
So,
The overall rate change = 1340 / 1020
= 440 / 330
= 4 / 3
So,
The new rate of change is representad by the function
f (t) = (4 / 3) v (t)
Hence,
The representation of the graph of v using the functions v (t) = 1020t and v (t) = 1360t is:

b. Before filling up the pool with a hose, a water truck adds 2000 gallons of water to the pool.
Answer:
We know that,
From part (a),
The function for filling up the pool with a hose is given as:
v (t) = 1020t
It is given that before filling up the pool with a hose, a water truck adds 2000 gallons of water to the pool
So,
v (t) = 1020t + 2000
Hence,
The representations of the functions before filling the hose and  a water truck adds 2000 gallons of water are:

Question 53.
ANALYZING RELATIONSHIPS
You have $50 to spend on fabric for a blanket. The amount m (in dollars) of money you have after buying y yards of fabric is given by the function m(y) = -9.98y + 50. How does the graph of m change in each situation?
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 168
a. You receive an additional $10 to spend on the fabric.
b. The fabric goes on sale, and each yard now costs $4.99.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.6 Question 53

Question 54.
THOUGHT-PROVOKING
Write a function g whose graph passes through the point (4, 2) and is a transformation of the graph of f(x) = x.
Answer:
It is given that a function g whose graph passes through the point (4, 2) and is a transformation of the graph of f(x) = x
We know that,
The equation that passes through (h, k) and it is a transformation of another function is:
y = fx – h) + k
So,
From (4, 2)
As the translation between the values of x and y in the given point is 2,
We can make the value of ‘h’ as -2
We can make the value of k as -4
Hence,
The equation that passes through (4, 2) is:
y = (x + 2 ) – 4
y = x – 2
Hence, from the above,
We can conclude that the function that passes through (4, 2) is:
g (x) = x – 2

In Exercises 55–60, graph f and g. Write g in terms of f. Describe the transformation from the graph of f to the graph of g.

Question 55.
f(x) = 2x – 5; g(x) = 2x – 8
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.6 Question 55

Question 56.
f(x) = 4x + 1; g(x) = -4x – 1
Answer:
The given equations are:
f (x) = 4x + 1
g (x) = -4x – 1
So,
g (x) = – (4x + 1)
g (x) = -f (x)
Hence,
The representations of f (x) and g (x) in the coordinate plane is:

Hence, from the above,
We can conclude that f (x) and g (x) are on the opposite quadrants

Question 57.
f(x) = 3x + 9; g(x) = 3x + 15
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.6 Question 57

Question 58.
f(x) = -x – 4 ; g(x) = x – 4
Answer:
The given equations are:
f (x) = -x – 4
g (x) = x – 4
Hence,
The representations of f (x) and g (x) in the coordinate plane are:

Hence, from the above,
We can conclude that f (x) and g (x) are perpendicular lines since only the slopes vary and the y-intercepts are the same

Question 59.
f(x) = x + 2; g(x) = \(\frac{2}{3}\)x + 2
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.6 Question 59

Question 60.
f(x) = x – 1; g(x) = 3x – 3
Answer:
The given equations are:
f (x) = x – 1
g (x) = 3x – 3
So,
g (x) = 3 (x – 1)
g (x) = 3 f (x)
Hence,
The representations of f (x) and g (x) in the coordinate plane are:

Hence, from the above,
We can conclude that g (x) translates 3 units away from f (x)

Question 61.
REASONING
The graph of f(x) = x + 5 is a vertical translation 5 units up of the graph of f(x) = x. How can you obtain the graph of f(x) = x + 5 from the graph of f(x) = x using a horizontal translation?
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.6 Question 61

Question 62.
HOW DO YOU SEE IT? Match each function with its graph. Explain your reasoning.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 169
a. a(x) = f(-x)
b. g(x) = f(x) – 4
c. h(x) = f(x) + 2
d. k(x) = f(3x)
Answer:

REASONING
In Exercises 63–66, find the value of r.

Question 63.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 170
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.6 Question 63

Question 64.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 171
Answer:
The given equations are:
f (x) = -3x + 5
g (x) = f (rx)
So,
g (x) = -3 (rx) + 5
Where,
r is the transformational factor
From the graph,
The x-intercept of g(x) is: 5
Now,
To find he x-intercept, put y = 0
So,
0 = -3 (rx) + 5
3 (rx) = 5
3r (5) = 5
3r = 5 / 5
3r = 1
r = \(\frac{1}{3}\)
Hence, from the above,
WE can conclude that the value of r is: \(\frac{1}{3}\)

Question 65.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 172
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.6 Question 65

Question 66.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 173
Answer:
The given equations are:
f (x) = \(\frac{1}{2}\)x + 8
g (x) = f (x) + r
So,
g (x) = \(\frac{1}{2}\)x + 8 + r
From the graph,
We can observe that f (x) and g (x) are both the parallel lines
So,
The slopes of f (x) and g (x) are the same and only the y-intercepts differ in f (x) and g (x)
From the graph,
We can observe that the y-intercept of g (x) is: 0
So,
g (x) = \(\frac{1}{2}\)x + 8 + 0
Hence, from the above,
We can conclude that the value of r is: 0

Question 67.
CRITICAL THINKING
When is the graph of y = f(x) + w the same as the graph of y = f(x + w) for linear functions? Explain your reasoning.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.6 Question 67

Maintaining Mathematical Proficiency

Solve the formula for the indicated variable.(Section 1.5)

Question 68.
Solve for h.
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 174
Answer:
The given formula from the given figure is:
V = πr²h
Divide by πr² into both sides
So,
\(\frac{V}{πr²}\) = h
Hence, from the above,
We can conclude that the formula for ‘h’ is:
h = \(\frac{V}{πr²}\)

Question 69.
Solve for w.
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 175
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.6 Question 69

Solve the inequality. Graph the solution, if possible. (Section 2.6)

Question 70.
| x – 3 | ≤ 14
Answer:
The given absolute value inequality is:
| x + 3 | ≤ 14
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
x + 3 ≤ 14 and x + 3 ≥ -14
x ≤ 14 – 3 and x ≥ -14 – 3
x ≤ 11 and x ≥ -17
Hence, from the above,
We can conclude that the solutions to the given absolute value inequality are:
x ≤11 and x ≥ -17
The representation of the solutions of the given inequality in the graph is:

Question 71.
| 2x + 4 | > 16
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.6 Question 71

Question 72.
5 | x + 7 | < 25
Answer:
The given absolute value inequality is:
5 | x + 7 | < 25
So,
| x + 7 | < 25 / 5
| x + 7 | < 5
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
x + 7 < 5 and x + 7 > -5
x < 5 – 7 and x > -5 – 7
x < -2 and x > -12
Hence, from the above,
We can conclude that the solutions of the given absolute value inequality are:
x < -2 and x > -12
The representations of the solutions of the given absolute value inequality in the graph is:

Question 73.
-2 | x + 1 | ≥ 18
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.6 Question 73

Lesson 3.7 Graphing Absolute Value Functions

Essential Question
How do the values of a, h, and k affect the graph of the absolute value function g(x) = a | x – h | + k? The parent absolute value function is f(x) = | x | Parent absolute value function
The graph of f is V-shaped.
Answer:
The given absolute value function is:
g (x) = a | x – h | + k
The parent function for the given absolute value function is:
f (x) = | x |
It is also given that the graph of f is v-shaped
Now,
The given absolute value function has the constants a, x, h, and k
Now,
“x – h” represents the translation of the x-axis where h is the translation value
Ex:
x – h =x – 3
Where,
x is the original function and h is the translation factor of x
Now,
“k” represents the y-intercept of the given absolute value function
The value of k affects the graph of the given absolute value function to translate on the y-axis up and down or shrink and stretch

EXPLORATION 1
Identifying Graphs of Absolute Value Functions
Work with a partner.
Match each absolute value function with its graph. Then use a graphing calculator to verify your answers.
a. g(x) = | x – 2 |
b. g(x) = | x – 2 | + 2
c. g(x) = | x + 2 | + 2
d. g(x) = – | x + 2 | + 2
e. g(x) = 2 | x – 2 |
f. g(x) = – | x + 2 | + 2
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 176
Answer:
The given absolute value functions are:
a. g(x) = | x – 2 |
b. g(x) = | x – 2 | + 2
c. g(x) = | x + 2 | + 2
d. g(x) = – | x + 2 | + 2
e. g(x) = 2 | x – 2 |
f. g(x) = – | x + 2 | + 2
We know that,
| x | =x for x > 0
| x | = -x for x < 0
Now,
a.
The given absolute value function is:
g(x) = | x – 2 |
So,
g (x) = x – 2 or g (x) = – (x – 2)
g (x) = x – 2 or g (x) = 2 – x
Hence,
The representation of the given absolute value function in the coordinate plane is:

Hence, fro the above,
We can conclude that graph F) matches the given absolute value function
b.
The given absolute value function is:
g(x) = | x – 2 | + 2
So,
g (x) = x – 2 + 2 or g (x) = – (x – 2) + 2
g (x) = x or g (x) = 2 – x + 2
g (x) = x or g (x) = 4 – x
Hence,
The representation of the given absolute value function in the coordinate plane is:

Hence, from the above,
We can conclude that graph C) matches the given absolute value function
c.
The given absolute value function is:
g(x) = | x + 2 | + 2
So,
g (x) = x + 2 + 2 or g (x) = – (x + 2 ) + 2
g (x) = x + 4 or  g (x) = -x – 2 + 2
g (x) = x + 4 or g (x) = -x
Hence,
The representation of the given absolute value function in the coordinate plane is:

Hence, from the above,
We can conclude that graph C) matches the given absolute value function
d.
The given absolute value function is:
g(x) = – | x + 2 | + 2
So,
g (x) = – (x + 2 )  + 2 or g (x) = – [- (x + 2 ) ] + 2
g (x) = -x – 2 + 2 or g (x) = x + 2 + 2
g (x) = x or g (x) =x + 4
Hence,
The representation of the given absolute value function in the coordinate plane is:

Hence, from the above,
We can conclude that graph B) matches the given absolute value function
e.
The given absolute value function is:
g(x) = 2 | x – 2 |
So,
g (x) = 2 (x – 2) or g (x) = 2 ( – (x – 2) )
g (x) = 2x – 4 or g (x) = 2 (-x + 2 )
g (x)= 2x – 4 or g (x) = -2x + 4
Hence,
The representation of the given absolue value functions in the coordinate plane is:

Hence, from the above,
We can conclude that the graph A) matches the given absolute value function
f.
The given absolute value function is:
g(x) = – | x + 2 | + 2
So,
g (x) = – (x + 2) + 2 or g (x) = – [- (x – 2} ] + 7
g (x) = x + 4 or g (x) = 2 – x + 7
g (x) = x + 4 or g (x) = -x + 9
Hence,
The representation of the given absolute value function in the coordinate plane is:

Hence, from the above,
We can conclude that graph E) matches the given absolute value function

Communicate Your Answer

Question 2.
How do the values of a, h, and k affect the graph of the absolute value function g(x) = a | x – h | + k?
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 176.1
Answer:
The given absolute value function is:
g (x) = a | x – h | + k
From the above absolute value function,
We can say that
a, h and k are constants
Now,
“a” represents the constant to multiply g (x)
“h” represents the translation factor on the x-axis
“k” represents the y-intercept of the given function
The value of “h” and “k” affect the graph in such a way that the graph can move anywhere in the graph

Question 3.
Write the equation of the absolute value function whose graph is shown. Use a graphing calculator to verify your equation.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 177
Answer:
The given graphing calculator is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 177
From the above graph,
We can observe that the plot is in downwqrd direction i.e., the absolute value fuction is negative and the plot passes through 2 in the x-axis
Hence,
The absolute value function representing the given graph is:
f (x) = -| x + 1 | + 2

3.7 Lesson

Monitoring Progress

Graph the function. Compare the graph to the graph of f(x) = | x |. Describe the domain and range.

Question 1.
h(x) = | x | – 1
Answer:
The given functions are:
f (x) = | x |
h (x) = | x | – 1
Hence,
The representation of f (x) and h (x) in  the same coordinate plane is:

Hence, from the above graph,
h (x) is translated vertically away from f (x) on the y-axis by 1unit
Now,
The domain of the given functions are: -10 ≤ x ≤ 10
The range of the given functions are: -1 ≤ y ≤ 10

Question 2.
n(x) = | x + 4 |
Answer:
The given functions are:
f (x) = | x |
h (x) = | x + 4 |
Hence,
The representation of f (x) and h (x) in the same coordinate plane is:

Hence, from the above graph,
h (x) is translated from f (x) 4 units away on the x-axis
Now,
The domain of f (x) is: -10 ≤ x ≤ 10
The domain of h (x) is: -10 ≤ x ≤6
The range of f (x) is: 0 ≤ y ≤ 10
The range of h (x) is: 0 ≤y ≤ 6

Graph the function. Compare the graph to the graph of f(x) = | x |. Describe the domain and range.

Question 3.
t(x) = -3| x |
Answer:
The given functions are:
f (x) = | x |
t (x) = -3 | x |
Hence,
The representation of f (x) and t (x) in the same coordinate plane is:

Hence, from the above graph,
t (x) is shrunk by 3units and rotated by 90° on the y-axis
Now,
The domain of f (x) is: -10 ≤ x ≤ 10
The domain of t (x) is: -3 ≤x ≤ 3
The range of f (x) is: 0 ≤ y ≤ 10
The range of t (x) is: -10 ≤ y ≤ 0

Question 4.
v(x) = \(\frac{1}{4}\)| x |
Answer:
The given functions are:
f (x) = | x |
v (x) = \(\frac{1}{4}\)| | x |
Hence,
The representation of f (x) and v (x) in the same coordinate plane is:

Hence, from the above graph,
t (x) translates \(\frac{1}{4}\)| units of f (x) on the y-axis
Now,
The domain of the given functions are: -10 ≤ x ≤ 10
The range of f (x) is: 0 ≤ y ≤ 10
The range of t (x) is: 0 ≤ y ≤ 2.5

Question 5.
Graph f(x) = | x – 1 | and g(x) = | \(\frac{1}{2}\)x – 1 |.
Compare the graph of g to the graph of f.
Answer:
The given functions are:
f (x) = | x – 1 |
g (x) = | \(\frac{1}{2}\)x – 1 |
Hence,
The representation of f (x) and g (x) in the same coordinate plane is:

Hence, from the above graph,
g (x) translates 1 unit away from f (x) on the x-axis
Now,
The domain of f (x) is: -8 ≤ x ≤  10
The domain of g (x) is: -10 ≤  x ≤  10
The range of f (x) is: 0 ≤  y ≤ 10
The range of g (x) is: 0 ≤  y ≤  6

Question 6.
Graph f(x) = | x + 2 | + 2 and g(x) = | -4x + 2 | + 2.
Compare the graph of g to the graph of f.
Answer:
The given functions are:
f (x) = | x + 2 | + 2
g (x) = | -4x + 2 | + 2
Hence,
The representation of f (x) and g(x) in the same coordinate plane is:

Hence, from the above graph,
g (x) translates \(\frac{3}{2}\) units away from f (x) on the x-axis
Now,
The domain of f (x) is: -10 ≤ x ≤ 6
The domain of g (x) is:-1.5 ≤ x ≤ 2.5
The range of f (x) is: 0 ≤ y ≤ 10
The range of g (x) is: 2 ≤y ≤ 10

Question 7.
Let g(x) = | –\(\frac{1}{2}\)x + 2 | + 1.
(a) Describe the transformations from the graph of f(x) = | x | to the graph of g.
(b) Graph g.
Answer:
a.
The given functions are:
f (x) = | x |
g (x) = | –\(\frac{1}{2}\)x + 2 | + 1
Hence,
The representation of f (x) and g (x) in the same coordinate plane is:

Hence, from the above graph,
We can observe that
g (x) translates 4 units away from f (x) on the positive x-axis
g (x) translates 2 units away from f (x) on the negative x-axis

b.
The function of g (x) is:
| –\(\frac{1}{2}\)x + 2 | + 1
We know that,
g can be written as g (x)
Hence,
The representation of g (x) in the coordinate plane is:

Graphing Absolute Value Functions 3.7 Exercises

Vocabulary and Core Concept Check

Question 1.
COMPLETE THE SENTENCE
The point (1, -4) is the _______ of the graph of f(x) = -3 | x – 1 | – 4.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.7 Question 1

Question 2.
USING STRUCTURE
How do you know whether the graph of f(x) = a | x – h | + k is a vertical stretch or a vertical shrink of the graph of f(x) = | x | ?
Answer:
The given function is:
f (x) = a | x – h | + k
Where,
a is the vertical stretch or vertical shrink
h is the horizontal translation
k is the vertical shift
Now,
‘a’ will be the vertical stretch if the value of a is an integer
‘a’ will be the vertical shrink if the value of ‘a’ is a fraction
Hence, from the above,
We can conclude that the given function is a vertical stretch

Question 3.
WRITING
Describe three different types of transformations of the graph of an absolute value function.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.7 Question 3

Question 4.
REASONING
The graph of which function has the same y-intercept as the graph of f(x) = | x – 2 | + 5? Explain.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 178
Answer:
The given absolute value functions are:
a. f (x) = | x – 2 | + 5
b. g (x) = | 3x – 2 | + 5
c. h (x) = 3 | x – 2 | + 5
We know that,
| x | = x for x > 0
| x | = -x for x < 0
Now,
a.
The given absolute value function is:
f (x) = | x – 2 | + 5
So,
f (x) = x – 2 + 5 or f (x) = -(x – 2) + 5
f (x) = x + 3 or f (x) = -x + 2 + 5
f (x) = x + 3 or f (x) = -x + 7

b.
The given absolute value function is:
g (x) = | 3x – 2 | + 5
So,
g (x) = 3x – 2 + 5 or g (x) = – (3x – 2 ) + 5
g (x) = 3x + 3 or g (x) = -3x + 2 + 5
g (x) = 3x + 3 or g (x) = -3x + 7

c.
The given absolute value function is:
h (x) = 3 | x – 2 | + 5
So,
h (x) = 3 (x – 2) + 5 or h (x) = -3 (x – 2) + 5
h (x) = 3x – 6 + 5 or h (x) = -3x + 6 + 5
h (x) = 3x – 1 or h (x) = -3x + 11
Compare the solutions of the given absolute value functions with
f (x) = mx + c
Where,
m is the slope
c is the y-intercept
Hence, from the above,
We can conclude that a) and b ) have the same y-intercepts

Monitoring Progress and Modeling with Mathematics

In Exercises 5–12, graph the function. Compare the graph to the graph of f(x) = | x |. Describe the domain and range.

Question 5.
d(x) = | x | – 4
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.7 Question 5

Question 6.
r(x) = | x | + 5
Answer:
The given functions are:
f (x) = | x |
r (x) = | x | + 5
Hence,
The representation of f (x) and r (x) in the coordinate plane is:

Hence, from the above,
We can conclude that r (x) is 5 units away from f (x)
Now,
The domain of the given absolute value function is: -5 ≤ x ≤ 5
The range of the given absolute value function is: 5 ≤y ≤10

Question 7.
m(x) = | x + 1 |
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.7 Question 7

Question 8.
v(x) = | x – 3 |
Answer:
The given absolute value functions are:
f (x) = | x |
v (x) = | x – 3 |
Hence,
The representation of f(x) and v (x) in the coordinate plane is:

Hence, from the above,
We can conclude that v (x) translates 3 units away from f (x)
Now,
The domain of the given absolute value function is: -6 ≤ x ≤ 10
The range of the given absolute value function is: 6 ≤ y ≤ 10

Question 9.
p(x) = \(\frac{1}{3}\) | x |
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.7 Question 9

Question 10.
j(x) = 3 | x |
Answer:
The given absolute value functions are:
f (x) = | x |
j (x) = 3 | x |
Hence,
The representation of f (x) and j (x) in the same coordinate plane is:

Hence, from the above,
We can conclude that j (x) vertically stretches 3 units away from f (x)
Now,
The domain of the given absolute value function is: -3.5 ≤ x ≤ 3.5
The range of the given absolute value function is: 0 ≤ y ≤ 10

Question 11.
a(x) = -5 | x |
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.7 Question 11

Question 12.
q(x) = – \(\frac{3}{2}\) | x |
Answer:
The given absolute value functions are:
f (x) = | x |
q (x) = –\(\frac{3}{2}\) | x |
Hence,
The representation of f (x) and q (x) in the same coordinate plane is:

Hence, from the above,
We can conclude that q (x) shrinks by \(\frac{3}{2}\) units from f (x)
Now,
The domain of the given absolute value function is: -6.5 ≤ x ≤ 6.5
The range of the given absolute value function is: -10 ≤ y ≤ 0

In Exercises 13–16, graph the function. Compare the graph to the graph of f(x) = | x − 6 |.

Question 13.
h(x) = | x – 6 | + 2
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.7 Question 13

Question 14.
n(x) = \(\frac{1}{2}\) | x – 6 |
Answer:
The given functions are:
f (x) = | x – 6 |
n (x) = \(\frac{1}{2}\) | x – 6 |
Hence,
The representation of f (x) and n (x) in the coordinate plane is:

Hence, from the above,
We can conclude that n (x) vertically stretches \(\frac{1}{2}\) units of f (x)

Question 15.
k(x) = -3 | x – 6 |
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.7 Question 15

Question 16.
g(x) = | x – 1 |
Answer:
The given functions are:
f (x) = | x – 6 |
g (x) = | x – 1 |
Hence,
The representation of f (x) and g (x) in the coordinate plane is:

Hence, from the above,
We can conclude that g (x) translates 4 units away from f (x) on the positive x-axis

In Exercises 17 and 18, graph the function. Compare the graph to the graph of f(x) = | x + 3 | − 2.

Question 17.
y(x) = | x + 4 | – 2
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.7 Question 17

Question 18.
b(x) = | x + 3 | + 3
Answer:
The given absolute value functions are:
f (x) = | x + 3 | – 2
b (x) = | x + 3 | + 3
Hence,
The representation of f (x) and b (x) in the coordinate plane is:

Hence, from the above,
We can conclude that f (x) and b (x) are parallel lines since the slopes are constant and there are different values of the y-intercepts

In Exercises 19–22, compare the graphs. Find the value of h, k, or a.

Question 19.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 179
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.7 Question 19

Question 20.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 180
Answer:
The given graph is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 180
From the above graph,
The given functions are:
f (x) = | x |
t (x) = | x – h |
Where,
h is the horizontal translation
From the graph,
We can observe that the translation occurs at (1, 0) of f (x)
Hence, from the above,
We can conclude that the value of ‘h’ is: 1

Question 21.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 181
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.7 Question 21

Question 22.
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 182
Answer:
The given graph is:
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 182
From the given graph,
The given graphs are:
f (x) = | x |
w (x) = a | x |
Where,
‘a’ is the vertical stretch
From the graph,
We can observe that the vertical stretch occurs at (0, 2)
Hence, from the above,
We can conclude that the value of a is: 2

In Exercises 23–26, write an equation that represents the given transformation(s) of the graph of g(x) = | x |.

Question 23.
vertical translation 7 units down
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.7 Question 23

Question 24.
horizontal translation 10 units left
Answer:
The given statement is:
Horizontal translation 10 units left
Let the absolute value function be y = | x |
So,
“Horizontal translation” means the moving of value on the x-axis either to the left side or the right side i.e., either on the negative or positive x-axis
So,
The given statement in the absolute value function form is:
y = | x | + 10

Question 25.
vertical shrink by a factor of \(\frac{1}{4}\)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.7 Question 25

Question 26.
vertical stretch by a factor of 3 and a reflection in the x-axis
Answer:
The given statement is:
Vertical stretch by a factor of 3 and a reflection in the x-axis
We know that,
The representation of the vertical stretch ‘a’ of the function is:
y = a.f (x)
The representation of the reflection of the function is:
y= f (-x)
Now,
Let the absolute value function be
y = x
Hence,
The representation of the given statement in the absolute value function form is:
Vertical stretch:
y = 3x
Reflection:
y = -3x

In Exercises 27–32, graph and compare the two functions.

Question 27.
f(x) = | x – 4 |; g(x) = | 3x – 4 |
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.7 Question 27

Question 28.
h(x) = | x + 5 |; t(x) = | 2x + 5 |
Answer:
The given functions are:
h (x) = | x + 5 |
t (x) = | 2x + 5 |
Hence,
The representation of h (x) and t (x) in the same coordinate plane is:

Hence, from the above,
We can conclude that t (x) translates 6 units away from h (x) in the negative y-axis

Question 29.
p(x) = | x + 1 | – 2; q(x) = | \(\frac{1}{4}\)x + 1 | – 2
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.7 Question 29

Question 30.
w(x) = | x – 3 | + 4; y(x) = | 5x – 3 | + 4
Answer:
The given absolute value functions are:
w (x) = | x – 3 | + 4
y (x) = | 5x – 3 | + 4
Hence,
The representation of w (x) and y (x) in the coordinate plane is:

Hence, from the above,
We can conclude that w (x) is 7 units away from y (x) in the positive y-axis

Question 31.
a(x) = | x + 2 | + 3; b(x) = | -4x + 2 | + 3
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.7 Question 31

Question 32.
u(x) = | x – 1 | + 2; v(x) = | –\(\frac{1}{2}\)x – 1 | + 2
Answer:
The given absolute value functions are:
u (x) = | x – 1 | + 2
v (x) = | –\(\frac{1}{2}\)x – 1 | + 2
Hence,
The representation of u (x) and v (x) in the coordinate plane is:

Hence, from the above,
We can conclude that u (x) is 1 unit away from v (x) in the positive y-axis

In Exercises 33–40, describe the transformations from the graph of f(x) = | x | to the graph of the given function. Then graph the given function.

Question 33.
r(x) = | x + 2 | – 6
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.7 Question 33

Question 34.
c(x) = | x + 4 | + 4
Answer:
The given absolute value function is:
c (x) = | x + 4 | + 4
Hence,
The representation of c (x) in the coordinate plane is:

Hence, from the above,
We can conclude that c (x) translates 4 units away in the positive x-axis and 4 units vertically in the y-axis

Question 35.
d(x) = – | x – 3 | + 5
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.7 Question 35

Question 36.
v(x) = -3| x + 1 | + 4
Answer:
The given absolute value function is:
v (x) = -3 | x + 1 | + 4
Hence,
The representation of v (x) in the coordinate plane is:

Hence, from the above,
We can conclude that v (x) translates 4 units vertically above and 2 units to the right side of the positive x-axis

Question 37.
m(x) = \(\frac{1}{2}\) | x + 4 | – 1
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.7 Question 37

Question 38.
s(x) = | 2x – 2 | – 3
Answer:
The given absolute value function is:
s (x) = | 2x – 2 | – 3
Hence,
The representation of s (x) in the coordinate plane is:

Hence, from the above,
We can conclude that s (x) translates 4 units away to the right side of the positive x-axis

Question 39.
j(x) = | -x + 1 | – 5
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.7 Question 39

Question 40.
n(x) = | –\(\frac{1}{3}\)x + 1 | + 2
Answer:
The given absolute value function is:
n (x) = | –\(\frac{1}{3}\)x + 1 | + 2
Hence,
The representation of n (x) in the coordinate plane is:

Hence, from the above,
We can conclude that n (x) translates 4 units away from the origin

Question 41.
MODELING WITH MATHEMATICS
The number of pairs of shoes sold s (in thousands) increases and then decreases as described by the function s(t) = -2 | t – 15 | + 50, where t is the time (in weeks).
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 183
a. Graph the function.
b. What is the greatest number of pairs of shoes sold in 1 week?
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.7 Question 41

Question 42.
MODELING WITH MATHEMATICS
On the pool table shown, you bank the five ball off the side represented by the x-axis. The path of the ball is described by the function p(x) = \(\frac{4}{3}\) | x – \(\frac{5}{4}\) |.
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 184
a. At what point does the five-ball bank off the side?
Answer:
It is given that on the pool table shown, you bank the five ball off the side represented by the x-axis. The path of the ball is described by the function
p(x) = \(\frac{4}{3}\) | x – \(\frac{5}{4}\) |
Now,
To find the point where the five-ball bank off the side, draw the plot of the given absolute value function
So,
The representation of the given absolute value function in the coordinate plane is:

Hence, from the above,
The point where the five-ball offside is: (1, 0)

b. Do you make the shot? Explain your reasoning.
Answer:
The given graph is:
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 184
From the graph,
We can observe that the point is accurately going through the hole i.e., at (5, 5)
Hence, from the above,
We can conclude that you make the shot

Question 43.
USING TRANSFORMATIONS
The points A (-\(\frac{1}{2}\), 3) , B(1, 0), and C(-4, -2) lie on the graph of the absolute value function f. Find the coordinates of the points corresponding to A, B, and C on the graph of each function.
a. g(x) = f(x) – 5
b. h(x) = f(x – 3)
c. j(x) = -f(x)
d. k(x) = 4f(x)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.7 Question 43.1
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.7 Question 43.2
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.7 Question 43.3

Question 44.
USING STRUCTURE
Explain how the graph of each function compares to the graph of y = | x | for positive and negative values of k, h, and a.
a.y = | x | + k
b. y = | x – h |
c. y = a| x |
d. y = | ax |
Answer:
The given absolute value functions are:
a.y = | x | + k
b. y = | x – h |
c. y = a| x |
d. y = | ax |
We know that,
| x | = x for x > 0
| x | = -x for x < 0
It is given that
The parent function is:
y = | x |
Now,
a.
The given absolute value function is:
y = | x | + k
Hence,
The given absolute value function translates from the parent function k units away in the positive y-axis vertically
b.
The given absolute value function is:
y = | x – h |
Hence,
The given absolute value function translates from the parent function h units away in the positive x-axis horizontally
c.
The given absolute value function is:
y = a| x |
Hence,
The given absolute value function reflects from the parent function ‘a’ units away
d.
The given absolute value function is:
y = | ax |
Hence,
The given absolute value function dilates from the parent function with the value of ‘a’ units

ERROR ANALYSIS
In Exercises 45 and 46, describe and correct the error in graphing the function.

Question 45.
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 185
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.7 Question 45

Question 46.
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 186
Answer:
The given absolute value function is:
y = -3 | x|
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
y = -3 (x) or y = -3 (-x)
So,
y = -3x or y = 3x
Hence,
The representation of the given absolute value function in the coordinate plane is:

Hence, from the above,
We can conclude that the given absolute value function in downward direction

MATHEMATICAL CONNECTIONS
In Exercises 47 and 48, write an absolute value function whose graph forms a square with the given graph.

Question 47.
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 187
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.7 Question 47

Question 48.
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 188
Answer:
The given graph is:
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 188
From the graph,
The given absolute value function is:
y = | x – 3 | + 1
From the given absolute value function,
We can say that the x-axis translates 3 units away in the positive x-axis and the y-axis is a vertical stretch with the value of k as 1

Question 49.
WRITING
Compare the graphs of p(x) = | x – 6 | and q(x) = | x | – 6.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.7 Question 49

Question 50.
HOW DO YOU SEE IT? The object of a computer game is to break bricks by deflecting a ball toward them using a paddle. The graph shows the current path of the ball and the location of the last brick.
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 189
a. You can move the paddle up, down, left, and right. At what coordinates should you place the paddle to break the last brick? Assume the ball deflects at a right angle.
Answer:
The given graph is:
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 189
It is given that the paddle moves at the right angle using a ball
So,
From the graph,
The coordinates that are at a right angle to break the brick using a paddle is:
(3, 8) and (14, 8)

b. You move the paddle to the coordinates in part (a), and the ball is deflected. How can you write an absolute value function that describes the path of the ball?
Answer:
From part (a),
The coordinates that are used to break the brick using a paddle is:
(3, 8) and (14, 8)
We know that,
For absolute value function,
The parent function will always be:
f (x) = | x |
From part (a),
The y-axis is constant and the x-axis is translating 3 units away and 14 units away
We know that,
The value of the absolute value function with the translation of x-value and  y-value as k-value
So,
f (x) = | x – h | + k
Hence,
The absolute value function that describes the path of the ball is:
f (x) = | x – 3 | + 8
f (x) = | x – 14 | + 8

In Exercises 51–54, graph the function. Then rewrite the absolute value function as two linear functions, one that has the domain x < 0 and one that has the domain x ≥ 0.

Question 51.
y = | x |
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.7 Question 51

Question 52.
y = | x | – 3
Answer:
The given absolute value function is:
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
y = x – 3 or y = -x – 3
Hence,
The representation of the given absolute value function in the coordinate plane is:

Question 53.
y = – | x | + 9
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.7 Question 53

Question 54.
y = – 4 | x |
Answer:
The given absolute value function is:
y = -4 | x |
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
y = -4 (x) or y = -4 (-x)
y = -4x or y = 4x
Hence,
The representation of the given absolute value function in the coordinate plane is:

In Exercises 55–58, graph and compare the two functions.

Question 55.
f(x) = | x – 1 | + 2; g(x) = 4 | x – 1 | + 8
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.7 Question 55

Question 56.
s(x) = | 2x – 5 | – 6; t(x) = \(\frac{1}{2}\) | 2x – 5 | – 3
Answer:
The given functions are:
s (x) = | 2x – 5 | – 6
t (x) = \(\frac{1}{2}\) | 2x – 5 | – 3
Hence,
The representation of s (x) and t (x) in the coordinate plane is:

Hence, from the above,
We can conclude that s (x) translates 60 units away from t (x)

Question 57.
v(x) = -2 | 3x + 1 | + 4; w(x) = 3 | 3x + 1 | – 6
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.7 Question 57

Question 58.
c(x) = 4 | x + 3 | – 1; d(x) = –\(\frac{4}{3}\) | x + 3 | + \(\frac{1}{3}\)
Answer:
The given functions are:
c (x) = 4 | x + 3 | – 1
d (x) =-\(\frac{4}{3}\) | x + 3 | + \(\frac{1}{3}\)
Hence,
The representation of c (x) and d (x) in the same coordinate plane is:

Hence, from the above,
We can conclude that d(x) and c (x) are opposite to each other

Question 59.
REASONING
Describe the transformations from the graph of g(x) = -2 | x + 1 | + 4 to the graph of h(x) =| x |. Explain your reasoning.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.7 Question 59

Question 60.
THOUGHT-PROVOKING
Graph an absolute value function f that represents the route a wide receiver runs in a football game. Let the x-axis represent the distance (in yards) across the field horizontally. Let the y-axis represent the distance (in yards) down the field. Be sure to limit the domain so the route is realistic.
Answer:
It is given that
Let the x-axis represent the distance (in yards) across the field horizontally
Let the y-axis represent the distance (in yards) down the field.
Let,
The distance across the field horizontally is: 50 yards
The distance down the field is: 25 yards
Remember we can take any value in the x-axis and the y-axis
So,
(x, y) = (50, 25)
So,
The absolute value function f that represents the route a wide receiver runs in a football game is:
f (x) = | x – 50 | + 25
Hence,
The representation of the above absolute value function in the coordinate plane is:

Hence, from the above,
The domain of the given absolute value function is: x ≥ 0
The range of the given absolute value function is: y > 0

Question 61.
SOLVING BY GRAPHING
Graph y = 2 | x + 2 | – 6 and y = -2 in the same coordinate plane. Use the graph to solve the equation 2 | x + 2 | – 6 = -2. Check your solutions.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.7 Question 61

Question 62.
MAKING AN ARGUMENT
Let p be a positive constant. Your friend says that because the graph of y =| x | + p is a positive vertical translation of the graph of y = | x |, the graph of y = | x + p | is a positive horizontal translation of the graph of y = | x |. Is your friend correct? Explain.
Answer:
It is given that your friend says that because the graph of y =| x | + p is a positive vertical translation of the graph of y = | x |, the graph of y = | x + p | is a positive horizontal translation of the graph of y = | x |.
Now,
The graph of
y = | x | + p is a positive vertical translation because the y-intercept in the given equation is positive
Now,
The graph of
y = | x + p | will be a horizontal translation with negative translation value
But, according to your friend,
y = | x + p | is a positive horizontal translation
Hence, from the above,
We can conclude that your friend is not correct

Question 63.
ABSTRACT REASONING
Write the vertex of the absolute value function f(x) = | ax – h | + k in terms of a, h, and k.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.7 Question 63

Maintaining Mathematical Proficiency

Solve the inequality. (Section 2.4)

Question 64.
8a – 7 ≤ 2(3a – 1)
Answer:
The given inequality is:
8a – 7 ≤ 2 (3a – 1)
So,
8a – 7 ≤ 2 (3a) – 2 (1)
8a – 7 ≤ 6a – 2
8a – 6a ≤ -2 + 7
2a ≤ 5
a ≤ \(\frac{5}{2}\)
Hence, from the above,
We can conclude that the solution to the given inequality is:
a ≤ \(\frac{5}{2}\)

Question 65.
-3(2p + 4) > -6p – 5
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.7 Question 65

Question 66.
4(3h + 1.5) ≥ 6(2h – 2)
Answer:
The given inequality is:
4 (3h + 1.5) ≥ 6 (2h – 2)
So,
4 (3h) + 4 (1.5) ≥ 6 (2h) – 6 (2)
12h + 6 ≥ 12h – 12
12h – 12h + 6 ≥ -12
6 ≥ -12
Since the above inequality is true,
The given inequality has infinite solutions

Question 67.
-4(x + 6) < 2(2x – 9)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.7 Question 67

Find the slope of the line. (Section 3.5)

Question 68.
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 190
Answer:
The given graph is:
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 190
From the above graph,
The points are: (0, 3), (-2, -2)
We know that,
The slope of the give line when (x1, y1), (x2, y2) are given is:
m= \(\frac{y2 – y1}{x2 – x1}\)
So,
(x1, y1) = (0, 3) and (x2, y2) = (-2, -2)
So,
m = \(\frac{-2 – 3}{-2 – 0}\)
m = \(\frac{-5}{-2}\)
m = \(\frac{5}{2}\)
Hence, from the above,
We can conclude that the slope of the given line is: \(\frac{5}{2}\)

Question 69
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 191
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.7 Question 69

Question 70.
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 192
Answer:
The given graph is:
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 192
From the above graph,
The given points are: (-3, 1), (1, -4)
We know that,
The slope of the give line when (x1, y1), (x2, y2) are given is:
m= \(\frac{y2 – y1}{x2 – x1}\)
So,
(x1, y1) = (-3, 1) and (x2, y2) = (1, -4)
So,
m = \(\frac{-4 – 1}{1 – [-3]}\)
m = \(\frac{-5}{1 + 3}\)
m = \(\frac{-5}{4}\)
Hence, from the above,
We can conclude that the slope of the given line is: –\(\frac{5}{4}\)

Graphing Linear Functions Performance Task: The Cost of a T-Shirt

3.4–3.7 What Did You Learn?

Core Vocabulary
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 193

Core Concepts
Section 3.4
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 194

Section 3.5
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 195

Section 3.6
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 196

Section 3.7
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 197

Mathematical Practices

Question 1.
Explain how you determined what units of measure to use for the horizontal and vertical axes in Exercise 37 on page 142.
Answer:
In Exercise 37 on page 142,
The given function is:
d (t)= (1/2)t + 6
We know that,
The horizontal axis represents the independent variable
The vertical axis represents the dependent variable
So,
The horizontal axis represents the time
The vertical axis represents the depth

Question 2.
Explain your plan for solving Exercise 48 on page 153.
Answer:
In Exercise 48 on page 153,
The profits of a school obtained by a school before playoffs and during the playoffs
So,
First, plot the 2 equations in a coordinate plane and compare the functions in the coordinate plane

Performance Task 

The Cost of a T-Shirt

You receive bids for making T-shirts for your class fundraiser from four companies. To present the pricing information, one company uses a table, one company uses a written description, one company uses an equation, and one company uses a graph. How will you compare the different representations and make the final choice? To explore the answers to this question and more, go to Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 198
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 199

Graphing Linear Functions Chapter Review

3.1 Functions (pp. 103 – 110)

Determine whether the relation is a function. Explain.
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 200
Answer:
The given table is:
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 200
We know that,
Every input has exactly one output.
x represents the input
y represents the output
Hence, from the table,
We can conclude that the given table is a function

Determine whether the relation is a function. Explain. 

Question 1.
(0, 1), (5, 6), (7, 9)
Answer:
The given ordered pairs are:
(0, 1), (5, 6), (7, 9)
We know that,
Every input has exactly one output
x represents the input
y represents the output
Hence, from the above,
We can conclude that the given relation is a function

Question 2.
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 201
Answer:
The given graph is:
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 201
We know that,
Every input has exactly one output
x represents the input
y represents the output
Now,
From the graph,
The input ‘2’ has 2 outputs i.e., (2, 0), and (2, 2)
Hence, from the above,
We can conclude that the given graph is not a function

Question 3.
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 202
Answer:
The given relation is:
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 202
We know that,
Every input has exactly one output
x represents the input
y represents the output
Hence, from the above,
We can conclude that the given relation is a function

Question 4.
The function y = 10x + 100 represents the amount y (in dollars) of money in your bank account after you babysit for x hours.
a. Identify the independent and dependent variables.
Answer:
The given function is:
y = 10x + 100
Where,
y represents the amount in dollars
x represents the number of hours
Now,
From the given function,
WE can say that
The independent variable of the given function is: x
The dependent variable of the given function is: y

b. You babysit for 4 hours. Find the domain and range of the function.
Answer:
The given function is:
y = 10x + 100
Where,
y represents the amount in dollars
x represents the number of hours
It is given that you babysit for 4 hours
So,
The given value of x is: 4
So,
The maximum value of y is:
y = 10 (4) + 100
y = 40 + 100
y = $140
We know that,
The amount and the time must not be the negative values
Hence,
The domain of the given function is: 0 ≤ x ≤ 4
The range of the give function is: 0 ≤ y ≤ 140

3.2 Linear Functions (pp. 111–120)

Does the table or equation represent a linear or nonlinear function? Explain.
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 203
Answer:
As x increases by 4, y increases by different amounts. The rate of change is not constant.
So,
The function is nonlinear.

b. y = 3x – 4
Answer:
The equation is in the form y = mx + b.
So,
The equation represents a linear function.

Does the table or graph represent a linear or nonlinear function? Explain. 

Question 5.
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 204
Answer:
The given table is:
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 204
From the given table,
The difference between all the values of x is: 5
The difference between all the values of y is: -3
Since there is a constant difference present between the values of x and the values of y,
The given table is a linear function

Question 6.
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 205
Answer:
The given graph is:
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 205
We know that,
The representation of a linear function in the graph is a “Straight line”
But,
We can say that the given graph is not a straight line from the graph
Hence, from the above,
We can conclude that the given graph is a non-linear function

Question 7.
The function y = 60 – 8x represents the amount y (in dollars) of money you have after buying x movie tickets.
(a) Find the domain of the function. Is the domain discrete or continuous? Explain.
Answer:
The given function is:
y = 60 – 8x
Where,
y represents the amount in dollars
x represents the number of movie tickets
We know that,
The domain is the set of all the values of x so that the given function will be satisfied
Since x represents the number of movie tickets, the value of x can’t be negative and the number of tickets will be infinity
Hence,
The domain of the given function is: 0 ≤ x ≤ ∞
Hence, from the above domain,
We can conclude that the domain is continuous

(b) Graph the function using its domain. Evaluate the function when x = -3, 0 and 5.
Answer:
The given function is:
y = 60 – 8x
Now,
The value of the given function when x = -3 is:
y = 60 – 8 (-3)
y = 60 + 24
y = 84
The value of the given function when x = 0 is:
y = 60 – 8 (0)
y = 60
The value of the given function when x = 5 is:
y = 60 – 8(5)
y = 60 – 40
y = 20
Now,
The representation of the given function along with its domain is:

Question 8.
f(x) = x + 8
Answer:
The given equation is:
f (x) = x + 8
We know that,
The standard representation of the function f (x) is: y
The standard representation of the linear equation is:
y = mx + c
Now,
y = x + 8
By comparing the given equation and the standard representation of the linear equation,
We can conclude that the given equation is a linear equation

Find the value of x, so that the function has the given value.

Question 9.
g(x) = 4 – 3x

Question 10.
k(x) = 7x; k(x) = 49
Answer:
The given function is:
k (x) = 7x with k (x) = 49
So,
49 = 7x
x = 49 / 7
x = 7
Hence, from the above,
We can conclude that the value of the given function is: 7

Question 11.
r(x) = -5x – 1; r(x) = 19
Answer:
The given function is:
r (x) = 5x – 1 with r (x) = 19
So,
19 = 5x – 1
5x = 19 + 1
5x = 20
x = 20 / 5
x = 4
Hence, from the above,
We can conclude that the value of the given function is: 4

Graph the linear function.

Question 12.
g(x) = -2x – 3
Answer:
The given function is:
g (x) = -2x – 3
Hence,
The representation of g (x) in the coordinate plane is:

Question 13.
h(x) = \(\frac{2}{3}\)x + 4
Answer:
The given function is:
h (x) = \(\frac{2}{3}\)x + 4
Hence,
The representation of h (x) in the coordinate plane is:

Question 14.
8x – 4y = 16
Answer:
The given function is:
8x – 4y = 16
4y = 8x – 16
y = \(\frac{8x  16}{4}\)
y = \(\frac{8x}{4}\) – \(\frac{16}{4}\)
y =2x – 4
Hence,
The representation of the given function in the coordinate plane is:

Question 15.
-12x – 3y = 36
Answer:
The given function is:
-12x – 3y = 36
Hence,
The representation of the given function in the coordinate plane is:

Question 16.
y = -5
Answer:
The given function is:
y = -5
Hence,
The representation of the given function in the coordinate plane is:

Question 17.
x = 6
Answer:
The given function is:
x = 6
Hence,
The representation of the given function in the coordinate plane is:

The points represented by the table lie on a line. Find the slope of the line. 

Question 18.
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 206
Answer:
The given table is:
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 206
We know that,
If the input and output values are given in the form of the table,
We can take any 2 pairs of the input and output values to find the slope
So,
The representation of ordered pairs to find the slope is:
(x1, y1) = (6, 9),
(x2, y2) = (11, 15)
Now,
We know that,
The slope of the line = \(\frac{y2 – y1}{x2 – x1}\)
= \(\frac{15 – 9}{11 – 6}\)
= \(\frac{6}{5}\)
Hence, from the above,
We can conclude that the slope of the given line is: \(\frac{6}{5}\)

Question 19.
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 207
Answer:

The given table is:
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 207
We know that,
If the input and output values are given in the form of the table,
We can take any 2 pairs of the input and output values to find the slope
So,
The representation of ordered pairs to find the slope is:
(x1, y1) = (3, -5),
(x2, y2) = (3, -2)
Now,
We know that,
The slope of the line = \(\frac{y2 – y1}{x2 – x1}\)
= \(\frac{-2 – [-5]}{3 – 3}\)
= \(\frac{5 – 2}{0}\)
= Undefined or ∞
Hence, from the above,
We can conclude that the slope of the given line is: Undefined or ∞

Question 20.
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 208
Answer:
The given table is:
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 208
We know that,
If the input and output values are given in the form of the table,
We can take any 2 pairs of the input and output values to find the slope
So,
The representation of ordered pairs to find the slope is:
(x1, y1) = (-4, -1),
(x2, y2) = (-3, -1)
Now,
We know that,
The slope of the line = \(\frac{y2 – y1}{x2 – x1}\)
= \(\frac{-1 – [-1]}{-3 – [-4]}\)
= \(\frac{-1 + 1}{-3 + 4}\)
= 0
Hence, from the above,
We can conclude that the slope of the given line is: 0

Graph the linear equation. Identify the x-intercept. 

Question 21.
y = 2x + 4
Answer:
The given linear equation is:
y = 2x + 4
To find the x-intercept, put y = 0
So,
2x + 4 = 0
2x = -4
x = -4 / 2
x = -2
Hence,
The representation of the given linear equation in the coordinate plane is:

Question 22.
-5x + y = -10
Answer:
The given linear equation is:
-5x + y = -10
To find the x-intercept, put y = 0
-5x + 0 = -10
-5x = -10
5x = 10
x = 10 / 5
x = 2
Hence,
The representation of the given linear equation in the coordinate plane is:

Question 23.
x + 3y = 9
Answer:
The given linear equation is:
x + 3y = 9
To find the x-intercept, put y = 0
So,
x + 0 = 9
x = 9
Hence,
The representation of the given linear equation in the coordinate plane is:

Question 24.
A linear function h models a relationship in which the dependent variable decreases 2 units for every 3 units the independent variable increases. Graph h when h(0) = 2. Identify the slope, y-intercept, and x-intercept of the graph. Let f(x) = 3x + 4. Graph f and h. Describe the transformation from the graph of f to the graph of h.
Answer:
It is given that a linear function h models a relationship in which the dependent variable decreases 2 units for every 3 units the independent variable increases.
We know that,
x represents the independent variable
y represents the dependent variable
So,
The given x value is: 3
The given y value is: -2
So,
The rate of change (or) slope (m) = \(\frac{y}{x}\)
= \(\frac{-2}{3}\)
= –\(\frac{2}{3}\)
It is also given that h (0) = 2
That means the y-intercept of h (x) is 2 when x is 0
We know that,
The representation of the standad form of the linear equation is:
y = mx + c
So,
h (x) =-\(\frac{2}{3}\)x + 2
The other given function is:
f (x) = 3x + 4
Hence,
The representation of f (x0 and h (x) in the same coordinate plane is:

Hence, from the above,
We can conclude that h (x) is 1.5 units away from f (x) on the x-axis

Let f(x) = 3x + 4.Graph f and h. Describe the transformation from the graph of f to the graph of h.

Question 25.
h(x) = f(x + 3)
ANswer:
The given functions are:
f (x) = 3x + 4
h (x) = f (x + 3)
So,
h (x) = 3 (x + 3) + 4
h (x) = 3 (x) + 3 (3) + 4
h (x) = 3x + 9 + 4
h (x) = 3x + 13
Hence,
The representation of f (x) and h (x) in the same coordinate plane is:

Hence, from the above,
We can conclude that f (x) and h (x) are parallel lines

Question 26.
h(x) = f(x) + 1
Answer:
The given functions are:
f (x) = 3x + 4
h (x) = f (x) + 1
So,
h (x) = 3x + 4 + 1
h (x) = 3x + 5
Hence,
The representation of f (x) and h (x) in the same coordinate plane is:

Hence, from the above,
We can conclude that f (x) and h (x) are the paralel lines

Question 27.
h(x) = f(-x)
Answer:
The given functions are:
f (x) = 3x + 4
h (x) = f (-x)
So,
h (x) = 3 (-x) + 4
h (x) = -3x + 4
Hence,
The representation of f (x) and h (x) in the same coordinate plane is:

Hence, from the above,
We can conclude that h (x) and f (x) are inversions of each other

Question 28.
h(x) = -f(x)
Answer:
The given functions are:
f (x) = 3x + 4
h (x) = -f (x)
So,
h (x) = – (3x + 4)
h (x) = -3x – 4
Hence,
The representation of f (x) and h (x) in the same coordinate plane is:

Hence, from the above,
We can conclude that h (x) and f (x) are reflections to each other

Question 29.
h(x) = 3f(x)
Answer:
The given function srae:
f (x) = 3x + 4
h (x) = 3 f (x)
So,
h (x) = 3 (3x + 4)
Hence,
The representation of f (x) and  h (x) in the same coordinate plane is:

Hence, from the above,
We can conclude that h (x) is a vertical stretch of f (x)

Question 30.
h(x) = f(6x)
Answer:
The given functions are:
f (x) = 3x + 4
h (x) = f (6x)
So,
h (x) = 3 (6x) + 4
h (x) = 18x + 4
Hence,
The representation of f (x) and h (x) in the same coordinate plane is:

Hence, from the above,
We can conclude that h (x) translates 2 units away from f (x) on the x-axis

Question 31.
Graph f(x) = x and g(x) = 5x + 1. Describe the transformations from the graph of f to the graph of g.
Answer:
The given functions are:
f (x) = x
g (x) = 5x + 1
Hence,
The representation of f (x) and g (x) in the same coordinate plane is:

Hence, from the above,
We can conclude that g (x) translates away 8 units away from f (x)  on the x-axis

Graph the function. Compare the graph to the graph of f(x) = | x |. Describe the domain and range.

Question 32.
m(x) = | x | + 6
Answer:
The given absolute value functions are:
f (x) = | x |
m (x) = | x | + 6
Hence,
The representation of f (x) and m (x) in the same coordinate plane is:

Hene, from the above,
We can conclude that m (x) translates 6 units up from the positive x-axis
Hence,
The domain of f (x) is: -10 ≤ x ≤ 10
The domain of m (x) is: -4 ≤ x ≤ 4
The range of m (x) is: 0 ≤y ≤ 10
The range of f (x) is: 0 ≤ y ≤10
The range of m (x) is: 6 ≤y ≤10

Question 33.
p(x) = | x – 4 |
Answer:
The given absolute value functions are:
f (x) = | x |
p (x) = | x – 4 |
Hence,
The representation of f (x) and p (x) in the same coordinate plane is:

Hene, from the above,
We can conclude that p (x) translates 1 unit up from the positive x-axis
Hence,
The domain of f (x) is: -10 ≤ x ≤ 10
The domain of p (x) is: -6 ≤ x ≤ 10
The range of f (x) is: 0 ≤ y ≤10
The range of p (x) is: 6 ≤y ≤10

Question 34.
q(x) = 4 | x |
Answer:
The given absolute value functions are:
f (x) = | x |
q (x) = 4 | x |
Hence,
The representation of f (x) and q (x) in the same coordinate plane is:

Hene, from the above,
We can conclude that q (x) translates 7 units up from the positive x-axis
Hence,
The domain of f (x) is: -10 ≤ x ≤ 10
The domain of q (x) is: -2.5 ≤ x ≤ 2.5
The range of q (x) is: 0 ≤y ≤ 10
The range of f (x) is: 0 ≤ y ≤10

Question 35.
r(x) = –\(-\frac{1}{4}\)| x |
Answer:
The given absolute value functions are:
f (x) = | x |
r (x) = –\(-\frac{1}{4}\) | x |
So,
r (x) = \(\frac{1}{4}\) | x |
Hence,
The representation of f (x) and r (x) in the same coordinate plane is:

Hene, from the above,
We can conclude that r (x) translates  units up from the positive x-axis
Hence,
The domain of f (x) is: -10 ≤ x ≤ 10
The domain of r (x) is: -10 ≤ x ≤ 10
The range of r (x) is: 0 ≤y ≤ 2
The range of f (x) is: 0 ≤ y ≤10

Question 36.
Graph f(x) = | x – 2 | + 4 and g(x) = | 3x – 2 | + 4. Compare the graph of g to the graph of f.
Answer:
The given functions are:
f (x) = | x – 2 | + 4
g (x) = | 3x – 2 | + 4
Hence,
The representation of f (x) and g (x) in the same coordinate plane is:

Hence, from the above,
We can conclude that g (x) translates 5 units away from f (x) on the x-axis

Question 37.
Let g(x) = \(\frac{1}{3}\) | x – 1 | – 2.
(a) Describe the transformations from the graph of f(x) = | x | to the graph of g.
Answer:
The given functions are:
f (x) = | x |
g (x) = \(\frac{1}{3}\) | x – 1 | – 2
Hence,
The representation of f (x) and g (x) in the coordinate plane is:

Hence, from the above,
We can conclude that g (x) translates 2 units away from f (x) on the y-axis
(b) Graph g.
Answer:
It is given that
g (x) = \(\frac{1}{3}\) | x – 1 | – 2
Hence,
The representation of g (x) in the coordinate plane is:

Graphing Linear Functions Chapter Test

Determine whether the relation is a function. If the relation is a function, determine whether the function is linear or nonlinear. Explain. 

Question 1.
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 209
Answer:
The given table is:
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 209
We know that,
Every input has exactly one output
So,
The given table is a function
Now,
The constant between all the values of x is: 1
There is no constant difference between all the values of y
Since there is no constant rate of change,
We can conclude that the obtained function is a non-linear function

Question 2.
y = -2x + 3
Answer:
The given equation is:
y = -2x + 3
We know that,
The standard form of the linear equation is:
y = mx + c
For a relation o be a function, every input has exactly only one output
Hence, from the above,
We can conclude that the given equation is a function and it is a linear function

Question 3.
x = -2
Answer:
The given equation is:
x = -2
We know that,
A relation is said to be a function if every input has exactly one output
So,
For x = -2
The input is: 2
The output is: 0
So,
The given equation is a function
For a given function to be a linear function, it will be in the form
y = mx + c
Hence, from the above,
We can conclude that the given function is a non-linear function

Graph the equation and identify the intercept(s). If the equation is linear, find the slope of the line. 

Question 4.
2x – 3y = 6
Answer:
The given equation is:
2x – 3y = 6
3y = 2x – 6
y = \(\frac{2x – 6}{3}\)
y = \(\frac{2x}{3}\) – \(\frac{6}{3}\)
y = \(\frac{2}{3}\)x – 2
Hence,
The above equation is in the form of
y = mx + c
So,
The given equation is a linear equation
So,
m = \(\frac{2}{3}\)
The y-intercept is: -2
To find the x-intercept, put y = 0
\(\frac{2}{3}\)x – 2 = 0
x = 3
Hence, from the above,
We can conclude that
Slope (m) = \(\frac{2}{3}\)
The x-intercept is: 3
The y-intercept is: -2

Question 5.
y = 4.5
Answer:
The given equation is:
y = 4.5
The given equation is not in the form of
y = mx + c
Hence,
The given equation is non-linear

Question 6.
y = | x – 1 | – 2
Answer:
The given equation is:
y = | x – 1 | – 2
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
y = x – 1 – 2 or y = -(x – 1) – 2
y = x – 3 or y = -x + 1 – 2
y = x – 3 or y = -x – 1
The above 2 equations are in the form of
y = mx + c
Hence,
The slope is: 1 or -1
The y-intercept is: -3 or -1
To find the x-intercept, put y =0
So,
0 = x – 3 or 0 = -x – 1
x = 3 or x = -1
Hence, from the above,
We can conclude that
The slope is: 1 or -1
The x-intercept is: 3 or -1
The y-intercept is: -3 or -1

Find the domain and range of the function represented by the graph. Determine whether the domain is discrete or continuous. Explain. 

Question 7.
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 210
Answer:
The given graph is:
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 210
From the graph,
We can observe that the graph is not a straight line
So,
We can say that the given graph is a non-linear function
Hence, from the above,
We can conclude that the domain is continuous by observing the graph

Question 8.
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 211
Answer:
The given graph is:
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 211
From the graph,
We can observe that the points form a straight line
So,
We can say that the given graph is a linear function
Hence, from the above,
We can conclude that the domain is continuous since all the points are connected

Graph f and g. Describe the transformations from the graph of f to the graph of g. 

Question 9.
f(x) = x; g(x) = -x + 3
Answer:
The given functions are:
f (x) = x
g (x) = -x + 3
Hence,
The representation of f (x) and g (x) in the coordinate plane is:

Hence, from the above,
We can conclude that f (x) and g (x) are perpendicular lines with a slope of -1

Question 10.
f(x) = | x | ; g(x) = | 2x + 4 |
Answer:
The given functions are:
f (x) = | x |
g (x) = | 2x + 4 |
Hence,
The representation of f (x) and g (x) in the coordinate plane is:

Hence, from the above,
We can conclude that g (x) is 2 units away from f (x) on the x-axis

Question 11.
Function A represents the amount of money in a jar based on the number of quarters in the jar. Function B represents your distance from home over time. Compare the domains.
Answer:
It is given that
Function A represents the amount of money in a jar based on the number of quarters in the jar
Function B represents your distance from home over time
We know that,
The distance should be greater than or equal to 0
The amount is greater than or equal to a quarter of the amount of the money
Hence,
The domain of function A is: \(\frac{1}{4}\) ≤ x ≤ ∞
The domain of function B is: 0 ≤ x ≤ ∞

Question 12.
A mountain climber is scaling a 500-foot cliff. The graph shows the elevation of the climber over time.
a. Find and interpret the slope and the y-intercept of the graph.
b. Explain two ways to find f(3). Then find f(3) and interpret its meaning.
c. How long does it take the climber to reach the top of the cliff? Justify your answer.
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 212
Answer:
a.
It is given that a mountain-climber is scaling a 500-foot cliff
The given graph is:
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 212
From the graph,
The equation that shows the elevation of the climber over tie is:
f (x) = 125x + 50
We know that,
The standard representation of the output for the function output is: y
So,
y = 125x + 50
Compare the above equation with
y = mx + c
So,
m = 125 and the y-intercept is: 50
Hence, from the above,
We can conclude that
The slope of the given equation is: 125
The y-intercept of the given equation is: 50

b.
From part (a),
The given equation is:
f (x) = 125x + 50
So,
f (3) = 125 (3) + 50
f (3) = 375 + 50
f (3) = 425
Hence, from the above,
We can conclude that for 3 hours, the climber climbs 425 feet

c.
The given graph is:
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 212
From the given graph,
The top of the hill is the maximum height i.e., the highest value on the y-axis
Hence, from the above,
We can conclude that it takes 4 hours to reach the top of the cliff

Question 13.
Without graphing, compare the slopes and the intercepts of the graphs of the functions f(x) = x + 1 and g(x) = f(2x).
Answer:
The given functons are:
f (x) = x + 1
g (x) = f (2x)
So,
g (x) = 2x + 1
Now,
Compare f (x) and g (x) with the standard linear equation
y = mx + c
So,
For f (x),
m = 1 and c = 1
Where,
c is the y-intercept
For g (x),
m = 2 and c = 1
Where,
c is the y-intercept

Question 14.
A rock band releases a new single. Weekly sales s (in thousands of dollars) increase and then decrease as described by the function s(t) = -2 | t – 20 | + 40, where t is the time (in weeks).
a. Identify the independent and dependent variables.
b. Graph s. Describe the transformations from the graph of f(x) = | x | to the graph of s.
Answer:
a.
It is given that a rock band releases a new single and weekly sales s (in thousands of dollars) increase and then decrease as described by the function
s(t) = -2 | t – 20 | + 40
where,
t is the time (in weeks)
Now,
The independent variable of the given function  is: t
The dependent variable of the given function is: s (t)

b.
The given absolute value functions are:
f (x) = | x |
s (t) = -2 | t – 20 | + 40
Hence,
The representation of f (x) and s (t) in the coordinate plane is:

Hence, from the above,
We can conclude that s (t) translates 5 units away from f (x) on the y-axis

Graphing Linear Functions Cumulative Assessment

Question 1.
You claim you can create a table of values that represents a linear function. Your friend claims he can create a table of values that represents a nonlinear function. Using the given numbers, what values can you use for x (the input) and y (the output) to support your claim? What values can your friend use?
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 213
Answer:
It is given that you claim you can create a table of values that represents a linear function. Your friend claims he can create a table of values that represents a nonlinear function.
Hence,
The values you and your friend use are:

Question 2.
A car rental company charges an initial fee of $42 and a daily fee of $12.
a. Use the numbers and symbols to write a function that represents this situation.
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 214
b. The bill is $138. How many days did you rent the car?
Answer:
a.
It is given that a car rental company charges an initial fee of $42 and a daily fee of $12
So,
The function that represents the situation is:
f (x) = (42 + 12) × x
f (x) = 52x
Where,
x is the number of days that you rent the car

b.
It is given that the bill is: $138
So,
138 = 52x
x = 138 ÷ 52
x = 3
Hence, from the above,
We can conclude that you can rent the car for 3 days

Question 3.
Fill in values for a and b so that each statement is true for the inequality ax − b> 0.
a. When a = _____ and b = _____, x > \(\frac{b}{a}\).
b. When a = _____ and b = _____, x < \(\frac{b}{a}\).
Answer:
a.
The given inequality is:
ax – b > 0
ax > b
x > b / a
So,
The value of a is less than b and the value of b is greater than a
b.
The value of a is greater than b and the value of b is less than a

Question 4.
Fill in the inequality with <, ≤, >, or ≥ so that the solution of the inequality is represented by the graph.
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 215
Answer:
The given number line is:
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 215
From the given number line,
The given inequality is:
-3 (x + 7) _____ -24
-3 (x) – 3 (7) _____ -24
-3x – 21 _____ -24
-3x ____ -24 + 21
-3x ____-3
3x ____ 3
x ____ 3 / 3
x ____ 1
From the number line
The marked line represented from 1 including 1 and continued till the right end of the number line
Hence,
x ≥ 1
Hence, from the above,
We can conclude that the symbol used for the given inequality is: ≥

Question 5.
Use the numbers to fill in the coefficients of ax + by = 40 so that when you graph the function, the x-intercept is -10 and the y-intercept is 8.
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 216
Answer:
The given function is:
ax + by = 40
It is given that the x-intercept and the y-intercept is: -10 and 8
Now,
To find the x-intercept, put y = 0
So,
ax = 40
a (-10) = 40
a = -40 / 10
a = -4
To find the y-intercept, put x = 0
So,
by = 40
b (8) = 40
b = 40 / 8
b = 5
Hence, from the above,
We can conclude that the values of a and b are: -4 and 5

Question 6.
Solve each equation. Then classify each equation based on the solution. Explain your reasoning.
a. 2x – 9 = 5x – 33
Answer:
The given expression is:
2x – 9 = 5x – 33
2x – 5x = -33 + 9
-3x = -24
x = 24 / 3
x = 8
Hence, from the above,
We can conclude that the given expression has 1 solution

b. 5x – 6 = 10x + 10
Answer:
The given expression is:
5x – 6 = 10x + 10
5x – 10x = 10 + 6
-5x = 16
x = -16 / 5
Hence, from the above,
We can conclude that the given expression has only 1 solution

c. 2(8x – 3) = 4(4x + 7)
Answer:
The given expression is:
2 (8x – 3 ) = 4 (4x + 7)
2 (8x) – 2 (3) = 4 (4x) + 4(7)
16x – 6 = 16x + 28
Hence, from the above,
We can conclude that the given expression has no solution

d. -7x + 5 = 2(x – 10.1)
Answer:
The given expression is:
-7x + 5 = 2 (x – 10.1)
-7x + 5 = 2x – 20.2
-7x – 2x = -20.2 – 5
-9x = -25.2
x = 25.9 / 9
Hence, from the above,
We can conclude that the given expression has only 1 solution

e. 6(2x + 4) = 4(4x + 10)
Answer:
The given expression is:
6 (2x + 4) = 4 (4x + 10)
12x + 24 = 16x + 40
12x – 16x = 40 – 24
-4x = 16
x = -16 / 4
x = -4
Hence, from the above,
We can conclude that the given expression has only 1 solution

f. 8(3x + 4) = 2(12x + 16)
Answer:
The given expression is:
8 (3x + 4) = 2 (12x + 16)
24x + 32 = 24x + 32
Hence, from the above,
We can conclude that the given expression has no solution

Question 7.
The table shows the cost of bologna at a deli. Plot the points represented by the table in a coordinate plane. Decide whether you should connect the points with a line. Explain your reasoning.
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 217
Answer:
The given table is:
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 217
Hence,
The representation of the points in the coordinate plane is:

Hence, from the above,
We can say that we can connect the points in the graph

Question 8.
The graph of g is a horizontal translation right, then a vertical stretch, then a vertical translation down of the graph of f(x) = x. Use the numbers and symbols to create g.
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 218
Answer:
The given function is:
f (x) = x
We know that,
f (x) can be re-written as f
g (x) can b ere-written as g
Now,
Let the operations performed on f (x) can be expressed in terms of g (x)
So,
Horizontal translation right:
g (x) = f (x) – 3 or g (x) = f (x) – 1 or g (x)  f (x) – (1/2)
Vertical stretch:
g (x) = 3 f (x)
Vertical translation down:
g (x) = f (x – 1)

Question 9.
What is the sum of the integer solutions of the compound inequality 2 | x – 5 | < 16?
A. 72
B. 75
C. 85
D. 88
Answer:
The given compound inequality is:
2 | x – 5 | < 16
So,
2 (x – 5) < 16 an d 2 (x – 5 ) > -16
2x – 10 < 16 and 2x – 10 > -16
2x < 26 and 2x >-6
x < 13 and x > -3
Hence,
The solution of the given compoud inequality is: -3 < x < 13
Now,
The sum of all the integers = -2 – 1 + 0 + 1 + 2 + 3 + 4 + 5+ 6 + 7 + 8 +9 + 10 + 11 + 12
= 75
Hence, from the above,
We can conclude that the sum of all the integers is: 75

Question 10.
Your bank offers a text alert service that notifies you when your checking account balance drops below a specific amount. You set it up so you are notified when your balance drops below $700. The balance is currently $3000. You only use your account for paying your rent (no other deposits or deductions occur). Your rent each month is $625.
a. Write an inequality that represents the number of months m you can pay your rent without receiving a text alert.
Answer:
It is given that your bank offers a text alert service that notifies you when your checking account balance drops below a specific amount. You set it up so you are notified when your balance drops below $700. The balance is currently $3000. You only use your account for paying your rent (no other deposits or deductions occur). Your rent each month is $625
Let m be the number of months
Hence,
The inequality that represents the number of months you can pay rent without receiving a text is:
3000 – 625x > 700

b. What is the maximum number of months you can pay your rent without receiving a text alert?
Answer:
From part (a),
The inequality that represents the number of months m without receiving a text alert is:
3000 – 625x > 700
3000 – 700 > 625x
2300 > 625x
2300 / 625 > x
3.68 > x
x < 4 months [ Since the number of months will not be in decimals]
Hence, from the above,
We can conclude that the maximum number of months you can pay your rent without receiving a text alert is: 4 months

c. Suppose you start paying rent in June. Select all the months you can pay your rent without making a deposit.
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 219
Answer:
From part (b),
The maximum number of months is:
x < 4 months
It is given that you start paying rent in June
So,
All the months you can pay your rent without making a deposit is:
June, July, August

Big Ideas Math Answers Grade 5 Chapter 10 Divide Fractions

Big Ideas Math Answers Grade 5 Chapter 10 Divide Fractions

Big Ideas Math Answers Grade 5 Chapter 10 Divide Fractions PDF is included here. We have given the solutions to all the questions in Big Ideas Math Answers Grade 5 Chapter 10 Divide Fractions in pdf format. Either you can learn BIM Grade 5 Chapter 10 Divide Fractions Answers online or offline using pdf. Big Ideas Math Answers Grade 5 Chapter 10 Divide Fractions Answer Key helps the students to enhance their knowledge. Also, it is the best source to get in-depth knowledge on the concept. Find out the real-time examples on Big Ideas Grade 5 Chapter 10 Divide Fractions Math Answers for better practice.

Big Ideas Grade 5 Chapter 10 Divide Fractions Math Book Answer Key

All the quick methods and tips are helpful for the students to save their time. Therefore, get the Big Ideas Math Answers Grade 5 Chapter 10 Divide Fractions and make use of it to have quick learning. Check answers for every question of every topic. The following sections have quick links where they have each topic with problems. Get the complete grip on the fractions by referring to our Big Ideas Grade 5 Math Answers Chapter 10 Divide Fractions.

Lesson: 1 Interpret Fractions as Division

Lesson: 2 Mixed Numbers as Quotients

Lesson: 3 Divide Whole Numbers by Unit Fractions

Lesson: 4 Divide Unit Fractions by Whole Numbers

Lesson: 5 Problem Solving: Fraction Division

Chapter: 10 – Divide Fractions

Lesson 10.1 Interpret Fractions as Division

Explore and Grow`

You share 4 sheets of construction paper equally among 8 people. Write a division expression that represents the situation. What fraction of a sheet of paper does each person get? Use a model to support your answer?
Answer:
The division expression that represents the fraction of a sheet of paper does each person get is:
4 ÷ 8 = \(\frac{1}{2}\)

Explanation:
It is given that you have 4 sheets of construction paper equally among 8 people.
Hence,
The division expression that represents the fraction of a sheet of paper is:
( The number of sheets of construction paper ) ÷ ( The number of people )
= 4 ÷ 8
= \(\frac{1}{2}\)
Hence, from the above,
We can conclude that the fraction of a sheet of paper does each person get is: \(\frac{1}{2}\)

Structure
How can you check your answer using multiplication?
Answer: We can check the answer by using the partial products method or by using the simplification method.

Think and Grow: Divide Whole Numbers
You can use models to divide whole numbers that have a fraction as the quotient.
Answer: 
From the above model,
The number of colored parts is: 4
The total number of parts are: 8
So,
The fraction of the colored part out of the total number of parts = 4 ÷ 8
= \(\frac{4}{8}\) = \(\frac{1}{2}\)
In \(\frac{1}{2}\),
1 represents the quotient
2 represents the remainder
Example
Find 2 ÷ 3.
One Way: Use a tape diagram. Show 2 wholes. Divide each whole into 3 equal parts.

Another Way: Use an area model. Show 2 wholes. Divide each whole into 3 equal parts. Then separate the parts into 3 equal groups.

Show and Grow

Divide. Use a model to help

Question 1.
2 ÷ 4 =0.5
Answer:
From the above model,
The number of colored parts is: 2
The number of total parts is: 4
So,
The fraction of the colored parts out of the total number of parts = 2 ÷ 4
= \(\frac{2}{4}\)
= \(\frac{1}{2}\)
So,
The fraction of the colored parts out of the total number of parts in the decimal form is: 0.5

Question 2.
1 ÷ 3 = 0.33
Answer:

From the above model,
The number of colored parts is: 1
The number of total parts is: 3
So,
The fraction of the colored parts out of the total number of parts = 1 ÷ 3
= \(\frac{1}{3}\)
So,
The fraction of the colored parts out of the total number of parts in the decimal form is: 0.33

Apply and Grow: Practice

Divide. Use a model to help.

Question 3.
1 ÷ 8 =0.018
Answer:
From the above model,
The number of colored parts is: 1
The number of total parts is: 8
So,
The fraction of the colored parts out of the total number of parts = 1 ÷ 8
= \(\frac{1}{8}\)
So,
The fraction of the colored parts out of the total number of parts in the decimal form is: 0.018

Question 4.
1 ÷ 4 =0.25
Answer:
From the above model,
The number of colored parts is: 1
The number of total parts is: 4
So,
The fraction of the colored parts out of the total number of parts = 1 ÷ 4
= \(\frac{1}{4}\)
So,
The fraction of the colored parts out of the total number of parts in the decimal form is: 0.25

Question 5.
2 ÷ 6 =0.33
Answer:
From the above model,
The number of colored parts is: 2
The number of total parts is: 6
So,
The fraction of the colored parts out of the total number of parts = 2 ÷ 6
= \(\frac{2}{6}\)
= \(\frac{1}{3}\)
So,
The fraction of the colored parts out of the total number of parts in the decimal form is: 0.33

Question 6.
2 ÷ 5 = 0.4
Answer:
From the above model,
The number of colored parts is: 2
The number of total parts is: 5
So,
The fraction of the colored parts out of the total number of parts = 2 ÷ 5
= \(\frac{2}{5}\)
So,
The fraction of the colored parts out of the total number of parts in the decimal form is: 0.4

Question 7.
3 ÷ 7 = 0.42
Answer: 
From the above model,
The number of colored parts is: 3
The number of total parts is: 7
So,
The fraction of the colored parts out of the total number of parts = 3 ÷ 7
= \(\frac{3}{7}\)
So,
The fraction of the colored parts out of the total number of parts in the decimal form is: 0.42

Question 8.
5 ÷ 6 = 0.83
Answer:
From the above model,
The number of colored parts is: 5
The number of total parts is: 6
So,
The fraction of the colored parts out of the total number of parts = 5 ÷ 6
= \(\frac{5}{6}\)
So,
The fraction of the colored parts out of the total number of parts in the decimal form is: 0.83

Question 9.
How many 6s are in 1?
Answer: There are six \(\frac{1}{6}\)s in 1

Explanation:
The number of 6s in 1 can be obtained by dividing 1 into 6 equal parts.
So,
The figure obtained will be like;

From the above model,
The number of colored parts is: 1
The number of total parts is: 6
So,
The fraction of the colored parts out of the total number of parts = 1 ÷ 6
= \(\frac{1}{6}\)
Hence, from the above,
We can conclude that there are six 6s in 1

Question 10.
How many 10s are in 9?
Answer: There are 9 \(\frac{9}{10}\)s in 9

Explanation:
The model for the number of 10s in 9 are:

From the above model,
The number of colored parts is: 9
The number of total parts is: 10
So,
The fraction of the colored parts out of the total number of parts = 9 ÷ 10
= \(\frac{9}{10}\)
Hence, from the above,
We can conclude that there are nine 9s in 10

Question 11.
Number Sense
For which equations does k = 8?
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 3
Answer: Let the equations named A), B), C), and D)
So,
The four equations are:
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 3
From the above equations,
The value ‘k’ must be in the numerator.
So,
In A), the value of the numerator is: 3
In B), the value of the numerator is: k
In C), the value of the numerator is: 2
In D) the value of the numerator is: 8
So,
From the above numerator values,
We can say that “k=8” holds good for Equation B)

Question 12.
Writing
Write and solve a real-life problem for 7 ÷ 12.
Answer:
From the above model,
The number of colored parts is: 7
The number of total parts is: 12
So,
The fraction of the colored parts out of the total number of parts = 7 ÷ 12
= \(\frac{7}{12}\)
Hence,
The fraction of the colored parts out of the total number of parts in the decimal form is: 0.58

Think and Grow: Modeling Real Life

Example
Three fruit bars are shared equally among 4 friends. What fraction of a fruit bar does each friend get?
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 4
Divide 3 by 4 to find what fraction of a fruit bar each friend gets.
Use an area model to find 3 ÷ 4. Show 3 whole fruit bars. Divide each fruit bar into 4 equal parts. Then separate the parts into 4 equal groups.

Show and Grow

Question 13.
You cut a 5-foot streamer into 6 pieces of equal size. What is the length of each piece in feet? in inches?
Answer: The length of each piece in feet is: \(\frac{5}{6}\)

Explanation:
It is given that you cut a 5-foot streamer into 6 equal pieces of equal size.
So,
The model representing the 6 equal pieces of the 5-foot streamer is:

From the above model,
We can see that each part in the model represents \(\frac{5}{6}\) of each part.
Hence, from the above,
We can conclude that the length of each piece of a 5-foot streamer in feet is: \(\frac{5}{6}\)

Question 14.
Four circular lemon slices are shared equally among 8 glasses of water. What fraction of a lemon slice does each glass get?
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 7
Answer: The fraction of a lemon slice does each glass get is: \(\frac{1}{2}\)

Explanation:
It is given that the four circular lemon slices are shared equally among 8 glasses of water.
So,
The model representing the portion that each glass get is:

From the above model,
We can say that each part represents \(\frac{1}{2}\) part
Hence, from the above,
We can conclude that the portion of a lemon slice does glass get is: \(\frac{1}{2}\)

Question 15.
You cut a 5-foot streamer into 6 pieces of equal size. What is the length of each piece in feet? in inches?
Answer: The length of each piece in feet is: \(\frac{5}{6}\)

Explanation:
It is given that you cut a 5-foot streamer into 6 equal pieces of equal size.
So,
The model representing the 6 equal pieces of the 5-foot streamer is:

From the above model,
We can see that each part in the model represents \(\frac{5}{6}\) of each part.
Hence, from the above,
We can conclude that the length of each piece of a 5-foot streamer in feet is: \(\frac{5}{6}\)

Question 16.
DIG DEEPER!
A fruit drink is made using \(\frac{7}{4}\) quarts of orange juice and \(\frac{5}{4}\) quarts of pineapple juice. The drink is shared equally among 12 guests. What fraction of a quart does each guest get?
Answer: The fraction of a quart does each guest get is: \(\frac{1}{4}\)

Explanation:
It is given that a fruit drink is made using \(\frac{7}{4}\) quarts of orange juice and \(\frac{5}{4}\) quarts of pineapple juice.
So,
The total amount of fruit juice= \(\frac{7}{4}\) + \(\frac{5}{4}\)
= \(\frac{ 7 + 5}{4}\)
= \(\frac{12}{4}\)
It is also given that the drink is shared equally among 12 guests
So,
The fraction of a quart does each gust get = \(\frac{12}{4}\) ÷ 12
= \(\frac{12}{4}\) ÷ \(\frac{12}{1}\)
= \(\frac{12}{4}\) × \(\frac{1}{12}\)
= \(\frac{1}{4}\)
Hence, from the above,
We can conclude that the fraction of a quart does each person get is: \(\frac{1}{4}\)

Interpret Fractions as Division Homework & Practice 10.1

Divide. Use a model to help.

Question 1.
1 ÷ 6 =0.16
Answer:


From the above model,
The number of colored parts is: 1
The number of total parts is: 6
So,
The fraction of the colored parts out of the total number of parts = 1 ÷ 6
= \(\frac{1}{6}\)
So,
The fraction of the colored parts out of the total number of parts in the decimal form is: 0.16

Question 2.
1 ÷ 7 =0.14
Answer:
From the above model,
The number of colored parts is: 1
The number of total parts is: 7
So,
The fraction of the colored parts out of the total number of parts = 1 ÷ 7
= \(\frac{1}{7}\)
So,
The fraction of the colored parts out of the total number of parts in the decimal form is: 0.14

Question 3.
1 ÷ 5 = 0.20
Answer:
From the above model,
The number of colored parts is: 1
The number of total parts is: 5
So,
The fraction of the colored parts out of the total number of parts = 1 ÷ 5
= \(\frac{1}{5}\)
So,
The fraction of the colored parts out of the total number of parts in the decimal form is: 0.20

Question 4.
3 ÷ 4 = 0.75
Answer:
From the above model,
The number of colored parts is: 3
The number of total parts is: 4
So,
The fraction of the colored parts out of the total number of parts = 3 ÷ 4
= \(\frac{3}{4}\)
So,
The fraction of the colored parts out of the total number of parts in the decimal form is: 0.75

Question 5.
6 ÷ 7 = 0.85
Answer:
From the above model,
The number of colored parts is: 6
The number of total parts is: 7
So,
The fraction of the colored parts out of the total number of parts = 6 ÷ 7
= \(\frac{6}{7}\)
So,
The fraction of the colored parts out of the total number of parts in the decimal form is: 0.85

Question 6.
5 ÷ 9 = 0.55
Answer:
From the above model,
The number of colored parts is: 5
The number of total parts is: 9
So,
The fraction of the colored parts out of the total number of parts = 5 ÷ 9
= \(\frac{5}{9}\)
So,
The fraction of the colored parts out of the total number of parts in the decimal form is: 0.55

Question 7.
YOU BE THE TEACHER
Your friend says \(\frac{5}{12}\) is equivalent to 12 ÷ 5. Is your friend correct? Explain.
Answer: No, your friend s not correct.

Explanation:
The given fraction is: \(\frac{5}{12}\)
From the given fraction,
The numerator is: 5
The denominator is: 12
We can write a fraction in the following form:
Fraction = \(\frac{Numerator}{Denominator}\)
So,
\(\frac{5}{12}\) is equivalent to 5 ÷ 12
But, according to your friend,
\(\frac{5}{12}\) is equivalent to 12 ÷ 5
Hence, from the above,
we can conclude that your friend is not correct.

Question 8.
Writing
Explain how fractions and division are related.

Question 9.
Structure
Write a division equation represented by the model.
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 8
Answer:
The division equation represented by the model is: 1 ÷ 4

Explanation:
The given model is:
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 8
From the given model,
The number of shaded parts is: 1
The total number of parts are: 4
So,
The division equation can be represented as:
Division equation = (The number of shaded parts) ÷ ( The total number of parts )
= 1 ÷ 4
= \(\frac{1}{4}\)

Question 10.
Number Sense
Eight friends share multiple vegetable pizzas, and each gets \(\frac{3}{8}\) of a pizza. How many pizzas do they share?
Answer: The total number of pizzas the eight friends shared are: 3 pizzas

Explanation:
It is given that the eight friends share multiple vegetable pizzas and each gets \(\frac{3}{8}\) of a pizza.
So,
The total number of pizzas shared by the eight friends = \(\frac{3}{8}\) × 8
= \(\frac{3}{8}\) × \(\frac{8}{1}\)
= \(\frac{3 × 8}{8 × 1}\)
= \(\frac{3}{1}\)
= 3
Hence, from the above,
We can conclude that the total number of pizzas shared by the eight friends is: 3 pizzas

Question 11.
Modeling Real Life
Seven friends each run an equal part of a 5-kilometer relay race. What fraction of a kilometer does each friend complete?
Answer: The fraction of a kilometer does each friend complete is: \(\frac{5}{7}\) kilometer

Explanation:
It is given that there are seven friends each run an equal part of a 5-kilometer relay race.
So,
The fraction that each friend run = \(\frac{The total distance} {The number of friends}\)
= \(\frac{5}{7}\)
Hence, from the above,
We can conclude that the fraction of a kilometer does each friend complete is: \(\frac{5}{7}\) kilometer

Question 12.
Modeling Real Life
A group of friends equally share 3 bags of pretzels. Each friend gets \(\frac{3}{5}\) of a bag of pretzels. How many friends are in the group?
Answer: The total number of friends in the group are: 5

Explanation:
It is given that a group of friends equally share 3 bags of pretzels and each friend gets \(\frac{3}{5}\) of a bag of pretzels.
So,
The total number of friends = \(\frac{The total number of bags}{The amount each friend gets}\)
= \(\frac{3}{1}\) × \(\frac{5}{3}\)
= \(\frac{5}{1}\)
= 5
Hence, from the above,
We can conclude that the total number of friends are: 5

Review & Refresh

Multiply.

Question 13.
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 9
Answer: 9 × \(\frac{2}{3}\) = 6

Explanation:
The given fractions are: \(\frac{9}{1}\) and \(\frac{2}{3}\)
So,
\(\frac{9}{1}\) × \(\frac{2}{3}\)
= \(\frac{9 × 2}{1 × 3}\)
= \(\frac{6}{1}\)
= 6
Hence,
9 × \(\frac{2}{3}\) = 6

Question 14.
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 10
Answer: 5 × \(\frac{7}{10}\) = \(\frac{7}{2}\)

Explanation:
The given fractions are: \(\frac{5}{1}\) and \(\frac{7}{10}\)
So,
\(\frac{5}{1}\) × \(\frac{7}{10}\)
= \(\frac{5 × 7}{1 × 10}\)
= \(\frac{7}{2}\)
Hence,
5 × \(\frac{7}{10}\) = \(\frac{7}{2}\)

Question 15.
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 11
Answer: 3 × \(\frac{5}{12}\) = \(\frac{5}{4}\)

Explanation:
The given fractions are: \(\frac{3}{1}\) and \(\frac{5}{12}\)
So,
\(\frac{3}{1}\) × \(\frac{5}{12}\)
= \(\frac{3 × 5}{1 × 12}\)
= \(\frac{5}{4}\)
Hence,
3 × \(\frac{5}{12}\) = \(\frac{5}{4}\)

Lesson 10.2 Mixed Numbers as Quotients

Explore and Grow

You share 6 sheets of construction paper equally among 4 people. Write a division expression that represents the situation. How much paper does each person get? Use a model to support your answer.
Answer:
The division expression representing the situation is: 6 ÷ 4

Explanation:
It s given that you have shared 6 sheets of construction paper equally among 4 people
So,
The division equation representing the sharing of construction papers is: 6 ÷ 4
Now,
6 ÷ 4 = \(\frac{6}{4}\)
So,
The above equation represents that 4 is divided into 6 parts.
So,
The model representing the situation is:

From the above model,
We can say that the amount of does each person get is: 1\(\frac{1}{2}\) or 1.5 or \(\frac{3}{2}\)

Precision
Does each person get less than or more than 1 sheet of paper? Use the dividend and divisor to explain why your answer makes sense.
Answer:
From the above problem,
We can say that each person gets more than 1 paper.
So,
The division equation of the above problem is: 6 ÷ 4
The equivalent form of 6 ÷ 4 is: \(\frac{6}{4}\)
Now,
The simplest form of \(\frac{6}{4}\) is: \(\frac{3}{2}\) ( The simplest form is the division of the numerator and the denominator with the common multiple if we can divide)
The mixed form of \(\frac{3}{2}\) is: 1\(\frac{1}{2}\)

Think and Grow: Divide Whole Numbers

You can use models to divide whole numbers that have a mixed number as the quotient.
Example
Find 3 ÷ 2.
One Way:
Use a tape diagram. Show 3 wholes. Divide each whole into 2 equal parts.

Another Way: Use an area model. Show 3 wholes. Divide each whole into 2 equal parts. Then separate the parts into 2 equal groups.

Show and Grow

Divide. Use a model to help

Question 1.
5 ÷ 3 = ___
Answer: 5 ÷ 3 = 1\(\frac{2}{3}\)

Explanation;
The given division equation is: 5 ÷ 3
The model representing the division equation is:

From the above model,
5 ÷ 3 = 3 ÷ 3
= 1 R 2
Hence,
We can say that each part is divided into 1\(\frac{2}{3}\) or \(\frac{5}{3}\)

Question 2.
7 ÷ 2 = ___

Answer: 7 ÷ 2 = 3\(\frac{1}{2}\)

Explanation;
The given division equation is: 7 ÷ 2
The model representing the division equation is:

From the above model,
7 ÷ 2 = 6 ÷ 2
= 3 R 1
Hence,
We can say that each part is divided into 3\(\frac{1}{2}\) or \(\frac{7}{2}\) or 3.5

Apply and Grow: Practice

Divide. Use a model to help.

Question 3.
12 ÷ 7 = ___

Answer: 12 ÷ 7 = 1\(\frac{5}{7}\)

Explanation;
The given division equation is: 12 ÷ 7
The model representing the division equation is:

From the above model,
12 ÷ 7 = 7 ÷ 7
= 1 R 5
Hence,
We can say that each part is divided into 1\(\frac{5}{7}\) or \(\frac{12}{7}\)

Question 4.
25 ÷ 20 = ___

Answer: 25 ÷ 20 = 1\(\frac{5}{20}\) = \(\frac{5}{4}\)

Explanation;
The given division equation is: 25 ÷ 20
The model representing the division equation is:

From the above model,
25 ÷ 20 = 20 ÷ 20
= 1 R 5
Hence,
We can say that each part is divided into 1\(\frac{5}{20}\) or \(\frac{5}{4}\)

Question 5.
15 ÷ 4 = ___

Answer: 15 ÷ 4 = 3\(\frac{3}{4}\)

Explanation;
The given division equation is: 15 ÷ 4
The model representing the division equation is:

From the above model,
15 ÷ 4 = 12 ÷ 4
= 3 R 3
Hence,
We can say that each part is divided into 3\(\frac{3}{4}\) or \(\frac{15}{4}\)

Question 6.
13 ÷ 6 = ___

Answer: 13 ÷ 6 = 2\(\frac{1}{6}\)

Explanation;
The given division equation is: 13÷ 6
The model representing the division equation is:

From the above model,
13 ÷ 6 = 12 ÷ 6
= 2 R 1
Hence,
We can say that each part is divided into 2\(\frac{1}{6}\) or \(\frac{13}{6}\)

Question 7.
16 ÷ 8 = ___

Answer: 16 ÷ 8 = 2

Explanation;
The given division equation is: 16÷ 8
The model representing the division equation is:

From the above model,
16 ÷ 8
= 2 R 0
Hence,
We can say that each part is divided into 2 equal parts

Question 8.
92 ÷ 50 = ___

Answer: 92 ÷ 50 = 1\(\frac{21}{25}\)

Explanation;
The given division equation is: 92÷ 50
So,
92 ÷ 50 = 50 ÷ 50
= 1 R 42
Hence,
We can say that each part is divided into 1\(\frac{42}{50}\) or 1\(\frac{21}{25}\)

Question 9.
How many 3s are in 7?
Answer: The number of 3 in 7 are: \(\frac{7}{3}\) or 2\(\frac{1}{3}\)

Explanation:
The division equation is: 7 ÷ 3
So,
The model for the given division equation is:

From the above model,
7 ÷ 3 = 6 ÷ 3
= 2 R 1
Hence, from the above,
We can conclude that there are 2\(\frac{1}{3}\) 3s in 7

Question 10.
How many 6s are in 21?
Answer: The number of 6s in 21 are: \(\frac{21}{6}\) or 3\(\frac{3}{6}\)

Explanation:
The division equation is: 21 ÷ 6
So,
The model for the given division equation is:

From the above model,
21 ÷ 6 = 18 ÷ 6
= 3 R 3
Hence, from the above,
We can conclude that there are 3\(\frac{3}{6}\) 3s in 21

Question 11.
YOU BE THE TEACHER
Your friend says that \(\frac{35}{6}\) is equivalent to 35 ÷ 6. Is your friend correct? Explain.
Answer: Yes, your friend is correct

Explanation:
It is given that \(\frac{35}{6}\)
We know that,
The decimal equation can be converted into a fraction as \(\frac{Numerator}{Denominator}\)
So,
\(\frac{35}{6}\) = 35 ÷ 6
Hence, from the above,
We can conclude that your friend is correct

Question 12.
Writing
Write and solve a real-life problem for 24 ÷ 5.
Answer: 24 ÷ 5 = 4\(\frac{4}{5}\)

Explanation;
The given division equation is: 24÷ 5
The model for the above division equation is:

From the above model,
24 ÷ 5 = 20 ÷ 5
= 4 R 4
Hence,
We can say that each part is divided into 4\(\frac{4}{5}\)

Think and Grow: Modeling Real Life

Example
You share 7 bales of hay equally among 3 horse stalls. How many whole bales are in each stall? What fractional amount of a bale is in each stall?
Divide 7 by 3 to find how many bales of hay are in each stall. Use an area model to help.

Show and Grow

Question 13.
Six muffins are shared equally among 4 friends. How many whole muffins does each friend get? What fractional amount of a muffin does each friend get?
Answer: Each friend will get 1 muffin and 2 muffins are leftovers
The fractional part of a muffin does each friend get is: \(\frac{1}{2}\)

Explanation:
It is given that there are six muffins are shared equally among 4 friends.
So,
The number of muffins each friend get = 6 ÷ 4
= 4 ÷ 4
= 1 R 2
Hence, from the above,
We can conclude that each friend gets 1 muffin each and the fraction of each muffin get is: \(\frac{1}{2}\)

Question 14.
A cyclist bikes 44 miles in 5 days. She bikes the same distance each day. Does she bike more than 8\(\frac{1}{2}\) miles each day? Explain.
Answer: She bikes more than 8\(\frac{1}{2}\) miles each day.

Explanation:
It is given that a cyclist bikes 44 miles in 5 days.
So,
The distance that she bikes each day = 44 ÷ 5
So,
44 ÷ 5 = 40 ÷ 5
= 8 R 4
= 8\(\frac{4}{5}\) miles
But, it is given that she bikes 8\(\frac{1}{2}\) miles each day
Hence, from the above,
We can conclude that she bikes more than 8\(\frac{1}{2}\) miles each day.

Question 15.
DIG DEEPER!
At Table A, 4 students share 7 packs of clay equally. At Table B, 5 students share 8 packs of clay equally. At which table does each student get a greater amount of clay? Explain.
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 15
Answer: At Table A, each student gets a greater amount of clay.

Explanation:
It is given that at Table A, 4 students share 7 packs of clay equally.
So,
The representation of clay at table A is: \(\frac{7}{4}\)
It is also given that at Table B, 5 students share 8 packs of clay equally.
So,
The representation of clay at table B is: \(\frac{8}{5}\)
So,
For comparison, equate the denominators.
So,
Multiply the first fraction at table A by \(\frac{5}{5}\) and the fraction at table B by \(\frac{4}{4}\)
So,
\(\frac{7}{4}\) × \(\frac{5}{5}\)
= \(\frac{35}{20}\)
So,
\(\frac{8}{5}\) × \(\frac{4}{4}\)
= \(\frac{32}{20}\)
Hence, from the above,
We can conclude that at table A, the students will get more amount of clay.

Mixed Numbers as Quotients Homework & Practice 10.2

Divide. Use a model to help.

Question 1.
5 ÷ 2 = ___
Answer: 5 ÷ 2 = 2\(\frac{1}{2}\)

Explanation;
The given division equation is: 5 ÷ 2
The model representing the division equation is:

From the above model,
5 ÷ 2 = 4 ÷ 2
= 2 R 1
Hence,
We can say that 5 ÷ 2 = 2\(\frac{1}{2}\) or 2.5 or \(\frac{5}{2}\)

Question 2.
10 ÷ 7 = ___
Answer: 10 ÷ 7 = 1\(\frac{3}{7}\) = \(\frac{10}{7}\)

Explanation;
The given division equation is: 10 ÷ 7
The model representing the division equation is:

From the above model,
10 ÷ 7 = 7 ÷ 7
= 1 R 3
Hence,
We can say that 10 ÷ 7 = 1\(\frac{3}{7}\) or \(\frac{10}{7}\)

Question 3.
3 ÷ 9 = ___
Answer: 3 ÷ 9 = \(\frac{1}{3}\)

Explanation;
The given division equation is: 3 ÷ 9
The model representing the division equation is:

From the above model,
3 and 9 are the multiples of 3.
So,
3 ÷ 9 = \(\frac{1}{3}\)
Hence,
We can say that 3 ÷ 9 = \(\frac{1}{3}\)

Question 4.
11 ÷ 4 = ___
Answer: 11 ÷ 4 = 2\(\frac{3}{4}\)

Explanation;
The given division equation is: 11 ÷ 4
The model representing the division equation is:

From the above model,
11 ÷ 4 = 8 ÷ 4
= 2 R 3
Hence,
We can say that 11 ÷ 4 = \(\frac{11}{4}\) or 2\(\frac{3}{4}\)

Question 5.
13 ÷ 6 = ___
Answer: 13 ÷ 6 = 2\(\frac{1}{6}\)

Explanation;
The given division equation is: 13 ÷ 6
The model representing the division equation is:

From the above model,
13 ÷ 6 = 12 ÷ 6
= 2 R 1
Hence,
We can say that 13 ÷ 6 = \(\frac{13}{6}\) or 2\(\frac{1}{6}\)

Question 6.
45 ÷ 8 = ___
Answer: 45 ÷ 8 = 5\(\frac{5}{8}\)

Explanation;
The given division equation is: 45 ÷ 8
The model representing the division equation is:

From the above model,
45 ÷ 8 = 40 ÷ 8
= 5 R 5
Hence,
We can say that 45 ÷ 8 = \(\frac{45}{8}\) or 5\(\frac{5}{8}\)

Question 7.
Number Sense
Between which two whole numbers is the quotient of 74 and 9?
Answer: The quotient of 74 and 9 is between 8 and 9

Explanation:
The given two numbers are 7 and 9
So,
By using the partial quotients method,
74 ÷ 9= 72 ÷ 9
= 8 R 2
So,
74 ÷ 9 = \(\frac{74}{9}\) or 8\(\frac{2}{9}\) or 8.3
Hence, from the above,
We can conclude that the quotient of 74 and 9 is between 8 and 9

Question 8.
Reasoning
Three friends want to share 22 baseball cards. For this situation, why does the quotient 7 R1 make more sense than the quotient 7\(\frac{1}{3}\)?
Answer:
It is given that three friends want to share 22 baseball cards.
So,
We have to find the number of baseball cards each friend possesses.
So,
It is sufficient to write the number of baseball cards possessed by each friend in the remainder form rather than the fraction form.
So,
The number of baseball cards possessed by each friend = \(\frac{The total number of baseball cards}{The number of friends}\)
= 22 ÷ 3
= 21 ÷ 3
= 7 R 1
Hence, from the above,
We can conclude that the remainder form is sufficient to find the number of baseball cars possessed by each friend rather than the fraction form.

Question 9.
DIG DEEPER!
Is \(\frac{2}{5}\) × 3 equivalent to 2 × 3 ÷ 5? Explain.
Answer: Yes, \(\frac{2}{5}\) × 3 equivalent to 2 × 3 ÷ 5

Explanation:
The given fraction and the number is: \(\frac{2}{5}\) and 3
So,
\(\frac{2}{5}\) × 3 = \(\frac{2}{5}\) × \(\frac{3}{1}\)
= \(\frac{2 × 3}{5}\)
= 2 × 3 ÷ 5
Hence, from the above,
We can conclude that \(\frac{2}{5}\) × 3 equivalent to 2 × 3 ÷ 5

Question 10.
Modeling Real Life
A bag of 4 balls weighs 6 pounds. Each ball weighs the same amount. What is the weight of each ball?
Answer: The weight of each ball is: \(\frac{3}{2}\) pounds or 1.5 pounds

Explanation:
It is given that a bag of 4 balls weighs 6 pounds
So,
The weight of each ball = \(\frac{The total weight of the balls}{The number of balls}\)
= 6 ÷ 4
Since 6 and 4 are the multiples of 2, divide the two numbers by 2
So,
6 ÷ 4 = 3 ÷ 2
So,
3 ÷ 2 = 2 ÷ 2
= 1 R 1
= 1\(\frac{1}{2}\) pounds
Hence, from the above,
We can conclude that the weight of each ball is: 1\(\frac{1}{2}\) pounds or 1.5 pounds

Question 11.
Modeling Real Life
Zookeepers order 600 pounds of bamboo for the pandas. The bamboo lasts 7 days. How many whole pounds of bamboo do the pandas eat each day? What fractional amount of a pound do the pandas eat each day?
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 17
Answer:
The amount of bamboos the pandas eat each day is around 85 pounds
The amount of bamboos the pandas eat each day in the fraction form is: 85\(\frac{5}{7}\)

Explanation:
It is given that zookeepers order 600 pounds of bamboo for the pandas and the bamboos last 7 days for the pandas
So,
The number of bamboos the pandas eat each day = 600 ÷ 7
So,
By using the partial quotients method,
600 ÷ 7 = ( 560 + 35 ) ÷ 7
= ( 560 ÷ 7 ) + ( 35 ÷ 7 )
= 80 + 5
= 85 R 5
Hence, from the above,
We can conclude that
The amount of bamboos the pandas eat each day is around 85 pounds
The amount of bamboos the pandas eat each day in the fraction form is: 85\(\frac{5}{7}\)

Question 12.
Modeling Real Life
A plumber has 20 feet of piping. He cuts the piping into 6 equal pieces. Is each piece greater than, less than, or equal to 3\(\frac{1}{2}\) feet?
Answer: Each piece is less than 3\(\frac{1}{2}\) feet

Explanation:
It is given that a plumber has 20 feet of piping and he cuts the piping into 6 equal pieces.
So,
The length of each piece = 20 ÷ 6
By using the partial quotients method,
20 ÷ 6 = 18 ÷ 6
= 3 R 2
So,
20 ÷ 6 = 3\(\frac{2}{6}\)
Now,
3\(\frac{1}{2}\) = \(\frac{7}{2}\)
3\(\frac{2}{6}\) = \(\frac{20}{6}\)
For comparison, we have to equate whether the denominators or the numerators.
So,
Multiply 3\(\frac{1}{2}\) with \(\frac{3}{3}\)
So,
3\(\frac{1}{2}\) = \(\frac{21}{6}\)
Hence, from the above,
We can conclude that each piece is less than 3\(\frac{1}{2}\) feet

Review & Refresh

Add.

Question 13.
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 18
Answer: \(\frac{2}{9}\) + \(\frac{2}{3}\) = \(\frac{8}{9}\)

Explanation:
The two given fractions are: \(\frac{2}{9}\) and \(\frac{2}{3}\)
So, in addition, we have to make either the numerators or the denominators equal
So,
Multiply \(\frac{2}{3}\)  with \(\frac{3}{3}\)
So,
\(\frac{2}{3}\)  = \(\frac{6}{9}\)
Hence, from the above,
We can conclude that \(\frac{2}{9}\) + \(\frac{2}{3}\) = \(\frac{8}{9}\)

Question 14.
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 19
Answer: \(\frac{1}{10}\) + \(\frac{3}{4}\) = \(\frac{34}{40}\)

Explanation:
The two given fractions are: \(\frac{1}{10}\) and \(\frac{3}{4}\)
So, in addition, we have to make either the numerators or the denominators equal
So,
Multiply \(\frac{1}{10}\)  with \(\frac{4}{4}\)
Multiply \(\frac{3}{4}\)  with \(\frac{10}{10}\)
So,
\(\frac{1}{10}\)  = \(\frac{4}{40}\)
\(\frac{3}{4}\)  = \(\frac{30}{40}\)
Hence, from the above,
We can conclude that \(\frac{1}{10}\) + \(\frac{3}{4}\) = \(\frac{34}{40}\)

Question 15.
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 20
Answer: \(\frac{3}{5}\) + \(\frac{5}{6}\) + \(\frac{1}{5}\) = \(\frac{49}{30}\)

Explanation:
The three given fractions are: \(\frac{3}{5}\) , \(\frac{5}{6}\) and \(\frac{1}{5}\)
So, in addition, we have to make either the numerators or the denominators equal
So,
Multiply \(\frac{3}{5}\)  with \(\frac{6}{6}\)
Multiply \(\frac{5}{6}\)  with \(\frac{5}{5}\)
Multiply \(\frac{1}{5}\)  with \(\frac{6}{6}\)
So,
\(\frac{3}{5}\)  = \(\frac{18}{30}\)
\(\frac{5}{6}\)  = \(\frac{25}{30}\)
\(\frac{1}{5}\)  = \(\frac{6}{30}\)
Hence, from the above,
We can conclude that \(\frac{3}{5}\) + \(\frac{5}{6}\) +\(\frac{1}{5}\)  = \(\frac{49}{30}\)

Lesson 10.3 Divide Whole Numbers by Unit Fractions

Explore and Grow

Write a real-life problem that can be represented by 6 ÷ \(\frac{1}{2}\)?
Answer:
Suppose, we have an apple and there are 6 children and we are giving each child half of the piece.
So,
Each child receives 6 ÷ \(\frac{1}{2}\) piece of the apple

What is the solution to the problem? Use a model to support your answer?
Answer:
The above problem is the division of an apple among the six children
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
So,
The amount each child receive from an apple = 6 ÷ \(\frac{1}{2}\)
= 6 × \(\frac{2}{1}\)
= \(\frac{6}{1}\) × \(\frac{2}{1}\)
= \(\frac{ 6 × 2}{1 × 1}\)
= 12

Structure
How can you check your answer using multiplication?
Answer:
We can check the answer using multiplication by the two rules regarding division and multiplication. They are:
A) a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
B) a= \(\frac{a}{1}\)

Think and Grow: Divide Whole Numbers by Unit Fractions

You can use models to divide whole numbers by unit fractions.
Example
Find 4 ÷ \(\frac{1}{3}\)
One Way:
Use a tape diagram to find how many \(\frac{1}{3}\)s are in 4. There are 4 wholes.
Divide each whole into 3 equal parts. Each part is \(\frac{1}{3}\).
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 21
Because there are 3 one-thirds in 1 whole, there are
4 × 3 equal parts = 12 one-thirds in 4 wholes.

Show and Grow

Divide. Use a model to help

Question 1.
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 23
Answer: 3 ÷ \(\frac{1}{2}\) = 6

Explanation:
The given numbers are: 3 and \(\frac{1}{2}\)
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
So,
3 ÷ \(\frac{1}{2}\)  = 3 × \(\frac{2}{1}\)
= \(\frac{3}{1}\) × \(\frac{2}{1}\)
= \(\frac{ 3 × 2}{1 × 1}\)
= 6
Hence,
3÷ \(\frac{1}{2}\) = 6

Question 2.
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 24
Answer: 2 ÷ \(\frac{1}{5}\) = 10

Explanation:
The given numbers are: 2 and \(\frac{1}{5}\)
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
So,
2 ÷ \(\frac{1}{5}\)  = 2 × \(\frac{5}{1}\)
= \(\frac{5}{1}\) × \(\frac{2}{1}\)
= \(\frac{ 5 × 2}{1 × 1}\)
= 10
Hence,
2÷ \(\frac{1}{5}\) = 10

Apply and Grow: Practice

Divide. Use a model to help.

Question 3.

Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 25
Answer: 1 ÷ \(\frac{1}{3}\) = 3

Explanation:
The given numbers are: 1 and \(\frac{1}{3}\)
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
So,
1 ÷ \(\frac{1}{3}\)  = 1 × \(\frac{3}{1}\)
= \(\frac{3}{1}\) × \(\frac{1}{1}\)
= \(\frac{ 3 × 1}{1 × 1}\)
= 3
Hence,
1÷ \(\frac{1}{3}\) = 3

Question 4.
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 26
Answer: 3 ÷ \(\frac{1}{5}\) = 15

Explanation:
The given numbers are: 3 and \(\frac{1}{5}\)
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
So,
3 ÷ \(\frac{1}{5}\)  = 3 × \(\frac{5}{1}\)
= \(\frac{3}{1}\) × \(\frac{5}{1}\)
= \(\frac{ 3 × 5}{1 × 1}\)
= 15
Hence,
3÷ \(\frac{1}{5}\) = 15

Question 5.
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 27
Answer: 5 ÷ \(\frac{1}{3}\) = 15

Explanation:
The given numbers are: 5 and \(\frac{1}{3}\)
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
So,
5 ÷ \(\frac{1}{3}\)  = 5 × \(\frac{3}{1}\)
= \(\frac{3}{1}\) × \(\frac{5}{1}\)
= \(\frac{ 3 × 5}{1 × 1}\)
= 15
Hence,
5÷ \(\frac{1}{3}\) = 15

Question 6.
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 28
Answer: 4 ÷ \(\frac{1}{4}\) = 16

Explanation:
The given numbers are: 4 and \(\frac{1}{4}\)
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
So,
4 ÷ \(\frac{1}{4}\)  = 4 × \(\frac{4}{1}\)
= \(\frac{4}{1}\) × \(\frac{4}{1}\)
= \(\frac{ 4 × 4}{1 × 1}\)
= 16
Hence,
4÷ \(\frac{1}{4}\) = 16

Question 7.
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 29
Answer: 7 ÷ \(\frac{1}{2}\) = 14

Explanation:
The given numbers are: 7 and \(\frac{1}{2}\)
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
So,
7 ÷ \(\frac{1}{2}\)  = 7 × \(\frac{2}{1}\)
= \(\frac{7}{1}\) × \(\frac{2}{1}\)
= \(\frac{ 7 × 2}{1 × 1}\)
= 14
Hence,
7÷ \(\frac{1}{2}\) = 14

Question 8.
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 30
Answer: 2 ÷ \(\frac{1}{7}\) = 14

Explanation:
The given numbers are: 2 and \(\frac{1}{7}\)
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
So,
2 ÷ \(\frac{1}{7}\)  = 2 × \(\frac{7}{1}\)
= \(\frac{2}{1}\) × \(\frac{7}{1}\)
= \(\frac{ 7 × 2}{1 × 1}\)
= 14
Hence,
2÷ \(\frac{1}{7}\) = 14

Question 9.
How many \(\frac{1}{4}\)s are in 5?
Answer: There are 20 \(\frac{1}{4}\)s in 5

Explanation:
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
Now,
We have to find the number of \(\frac{1}{4}\)s in 5
So,
5 ÷ \(\frac{1}{4}\)  = 5 × \(\frac{4}{1}\)
= \(\frac{5}{1}\) × \(\frac{4}{1}\)
= \(\frac{ 5 × 4}{1 × 1}\)
= 20
Hence, from the above,
We can conclude that there are 20 \(\frac{1}{4}\)s in 5.

Question 10.
How many \(\frac{1}{6}\)s are in 2?
Answer: There are 12 \(\frac{1}{6}\)s in 2

Explanation:
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
Now,
We have to find the number of \(\frac{1}{6}\)s in 2
So,
2 ÷ \(\frac{1}{6}\)  = 2 × \(\frac{6}{1}\)
= \(\frac{2}{1}\) × \(\frac{6}{1}\)
= \(\frac{ 2 × 6}{1 × 1}\)
= 12
Hence, from the above,
We can conclude that there are 12 \(\frac{1}{6}\)s in 2.

Question 11.
YOU BE THE TEACHER
Newton finds 6 ÷ \(\frac{1}{3}\). Is he correct? Explain.
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 31
Answer: No, Newton is not correct

Explanation:
The given division equation is: 6 ÷ \(\frac{1}{3}\)
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
So,
6 ÷ \(\frac{1}{3}\)  = 6 × \(\frac{3}{1}\)
= \(\frac{6}{1}\) × \(\frac{3}{1}\)
= \(\frac{ 3 × 6}{1 × 1}\)
= 18
But, according to Newton,
6 ÷ \(\frac{1}{3}\) = 2
Hence, from the above,
We can conclude that Newton is not correct.

Question 12.
Writing
Write and solve a real-life problem for 4 ÷ \(\frac{1}{2}\).
Answer:
Suppose we have 4 bags of wheat and we have to distribute the 4 bags by dividing each bag of wheat in half
So,
Each person receives 4 ÷ \(\frac{1}{2}\) bag of wheat
Now,
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
So,
4 ÷ \(\frac{1}{2}\)  = 4 × \(\frac{2}{1}\)
= \(\frac{4}{1}\) × \(\frac{2}{1}\)
= \(\frac{ 4 × 2}{1 × 1}\)
= 8
Hence, from the above,
We can conclude that there are 8 bags of wheat when divide the 4 bags of wheat in half.

Think and Grow: Modeling Real Life

Example
A chef makes 3 cups of salsa. A serving of salsa is \(\frac{1}{8}\) cup. How many servings does the chef make?
To find the number of servings, find the number of \(\frac{1}{8}\) cups in 3 cups.
Use an area model to find 3 ÷ \(\frac{1}{8}\). Divide each cup into 8 equal parts.
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 32

Show and Grow

Question 13.
A litter of kittens weighs a total of 2 pounds. Each newborn kitten weighs \(\frac{1}{4}\) pound. How many kittens are in the litter?
Answer: The number of kittens in the litter are: 8 kittens

Explanation:
It is given that a litter of kittens weighs a total of 2 pounds and each newborn kitten weighs \(\frac{1}{4}\) pound.
So,
The number of kittens in the litter = \(\frac{The total weight of litter}{The weight of each newborn kitten}\)
= 2 ÷ \(\frac{1}{4}\)
Now,
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
So,
2 ÷ \(\frac{1}{4}\)  = 2 × \(\frac{4}{1}\)
= \(\frac{4}{1}\) × \(\frac{2}{1}\)
= \(\frac{ 4 × 2}{1 × 1}\)
= 8
Hence, from the above,
We can conclude that the number of kittens in the litter are: 8 kittens

Question 14.
You put signs on a walking trail that is 7 miles long. You put a sign at the start and at the end of the trail. You also put a sign every \(\frac{1}{10}\) mile. How many signs do you put on the trail?
Answer: The total number of signs you put on the trail is: 72

Explanation:
It is given that you put signs on a walking trail that is 7 miles long and you put a sign at the start and at the end of the trail.
It is also given that you put a sign every \(\frac{1}{10}\) mile.
So,
The total number of signs you put on the trail = The sign at the start of the trail + The sign at the end of the trail + The total number of signs for \(\frac{1}{10}\) mile
Now,
The total number of signs for \(\frac{1}{10}\) mile = 7 ÷ \(\frac{1}{10}\)
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
So,
7 ÷ \(\frac{1}{10}\)  = 7 × \(\frac{10}{1}\)
= \(\frac{7}{1}\) × \(\frac{10}{1}\)
= \(\frac{ 7 × 10}{1 × 1}\)
= 70
So,
The total number of signs you put on the trail = 1 + 1 + 70
= 72
hence, from the above,
We can conclude that there are 72 signs that you put on the trail

Question 15.
DIG DEEPER!
You have 2 boards that are each 8 feet long. You cut \(\frac{1}{2}\)– foot pieces to make square picture frames. How many picture frames can you make?
Answer: The number of picture frames you can make is: 32

Explanation:
It is given that you have 2 boards that are each 8 feet long.
So,
The total length of 2 boards = 2 × 8 = 16 feet
It is also given that you cut \(\frac{1}{2}\)– foot pieces to make square picture frames.
So,
The total number of picture frames = \(\frac{The total length of 2 boards}{The length of each square frame}\)
= 16 ÷ \(\frac{1}{2}\)
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
So,
16 ÷ \(\frac{1}{2}\)  = 16 × \(\frac{2}{1}\)
= \(\frac{16}{1}\) × \(\frac{2}{1}\)
= \(\frac{ 16 × 2}{1 × 1}\)
= 32
Hence, from the above,
We can conclude that we can make 32 picture frames.

Divide Whole Numbers by Unit Fractions Homework & Practice 10.3

Divide. Use a model to help.

Question 1.
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 33
Answer: 1 ÷ \(\frac{1}{9}\) = 9

Explanation:
The given numbers are: 1 and \(\frac{1}{9}\)
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
So,
1 ÷ \(\frac{1}{9}\)  = 1 × \(\frac{9}{1}\)
= \(\frac{1}{1}\) × \(\frac{9}{1}\)
= \(\frac{ 1 × 9}{1 × 1}\)
= 9
Hence,
1÷ \(\frac{1}{9}\) = 9

Question 2.
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 34
Answer: 2 ÷ \(\frac{1}{3}\) = 6

Explanation:
The given numbers are: 2 and \(\frac{1}{3}\)
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
So,
2 ÷ \(\frac{1}{3}\)  = 2 × \(\frac{3}{1}\)
= \(\frac{3}{1}\) × \(\frac{2}{1}\)
= \(\frac{ 3 × 2}{1 × 1}\)
= 6
Hence,
2÷ \(\frac{1}{3}\) = 6

Question 3.
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 35
Answer: 5 ÷ \(\frac{1}{2}\) = 10

Explanation:
The given numbers are: 5 and \(\frac{1}{2}\)
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
So,
5 ÷ \(\frac{1}{2}\)  = 5 × \(\frac{2}{1}\)
= \(\frac{5}{1}\) × \(\frac{2}{1}\)
= \(\frac{ 5 × 2}{1 × 1}\)
= 10
Hence,
5÷ \(\frac{1}{2}\) = 10

Divide. Use a model to help.

Question 4.
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 36
Answer: 9 ÷ \(\frac{1}{4}\) = 36

Explanation:
The given numbers are: 9 and \(\frac{1}{4}\)
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
So,
9 ÷ \(\frac{1}{4}\)  = 9 × \(\frac{4}{1}\)
= \(\frac{9}{1}\) × \(\frac{4}{1}\)
= \(\frac{ 9 × 4}{1 × 1}\)
= 36
Hence,
9÷ \(\frac{1}{4}\) = 36

Question 5.
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 37
Answer: 7 ÷ \(\frac{1}{3}\) = 21

Explanation:
The given numbers are: 7 and \(\frac{1}{3}\)
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
So,
7 ÷ \(\frac{1}{3}\)  = 7 × \(\frac{3}{1}\)
= \(\frac{3}{1}\) × \(\frac{7}{1}\)
= \(\frac{ 3 × 7}{1 × 1}\)
= 21
Hence,
7÷ \(\frac{1}{3}\) = 21

Question 6.
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 38
Answer: 8 ÷ \(\frac{1}{5}\) = 40

Explanation:
The given numbers are: 8 and \(\frac{1}{5}\)
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
So,
8 ÷ \(\frac{1}{5}\)  = 8 × \(\frac{5}{1}\)
= \(\frac{8}{1}\) × \(\frac{5}{1}\)
= \(\frac{ 8 × 5}{1 × 1}\)
= 40
Hence,
8÷ \(\frac{1}{5}\) = 40

Question 7.
Number Sense
Explain how you can check your answer for Exercise 6.
Answer:
We can check the answer for exercise 6 by using the below model:

From the above model,
Each part represents \(\frac{8}{5}\)
So,
The total value of the 5 parts is: \(\frac{40}{5}\)
Hence,
In the above way, we can say that we check the answer

Question 8.
YOU BE THE TEACHER
Descartes finds 5 ÷ \(\frac{1}{4}\). Is he correct? Explain.
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 38.1
Answer: Yes, he is correct

Explanation:
We can write 5 as \(\frac{20}{4}\) or \(\frac{5}{1}\)
But, we only take \(\frac{20}{4}\) because the divided number given is 4
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
So,
\(\frac{20}{4}\) ÷ \(\frac{1}{4}\)
= \(\frac{20}{4}\) × \(\frac{4}{1}\)
= \(\frac{ 20 × 4}{4 × 1}\)
= 20
Hence, from the above,
We can conclude that Descartes is correct.

Question 9.
Modeling Real Life
You need \(\frac{1}{2}\) pound of clay to make a pinch pot. How many pinch pots can you make with 12 pounds of clay?
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 42
Answer: You can make 24 pinch pots with 12 pounds of clay

Explanation:
It is given that you need \(\frac{1}{2}\) pound of clay to make a pinch pot.
It is also given that you have 12 pounds of clay
So,
The number of pinch pots you can make by using 12 pounds of clay = \(\frac{The total amount of clay}{The amount of clay used to make each pinch pot}\)
= 12 ÷ \(\frac{1}{2}\)
Now,
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
So,
12 ÷ \(\frac{1}{2}\)
= \(\frac{12}{1}\) × \(\frac{2}{1}\)
= \(\frac{ 12 × 2}{4 × 1}\)
= 24
Hence, from the above,
We can conclude that we can make 24 pinch pots by using 12 pounds of clay.

Question 10.
Modeling Real Life
Your art teacher has 5 yards of yellow string and 4 yards of green string. She cuts both colors \(\frac{1}{3}\)-yard pieces to hang of string into student artwork. How many pieces of student artwork can she hang?
Answer: The number of pieces of student artwork she can hang is: 27

Explanation:
It is given that your art teacher has 5 yards of yellow string and 4 yards of green string.
So,
The total number of yards of string = 5 + 4 = 9 yards of string
It is also given that she cuts both colors \(\frac{1}{3}\)-yard pieces to hang of string into student artwork.
So,
The number of pieces of student artwork she can hang = \(\frac{The total number of yards of strings}{The length of each yard f string}\)
= 9 ÷ \(\frac{1}{3}\)
Now,
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
So,
9 ÷ \(\frac{1}{3}\)
= \(\frac{9}{1}\) × \(\frac{3}{1}\)
= \(\frac{ 9 × 3}{1 × 1}\)
= 27
Hence, from the above,
We can conclude that there are 27 pieces of student artwork that she can hang.

Review & Refresh

Question 11.
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 43
Answer: \(\frac{2}{5}\) × \(\frac{3}{4}\) = \(\frac{6}{20}\)

Explanation:
The given fractions are: \(\frac{3}{4}\) and \(\frac{2}{5}\)
So,
\(\frac{2}{5}\) × \(\frac{3}{4}\)
= \(\frac{2 × 3}{5 × 4}\)
= \(\frac{6}{20}\)
Hence,
\(\frac{2}{5}\) × \(\frac{3}{4}\) = \(\frac{6}{20}\)

Question 12.
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 44
Answer: \(\frac{1}{8}\) × \(\frac{5}{8}\) = \(\frac{5}{64}\)

Explanation:
The given fractions are: \(\frac{1}{8}\) and \(\frac{5}{8}\)
So,
\(\frac{1}{8}\) × \(\frac{5}{8}\)
= \(\frac{1 × 5}{8 × 8}\)
= \(\frac{5}{64}\)
Hence,
\(\frac{1}{8}\) × \(\frac{5}{8}\) = \(\frac{5}{64}\)

Question 13.
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 45
Answer: \(\frac{4}{9}\) × \(\frac{2}{7}\) = \(\frac{8}{63}\)

Explanation:
The given fractions are: \(\frac{4}{9}\) and \(\frac{2}{7}\)
So,
\(\frac{4}{9}\) × \(\frac{2}{7}\)
= \(\frac{2 × 4}{7 × 9}\)
= \(\frac{8}{63}\)
Hence,
\(\frac{4}{9}\) × \(\frac{2}{7}\) = \(\frac{8}{63}\)

Lesson 10.4 Divide Unit Fractions by Whole Numbers

Write a real-life problem that can be represented by \(\frac{1}{2}\) ÷ 3?
Answer:
Suppose we have 3 people and those 3 people each has to share \(\frac{1}{2}\) of the apple

What is the solution to the problem? Use a model to support your answer?
Answer:
The above problem is: We have to share \(\frac{1}{2}\) each for the 3 people
Now,
We know that,
\(\frac{a}{b}\) ÷ a  = \(\frac{a}{b}\) × \(\frac{1}{a}\)
a= \(\frac{a}{1}\)
So,
\(\frac{1}{2}\) ÷ 3
= \(\frac{1}{2}\) × \(\frac{1}{3}\)
= \(\frac{ 1 × 1}{2 × 3}\)
= \(\frac{1}{6}\)
Hence,
\(\frac{1}{6}\) is the solution to the above problem.

Precision
Is the answer greater than or less than 1? Explain?
Answer: The answer is less than 1

Explanation:
The answer for the problem is: \(\frac{1}{6}\)
So,
For the comparison of \(\frac{1}{6}\) with 1, we have to see whether the numerators or the denominators are equal or not
So, in this case, the numerators are equal
So, compare the denominators
So,
1 < 6
Hence, from the above,
We can conclude that \(\frac{1}{6}\) is less than 1

Think and Grow: Divide Unit Fractions by Whole Numbers

You can use models to divide unit fractions by whole numbers.

Show and Grow

Divide. Use a model to help.

Question 1.
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 48
Answer: \(\frac{1}{4}\) ÷ 2 = \(\frac{1}{8}\)

Explanation:
The given numbers are: \(\frac{1}{4}\) and 2
We know that,
\(\frac{a}{b}\) ÷ a  = \(\frac{a}{b}\) × \(\frac{1}{a}\)
a= \(\frac{a}{1}\)
So,
\(\frac{1}{4}\) ÷ 2
= \(\frac{1}{4}\) × \(\frac{1}{2}\)
= \(\frac{ 1 × 1}{2 × 4}\)
= \(\frac{1}{8}\)
Hence,
\(\frac{1}{4}\) ÷ 2 = \(\frac{1}{8}\)

Question 2.
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 49
Answer: \(\frac{1}{2}\) ÷ 5 = \(\frac{1}{10}\)

Explanation:
The given numbers are: \(\frac{1}{2}\) and 5
We know that,
\(\frac{a}{b}\) ÷ a  = \(\frac{a}{b}\) × \(\frac{1}{a}\)
a= \(\frac{a}{1}\)
So,
\(\frac{1}{2}\) ÷ 5
= \(\frac{1}{2}\) × \(\frac{1}{5}\)
= \(\frac{ 1 × 1}{2 × 5}\)
= \(\frac{1}{10}\)
Hence,
\(\frac{1}{2}\) ÷ 5 = \(\frac{1}{10}\)

Apply and Grow: Practice

Divide. Use a model to help.

Question 3.
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 50
Answer: \(\frac{1}{5}\) ÷ 3 = \(\frac{1}{15}\)

Explanation:
The given numbers are: \(\frac{1}{5}\) and 3
We know that,
\(\frac{a}{b}\) ÷ a  = \(\frac{a}{b}\) × \(\frac{1}{a}\)
a= \(\frac{a}{1}\)
So,
\(\frac{1}{5}\) ÷ 3
= \(\frac{1}{5}\) × \(\frac{1}{3}\)
= \(\frac{ 1 × 1}{5 × 3}\)
= \(\frac{1}{15}\)
Hence,
\(\frac{1}{5}\) ÷ 3 = \(\frac{1}{15}\)

Question 4.
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 51
Answer: \(\frac{1}{6}\) ÷ 2 = \(\frac{1}{12}\)

Explanation:
The given numbers are: \(\frac{1}{6}\) and 2
We know that,
\(\frac{a}{b}\) ÷ a  = \(\frac{a}{b}\) × \(\frac{1}{a}\)
a= \(\frac{a}{1}\)
So,
\(\frac{1}{6}\) ÷ 2
= \(\frac{1}{6}\) × \(\frac{1}{2}\)
= \(\frac{ 1 × 1}{2 × 6}\)
= \(\frac{1}{12}\)
Hence,
\(\frac{1}{6}\) ÷ 2 = \(\frac{1}{12}\)

Question 5.
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 52
Answer: \(\frac{1}{3}\) ÷ 5 = \(\frac{1}{15}\)

Explanation:
The given numbers are: \(\frac{1}{3}\) and 5
We know that,
\(\frac{a}{b}\) ÷ a  = \(\frac{a}{b}\) × \(\frac{1}{a}\)
a= \(\frac{a}{1}\)
So,
\(\frac{1}{3}\) ÷ 5
= \(\frac{1}{3}\) × \(\frac{1}{5}\)
= \(\frac{ 1 × 1}{3 × 5}\)
= \(\frac{1}{15}\)
Hence,
\(\frac{1}{3}\) ÷ 5 = \(\frac{1}{15}\)

Question 6.
Big Ideas Math Answers 5th Grade Chapter 10 Divide Fractions 53
Answer: \(\frac{1}{5}\) ÷ 4 = \(\frac{1}{20}\)

Explanation:
The given numbers are: \(\frac{1}{5}\) and 4
We know that,
\(\frac{a}{b}\) ÷ a  = \(\frac{a}{b}\) × \(\frac{1}{a}\)
a= \(\frac{a}{1}\)
So,
\(\frac{1}{5}\) ÷ 4
= \(\frac{1}{5}\) × \(\frac{1}{4}\)
= \(\frac{ 1 × 1}{5 × 4}\)
= \(\frac{1}{20}\)
Hence,
\(\frac{1}{5}\) ÷ 4 = \(\frac{1}{20}\)

Question 7.
Big Ideas Math Answers 5th Grade Chapter 10 Divide Fractions 54
Answer: \(\frac{1}{3}\) ÷ 3 = \(\frac{1}{9}\)

Explanation:
The given numbers are: \(\frac{1}{3}\) and 3
We know that,
\(\frac{a}{b}\) ÷ a  = \(\frac{a}{b}\) × \(\frac{1}{a}\)
a= \(\frac{a}{1}\)
So,
\(\frac{1}{3}\) ÷ 3
= \(\frac{1}{3}\) × \(\frac{1}{3}\)
= \(\frac{ 1 × 1}{3 × 3}\)
= \(\frac{1}{9}\)
Hence,
\(\frac{1}{3}\) ÷ 3 = \(\frac{1}{9}\)

Question 8.
Big Ideas Math Answers 5th Grade Chapter 10 Divide Fractions 55
Answer: \(\frac{1}{8}\) ÷ 2 = \(\frac{1}{16}\)

Explanation:
The given numbers are: \(\frac{1}{8}\) and 2
We know that,
\(\frac{a}{b}\) ÷ a  = \(\frac{a}{b}\) × \(\frac{1}{a}\)
a= \(\frac{a}{1}\)
So,
\(\frac{1}{8}\) ÷ 2
= \(\frac{1}{8}\) × \(\frac{1}{2}\)
= \(\frac{ 1 × 1}{2 × 8}\)
= \(\frac{1}{16}\)
Hence,
\(\frac{1}{8}\) ÷ 2 = \(\frac{1}{16}\)

Question 9.
How many 6s are in \(\frac{1}{2}\)?
Answer: There are \(\frac{1}{12}\) 6s in \(\frac{1}{2}\)

Explanation:
We know that,
\(\frac{a}{b}\) ÷ a  = \(\frac{a}{b}\) × \(\frac{1}{a}\)
a= \(\frac{a}{1}\)
Now,
We have to find the number of 6s in \(\frac{1}{2}\)
So,
\(\frac{1}{2}\) ÷ 6
= \(\frac{1}{2}\) × \(\frac{1}{6}\)
= \(\frac{ 1 × 1}{2 × 6}\)
= \(\frac{1}{12}\)
Hence, from the above,
We can conclude that there are \(\frac{1}{12}\) 6s in \(\frac{1}{2}\)

Question 10.
How many 2s are in \(\frac{1}{3}\) ?
Answer: There are \(\frac{1}{6}\) 2s in \(\frac{1}{3}\)

Explanation:
We know that,
\(\frac{a}{b}\) ÷ a  = \(\frac{a}{b}\) × \(\frac{1}{a}\)
a= \(\frac{a}{1}\)
Now,
We have to find the number of 2s in \(\frac{1}{3}\)
So,
\(\frac{1}{3}\) ÷ 2
= \(\frac{1}{3}\) × \(\frac{1}{2}\)
= \(\frac{ 1 × 1}{2 × 3}\)
= \(\frac{1}{6}\)
Hence, from the above,
We can conclude that there are \(\frac{1}{6}\) 2s in \(\frac{1}{2}\)

Question 11.
Writing
Write and solve a real-life problem for
Big Ideas Math Answers 5th Grade Chapter 10 Divide Fractions 56
Answer:
Suppose a box has 7 chocolates. We have to divide these seven chocolates into further \(\frac{1}{2}\) parts so that the chocolates can be distributed to more people
So,
The each part of chocolate we can get = \(\frac{1}{2}\) ÷ 7
Now,
We know that,
\(\frac{a}{b}\) ÷ a  = \(\frac{a}{b}\) × \(\frac{1}{a}\)
a= \(\frac{a}{1}\)
So,
\(\frac{1}{2}\) ÷ 7
= \(\frac{1}{2}\) × \(\frac{1}{7}\)
= \(\frac{ 1 × 1}{2 × 7}\)
= \(\frac{1}{14}\)
Hence, from the above,
We can conclude that we can get \(\frac{1}{14}\) part of each chocolate.

Question 12.
Reasoning
Complete the statements.

Think and Grow: Modeling Real Life

You melt \(\frac{1}{4}\) quart of soap. You pour the soap into 4 of the same-sized molds. What fraction of a quart of soap does each mold hold?
You are dividing \(\frac{1}{4}\) quart into 4 equal parts, so you need to find \(\frac{1}{4}\) ÷ 4.

Show and Grow

Question 13.
You buy \(\frac{1}{2}\) pound of grapes. You equally divide the grapes into 2 bags. What fraction of a pound of grapes do you put into each bag?
Answer: The fraction of a pound of grapes you put into each bag is: \(\frac{1}{8}\) pound

Explanation:
It is given that you buy \(\frac{1}{2}\) pound of grapes.
It is also given that you equally divide the grapes into 2 bags.
So,
The number of grapes in each bag = \(\frac{1}{2}\) ÷ 2
Now,
The fraction of pound of grapes you put into each bag = \(\frac{The number of grapes in each bag}{2}\)
=  ( \(\frac{1}{2}\) ÷ 2 ) ÷ 2
Now,
We know that,
\(\frac{a}{b}\) ÷ a  = \(\frac{a}{b}\) × \(\frac{1}{a}\)
a= \(\frac{a}{1}\)
So,
( \(\frac{1}{2}\) ÷ 2 ) ÷ 2
= ( \(\frac{1}{2}\) × \(\frac{1}{2}\) ) × \(\frac{1}{2}\)
= \(\frac{ 1 × 1}{2 × 2}\) × \(\frac{1}{2}\)
= \(\frac{1}{4}\) × \(\frac{1}{2}\)
= \(\frac{ 1 × 1}{2 × 4}\)
= \(\frac{1}{8}\)
Hence, from the above
We can conclude that the fraction of pound of grapes in each bag is: \(\frac{1}{8}\) pound

Question 14.
You have \(\frac{1}{8}\) cup of red sand, \(\frac{1}{4}\) cup of blue sand, and \(\frac{1}{2}\) cup of white sand. You equally divide the sand into 3 containers. What fraction of a cup of sand do you pour into each container?
Answer: The fraction of a cup of sand you pour into each container is: \(\frac{7}{24}\)

Explanation:
It is given that you have \(\frac{1}{8}\) cup of red sand, \(\frac{1}{4}\) cup of blue sand, and \(\frac{1}{2}\) cup of white sand.
So,
The total amount of sand = \(\frac{1}{8}\) cup of red sand + \(\frac{1}{4}\) cup of blue sand + \(\frac{1}{2}\) cup of white sand
In addition, we have to see either the numerators are equal or the denominators are equal.
If the numerators are equal we have to ake the denominators also equal.
So,
\(\frac{1}{4}\) is multplied by \(\frac{2}{2}\)
\(\frac{1}{2}\) is multiplied by \(\frac{4}{4}\)
So,
\(\frac{1}{4}\) = \(\frac{2}{8}\)
\(\frac{1}{2}\) = \(\frac{4}{8}\)
So,
\(\frac{1}{8}\) + \(\frac{2}{8}\) + \(\frac{4}{8}\) = \(\frac{7}{8}\)
It is also given that all the sand is equally distributed into 3 containers
So,
The amount of sand in each container = \(\frac{7}{8}\) ÷ 3
Now,
We know that,
\(\frac{a}{b}\) ÷ a  = \(\frac{a}{b}\) × \(\frac{1}{a}\)
a= \(\frac{a}{1}\)
So,
\(\frac{7}{8}\) ÷ 3
= \(\frac{7}{8}\) × \(\frac{1}{3}\)
= \(\frac{ 7 × 1}{8 × 3}\)
= \(\frac{7}{24}\)
Hence, from the above,
We can conclude that the amount of sand in each container is: \(\frac{7}{24}\) cup.

Question 15.
DIG DEEPER!
You, your friend, and your cousin share \(\frac{1}{2}\) of a vegetable pizza and \(\frac{1}{4}\) of a cheese share pizza. The pizzas are the same size. What fraction of a pizza do you get in all?
Big Ideas Math Answers 5th Grade Chapter 10 Divide Fractions 58

Divide. Use a model to help
Answer: The fraction of a pizza you got is: \(\frac{3}{12}\)

Explanation:
It is given that you, your friend, and your cousin share \(\frac{1}{2}\) of a vegetable pizza and \(\frac{1}{4}\) of a cheese share pizza.
So,
The total amount of pizza = \(\frac{1}{2}\) of a vegetable pizza + \(\frac{1}{4}\) of a cheese share pizza
In addition, we have to see either the numerators are equal or the denominators are equal.
If the numerators are equal we have to ake the denominators also equal.
So,
\(\frac{1}{2}\) is multplied by \(\frac{2}{2}\)
So,
\(\frac{1}{2}\) = \(\frac{2}{4}\)
So,
\(\frac{2}{4}\) + \(\frac{1}{4}\) = \(\frac{3}{4}\)
So,
The fraction of pizza each get = \(\frac{The total amount of pizza}{3}\)
= \(\frac{3}{4}\) ÷ 3
Now,
We know that,
\(\frac{a}{b}\) ÷ a  = \(\frac{a}{b}\) × \(\frac{1}{a}\)
a= \(\frac{a}{1}\)
So,
\(\frac{3}{4}\) ÷ 3
= \(\frac{3}{4}\) × \(\frac{1}{3}\)
= \(\frac{ 3 × 1}{4 × 3}\)
= \(\frac{3}{12}\)
Hence, from the above,
We can conclude that the fraction of pizza each get is: \(\frac{3}{12}\)

Divide Unit Fractions by Whole Numbers Homework & Practice 10.4

Question 1.
Big Ideas Math Answers 5th Grade Chapter 10 Divide Fractions 59
Answer: \(\frac{1}{3}\) ÷ 4 = \(\frac{1}{12}\)

Explanation:
The given numbers are: \(\frac{1}{3}\) and 4
We know that,
\(\frac{a}{b}\) ÷ a  = \(\frac{a}{b}\) × \(\frac{1}{a}\)
a= \(\frac{a}{1}\)
So,
\(\frac{1}{3}\) ÷ 4
= \(\frac{1}{3}\) × \(\frac{1}{4}\)
= \(\frac{ 1 × 1}{3 × 4}\)
= \(\frac{1}{12}\)
Hence,
\(\frac{1}{3}\) ÷ 4 = \(\frac{1}{12}\)

Question 2.
Big Ideas Math Answers 5th Grade Chapter 10 Divide Fractions 60
Answer: \(\frac{1}{6}\) ÷ 3 = \(\frac{1}{18}\)

Explanation:
The given numbers are: \(\frac{1}{6}\) and 3
We know that,
\(\frac{a}{b}\) ÷ a  = \(\frac{a}{b}\) × \(\frac{1}{a}\)
a= \(\frac{a}{1}\)
So,
\(\frac{1}{6}\) ÷ 3
= \(\frac{1}{6}\) × \(\frac{1}{3}\)
= \(\frac{ 1 × 1}{6 × 3}\)
= \(\frac{1}{18}\)
Hence,
\(\frac{1}{6}\) ÷ 3 = \(\frac{1}{18}\)

Question 3.
Big Ideas Math Answers 5th Grade Chapter 10 Divide Fractions 61
Answer: \(\frac{1}{4}\) ÷ 5 = \(\frac{1}{20}\)

Explanation:
The given numbers are: \(\frac{1}{4}\) and 5
We know that,
\(\frac{a}{b}\) ÷ a  = \(\frac{a}{b}\) × \(\frac{1}{a}\)
a= \(\frac{a}{1}\)
So,
\(\frac{1}{4}\) ÷ 5
= \(\frac{1}{4}\) × \(\frac{1}{5}\)
= \(\frac{ 1 × 1}{5 × 4}\)
= \(\frac{1}{20}\)
Hence,
\(\frac{1}{4}\) ÷ 5 = \(\frac{1}{20}\)

Divide. Use a model to help.

Question 4.
Big Ideas Math Answers 5th Grade Chapter 10 Divide Fractions 62
Answer: \(\frac{1}{5}\) ÷ 9 = \(\frac{1}{45}\)

Explanation:
The given numbers are: \(\frac{1}{5}\) and 9
We know that,
\(\frac{a}{b}\) ÷ a  = \(\frac{a}{b}\) × \(\frac{1}{a}\)
a= \(\frac{a}{1}\)
So,
\(\frac{1}{5}\) ÷ 9
= \(\frac{1}{5}\) × \(\frac{1}{9}\)
= \(\frac{ 1 × 1}{5 × 9}\)
= \(\frac{1}{45}\)
Hence,
\(\frac{1}{5}\) ÷ 9 = \(\frac{1}{45}\)

Question 5.
Big Ideas Math Answers 5th Grade Chapter 10 Divide Fractions 63
Answer: \(\frac{1}{8}\) ÷ 6 = \(\frac{1}{48}\)

Explanation:
The given numbers are: \(\frac{1}{8}\) and 6
We know that,
\(\frac{a}{b}\) ÷ a  = \(\frac{a}{b}\) × \(\frac{1}{a}\)
a= \(\frac{a}{1}\)
So,
\(\frac{1}{8}\) ÷ 6
= \(\frac{1}{8}\) × \(\frac{1}{6}\)
= \(\frac{ 1 × 1}{8 × 6}\)
= \(\frac{1}{48}\)
Hence,
\(\frac{1}{8}\) ÷ 6 = \(\frac{1}{48}\)

Question 6.
Big Ideas Math Answers 5th Grade Chapter 10 Divide Fractions 64
Answer: \(\frac{1}{7}\) ÷ 4 = \(\frac{1}{28}\)

Explanation:
The given numbers are: \(\frac{1}{7}\) and 4
We know that,
\(\frac{a}{b}\) ÷ a  = \(\frac{a}{b}\) × \(\frac{1}{a}\)
a= \(\frac{a}{1}\)
So,
\(\frac{1}{7}\) ÷ 4
= \(\frac{1}{7}\) × \(\frac{1}{4}\)
= \(\frac{ 1 × 1}{7 × 4}\)
= \(\frac{1}{28}\)
Hence,
\(\frac{1}{7}\) ÷ 4 = \(\frac{1}{28}\)

Question 7.
YOU BE THE TEACHER
Your friend divides \(\frac{1}{3}\) by 7 to get \(\frac{1}{21}\). He checks his answer by multiplying \(\frac{1}{21}\) × \(\frac{1}{3}\). Does your friend check his answer correctly? Explain.
Answer: No, your friend does not check his answer correctly

Explanation:
It is given that your friend divides \(\frac{1}{3}\) by 7 to get \(\frac{1}{21}\).
We know that,
\(\frac{a}{b}\) ÷ a  = \(\frac{a}{b}\) × \(\frac{1}{a}\)
a= \(\frac{a}{1}\)
So,
\(\frac{1}{3}\) ÷ 7
= \(\frac{1}{3}\) × \(\frac{1}{7}\)
= \(\frac{ 1 × 1}{7 × 3}\)
= \(\frac{1}{21}\)
It is also given that your friend checks his answer by multiplying \(\frac{1}{21}\) × \(\frac{1}{3}\).
Now,
\(\frac{1}{21}\) × \(\frac{1}{3}\)
= \(\frac{1 × 1}{21 × 3}\)
= \(\frac{1}{63}\)
But, your friend wanted to check whether \(\frac{1}{21}\) × \(\frac{1}{3}\) = \(\frac{1}{7}\)
But, the value becomes \(\frac{1}{63}\)
Hence, from the above,
We can conclude that your friend does not check the answer correctly.

Question 8.
Logic
Find the missing numbers.

Question 9.
Modeling Real Life
You win tickets that you can exchange for prizes. You exchange \(\frac{1}{5}\) of your tickets and then divide them equally among 3 prizes. What fraction of your tickets do you spend on each prize?
Answer: The fraction of your tickets you spend on each prize is: \(\frac{1}{15}\)

Explanation:
It is given that you win tickets that you can exchange for prizes.
It is also given that you exchange \(\frac{1}{5}\) of your tickets and then divide them equally among 3 prizes
So,
The fraction of the tickets spent on each prize = \(\frac{The value of Exchange}{The number of prizes}\)
= \(\frac{1}{5}\) ÷ 3
Now,
We know that,
\(\frac{a}{b}\) ÷ a  = \(\frac{a}{b}\) × \(\frac{1}{a}\)
a= \(\frac{a}{1}\)
So,
\(\frac{1}{5}\) ÷ 3
= \(\frac{1}{5}\) × \(\frac{1}{3}\)
= \(\frac{ 1 × 1}{5 × 3}\)
= \(\frac{1}{15}\)
Hence, from the above,
We can conlude that the fraction of tickets you spend on each prize is: \(\frac{1}{15}\)

Question 10.
DIG DEEPER!
You have \(\frac{1}{8}\) gallon of melted crayon wax. You pour the wax equally into 8 different molds to make new crayons. What fraction of a cup of melted wax is in each mold? Think: 1 gallon is 16 cups.
Big Ideas Math Answers 5th Grade Chapter 10 Divide Fractions 66
Answer: The fraction of a cup of melted wax in each mold is: \(\frac{1}{4}\)

Explanation:
It is given that you have \(\frac{1}{8}\) gallon of melted crayon wax.
It is also given that you pour the wax equally into 8 different molds to make new crayons.
So,
The fraction of melted crayon wax in each mold in gallons = \(\frac{The total amount of melted crayon wax }{The number of molds}\)
= \(\frac{1}{8}\) ÷ 8
Now,
We know that,
\(\frac{a}{b}\) ÷ a  = \(\frac{a}{b}\) × \(\frac{1}{a}\)
a= \(\frac{a}{1}\)
So,
\(\frac{1}{8}\) ÷ 8
= \(\frac{1}{8}\) × \(\frac{1}{8}\)
= \(\frac{ 1 × 1}{8 × 8}\)
= \(\frac{1}{64}\) gallons
But, it is given that
1 gallon = 16 cups
So,
The total number of cups that the melted crayon wax contained = \(\frac{1}{64}\) × \(\frac{16}{1}\)
= \(\frac{1 × 16 }{64 × 1}\)
= \(\frac{1}{4}\)
Hence, from the above,
We can conclude that there are \(\frac{1}{4}\) cups of melted crayon wax in each mold.

Review & Refresh

Question 11.
0.9 ÷ 0.1 = ___
Answer: 0.9 ÷ 0.1 = 9

Explanation:
The given decimal numbers are: 0.9 and 0.1
The representation of the decimal numbers in the fraction form is: \(\frac{9}{10}\) and \(\frac{1}{10}\)
Now,
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
So,
\(\frac{9}{10}\) ÷ \(\frac{1}{10}\)  = \(\frac{9}{10}\) × \(\frac{10}{1}\)
= \(\frac{ 9 × 10}{10 × 1}\)
= 9
Hence, 0.9 ÷ 0.1 = 9

Question 12.
38.6 ÷ 100 = ___

Answer: 38.6 ÷ 100 = 0.386

Explanation:
The given numbers are: 38.6 and 100
The representation of the numbers in the fraction form is: \(\frac{386}{10}\) and \(\frac{100}{1}\)
Now,
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
So,
\(\frac{386}{10}\) ÷ \(\frac{100}{1}\)  = \(\frac{386}{10}\) × \(\frac{1}{100}\)
= \(\frac{ 386 × 1}{100 × 10}\)
= \(\frac{386}{1000}\)
= 0.386
Hence, 38.6 ÷ 100 = 0.386

Question 13.
2.57 ÷ 0.01 = ___
Answer: 2.57 ÷ 0.01 = 257

Explanation:
The given decimal numbers are: 2.57 and 0.01
The representation of the decimal numbers in the fraction form is: \(\frac{257}{100}\) and \(\frac{1}{100}\)
Now,
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
So,
\(\frac{257}{100}\) ÷ \(\frac{1}{100}\)  = \(\frac{257}{100}\) × \(\frac{100}{1}\)
= \(\frac{ 257 × 100}{100 × 1}\)
= 257
Hence, 2.57 ÷ 0.01 = 257

Lesson 10.5 Problem Solving: Fraction Division

Explore and Grow

You want to make a \(\frac{1}{3}\) batch of the recipe. How you can use division to find the amount of each ingredient you need?
Answer:
It is given that you want to make a \(\frac{1}{3}\) batch of the recipe.
So,
From \(\frac{1}{3}\),
1 represents a batch of the recipe
3 represents the total number of ingredients in a batch
So,
The amount of each ingredient you need = \(\frac{The amount of the batch of the recipe }{The total number of ingredients}\)
= \(\frac{1}{3}\) ÷ 3
Now,
We know that,
\(\frac{a}{b}\) ÷ a  = \(\frac{a}{b}\) × \(\frac{1}{a}\)
a= \(\frac{a}{1}\)
So,
\(\frac{1}{3}\) ÷ 3
= \(\frac{1}{3}\) × \(\frac{1}{3}\)
= \(\frac{ 1 × 1}{3 × 3}\)
= \(\frac{1}{9}\)
Hence, from the above,
We can conclude that the amount of each ingredient you need is: \(\frac{1}{9}\)

Reasoning
Without calculating, explain how you can tell whether you need more than or less than 1 tablespoon of olive oil?
Answer: You need less than 1 tablespoon of olive oil

Explanation:
From the above problem,
The amount of each ingredient is: \(\frac{1}{9}\)
Since the amount of each ingredient is less than 1, you need less than 1 tablespoon of olive oil

Think and Grow: Problem Solving: Fraction Division

Example
You have 4 cups of yellow paint and 3 cups of blue paint. How many batches of green paint can you make?
Big Ideas Math Answers 5th Grade Chapter 10 Divide Fractions 67

Understand the Problem

What do you know?

  • You have 4 cups of yellow paint and 3 cups of blue paint.
  • One batch of green paint is made of \(\frac{1}{2}\) cup of yellow and \(\frac{1}{3}\) cup of blue.

What do you need to find?

  • You need to find how many batches of green paint you can make.

Make a Plan
How will you solve?

  • Find how many batches are possible from yellow, and how many from blue.
  • Choose the lesser number of batches.

Solve

So, you can make 8 batches of green paint.

Show and Grow

Question 1.
In the example, explain why you choose the fewer number of batches.
Answer: In the above example, the yellow paint has the less number of batches as the amount of each batch of yellow paint-filled is more than the batch of green paint
Hence,
We choose the fewer number of batches of yellow paint

Apply and Grow: Practice

Understand the problem. What do you know? What do you need to find? Explain.

Question 2.
A landowner donates 3 acres of land to a city. The mayor of the city uses 1 acre of the land for a playground and the rest of the land for community garden plots. Each garden plot is \(\frac{1}{3}\) acre. How many plots are there?
Understand the problem. Then make a plan. How will you solve it? Explain?
Answer: The number of plots in the community is: 6

Explanation:
It is given that a landowner donates 3 acres of land to a city and the mayor of the city uses 1 acre of the land for a playground and the rest of the land for community garden plots.
So,
The portion of the land used for community garden plots is: 2 acres
It is also given that each garden plot is \(\frac{1}{3}\) acre.
So,
The number of plots = \(\frac{The portion of the land used for community garden plots}{The area of each garden plot}\)
= 2 ÷ \(\frac{1}{3}\)
= 2 × \(\frac{3}{1}\)
= \(\frac{2}{1}\) × \(\frac{3}{1}\)
= 6
Hence, from the above,
We can conclude that there are 6 plots

Question 3.
A craftsman uses \(\frac{3}{4}\) gallon of paint to paint 4 identical dressers. He uses the same amount of paint on each dresser. How much paint does he use to paint 7 of the same dressers?
Big Ideas Math Answers 5th Grade Chapter 10 Divide Fractions 69
Answer: The paint used by the craftsman to paint 7 of the same dressers is: \(\frac{21}{16}\)

Explanation:
It is given that a craftsman uses \(\frac{3}{4}\) gallon of paint to paint 4 identical dressers.
So,
The paint used to paint each dresser = \(\frac{3}{4}\) ÷ 4
= \(\frac{3}{4}\) × \(\frac{1}{4}\)
= \(\frac{3}{16}\) gallon
So,
The amount of paint used to paint the 7 identical dressers = \(\frac{The paint used to paint each dresser}{1}\) × 7
= \(\frac{3}{16}\) × \(\frac{7}{1}\)
= \(\frac{3 × 7}{16 × 1}\)
= \(\frac{21}{16}\) gallon
Hence, from the above,
We can conclude that the paint used to paint 7 identical dressers is: \(\frac{21}{16}\) gallon

Question 4.
An airplane travels 125 miles in \(\frac{1}{4}\) hour. It travels the same number of miles each hour. How many miles does the plane travel in 5 hours?
Answer: The number of miles the plane travel in 5 hours is: 2,500 miles

Explanation:
It is given that an airplane travels 125 miles in \(\frac{1}{4}\) hour
So,
The number of miles traveled by plane in 1 hour = 125 ÷ \(\frac{1}{4}\)
= 125 × \(\frac{4}{1}\)
= 125 × 4
= 500 miles
So,
The number of miles traveled by plane in 5 hours = ( The number of miles traveled by plane in 1 hour ) × 5
= 500 × 5
= 2,500 miles
Hence, from the above,
We can conclude that the number of miles traveled by plane in 5 hours is: 2,500 miles

Question 5.
You make bows for gifts using \(\frac{2}{3}\) yard of ribbon for each bow. You have 4 feet of red ribbon and 5 feet of green ribbon. How many bows can you make?
Answer: The number of bows you can make is: 2 bows

Explanation:
It is given that you make bows for gifts using \(\frac{2}{3}\) yard of ribbon for each bow.
It is also given that you have 4 feet of red ribbon and 5 feet of green ribbon
So,
The total length of ribbon = 5 + 4 = 9 feet
we know that,
1 foot = \(\frac{1}{3}\) yards
So,
9 feet = 9 × \(\frac{1}{3}\) yards
= \(\frac{9}{1}\) yards × \(\frac{1}{3}\) yards
= 3 yards
So,
The number of bows you can make = \(\frac{2}{3}\) yards × 3
= 2 bows
Hence, from the above,
We can conclude that the number of bows we can make is: 2

Question 6.
A landscaper buys 1 gallon of plant fertilizer. He uses \(\frac{1}{5}\) of the fertilizer, and then divides the rest into 3 smaller bottles. How many gallons does he put into each bottle?
Answer: The number of gallons he put into each bottle is: \(\frac{4}{15}\)

Explanation:
It is given that a landscaper buys 1 gallon of plant fertilizer and he uses \(\frac{1}{5}\) of the fertilizer
So,
The remaining amount of the fertilizer = 1 – \(\frac{1}{5}\)
= \(\frac{4}{5}\) gallons
It is also given that he divided the remaining amount of fertilizer into 3 smaller bottles.
So,
The amount of fertilizer put into each bottle = \(\frac{The remaining amount of the fertilizer}{The total number of bottles}\)
= \(\frac{4}{5}\) ÷ 3
= \(\frac{4}{5}\) × \(\frac{1}{3}\)
= \(\frac{4 × 1}{5 × 3}\)
= \(\frac{4}{15}\) gallons
hence, from the above,
We can conclude that the amount of remaining fertilizer put into each bottle is: \(\frac{4}{15}\) gallons

Think and Grow: Modeling Real Life

Example
A sponsor donates $0.10 to a charity for every \(\frac{1}{4}\) kilometer of the triathlon an athlete completes. The athlete completes the entire triathlon. How much money does the sponsor donate?
Big Ideas Math Answers 5th Grade Chapter 10 Divide Fractions 70
Think: What do you know? What do you need to find? How will you solve?
Write and solve an equation.
Add 1.9, 90, and 21.1 to find how many kilometers the athlete completes.
Divide the sum by \(\frac{1}{4}\) to find how many \(\frac{1}{4}\) kilometers the athlete completes.
Multiply the quotient by $0.10 to find how much money the sponsor donates.
Big Ideas Math Answers 5th Grade Chapter 10 Divide Fractions 71
Let m represent the total amount of money donated.

Show and Grow

Question 7.
You earn $5 for every \(\frac{1}{2}\) hour you do yard work. How much money do you earn in 1 week?
Big Ideas Math Answers 5th Grade Chapter 10 Divide Fractions 73
Answer: The amount you earn in 1 week is: $700

Explanation:
The given table is:
Big Ideas Math Answers 5th Grade Chapter 10 Divide Fractions 73
From the above table,
The total amount of time = 5\(\frac{1}{2}\) + 3 + 1\(\frac{1}{2}\)
= \(\frac{11}{2}\) + 3 + \(\frac{3}{2}\)
= \(\frac{11 + 3}{2}\) + 3
= 7 + 3
= 10 hours
It is given that you earn $5 for every \(\frac{1}{2}\) hour you do yard work
So,
The amount earned in 10 hours in a day = 10 ÷\(\frac{1}{2}\) × 5 ( Since we have the time in hours but the money earned is given in half an hour basis )
= 20 × 5
= $100
We know that 1 week = 7 days
So,
The amount earned in 1 week = 100 × 7 = $700
hence, from the above,
We can conclude that we can earn $700 in a week

Problem Solving: Fraction Division Homework &  10.5

Understand the problem. Then make a plan. How will you solve? Explain.

Question 1.
A train travels 75 miles in \(\frac{1}{2}\) hour. How many miles does the train travel in 8 hours?
Answer: The number of miles the train travel in  hours is: 1,200 miles

Explanation:
It is given that a train travels 75 miles in \(\frac{1}{2}\) hour.
So,
The number of miles the train travel in 1 hour = 75 ÷ \(\frac{1}{2}\)
= 75 × 2
= 150 miles
So,
The number of miles the train travel in 8 hours = The number of miles traveled by train in 1 hour × 8
= 150 × 8
= 1,200 miles
Hence, from the above,
We can conclude that the train travels 1,200 miles in 8 hours.

Question 2.
You need \(\frac{2}{3}\) yard of fabric to create a headband. You have 12 feet of blue fabric and 4 feet of yellow fabric. How many headbands can you make with all of the fabric?
Answer: The number of headbands you can make with all of the fabric is: 8 headbands

Explanation:
It is given that you need \(\frac{2}{3}\) yard of fabric to create a headband.
It is also given that you have 12 feet of blue fabric and 4 feet of yellow fabric.
So,
The total length of the fabric = 12 + 4 = 16 feet
We know that
1 foot = \(\frac{1}{3}\) yards
So,
16 feet = \(\frac{16}{3}\) yards
So,
The number of headbands you can create with all the fabric = \(\frac{The total length of the fabric}{The length of each fabric}\)
= \(\frac{16}{3}\) ÷ \(\frac{2}{3}\)
= \(\frac{16}{3}\) × \(\frac{3}{2}\)
= \(\frac{16 × 3}{3 × 2}\)
= 8 headbands
Hence, from the above,
We can conclude that we can create 8 headbands with all the fabric.

Question 3.
An art teacher has 8 gallons of paint. Her class uses \(\frac{3}{4}\) of the paint. The teacher divides the rest of the paint into 4 bottles. How much paint is in each bottle?
Answer: The amount of paint in each bottle is: \(\frac{1}{2}\)

Explanation:
It is given that an art teacher has 8 gallons of paint and her class uses \(\frac{3}{4}\) of the paint.
So,
The remaining amount of paint = \(\frac{1}{4}\) × 8
= \(\frac{1}{4}\) × \(\frac{8}{1}\)
=\(\frac{1 × 8}{4 × 1}\)
= 2 gallons
It is also given that the remaining amount of the paint divided into 4 bottles by the teacher
So,
The amount of paint present in each bottle = 2 ÷ 4
= \(\frac{1}{2}\) gallons
Hence, from the above,
We can conclude that the amout of paint present in each bottle is: \(\frac{1}{2}\) gallons

Question 4.
You mix 3\(\frac{1}{4}\) cups of frozen strawberries and 4\(\frac{1}{2}\) cups of frozen blueberries in a bowl. A smoothie requires \(\frac{1}{2}\) cup of your berry mix. How many smoothies can you make?
Answer: The number of smoothies you can make is:

Explanation:
It is given that you mix 3\(\frac{1}{4}\) cups of frozen strawberries and 4\(\frac{1}{2}\) cups of frozen blueberries in a bowl.
So,
The amount of berry mix = 3\(\frac{1}{4}\) cups of frozen strawberries + 4\(\frac{1}{2}\) cups of frozen blueberries
= 3\(\frac{1}{4}\) + 4\(\frac{1}{2}\)
= \(\frac{13}{4}\) + \(\frac{9}{2}\)
In addition, equate the denominators
So,
Multiply \(\frac{9}{2}\) with \(\frac{2}{2}\)
So,
\(\frac{9}{2}\) = \(\frac{18}{4}\)
So,
The amount of berry mix = \(\frac{13}{4}\) + \(\frac{18}{4}\)
= \(\frac{31}{4}\)
Now,
It is also given that the smoothie requires \(\frac{1}{2}\) cup of your berry mix.
So,
The number of smoothies = \(\frac{31}{4}\) ÷ \(\frac{1}{2}\)
= \(\frac{31}{4}\) × \(\frac{2}{1}\)
= \(\frac{31 × 2}{4 × 1}\)
= \(\frac{31}{2}\)
Hence, from the above,
We can conclude that the number of smoothies we can make are: \(\frac{31}{2}\)

Question 5.
Modeling Real Life
A sponsor donates $0.10 for every \(\frac{1}{4}\) dollar donated at the locations shown. How much money does the sponsor donate?
Big Ideas Math Answers 5th Grade Chapter 10 Divide Fractions 74
Answer: The amount of money the sponsor donates is: $40.4

Explanation:
The given table is:
Big Ideas Math Answers 5th Grade Chapter 10 Divide Fractions 74
From the above table,
The total amount of money collected = 25.25 + 12.50 + 63.25
= $101
It is given that a sponsor donates $0.10 for every \(\frac{1}{4}\) dollar
So,
the total amount of donated = The total amount of money collected ÷ \(\frac{1}{4}\) × 0.10
= 101 ÷ \(\frac{1}{4}\) × 0.10
= 101 × 4 × 0.10
= 04 × 0.10
= $40.4
Hence, from the above,
We can conclude that the amount of money donated by a sponsor is: $40.4

Question 6.
DIG DEEPER!
A nurse earns $16 for every \(\frac{1}{2}\) hour at work. How much money does she earn in 5 days?
Big Ideas Math Answers 5th Grade Chapter 10 Divide Fractions 75
Answer: The money she earns in 5 days is: $1,280

Explanation:
The given table is:
Big Ideas Math Answers 5th Grade Chapter 10 Divide Fractions 75
From the above table,
The total amount of time = 6\(\frac{3}{4}\) + 1 + \(\frac{1}{4}\)
= \(\frac{27}{4}\) + 1 + \(\frac{1}{4}\)
= \(\frac{27 + 1}{4}\) + 1
= 7 + 1
= 8 hours
It is given that a nurse earns $16 for every \(\frac{1}{2}\) hour at work.
So,
The money she earned for 1 hour = 16 ÷ \(\frac{1}{2}\)
= 16 × 2 = $32
So,
The money earned for 8 hours = The money earned in 1 hour × 8
= 32 × 8 = $256
So,
The money earned in 5 days = The money earned in 1 day × 5
= 256 × 5 = $1,280
hence, from the above,
we can conclude that she can earn $1,280 in 5 days.

Review & Refresh

Find the quotient. Then check your answer.

Question 7.
Big Ideas Math Answers 5th Grade Chapter 10 Divide Fractions 76
Answer:  186 ÷ 12 = 1 R 4

Explanation:
Let 185.88 be rounded to 186
So,
By using the partial quotients method,
186 ÷ 12 = ( 120 + 36 + 24 ) ÷ 12
= ( 120 ÷ 12 ) + ( 36 ÷ 12 ) + ( 24 ÷ 12 )
= 10 + 3 + 2
= 17 R 4
Hence, 186 ÷ 12 = 17 R 4

Question 8.
Big Ideas Math Answers 5th Grade Chapter 10 Divide Fractions 77
Answer: 74 ÷ 24 = 3 R 2

Explanation:
Let 74.4 be rounded to 74
So,
By using the partial quotients method,
74 ÷ 24 = 72 ÷ 24
= 3 R 2
Hence, 74 ÷ 24 = 3 R 2

Question 9.
Big Ideas Math Answers 5th Grade Chapter 10 Divide Fractions 78
Answer: 42 ÷ 46 = 0.9

Explanation:
Let 42.32 be rounded to 42
So,
By using the partial quotients method,
42 ÷ 46 = 0.9
Hence,
42 ÷ 46 = 0.9

Divide Fractions Performance Task 10

Your city has a robotics competition. Each team makes a robot that travels through a maze. The time each robot spends in the maze is used to find the team’s score.
1. One-third of the students in your grade participate in the competition. The number of participating students is divided into 12 teams.
Big Ideas Math Answers 5th Grade Chapter 10 Divide Fractions 79
a. What fraction of the total number of students in your grade is on each team?
Answer: The fraction of the total number of students in your grade is: \(\frac{1}{36}\)

Explanation:
It is given that there are \(\frac{1}{3}\) of the students in your grade are participating in the competition.
It is also given that the participating students are divided into 12 teams.
So,
The fraction of the total number of students in each team = \(\frac{The number of participating students}{Th total number of teams}\)
= \(\frac{1}{3}\) ÷ 12
= \(\frac{1}{3}\) ÷ \(\frac{12}{1}\)
= \(\frac{1}{3}\) × \(\frac{1}{12}\)
= \(\frac{1 × 1}{12 × 3}\)
= \(\frac{1}{36}\)
Hence, from the above,
We can conclude that the fraction of students that are in each team is: \(\frac{1}{36}\)

b. There are 3 students on each team. How many students are in your grade?
Answer: The number of students in your grade is: 36

Explanation:
It is given that the number of students is divided into 12 teams
It is also given that there are 3 students on each team
So,
The total number of students = The number of teams × The number of students in each team
= 12 × 3 = 36 students
Hence, from the above,
We can conclude that there are 36 students in your grade

Question 2.
The maze for the competition is shown.
a. Write the length of the maze in feet.
Big Ideas Math Answers 5th Grade Chapter 10 Divide Fractions 80
Answer:
The given maze for the competition is:
Big Ideas Math Answers 5th Grade Chapter 10 Divide Fractions 80
From the above maze,
The total length of the maze is: 8 feet 6 inches
We know that,
1 foot = 12 inches
Hence,
1 inch = \(\frac{1}{12}\) feet
So,
6 inches = 6 × \(\frac{1}{12}\)
= \(\frac{1}{12}\) × \(\frac{6}{1}\)
= \(\frac{1}{2}\) feet
So,
The total length of the maze in feet = 8 feet + \(\frac{1}{2}\) feet
= 8.5 feet or 8\(\frac{1}{2}\) feet

b. The length of the maze is divided into 6 equal sections. What is the length of each section of the maze?
Answer: The length of each section of the maze is: \(\frac{17}{12}\) feet

Explanation:
From the above Exercise,
The total length of the maze in feet is: 8.5 feet or 8\(\frac{1}{2}\) feet
It is given that the length of the maze is divided into 6 equal sections
So,
The length of each section of the maze = 8\(\frac{1}{2}\) ÷ 6
= 8\(\frac{1}{2}\) ÷ \(\frac{6}{1}\)
= 8\(\frac{1}{2}\) × \(\frac{1}{6}\)
= \(\frac{17}{2}\) × \(\frac{1}{6}\)
= \(\frac{17}{12}\) feet
Hence, from the above,
We can conclude that the length of each section in a maze is: \(\frac{17}{12}\) feet

Question 3.
Each team has 200 seconds to complete the maze. The rules require judges to use the expression (200 – x) ÷ \(\frac{1}{5}\), where x is the total number of seconds, to find a team’s total score.
a. Your robot completes the maze in 3 minutes 5 seconds. How many points does your team earn?
Answer: The number of points your team earn is: 75 points

Explanation:
It is given that each team has 200 seconds to complete the maze and the rules require judges to use the expression (200 – x) ÷ \(\frac{1}{5}\) where x is the total number of seconds
It is also given that your robot completes the maze in 3 minutes 5 seconds
We know that,
1 minute = 60 seconds
So,
The time taken by the robot to complete the maze in seconds = ( 3 × 60 ) + 5
= 185 seconds
So,
x= 185
So,
200 – x = 200 – 185 = 15
So,
The number of points the team earned = ( 200 – x ) ÷ \(\frac{1}{5}\)
= 15 ÷ \(\frac{1}{5}\)
= 15 × 5
= 75 points
Hence, from the above,
We can conclude that the number of points earned by the team is: 75 points

b. Do you think the team with the most points or the fewest points wins? Use an example to justify your answer.
Answer: The team with the most points wins the competition because
Reason:
Suppose team A takes 2 minutes and team B takes 3 minutes to complete the competition
So,
The time is taken by team A in seconds = 120 seconds
So,
x= 120
So,
200 – x = 200 – 120 = 80
Now,
The time is taken by team B in seconds = 180 seconds
So,
x= 180
So,
200 – x = 200 – 180 = 20
Now,
The number of points earned by team A = 80 ÷ \(\frac{1}{5}\)
= 400 points
The number of points earned by team B = 20 ÷ \(\frac{1}{5}\)
= 100 points
Hence, from the above,
We can conclude that the team with more points wins the competition

Divide Fractions Activity

Fraction Connection: Division

Directions:

  1. Players take turns rolling three dice.
  2. On your turn, evaluate the expression indicated by your roll and cover the answer.
  3. The first player to get four in a row, horizontally, vertically, or diagonally, wins!

Big Ideas Math Answers 5th Grade Chapter 10 Divide Fractions 81
Big Ideas Math Answers 5th Grade Chapter 10 Divide Fractions 82

Divide Fractions Chapter Practice 10

10.1 Interpret Fractions as Division

Divide. Use a model to help.

Question 1.
1 ÷ 2 = ___
Answer: 1 ÷ 2 = \(\frac{1}{2}\)

Explanation:
We know that,
a ÷ b = \(\frac{a}{b}\)
Hence,
1 ÷ 2 = \(\frac{1}{2}\)

Question 2.
3 ÷ 10 = __
Answer: 3 ÷ 10 = \(\frac{3}{10}\)

Explanation:
We know that,
a ÷ b = \(\frac{a}{b}\)
Hence,
3 ÷ 10 = \(\frac{3}{10}\)

Question 3.
4 ÷ 7 = __
Answer: 4 ÷ 7 = \(\frac{4}{7}\)

Explanation:
We know that,
a ÷ b = \(\frac{a}{b}\)
Hence,
4 ÷ 7 = \(\frac{4}{7}\)

Question 4.
11 ÷ 15 = ___
Answer: 11 ÷ 15 = \(\frac{11}{15}\)

Explanation:
We know that,
a ÷ b = \(\frac{a}{b}\)
Hence,
11 ÷ 15 = \(\frac{11}{15}\)

Question 5.
8 ÷ 9 = ___
Answer: 8 ÷ 9 = \(\frac{8}{9}\)

Explanation:
We know that,
a ÷ b = \(\frac{a}{b}\)
Hence,
8 ÷ 9 = \(\frac{8}{9}\)

Question 6.
13 ÷ 20 = ___
Answer: 13 ÷ 20 = \(\frac{13}{20}\)

Explanation:
We know that,
a ÷ b = \(\frac{a}{b}\)
Hence,
13 ÷ 20 = \(\frac{13}{20}\)

Question 7.
Modeling Real Life
Nine friends equally share 12 apples. What fraction of an apple does each friend get?
Answer: The fraction of an apple each friend get is: \(\frac{9}{12}\)

Explanation:
It is given that nine friends equally share 12 apples.
So,
The fraction of an apple each friend get = \(\frac{The number of friends}{The number of apples}\)
= \(\frac{9}{12}\)
Hence, from the above,
We can conclude that the fraction of an apple each friend get is: \(\frac{9}{12}\)

10.2 Mixed Numbers as Quotients

Divide. Use a model to help

Question 8.
8 ÷ 3 = ___

Answer: 8 ÷ 3 = 2\(\frac{2}{3}\)

Explanation:
We know that,
a ÷ b = \(\frac{a}{b}\)
a\(\frac{b}{c}\) = \(\frac{ac + b}{c}\)
\(\frac{a}{b}\) = Remainder\(\frac{Quotient}{Divisor}\)
So,
8 ÷ 3 = \(\frac{8}{3}\)
By using the partial quotients method,
8 ÷ 3 = 6 ÷ 3
= 2 R 2
Hence,
8 ÷ 3 = 2\(\frac{2}{3}\)

Question 9.
6 ÷ 5 = ___
Answer: 6 ÷ 5 = 1\(\frac{1}{5}\)

Explanation:
We know that,
a ÷ b = \(\frac{a}{b}\)
a\(\frac{b}{c}\) = \(\frac{ac + b}{c}\)
\(\frac{a}{b}\) = Remainder\(\frac{Quotient}{Divisor}\)
So,
6 ÷ 5 = \(\frac{6}{5}\)
By using the partial quotients method,
6 ÷ 5 = 5 ÷ 5
= 1 R 1
Hence,
6 ÷ 5 = 1\(\frac{1}{5}\)

Question 10.

10 ÷ 4 = ___
Answer: 10 ÷ 4 = 2\(\frac{2}{4}\)

Explanation:
We know that,
a ÷ b = \(\frac{a}{b}\)
a\(\frac{b}{c}\) = \(\frac{ac + b}{c}\)
\(\frac{a}{b}\) = Remainder\(\frac{Quotient}{Divisor}\)
So,
10 ÷ 4 = \(\frac{10}{4}\)
By using the partial quotients method,
10 ÷ 4 = 8 ÷ 4
= 2 R 2
Hence,
10 ÷ 4 = 2\(\frac{2}{4}\)

Question 11.

20 ÷ 11 = __
Answer: 20 ÷ 11 = 1\(\frac{9}{11}\)

Explanation:
We know that,
a ÷ b = \(\frac{a}{b}\)
a\(\frac{b}{c}\) = \(\frac{ac + b}{c}\)
\(\frac{a}{b}\) = Remainder\(\frac{Quotient}{Divisor}\)
So,
20 ÷ 11 = \(\frac{20}{11}\)
By using the partial quotients method,
20 ÷ 11 = 11 ÷ 11
= 1 R 9
Hence,
20 ÷ 11 = 1\(\frac{9}{11}\)

Question 12.

25 ÷ 2 = ___
Answer: 25 ÷ 2 = 12\(\frac{1}{2}\)

Explanation:
We know that,
a ÷ b = \(\frac{a}{b}\)
a\(\frac{b}{c}\) = \(\frac{ac + b}{c}\)
\(\frac{a}{b}\) = Remainder\(\frac{Quotient}{Divisor}\)
So,
25 ÷ 2 = \(\frac{25}{2}\)
By using the partial quotients method,
25 ÷ 2 = 24 ÷ 2
= 12 R 1
Hence,
25 ÷ 2 = 12\(\frac{1}{2}\)

Question 13.

64 ÷ 9 = ___
Answer: 64 ÷ 9 = 7\(\frac{1}{9}\)

Explanation:
We know that,
a ÷ b = \(\frac{a}{b}\)
a\(\frac{b}{c}\) = \(\frac{ac + b}{c}\)
\(\frac{a}{b}\) = Remainder\(\frac{Quotient}{Divisor}\)
So,
64 ÷ 9 = \(\frac{64}{9}\)
By using the partial quotients method,
64 ÷ 9 = 63 ÷ 9
= 7 R 1
Hence,
64 ÷ 9 = 7\(\frac{1}{9}\)

10.3 Divide Whole Numbers by Unit Fractions

Divide. Use a model to help.

Question 14.
Big Ideas Math Answers Grade 5 Chapter 10 Divide Fractions 83
Answer: 4 ÷ \(\frac{1}{2}\) = 8

Explanation:
The given numbers are: 4 and \(\frac{1}{2}\)
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
So,
4 ÷ \(\frac{1}{2}\)  = 4 × \(\frac{2}{1}\)
= \(\frac{4}{1}\) × \(\frac{2}{1}\)
= \(\frac{ 2 × 4}{1 × 1}\)
= 8
Hence,
4÷ \(\frac{1}{2}\) = 8

Question 15.
Big Ideas Math Answers Grade 5 Chapter 10 Divide Fractions 84
Answer: 6 ÷ \(\frac{1}{5}\) = 30

Explanation:
The given numbers are: 6 and \(\frac{1}{5}\)
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
So,
6 ÷ \(\frac{1}{5}\)  = 6 × \(\frac{5}{1}\)
= \(\frac{6}{1}\) × \(\frac{5}{1}\)
= \(\frac{ 6 × 5}{1 × 1}\)
= 30
Hence,
6÷ \(\frac{1}{5}\) = 30

Question 16.
Big Ideas Math Answers Grade 5 Chapter 10 Divide Fractions 85
Answer: 7 ÷ \(\frac{1}{4}\) = 28

Explanation:
The given numbers are: 7 and \(\frac{1}{4}\)
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
So,
7 ÷ \(\frac{1}{4}\)  = 7 × \(\frac{4}{1}\)
= \(\frac{7}{1}\) × \(\frac{4}{1}\)
= \(\frac{ 7 × 4}{1 × 1}\)
= 28
Hence,
7÷ \(\frac{1}{4}\) = 36

Question 17.
Big Ideas Math Answers Grade 5 Chapter 10 Divide Fractions 86
Answer: 8 ÷ \(\frac{1}{3}\) = 24

Explanation:
The given numbers are: 8 and \(\frac{1}{3}\)
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
So,
8 ÷ \(\frac{1}{3}\)  = 8 × \(\frac{3}{1}\)
= \(\frac{8}{1}\) × \(\frac{3}{1}\)
= \(\frac{ 8 × 3}{1 × 1}\)
= 24
Hence,
8÷ \(\frac{1}{3}\) = 24

Question 18.
Big Ideas Math Answers Grade 5 Chapter 10 Divide Fractions 87
Answer: 9 ÷ \(\frac{1}{2}\) = 18

Explanation:
The given numbers are: 9 and \(\frac{1}{2}\)
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
So,
9 ÷ \(\frac{1}{2}\)  = 9 × \(\frac{2}{1}\)
= \(\frac{9}{1}\) × \(\frac{2}{1}\)
= \(\frac{ 9 × 2}{1 × 1}\)
= 18
Hence,
9÷ \(\frac{1}{2}\) = 18

Question 19.
Big Ideas Math Answers Grade 5 Chapter 10 Divide Fractions 88
Answer: 2 ÷ \(\frac{1}{10}\) = 20

Explanation:
The given numbers are: 2 and \(\frac{1}{10}\)
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
So,
2 ÷ \(\frac{1}{10}\)  = 2 × \(\frac{10}{1}\)
= \(\frac{2}{1}\) × \(\frac{10}{1}\)
= \(\frac{ 2 × 10}{1 × 1}\)
= 20
Hence,
2÷ \(\frac{1}{10}\) = 20

10.4 Divide Unit Fractions by Whole Numbers

Divide. Use a model to help.

Question 20.
Big Ideas Math Answers Grade 5 Chapter 10 Divide Fractions 89
Answer: \(\frac{1}{7}\) ÷ 2 = \(\frac{1}{14}\)

Explanation:
The given numbers are: \(\frac{1}{7}\) and 2
We know that,
\(\frac{a}{b}\) ÷ a  = \(\frac{a}{b}\) × \(\frac{1}{a}\)
a= \(\frac{a}{1}\)
So,
\(\frac{1}{7}\) ÷ 2
= \(\frac{1}{7}\) × \(\frac{1}{2}\)
= \(\frac{ 1 × 1}{7 × 2}\)
= \(\frac{1}{14}\)
Hence,
\(\frac{1}{7}\) ÷ 2 = \(\frac{1}{14}\)

Question 21.
Big Ideas Math Answers Grade 5 Chapter 10 Divide Fractions 90
Answer: \(\frac{1}{2}\) ÷ 9 = \(\frac{1}{18}\)

Explanation:
The given numbers are: \(\frac{1}{2}\) and 9
We know that,
\(\frac{a}{b}\) ÷ a  = \(\frac{a}{b}\) × \(\frac{1}{a}\)
a= \(\frac{a}{1}\)
So,
\(\frac{1}{2}\) ÷ 9
= \(\frac{1}{2}\) × \(\frac{1}{9}\)
= \(\frac{ 1 × 1}{2 × 9}\)
= \(\frac{1}{18}\)
Hence,
\(\frac{1}{2}\) ÷ 9 = \(\frac{1}{18}\)

Question 22.
Big Ideas Math Solutions Grade 5 Chapter 10 Divide Fractions 91
Answer: \(\frac{1}{3}\) ÷ 7 = \(\frac{1}{21}\)

Explanation:
The given numbers are: \(\frac{1}{3}\) and 7
We know that,
\(\frac{a}{b}\) ÷ a  = \(\frac{a}{b}\) × \(\frac{1}{a}\)
a= \(\frac{a}{1}\)
So,
\(\frac{1}{3}\) ÷ 7
= \(\frac{1}{3}\) × \(\frac{1}{7}\)
= \(\frac{ 1 × 1}{7 × 3}\)
= \(\frac{1}{21}\)
Hence,
\(\frac{1}{3}\) ÷ 7 = \(\frac{1}{21}\)

Question 23.
Big Ideas Math Solutions Grade 5 Chapter 10 Divide Fractions 92
Answer: \(\frac{1}{6}\) ÷ 5 = \(\frac{1}{30}\)

Explanation:
The given numbers are: \(\frac{1}{6}\) and 5
We know that,
\(\frac{a}{b}\) ÷ a  = \(\frac{a}{b}\) × \(\frac{1}{a}\)
a= \(\frac{a}{1}\)
So,
\(\frac{1}{6}\) ÷ 5
= \(\frac{1}{6}\) × \(\frac{1}{5}\)
= \(\frac{ 1 × 1}{6 × 5}\)
= \(\frac{1}{30}\)
Hence,
\(\frac{1}{6}\) ÷ 5 = \(\frac{1}{30}\)

Question 24.
Big Ideas Math Solutions Grade 5 Chapter 10 Divide Fractions 93
Answer: \(\frac{1}{7}\) ÷ 3 = \(\frac{1}{21}\)

Explanation:
The given numbers are: \(\frac{1}{7}\) and 3
We know that,
\(\frac{a}{b}\) ÷ a  = \(\frac{a}{b}\) × \(\frac{1}{a}\)
a= \(\frac{a}{1}\)
So,
\(\frac{1}{7}\) ÷ 3
= \(\frac{1}{7}\) × \(\frac{1}{3}\)
= \(\frac{ 1 × 1}{7 × 3}\)
= \(\frac{1}{21}\)
Hence,
\(\frac{1}{7}\) ÷ 3 = \(\frac{1}{21}\)

Question 25.
Big Ideas Math Solutions Grade 5 Chapter 10 Divide Fractions 94

Answer: \(\frac{1}{8}\) ÷ 4 = \(\frac{1}{32}\)

Explanation:
The given numbers are: \(\frac{1}{8}\) and 4
We know that,
\(\frac{a}{b}\) ÷ a  = \(\frac{a}{b}\) × \(\frac{1}{a}\)
a= \(\frac{a}{1}\)
So,
\(\frac{1}{8}\) ÷ 4
= \(\frac{1}{8}\) × \(\frac{1}{4}\)
= \(\frac{ 1 × 1}{8 × 4}\)
= \(\frac{1}{32}\)
Hence,
\(\frac{1}{8}\) ÷ 4 = \(\frac{1}{32}\)

10.5 Problem Solving: Fraction Division

Question 26.
A mechanic buys 1 gallon of oil. She uses \(\frac{1}{6}\) of the oil, and then divides the rest into 4 smaller bottles. How much does she put into each bottle?
Big Ideas Math Solutions Grade 5 Chapter 10 Divide Fractions 95
Answer: The amount of oil she put into each bottle is: \(\frac{5}{24}\)

Explanation:
It is given that a mechanic buys 1 gallon of oil and she uses \(\frac{1}{6}\) of the oil
So,
the remaining part of the oil = 1 –  \(\frac{1}{6}\)
= \(\frac{5}{6}\)
It is also given that she divides the rest of the oil into 4 smaller bottles.
So,
The amount of oil in each bottle = \(\frac{The remaining part of the oil}{The number of bottles}\)
= \(\frac{5}{6}\) ÷ 4
= \(\frac{5}{6}\) × \(\frac{1}{4}\)
= \(\frac{5 × 1}{6 × 4}\)
= \(\frac{5}{24}\)
Hence, from the above,
We can conclude that the amount of oil in each bottle is: \(\frac{5}{24}\)

Conclusion:

Make use of the quick links and try to solve the problems in a simple manner. Redefine your true self with the BIM Answer Key for Grade 5 curated by subject experts. Test your knowledge by solving the questions which are given at the end of the chapter.

Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities

Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities

Clearing the math queries of algebra 1 concepts can be tricky for educators but we have shared an excellent guide for resolving all student’s issues while learning the concepts. The perfect study resource for students is Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities where it will help all of them score great marks in the exams. Using the best guide of Big ideas Math Book Answer Key Ch 2 Solving linear Functions, students can easily improve their math skills and problem-solving skills. Go through the below sections and find the BIM math book and observe Algebra 1 Ch 2  Solving Linear Inequalities thoroughly for good results in exams.

Big Ideas Math Book Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities

Guys who want to score the highest marks in the examinations should refer to this Topic-wise BIM Math Book Algebra 1 Ch 2 Answers from the below links. In Big Ideas Math Book Algebra 1 Solution Key Chapter 2 Solving Linear Inequalities, students can practice various questions covered in Exercises, Chapter tests, performance tasks, vocabulary, practice tests, cumulative assessments, and others.

Take a look at the comprehensive collection of Chapter 2 questions included in the Big Ideas Math Algebra 1 Solutions and become a master in maths. Also, you all can identify your weak areas by practicing the Big Ideas Math Answers Algebra 1 Ch 2. Download the common core curriculum standards provided Topic-wise BIM Algebra 1 Textbook Solutions Pdf through the below links and learn efficiently.

Solving Linear Inequalities Maintaining Mathematical Proficiency

Graph the number.

Question 1.
6
Answer:
The number line representing 6 is:

Question 2.
| 2 |
Answer:
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
The number line representing | 2 | is:

Question 3.
| -1 |
Answer:
We know that,
| -x | = x for x > 0
| -x | = -x for x < 0
So,
The number line representing | -1 | is:

Question 4.
2 + | -2 |
Answer:
The given absolute value expression is:
2 + | -2 |
We know that,
| -x | =x for x > 0
| -x | = -x for x < 0
So,
2 + | -2 | = 2 + 2                                                 2 + | – 2 | = 2 – 2
= 4                                                                       = 0
Hence,
The number line representing the values of 2 + | – 2 | is:

Question 5.
1 – | -4 |
Answer:
The given absolute value expression is:
1 – | – 4 |
We know that,
| -x | = x for x > 0
| – x | = -x for x < 0
So,
1 – | – 4 | = 1 – 4                                      1 – | – 4 | = 1 – ( -4 )
= -3                                                          = 5
Hence,
The number line representing the values of 1 – | – 4 | is:

Question 6.
-5 + | 3 |
Answer:
The given absolute value expression is:
-5 + | 3 |
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
-5 + | 3 | = -5 + 3                                 -5 + | 3 | = -5 -3
= -2                                                        = -8
Hence,
The number line representing the values of -5 + | 3 | is:

Complete the statement with <, >, or =.

Question 7.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 1
Answer:
We know that,
The order of the numbers in the ascending order is:
-9, -8, -7, -6, -5, -4 ,-3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
Hence, from the above,
We can conclude that
2 < 9

Question 8.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 2
Answer:
The order of the numbers in the ascending order is:
-9, -8, -7, -6, -5, -4 ,-3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
Hence, from the above,
We can conclude that
-6 < 5

Question 9.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 3
Answer:
The order of the numbers in the ascending order is:
-9, -8, -7, -6, -5, -4 ,-3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
We know that,
The greater number in the positive integers is the lesser number in the negative integers
Hence, from the above,
We can conclude that
-12 < -4

Question 10.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 4
Answer:
The order of the numbers in the ascending order is:
-9, -8, -7, -6, -5, -4 ,-3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
We know that,
The greater number in the positive integers is the lesser number in the negative integers
Hence, from the above,
We can conclude that
-7 > -13

Question 11.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 5
Answer:
We know that,
| -x | = x for x > 0
| -x | = -x for x < 0
The order of the numbers in the ascending order is:
-9, -8, -7, -6, -5, -4 ,-3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
We know that,
The greater number in the positive integers is the lesser number in the negative integers
Hence, from the above,
We can conclude that
8 = 8 ( or ) -8 = -8

Question 12.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 6
Answer:
We know that,
| -x | = x for x > 0
| -x | = -x for x < 0
The order of the numbers in the ascending order is:
-9, -8, -7, -6, -5, -4 ,-3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
We know that,
The greater number in the positive integers is the lesser number in the negative integers
Hence, from the above,
We can conclude that
-10 < 18    ( or ) -10 > -18

Question 13.
ABSTRACT REASONING
A number a is to the left of a number b on the number line. How do the numbers -a and -b compare?
Answer:
It is given that a number a is to the left of a number b on the number line.
So,
The representation of a and b on the number line is:

Hence, from the above number line
We can conclude that
-a > -b

Solving Linear Inequalities Mathematical Practices

Monitoring Progress

Use a graphing calculator to solve the inequality.

Question 1.
2x + 3 < x – 1
Answer:
The given inequality is:
2x + 3 < x – 1
We know that,
When we convert the mathematical symbols from LHS to RHS, the sign of that mathematical symbol change. i.e.,
+ will be converted into – and vice – versa
× will be converted into ÷ and vice-versa
So,
2x – x < -1 – 3
x < – 4
Hence,
The representation of the solved inequality in the number line is:

Question 2.
-x – 1 > -2x + 2
Answer:
The given inequality is:
-x – 1 > -2x + 2
We know that,
When we convert the mathematical symbols from LHS to RHS, the sign of that mathematical symbol change. i.e.,
+ will be converted into – and vice – versa
× will be converted into ÷ and vice-versa
So,
-x + 2x > 2 + 1
x > 3
Hence,
The representation of the solved inequality in the number line is:

Question 3.
\(\frac{1}{2}\)x + 1 > \(\frac{3}{2}\)x + 3
Answer:
The given inequality is:
\(\frac{1}{2}\)x + 1 > \(\frac{3}{2}\)x + 3
We know that,
When we convert the mathematical symbols from LHS to RHS, the sign of that mathematical symbol change. i.e.,
+ will be converted into – and vice – versa
× will be converted into ÷ and vice-versa
So,
\(\frac{1}{2}\)x – \(\frac{3}{2}\)x > 3 – 1
\(\frac{1 – 3}{2}\)x > 2
\(\frac{-2}{2}\)x > 2
\(\frac{-1}{1}\)x > 2
-x > 2
Multiply with –  both sides
So,
x > -2
Hence,
The representation of the solved inequality in the number line is:

Lesson 2.1 Writing and Graphing Inequalities

Essential Question

How can you use an inequality to describe a real-life statement?
Answer:
Inequality is just a relationship between two amounts, in which the amounts are not the same.
Example:
She completed her work before anyone else. i.e., her finish time was less than everyone else

EXPLORATION 1
Writing and Graphing Inequalities

Work with a partner. Write an inequality for each statement. Then sketch the graph of the numbers that make each inequality true.
a. Statement The temperature t in Sweden is at least -10°C.
Inequality Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 7
Graph Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 8
Answer:
The given graph is:
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 8
It s given that the temperature t in Sweden is at least -10°C
The meaning of the above statement is that the temperature of Sweden is -10°C or greater than -10°C
Hence,
The representation of the temperature of Sweden in the graph is:

b. Statement The elevation e of Alabama is at most 2407 feet.
Inequality Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 9
Graph Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 10
Answer:
The given graph is:
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 10
It is given that the elevation ‘ e ‘ of Alabama is at most 2,407 feet
From the given graph,
We can say that the gap between each elevation is 1,000 feet
So,
The elevation of Alabama will come between 2000 and 3000 in the number line
Hence,
The representation of the elevation of Alabama in the given graph is:

Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 11
Answer:
EXPLORATION 2
Writing Inequalities
Work with a partner. Write an inequality for each graph. Then, in other words, describe all the values of x that make each inequality true.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 12
Answer:
The given number lines are:
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 12
a)
From the given number line,
The value of x marked from 1 to 4
Hence,
The inequality representing the given number line is:
x ≥ 1
b)
From the given number line,
The value of x marked after 1 till the end of the number line
Hence,
The inequality representing the given number line is:
x > 1
c)
From the given number line,
The value of x marked from 1 till the end of the left side of the number line.
Hence,
The inequality representing the given number line is:
x ≤ 1
d)
From the given number line,
The value of x marked before 1 till the end of the left side of the number line
Hence,
The inequality representing the given number lie is:
x < 1

Communicate Your Answer

Question 3.
How can you use an inequality to describe a real-life statement?
Answer:
Inequality is just a relationship between two amounts, in which the amounts are not the same.
Example:
A ball in the net is worth two in the bush. i.e., the value of a single ball in the net is greater than the value of a single ball in the bush.

Question 4.
Write a real-life statement that involves each inequality.
a. x < 3.5
b. x ≤ 6
c. x > -2
d. x ≥ 10
Answer:
a)
The given inequality is:
x < 3.5
The real-life situation that involves the given inequality is:
The number of birds that are less than the height of 3.5 inches
b)
The given inequality is:
x ≤ 6
The real- life situation that involves the given inequality is:
The number of people that are buying the mobiles from a mobile store
c)
The given inequality is:
x > -2
The real-life situation that involves the given inequality is:
The number of integers that are greater than -2
d)
The given inequality is:
x ≥ 10
The real-life situation that involves the given inequality is:
The number of chocolates that are distributed greater than or equal to 10 children

2.1 Lesson

Monitoring Progress

Write the sentence as an inequality.

Question 1.
A number b is fewer than 30.4.
Answer:
The given sentence is:
A number b is fewer than 30.4
Hence,
The representation of the given sentence in the form of inequality is:
b < 30.4

Question 2.
–\(\frac{7}{10}\) is at least twice a number k minus 4.
Answer:
The given sentence is:
–\(\frac{7}{10}\) is at least twice a number k minus 4.
Hence,
The representation of the given sentence in the form of inequality is:
–\(\frac{7}{10}\) = 2k – 4

Tell whether −6 is a solution to the inequality.

Question 3.
c + 4 < -1
Answer:
-6 is not a solution to the given inequality

Explanation:
The given equation is:
c + 4 = -1
So,
c = -4 – 1
c = -5
Hence, from the above,
We can conclude that -6 is not a solution to the given inequality

Question 4.
10 ≤ 3 – m
Answer:
-6 is not a solution to the given inequality

Explanation:
The given inequality is:
10 ≤ 3 – m
So,
-m ≤ 10 – 3
-m ≤ 7
Multiply by – on both sides
m ≤ -7
Hence, from the above,
We can conclude that -6 is not a solution to the given inequality

Question 5.
21 ÷ x ≥ -3.5
Answer:
-6 is a solution to the given inequality

Explanation:
The given inequality is:
21 ÷ x ≥ -3.5
So,
-21 ÷ 3.5 ≥ x
-210 ÷ 35 ≥ x
-6 ≥ x
x ≤ -6
Hence, from the above,
We can conclude that -6 is a solution to the given inequality

Question 6.
4x – 25 > -2
Answer:
-6 is not a solution to the given inequality

Explanation:
The given inequality is:
4x – 25 > -2
So,
4x > -2 + 25
4x > 23
x > 23 ÷ 4
x > 5.75
Hence, from the above,
We can conclude that -6 is not a solution to the given inequality

Graph the inequality.

Question 7.
b > -8
Answer:
The given inequality is:
b > -8
Hence,
The representation of the given inequality in the number line is:

Question 8.
1.4 ≥ g
Answer:
The given inequality is:
1.4 ≥ g
So,
g ≤ 1.4
Hence,
The representation of the given inequality in the number line is:

Question 9.
r < \(\frac{1}{2}\)
Answer:
The given inequality is:
r < \(\frac{1}{2}\)
So,
r < 0.5
Hence,
The representation of the given inequality in the number line is:

Question 10.
v ≥ \(\sqrt{36}\)
Answer:
The given inequality is:
v > \(\sqrt{36}\)
So,
v > 6
Hence,
The representation of the given inequality in the number line is:

Question 11.
Write an inequality that represents the graph.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 13
Answer:
The given graph is:
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 13
From the given graph,
We can say that the marked line is from -6 and continued after -6 till the last number on the right side of the number line.
Hence,
The representation of the given number line in the form of inequality is:
x ≥ -6

Writing and Graphing Inequalities 2.1 Exercises

Vocabulary and Core Concept Check

Question 1.
COMPLETE THE SENTENCE
A mathematical sentence using the symbols <, >, ≤, or ≥ is called a(n)_______.
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.1 Question 1

Question 2.
VOCABULARY
Is 5 in the solution set of x + 3 > 8? Explain.
Answer:
5 is not in the solution of the given inequality
x + 3 > 8

Explanation:
The given inequality is:
x + 3 > 8
So,
x > 8 – 3
x > 5
Hence, from the above,
We can conclude that 5 is not in the solution of the given inequality

Question 3.
ATTENDING TO PRECISION
Describe how to graph an inequality.
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.1 Question 3

Question 4.
DIFFERENT WORDS, SAME QUESTION
Which is different? Write “both” inequalities.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 14
Answer:
The given inequalities in the worded form are:
a) w is greater than or equal to -7
b) w is not less than -7
c) w is no more than -7
d) w is at least -7
Now,
a)
The given worded form in the form of inequality is:
w ≥ -7
b)
The given worded form in the form of inequality is:
w > -7
c)
The given worded form in the form of inequality is:
w ≥ -7
d)
The given worded form in the form of inequality is:
w ≥ -7

Monitoring Progress and Modeling with Mathematics

In Exercises 5–12, write the sentence as an inequality.

Question 5.
A number x is greater than 3.
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.1 Question 5

Question 6.
A number n plus 7 is less than or equal to 9.
Answer:
The given worded form is:
A number n plus 7 is less than or equal to 9
Hence,
The representation of the given worded form in the form of inequality is:
n + 7 ≤ 9

Question 7.
Fifteen is no more than a number t divided by 5.
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.1 Question 7

Question 8.
Three times a number w is less than 18.
Answer:
The given worded form is:
Three times a number w is less than 18
Hence,
The representation of the given worded form in the form of inequality is:
3w < 18

Question 9.
One-half of a number y is more than 22.
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.1 Question 9

Question 10.
Three is less than the sum of a number s and 4.
Answer:
The given worded form is:
Three is less than the sum of a number s and 4
Hence,
The representation of the given worded form in the form of inequality is:
3 < s + 4

Question 11.
Thirteen is at least the difference between a number v and 1.
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.1 Question 11

Question 12.
Four is no less than the quotient of a number x and 2.
Answer:
The given worded form is:
Four is no less than the quotient of a number x and 2
Hence,
The representation of the given worded form in the form of inequality is:
4 > x ÷ 2

Question 13.
MODELING WITH MATHEMATICS
On a fishing trip, you catch two fish. The weight of the first fish is shown. The second fish weighs at least 0.5 pound more than the first fish. Write an inequality that represents the possible weights of the second fish.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 15
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.1 Question 13

Question 14.
MODELING WITH MATHEMATICS
There are 430 people in a wave pool. Write an inequality that represents how many more people can enter the pool.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 16
Answer:
The inequality that represents the number of more people that can enter the pool is:
430 + x = 600
Where,
x is the number of more people that can enter the pool

Explanation:
It is given that there are 430 people in a wave pool
It is also given that the maximum capacity in a pool is: 600
Let,
x be the number of more people that can enter the pool
Hence,
The inequality that represents the number of more people that can enter the pool is:
430 + x = 600

In Exercises 15–24, tell whether the value is a solution to the inequality.

Question 15.
r + 4 > 8; r = 2
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.1 Question 15

Question 16.
5 – x < 8; x = -3
Answer:
-3 is not a solution of the given inequality

Explanation:
The given inequality is:
5 – x < 8
It is given that x = -3
So,
5 – ( -3 ) < 8
5 + 3 < 8
8 < 8
Hence, from the above,
We can conclude that x = -3 is not a solution to the given inequality

Question 17.
3s ≤ 19; s = -6
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.1 Question 17

Question 18.
17 ≥ 2y ; y = 7
Answer:
y = 7 is a solution of the given inequality

Explanation:
The given inequality is:
17 ≥ 2y
It is given that y = 7
So,
17 ≥ 2 ( 7 )
17 ≥ 14
Hence, from the above,
We can conclude that y = 7 is a solution of the given inequality

Question 19.
-1 > –\(\frac{x}{2}\); x = 3
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.1 Question 19

Question 20.
–\(\frac{4}{z}\) ≥ 3; z = 2
Answer:
z = 2 is not a solution of the given inequality

Explanation;
The given inequality is:
–\(\frac{4}{z}\) ≥ 3
It is given that z = 2
So,
–\(\frac{4}{2}\) ≥ 3
-2 ≥ 3
Hence, from the above,
We can conclude that z = 2 is not a solution of the given inequality

Question 21.
14 ≥ -2n + 4; n = -5
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.1 Question 21

Question 22.
-5 ÷ (2s) < -1; s = 10
Answer:
s = 10 is a solution of the given inequality

Explaantion:
The given inequality is:
-5 ÷ ( 2s ) < -1
It is given that s = 10
So,
-5 ÷ 2 ( 10 ) < -1
-5 ÷ 20 < -1
–\(\frac{1}{4}\) < -1
\(\frac{1}{4}\) < 1
1 < 4
Hence, from the above,
We can conclude that s = 10 is a solution of the given inequality

Question 23.
20 ≤ \(\frac{10}{2z}\) + 20; z = 5
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.1 Question 23

Question 24.
\(\frac{3m}{6}\) – 2 > 3; m = 8
Answer:
m = 8 is not a solution of the given inequality

Explanation:
The given inequality is:
\(\frac{3m}{6}\) – 2 > 3
It is given that m = 8
So,
\(\frac{3 × 8}{6}\) > 3 + 2
4  3 + 2
4 > 5
Hence, from the above,
We can conclude that m = 8 is not a solution of the given inequality

Question 25.
MODELING WITH MATHEMATICS
The tallest person who ever lived was approximately 8 feet 11 inches tall.
a. Write an inequality that represents the heights of every other person who has ever lived.
b. Is 9 feet a solution of the inequality? Explain.
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.1 Question 25

Question 26.
DRAWING CONCLUSIONS
The winner of a weight-lifting competition bench-pressed 400 pounds. The other competitors all bench-pressed at least 23 pounds less.
a. Write an inequality that represents the weights that the other competitors bench-pressed.
Answer:
It is given that the winner of a weight-lifting competition bench pressed 400 pounds whereas the other competitors all bench-pressed at least 23 pounds less.
Let,
x be the number of pounds that all the other competitors’ bench-pressed
So,
The inequality that represents the weights that the other competitors bench-pressed is:
x  + 23 ≤ 400

b. Was one of the other competitors able to bench-press 379 pounds? Explain.
Answer:
No, one of the other competitors won’t be able to bench-press 379 pounds

Explanation:
From part (a),
The inequality that represents the weights that the other competitors bench-pressed is:
x + 23 ≤ 400
x ≤ 400 – 23
x ≤ 377 pounds
Hence, from the above,
We can conclude that one of the other competitors won’t be able to bench-press 379 pounds

ERROR ANALYSIS
In Exercises 27 and 28, describe and correct the error in determining whether 8 is in the solution set of the inequality.

Question 27.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 17
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.1 Question 27

Question 28.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 18
Answer:
8 is a solution set of the given inequality

Explanation:
The given inequality is:
\(\frac{1}{2}\)x + 2 ≤ 6
It is given that x = 8
So,
\(\frac{1}{2}\) × 8 ≤ 6 – 2
\(\frac{8}{2}\) ≤ 4
4 ≤ 4
Hence, from the above,
We can conclude that 8 is a solution set of the given inequality

In Exercises 29–36, graph the inequality.

Question 29.
x ≥ 2
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.1 Question 29

Question 30.
z ≤ 5
Answer:
The given inequality is:
z ≤ 5
Hence,
The representation of the given inequality in the number line is:

Question 31.
-1 > t
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.1 Question 31

Question 32.
-2 < w
Answer:
The given inequality is:
-2 < w
So,
w > -2
Hence,
The representation of the given inequality in the number line is:

Question 33.
v ≤ -4
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.1 Question 33

Question 34.
s < 1
Answer:
The given inequality is:
s < 1
Hence,
The representation of the given inequality in the number line is:

Question 35.
\(\frac{1}{4}\) < p
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.1 Question 35

Question 36.
r ≥ -| 5 |
Answer:
The given inequality is:
r ≥ – | 5 |
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
r ≥ -5   ( or ) r ≥ 5
Hence,
The representation of the given inequalities in the number line is:

In Exercises 37–40, write and graph an inequality for the given solution set.

Question 37.
{x | x < 7}
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.1 Question 37

Question 38.
{n | n ≥ -2}
Answer:
The given inequality is:
{n | n ≥ -2}
The given inequality can be rewritten as:
n ≥ -2
Hence,
The representation of the given inequality in the number line is:

Question 39.
{z | 1.3 ≤ z}
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.1 Question 39

Question 40.
{w | 5.2 > w}
Answer:
The given inequality is:
{w | 5.2 > w}
The given inequality can be rewritten as:
5.2 > w
So,
w < 5.2
From the above value,
We can say that 5.2 lies between 5 and 6
Hence,
The representation of the given inequality in the number line is:

In Exercises 41–44, write an inequality that represents the graph.

Question 41.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 19
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.1 Question 41

Question 42.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 21
Answer:
The given number line is:
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 21
From the above number line,
We can say that the marked line started from -2 and ended at the last value on the right side of the number line.
Hence,
The representation of the inequality for the given number line is:
x ≥ -2

Question 43.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 22
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.1 Question 43

Question 44.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 23
Answer:
The given number line is:
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 23
From the above number line,
We can say that the marked line started from -1 excluding – and continued till the end of the number line in the left side of the number line
Hence,
The representation of the inequality for the given number line is:
x < -1

Question 45.
ANALYZING RELATIONSHIPS
The water temperature of a swimming pool must be no less than 76°F. The temperature is currently 74°F. Which graph correctly shows how much the temperature needs to increase? Explain your reasoning.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 24
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.1 Question 45

Question 46.
MODELING WITH MATHEMATICS
According to state law for vehicles traveling on state roads, the maximum total weight of a vehicle and its contents depends on the number of axles on the vehicle. For each type of vehicle, write and graph an inequality that represents the possible total weights w (in pounds) of the vehicle and its contents.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 25
Answer:
It is given that the maximum total weight of a vehicle and its contents depends on the number of axles on the vehicle.
Now,
Let,
w be the total weight of the vehicle and its contents
For the 2 axles vehicle, if the maximum weight is 40,000 lb and w is the possible total weight of the vehicle and its contents, then
w < 40,000
The representation of the given inequality in the number line is:

For the 3 axles vehicle, if the maximum weight is 60,000 lb and w is the possible total weight of the vehicle and its contents, then
w < 60,000
The representation of the given inequality in the number line is:

For the 4 axles vehicle, if the maximum weight is 80,000 lb and w is the possible total weight of the vehicle and its contents, then
w < 80,000
The representation of the given inequality in the number line is:

Question 47.
PROBLEM-SOLVING
The Xianren Bridge is located in Guangxi Province, China. This arch is the world’s longest natural arch, with a length of 400 feet. Write and graph an inequality that represents the lengths ℓ (in inches) of all other natural arches.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 26
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.1 Question 47

Question 48.
THOUGHT-PROVOKING
A student works no more than 25 hours each week at a part-time job. Write an inequality that represents how many hours the student can work each day.
Answer:
It is given that a student works no more than 25 hours each week at a part-time job
Let the number of days that students work each week be x
So,
The inequality that represents the number of hours the students work each week is:
x < 25
We know that,
1 week = 7 days
So,
7x < 25
Hence,
The inequality that represents the number of hours the students work each day is:
x < \(\frac{25}{7}\)
x < 3.5
Hence,
The representation of the inequality in the number line is:

Question 49.
WRITING
Describe a real-life situation modeled by the inequality 23 + x ≤ 31.
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.1 Question 49

Question 50.
HOW DO YOU SEE IT?
The graph represents the known melting points of all metallic elements (in degrees Celsius). Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 27
a. Write an inequality represented by the graph.
Answer:
The given graph is:
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 27
From the above graph,
We can say that the marked line started from -38.87 and continued till the end of the right line of the number line
Hence,
The inequality that represents the given number line is:
x ≥ -38.87

b. Is it possible for a metallic element to have a melting point of -38.87°C? Explain.
Answer:
Yes, it is possible for a metallic element to have a melting point of -38.87°C.

Explanation:
The given graph is:
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 27
From the above graph,
The starting point is -38.87
Hence, from the above,
We can conclude that it is possible for a metallic element to have a melting point of -38.87°C

Question 51.
DRAWING CONCLUSIONS
A one-way ride on a subway costs $0.90. A monthly pass costs $24. Write an inequality that represents how many one-way rides you can buy before it is cheaper to buy the monthly pass. Is it cheaper to pay the one-way fare for 25 rides? Explain.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 27.1
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.1 Question 51

Question 52.
MAKING AN ARGUMENT
The inequality x ≤ 1324 represents the weights (in pounds) of all mako sharks ever caught using a rod and reel. Your friend says this means no one using a rod and reel has ever caught a mako shark that weighs 1324 pounds. Your cousin says this means someone using a rod and reel has caught a mako shark that weighs 1324 pounds. Who is correct? Explain your reasoning.
Answer:
Your cousin is correct

Explanation:
It is given that the inequality x ≤ 1324 represents the weights (in pounds) of all mako sharks ever caught using a rod and reel. Your friend says this means no one using a rod and reel has ever caught a mako shark that weighs 1324 pounds. Your cousin says this means someone using a rod and reel has caught a mako shark that weighs 1324 pounds.
Now,
The given inequality is:
x ≤ 1324
The meaning of inequality is the value of x is less than or equal to 1324 pounds
Now,
Your friend says this means no one using a rod and reel has ever caught a mako shark that weighs 1324 pounds
But, from the given inequality,
The statement of your friend is wrong
Your cousin says this means someone using a rod and reel has caught a mako shark that weighs 1324 pounds
But from the given inequality,
The statement of your cousin is correct.
Hence, from the above,
We can conclude that your cousin is correct

Question 53.
CRITICAL THINKING
Describe a real-life situation that can be modeled by more than one inequality.
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.1 Question 53

Question 54.
MODELING WITH MATHEMATICS
In 1997, Superman’s cape from the 1978 movie Superman was sold at an auction. The winning bid was $17,000. Write and graph an inequality that represents the amounts all the losing bids.
Answer:
It is given that in 1997, Superman’s cape from the 1978 movie Superman was sold at an auction. The winning bid was $17,000.
So,
The bid must not be less than $17,000, otherwise, the bid will lose
Hence,
The inequality that represents the amounts of all losing bids is:
x < 17,000
The representation of the inequality in the number line is:

MATHEMATICAL CONNECTIONS
In Exercises 55–58, write an inequality that represents the missing dimension x.

Question 55.
The area is less than 42 square meters.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 28
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.1 Question 55

Question 56.
The area is greater than or equal to 8 square feet.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 29
Answer:
The inequality that represents the value of x is:
x ≥ \(\frac{8}{5}\)

Explanation:
The given figure is:
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 29
From the above figure,
We can say that the given figure is a Right-angled triangle
We know that,
The area of the triangle = \(\frac{1}{2}\) × Base × Height
= \(\frac{1}{2}\) × x × 10
= 5x
It is given that the area of the triangle is greater than or equal to 8 square feet
So,
5x ≥ 8 square feet
x ≥ \(\frac{8}{5}\)
Hence, from the above,
We can conclude that the inequality that represents x is:
x ≥ \(\frac{8}{5}\)

Question 57.
The area is less than 18 square centimeters.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 30
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.1 Question 57

Question 58.
The area is greater than 12 square inches.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 31
Answer:
The inequality that represents the value of x is:
x > 6

Explanation:
The given figure is:
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 31
From the above figure,
We can say that the given figure is the rectangle.
We know that,
The area of the rectangle = Length × Width
= 2 × x
It is given that the area of the rectangle is greater than 12 square inches
So,
2x > 12
x > 12 / 2
x > 6
Hence, from the above,
We can conclude that the inequality that represents the value of x is:
x > 6

Question 59.
WRITING
A runner finishes a 200-meter dash in 35 seconds. Let r represent any speed (in meters per second) faster than the runner’s speed.
a. Write an inequality that represents r. Then graph the inequality.
b. Every point on the graph represents a speed faster than the runner’s speed. Do you think every point could represent the speed of a runner? Explain.
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.1 Question 59

Maintaining Mathematical Proficiency

Solve the equation. Check your solution.(Section 1.1)

Question 60.
x + 2 = 3
Answer:
The value of x is: 1

Explanation:
The given equation is:
x + 2 = 3
So,
x = 3 – 2
x = 1
Hence, from the above
We can conclude that the value of x is: 1

Question 61.
y – 9 = 5
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.1 Question 61

Question 62.
6 = 4 + y
Answer:
The value of y is: 2

Explanation:
The given equation is:
6 = 4 + y
So,
6 – 4 = y
2 = y
y = 2
Hence, from the above,
We can conclude that the value of y is: 2

Question 63.
-12 = y – 11
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.1 Question 63

Solve the literal equation for x.(Section 1.5)

Question 64.
v = x • y • z
Answer:
The value of x is: \(\frac{v}{yz}\)

Explanation:
The given equation is:
v = x ⋅ y ⋅ z
x = \(\frac{v}{yz}\)
Hence, from the above,
We can conclude that the value of x is: \(\frac{v}{yz}\)

Question 65.
s = 2r + 3x
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.1 Question 65

Question 66.
w = 5 + 3(x – 1)
Answer:
The value of x is: \(\frac{w – 2}{3}\)

Explanation:
The given equation is:
w = 5 + 3 ( x – 1 )
So,
w = 5 + 3 ( x ) – 3 ( 1 )
w = 5 + 3x – 3
w = 3x + 2
3x = w – 2
x = \(\frac{w – 2}{3}\)
Hence, from the above,
We can conclude that the value of x is: \(\frac{w – 2}{3}\)

Question 67.
n = \(\frac{2x + 1}{2}\)
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.1 Question 67

Lesson 2.2 Solving Inequalities Using Addition or Subtraction

Essential Question
How can you use addition or subtraction to solve an inequality?
EXPLORATION 1
Quarterback Passing Efficiency
Work with a partner.
The National Collegiate Athletic Association (NCAA) uses the following formula to rank the passing efficiencies P of quarterbacks.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 32
Answer:
The formula used to rank the passing efficiencies P of the quarterbacks is:

By comparing the coefficients, we get
Y = 8.4
C = 100
T = 330
N = 200
A = 1
Now,
a) T < C
= 330 < 100
Hence, from the above,
We can conclude that the given inequality is not true
b) C + N ≤ A
= ( 100 + 200 ) ≤ 1
= 300 ≤ 1
Hence, from the above,
We can conclude that the given inequality is not true
c) N < A
= 200 < 1
Hence, from the above,
We can conclude that the given inequality is not true
d) A – C ≥ M
We know that,
M = C – N
So,
A-C ≥ C – N
( 1 – 100 ) ≥ ( 100 – 200 )
= -99 ≥ -100
= 99 ≥ 100
Hence, from the above,
We can conclude that the given inequality is not true

EXPLORATION 2
Finding Solutions of Inequalities
Work with a partner.
Use the passing efficiency formula to create a passing record that makes each inequality true. Record your results in the table. Then describe the values of P that make each inequality true.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 33
a. P < 0
b. P + 100 ≥ 250
c. P – 250 > -80
Answer:
Using the data from Exploration 1,
The completed table is:

Now,
The formula for passing efficiency is:

a) P < 0
To make P < 0,
The value of A or any of the values of the numerator must be less than 0
b) P + 100 ≥ 250
P ≥ 250 – 100
P ≥ 150
Hence, from the above,
We can conclude that the value of P must be greater than or equal to 150
c ) P – 250 > -80
P > -80 + 250
P > 170
Hence, from the above,
We can conclude that the value of P must be greater than 170

Communicate Your Answer

Question 3.
How can you use addition or subtraction to solve an inequality?
Answer:
If you want to add or subtract from one side of the equation, you must perform the same operation to the other side of the equation. When solving inequalities by adding or subtracting, our goal is to have the variable on its own

Question 4.
Solve each inequality.
a. x + 3 < 4 b. x – 3 ≥ 5 c. 4 > x – 2
d. -2 ≤ x + 1
Answer:
The given inequalities are:
a) x + 3 < 4
b) x – 3 ≥ 5
c) 4 > x – 2
d) -2 ≤ x + 1
Now,
a)
The given inequality is:
x + 3 < 4
So,
x < 4 – 3
x < 1
Hence, from the above,
We can conclude that x < 1
b)
The given inequality is:
x – 3 ≥ 5
x ≥ 5 + 3
x ≥ 8
Hence, from the above,
We can conclude that x ≥ 8
c)
The given inequality is:
4 > x – 2
So,
4 + 2 > x
6 > x
x < 6
Hence, from the above,
We can conclude that x < 6
d)
The given inequality is:
-2 ≤ x + 1
So,
-2 – 1 ≤ x
-3 ≤ x
x ≥ -3
Hence, from the above,
We can conclude that x ≥ -3

2.2 Lesson

Monitoring Progress

Solve the inequality. Graph the solution.

Question 1.
b – 2 > -9
Answer:
The given inequality is:
b – 2 > -9
So,
b > -9 + 2
b > -7
Hence, from the above,
We can conclude that the value of b is greater than -7
The representation of the inequality in the number line is:

Question 2.
m – 3 ≤ 5
Answer:
The given inequality is:
m – 3 ≤ 5
So,
m ≤ 5 + 3
m ≤ 8
Hence, from the above,
We can conclude that the value of m is less than or equal to 8
The representation of the inequality in the number line is:

Question 3.
\(\frac{1}{4}\) > y – \(\frac{1}{4}\)
Answer:
The given inequality is:
\(\frac{1}{4}\) > y – \(\frac{1}{4}\)
So,
\(\frac{1}{4}\) + \(\frac{1}{4}\) > y
\(\frac{1 + 1}{4}\) > y
\(\frac{2}{4}\) > y
\(\frac{1}{2}\) > y
y < \(\frac{1}{2}\)
Hence, from the above,
We can conclude that the value of y is less than \(\frac{1}{2}\)
The representation of the inequality in the number line is:

Solve the inequality. Graph the solution.

Question 4.
k + 5 ≤ -3
Answer:
The given inequality is:
k + 5 ≤ -3
So,
k ≤ -3 – 5
k ≤ -8
Hence, from the above,
We can conclude that the value of k is less than or equal to -8
The representation of the inequality in the number line is:

Question 5.
\(\frac{5}{6}\) ≤ z + \(\frac{1}{6}\)
Answer:
The given inequality is:
\(\frac{5}{6}\) ≤ z + \(\frac{1}{6}\)
So,
\(\frac{5}{6}\) – \(\frac{1}{6}\) ≤ z
\(\frac{5 – 1}{6}\) ≤ z
\(\frac{4}{6}\) ≤ z
\(\frac{2}{3}\) ≤ z
z ≥ \(\frac{2}{3}\)
z ≥ 0.6
z ≥ 1 ( Approx. )
Hence, from the above,
We can conclude that the value of z is approximately greater than or equal to 1
The representation of the inequality in the number line is:

Question 6.
p + 0.7 > -2.3
Answer:
The given inequality is:
p + 0.7 > -2.3
So,
p > -2.3 – 0.7
p > -3
Hence, from the above,
We can conclude that the value of p is greater than -3
The representation of the inequality in the number line is:

Monitoring Progress

Question 7.
The microwave oven uses only 1000 watts of electricity. Does this allow you to have both the microwave oven and the toaster plugged into the circuit at the same time? Explain your reasoning.
Answer:
Yes, this allows you to have both the microwave oven and the toaster plugged into the circuit at the same time

Explanation:
It is given that the microwave oven uses only 1000 watts of electricity.
We know that,
The toaster consumes less electricity than the microwave oven
Hence, from the above,
We can conclude that 1000 watts of electricity allow you to have both the microwave oven and the toaster plugged into the circuit at the same time

Solving Inequalities Using Addition or Subtraction 2.2 Exercises

In Exercises 3−6, tell which number you would add to or subtract from each side of the inequality to solve it.

Question 1.
VOCABULARY
why is the inequality x ≤ 6 equivalent to the inequality x – 5 ≤ 6 – 5 ?
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.2 Question 1

Question 2.
WRITING
Compare solving equations using addition with solving inequalities using addition.
Answer:
Solving equations and inequalities using addition is very similar. You have to add the same quantity on every side and the sign between the two sides does not change.
It remains the equal sign  “=”  in the case of equations and the inequality sign in the case of inequalities

Question 3.
k + 11 < -3
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.2 Question 3

Question 4.
v – 2 > 14
Answer:
The value of v is: 16

Explanation:
The given equation is:
v – 2 > 14
So,
v  – 2 + 2 > 14 + 2
v > 16
Hence, from the above,
We can conclude that we have to add 2 on both sides

Question 5.
-1 ≥ b – 9
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.2 Question 5

Question 6.
-6 ≤ 17 + p
Answer:
The value of p is: p ≥ -23

Explanation:
The given inequality is:
-6 ≤ 17 + p
So,
-6 + 6 ≤ 17 + 6 + p
0 ≤ 23 + p
-23 ≤ p
p ≥ -23
Hence, from the above,
We can conclude that we have to add 6 on both sides

In Exercises 7−20, solve the inequality. Graph the solution.

Question 7.
x – 4 < -5
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.2 Question 7

Question 8.
1 ≤ s – 8
Answer:
The given inequality is:
1 ≤ s – 8
1 + 8 ≤ s – 8 + 8
9 ≤ s
s ≥ 9
Hence,
The solution to the given inequality is:
s ≥ 9
The representation of the inequality in the number line is:

Question 9.
6 ≥ m – 1
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.2 Question 9

Question 10.
c – 12 > -4
Answer:
The given inequality is:
c – 12 > -4
So,
c – 12 + 4 > -4 + 4
c – 8 > 0
c > 0 + 8
c > 8
Hence, from the above,
We can conclude that the solution to the given inequality is:
c > 8
The representation of the inequality in the number line is:

Question 11.
r + 4 < 5
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.2 Question 11

Question 12.
-8 ≤ 8 + y
Answer:
The given inequality is:
-8 ≤ 8 + y
So,
-8 + 8 ≤ 8 + 8 + y
0 ≤ 16 + y
-16 ≤ y
y ≥ -16
Hence, from the above,
We can conclude that the solution to the given inequality is:
y ≥ -16
The representation of the inequality in the number line is:

Question 13.
9 + w > 7
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.2 Question 13

Question 14.
15 ≥ q + 3
Answer:
The given inequality is:
15 ≥ q + 3
So,
15 – 15 ≥ q + 3 – 15
0 ≥ q – 12
12 ≥ q
q ≤ 12
Hence, from the above,
We can conclude that the solution to the given inequality is:
q ≤ 12
The representation of the inequality in the number line is:

Question 15.
h – (-2) ≥ 10
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.2 Question 15

Question 16.
-6 > t – (-13)
Answer:
The given inequality is:
-6 > t – ( -13 )
So,
-6 + 6 > t + 13 + 6
0 > t + 19
-19 > t
t < -19
Hence, from the above,
We can conclude that the solution to the given inequality is:
t < -19
The representation of the inequality in the number line is:

Question 17.
j + 9 – 3 < 8
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.2 Question 17

Question 18.
1 – 12 + y ≥ -5
Answer:
The given inequality is:
1 – 12 + y ≥ -5
So,
-11 + y ≥ -5
-11 + 5 + y ≥ -5 + 5
6 + y ≥ 0
y ≥ -6
Hence, from the above,
We can conclude that the solution to the given inequality is:
y ≥ -6
The representation of the inequality in the number line is:

Question 19.
10 ≥ 3p – 2p – 7
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.2 Question 19

Question 20.
18 – 5z + 6z > 3 + 6
Answer:
The given inequality is:
18 – 5z + 6z > 3 + 6
So,
18 + z > 9
18 – 9 + z > 9 – 9
z + 9 > 0
z > -9
Hence, from the above,
We can conclude that the solution to the given inequality is:
z > -9
The representation of the inequality in the number line is:

In Exercises 21−24, write the sentence as an inequality. Then solve the inequality.

Question 21.
A number plus 8 is greater than 11.
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.2 Question 21

Question 22.
A number minus 3 is at least -5.
Answer:
The given worded form is:
A number minus 3 is at least -5
Let the number be x
So,
The representation of the given worded form in the form of inequality is:
x – 3 ≥ -5
So,
x – 3 + 5 ≥ -5 + 5
x + 2 ≥ 0
x ≥ -2
Hence, the solution to the given worded form inequality is:
x ≥ -2

Question 23.
The difference of a number and 9 is fewer than 4.
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.2 Question 23

Question 24.
Six is less than or equal to the sum of a number and 15.
Answer:
The given worded form is:
Six is less than or equal to the sum of a number and 15
Let the number be x
So,
The representation of the given worded form in the form of inequality is:
6 ≤ x + 15
6 – 6 ≤ x + 15 – 6
0 ≤ x + 9
-9 ≤ x
x ≥ -9
Hence, from the above,
We can conclude that the solution to the given worded form of the inequality is:
x ≥ -9

Question 25.
MODELING WITH MATHEMATICS
You are riding a train. Your carry-on bag can weigh no more than 50 pounds. Your bag weighs 38 pounds.
a. Write and solve an inequality that represents how much weight you can add to your bag.
b. Can you add both a 9-pound laptop and a 5-pound pair of boots to your bag without going over the weight limit? Explain.
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.2 Question 25

Question 26.
MODELING WITH MATHEMATICS
You order the hardcover book shown from a website that offers free shipping on orders of $25 or more. Write and solve an inequality that represents how much more you must spend to get free shipping.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 34
Answer:
The inequality that represents the more money you must spend to get free shipping is:

Explanation:
It is given that you order the hardcover book shown from a website that offers free shipping on orders of $25 or more.
From the figure,
The cost of a hardcover book is: $19.76
Let
The amount more money you must spend to get free shipping to be x
So,
x + 19.76 ≥ 25
x + 19.76 – 25 ≥ 25 – 25
x – 5.24 ≥ 0
x ≥ $5.24
Hence, from the above,
We can conclude that the amount more money you must spend to get free shipping is: $5.24
ERROR ANALYSIS
In Exercises 27 and 28, describe and correct the error in solving the inequality or graphing the solution.

Question 27.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 35
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.2 Question 27

Question 28.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 36
Answer:
The given inequality is:
-10 + x ≥ -9
-10 + 9 + x ≥ -9 + 9
-1 + x ≥ 0
x ≥ 1
Hence, from the above,
We can conclude that the solution to the given inequality is:
x ≥ 1
The representation of the inequality in the number line is:

Question 29.
PROBLEM-SOLVING
An NHL hockey player has 59 goals so far in a season. What are the possible numbers of additional goals the player can score to match or break the NHL record of 92 goals in a season?
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.2 Question 29

Question 30.
MAKING AN ARGUMENT
In an aerial ski competition, you perform two acrobatic ski jumps. The scores on the two jumps are then added together.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 37

a. Describe the score that you must earn on your second jump to beat your competitor.
Answer:
The given table is:
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 37
Let the score of your second acrobatic ski jump be x
Now,
From the above table,
We can say that your score must be greater than your competitor’s score to beat your competitor
Hence,
The inequality that represents the score you must earn on your second jump to beat your competitor is:
x > 119.8

b. Your coach says that you will beat your competitor if you score 118.4 points. A teammate says that you only need 117.5 points. Who is correct? Explain.
Answer:
Your coach and your teammate both are correct

Explanation:
The given table is:
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 37
From the above table,
In the 1st acrobatic ski jump,
Your competitor’s score is: 117.1
Your score is: 119.5
If you want to beat your competitor, then you have to score more than 117.1
So,
x > 117.1
It is given that your coach says that you will beat your competitor if you score 118.4 points. A teammate says that you only need 117.5 points
So,
According to your coach,
You have to score 118.4 points to beat your competitor
So,
118.4 > 117.1
According to a teammate,
You have to score 117.5 points to beat your competitor
So,
117.5 > 117.1
Hence, from the above,
We can conclude that your coach and your teammate both are correct

Question 31.
REASONING
Which of the following inequalities are equivalent to the inequality x – b < 3, where b is a constant? Justify your answer.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 38
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.2 Question 31

MATHEMATICAL CONNECTIONS
In Exercises 32 and 33, write and solve an inequality to find the possible values of x.

Question 32.
Perimeter < 51.3 inches
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 39
Answer:
The value of x is greater than 21.6 inches

Explanation:
The given figure is:
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 39
From the above figure,
We can say that the figure is the triangle
We know that,
The perimeter is the sum of all of the sides
So,
The perimeter of the triangle = 15.5 + 14.2 + x
It is given that
Perimeter < 51.3 inches
So,
51.3 < 29.7 + x
51.3 – 29.7 < x
21.6 < x
x > 21.6
Hence, from the above,
We can conclude that the value of x is greater than 21.6 inches

Question 33.
Perimeter ≤ 18.7 feet
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 39.1
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.2 Question 33

Question 34.
THOUGHT-PROVOKING
Write an inequality that has the solution shown in the graph. Describe a real-life situation that can be modeled by inequality.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 40
Answer:
The given number line is:
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 40
From the above number line,
We can observe that the marked line is starting from 16 and continued till the left end of the number line
So,
The inequality that represents the given number line is:
x ≤ 16
The real-life situation that can be modeled by inequality is:
A lift carrying no more than 16 people

Question 35.
WRITING
Is it possible to check all the numbers in the solution set of an inequality? When you solve the inequality x – 11 ≥ -3, which numbers can you check to verify your solution? Explain your reasoning.
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.2 Question 35

Question 36.
HOW DO YOU SEE IT?
The diagram represents the numbers of students in a school with brown eyes, brown hair, or both.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 41
Determine whether each inequality must be true. Explain your reasoning.
a. H ≥ E
b. H + 10 ≥ E
c. H ≥ X
d. H + 10 ≥ X
e. H > X
f. H + 10 > X
Answer:
The given diagram is:
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 41
From the above diagram,
We can observe that,
The spaces occupied by H and E are equal
The spaces occupied by H and E are greater than X
So,
We can say that
H = E; H > X; E > X
Now,
a. H ≥ E
From the above diagram,
The given inequality is not true
b. H + 10 ≥ E
From the above diagram,
The given inequality is not true
c. H ≥ X
From the above diagram,
The given inequality is not true
d. H + 10 ≥ X
From the above diagram,
The given inequality is not true
e. H > X
From the above diagram,
The given inequality is true
f. H + 10 > X
From the above diagram,
The given inequality is true

Question 37.
REASONING
Write and graph an inequality that represents the numbers that are not solutions to each inequality.
a. x + 8 < 14
b. x – 12 ≥ 5.7
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.2 Question 37

Question 38.
PROBLEM-SOLVING
Use the inequalities c – 3 ≥ d, b + 4 < a + 1, and a – 2 ≤ d – 7 to order a, b, c, and d from least to greatest.
Answer:
The order of a, b, c, and d from least to greatest is:
b, a, c, and d

Explanation:
The given inequalities are:
A) c – 3 ≥ d
B) b + 4 < a + 1
C) a – 2 ≤ d – 7
Now,
A)
The given inequality is:
c – 3 ≥ d
c ≥ d + 3
B)
The given inequality is:
b + 4 < a + 1
b < a + 1 – 4
b < a – 3
b + 3 < a
a > b + 3
C)
The given inequality is:
a – 2 ≤ d – 7
a ≤ d – 7 + 2
a ≤ d – 5
a + 5 ≤ d
d ≥ a + 5
Hence, from the above,
We can conclude that the order of a, b, c, d from least to greatest is:
b, a, c, and d

Maintaining Mathematical Proficiency

Find the product or quotient.

Question 39.
7 • (-9)
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.2 Question 39

Question 40.
-11 • (-12)
Answer:
The given expression is:
-11 ⋅ ( -12 )
= 11 ⋅ 12 [ Since we know that – × – = + ]
= 132
Hence, from the above,
We can conclude that the product of the given expression is: 132

Question 41.
-27 ÷ (-3)
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.2 Question 41

Question 42.
20 ÷ (-5)
Answer:
The given expression is:
20 ÷ ( -5 )
= -20 ÷ 5 [ Since we know that + ÷ – = – ]
= -4
Hence, from the above,
We can conclude that the product of the given expression is: -4

Solve the equation. Check your solution.(Section 1.1)

Question 43.
6x = 24
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.2 Question 43

Question 44.
-3y = -18
Answer:
The given equation is:
-3y = -18
y = -18 ÷ ( -3 )
y = 6 [ Sice we know that – ÷ – = + ]
Hence, fro the above,
We can conclude that the value of y in the given equation is: 6

Question 45.
\(\frac{s}{-8}\) = 13
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.2 Question 45

Question 46.
\(\frac{n}{4}\) = -7.3
Answer:
The given equation is:
\(\frac{n}{4}\) = -7.3
n = -7.3 × 4
n = -29.2
Hence, from the above,
We can conclude that the value of n in the given equation is: -29.2

Lesson 2.3 Solving Inequalities Using Multiplication or Division

Essential Question

How can you use division to solve inequality?

EXPLORATION 1
Writing a Rule
Work with a partner.
a. Copy and complete the table. Decide which graph represents the solution of the inequality 6 < 3x. Write the solution to the inequality.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 42
Answer:
a)
The completed table is:

The given number lines are:

From the above two number lines,
We can say that the expression 6 < 3x will be true if the value of x is greater than 2
From the 2nd number line,
We can observe that the value of x is greater than 2
b)
Rule:
Division Property of Inequality:
If you divide one side of an inequality by a number, you can divide the other side of the inequality by the same number.The given inequality is:
6 < 3x
3x > 6
x > 6 ÷ 3
x > 2
So,
From the table,
x > 2 means x = 3, 4, 5
So,
The value of x is: 3, 4, 5
Now,
i. 2x < 4
Answer:
The given inequality is:
2x < 4
x < 4 / 2
x < 2
Now,
From the above table,
We can observe that x < 2 will be held true if the value of x is -1, 0, 1

ii. 3 ≥ 3x
Answer:
The given inequality is:
3 ≥ 3x
1 ≥ x
x ≤ 1
Now,
From the above table,
We can observe that x ≤ 1 will be held true if the value of the x is -1, 0, 1

iii. 2x < 8
Answer:
The given inequality is:
2x < 8
x < 8 / 2
x < 4
Now,
From the above table,
We can observe that x < 4 will be held true if the value of x is -1, 0, 1, 2, 3

iv. 6 ≥ 3x
Answer:
The given inequality is:
6 ≥ 3x
So,
6 / 3 ≥ x
2 ≥ x
x ≤ 2
Now,
from the above table,
We can observe that x ≤ 2 will be held true if the value of x is -1, 0, 1, 2

EXPLORATION 2
Writing a Rule
Work with a partner.
a. Copy and complete the table. Decide which graph represents the solution of the inequality 6 < -3x. Write the solution of the inequality.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 43
Answer:
The complete table is:

From the given number lines,
The first graph indicates the inequality
x < -2
The second graph indicates the inequality
x > -2
Hence, from the table and the number line,
The inequality that represents the table is:
x < -2

b. Use a table to solve each inequality. Then write a rule that describes how to use division to solve the inequalities.
Rule:
Division Property of Inequality:
If you divide one side of an inequality by a number, you can divide the other side of the inequality by the same number.
i. -2x < 4
Answer:
The given inequality is:
-2x < 4
-x < 4 / 2
-x < 2
x < -2
Hence,
From the above table,
We can observe that the values of x for the inequality x < -2 is -5, -4, -3

ii. 3 ≥ -3x
Answer:
The given inequality is:
3 ≥ -3x
1 ≥ -x
-1 ≥ x
x ≤ -1
Hence,
From the above table,
We can observe that the values of x for the inequality x ≤ -1 is -5, -4, -3, -2, -1

iii. -2x < 8
Answer:
The given inequality is:
-2x < 8
x < -8 /2
x < -4
Hence,
From the above table,
We can observe that the values of x for the inequality x < -4 is  -5

iv.6 ≥ -3x
Answer:
The given inequality is:
6 ≥ -3x
6 / ( -3 ) ≥ x
-2 ≥ x
x ≤ -2
Hence,
From the above table,
We can observe that the values of x for the inequality x ≤ -2 is -5, -4, -3, -2

Communicate Your Answer

Question 3.
How can you use division to solve inequality?
Answer:
You can use division to solve the inequality by using the Division Property of Inequality
Division Property of Inequality:
If you divide one side of an inequality by a number, you can divide the other side of the inequality by the same number.

Question 4.
Use the rules you wrote in Explorations 1(b) and 2(b) to solve each inequality.
a. 7x < -21
b. 12 ≤ 4x
c. 10 < -5x d. -3x ≤ 0
Answer:
Rule:
Division Property of Inequality:
If you divide one side of an inequality by a number, you can divide the other side of the inequality by the same number.
By using the above rule, solve the given inequalities
Now,
a. 7x < -21
Answer:
The given inequality is:
7x < -21
x < -21 /
x < -3
Hence, from the above,
We can conclude that the solution to the given inequality is x < -3

b. 12 ≤ 4x
Answer:
The given inequality is:
12 ≤ 4x
12 / 4 ≤ x
3 ≤ x
x ≥ 3
Hence, from the above,
We can conclude that the solution to the given inequality is x ≥ 3

c. 10 < -5x
Answer:
The given inequality is:
10 < -5x
10 / -5 < x
2 < x
x > 2
Hence, from the above,
We can conclude that the solution to the given inequality is x > 2

d. -3x ≤ 0
Answer:
The given inequality is:
-3x ≤ 0
x ≤ 0 / -3
x ≤ 0
Hence, from the above,
We can conclude that the solution to the given inequality is x ≤ 0

2.3 Lesson 

Monitoring Progress Solve the inequality. Graph the solution. 

Question 1.
\(\frac{n}{7}\) ≥ -1
Answer:
The given inequality is:
\(\frac{n}{7}\) ≥ -1
n ≥ -1 ( 7 )
n ≥ -7
Hence, from the above,
We can conclude that the solution o the inequality is n ≥ -7
The representation of the inequality in the number line is:

Question 2.
-6.4 ≥ \(\frac{1}{5}\)w
Answer:
The given inequality is:
-6.4 ≥ \(\frac{1}{5}\)w
-6.4 ( 5 ) ≥ w
-32 ≥ w
w ≤ -32
Hence, from the above,
We can conclude that the solution to the given inequality is w ≤ -32
The representation of the inequality in the number line is:

Question 3.
4b ≥ 36
Answer:
The given inequality is:
4b ≥ 36
b ≥ 36 / 4
b ≥ 9
Hence, from the above,
We can conclude that the solution to the given inequality is b ≥ 9
The representation of the inequality in the number line is:

Question 4.
-18 > 1.5q
Answer:
The given inequality is:
-18 > 1.5q
-18 / 1.5 > q
-180 / 15 > q
-12 > q
q < -12
Hence, from the above,
We can conclude that the solution to the given inequality is q < -12
The representation of the inequality in the number line is:

Monitoring Progress

Solve the inequality. Graph the solution.

Question 5.
\(\frac{p}{-4}\) < 7
Answer:
The given inequality is:
\(\frac{p}{-4}\) < 7
–\(\frac{p}{4}\) < 7
-p < 7 ( 4 )
-p < 28
p < -28
Hence, from the above,
We can conclude that the solution to the given inequality is p < -28
The representation of the inequality in the number line is:

Question 6.
\(\frac{x}{-5}\) ≤ -5
Answer:
The given inequality is:
\(\frac{x}{-5}\) ≤ -5
–\(\frac{x}{5}\) ≤ -5
\(\frac{x}{5}\) ≤ 5
x ≤ 5 ( 5 )
x ≤ 25
Hence, from the above,
We can conclude that the solution to the given inequality is x ≤ 25
The representation of the inequality in the number line is:

Question 7.
-1 ≥ –\(\frac{1}{10}\)z
Answer:
The given inequality is:
-1 ≥ –\(\frac{1}{10}\)z
1 ≥ \(\frac{1}{10}\)z
1 ( 10 ) ≥ z
10 ≥ z
z ≤ 10
Hence, from the above,
We can conclude that the solution to the given inequality is z ≤ 10
The representation of the inequality in the number line is:

Question 8.
-9m > 63
Answer:
The given inequality is:
-9m > 63
m > -63 / 9
m > -7
Hence, from the above
We can conclude that the solution to the given inequality is m > -7
The representation of the inequality in the number line is:

Question 9.
-2r ≥ -22
Answer:
The given inequality is:
-2r ≥ -22
2r ≥ 22
r ≥ 22 / 2
r ≥ 11
Hence, from the above
We can conclude that the solution to the given inequality is r ≥ 11
The representation of the inequality in the number line is:

Question 10.
-0.4y ≥ -12
Answer:
The given inequality is:
-0.4y ≥ -12
0.4y ≥ 12
y ≥ 12 / 0.4
y ≥ 120 / 4
y ≥ 30
Hence, from the above
We can conclude that the solution to the given inequality is y ≥ 30
The representation of the inequality in the number line is:

Question 11.
You have at most $3.65 to make copies. Each copy costs $0.25. Write and solve an inequality that represents the number of copies you can make.
Answer:
The inequality that represents the number of copies you can make is:
x + 0.25 ≤ 3.65

Explanation:
It is given that you have at most $3.65 i.e., you have a maximum of only $3.65 to make copies and it is also given that each copy costs $0.25.
Let,
The additional cost to make the copies be $x
So,
The total cost = x + 0.25
But,
The total cost won’t be greater than 3.65
Hence,
The inequality that represents the number of copies you can make is:
x + 0.25 ≤ 3.65

Question 12.
The maximum speed limit for a school bus is 55 miles per hour. Write and solve an inequality that represents the number of hours it takes to travel 165 miles in a school bus.
Answer:
The inequality that represents the number of hours it takes to travel 165 miles in a school bus is:
x ≥ 3

Explanation:
It is given that the maximum speed limit for a school bus is 55 miles per hour.
Let,
x be the number of hours
So,
The inequality that represents the number of hours it takes to travel 165 miles in a school bus is:
55x ≥ 165
x ≥ 165 / 55
x ≥ 3
Hence, from the above,
We can conclude that the inequality that represents the number of hours it takes to travel 165 miles in a school bus is:
x ≥ 3

Solving Inequalities Using Multiplication or Division 2.3 Exercises

In Exercises 3–10, solve the inequality. Graph the solution.

Vocabulary and Core Concept Check

Question 1.
WRITING
Explain how solving 2x < -8 is different from solving -2x < 8
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.3 Question 1

Question 2.
OPEN-ENDED
Write an inequality that is solved using the Division property of Inequality where the inequality symbol needs to be reversed.
Answer:
The required inequality is:
-6 < 3x
Using the Division Property of Inequality,
-6 / 3 < ( 3 / 3 )x
-2 < x
x > -2
Hence, from the above,
We can conclude that the inequality is solved using the Division Property of Inequality wheer the inequality symbol needs to be reversed

Question 3.
4x < 8
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.3 Question 3

Question 4.
3y ≤ -95.
Answer:
The given inequality is:
3y ≤ -95
y ≤ -95 / 3
y ≤ -31.6
y ≤ -32 [ Approximate value ]
Hence, from the above,
We can conclude that the solution to the given inequality is y ≤ -32
The representation of the inequality in the number line is:

Question 5.
-20 ≤ 10n
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.3 Question 5

Question 6.
35 < 7t
Answer:
The given inequality is:
35 < 7t
35 / 7 < t
5 < t
t > 5
Hence, from the above,
We can conclude that the solution to the given inequality is t > 5
The representation of the inequality in the number line is:

Question 7.
\(\frac{x}{2}\) > -2
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.3 Question 7

Question 8.
\(\frac{a}{4}\) < 10.2
Answer:
The given inequality is:
\(\frac{a}{4}\) < 10.2
a < 10.2 ( 4 )
a < 40.8
a < 40 [ Approximate value ]
Hence, from the above,
We can conclude that the solution to the given inequality is a < 40
The representation of the inequality in the number line is:

Question 9.
20 ≥ \(\frac{4}{5}\)w
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.3 Question 9

Question 10.
-16 ≤ \(\frac{8}{3}\)t
Answer:
The given inequality is:
-16 ≤ \(\frac{8}{3}\)t
-16 ( 3) ≤ 8t
-48 ≤ 8t
-48 / 8 ≤ t
-6 ≤ t
t ≥ -6
Hence, from the above,
We can conclude that the solution to the given inequality is t ≥ -6
The representation of the inequality in the number line is:

In Exercises 11–18, solve the inequality. Graph the solution.

Question 11.
-6t < 12
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.3 Question 11

Question 12.
-9y > 9
Answer:
The given inequality is:
-9y > 9
y > -9 / 9
y > -1
Hence, from the above,
We can conclude that the solution to the given inequality is y > -1
The representation of the inequality in the number line is:

Question 13.
-10 ≥ -2z
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.3 Question 13

Question 14.
-15 ≤ -3c
Answer:
The given inequality is:
-15 ≤ -3c
15 ≤ 3c
15 / 3 ≤ c
5 ≤ c
c ≥ 5
Hence, from the above,
We can conclude that the solution to the given inequality is c ≥ 5
The representation of the inequality in the number line is:

Question 15.
\(\frac{n}{-3}\) ≥ 1
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.3 Question 15

Question 16.
\(\frac{w}{-5}\) ≤ 16
Answer:
The given inequality is:
\(\frac{w}{-5}\) ≤ 16
w ≤ -16 ( 5 )
w ≤ -80
Hence, from the above,
We can conclude that the solution to the given inequality is w ≤ -80
The representation of the inequality in the number line is:

Question 17.
-8 < –\(\frac{1}{4}\)m
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.3 Question 17

Question 18.
-6y > –\(\frac{2}{3}\)
Answer:
The given inequality is:
-6 > –\(\frac{2}{3}\)y
6  > \(\frac{2}{3}\)y
6 ( 3 ) > 2y
18 > 2y
18 / 2 > y
9 > y
y < 9
Hence, from the above,
We can conclude that the solution to the given inequality is y < 9
The representation of the inequality in the number line is:

Question 19.
MODELING WITH MATHEMATICS
You have $12 to buy five goldfish for your new fish tank. Write and solve an inequality that represents the prices you can pay per fish.
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.3 Question 19

Question 20.
MODELING WITH MATHEMATICS
A weather forecaster predicts that the temperature in Antarctica will decrease by 8°F each hour for the next 6 hours. Write and solve an inequality to determine how many hours it will take for the temperature to drop at least 36°F.
Answer:
The inequality to determine the number of hours it will take for the temperature to drop at least 36° F is:
x ≥ 4.5

Explanation:
It is given that a weather forecaster predicts that the temperature in Antarctica will decrease by 8°F each hour for the next 6 hours.
Now,
Let the number of hours that will take to drop the temperature be x
So,
8x ≥ 36
x ≥ 36 / 8
x ≥ 4.5
Hence, from the above,
We can conclude that the inequality to determine the number of hours it will take for the temperature to drop at least 36° F is:
x ≥ 4.5 hours

USING TOOLS
In Exercises 21–26, solve the inequality. Use a graphing calculator to verify your answer.

Question 21.
36 < 3y
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.3 Question 21

Question 22.
17v ≥ 51
Answer:
The given inequality is:
17v ≥ 51
v ≥ 51 / 17
v ≥ 3
Hence, from the above,
We can conclude that the solution to the given inequality is v ≥ 3
The representation of the inequality in the number line is:

Question 23.
2 ≤ –\(\frac{2}{9}\)x
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.3 Question 23

Question 24.
4 > \(\frac{n}{-4}\)
Answer:
The given inequality is:
4 > \(\frac{n}{-4}\)
4 ( -4 ) > n
-16 > n
n < -16
Hence, from the above,
We can conclude that the solution to the given inequality is n < -16
The representation of the inequality in the number line is:

Question 25.
2x > \(\frac{3}{4}\)
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.3 Question 25

Question 26.
1.1y < 4.4
Answer:
The given inequality is:
1.1y < 4.4
\(\frac{11}{10}\)y > \(\frac{44}{10}\)
y > \(\frac{10 × 44}{11 × 10}\)
y > 4
Hence, from the above,
We can conclude that the solution to the given inequality is y > 4
The representation of the inequality in the number line is:

ERROR ANALYSIS
In Exercises 27 and 28, describe and correct the error in solving the inequality.

Question 27.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 45
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.3 Question 27

Question 28.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 46
Answer:
The given inequality is:
-4y ≤ -32
4y ≤ 32
y ≤ 32 /4
y ≤ 8
Hence, from the above,
We can conclude that the solution to the given inequality is:
y ≤ 8

Question 29.
ATTENDING TO PRECISION
You have $700 to buy a new carpet for your bedroom. Write and solve an inequality that represents the costs per square foot that you can pay for the new carpet. Specify the units of measure in each step.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 47
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.3 Question 29

Question 30.
HOW DO YOU SEE IT?
Let m > 0. Match each inequality with its graph. Explain your reasoning.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 48
Answer:
The given inequalities are:
a. \(\frac{x}{m}\) < -1
b. \(\frac{x}{m}\) > 1
c. \(\frac{x}{m}\) < 1
d. –\(\frac{x}{m}\) < 1
It is given that m > 0 i.e., m is a positive number
Let,
The value of x is: 1
Now,
a.
The given inequality is:
\(\frac{x}{m}\) < -1
\(\frac{1}{m}\) < -1
1 < -m
-1 < m
m > -1
Hence,,
From the number lines, D) represents the required number line  for this inequality
b.
The given inequality is:
\(\frac{x}{m}\) > 1
x > m
1 > m
m < 1
Hence,
From the number lines, B) represents the required number line for this inequality
c.
The given inequality is:
\(\frac{x}{m}\) < 1
x < m
1 < m
m > 1
Hence,
From the number lines, A) represents the required number line for this inequality
d.
The given inequality is:
–\(\frac{x}{m}\) < 1
-x < -m
x < m
1 < m
m > 1
Hence,
From the number lines, A) represents the required number line for this inequality

Question 31.
MAKING AN ARGUMENT
You run for 2 hours at a speed no faster than 6.3 miles per hour.
a. Write and solve an inequality that represents the possible numbers of miles you run.
b. A marathon is approximately 26.2 miles. Your friend says that if you continue to run at this speed, you will not be able to complete a marathon in less than 4 hours. Is your friend correct? Explain.
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.3 Question 31

Question 32.
THOUGHT-PROVOKING
The inequality
\(\frac{x}{4}\) ≤ 5 has a solution of x = p. Write a second inequality that also has a solution of x = p.
Answer:
The second inequality that also has a solution x = p is:
x ≤ 20

Explanation:
The given inequality is:
\(\frac{x}{4}\) ≤ 5
It is given that the given inequality has a solution of x = p
So,
The second inequality that also has a second solution of x = p is:
\(\frac{x}{4}\) ≤ 5
x ≤ 5 ( 4 )
x ≤ 20
Hence, from the above,
We can conclude that the second inequality that also has a solution of x = p is:
x ≤ 20

Question 33.
PROBLEM-SOLVING
The U.S. Mint pays $0.02 to produce every penny. How many pennies are produced when the U.S. Mint pays more than $6 million in production costs?
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.3 Question 33

Question 34.
REASONING
Are x ≤ \(\frac{2}{3}\) and -3x ≤ -2 equivalent? Explain your reasoning.
Answer:
Yes,
x ≤ \(\frac{2}{3}\) and -3x ≤ -2 are equivalent

Explanation:
The given inequalities are -3x ≤ -2 and x ≤ \(\frac{2}{3}\)
Now,
x ≤ \(\frac{2}{3}\)
3x ≤ 2
Now,
-3x ≤ -2
Multiply with ‘-‘ on both sides
We know that,
– × – = +
So,
3x ≤ 2
Hence, from the above,
We can conclude that the given two inequalities are equivalent

Question 35.
ANALYZING RELATIONSHIPS
Consider the number line shown.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 49
a. Write an inequality relating A and B.
b. Write an inequality relating -A and -B.
c. Use the results from parts (a) and (b) to explain why the direction of the inequality symbol must be reversed when multiplying or dividing each side of an inequality by the same negative number.
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.3 Question 35

Question 36.
REASONING
Why might solving the inequality \(\frac{4}{x}\) ≥ 2 by multiplying each side by x lead to an error? (Hint: Consider x > 0 and x < 0.)
Answer:
The given inequality is:
\(\frac{4}{x}\) ≥ 2
4 ≥ 2x
4 / 2 ≥ x
2 ≥ x
x ≤ 2
It is given that to consider x > 0 and x < 0 i.e., x as a positive number and a negative number
Now,
Multiply the given inequality by x on both sides and consider x as positive
4x ≥ 2x²
4x / 2 ≥ x²
2x / x ≥ x
2 ≥ x
x ≤ 2
Now,
Multiply the given inequality by x on both sides and consider x as negative
-4x ≥ -2x²
4x / 2 ≥ x²
2x ≥ x²
2x / x ≥ x
2 ≥ x
x ≤ 2
Hence, from the above,
We can conclude that there will be no error even when we multiply the given inequality with x on both sides

Question 37.
MATHEMATICAL CONNECTIONS
The radius of a circle is represented by the formula r = \(\frac{C}{2π}\). Write and solve an inequality that represents the possible circumferences C of the circle.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 50
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.3 Question 37

Question 38.
CRITICAL THINKING
A water-skiing instructor recommends that a boat pulling a beginning skier has a speed less than 18 miles per hour. Write and solve an inequality that represents the possible distances d (in miles) that a beginner can travel in 45 minutes of practice time.
Answer:
The inequality that represents the possible distances that a beginner can travel in 45 minutes of practice time is:
d < 13.5 miles

Explanation:
It is given that a water-skiing instructor recommends that a boat pulling a beginning skier has a speed less than 18 miles per hour
The given speed is in terms of miles per hour
So,
The time should also be in hours
But the given time is in minutes
So,
We know that,
60 minutes = 1 hour
So,
45 minutes = \(\frac{45}{60}\) hours
= \(\frac{3}{4}\) hours
It is also given that the distance is d
We know that,
Speed = \(\frac{Distance}{Time}\)
So,
Distance = Speed × Time
Hence,
The inequality that represents the possible distances that a beginner can travel in 45 minutes of practice time is:
d < 18 ( \(\frac{3}{4}\) )
d < \(\frac{18 × 3}{4}\)
d < \(\frac{54}{4}\)
d < \(\frac{27}{2}\)
d < 13.5 miles
Hence, from the above,
We can conclude that the inequality that represents the possible distances that a beginner can travel in 45 minutes of practice time is:
d < 13.5 miles

Question 39.
CRITICAL THINKING
A local zoo employs 36 people to take care of the animals each day. At most, 24 of the employees work full time. Write and solve an inequality that represents the fraction of employees who work part-time. Graph the solution. Maintaining Mathematical Proficiency Solve the equation. Check your solution. (Section 1.2 and Section 1.3)
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.3 Question 39

Question 40.
5x + 3 = 13
Answer:
The given equaltion is:
5x + 3 = 13
5x = 13 – 3
5x = 10
x = 10 / 2
x = 5
Hence, from the above,
We can conclude that the solution to the given equaltion is x = 5

Question 41.
\(\frac{1}{2}\)y – 8 = -10
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.3 Question 41

Question 42.
-3n + 2 = 2n – 3
Answer:
The given equation is:
-3n + 2 = 2n – 3
-3n – 2n = -3 – 2
-5n = -5
5n = 5
n = 5 / 5
n = 1
Hence, from the above,
We can conclud ethat the solution to the given equation is n = 1

Question 43.
\(\frac{1}{2}\)z + 4 = \(\frac{5}{2}\)z – 8 Tell which number is greater.
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.3 Question 43

Question 44.
0.8, 85%
Answer:
The relation between the given numbers are:
0.8 < 85%

Explanation:
The given numbers are: 0.8 and 85%
We know that,
x% = \(\frac{x}{100}\)
We can rewrite 0.8 as 0.80 since both 0.8 and 0.80 are the same
So,
85% = \(\frac{85}{100}\)
0.8 = \(\frac{80}{100}\)
By comparison, we can get
0.8 is less than 85%
Hence, from the above,
We can conclude that
0.8 < 85%

Question 45.
\(\frac{16}{30}\), 50%
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.3 Question 45

Question 46.
120%, 0.12
Answer:
The relation between the given numbers are:
120% > 0.12

Explanation:
The given numbers are: 120% and 0.12
We know that,
x% = \(\frac{x}{100}\)
Now,
120% = \(\frac{120}{100}\)
0.12 = \(\frac{12}{100}\)
By comparison, we can get
120% is greater than 0.12
Hence, from the above,
We can conclude that
120% > 0.12

Question 47.
60%, \(\frac{2}{3}\)
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.3 Question 47

Lesson 2.4 Solving Multi-step Inequalities

Essential Question
How can you solve a multi-step inequality?

EXPLORATION 1
Solving a Multi-Step Inequality
Work with a partner.

• Use what you already know about solving equations and inequalities to solve each multi-step inequality. Justify each step.
• Match each inequality with its graph. Use a graphing calculator to check your answer.
a. 2x + 3 ≤ x + 5
b. -2x + 3 > x + 9
c. 27 ≥ 5x + 4x
d. -8x + 2x – 16 < -5x + 7x
e. 3(x – 3) – 5x > -3x – 6
f. -5x – 6x ≤ 8 – 8x – x
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 51
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 51.1
Answer:
The given inequalities are:
a. 2x + 3 ≤ x + 5
b. -2x + 3 > x + 9
c. 27 ≥ 5x + 4x
d. -8x + 2x – 16 < -5x + 7x
e. 3(x – 3) – 5x > -3x – 6
f. -5x – 6x ≤ 8 – 8x – x
The given graphing calculators are:
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 51.1
Now,
From the graphing calculators, we can observe that the graph is divided into 4 parts.
The first part indicates +x
The second part indicates -x
Now,
a.
The given inequality is:
2x + 3 ≤ x + 5
So,
2x – x ≤ 5 – 3
x ≤ 2
Hence, from the above,
We can conclude that the solution to the given inequality is x ≤ 2
The graph B) matches the solution of the given inequality

b.
The given inequality is:
-2x + 3 > x + 9
So,
-2x – x > 9  -3
-3x > 6
x > -6 / 3
x > -2
Hence, from the above,
We can conclude that the solution to the given inequality is x > -2
The graph C) matches the solution of the given inequality

c.
The given inequality is:
27 ≥ 5x + 4x
So,
27 ≥ 9x
27 / 9 ≥ x
3 ≥ x
x ≤ 3
Hence, from the above,
We can conclude that the solution to the given inequality is x ≤ 3
The graph E) matches the solution of the given inequality

d.
The given inequality is:
-8x + 2x – 16 < -5x + 7x
So,
-6x – 16 < 2x
-6x – 2x < 16
-8x < 16
x < -16 / 8
x < -2
Hence, from the above,
We can conclude that the solution to the given inequality is x < -2
Graph A) matches the solution of the given inequality

e.
The given inequality is:
3(x – 3) – 5x > -3x – 6
So,
3 ( x ) – 3 ( 3 ) – 5x > -3x – 6
3x – 9 – 5x > -3x – 6
-2x – 9 > -3x – 6
-2x + 3x > -6 + 9
x > 3
Hence, from the above,
We can conclude that the solution to the given inequality is x > 3
The graph D) matches the solution of the given inequality

f.
The given inequality is:
-5x – 6x ≥ 8 – 8x – x
So,
-11x ≥ 8 – 9x
-11x + 9x ≥ 8
-2x ≥ 8
x ≥ -8 / 2
x ≥ -4
Hence, from the above,
We can conclude that the solution to the given inequality is x ≥ -4
The graph F) matches the solution of the given inequality

Question 2.
How can you solve a multi-step inequality?
Answer:
The general procedure for solving multi-step inequality is as follows:
a) Clear parenthesis i.e., Brackets on both sides of the inequality and collect like terms
b) Addor subtract terms so the variable is on one side and the constant is on another side of the inequality sign

Question 3.
Write two different multi-step inequalities whose solutions are represented by the graph.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 52
Answer:
The given graph is:
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 52
From the given graph,
We can observe that the marking started from -1 but by excluding -1 and continued till the left end of the graph
So,
The inequality represented by the graph is given as:
x < -1

2.4 Lesson

Monitoring Progress
Solve the inequality. Graph the solution.

Question 1.
4b – 1 < 7
Answer:
The given inequality is:
4b – 1 < 7
So,
4b < 7 + 1
4b < 8
b < 8 / 4
b < 2
Hence, from the above,
We can conclude that the solution to the given inequality is b < 2
The representation of the inequality in the number line is:

Question 2.
8 – 9c ≥ -28
Answer:
The given inequality is:
8 – 9c ≥ -28
-9c ≥ -28 – 8
-9c ≥ -36
9c ≥ 36
c ≥ 36 / 9
c ≥ 4
Hence, from the above,
We can conclude that the solution to the given inequality is c ≥ 4
The representation of the inequality in the number line is:

Question 3.
\(\frac{n}{-2}\) + 11 > 12
Answer:
The given inequality is:
\(\frac{n}{-2}\) + 11 > 12
\(\frac{n}{-2}\) > 12 – 11
\(\frac{n}{-2}\) > 1
n > 1 (-2)
n > -2
Hence, from the above,
We can conclude that the solution to the given inequality is n > -2
The representation of the inequality in the number line is:

Question 4.
6 ≥ 5 – \(\frac{v}{3}\)
Answer:
The given inequality is:
6 ≥ 5 – \(\frac{v}{3}\)
6 – 5 ≥ – \(\frac{v}{3}\)
-1 ≥ \(\frac{v}{3}\)
-3 ≥ v
v ≤ -3
Hence, from the above,
We can conclude that the solution to the given inequality is v ≤ -3
The representation of the inequality in the number line is:

Solve the inequality.

Question 5.
5x – 12 ≤ 3x – 4
Answer:
The given inequality is:
5x – 12 ≤ 3x – 4
5x – 3x ≤ 12 – 4
2x ≤ 8
x ≤ 8 / 2
x ≤ 4
Hence, from the above,
We can conclude that the solution to the given inequality is x ≤ 4

Question 6.
2(k – 5) < 2k + 5
Answer:
The given inequality is:
2 ( k – 5 ) < 2k + 5
So,
2 ( k ) – 2 ( 5 ) < 2k + 5
2k – 10 < 2k + 5
2k – 2k < 5 + 10
-10 < 5
Hence, from the above,
We can conclude that there is no solution for the given inequality

Question 7.
-4(3n – 1) > -12n + 5.2
Answer:
The given inequality is:
-4 ( 3n – 1 ) > -12n + 5.2
So,
-4 ( 3n ) – 4 ( -1 ) > -12n + 5.2
-12n + 4 > -12n + 5.2
-12n + 12n + 4 > 5.2
4 > 5.2
Hence, from the above,
We can conclude that there is no solution for the given inequality

Question 8.
3(2a – 1) ≥ 10a – 11
Answer:
The given inequality is:
3 ( 2a – 1 ) ≥ 10a – 11
So,
3 ( 2a ) – 3 ( 1 ) ≥ 10a – 11
6a – 3 ≥ 10a – 11
6a – 10a ≥ -11 + 3
-4a ≥ -8
4a ≥ 8
a ≥ 8 / 4
a ≥ 2
Hence, from the above,
We can conclude that the solution to the given inequality is a ≥ 2

Question 9.
WHAT IF?
You need a mean score of at least 85 points to advance to the next round. What scores in the fifth game will allow you to advance?
Answer:

Solving Multi-step Inequalities 2.4 Exercises

Vocabulary and Core Concept Check

Question 1.
WRITING
Compare solving multi-step inequalities and solving multi-step equations.
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.4 Question 1

Question 2.
WRITING
Without solving, how can you tell that the inequality 4x + 8 ≤ 4x – 3 has no solution?
Answer:
The given inequality is:
4x + 8  ≤ 4x – 3
Rearrange the variables ad constant terms
So,
4x – 4x ≤ -3 – 8
0 + 8 ≤ -3
8 ≤ -3
Hence, from the above,
We can observe that there is no variable x
Hence,
We can conclude that there is no solution for the given inequality

In Exercises 3–6, match the inequality with its graph.

Question 3.
7b – 4 ≤ 10
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.4 Question 3

Question 4.
4p + 4 ≥ 12
Answer:
The given inequality is:
4p + 4 ≥ 12
So,
4p ≥ 12 – 4
4p ≥ 8
p ≥ 8 / 4
p ≥ 2
Hence, from the above,
We can conclude that the solution to the given inequality is p ≥ 2
The graph represented for the solution is A) [ Se the graphs from Exercise 6 ]

Question 5.
-6g + 2 ≥ 20
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.4 Question 5

Question 6.
3(2 – f) ≤ 15

Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 53
Answer:
The given inequality is:
3 (2 – f) ≤  15
So,
3 (2) – 3 (f) ≤  15
6 – 3f ≤  15
-3f ≤  15 – 6
-3f ≤  9
f ≤  -9 / 3
f ≤  -3
Hence, from the above,
We can conclude that the solution to the given inequality is f ≤  -3
The graph C) represents the solution of the given inequality

In Exercises 7–16, solve the inequality. Graph the solution.

Question 7.
2x – 3 > 7
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.4 Question 7

Question 8.
5y + 9 ≤ 4
Answer:
The given inequality is:
5y + 9 ≤ 4
So,
5y ≤  4  -9
5y ≤  -5
y ≤  -5 / 5
y ≤  -1
Hence, from the above,
We can conclude that the solution to the given inequality is y ≤  -1
The representation of the solution in the graph is:

Question 9.
-9 ≤ 7 – 8v
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.4 Question 9

Question 10.
2 > -3t – 10
Answer:
The given inequality is:
2 > -3t – 10
So,
2 + 10 > -3t
12 > -3t
-12 / 3 > t
-4 > t
t < -4
Hence, from the above,
We can conclude that the solution to the given inequality is t < -4
The representation of the solution of the inequality in the graph is:

Question 11.
\(\frac{w}{2}\) + 4 > 5
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.4 Question 11

Question 12.
1 + \(\frac{m}{3}\) ≤ 6
Answer:
The given inequality is:
1 + \(\frac{m}{3}\) ≤ 6
So,
\(\frac{m}{3}\) ≤ 6 – 1
\(\frac{m}{3}\) ≤ 5
m ≤ 5 (3)
m ≤ 15
Hence, from the above,
We can conclude that the solution to the given inequality is m ≤ 15
The representation of the solution of the inequality in the graph is:

Question 13.
\(\frac{p}{-8}\) + 9 > 13
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.4 Question 13

Question 14.
3 + \(\frac{r}{-4}\) ≤ 6
Answer:
The given inequality is:
3 + \(\frac{r}{-4}\) ≤ 6
\(\frac{r}{-4}\) ≤ 6 – 3
\(\frac{r}{-4}\) ≤ 3
r ≤ 3 (-4)
r ≤ -12
Hence, from the above,
We can conclude that the solution to the given inequality is r ≤ -12
The representation of the solution of the given inequality in the graph is:

Question 15.
6 ≥ -6(a + 2)
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.4 Question 15

Question 16.
18 ≤ 3(b – 4)
Answer:
The given inequality is:
18 ≤ 3 (b – 4)
So,
18 ≤ 3 (b) – 3 (4)
18 ≤ 3b – 12
18 + 12 ≤ 3b
30 ≤ 3b
30 / 3 ≤ b
10 ≤ b
b ≥ 10
Hence, from the above,
We can conclude that the solution to the given inequality is b ≥ 10
The representation of the solution of the given inequality in the graph is:

In Exercises 17–28, solve the inequality.

Question 17.
4 – 2m > 7 – 3m
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.4 Question 17

Question 18.
8n+ 2 ≤ 8n – 9
Answer:
The given nequality is:
8n + 2 ≤ 8n – 9
So,
8n – 8n + 2 ≤ -9
2 ≤ 9
Hence, from the above,
We can conclude that there is no solution for the given inequality

Question 19.
-2d – 2 < 3d + 8
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.4 Question 19

Question 20.
8 + 10f > 14 – 2f
Answer:
The given inequality is:
8 + 10f > 14 – 2f
10f + 2f > 14 – 8
12f > 6
f > 6 / 12
f > \(\frac{1}{2}\)
Hence, from the above,
We can conclude that the solution to the given inequality is f > \(\frac{1}{2}\)

Question 21.
8g – 5g – 4 ≤ -3 + 3g
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.4 Question 21

Question 22.
3w – 5 > 2w + w – 7
Answer:
The given inequality is:
3w – 5 > 2w + w – 7
So,
3w – 5 > 3w – 7
3w – 5 – 3w > -7
-5 > -7
5 > 7
Hence, from the above,
We can conclude that there is no solution for the given inequality

Question 23.
6(ℓ + 3) < 3(2ℓ + 6)
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.4 Question 23

Question 24.
2(5c – 7) ≥ 10(c – 3)
Answer:
The given inequality is:
2 (5c – 7) ≥ 10 (c – 3)
So,
2 (5c) – 2 (7) ≥ 10 (c) – 10 (3)
10c – 14 ≥ 10c – 30
10c – 10c – 14 ≥ -30
-14 ≥ -30
14 ≥ 30
Hence, from the above,
We can conclude that there is no solution for the given inequality

Question 25.
4 (\(\frac{1}{2}\)t – 2 )> 2(t – 3)
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.4 Question 25

Question 26.
15 (\(\frac{1}{3}\)b + 3 ) ≤ 6(b + 9)
Answer:
The given inequality is:
15 ( (\(\frac{1}{3}\)b + 3 ) ) ≤ 6 (b + 9)
So,
15 ( (\(\frac{1}{3}\)b ) ) + 15 (3) ≤ 6 (b) + 6 (9)
5b + 45 ≤ 6b + 54
5b – 6b ≤ 54 – 45
-b ≤ 9
b ≤ -9
Hence, from the above,
We can conclude that the solution to the given inequality is b ≤ -9

Question 27.
9j – 6 + 6j ≥ 3(5j – 2)
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.4 Question 27

Question 28.
6h – 6 + 2h < 2(4h – 3)
Answer:
The given inequality is:
6h – 6 + 2h < 2 (4h – 3)
So,
8h – 6 < 2 (4h) – 2 (3)
8h – 6 < 8h – 6
8h – 8h < -6 + 6
0 < 0
Hence, from the above,
We can conclude that there is no solution for the given inequality

ERROR ANALYSIS
In Exercises 29 and 30, describe and correct the error in solving the inequality.

Question 29.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 54
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.4 Question 29

Question 30.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 55
Answer:
The given inequality is:
-2 (1 – x) ≤ 2x – 7
-2 (1) + 2 (x) ≤ 2x – 7
-2 + 2x ≤ 2x – 7
-2 + 2x – 2x ≤ -7
-2 ≤ -7
2 ≤  7
Hence, from the above,
We can conclude that the given inequality has no solution

Question 31.
MODELING WITH MATHEMATICS
Write and solve an inequality that represents how many $20 bills you can withdraw from the account without going below the minimum balance.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 56
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.4 Question 31

Question 32.
MODELING WITH MATHEMATICS
A woodworker wants to earn at least $25 an hour making and selling cabinets. He pays $125 for materials. Write and solve an inequality that represents how many hours the woodworker can spend building the cabinet.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 57
Answer:
The inequality that represents the number of hours the woodworker can spend build the cabinet is:
x ≥ 5

Explanation:
It is given that a woodworker wants to earn at least $25 an hour making and selling cabinets. He pays $125 for materials.
Let x be the number of hours taken by the woodworker to build the cabinet
So,
The inequality that represents the number of hours the woodworker can spend build the cabinet is:
25 (x) ≥ 125
x ≥ 125 / 25
x ≥ 5
Hence, from the above,
We can conclude that the inequality that represents the number of hours the woodworker can spnd build the cabinet is x ≥ 5

Question 33.
MATHEMATICAL CONNECTIONS
The area of the rectangle is greater than 60 square feet. Write and solve an inequality to find the possible values of x.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 58
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.4 Question 33

Question 34.
MAKING AN ARGUMENT
Forest Park Campgrounds charges a $100 membership fee plus $35 per night. Woodland Campgrounds charges a $20 membership fee plus $55 per night. Your friend says that if you plan to camp for four or more nights, then you should choose Woodland Campgrounds. Is your friend correct? Explain.
Answer:
Yes, your friend is correct

Explanation:
It is given that the Forest Park Campgrounds charges a $100 membership fee plus $35 per night and the Woodland Campgrounds charges a $20 membership fee plus $55 per night.
It is also given that your friend says that if you plan to camp for four or more nights, then you should choose Woodland Campgrounds.
So,
Let x be the number of nights you spend
Now,
The charge of the Forest Park Campgrounds for x nights = 100 + 35x
The charge of the Woodland Campgrounds = 20 + 55x
Now,
It is given that for four or more nights i.e., x ≥ 4, you will choose Woodland Campgrounds
So,
We have to verify the above statement
So,
20 + 55x ≥ 100 + 35x
55x – 35x ≥ 100 – 20
20x ≥ 80
x ≥ 80 / 20
x ≥ 4
Hence, from the above,
We can conclude that your friend is correct.

Question 35.
PROBLEM-SOLVING
The height of one story of a building is about 10 feet. The bottom of the ladder on the truck must be at least 24 feet away from the building. How many stories can the ladder reach? Justify your answer.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 59
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.4 Question 35.1
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.4 Question 35.2

Question 36.
HOW DO YOU SEE IT?
The graph shows your budget and the total cost of x gallons of gasoline and a car wash. You want to determine the possible amounts (in gallons) of gasoline you can buy within your budget.
a. What is your budget?
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 60
Answer:
Your budget is: $40

Explanation:
The given graph is:
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 60
From the graph,
We can observe that x-axis represents the amount of gasoline ( gallons ) and the y-axis represents the budget
From the graph,
The line is represented parallel to the x-axis i.e., Budget
So,
From the graph,
The given equation is:
y = 40
Hence, from the above,
We can conclude that the budget is: $40

b. How much does a gallon of gasoline cost? How much does a car wash cost?
Answer:
The cost of a gallon of gasoline is: $3.55
The cost of a car wash is: $8

Explanation:
It is given that the graph shows your budget and the total cost of x gallons of gasoline and a car wash.
From the graph,
We know that,
The x-axis represents the amount of gasoline ( gallons )
So,
The equation represented by the x-axis is:
y = 3.55x + 8
So,
From the equation,
We know that,
The coefficient of x represents the cost of x gallons of gasoline and the constant term represents the cost of a car wash
Hence, from the above,
We can conclude that
The cost of a gallon of gasoline is: $3.55
The cost of a car wash is: $8

c. Write an inequality that represents the possible amounts of gasoline you can buy.
Answer:
The inequality that represents the possible amounts of gasoline you can buy is:
x ≤ 9 gallons

Explanation:
From the graph,
We know that,
The x-axis represents the amounts of gasoline you can buy
We know that,
y = 3.55x + 8
We know that,
The budget will be less than or equal to $40
So,
3.55x + 8 ≤ 40
3.55x ≤ 40 – 8
3.55x ≤ 32
x ≤ 32 / 3.55
x ≤ 9 [ Approx. ]
Hence, from the above,
We can conclude that the inequality that represents the possible amounts of gasoline you can buy is:
x ≤ 9

d. Use the graph to estimate the solution of your inequality in part (c).
Answer:
From part (c),
The required inequality is:
3.55x + 8 ≤ 40
From the graph,
We can observe the x-axis and y-axis intersected at one particular point.
The particular point is the estimated solution of the inequality
So,
The particular point = (x , y ) = ( 9 , 40 )
Hence, from the above,
We can conclude that the estimated solution of your inequality in part (c) is ( 9, 40 )

Question 37.
PROBLEM-SOLVING
For what r values of r will the area of the shaded region be greater than or equal to 9(π – 2)?
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.4 Question 37

Question 38.
THOUGHT-PROVOKING
A runner’s times (in minutes) in the four races he has completed are 25.5, 24.3, 24.8, and 23.5. The runner plans to run at least one more race and wants to have an average time of less than 24 minutes. Write and solve an inequality to show how the runner can achieve his goal.
Answer:
The inequality that shows the runner can achieve his goal is:
x < 21.9 minutes

Explanation:
It is given that a runner’s times (in minutes) in the four races he has completed are 25.5, 24.3, 24.8, and 23.5.
It is also given that the runner plans to run at least one more race and wants to have an average time of less than 24 minutes.
Now,
Let x be the time in the fifth race [ Since it is given that the runner plans to run at least one more race ]
We know that,
Average = \(\frac{The sum of all the items}{The number of items}\)
Now,
Average = \(\frac{25.5 + 24.3 + 24.8 + 23.5 + x}{5}\)
Average = \(\frac{98.1 + x}{5}\)
It is given that the average time will be less than 24 minutes
So,
\(\frac{98.1 + x}{5}\) < 24
98.1 + x < 24 (5)
98.1 + x < 120
x < 120 – 98.1
x < 21.9 minutes
Hence, from the above,
We can conclude that the inequality that shows the runner can achieve his goal is x < 21.9 minutes

REASONING
In Exercises 39 and 40, find the value of a for which the solution of the inequality is all real numbers.

Question 39.
a(x + 3) < 5x + 15 – x
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.4 Question 39

Question 40.
3x + 8 + 2ax ≥ 3ax – 4a
Answer:
The value of a is: 3

Explanation:
The given inequality is:
3x + 8 + 2ax ≥ 3ax – 4a
So,
3x + 2ax – 3ax ≥ -4a – 8
3x – ax ≥ -4a – 8
It is given that the solution of the given inequality is all real numbers
So,
x = 0
So,
To make x = 0,
We can observe from the inequality that ‘a’ must be equal to 3
Hence, from the above,
We can conclude that the value of a is: 3

Maintaining Mathematical Proficiency
Write the sentence as an inequality. (Section 2.1)

Question 41.
Six times a number y is less than or equal to 10.
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.4 Question 41

Question 42.
A number p plus 7 is greater than 24.
Answer:
The given worded form is:
A number p plus 7 is greater than 24
Hence,
The representation of the worded form in the form of inequality is:
p + 7 > 24

Question 43.
The quotient of a number r and 7 is no more than 18.
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.4 Question 43

Solving Linear Inequalities Study Skills: Analyzing Your Errors

2.1–2.4 What Did You Learn?

Core Vocabulary
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 61

Core Concepts
Section 2.1
Representing Linear Inequalities, p. 57

Section 2.2
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 62

Section 2.3
Multiplication and Division Properties of Inequality (c > 0), p. 68
Multiplication and Division Properties of Inequality (c < 0), p. 69

Section 2.4
Solving Multi-Step Inequalities, p. 74
Special Solutions of Linear Inequalities, p. 75

Mathematical Practices

Question 1.
Explain the meaning of the inequality symbol in your answer to Exercise 47 on page 59. How did you know which symbol to use?
Answer:
In Exercise 47 on page 59,
The inequality symbol we used is: ≤
The meaning of ≤ is ” Less than or equal to ”
In Exercise 47,
It is given that the Xianren bridge arch is the longest natural arch with a length of 400 feet i.e., there is no arch longer than the Xianren bridge arch and the remaining natural arches are shorter than the Xianren arch
Hence,
The lengths of all the arches including the Xianren arch will be represented by the inequality symbol “≤”

Question 2.
In Exercise 30 on page 66, why is it important to check the reasonableness of your answer in part (a) before answering part (b)?
Answer:
In part (a), it is given that you have to beat your competitor with your score.
So,
Your score must be greater than your competitor
Then only you can solve part (b).

Question 3.
Explain how considering the units involved in Exercise 29 on page 71 helped you answer the question.
Answer:
In Exercise 29 on page 71,
The mat that is given is in the form of the square since its length and width both are the same
So,
We know that,
Area of the square = Side²
Let the cost per square foot be x
So,
The inequality formed will be like:
( Area of the square ) ⋅ x ≤ $700

Study Skills

Analyzing Your Errors

Application Errors
What Happens: You can do numerical problems, but you struggle with problems that have context.
How to Avoid This Error: Do not just mimic the steps of solving an application problem. Explain out loud what the question is asking and why you are doing each step. After solving the problem, ask yourself, “Does my solution make sense?”
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 63

Solving Linear Inequalities Maintaining 2.1 – 2.4 Quiz

Write the sentence as an inequality. (Section 2.1)

Question 1.
A number z minus 6 is greater than or equal to 11.
Answer:
The given worded form is:
A number z minus 6 is greater than or equal to 11
Hence,
The representation of the worded form in the form of inequality is:
z – 6 ≥ 11

Question 2.
Twelve is no more than the sum of -1.5 times a number w and 4.
Answer:
The given worded form is:
Twelve is no more than the sum of -1.5 times a number w and 4
Hence,
The representation of the worded form in the form of inequality is:
12 ≤ -1.5w + 4

Write an inequality that represents the graph.(Section 2.1)

Question 3.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 64
Answer:
The inequality that represents the given graph is:
x < 0

Explanation:
The given graph is:
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 64
From the above number line,
We can observe that the marked line started from 0 that excludes 0 and continued till the left end of the number line
Hence,
The inequality that represents the given graph is:
x < 0

Question 4.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 65
Answer:
The inequality that represents the given graph is:
x ≥ 8

Explanation:
The given graph is:
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 65
From the above graph,
The marked line started from 8 that includes 8 and continued till the right end of the number line.
Hence,
The inequality that represents the given graph is:
x ≥ 8

Solve the inequality. Graph the solution. (Section 2.2 and Section 2.3)

Question 5.
9 + q ≤ 15
Answer:
The solution to the given inequality is:
q ≤ 6

Explanation:
The given inequality is:
9 + q ≤ 15
So,
q ≤ 15 – 9
q ≤ 6
Hence, from the above,
We can conclude that the solution to the given inequality is q ≤ 6
The representation of the solution of the inequality in the graph is:

Question 6.
z – (-7) < 5
Answer:
The solution to the given inequality is:
z < -2

Explanation:
The given inequality is:
z – (-7) < 5
So,
z + 7  < 5
z < 5 – 7
z < -2
Hence, from the above,
We can conclude that the solution to the given inequality is z < -2
The representation of the solution of the inequality in the graph is:

Question 7.
-3 < y – 4
Answer:
The solution to the given inequality is:
y > 1

Explanation:
The given inequality is:
-3 < y – 4
So,
-3 + 4 < y
1 < y
y > 1
Hence, from the above,
We can conclude that the solution to the given inequality is y > 1
The representation of the solution of inequality in the graph is:

Question 8.
3p ≥ 18
Answer:
The solution to the given inequality is:
p ≥ 6

Explanation:
The given inequality is:
3p ≥ 18
p ≥ 18 / 3
p ≥ 6
Hence, from the above,
We can conclude that the solution to the given inequality is p ≥ 6
The representation of the solution of the inequality in the graph is:

Question 9.
6 > \(\frac{w}{-2}\)
Answer:
The solution to the given inequality is:
w < -12

Explanation:
The given inequality is:
6 > \(\frac{w}{-2}\)
So,
6 (-2) > w
-12 > w
w < -12
Hence, from the above,
We can conclude that the solution to the given inequality is w < -12
The representation of the solution of the inequality in the graph is:

Question 10.
-20x > 5
Answer:
The solution to the given inequality is:
x > – \(\frac{1}{4}\)
So,
x > -0.25

Explanation:
The given inequality is:
-20x > 5
So,
x > -5 / 20
x > – \(\frac{1}{4}\)
Hence, from the above,
We can conclude that the solution to the given inequality is x > –\(\frac{1}{4}\)
The representation of the solution of the inequality in the graph is:

Solve the inequality. (Section 2.4)

Question 11.
3y – 7 ≥ 17
Answer:
The solution to the given inequality is:
y ≥ 8

Explanation:
The given inequality is:
3y – 7 ≥ 17
So,
3y ≥ 17 + 7
3y ≥ 24
y ≥ 24 / 3
y ≥ 8
Hence, from the above,
We can conclude that the solution to the given inequality is y ≥ 8

Question 12.
8(3g – 2) ≤ 12(2g + 1)
Answer:
There is no soution for the given inequality

Explanation:
The given inequality is:
8 (3g – 2) ≤ 12 (2g + 1)
So,
8 (3g) – 8 (2) ≤ 12 (2g) + 12 (1)
24g – 16 ≤ 24g + 12
24g – 24g – 16 ≤ 12
-16 ≤ 12
Hence, from the above,
We can conclude that there is no solution for the given inequality

Question 13.
6(2x – 1) ≥ 3(4x + 1)
Answer:
There is no solution for the given inequality

Explanation:
The given inequality is:
6 (2x – 1)≥ 3 (4x + 1)
So,
6 (2x) – 6 (1) ≥ 3 (4x) + 3 (1)
12x – 6 ≤ 12x + 3
12x – 6 – 12x ≤ 3
-6 ≤ 3
Hence, from the above,
We can conclude that there is no solution for the given inequality

Question 14.
Three requirements for a lifeguard training course are shown. (Section 2.1)
a. Write and graph three inequalities that represent the requirements.
b. You can swim 250 feet, tread water for 6 minutes, and swim 35 feet underwater without taking a breath. Do you satisfy the requirements of the course? Explain.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 66
Answer:
a.
The given requirements are:
1. Swim at least 100 yards
2. Tread water for at least 5 minutes
3. Swim 10 yards or more underwater without taking a breath
Now,
1.
The inequality that represents the first requirement is:
x ≥ 100 yards
2.
The inequality that represents the second requirement is:
x ≥ 5 minutes
3.
The inequality that represents the third requirement is:
x ≥ 10 yards

b.
It is given that you can swim 250 feet, tread water for 6 minutes, and swim 35 feet underwater without taking a breath.
From part (a),
The inequalities for the three requirements are:
x ≥ 100 yards
x ≥ 5 minutes
x ≥ 10 yards
Hence, from the above,
We can conclude that you satisfied the requirements of the course

Question 15.
The maximum volume of an American white pelican’s bill is about 700 cubic inches. A pelican scoops up 100 cubic inches of water. Write and solve an inequality that represents the additional volumes the pelican’s bill can contain. (Section 2.2)
Answer:
The inequality that represents the additional volumes of the pelican’s bill can contain is:
x  ≤ 600

Explanation:
It is given that the maximum volume of an American white pelican’s bill is about700 cubic inches. A pelican scoops up 100 cubic inches of water.
Let x be the additional volumes the pelican’s bill can contain
It is also given that the maximum volume is 700 cubic inches and a pelican’s bill scoops up 100 cubic inches of water
Hence,
The inequality that represents the additional volumes of the pelican’s bill can contain is:
x + 100 ≤ 700
x ≤ 700 – 100
x ≤ 600

Question 16.
You save $15 per week to purchase one of the bikes shown. (Section 2.3and Section 2.4)
a. Write and solve an inequality to find the numbers of weeks you need to save to purchase a bike.
b. Your parents give you $65 to help you buy the new bike. How does this affect your answer in part (a)? Use an inequality to justify your answer.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 67
Answer:
a.
It is given that you save $15 per week to purchase one of the bikes
From the fig,
It is given that the starting price of the bike is $120
Now,
Let x be the number of weeks
Hence,
The inequality that representing the number of weeks you need to save to purchase a bike is:
15 ⋅ x ≥ 120
15x ≥ 120
x ≥ 120 / 15
x ≥ 8
Hence, from the above,
We can conclude that the number of weeks you need to save to purchase a bike is 8 weeks

b.
It is given that your parents give you $65 to help you buy the new bike.
We know that,
The starting price of the bike is: $120
So,
The remaining amount that needs to purchase a bike = 120 – 65
= $55

Lesson 2.5 Solving Compound Inequalities

Essential Question
How can you use inequalities to describe intervals on the real number line?
EXPLORATION 1
Describing Intervals on the Real Number
Work with a partner.
In parts (a)–(d), use two inequalities to describe the interval.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 68
e. Do you use “and” or “or” to connect the two inequalities in parts (a)–(d)? Explain.
Answer:
(a) – (d):
The given graphs are:
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 68
From the given graphs,
We can observe that the marked lines are from a certain number to a certain number
Now,
a.
In the given graph,
The marked line starting from -3 excluding -3 and continued till -6 including -6
So,
The inequality that represents the given interval is:
x ≥ -6 and x < -3
Hence,
The compound inequality that represents the given interval is:
-6 ≤ x < -3

b.
In the given graph,
The marked line starting from -5 excluding -5 and continued till 4 including 4
So,
The inequality that represents the given interval is:
x > -5 and x < 4
Hence,
The compound inequality that represents the given interval is:
-5 < x < 4

c.
In the given graph,
The marked line starting from -4 including -4 and continued till 5 including 5
So,
The inequality that represents the given interval is:
x ≥ -4 and x ≤ 5
Hence,
The compound inequality that represents the given interval is:
-4 ≤ x ≤ 5

d.
In the given graph,
The marked line starting from -3 excluding -3 and continued till 6 including 6
So,
The inequality that represents the given interval is:
x > -3 and x < 6
Hence,
The compound inequality that represents the given interval is:
-3 < x < 6

e.
We use “and” to connect the two inequalities in parts (a) – (d)
Now,
From (a) – (d),
We can observe that,
For Half-open interval,
We use ≥ and <
For closed interval,
We use ≥ and ≤
For open interval,
We use > and <

EXPLORATION 2
Describing Two Infinite Intervals
Work with a partner. In parts (a)–(d), use two inequalities to describe the interval.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 69
e. Do you use “and” or “or” to connect the two inequalities in parts (a)–(d)? Explain.
Answer:
The given graphs are:
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 69
(a) – (d):
a.
In the given graph,
The first marked line starting from -6 including -6 and continued till the end of the left side of the graph
The second marked line starting from 3 excluding 3 and continued till the end of the right side of the graph
So,
The inequality that represents the given intervals are:
x ≤ -6 and x > 3
Hence,
The compound inequality that represents the given interval is:
-6 ≤ x > 3

b.
In the given graph,
The first marked line starting from -5 excluding -5 and continued till the end of the left side of the graph
The second marked line starting from 4 including 4 and continued till the end of the right side of the graph
So,
The inequality that represents the given intervals are:
x < -5 and x ≥ 4
Hence,
The compound inequality that represents the given interval is:
-5 < x ≥ 4

c.
In the given graph,
The first marked line starting from -4 including -4 and continued till the end of the left side of the graph
The second marked line starting from 5 including 5 and continued till the end of the right side of the graph
So,
The inequality that represents the given intervals are:
x ≤ -4 and x ≥ 5
Hence,
The compound inequality that represents the given interval is:
-4 ≤ x ≥ 5

d.
In the given graph,
The first marked line starting from -3 excluding -3 and continued till the end of the left side of the graph
The second marked line starting from 6 excluding 6 and continued till the end of the right side of the graph
So,
The inequality that represents the given intervals are:
x < -3 and x > 6
Hence,
The compound inequality that represents the given interval is:
-3 < x > 6

e.
We use “and” to connect the two inequalities in parts (a) – (d)
Now,
From (a) – (d),
We can observe that,
For Half-open interval,
We use ≥ and <
For closed interval,
We use ≥ and ≤
For open interval,
We use > and <

Communicate Your Answer

Question 3.
How can you use inequalities to describe intervals on the real number line?
Answer:
An inequality that includes the boundary point indicated by the “or equal” part of the symbols ≤ and ≥ with a closed dot on the number line. The symbol (∞) indicates the interval is unbounded to the right. Express ordering relationships using the symbol < for “less than” and > for “greater than.”

2.5 Lesson

Monitoring Progress

Write the sentence as an inequality. Graph the inequality.

Question 1.
A number d is more than 0 and less than 10.
Answer:
The given worded form is:
A number d is more than 0 and less than 10
Hence,
The representation of the given worded form in the form of inequality is:
d > 0 and d < 10
The representation of the inequalities in the form of compound inequality is:
0 < d < 10
The representation of the compound inequality in the graph is:

Question 2.
A number a is fewer than -6 or no less than -3.
Answer:
The given worded form is:
A number a is fewer than -6 or no less than -3
Hence,
The representation of the given worded form in the form of inequality is:
a < -6 or a > -3
The representation of the inequalities in the graph is:

Solve the inequality. Graph the solution.

Question 3.
5 ≤ m + 4 < 10
Answer:
The given inequality is:
5 ≤ m + 4 < 10
Subtract by 4 on both sides
5 – 4 ≤ m + 4 – 4 < 10 – 4
1 ≤ m < 6
Hence, from the above,
We can conclude that the compound inequality for the given inequality is: 1 ≤ m < 6
The representation of the compound inequality in the graph is:

Question 4.
-3 < 2k – 5 < 7
Answer:
The given inequality is:
-3 < 2k – 5 < 7
Add 5 on both sides
-3 + 5 < 2k – 5 + 5 < 7 + 5
2 < 2k < 12
Divide by 2 on both sides
1 < k < 6
Hence, from the avove,
We can conclude that the compound inequality for the given inequality is: 1 < k < 6
The representation of the compound inequality in the graph is:

Question 5.
4c + 3 ≤ -5 or c – 8 > -1
Answer:
The given inequality is:
4c + 3 ≤ -5 or c – 8 > -1
So,
4c ≤ -5 – 3 or c > -1 + 8
4c ≤ -8 or c > 7
c ≤ -8 / 4 or c > 7
c ≤ -2 or c > 7
Hence, from the above,
We can conclude that the solutions to the given inequality are: c ≤ -2 or c > 7
The representations of the solutions in the graph are:

Question 6.
2p + 1 < -7 or 3 – 2p ≤ -1
Answer:
The given inequality is:
2p + 1 < -7 or 3 – 2p ≤ -1
So,
2p < -7 – 1 or -2p ≤ -1 – 3
2p < -8 or -2p ≤ -4
p < -8 / 2 or p ≤ -4 / (-2)
p < -4 or p ≤ 2
Hence, from the above,
We can conclude that the solutions to the given inequality are: p < -4 or p ≤ 2
The representations of the solutions in the graph are:

Question 7.
Write and solve a compound inequality that represents the temperature rating (in degrees Fahrenheit) of the winter boots.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 70
Answer:
The range of the temperature rating ( In °C ) of the winter boots is: -40°C to 15°C
We know that,
°F = ( °C × \(\frac{9}{5}\) ) + 32
Now,
To convert -40°C into °F,
°F = ( -40°C × \(\frac{9}{5}\) ) + 32
°F = -72 + 32
°F = – 40°F
Now,
To convert 15°C into °F,
°F = ( 15 × \(\frac{9}{5}\) ) + 32
°F = 27 + 32
°F = 59°F
Let t be the temperature in °F
So,
The representation of the temperature ranges in the form of inequalities is:
t > -40°F and t < 59°F
Hence,
The representation of the temperature ranges in the form of compound inequality is:
-40°F < t < 59°F

Solving Compound Inequalities 2.5 Exercises

Vocabulary and Core Concept Check

Question 1.
WRITING
Compare the graph of -6 ≤ x ≤ -4 with the graph of x ≤ -6 or x ≥ -4.
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.5 Question 1

Question 2.
WHICH ONE do DOESN’T BELONG?
Which compound inequality does not belong with the other three? Explain your reasoning.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 70.1
Answer:
The given inequalities are:
a. a > 4 or a < -3
b. a< -2 or a > 8
c. a > 7 or a < -5
d. a < 6 or a > -9
Now,
Represent all the inequalities in the graph
So,
a.
The given inequality is:
a > 4 or a < -3
The representation of the inequalities in the graph is:

b.
The given inequality is:
a < -2 or a > 8
The representation of the inequalities in the graph is:

c.
The given inequality is:
a > 7 or a < -5
The representation of the inequalities in the graph is:

d.
The given inequality is:
a < 6  or a > -9
The representation of the inequalities in the graph is:

From the above,
We can conclude that inequality c) does not belong with the other three inequalities

In Exercises 3–6, write a compound inequality that is represented by the graph.

Question 3.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 71
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.5 Question 3

Question 4.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 72
Answer:
The given graph is:
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 72
From the above graph,
We can observe that the marked line started from 7 excluding 7 and continued till 14 excluding 14
Hence,
The representation of the inequalities from the given graph is:
x > 7 and x < 14
The representation of the inequalities in the form of compound inequality is:
7 < x < 14

Question 5.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 73
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.5 Question 5

Question 6.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 74
Answer:
The given graph is:
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 74
From the above graph,
We can observe that
The first marked line started from 4 including 4 and continued till the end of the left end of the graph
The second marked line started from 6 excluding 6 and continued till the end of the right end of the graph
Hence,
The representation of the given graph in the form of inequality is:
x ≤ 4 or x > 6

In Exercises 7–10, write the sentence as an inequality. Graph the inequality.

Question 7.
A number p is less than 6 and greater than 2.
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.5 Question 7

Question 8.
A number n is less than or equal to -7 or greater than 12.
Answer:
The given worded form is:
A number n is less than or equal to -7 or greater than 12
Hence,
The representation of the given worded form in the form of inequality is:
n ≤ -7 or n > 12
The representation of the inequalities in the graph is:

Question 9.
A number m is more than -7\(\frac{2}{3}\) or at most -10.
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.5 Question 9

Question 10.
A number r is no less than -1.5 and fewer than 9.5.
Answer:
The given worded form is:
A number r is no less than -1.5 and fewer than 9.5
Hence,
The representation of the given worded form in the form of inequality is:
r > -1.5 and r < 9.5
Hence,
The representation of the given worded form in the form of compound inequality is:
-1.5 < r < 9.5
The representation of the compound inequality in the graph is:

Question 11.
MODELING WITH MATHEMATICS
Slitsnails are large mollusks that live in deep waters. They have been found in the range of elevations shown. Write and graph a compound inequality that represents this range.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 75
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.5 Question 11

Question 12.
MODELING WITH MATHEMATICS
The life zones on Mount Rainier, a mountain in Washington, can be approximately classified by elevation, as follows.
Low-elevation forest: above 1700 feet to 2500 feet mid-elevation forest: above 2500 feet to 4000 feet Subalpine: above 4000 feet to 6500 feet
Alpine: above 6500 feet to the summit
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 76
Write a compound inequality that represents the elevation range for each type of plant life.
a. trees in the low-elevation forest zone
b. flowers in the subalpine and alpine zones
Answer:
It is given that the elevations of the life zones on Mount Rainier are as follows:
Low-elevation forest: above 1700 feet to 2500 feet
Mid-elevation forest: above 2500 feet to 4000 feet
Subalpine: above 4000 feet to 6500 feet
Alpine: above 6500 feet to the summit
Let the elevation range for each type of plant life be x
So,
a.
The inequality that represents the elevation range of trees in the low-elevation forest zone is:
x > 1700 feet and x < 2500 feet
Hence,
The representation of the elevation range of trees in the low-elevation forest zone in the form of compound inequality is:
1700 feet < x < 2500 feet
b.
The inequalities that represent the elevation range of trees in the subalpine and alpine zones is:
Subalpine zone: x > 4000 feet and x < 6500 feet
Alpine zone: x > 6500 feet
Hence,
The representation of the elevation range of trees in the Subalpine zone and Alpine zone in the form of compound inequality is:
Subalpine zone: 4000 feet < x < 6500 feet
Alpine zone: x > 6500 feet

In Exercises 13–20, solve the inequality. Graph the solution.

Question 13.
6 < x + 5 ≤ 11
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.5 Question 13

Question 14.
24 > -3r ≥ -9
Answer:
The given inequality is:
24 > -3r ≥ -9
Divide by 3 on both sides
( 24 / 3 ) > ( -3r / 3 ) ≥ ( -9 / 3 )
8 > -r ≥ -3
Multiply with ‘-‘ on both sides
-8 < r ≤ 3 [ Because when we multiply with ‘-‘, we have to change the inequality signs also ]
Hence,
The representation of the solution of the given inequality in the graph is:

Question 15.
v + 8 < 3 or -8v < -40
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.5 Question 15

Question 16.
-14 > w + 3 or 3w ≥ -27
Answer:
The given inequality is:
-14 > w + 3 or 3w ≥ -27
So,
-14 – 3 > w or w ≥ -27 / 3
-17 > w or w ≥ -9
w < -17 or w ≥ -9
Hence,
The representation of the solutions of the given inequality in the graph is:

Question 17.
2r + 3 < 7 or -r + 9 ≤ 2
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.5 Question 17

Question 18.
-6 < 3n + 9 < 21
Answer:
The given inequality is:
-6 < 3n + 9 < 21
Subtract with 9 on both sides
-6 – 9 < 3n + 9 – 9 < 21 – 9
-15 < 3n < 12
Divide by 3 on both sides
( -15 / 3 ) < ( 3n / 3 ) < ( 12 / 3 )
-5 < n < 4
Hence,
The representation of the solution of the inequality in the graph is:

Question 19.
-12 < \(\frac{1}{2}\)(4x + 16) < 18
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.5 Question 19

Question 20.
35 < 7(2 – b) or \(\frac{1}{3}\)(15b – 12) ≥ 21
Answer:
The given inequality is:
35 < 7 (2 – b) or \(\frac{1}{3}\) ( 15b – 12 ) ≥ 21
So,
35 < 7 (2) – 7 (b) or 15b – 12 ≥ 21 (3)
35 < 14 – 7b or 15b – 12 ≥ 63
7b < 14 – 35 or 15b ≥ 63 + 12
7b < -21 or 15b ≥ 75
b < -21 / 7 or b ≥ 75 / 15
b < -3 or b ≥ 5
Hence,
The representation of the solutions of the given inequality in the graph is:

ERROR ANALYSIS
In Exercises 21 and 22, describe and correct the error in solving the inequality or graphing the solution.

Question 21.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 77
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.5 Question 21

Question 22.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 78
Answer:
The given inequality is:
x – 2 > 3 or x + 8 < -2
So,
x > 3 + 2 or x < -2 – 8
x > 5 or x < -10
Hence,
The representation of the solutions of the inequality in the graph is:

Question 23.
MODELING WITH MATHEMATICS
Write and solve a compound inequality that represents the possible temperatures (in degrees Fahrenheit) of the interior of the iceberg.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 79
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.5 Question 23

Question 24.
PROBLEM-SOLVING
A ski shop sells skis with lengths ranging from 150 centimeters to 220 centimeters. The shop says the length of the skis should be about 1.16 times a skier’s height (in centimeters). Write and solve a compound inequality that represents the heights of skiers the shop does not provide skis for.
Answer:
The compound inequality that represents the heights of skiers the shop does not provide skis for is:
129.3 centimeters > h > 189.7 centimeters

Explanation:
It is given that a ski shop sells skis with lengths ranging from 150 centimeters to 220 centimeters.
It is also given that the shop says the length of the skis should be about 1.16 times a skier’s height (in centimeters).
Now,
Let x be the possible length of the skis
Let h be the possible height of the skiers for whom the shop does not provide skis for
So,
The inequality that represents the lengths of the skis is:
x < 150 centimeters or x > 220 centimeters
So,
The compound inequality that represents the length of the skis is:
150 centimeters > x > 220 centimeters
But,
x = 1.16h
So,
150 centimeters > 1.16h > 220 centimeters
(150 / 1.16) centimeters > h > (220 / 1.16) centimeters
129.3 centimeters > h > 189.7 centimeters
Hence, from the above,
We can conclude that the compound inequality that represents the heights of skiers the shop does not provide skis for is:
129.3 centimeters > h > 189.7 centimeters

In Exercises 25–30, solve the inequality. Graph the solution, if possible.

Question 25.
22 < -3c + 4 < 14
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.5 Question 25

Question 26.
2m – 1 ≥ 5 or 5m > -25
Answer:
The given inequality is:
2m – 1≥ 5 or 5m > -25
So,
2m ≥ 5 + 1 or m > -25 / 5
2m ≥ 6 or m > -5
m ≥ 6 / 2 or m > -5
m ≥ 3 or m > -5
Hence, from the above
We can conclude that the solutions to the given inequality are:
m ≥ 3 or m > -5
The compound inequality of the solutions of the given inequality is:
m > -5
The representation of the solutions of the given inequality in the graph is:

Question 27.
-y + 3 ≤ 8 and y + 2 > 9
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.5 Question 27

Question 28.
x – 8 ≤ 4 or 2x + 3 > 9
Answer:
The given inequality is:
x – 8 ≤ 4 or 2x + 3 > 9
So,
x ≤ 4 + 8 or 2x >  9 – 3
x ≤ 12 or 2x > 6
x ≤ 12 or x > 6 / 2
x ≤ 12 or x > 3
Hence, from the above,
We can conclude that the solutions to the given inequality are:
x ≤ 12 or x > 3
The compound inequality of the solutions of the given inequality is:
3 < x ≤ 12
The representation of the solutions of the inequalities in the graph is:

Question 29.
2n + 19 ≤ 10 + n or -3n + 3 < -2n + 33
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.5 Question 29

Question 30.
3x – 18 < 4x – 23 and x – 16 < -22
Answer:
The given inequality is:
3x – 18 < 4x – 23 and x – 16 < -22
So,
3x – 4x < -23 + 18 and x < -22 + 16
-x < -5 and x < – 6
x < 5 and x < -6
Hence, from the above,
We can conclude that the solutions to the given inequality are:
x < 5 and x < -6
The compound inequality that represents the solutions of the given inequality is:
x < 5
The representation of the solutions of the given inequality in the graph is:

Question 31.
REASONING
Fill in the compound inequality Big Ideas Math Answer Key Algebra 1 Chapter 2 Solving Linear Inequalities 110 and 5(x + 2) ≥ 2(x + 8) with <, ≤, >, or ≥ so that the solution is only one value.
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.5 Question 31

Question 32.
THOUGHT-PROVOKING
Write a real-life story that can be modeled by the graph.
Answer:
The real-life situation that can be modeled by the graph is:
Suppose we have 100 children in a school. Out of the 100 children, there are 60 boys and the rest are girls.
The attendance of all the boys in a particular month is 50% and all of the girls is 70%. Mark the percentage attendance that is between the attendance of all the boys and girls?
The graph representing the above real-life situation is:

Question 33.
MAKING AN ARGUMENT
The sum of the lengths of any two sides of a triangle is greater than the length of the third side. Use the triangle shown to write and solve three inequalities. Your friend claims the value of x can be 1. Is your friend correct? Explain.
Big Ideas Math Answer Key Algebra 1 Chapter 2 Solving Linear Inequalities 111
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.5 Question 33

Question 34.
HOW DO YOU SEE IT?
The graph shows the annual profits of a company from 2006 to 2013.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 80
a. Write a compound inequality that represents the annual profits from 2006 to 2013.
Answer:
The given graph is:
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 80
From the graph,
We can observe that the annual profits from 2006 to 2013
So,
The minimum value in the graph is: $65 million
The highest value in the graph is: $100 million
Let x be the annual profit
Hence,
The inequalities that represent the annual profits from 2006 to 2013 are:
x ≥ $65 million and x ≤ $100 million
So,
The compound inequality that represents the annual profits from 2006 to 2013 is:
$65 million ≤ x ≤ $100 million

b. You can use the formula P = R – C to find the profit P, where R is the revenue and C is the cost. From 2006 to 2013, the company’s annual cost was about $125 million. Is it possible the company had annual revenue of $160 million from 2006 to 2013? Explain.
Answer:
Yes, it is possible that the company had annual revenue of $160 million from 2006 to 2013

Explanation:
From part (a),
The total profit from 2006 to 2013 is:
P = 65 + 70 + 85 + 50 + 65 + 70 + 90
= $575 million
It is given that you can use the formula P = R – C to find the profit P, where R is the revenue and C is the cost.
It is also given that from 2006 to 2013, the company’s annual cost was about $125 million.
So,
P = R – C
R = P + C
R = 575 + 125
= $700 million
Hence, from the above,
We can conclude that it is possible that the company had annual revenue of $160 million from 2006 to 2013

Maintaining Mathematical Proficiency

Solve the equation. Graph the solutions, if possible.

Question 35.
| \(\frac{d}{9}\) | = 6
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.5 Question 35

Question 36.
7 | 5p – 7 | = -21
Answer:
The given absolute value equation is:
7 | 5p – 7 | = 21
So,
| 5p – 7 | = -21 / 7
| 5p – 7 | = -3
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
5p – 7 = -3 or 5p – 7 = – (-3)
5p = -3 + 7 or 5p = 3 + 7
5p = 4 or 5p = 10
p = \(\frac{4}{5}\) or p = 2
p = 0.8 or p = 2
p = 1 [Approx. value] or p = 2
Hence, from the above,
We can conclude that the solutions to the given absolute value equation are:
p = 1 or p = 2
The representation of the solutions of the given absolute value equation in the graph is:

Question 37.
| r + 2 | = | 3r – 4 |
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.5 Question 37.1
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.5 Question 37.2

Question 38.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 81
Answer:
The given absolute value equation is:
| \(\frac{1}{2}\)w – 6 | = | w + 7 |
So,
\(\frac{1}{2}\)w – 6 = w + 7
\(\frac{1}{2}\)w – w = 7 + 6
–\(\frac{1}{2}\)w = 13
w = 13 (-2)
w = -26
Hence, from the above,
We can conclude that the solution to the given absolute value equation is:
w = -26
The representation of the solution of the given absolute value equation in the graph is:

Find and interpret the mean absolute deviation of the data.

Question 39.
1, 1, 2, 5, 6, 8, 10, 12, 12, 13
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.5 Question 39

Question 40.
24, 26, 28, 28, 30, 30, 32, 32, 34, 36
Answer:
The given numbers are:
24, 26, 28, 28, 30, 30, 32, 32, 34, 36
We know that,
Mean = \(\frac{The sum of the given numbers}{The total number of numbers}\)
= \(\frac{24 + 26+ 28 + 28 + 30 + 30 + 32 + 32 + 34 + 36}{10}\)
= \(\frac{300}{10}\)
= 30
Now,
We know that,
Absolute deviation = ( The mean ) – ( Given number )
So,
Absolute deviations are:
30 – 24, 30 – 26, 30 – 28, 30 – 28, 30 – 30, 30 – 30, 30 – 32, 30 – 32, 30 – 34, 30 – 36
= 6, 4, 2, 2, 0, 0, -2, -2, -6, -4
So,
Absolute Mean deviation = \(\frac{Sum of Absolute deviations}{The numebr of absolute deviations}\)
= \(\frac{6 + 4 + 2 + 2 + 0 + 0 – 2 – 2 – 6 – 4}{10}\)
= \(\frac{0}{10}\)
= 0
Hence, from the above,
We can conclude that there is no deviation

Lesson 2.6 Solving Absolute Value Inequalities

Essential Question
How can you solve an absolute value inequality? Solving an Absolute Value Inequality Algebraically
EXPLORATION 1
Solving an Absolute Value Inequality Algebraically
Work with a partner.

Consider the absolute value inequality | x + 2 | ≤ 3.
a. Describe the values of x + 2 that make the inequality true. Use your description to write two linear inequalities that represent the solutions of the absolute value inequality.
b. Use the linear inequalities you wrote in part (a) to find the solutions of the absolute value inequality.
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 82
c. How can you use linear inequalities to solve an absolute value inequality?
Answer:
The given absolute value equation is:
| x + 2 | ≤ 3
a.
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
x + 2 ≤ 3 or x + 2 ≤ -3
x ≤ 3 – 2 or x ≤ -3 – 2
x ≤ 1 or x ≤ -5
Hence, from the above,
We can conclude that the solutions to the given absolute value equation is:
x ≤ 1 or x ≤ -5

b.
The given absolute value equation is:
| x + 2 | ≤ 3
From part (a),
The solutions of the given absolute value equation are:
x ≤ 1 or x ≤ -5
We know that,
| x | = x for x > 0
| x | = -x for x < 0
Now,
For x = 1,
| 1 + 2 | ≤ 3
| 3 | ≤ 3
3 ≤ 3
For x = -5,
| -5 + 2 | ≤ 3
| -3 | ≤ 3
3 ≤ 3

c.
The following are the steps to solve the absolute value inequality:
A) Isolate the absolute value expression on the left side of the inequality
B) If the number on the other side of the inequality sign is negative, then your equation either has no solution or all real numbers as solutions
C) Remove the absolute value bars by setting up a compound inequality
D) Solve the inequalities

EXPLORATION 2
Solving an Absolute Value Inequality Graphically
Work with a partner.
Consider the absolute value inequality
| x + 2 | ≤ 3.
a. On a real number line, locate the point for which | x + 2 | ≤ 3
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 83
Answer:
The given absolute value inequality is:
| x + 2 | ≤ 3
From Exploration 1,
The solutions of the absolute value inequality | x + 2 | ≤ 3 are:
x ≤ 1 and x ≤ -5
The compound inequality of the solution of the given absolute value inequality is:
x ≤ 1 ( Since the absolute value inequality can never be negative )
Now,
The given graph is:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 83
Hence,
The representation of the solutions of the absolute value inequality is:

b. Locate the points that are within 3 units from the point you found in part (a). What do you notice about these points?
Answer:
From part (a),
The compound inequality that represents the solution of the given absolute value inequality is:
x ≤ 1
The points that are within 3 units from the point x ≤ 1, i.e., the given point + 3
So,
1 + 3 = 4
Hence,
The point we have to locate in the graph is:
x ≤ 4
Hence,
The representation of the compound inequality of the given absolute value inequality in the graph is:

c. How can you use a number line to solve an absolute value inequality?
Answer:
You begin the marking of the points on the number line by making it into separate equations and then solving them separately. An absolute value equation has no solution if the absolute value expression equals a negative number since an absolute value can never be negative. You can write an absolute value inequality as a compound value inequality

EXPLORATION 3
Solving an Absolute Value Inequality Numerically
Work with a partner.

Solving an Absolute Value Inequality Numerically
Work with a partner. Consider the absolute value inequality | x + 2 | ≤ 3.
a. Use a spreadsheet, as shown, to solve the absolute value inequality.
b. Compare the solutions you found using the spreadsheet with those you found in Explorations 1 and 2. What do you notice?
c. How can you use a spreadsheet to solve an absolute value inequality?
Answer:

Communicate Your Answer

Question 4.
How can you solve an absolute value inequality?
Answer:
Isolate the absolute value expression on the left side of the inequality. If the number on the other side of the inequality sign is negative, your equation either has no solution or all real numbers as solutions. Use the sign of each side of your inequality to decide which of these cases holds.

Question 5.
What do you like or dislike about the algebraic, graphical, and numerical methods for solving an absolute value inequality? Give reasons for your answers.
Answer:
We have to use algebraic, graphical, and numerical methods for solving an absolute value inequality depending on the situation
The algebraic and graphical methods for solving an absolute value inequality are useful when the size of the inequality is small i.e., have only 1 variable

2.6 Lesson

Monitoring Progress

Solve the inequality. Graph the solution, if possible

Question 1.
| x | ≤ 3.5
Answer:
The given absolute value inequality is:
| x | ≤ 3.5
We know that,
| x | =x for x > 0
| x | = -x for x < 0
So,
x ≤ 3.5
x ≤ 4 [Approximate value}
Hence, from the above,
We can conclude that the solution of the given absolute value inequality is:
x ≤ 4 [Approximate value]
The representation of the solution of the given absolute value inequality in the graph is:

Question 2.
| k – 3 | < -1
Answer:
The given absolute value inequality is:
| k – 3 | < -1
We know that,
If the number on the other side of the absolute value expression is negative, then the given absolute value equation has no solution or has real numbers as a solution
Hence, from the above,
We can conclude that the given absolute value inequality has no solution

Question 3.
| 2w – 1 | < 11
Answer:
The given absolute value inequality is:
| 2w – 1 | < 11
We know that,
| x | =x for x > 0
| x | = -x for x < 0
So,
2w – 1 < 11
2w < 11 + 1
2w < 12
w < 12 /  2
w < 6
Hence, from the above,
We can conclude that the solution of the given absolute value inequality is:
w < 6
The representation of the solution of the given absolute value inequality in the graph is:

Question 4.
| x + 3 | > 8
Answer:
The given absolute value inquality is:
| x + 3 | > 8
We know that,
| x | = xfor x > 0
| x | = -x for x < 0
So,
x + 3 > 8
x > 8 – 3
x > 5
Hence, from the above,
We can conclude that the solution of the given absolute value inequality is:
x > 5
The representation of the solution of the given absolute value inequality in the graph is:

Question 5.
| n + 2 | – 3 ≥ -6
Answer:
The given absolute value inequality is:
| n + 2 | – 3 ≥ -6
So,
| n + 2 | ≥ -6 + 3
| n + 2 | ≥ -3
We know that,
If the number on the other side of the absolute value expression is negative, then the given absolute value equation has no solution or has real numbers as a solution
Hence, from the above,
We can conclude that the given absolute value inequality has no solution

Question 6.
3 | d + 1 | – 7 ≥ -1
Answer:
The given absolute value inequality is:
3 | d + 1 | – 7 ≥ -1
So,
3 | d + 1 | ≥ -1 + 7
3 | d + 1 | ≥ 6
| d + 1 | ≥ 6 / 3
| d + 1 | ≥ 2
We know that,
| x | = x for x > 0
| x | = -x for x < 0
Hence,
d + 1 ≥ 2
d ≥ 2 – 1
d ≥ 1
Hence, from the above,
We can conclude that the solution of the given absolute value inequality is:
d ≥ 1
The representation of the solution of the given inequality in the graph is:

Question 7.
WHAT IF?
You are willing to pay the mean price with an absolute deviation of at most $75. How many of the computer prices meet your condition?
Answer:

Solving Absolute Value Inequalities 2.6 Exercises

Question 1.
REASONING
Can you determine the solution of | 4x – 2 | ≥ -6 without solving it? Explain
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.6 Question 1

Question 2.
WRITING
Describe how solving | w – 9 | ≤ 2 is different from solving | w – 9 | ≥ 2.
Answer:
The given absolute value inequalities are:
| w – 9 | ≤ 2 and | w – 9 | ≥ 2
Now,
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
w – 9 ≤ 2 and w – 9 ≥ 2
w ≤ 2 + 9 and w ≥ 2 + 9
w ≤ 11 and w ≥ 11
w – 9 ≤ -2 and w – 9 ≥ -2
w ≤ -2 + 9 and w ≥ -2 + 9
w ≤ 7 and w ≥ 7
Hence, from the above,
We can conclude that the given absolute value inequalities are different by observing the above values for the 2 absolute value inequalities

Monitoring Progress and Modeling with Mathematics 

In Exercises 3–18, solve the inequality. Graph the solution, if possible.

Question 3.
| x | < 3
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.6 Question 3

Question 4.
| y | ≥ 4.
Answer:
The given absolute value inequality is:
| y | ≥ 4
We know that,
| x | = x for x > 0
| x | = -x for x < 0
Hence,
The solutions for the given absolute value inequality is:
y ≥ 4 or y ≤ -4
The representation of the compound inequality solution of the given absolute value inequality is:

Question 5.
| d + 9 | > 3
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.6 Question 5

Question 6.
| h – 5 | ≤ 10
Answer:
The given absolute value inequality is:
| h – 5 | ≤ 10
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
h – 5 ≤ 10
h ≤ 10 + 5 and h ≥ -10 + 5
h ≤ 15 and h ≥ -5
Hence,
The compound inequality of the solutions of the given absolute value inequality are:
-5 ≤  h ≤  15
The representation of the compound inequality in the graph is:

Question 7.
| 2s – 7 | ≥ -1
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.6 Question 7

Question 8.
| 4c + 5 | > 7
Answer:
The given absolute value inequality is:
| 4c + 5 | > 7
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
4c + 5 > 7 or 4c + 5 < -7
4c > 7 – 5 or 4c < -7 – 5
4c > 2 or 4c < -12
c > \(\frac{2}{4}\) or c < –\(\frac{4}{12}\)
c > \(\frac{1}{2}\) or c < –\(\frac{1}{3}\)
Hence,
The solutions of the given absolute value equation are:
c > \(\frac{1}{2}\) or c < –\(\frac{1}{3}\)
c > 1 or c < 0 [ Approximate values of the inequality ]
The representation of the solutions of the given absolute value inequality in the graph is:

Question 9.
| 5p + 2 | < -4
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.6 Question 9

Question 10.
| 9 – 4n | < 5
Answer:
The given absolute value inequality is:
| 9 – 4n | < 5
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
9 – 4n < 5 and 9 – 4n > -5
9 – 5 < 4n and 9 + 5 > 4n
4 < 4n and 14 > 4n
4n > 4 and 4n < 14
n > 4/ 4 and n < 14 / 4
n > 1 and n < 7 / 2
n > 1 and n < 3.5
Hence,
The solutions of the given absolute value inequality are:
n > 1 and n < 3.5
The compound solution of the solutions of the given absolute value inequality is:
1 < n < 3.5
The representation of the compound inequality of the solutions of the given inequality in the graph is:

Question 11.
| 6t – 7 | – 8 ≥ 3
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.6 Question 11

Question 12.
| 3j – 1 | + 6 > 0
Answer:
The given absolute value inequality is:
| 3j – 1 | + 6 > 0
So,
| 3j – 1 | > 0 – 6
| 3j – 1 | > -6
We know that,
By definition of absolute values, the absolute value of an expression must be greater than or equal to 0
So,
The expression | 3j – 1 |  will always be greater than -6
Hence,
For the given absolute value inequality, all the real numbers are solutions

Question 13.
3 | 14 – m | > 18
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.6 Question 13

Question 14.
-4 | 6b – 8 | ≤ 12
Answer:
The given absolute value inequality is:
-4 | 6b – 8 | ≤ 12
So,
| 6b – 8 | ≤ -3
We know that,
By definition of absolute values, the absolute value of an expression must be greater than or equal to 0.
So,
The expression | 6b – 8 | must not be less than -3
Hence,
For the given absolute value inequality, there is no solution

Question 15.
2 | 3w + 8 | ≥ 13
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.6 Question 15

Question 16.
3 | 2 – 4u | + 5 < -13
Answer:
The given absolute value inequality is:
3 | 2 – 4u | + 5 < -13
So,
3 | 2 – 4u | < -13 – 5
3 | 2 – 4u | < -18
| 2 – 4u | < -6
We know that,
By definition of absolute values, the absolute value of an expression must be greater than or equal to 0.
So,
The expression | 2 – 4u | must not be less than -6
Hence,
For the given absolute value inequality, there is no solution

Question 17.
6 | -f + 3 | + 7 > 7
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.6 Question 17

Question 18.
\(\frac{2}{3}\) |4v + 6 | – 2 ≤ 10
Answer:
The given absolute value inequality is:
\(\frac{2}{3}\) | 4v + 6 | – 2 ≤ 10
\(\frac{2}{3}\) | 4v + 6 | ≤ 10 + 2
\(\frac{2}{3}\) | 4v + 6 | ≤ 12
| 4v + 6 | ≤ 12 × \(\frac{3}{2}\)
| 4v + 6 | ≤ 18
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
4v + 6 ≤ 18 and 4v + 6 ≥ -18
4v ≤ 18 – 6 and 4v ≥ -18 – 6
4v ≤ 12 and 4v ≥ -24
v ≤ 12 / 4 and v ≥ -24 / 4
v ≤ 3 and v ≥ -6
Hence, from the above,
We can conclude that the solutions of the given absolute value inequality are:
v ≤ 3 and v ≥ -6
The representation of the compound inequality of the solutions of the given absolute value inequality is:
-6 ≤ v ≤ 3
The representation of the compound inequality in the graph is:

Question 19.
MODELING WITH MATHEMATICS
The rules for an essay contest say that entries can have 500 words with an absolute deviation of at most 30 words. Write and solve an absolute value inequality that represents the acceptable numbers of words.
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.6 Question 19

Question 20.
MODELING WITH MATHEMATICS
The normal body temperature of a camel is 37°C. This temperature varies by up to 3°C throughout the day. Write and solve an absolute value inequality that represents the range of normal body temperatures (in degrees Celsius) of a camel throughout the day.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 84
Answer:
The absolute value inequality that represents the range of normal body temperatures of a camel throughout the day is:
34°C ≤ t ≤ 40°C

Explanation:
It is given that the normal body temperature of a camel is 37°C and this temperature varies by up to 3°C throughout the day.
Let t be the temperature in °C
So,
The absolute value inequality that represents the range of normal body temperatures of a camel throughout the day is:
37 – 3 ≤ t ≤ 37 + 3
34°C ≤ t ≤ 40°C

ERROR ANALYSIS
In Exercises 21 and 22, describe and correct the error in solving the absolute value inequality.

Question 21.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 85
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.6 Question 21

Question 22.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 86
Answer:
The given absolute value inequality is:
| x + 4 | > 13
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
x + 4 > 13 or x + 4 < -13
x > 13 – 4 or x < -13 – 4
x > 9 or x < -17
Hence, from the above,
We can conclude that the solutions of the given absolute value inequality are:
x > 9 or x < -17

In Exercises 23–26, write the sentence as an absolute value inequality. Then solve the inequality.

Question 23.
A number is less than 6 units from 0.
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.6 Question 23

Question 24.
A number is more than 9 units from 3.
Answer:
The given worded form is:
A number is more than 9 units from 3
Let n be the number
Hence,
The representation of the given worded form in the form of absolute value inequality is:
| n – 3 | > 9
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
n – 3 > 9 or n – 3 < -9
n > 9 + 3 or n < -9 + 3
n > 12 or n < -6
Hence, from the above,
We can conclude that the solutions o the given absolute value inequality is:
n > 12 or n < -6

Question 25.
Half of a number is at most 5 units from 14.
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.6 Question 25

Question 26.
Twice a number is no less than 10 units from -1.
Answer:
The given worded form is:
Twice a number is no less than 10 units from -1
Let x be the number
So,
The representation of the given worded form in the form of absolute value inequality is:
| 2x – (-1) | > 10
So,
| 2x + 1 | > 10
We know that,
| x | = x for x > 0
| x | = -x for x < 0
Hence,
2x + 1 > 10 or 2x + 1 < -10
2x > 10 – 1 or 2x < -10 – 1
2x > 9 or 2x < -11
x > \(\frac{9}{2}\) or x < –\(\frac{11}{2}\)
Hence, from the above,
We can conclude that the solutions of the given absolute value inequality is:
x > \(\frac{9}{2}\) or x < –\(\frac{11}{2}\)

Question 27.
PROBLEM SOLVING
An auto parts manufacturer throws out gaskets with weights that are not within 0.06 pound of the mean weight of the batch. The weights (in pounds) of the gaskets in a batch are 0.58, 0.63, 0.65, 0.53, and 0.61. Which gasket(s) should be thrown out?
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.6 Question 27

Question 28.
PROBLEM-SOLVING
Six students measure the acceleration (in meters per second ) of an object in free fall. The measured values are shown. The students want to state that the absolute deviation of each measured value x from the mean is at most d. Find the value of d.
10.56, 9.52, 9.73, 9.80, 9.78, 10.91
Answer:
The value of d is: 0.86

Explanation:
The given numbers are:
10.56, 9.52, 9.73, 9.80, 9.78, 10.91
We know that,
Mean = \(\frac{Sum of the given observations}{Number of observations}\)
= \(\frac{10.56 + 9.52 + 9.73 + 9.80 + 9.78 + 10.91}{6}\)
= \(\frac{60.3}{6}\)
= 10.05
Now,
We know that,
Absolute deviation = Mean – ( Given observation)
It is given that that the measured value is x
Hence,
The representation of the absolute deviation in the form of absolute value inequality is:
| x – 10.05 | ≤ d
So,
| 10.56 – 10.05 | ≤ d
| 9.52 – 10.05 | ≤ d
| 9.73 – 10.05 | ≤ d
| 9.80 – 10.05 | ≤ d
| 9.78 – 10.05 | ≤ d
| 10.91 – 10.05 | ≤ d
So,
d ≥ 0.51
d ≥ 0.53
d ≥ 0.32
d ≥ 0.25
d ≥ 0.27
d ≥ 0.86
Hence, from the above,
We can conclude that the value of d will be the highest deviation from the mean value
The value of d is: 0.86

MATHEMATICAL CONNECTIONS
In Exercises 29 and 30, write an absolute value inequality that represents the situation. Then solve the inequality.

Question 29.
The difference between the areas of the figures is less than 2.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 86.1
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.6 Question 29

Question 30.
The difference between the perimeters of the figures is less than or equal to 3.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 87
Answer:
The given figures are:
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 87
From the above,
The given figures are: Rectangle and square
We know that,
The perimeter of a rectangle = 2 ( Length + Width )
The perimeter of a square = 4 ( Side )
So,
The perimeter of a rectangle = 2 ( 3 + x + 1 ) = 2 ( x + 4 )
The perimeter of a square = 4 (x) = 4x
It is given that the difference of the perimeters of the given figures is less than or equal to 3
So,
| 2 ( x + 4 ) – 4x | ≤ 3
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
2 (x + 4) – 4x ≤3 and 2 ( x + 4 ) – 4x ≥ -3
2x + 8 – 4x ≤ 3 and 2x + 8 – 4x ≥ -3
8 – 2x ≤ 3 and 8- 2x ≥ -3
-2x ≤ 3 – 8 and -2x ≥ -3 – 8
-2x ≤ -5 and -2x ≥ -11
2x ≤ 5 and 2x ≥ 11
x ≤ \(\frac{5}{2}\) and x ≥ \(\frac{11}{2}\)
Hence, from the above,
We can conclude that the compound inequality for the solutions of the given absolute value inequality is:
\(\frac{11}{2}\) ≤ x ≤ \(\frac{5}{2}\)

REASONING
In Exercises 31–34, tell whether the statement is true or false. If it is false, explain why.

question 31.
If a is a solution of | x + 3 | ≤ 8, then a is also a solution of x + 3 ≥ -8.
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.6 Question 31

Question 32.
If a is a solution of | x + 3 | > 8, then a is also a solution of x + 3 > 8.
Answer:
The given statement is true

Explanation:
The given statement is:
If a is a solution of | x + 3 | > 8, then a is also a solution of x + 3 > 8
Now,
The given inequality is:
| x + 3 | > 8
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
x + 3 > 8 or x + 3 < -8
We know that,
The value of the absolute value expression must be greater than or equal to 0
So,
x + 3 < -8 has no solution
So,
a is a solution of x + 3 > 8
Hence, from the above,
We can conclude that the given statement is true

Question 33.
If a is a solution of | x + 3 | ≥ 8, then a is also a solution of x + 3 ≥ -8.
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.6 Question 33

Question 34.
If a is a solution of x + 3 ≤  -8, then a is also a solution of | x + 3 | ≥ 8.
Answer:
The given statement is true

Explanation:
The given statement is:
If a is a solution of x + 3 ≤ -8, then a is a solution of | x + 3 | ≥ 8
Now,
The given absolute value inequality is:
| x + 3 | ≥ 8
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
x + 3 ≥ 8 or x + 3 ≤ -8
If a is a solution of x + 3  ≥ 8, then a is also a solution of x + 3 ≤ -8
Hence, from the above,
We can conclude that the given statement is true

Question 35.
MAKING AN ARGUMENT
One of your classmates claims that the solution of | n | > 0 is all real numbers. Is your classmate correct? Explain your reasoning.
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.6 Question 35

Question 36.
THOUGHT-PROVOKING
Draw and label a geometric figure so that the perimeter P of the figure is a solution to the inequality | P – 60 | ≤ 12.
Answer:
The given absolute value inequality is:
| p – 60 | ≤ 12
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
p – 60 ≤ 12 and p – 60 ≥ -12
p ≤ 12 + 60 and p ≥ -12 + 60
p ≤ 72 and p ≥ 48
Hence,
The representation of the compound inequality of the perimeter of the figure is:
48 ≤ p ≤ 72
Now, find the value of 4p i.e., the perimeter of the square
So,
192 ≤ 4p ≤ 288
Hence, from the above,
We can conclude that the perimeter is represented for the perimeter of the square

Question 37.
REASONING
What is the solution of the inequality | ax + b | < c, where c < 0? What is the solution of the inequality | ax + b | > c, where c < 0? Explain.
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.6 Question 37

Question 38.
HOW DO YOU SEE IT?
Write an absolute value inequality for each graph.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 88
How did you decide which inequality symbol to use for each inequality?
Answer:
The given graphs are:
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 88
a.
From the given graph,
The 1st marked line starts from -1 including -1 and continued till the left end of the number line
The 2nd marked line starts from 5 including from 5 and continued till the right end of the number line
Hence,
The representation of the marked lines in the form of absolute value inequality is:
| x + 1 | ≤ 3 and | x – 5 | ≥ 1

b.
From the given graph,
The marked line starts from -1 excluding -1 and continued till 5 on the right end of the number line
Hence,
The representation of the marked line in the form of absolute value inequality is:
| x + 1 | > 6
c.
From the given graph,
The marked line starts from -1 including -1 and continued till 5 on the right end of the number line
Hence,
The representation of the marked line in the form of absolute value inequality is:
| x + 1 | ≥ 6
d.
Fro the given graph,
The 1st marked line starts from -1 excluding -1 and continued till the left end of the number line
The 2nd marked line starts from 5 excluding 5 and continued till the right end of the number line
Hence,
The representation of the marked lines in the form of absolute value inequality is:
| x + 1 | < 3 and | x – 5 | > 1
If the marked point is indicated on the number line and if that marked line continued till the right end of the number line, then we will use ≥
If the marked point is not indicated on the number line and if that marked line continued till the right end of the number line, then we will use >
If the marked point is indicated on the number line and if that marked line continued till the left end of the number line, then we will use ≤
If the marked point is not indicated on the number line and if that marked line continued till the left end of the number line, then we will use <

Question 39.
WRITING
Explain why the solution set of the inequality | x | < 5 is the intersection of two sets, while the solution set of the inequality | x | > 5 is the union of two sets.
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.6 Question 39

Question 40.
PROBLEM-SOLVING
Solve the compound inequality below. Describe your steps.
| x – 3 | < 4 and | x + 2 | > 8
Answer:
The given inequalities are:
| x + 3 | < 4 and | x + 2  | > 8
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
x + 3 < 4 or x + 3 > -4 and x + 2 > 8 or x + 2 < -8
x < 4 – 3 or x > -4 – 3 and x > 8 – 2 or x < -8 – 2
x < 1 or x > -7 and x > 6 or x < -10
Hence,
The compound inequality of the solutions of the given absolute value inequalities are:
-7 < x < 1 and -10 > x > 6

Maintaining Mathematical Proficiency

Plot the ordered pair in a coordinate plane. Describe the location of the point.

Question 41.
A(1, 3)
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.6 Question 41

Question 42.
B(0, -3)
Answer:
In the graph,
X-axis:
The right side of the x-axis will be: +ve
The left side of the x-axis will be: -ve
Y-axis:
The top side of the y-axis will be: +ve
The downside of the y-axis will be: -ve
We know that,
The graph is divided into 4 quadrants. They are:
1st Quadrant: x: +ve and y: -ve
2nd Quadrant: x: -ve and y: +ve
3rd Quadrant: x: -ve and y: -ve
4th Quadrant: x: +ve and y: -ve
Hence,
B(0, -3) will be in the 4th Quadrant
The representation of point B in the graph is:

Question 43.
C(-4, -2)
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.6 Question 43

Question 44.
D(-1, 2)
Answer:
In the graph,
X-axis:
The right side of the x-axis will be: +ve
The left side of the x-axis will be: -ve
Y-axis:
The top side of the y-axis will be: +ve
The downside of the y-axis will be: -ve
We know that,
The graph is divided into 4 quadrants. They are:
1st Quadrant: x: +ve and y: -ve
2nd Quadrant: x: -ve and y: +ve
3rd Quadrant: x: -ve and y: -ve
4th Quadrant: x: +ve and y: -ve
Hence,
D(-1, 2) will be in the 2nd Quadrant
The representation of point D in the graph is:

Copy and complete the table.

Question 45.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 89
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.6 Question 45

Question 46.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 90
Answer:
The completed table is:

Solving Linear Inequalities Performance Task: Grading Calculations

2.5–2.6 What Did You Learn?

Core Vocabulary
compound inequality, p. 82
absolute value inequality, p. 88
absolute deviation, p. 90

Core Concepts
Section 2.5
Writing and Graphing Compound Inequalities, p. 82
Solving Compound Inequalities, p. 83

Section 2.6
Solving Absolute Value Inequalities, p. 88

Mathematical Practices

Question 1.
How can you use a diagram to help you solve Exercise 12 on page 85?
Answer:
In Exercise 12 on page 85,
The given diagram represents the maximum elevation of Mount Rainier
In the diagram,
The given maximum elevation of Mount Rainier is: 14,410 ft
So,
This 14,410 ft is divided into different types of elevation in Exercise 12

Question 2.
In Exercises 13 and 14 on page 85, how can you use structure to break down the compound inequality into two inequalities?
Answer:
In Exercises 13 and 14 on page 85,
First, perform the necessary mathematical operations so that only the variable must be present
So,
After performing the necessary mathematical operations, the structure of the compound inequality will be like
a < x < b or a > x > b
The breakdown of the compound inequality a < x < b will be like x >a and x < b
The breakdown of the compound inequality a > x > b will be like x < a and x > b
We can put the different types of inequality symbols in the breakdown of inequalities

Question 3.
Describe the given information and the overall goal of Exercise 27 on page 91.
Answer:
The given information is:
An auto parts manufacturer throws out gaskets with weights that are not within 0.06 pounds of the mean weight of the batch. The weights (in pounds) of the gaskets in a batch are 0.58, 0.63, 0.65, 0.53, and 0.61
The overall goal of Exercise 27:
Find the mean of the given weights and find the absolute deviation of the gaskets
The absolute deviation of the gaskets that are less than 0.0 pounds of the man weight should be thrown out

Question 4.
For false statements in Exercises 31–34 on page 92, use examples to show the statements are false.
Answer:
In Exercises 31 – 34 on page 92,
Exercises 33 and 34 are false
The Exercise 33 is: | x + 3 | ≥ 8
The Exercise 34 is: | x + 3 | ≥ 8
So,
In Exercise 33,
a is a solution for x + 3 ≥ 8 and x + 3 ≤ -8
In Exercise 34,
a is a solution of x + 3 ≥ 8 and x + 3 ≤ -8

Performance Task

Grading Calculations

You are not doing as well as you had hoped in one of your classes. So, you want to figure out the minimum grade you need on the final exam to receive the semester grade that you want. Is it still possible to get an A? How would you explain your calculations to a classmate?
To explore the answers to this question and more, go to Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 91
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 91.1

Solving Linear Inequalities Chapter Review

2.1 Writing and Graphing Inequalities (pp. 53–60)
a. A number x plus 36 is no more than 40. Write this sentence as an inequality.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 91.2
b. Graph w > −3.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 92

Write the sentence as an inequality.

Question 1.
A number d minus 2 is less than -1.
Answer:
The given worded form is:
A number d minus 2 is less than -1
Hence,
The representation of the given worded form in the form of inequality is:
d – 2 < -1

Question 2.
Ten is at least the product of a number h and 5.
Answer:
The given worded form is:
Ten is at least the product of a number h and 5
Hence,
The representation of the given worded form in the form of inequality is:
10 ≥ h(5)
10 ≥ 5h
10 / 5 ≥ h
2 ≥ h
h ≤ 2
Hence, from the above,
We can conclude that the representation of the given worded form in the form of inequality is:
h ≤ 2

Graph the inequality.

Question 3.
x > 4
Answer:
The given inequality is:
x > 4
Hence,
The representation of the given inequality in the graph is:

Question 4.
y ≤ 2
Answer:
The given inequality is:
y ≤ 2
Hence,
The representation of the given inequality in the graph is:

Question 5.
-1 ≥ z
Answer:
The given inequality is:
-1 ≥ z
z ≤ -1
Hence,
The representation of the given inequality in the graph is:

2.2 Solving Inequalities Using Addition or Subtraction (pp. 61–66)

Solve x + 2.5 ≤ −6. Graph the solution.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 93

Solve the inequality. Graph the solution.

Question 6.
p + 4 < 10
Answer:
The given inequality is:
p + 4 < 10
So,
p < 10 – 4
p < 6
Hence, from the above,
We can conclude that the solution to the given inequality is p < 6
The representation of the solution of the given inequality in the graph is:

Question 7.
r – 4 < -6
Answer:
The given inequality is:
r – 4 < -6
So,
r < -6 + 4
r < -2
Hence, from the above,
We can conclude that the solution to the given inequality is r < -2
The representation of the solution of the given inequality in the graph is:

Question 8.
2.1 ≥ m – 6.7
Answer:
The given inequality is:
2.1 ≥ m – 6.7
So,
2.1 + 6.7 ≥ m
8.8 ≥ m
9 ≥ m [Approximate value]
m ≤ 9
Hence, from the above,
We can conclude that the solution to the given inequality is m ≤ 9
The representation of the solution of the given inequality in the graph is:

2.3 Solving Inequalities. Using Multiplication or Division

Solve \(\frac{n}{-10}\) > 5. Graph the solution.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 94

Solve the inequality. Graph the solution.

Question 9.
3x > -21
Answer:
The given inequality is:
3x > -21
So,
x > -21 / 3
x > -7
Hence, from the above,
We can conclude that the solution to the given inequality is x > -7
The representation of the solution of the given inequality in the graph is:

Question 10.
-4 ≤ \(\frac{g}{5}\)
Answer:
The given inequality is:
-4 ≤ \(\frac{g}{5}\)
-4 (5) ≤ g
-20 ≤ g
g ≥ -20
Hence, from the above,
We can conclude that the solution to the given inequality is g ≥ -20
The representation of the solution of the given inequality in the graph is:

Question 11.
–\(\frac{3}{4}\)n ≤ 3
Answer:
The given inequality is:
–\(\frac{3}{4}\)n ≤ 3
n ≤ 3 × –\(\frac{4}{3}\)
n ≤ —\(\frac{3 × 4}{3}\)
n ≤ -4
Hence, from the above,
We can conclude that the solution to the given inequality is n ≤ -4
The representation of the solution of the given inequality in the graph is:

Question 12.
\(\frac{s}{-8}\) ≥ 11
Answer:
The given inequality is:
\(\frac{s}{-8}\) ≥ 11
s ≥ 11 (-8)
s ≥ -88
Hence, from the above,
We can conclude that the solution to the given inequality is s ≥ -88
The representation of the solution of the given inequality in the graph is:

Question 13.
36 < 2q
Answer:
The given inequality is:
36 < 2q
So,
36 / 2 < q
18 < q
q > 18
Hence, from the above,
We can conclude that the solution to the given inequality is q > 18
The representation of the solution of the given inequality in the graph is:

Question 14.
-1.2k > 6
Answer:
The given inequality is:
-1.2k > 6
So,
k > 6 / -1.2
k > -60 / 12
k > -5
Hence, from the above,
We can conclude that the solution to the given inequality is k > -5
The representation of the solution of the given inequality in the graph is:

2.4 Solving Multi-step Inequalities

Solve 22 + 3y ≥ 4. Graph the solution.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 95

Solve the inequality. Graph the solution, if possible.

Question 15.
3x – 4 > 11
Answer:
The given inequality is:
3x – 4 > 11
So,
3x > 11 + 4
3x > 15
x > 15 / 3
x > 5
Hence, from the above,
We can conclude that the solution to the given inequality is x > 5
The representation of the solution of the given inequality in the graph is:

Question 16.
-4 < \(\frac{b}{2}\) + 9
Answer:
The given inequality is:
-4 < \(\frac{b}{2}\) + 9
-4 – 9 < \(\frac{b}{2}\)
-13 < \(\frac{b}{2}\)
-13 (2) < b
-26 < b
b > -26
Hence, from the above,
We can conclude that the solution of the given inequality is b > -26
The representation of the solution of the given inequality in the graph is:

Question 17.
7 – 3n ≤ n + 3
Answer:
The given inequality is:
7 – 3n ≤ n + 3
-3n – n ≤ 3 – 7
-4n ≤ -4
4n ≤ 4
n ≤ 4 / 4
n ≤ 1
Hence, from the above,
We can conclude that the solution to the given inequality is n ≤ 1
The representation of the solution of the given inequality in the graph is:

Question 18.
2(-4s + 2) ≥ -5s – 10
Answer:
The given inequality is:
2 (-4s +2) ≥ -5s – 10
So,
2 (-4s) + 2 (2) ≥ -5s – 10
-8s + 4 ≥ -5s – 10
-8s + 5s ≥ -10 –  4
-3s ≥ -14
3s ≥ 14
s ≥ 14 / 3
s ≥ 5 [ Approximate value }
Hence, from the above,
We can conclude that the solution of the given inequality is s ≥ 5
The representation of the solution of the given inequality in the graph is:

Question 19.
6(2t + 9) ≤ 12t – 1
Answer:
The given inequality is:
6(2t + 9) ≤ 12t – 1
So,
6 (2t) + 6 (9) ≤ 12t – 1
12t + 54 ≤ 12t – 1
1t – 12t + 54 ≤ -1
54 ≤ -1
Hence, from the above,
We can conclude that there is no solution for the given inequality

Question 20.
3r – 8 > 3(r – 6)
Answer:
The given inequality is:
3r – 8 > 3 ( r – 6 )
So,
3r – 8 > 3 (r) – 3 (6)
3r – 8 > 3r – 18
3r – 8 – 3r > -18
-8 > -18
8 > 18
Hence, from the above,
We can conclude that there is no solution to the given inequality

2.5 Solving Compound Inequalities

Solve −1 ≤ −2d + 7 ≤ 9. Graph the solution.
Big Ideas Math Algebra 1 Solutions Chapter 2 Solving Linear Inequalities 96

Question 21.
A number x is more than -6 and at most 8. Write this sentence as an inequality. Graph the inequality.
Answer:
The given worded form is:
A number x is more than -6 and at most 8
The representation of the given worded form in the form of inequality is:
x > -6 and x ≤ 8
Hence,
The representation of the solutions of the given worded form in the form of compound inequality is:
-6 < x ≤ 8
The representation of the compound inequality in the graph is:

Solve the inequality. Graph the solution.

Question 22.
19 ≥ 3z + 1 ≥ -5
Answer:
The given inequality is:
19 ≥ 3z + 1 ≥ -5
Subtract with -1 on both sides
So,
19 – 1 ≥ 3z + 1 – 1 ≥ -5 – 1
18 ≥ 3z ≥ -6
Divide by 3 on both sides
So,
(18 / 3) ≥ (3z / 3) ≥ (-6 / 3)
6 ≥ z ≥ -2
-2 ≤ z ≤ 6
Hence, from the above,
We can conclude that the compound inequality solution to the given inequality is:
-2 ≤ z ≤ 6
The representation of the compound inequality in the graph is:

Question 23.
\(\frac{r}{4}\) < -5 or -2r – 7 ≤ 3
Big Ideas Math Algebra 1 Solutions Chapter 2 Solving Linear Inequalities 97
Answer:
The given inequality is:
\(\frac{r}{4}\) < -5 or -2r – 7 ≤ 3
r < -5 (4) or -2r ≤ 3 + 7
r < -20 or -2r ≤ 10
r < -20 or r ≤ -5
Hence, from the above,
We can conclude that the solutions to the given inequality are:
r < -20 or r ≤ -5
The representations of the solutions of the given inequality in the graph is:

2.6 Solving Absolute Value Inequalities

Solve the inequality. Graph the solution, if possible.

Question 24.
| m | ≥ 10
Answer:
The given absolute value inequality is:
| m | ≥ 10
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
m ≥ 10 or m ≤ -10
We know that,
The value of absolute value expression must be greater than or equal to 0
So,
m ≥ 10
Hence, from the above,
We can conclude that the solution of the given absolute value inequality is m ≥ 10
The representation of the solution of the given absolute value inequality in the graph is:

Question 25.
| k – 9 | < -4
Answer:
The given absolute value inequality is:
| k – 9 | < 4
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
k – 9 < 4 and k – 9 > -4
k < 4 + 9 and k > -4 + 9
k < 13 and k > 5
Hence, from the above,
We can conclude that the solutions of the given absolute value inequality are:
k < 13 and k > 5
The representation of the compound inequality of the solutions of the given absolute inequality is:
5 < k < 13
The representation of the compound inequality in the graph is:

Question 26.
4 | f – 6 | ≤ 12
Answer:
The given absolute value inequality is:
4 | f – 6 | ≤ 12
| f – 6 | ≤ 3
We know that,
| x | = x for x > 0
| X | = -x for x < 0
So,
f – 6 ≤ 3 and f – 6 ≥ -3
f ≤ 3 + 6 and f ≥ -3 + 6
f ≤ 9 and f ≥ 3
Hence, from the above,
We can conclude that the solutions of the given absolute value inequality are:
f ≤ 9 and f ≥ 3
The representation of the compound inequality of the solutions of the given absolute value inequality is:
3 ≤ f ≤ 9
The representation of the compound inequality in the graph is:

Question 27.
5 | b + 8 | – 7 > 13
Answer:
The given absolute value inequality is:
5 | b + 8 | – 7 > 13
5 | b + 8 | > 13 + 7
5 | b + 8 | > 20
| b + 8 | > 4
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
b + 8  > 4 or b + 8 < -4
b > 4 + 8 or b < -4 – 8
b > 12 or b < -12
We know that,
The value of absolute value expression must be greater than or equal to 0
So,
b > 12
Hence, from the above,
We can conclude that the solution to the given absolute value inequality is b > 12
The representation of the solution of the given absolute value inequality in the graph is:

Question 28.
| -3g – 2 | + 1 < 6
Answer:
The given absolute value inequality is:
| -3g – 2 | + 1 < 6
– | 3g + 2 | < 6 – 1
– | 3g + 2 | < 5
| 3g + 2 | < -5
We know that,
The value of absolute value expression must be greater than or equal to 0
Hence, from the above,
We can conclude that there is no solution to the given absolute value inequality

Question 29.
| 9 – 2j | + 10 ≥ 2
Answer:
The given absolute value inequality is:
| 9 – 2j | + 10 ≥  2
| 9 – 2j | ≥ 2 – 10
| 9 – 2j | ≥  -8
We know that,
The value of absolute value expression must be greater than or equal to 0
Hence, from the above,
We can conclude that there is no solution to the given absolute value inequality

Question 30.
A safety regulation states that the height of a guardrail should be 106 centimeters with an absolute deviation of no more than 7 centimeters. Write and solve an absolute value inequality that represents the acceptable heights of a guardrail.
Answer:
The absolute value inequality that represents the acceptable heights of a guardrail is:
| x – 106 | ≤ 7 centimeters

Explanation:
It is given that a safety regulation states that the height of a guardrail should be 106 centimeters with an absolute deviation of no more than 7 centimeters.
Now,
We know that,
Absolute deviation = ( Mean ) – ( The given observation )
Let the Mean be x
So,
Absolute deviation = x – 106
Hence,
The absolute value inequality that represents the acceptable heights of a guardrail is:
| x – 106 | ≤ 7
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
x – 106 ≤ 7 and x – 106 ≥ -7
x ≤ 7 + 106 and x ≥ -7 + 106
x ≤ 113 and x ≥ 99
Hence, from the above,
We can conclude that the absolute value inequality that represents the acceptable heights of a guardrail is:
| x – 106 | ≤ 7

Solving Linear Inequalities Chapter Test

Write the sentence as an inequality.

Question 1.
The sum of a number y and 9 is at least -1.
Answer:
The given worded form is:
The sum of a number y and 9 is at least -1
Hence,
The representation of the given worded form in the form of inequality is:
y + 9 ≥ -1

Question 2.
A number r is more than 0 or less than or equal to -8.
Answer:
The given worded form is:
A number r is more than 0 or less than or equal to -8
Hence,
The representation of the given worded form in the form of inequality is:
r > 0 or r ≤ -8

Question 3.
A number k is less than 3 units from 10.
Answer:
The given worded form is:
A number k is less than 3 units from 10
Hence,
The representation of the given worded form in the form of inequality is:
k – 10 < 3

Solve the inequality. Graph the solution, if possible.

Question 4.
\(\frac{x}{2}\) – 5 ≥ -9
Answer:
The given inequality is:
\(\frac{x}{2}\) – 5 ≥ -9
\(\frac{x}{2}\) ≥ -9 + 5
\(\frac{x}{2}\) ≥ -4
x ≥ -4 (2)
x ≥ -8
Hence, from the above,
We can conclude that the solution to the given inequality is x ≥ -8
The representation of the solution of the given inequality in the graph is:

Question 5.
-4s < 6s + 1
Answer:
The given inequality is:
-4s < 6s + 1
So,
-4s – 6s < 1
-10s < 1
10s < -1
s < –\(\frac{1}{10}\)
s < -0.1
Hence, from the above,
We can conclude that the solution to the given inequality is s < -0.1
The representation of the solution of the given inequality in the graph is:

Question 6.
4p + 3 ≥ 2(2p + 1)
Answer:
The given inequality is:
4p + 3 ≥ 2 (2p + 1)
So,
4p + 3 ≥ 2 (2p) + 2 (1)
4p + 3 ≥ 4p + 2
4p – 4p + 3 ≥ 2
3 ≥ 2
Hence, from the above,
We can conclude that there is no solution for the given inequality

Question 7.
-7 < 2c – 1 < 10
Answer:
The given inequality is:
-7 < 2c – 1 < 10
Add 1 on both sides
So,
-7 + 1 < 2c – 1 + 1 < 10 + 1
-6 < 2c < 11
Divide by 3 on both sides
(-6 / 2) < (2c / 2) < (11 / 2)
-3 < c < 5.5
-3 < c < 6 [Approximate inequality]
Hence, from the above,
We can conclude that the solution to the given inequality is -3 < c < 6
The representation of the solution of the given inequality in the graph is:

Question 8.
-2 ≤ 4 – 3a ≤ 13
Answer:
The given inequality is:
-2 ≤ 4 – 3a ≤ 13
Subtract 4 on both sides
-2 – 4 ≤ 4 – 4 – 3a ≤ 13 – 4
-6 ≤ -3a ≤ 9
Divide by – on both sides
6 ≥ 3a ≥ -9
Divide by 3 on both sides
(6 / 3) ≥ (3a / 3) ≥ (-9 / 3)
2 ≥ a ≥ -3
-3 ≤ a ≤ 2
Hence, from the above,
We can conclude that the solution to the given inequality is -3 ≤ a ≤ 2
The representation of the solution of the inequality in the graph is:

Question 9.
-5 < 2 – h or 6h + 5 > 71
Answer:
The given inequality is:
-5 < 2 – h or 6h + 5 > 71
-5 – 2 < h or 6h > 71 – 5
-7 < -h or 6h > 66
7 < h or h > 66 / 6
h > 7 or h > 11
Hence, from the above,
We can conclude that the solution to the given inequality is h > 7 [Since h > 11 is before  h > 7]
The representation of the solution of the given inequality in the graph is:

Question 10.
| 2q + 8 | > 4
Answer:
The given absolute value inequality is:
| 2q + 8 | > 4
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
2q + 8 > 4 or 2q + 8 < -4
2q > 4 – 8 or 2q < -4 – 8
2q > -4 or 2q < -12
We know that,
The value of the absolute value expression must be greater than or equal to 0
Hence, from the above,
We can conclude that the given absolute value inequality has no solution

Question 11.
-2 | y – 3 | – 5 ≥ -4
Answer:
The given absolute value inequality is:
-2 | y – 3 | – 5 ≥ -4
So,
-2 | y – 3 | ≥ -4 + 5
-2 | y – 3 | ≥1
| y – 3 | ≥ –\(\frac{1}{2}\)
We know that,
The value of the absolute value expression must be greater than or equal to 0
Hence, from the above,
We can conclude that the given absolute inequality has no solution

Question 12.
4 | -3b + 5 | – 9 < 7
Answer:
The given absolute value inequality is:
4 | -3b + 5 | – 9 < 7
So,
4 | -3b + 5 | < 7 + 9
4 | -3b + 5 | < 16
| -3b + 5 | < 16 / 4
| -3b + 5 | < 4
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
-3b + 5 < 4 and -3b + 5 > -4
-3b < 4 – 5 and -3b > -4 – 5
-3b < -1 and -3b > -9
3b < 1 and b > 3
b < \(\frac{1}{3}\) and b > 3
b < 0 and b > 3 [ Approximate value of inequality ]
Hence, from the above,
We can conclude that the solutions of the given inequality are b < 0 and b > 3
The representation of the solutions of the given inequality in the graph is:

Question 13.
You start a small baking business, and you want to earn a profit of at least $250 in the first month. The expenses in the first month are $155. What are the possible revenues that you need to earn to meet the profit goal?
Answer:
The possible revenues that you need to earn to meet the profit goal is:
R ≥ $405

Explanation:
It is given that you start a small baking business, and you want to earn a profit of at least $250 in the first month. The expenses in the first month are $155.
Now,
Let
P is the profit
R is the Revenue
E is the expenses
We know that,
P = R – E
It is given that P ≥ $250
So,
R – E ≥250
R – 155 ≥ 250
R ≥ 250 + 155
R ≥ 405
Hence, from the above,
We can conclude that the possible revenues that you need to earn to meet the profit goal is:
R ≥ $405

Question 14.
A manufacturer of bicycle parts requires that a bicycle chain have a width of 0.3 inches with an absolute deviation of at most 0.0003 inches. Write and solve an absolute value inequality that represents the acceptable widths.
Big Ideas Math Algebra 1 Solutions Chapter 2 Solving Linear Inequalities 96.1
Answer:
The absolute value inequality that represents the acceptable widths is:
0.2997 inches ≤ w ≤0.3003 inches

Explanation:
It is given that a manufacturer of bicycle parts requires that a bicycle chain have a width of 0.3 inches with an absolute deviation of at most 0.0003 inches.
We know that,
Absolute deviation = Mean – Observation
Let the Mean be x
So,
x – 0.3 ≤ 0.0003
Hence,
The absolute value inequality that represents the acceptable widths is:
| x – 0.3 | ≤ 0.0003
-0.0003 ≤ x – 0.3 ≤ 0.0003
Add with 0.3 on both sides
-0.0003 + 0.3 ≤ x – 0.3 + 0.3 ≤ 0.0003 + 0.3
0.2997 ≤ x ≤ 0.3003
Hence, from the above,
We can conclude that the absolute value inequality that represents the acceptable widths is:
0.2997 inches ≤ w ≤0.3003 inches

Question 15.
Let a, b, c, and d be constants. Describe the possible solution sets of the inequality ax + b < cx + d.
Answer:

Write and graph a compound inequality that represents the numbers that are not solutions to the inequality represented by the graph shown. Explain your reasoning.

Question 16.
Big Ideas Math Algebra 1 Solutions Chapter 2 Solving Linear Inequalities 97.1
Answer:
The given graph is:
Big Ideas Math Algebra 1 Solutions Chapter 2 Solving Linear Inequalities 97.1
From the graph,
The 1st marked line starts from -3 including -3 and continued till the left end of the number line
The 2nd marked line starts from 2 excluding 2 and continued till the right end of the number line
Hence,
The representation of the inequalities from the graph is:
x ≤ -3 and x > 2

Question 17.
Big Ideas Math Algebra 1 Solutions Chapter 2 Solving Linear Inequalities 98
Answer:
The given graph is:
Big Ideas Math Algebra 1 Solutions Chapter 2 Solving Linear Inequalities 98
From the graph,
The marked line starts from -4 including -4 and continued till -1 including -1 to the right end of the number line
Hence,
The representation of the inequality from the graph is:
x ≥ -4 and x ≤ -1
Hence,
The representation of the inequalities in the form of compound inequality is:
–4 ≤ x ≤ -1

Question 18.
A state imposes a sales tax on items of clothing that cost more than $175. The tax applies only to the difference in the price of the item and $175.
a. Use the receipt shown to find the tax rate (as a percent).
b. A shopper has $430 to spend on a winter coat. Write and solve an inequality to find the prices p of coats that the shopper can afford. Assume that p ≥ 175.
c. Another state imposes a 5% sales tax on the entire price of an item of clothing. For which prices would paying the 5% tax to be cheaper than paying the tax described above? Write and solve an inequality to find your answer and list three prices that are solutions.
Big Ideas Math Algebra 1 Solutions Chapter 2 Solving Linear Inequalities 99
Answer:
a.
From the given receipt,
The price is: $295
The total price is: $302.50
It is given that
Tax = Product price – $175
Now,
Let R be the tax rate
So,
R (product price – 175) = 7.5
R ( 295 – 175 ) = 7.5
R ( 120 ) = 7.5
%R = \(\frac{7.5}{120}\) × 100
%R = \(\frac{75}{1200}\) × 100
%R = 0.0625 × 100
%R = 6.25%
Hence, from the above,
We can conclude that the tax rate using the receipt is: 6.25%

b.
It is given that a shopper has $430 to spend on a winter coat
Let p be the number of coats
It is also given that to assume p ≥ 175
We know that,
The total price = The given price + Tax
So,
The representation of the inequality that represents the price of coats is:
p + R ( p – 175 ) ≤ 430
p + 0.0625 (p – 175)  ≤ 430
1.0625p – 10.9375 ≤ 430
1.0625p ≤ 430 + 10.9375
1.0625p ≤ 440.9375
p ≤ 440.9375 / 1.0625
p ≤ 415
Hence, from the above,
We can conclude that the prices of the coats that the shopper can afford are:
p ≤ $415

c.
It is given that another state imposes a 5% sales tax on the entire price of an item of clothing.
Now,
Let t be the tax rate that another state-imposed
Let x be the price
So,
(1+0.05) × t ≤ t + 0.0625 (t – 175)
1.05t ≤ 1.0625t – 10.9375
10.9375 ≤ 1.0625t – 1.05t
10.9375 ≤ 0.0125t
10.9375 / 0.0125 ≤ t
875 ≤ t
t ≥ 875
hence, from the above,
We can conclude that $875 is cheaper

Solving Linear Inequalities Maintaining Cumulative Assessment

Question 1.
The expected attendance at a school event is 65 people. The actual attendance can vary by up to 30 people. Which equation can you use to find the minimum and maximum attendances?
A. | x – 65 | = 30
B. | x + 65 | = 30
C. | x – 30 | = 65
D. | x + 30 | = 65
Answer:
It is given that the expected attendance at a school event is 65 people. The actual attendance can vary by up to 30 people.
So,
The absolute value equation that represents the maximum and minimum attendances are:
| x – Expected maximum attendance | ≤ 30
Hence,
| x – 65 | ≤ 30
| x – 65 | = 30
Hence, from the above,
We can conclude that option A represents the minimum and maximum attendances at a school event

Question 2.
Fill in values for a and b so that each statement is true for the inequality ax + 4 ≤ 3x + b.
a. When a = 5 and b = _____, x ≤ -3.
b. When a = _____ and b = _____, the solution of the inequality is all real numbers.
c. When a = _____ and b = _____, the inequality has no solution.
Answer:
The given inequality is:
ax + 4  ≤ 3x + b
a.
When a = 5,
5x + 4 ≤ 3x + b
5x – 3x + 4 ≤ b
2x + 4 ≤ b
2x ≤ b – 4
x ≤ ( b – 4 ) / 2
So,
The solution x ≤ -3 will be possible only if b = -2

b.
The given inequality is:
ax + 4 ≤ 3x + b
Compare the like coefficients
Hence,
The values of a and b so that the solution of an inequality is all real numbers are:
a = 3, b ≥ 4

c.
The given inequality is:
ax + 4 ≤ 3x + b
Hence,
The values of a and b so that the inequality has no solution are:
a = 3 and b < 4

Question 3.
Place each inequality into one of the two categories.
Big Ideas Math Algebra 1 Solutions Chapter 2 Solving Linear Inequalities 101
Answer:
The inequalities that have at least one integer solution must have x variable
The inequalities that have no integer solution must not have x variable
So,
The given inequalities are:
a. 5x – 6 + x ≥ 2x – 8
b. x – 8 + 4x ≤ 3 (x – 3) + 2x
c. 2 (3x + 8) > 3 (2x + 6)
d. 9x – 3 < 12 or 6x + 2 > -10
e. 17 < 4x + 5 < 21
f. 5 (x – 1) ≤ 5x – 3
Hence, from the above,
We can conclude that
The inequalities that have at least one integer solution are: a, d, e
The inequalities that have no integer solution are: b, c, f

Question 4.
Admission to a play costs $25. A season pass costs $180.
a. Write an inequality that represents the number x of plays you must attend for the season pass to be a better deal.
b. Select the number of plays for which the season pass is not a better deal.
Big Ideas Math Answer Key Algebra 1 Chapter 2 Solving Linear Inequalities 104
Answer:
a.
It is given that admission to a play costs $25 and a season pass costs $180
Let the number of plays be x
So,
The total number of pays = 25x
Hence,
The inequality that represents the number x of plays you must attend for the season pass to be a  better deal is:
25x > 180
x > 180 / 25
x > 7.2
Hence, from the above,
We can conclude that the inequality that represents the number x of plays you must attend for the season pass to be a better deal is:
x > 7.2

b.
From part (a),
The inequality that represents the season pas to be a better deal is:
x > 7.2
So,
The inequality that represents the season pass not to be a better deal is:
x < 7.2
Hence, from the given numbers,
The number of plays that the season pass is not a better deal is: 0, 1, 2, 3, 4, 5, 6, and 7

Question 5.
Select the values of a that make the solution of the equation 3(2x – 4) = 4(ax – 2) positive.
Big Ideas Math Answer Key Algebra 1 Chapter 2 Solving Linear Inequalities 105
Answer:
The given equation is:
3 (2x – 4) = 4 (ax – 2)
So,
3 (2x) – 3 (4) = 4 (ax) – 4 (2)
6x – 12 = 4ax – 8
6x – 12 + 8 = 4ax
6x – 4 = 4ax
6x – 4ax = 4
x (6 – 4a) = 4
Now,
The given numbers are: -2, -1, 0, 1, 2, 3, 4, 5
So,
x (6 – 4(-2) ) = 4
x ( 6 – 4 ( -1) ) = 4
x ( 6 – 4(0) ) = 4
x ( 6 – 4(1) ) = 4
x ( 6 – 4 (2) ) = 4
x ( 6 – 4 (3) ) = 4
x ( 6 – 4 (4) ) = 4
x ( 6 – 4 (5) ) = 4
Hence, from the above,
We can conclude that the values of a so that the solution is positive are: -2, -1, 0, 1

Question 6.
Fill in the compound inequality with <, ≤, , ≥, or > so the solution is shown in the graph.
Big Ideas Math Answer Key Algebra 1 Chapter 2 Solving Linear Inequalities 107
Answer:
From the graph,
The representation of the inequality is:
-2 ≤ x < 3
Now,
The given inequality is:
4x – 18 ____ -x – 3 and -3x – 9 _____ -3
4x + x ____ -3 + 18 and -3x ___ -3 + 9
5x ____ 15 and -3x ___ 6
x ____ 15 / 5 and x ____ -3 / 6
x ____ 3 and x ____ -2
Hence,
The blanks that can fill the inequality are:
x < 3 and x ≥ -2

Question 7.
You have a $250 gift card to use at a sporting goods store.
Big Ideas Math Answer Key Algebra 1 Chapter 2 Solving Linear Inequalities 108
a. Write an inequality that represents the possible numbers x of pairs of socks you can buy when you buy 2 pairs of sneakers. Can you buy 8 pairs of socks? Explain.
b. Describe what the inequality 60 + 80x ≤ 250 represents in this context.
Answer:
a.
It is given that
The cost of a pair of sneakers is: $80
The cost of a pair of socks is: $12
It is given that there are 2 pairs of sneakers and x pairs of socks
Hence,
The inequality that represents the possible number of x pairs of socks you can buy when you buy 2 pairs of sneakers is:
2 (80) + 12x ≤ 250
160 + 12x ≤ 250
12x ≤ 250 – 160
12x ≤ 90
x ≤ 90 / 12
x ≤ 7.5
Hence, from the above,
We can conclude that you can buy only 7 pairs of socks

b.
The given inequality is:
60 + 8x ≤ 250
12 (5) + 8x ≤ 250
Hence, from the above,
We can conclude that you can buy 5 pairs of sneakers in this context

Question 8.
Consider the equation shown, where a, b, c, and d are integers.
ax + b = cx + d
Student A claims the equation will always have one solution. Student B claims the equation will always have no solution. Use the numbers shown to answer parts (a)–(c).
Big Ideas Math Answer Key Algebra 1 Chapter 2 Solving Linear Inequalities 109
a. Select values for a, b, c, and d to create an equation that supports Student A’s claim.
Answer:
The given equation is:
ax + b = cx + d
Student A claims the equation will have always one solution
Hence,
From the given numbers,
The values of a, b, c, and d are:
a = -1, b = 2, c = -1, d = 4
So,
The value of x is:
-x + 2 = x + 4
-x – x = 4 – 2
-2x = 2
x = -2 /2
x = -1
Hence, from the above,
We can conclude that according to Student A’s claim,
The values of a, b, c, and d are:
a = -1, b = 2, c = 1, d = 4

b. Select values for a, b, c, and d to create an equation that supports Student B’s claim.
Answer:
The given equation is:
ax + b = cx + d
Student B claims that the equation will have no solution
So,
The value of x must be equal to 0
So,
a must be equal to c
Hence,
The values of a, b, c, and d are: a = c and the values of b and d will be any number from the given numbers

c. Select values for a, b, c, and d to create an equation that shows both Student A and Student B are incorrect.
Answer:
The given equation is:
ax + b = cx + d
For the claims of both Student A and Student B to be false,
The given equation must have the real numbers as a solution
So,
To have the real numbers as a solution, the values of a and b will be arbitrary from the given numbers

Big Ideas Math Answers Grade 4 Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers

Big Ideas Math Answers Grade 4 Chapter 5

Big Ideas Math Answers Grade 4 Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers are included here to help the students to improve their preparation level. You must practice every question given on Big Ideas Math Answers Grade 4 to aware of the different questions impose in the exam. Practice all the problems and verify the solution and explanation to find out the easy method to solve problems. Students are suggested to learn the in-depth concept of Divide Multi-Digit Numbers into One-Digit Numbers and start practicing the below questions.

Big Ideas Grade 4 Answer Key Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers

Learn from basics using Big Ideas Grade 4 Solution Key pdf. Click on the below attached links and start solving all the problems. Tips and tricks are also given to help the students to remember the process to solve different problems. So, by referring to our bigideasmathanswers.com students can get a complete grip on the concepts as well as they will become math experts easily. Find out various methods to solve problems and choose the best out of one that makes your learning easy.

Lesson: 1 Divide Tens, Hundreds, and Thousands

Lesson: 2 Estimate Quotients

Lesson: 3 Understand Division and Remainder

Lesson: 4 Use Partial Quotients

Lesson: 5 Use Partial Quotients with a Remainder

Lesson: 6 Divide Two-Digit Numbers by One-Digit Numbers

Lesson: 7 Divide Multi-Digit Numbers by One-Digit Numbers

Lesson: 8 Divide by One-Digit Numbers

Lesson: 9 Problem Solving: Division

Performance Task

Lesson 5.1 Divide Tens, Hundreds, and Thousands


Explore and Grow

Use a model to find each missing factor. Draw each model. Then write the related division equation.
Big Ideas Math Answer Key Grade 4 Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.1 1

What pattern do you notice?

Answer: The pattern follows the multiplication of 4 with ones, tens, hundreds, and thousands.

Explanation:
Let the missing number be X.
Now,
The given Expressions are:

A) X × 2 = 8
So, X can be calculated by
X= 8 ÷ 2 = 4

B) X × 2 = 80
So, X can be calculated by
X= 80 ÷ 2 = 40

C) X × 2 = 800
So, X can be calculated by
X= 800 ÷ 2 = 400

D) X × 2 = 8,000
So, X can be calculated by
X= 8,000 ÷ 2 = 4,000
Hence, from the above,
We can conclude that the pattern follows the multiplication of 4 with ones, tens, hundreds, and thousands.

Repeated Reasoning

Explain how 12 ÷ 4 can help you find 1,200 ÷ 4

Answer: 1,200 ÷ 4 is 100 times the value of 12 ÷ 4.

Explanation:

The value of 12 ÷ 4 = 3
Now,
we know that 1,200 times is 100 times the value of 12. ( From the place-value Concept)
So,
1,200 ÷ 4 = 120 tens ÷ 4
=30 tens
= 300
Hence, from the values of 3 and 300,
We can conclude that the value of 300 is 100 times the value of 3.

Think and Grow: Divide Tens, Hundreds and Thousands

You can use place value and basic division facts to divide tens, hundreds, or thousands by one-digit numbers.
Big Ideas Math Answer Key Grade 4 Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.1 2

Example
Find 270 ÷ 9
Think: 27 ÷ 9
270 ÷ 9 = 27 tens ÷ 9
= 3 tens
= 30
So, 270 ÷ 9 = 30

Big Ideas Math Answer Key Grade 4 Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.1 3

Example
Find 5,600 ÷ 8.
Think: 56 ÷ 8
5,600 ÷ 8 = 56 hundreds ÷ 8
= 7 hundreds
= 700
So, 5,600 ÷ 8 = 700

Show and Grow

Question 1.
Find 2,400 ÷ 6.
Think: 24 ÷ 6 = 3
2,400 ÷ 6 = 24 hundreds ÷ 6
= 3 hundreds
=300
So, 2,400 ÷ 6 = 300.

Question 2.
Find each quotient.
49 ÷ 7 =_____
490 ÷ 7 = ______
4,900 ÷ 7 = ______

Answer: Let the Expressions be A), B) and C)
So,
The quotients of A), B) and C) are:
A) 7
B) 70
C) 700

Explanation:
Let the given Expressions be A), B) and C)
So, the given Expressions are:
A) 49 ÷ 7
B) 490 ÷ 7
C) 4,900 ÷ 7
So,
A) 49 ÷ 7 = 7

B) 490 ÷ 7 = 49 tens ÷ 7
= 7 tens
=70
So, 490 ÷ 7 = 70

C) 4,900 ÷ 7 =  49 hundreds ÷ 7
= 7 hundred
= 700
So, 4,900 ÷ 7 = 700

Note: The “Quotient” of an Expression is defined as a number that divides the Dividend and the quotient should be an Integer.

Apply and Grow: Practice

Find the quotient
Question 3.
50 ÷ 5 = _____

Answer: 10

Explanation:
The given Expression is:
50 ÷ 5 = 5 tens ÷ 5
= 1 ten
=10
So, 50 ÷ 5 = 10

Note: The “Quotient” of an Expression is defined as a number that divides the Dividend and the quotient should be an Integer.

Question 4.
360 ÷ 6 = _____

Answer: 60

Explanation:
The given Expression is:
360 ÷ 6 = 36 tens ÷ 6
= 6 tens
=60
So, 360 ÷ 6 = 60

Note: The “Quotient” of an Expression is defined as a number that divides the Dividend and the quotient should be an Integer.

Question 5.
7,200 ÷ 8 = ______

Answer: 900

Explanation:
The given Expression is:
7,200 ÷ 8 = 72 hundreds ÷ 8
= 9 hundred
=900
So, 7,200 ÷ 8 = 900

Note: The “Quotient” of an Expression is defined as a number that divides the Dividend and the quotient should be an Integer.

Question 6.
180 ÷ 2 = ______

Answer: 90

Explanation:
The given Expression is:
180 ÷ 2 = 18 tens ÷ 2
= 9 tens
=90
So, 180 ÷ 2 = 90

Note: The “Quotient” of an Expression is defined as a number that divides the Dividend and the quotient should be an Integer.

Question 7.
4,200 ÷ 7 = _____

Answer: 600

Explanation:
The given Expression is:
4,200 ÷ 7 = 42 hundreds ÷ 7
= 6 hundred
=600
So, 4,200 ÷ 7 = 600

Note: The “Quotient” of an Expression is defined as a number that divides the Dividend and the quotient should be an Integer.

Question 8.
20 ÷ 2 = ____

Answer: 10

Explanation:
The given Expression is:
20 ÷ 2 = 2 tens ÷ 2
= 1 ten
=10
So, 20 ÷ 2 = 10

Note: The “Quotient” of an Expression is defined as a number that divides the Dividend and the quotient should be an Integer.

Question 9.
2,000 ÷ 5 = _____

Answer: 400

Explanation:
The given Expression is:
2,000 ÷ 5 = 20 hundreds ÷ 5
= 4 hundred
=400
So, 2,000 ÷ 5 = 400

Note: The “Quotient” of an Expression is defined as a number that divides the Dividend and the quotient should be an Integer.

Question 10.
30 ÷ 3 = _____

Answer: 10

Explanation:
The given Expression is:
30 ÷ 3 = 3 tens ÷ 3
= 1 ten
=10
So, 30 ÷ 3 = 10

Note: The “Quotient” of an Expression is defined as a number that divides the Dividend and the quotient should be an Integer.

Question 11.
320 ÷ 4 = ______

Answer: 80

Explanation:
The given Expression is:
320 ÷ 4 = 32 tens ÷ 4
= 8 tens
=80
So, 320 ÷ 4 = 80

Note: The “Quotient” of an Expression is defined as a number that divides the Dividend and the quotient should be an Integer.

Question 12.
140 ÷ 2 = _____

Answer: 70

Explanation:
The given Expression is:
140 ÷ 2 = 14 tens ÷ 2
= 7 tens
=70
So, 140 ÷ 2 = 70

Note: The “Quotient” of an Expression is defined as a number that divides the Dividend and the quotient should be an Integer.

Question 13.
5,400 ÷ 9 = ____

Answer: 600

Explanation:
The given Expression is:
5,400 ÷ 9 = 54 hundreds ÷ 9
= 6 hundred
=600
So, 5,400 ÷ 9 = 600

Note: The “Quotient” of an Expression is defined as a number that divides the Dividend and the quotient should be an Integer.

Question 14.
180 ÷ 6 = _____

Answer: 30

Explanation:
The given Expression is:
180 ÷ 6 = 18 tens ÷ 6
= 3 tens
=30
So, 180 ÷ 6 = 30

Note: The “Quotient” of an Expression is defined as a number that divides the Dividend and the quotient should be an Integer.

DIG DEEPER!

Find the missing number.
Question 15.
70 ÷ ____ = 10

Answer: The missing number is: 7

Explanation:
Let the missing number be X.
The given Expression is:
70 ÷ X = 10
So, X can be calculated by
X= 70 ÷ 10 = 70

Question 16.
4,000 ÷ _____ = 800

Answer: The missing number is: 20

Explanation:
Let the missing number be X.
The given Expression is:
4,000 ÷ X = 800
So, X can be calculated by
X= 4,000 ÷ 800 = 20

Question 17.
160 ÷ _____ = 40

Answer: The missing number is: 4

Explanation:
Let the missing number be X.
The given Expression is:
160 ÷ X = 40
So, X can be calculated by
X= 160 ÷ 40 = 4

Question 18.
_____ ÷ 7 = 300

Answer: The missing number is: 2,100

Explanation:
Let the missing number be X.
The given Expression is:
X ÷ 7 = 300
So, X can be calculated by
X= 70 × 300 = 2,100

Question 19.
_____ ÷ 5 = 70

Answer: The missing number is: 350

Explanation:
Let the missing number be X.
The given Expression is:
X ÷ 5 = 70
So, X can be calculated by
X= 70 × 5 = 350

Question 20.
_____ ÷ 6 = 10

Answer: The missing number is: 60

Explanation:
Let the missing number be X.
The given Expression is:
X ÷ 6 = 10
So, X can be calculated by
X= 6 × 10 = 60

Compare
Question 21.
Big Ideas Math Answer Key Grade 4 Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.1 4
Answer: 10 is equal to 10

Explanation:
Let the given Expressions be A) and B)
The given Expressions are:
A) 40 ÷ 4 = 4 tens ÷ 4
= 1 ten
= 10
B) 1 × 10 = 10
So, from the above,
We can conclude that 10 is equal to 10

Question 22.
Big Ideas Math Answer Key Grade 4 Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.1 5

Answer: 20 is less than 200

Explanation:
Let the given Expressions be A) and B)
The given Expressions are:
A) 160 ÷ 8 = 16 tens ÷ 8
= 2 tens
= 20
B) 2 × 100 = 200
So, from the above,
We can conclude that 20 is less than 200

Question 23.
Big Ideas Math Answer Key Grade 4 Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.1 6

Answer: 900 is greater than 90

Explanation:
Let the given Expressions be A) and B)
The given Expressions are:
A) 8,100 ÷ 9 = 81 hundreds ÷ 9
= 9 hundreds
= 900
B) 9 × 10 = 90
So, from the above,
We can conclude that 900 is equal to 90

Question 24.
There are 240 students visiting a fair. They are divided equally among 8 barns. How many students are in each barn?
Big Ideas Math Answer Key Grade 4 Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.1 7
Answer: There are 30 students in each barn.

Explanation:
Given that there are 240 students visiting a fair and these students are divided equally among 8 barns.
So,
The number of students in each barn can be calculated by dividing the number of students by the number of barns.
Hence,
The number of students in each barn is:
240 ÷ 8 = 24 tens ÷ 8
= 3 tens
= 30
So, 240 ÷ 8 = 30 students

Question 25.

YOU BE THE TEACHER
Is Descartes correct? Explain.
Big Ideas Math Answer Key Grade 4 Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.1 8

Answer: Descartes is not correct.

Explanation:
Given Expression is 2,500 ÷ 5
The value of 2,500 ÷ 5 is:
2,500 ÷ 5  = 25 hundreds ÷ 5
= 5 hundreds
= 500
So, 2,500 ÷ 5 = 500
But, According to Descartes,
2,500 ÷ 5 = 5,000
But, up on calculation, we get the result as 500.
So, Descartes is not correct.

Think and Grow: Modeling Real Life

Example
A lobster lays 5,400 eggs. It lays 9 times as many eggs as a seahorse. How many eggs does the seahorse lay?

Big Ideas Math Answer Key Grade 4 Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.1 9.1
Draw a model.

Show and Grow

Question 26.
A coach has 350 career wins. He has 7 times as many careers wins as wins this season. How many wins does the coach have this season?

Big Ideas Math Answer Key Grade 4 Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.1 10

Answer: 50 wins

Explanation:
Given that a coach has 350 career wins and he has 7 times as many careers wins as this season.
So,
The number of wins that the coach has in this season can be calculated by dividing the total career wins by the number of times the career wins of this season.
Hence,
The number of career wins in this season is:
350 ÷ 7 = 35 tens ÷ 7
= 5 tens
= 50
So, from the above,
We can conclude that the career wins of this season are: 50

Question 27.
There are 160 shareable bicycles in a city. There are 8 bicycle-sharing stations. Each station has the same number of bicycles. How many bicycles are at each station?

Answer: 20 bicycles are at each station.

Explanation:
Given that there are 160 shareable bicycles in a city and there are 8 bicycle-sharing stations.
It is also given that each station has the same number of bicycles.
So, the number of bicycles in each station are:
160 ÷ 8 = 16 tens ÷ 8
= 2 tens
= 20
So, 160 ÷ 8 = 20
Hence, from the above,
We can conclude that there are 20 bicycles at each station.

Question 28.
A charity has 637 adult volunteers and 563 teenage volunteers. All of the volunteers are divided, into 6 equal groups. How many volunteers are in each group?

Answer: There are 200 volunteers in each group.

Explanation;
Given that a charity has 637 adult volunteers and 563 teenage volunteers.
So,
The total number of Volunteers = Number of adult volunteers + Number of teenage volunteers
= 637 + 563
= 1,200 volunteers
It is also given that the volunteers are divided into 6 equal groups.
So, the number of volunteers in each group are 1,200 ÷ 6.
Now,
1,200 ÷ 6 = 12 hundreds ÷ 6
= 2 hundred
= 200
Hence, from the above,
We can conclude that there are 200 volunteers in each group.

Divide Tens, Hundreds and Thousands Homework & Practice 5.1

Question 1.
Find 150 ÷ 3.
Think: 15 ÷ 3 = 5
150 ÷ 3 = 15 tens ÷ 3
= 5 tens
= 50
So, 150 ÷ 3 = 50

Question 2.
Find 6,300 ÷ 7
Think: 63 ÷ 7 = 9
6,300 ÷ 7 = 63 hundreds ÷ 7
= 9  hundreds
= 900
So, 6300 ÷ 7 = 900

Find the quotient.

Question 3.
12 ÷ 2 = _____
120 ÷ 2 = ______
1,200 ÷ 2 = ______
Answer:
Let the Expressions be A), B) and C)
So,
The quotients of A), B) and C) are:
A) 6
B) 60
C) 600

Explanation:
Let the given Expressions be A), B) and C)
So, the given Expressions are:
A) 12 ÷ 2
B) 120 ÷ 2
C) 1,200 ÷ 2
So,
A) 12 ÷ 2 = 6

B) 120 ÷ 2 = 12 tens ÷ 2
= 6 tens
=60
So, 120 ÷ 2 = 60

C) 1,200 ÷ 2 =  12 hundreds ÷ 2
= 6 hundred
= 600
So, 1,200 ÷ 2 = 600

Note: The “Quotient” of an Expression is defined as a number that divides the Dividend and the quotient should be an Integer.

Question 4.
40 ÷ 8 = _____
400 ÷ 8 = ______
4,000 ÷ 8 = ______

Answer:
Let the Expressions be A), B) and C)
So,
The quotients of A), B) and C) are:
A) 5
B) 50
C) 500

Explanation:
Let the given Expressions be A), B) and C)
So, the given Expressions are:
A) 40 ÷ 8
B) 400 ÷ 8
C) 4,000 ÷ 8
So,
A) 40 ÷ 8 = 5

B) 400 ÷ 8 = 40 tens ÷ 8
= 5 tens
=50
So, 400 ÷ 8 = 50

C) 4,000 ÷ 8 =  40 hundreds ÷ 8
= 5 hundred
= 500
So, 4,000 ÷ 8 = 500

Note: The “Quotient” of an Expression is defined as a number that divides the Dividend and the quotient should be an Integer.

Find the quotient.
Question 5.
80 ÷ 8 = ______

Answer: 10

Explanation:
The given Expression is:
80 ÷ 8 = 8 tens ÷ 8
= 1 ten
=10
So, 80 ÷ 8 = 10

Note: The “Quotient” of an Expression is defined as a number that divides the Dividend and the quotient should be an Integer.

Question 6.
300 ÷ 6 = _____

Answer: 50

Explanation:
The given Expression is:
300 ÷ 6 = 30 tens ÷ 6
= 5 tens
=50
So, 300 ÷ 6 = 50

Note: The “Quotient” of an Expression is defined as a number that divides the Dividend and the quotient should be an Integer.

Question 7.
1,000 ÷ 5 = _____

Answer: 200

Explanation:
The given Expression is:
1,000 ÷ 5 = 10 hundreds ÷ 5
= 2 hundred
=200
So, 1,000 ÷ 5 = 200

Note: The “Quotient” of an Expression is defined as a number that divides the Dividend and the quotient should be an Integer.

Find the quotient.
Question 8.
40 ÷ 8 = ______

Answer: 5

Explanation:
The given Expression is:
40 ÷ 8 = 4 tens ÷ 8
=5
So, 40 ÷ 8 = 5

Note: The “Quotient” of an Expression is defined as a number that divides the Dividend and the quotient should be an Integer.

Question 9.
6,400 ÷ 8 = _____

Answer: 800

Explanation:
The given Expression is:
6,400 ÷ 8 = 64 hundred ÷ 8
= 8 hundred
=800
So, 6,400 ÷ 8 = 800

Note: The “Quotient” of an Expression is defined as a number that divides the Dividend and the quotient should be an Integer.

Question 10.
350 ÷ 5 = ______

Answer: 70

Explanation:
The given Expression is:
350 ÷ 5 = 35 tens ÷ 5
= 7 tens
=70
So, 350 ÷ 5 = 70

Note: The “Quotient” of an Expression is defined as a number that divides the Dividend and the quotient should be an Integer.

Question 11.
2,100 ÷ 7 = ____

Answer: 300

Explanation:
The given Expression is:
2,100 ÷ 7 = 21 hundreds ÷ 7
= 3 hundred
=300
So, 2,100 ÷ 7 = 300

Note: The “Quotient” of an Expression is defined as a number that divides the Dividend and the quotient should be an Integer.

Question 12.
240 ÷ 3 = ______

Answer: 80

Explanation:
The given Expression is:
240 ÷ 3 = 24 tens ÷ 3
= 8 tens
=80
So, 240 ÷ 3 = 80

Note: The “Quotient” of an Expression is defined as a number that divides the Dividend and the quotient should be an Integer.

Question 13.
90 ÷ 9 = _____

Answer: 10

Explanation:
The given Expression is:
90 ÷ 9 = 9 tens ÷ 9
= 1 ten
=10
So, 90 ÷ 9 = 10

Note: The “Quotient” of an Expression is defined as a number that divides the Dividend and the quotient should be an Integer.

DIG DEEPER!

Find the missing number.

Question 14.
50 ÷ ______ = 10

Answer: The missing number is:  5

Explanation:
Let the missing Expression be X.
The given Expression is:
50 ÷ X = 10
So, X can be calculated by
X= 160 ÷ 40 = 4

Question 15.
_____ ÷ 7 = 600

Answer: The missing number is: 4,200

Explanation:
Let the missing Expression be X.
The given Expression is:
X ÷ 7 = 600
So, X can be calculated by
X= 600 × 7 = 4,200

Question 16.
320 ÷ ______ = 40

Answer: The missing number is: 8

Explanation:
Let the missing number be X.
The given Expression is:
320 ÷ X = 40
So, X can be calculated by
X= 3200 ÷ 40 = 8

Compare
Question 17.
Big Ideas Math Answer Key Grade 4 Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.1 11

Answer: 10 is equal to 10

Explanation:
Let the given Expressions be A) and B)
The given Expressions are:
A) 30 ÷ 3 = 3 tens ÷ 3
= 1 ten
= 10
B) 1 × 10 = 10
So, from the above,
We can conclude that 10 is equal to 10

Question 18.
Big Ideas Math Answer Key Grade 4 Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.1 12
Answer: 70 is less than 700

Explanation:
Let the given Expressions be A) and B)
The given Expressions are:
A)560 ÷ 8 = 56 tens ÷ 8
= 7 tens
= 70
B) 9 × 100 = 700
So, from the above,
We can conclude that 70 is less than 700

Question 19.
Big Ideas Math Answer Key Grade 4 Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.1 13
Answer: 700 is less than 7,000

Explanation:
Let the given Expressions be A) and B)
The given Expressions are:
A) 4,900 ÷ 7 = 49 hundreds ÷ 7
= 7 hundreds
= 700
B) 7 × 1,000 = 7,000
So, from the above,
We can conclude that 700 is equal to 7,000

Question 20.
A movie theater has 180 seats. The seats are divided into 9 equal rows. How many seats are in each row?

Answer: There are 20 seats in each row.

Explanation:
Given that a model theater has 180 seats and the seats are divided into 9 equal rows.
Now, to find the number of seats in each row, we have to divide the total number of seats by the number of rows.
So,
180 ÷ 9 = 18 tens ÷ 9
= 2 tens
= 20 seats
So, from the above,
We can conclude that the number of seats in each row is: 20 seats

Question 21.
Number Sense
What is Newton’s number?

Big Ideas Math Answer Key Grade 4 Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.1 14

Answer:

Question 22.
Modeling Real Life
A gorilla understands 2,000 words. She understands 4 times as many words as a toddler. How many words does the toddler understand?
Big Ideas Math Answer Key Grade 4 Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.1 15

Answer: The toddler understands 500 words.

Explanation:
Given that a gorilla understands 2,000 words.
It is also given that the gorilla understands 4 times as many words as a toddler.
So,
The words understand by a toddler = The words understand by a gorilla ÷ 4
So,
2,000 ÷ 4 = 20 hundreds ÷ 4
= 5 hundred
= 500
Hence, from the above,
We can conclude that the words understand by a toddler are: 500 words

Review & Refresh

Compare

Question 23.
Big Ideas Math Answer Key Grade 4 Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.1 16
Answer: 1,834 is greater than 1,796

Explanation:
Given numbers are 1,834 and 1,796.
Hence, from these 2 numbers,
We can conclude that 1,834 is greater than 1,796

Question 24.
Big Ideas Math Answer Key Grade 4 Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.1 17
Answer: 62,905 is greater than 62,081

Explanation:
Given numbers are 62,905 and 62,081.
Hence, from these 2 numbers,
We can conclude that 62,905 is greater than 62,081

Question 25.
Big Ideas Math Answer Key Grade 4 Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.1 18
Answer: 9,142 is less than 9,146

Explanation:
Given numbers are 9,142 and 9,146
Hence, from these 2 numbers,
We can conclude that 9,142 is less than 9,146.

Question 26.
Big Ideas Math Answer Key Grade 4 Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.1 19
Answer: 52,048 is less than 52,071

Explanation:
Given numbers are 52,048 and 52,071
Hence, from these 2 numbers,
We can conclude that 52,048 is less than 52,071

Question 27.
Big Ideas Math Answer Key Grade 4 Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.1 20
Answer: 402,157 is equal to 402,157

Explanation:
Given numbers are 402,157 and 402,157
Hence, from these 2 numbers,
We can conclude that 402,157 is equal to 402,157

Question 28.
Big Ideas Math Answer Key Grade 4 Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.1 21
Answer: 387,402 is less than 384,927

Explanation:
Given numbers are 387,402 and 384,927
Hence, from these 2 numbers,
We can conclude that 387,402 is less than 384,927

Lesson 5.2 Estimate Quotients

Explore and Grow

Explain how you can use the table to estimate 740 ÷ 8.

740 ÷ 8 is about ______.

Answer: 90

Explanation:
Let 740 be Estimated to 720.
So, now we have to find the value of 720 ÷ 8
Now,
720 ÷ 8 = 72 tens ÷ 8
= 9 tens
= 90
Hence, from the above,
We can conclude that 740 ÷ 8 is about 90.

Reasoning
Why did you choose your estimate? Compare your results with your partner.

Answer: We choose the Estimate to round off the Result because the quotient must be the Integer.

Think and Grow: Estimate Quotients

You can use division facts and compatible numbers to estimate a quotient.
Example
Estimate 154 ÷ 4.

Big Ideas Math Answers 4th Grade Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.2 2

Look at the first two digits of the dividend and use basic division facts. and are close to the actual numbers.
Think: What number close to 154 is easily divided by 4?
Try 120. 12 ÷ 4 = 3, so 120 ÷ 4 =30.
Try 160. 16 ÷ 4 , = 4, so 160 ÷ 4 = 140.
Choose 160 because 154 is closer to 160.
So, 154 ÷ 4 is about 40.

When solving division problems, you can check whether an answer is reasonable by finding two numbers that a quotient is between.
Example
Find two numbers that the quotient 6,427 ÷ 7 is between.

Big Ideas Math Answers 4th Grade Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.2 3
Think: What numbers close to 6,427 are easily divided by7?
Use 6,300. 63 ÷ 7 = 9, so 6,300 ÷ 7 = 900.
Use 7,000. 70 ÷ 7 = 10, so 7,000 ÷ 7 = 1,000 .
6,427 is between 6,300 and 7,000.
So, the quotient 6,427 ÷ 7 is between 900 and 1,000.

Show and Grow

Estimate the quotient.
Question 1.
61 ÷ 3
Answer: 20

Explanation;
Let 61 be rounded off to 60.
So, now we have to find 60 ÷ 3
Now,
60 ÷ 3 = 6 tens ÷ 3
= 2 tens
= 20
Hence, from the above,
We can conclude that 61 ÷ 3 can be rounded off to 20.

Question 2.
465 ÷ 9
Answer: 50

Explanation:
Let 465 be rounded off to 450.
So, now we have to find 450 ÷ 9
Now,
450 ÷ 9 = 45 tens ÷ 9
= 5 tens
= 50
Hence, from the above,
We can conclude that 465 ÷ 9 can be rounded off to 50..

Find two numbers that the quotient is between.
Question 3.
477 ÷ 8
Answer: The Quotient of 477 ÷ 8 is between 50 and 60.

Explanation:
Use 400. 40 ÷ 8 = 5, so 400 ÷ 8 = 50.
Use 480. 48 ÷ 8 = 6, so 480 ÷ 8 = 60 .
477 is between 400 and 480.
So, the quotient of 477 ÷ 8 is between 50 and 60.

Question 4.
5,194 ÷ 6
Answer: The Quotient of 5,194 ÷ 6 is between 800 and 900

Explanation:
Use 4,800. 48 ÷ 6 = 8, so 4,800 ÷ 6 = 800.
Use 5,400. 54 ÷ 6 = 9, so 5,400 ÷ 6 = 900 .
5,194 is between 4,800 and 5,400.
So, the quotient of 5,194 ÷ 6 is between 800 and 900

Apply and Grow: Practice

Estimate the quotient
Question 5.
29 ÷ 5
Answer: 6

Explanation;
Let 29 be rounded off to 30.
So, now we have to find 30 ÷ 5
Now,
30 ÷ 5 = 3 tens ÷ 5
= 6
Hence, from the above,
We can conclude that 29 ÷ 5 can be rounded off to 6.

Question 6.
571 ÷ 8
Answer: 70

Explanation;
Let 571 be rounded off to 560.
So, now we have to find 560 ÷ 8
Now,
560 ÷ 8 = 56 tens ÷ 8
= 7 tens
= 70
Hence, from the above,
We can conclude that 571 ÷ 8 can be rounded off to 70.

Question 7.
202 ÷ 6
Answer: 30

Explanation;
Let 202 be rounded off to 180.
So, now we have to find 180 ÷ 6
Now,
180 ÷ 6 = 18 tens ÷ 6
= 3 tens
= 30
Hence, from the above,
We can conclude that 202 ÷ 6 can be rounded off to 30.

Question 8.
3,384 ÷ 7
Answer: 500

Explanation;
Let 3,384 be rounded off to 3,500.
So, now we have to find 3,500 ÷ 7
Now,
3,500 ÷ 7 = 35 hundreds ÷ 7
= 5 hundred
= 500
Hence, from the above,
We can conclude that 3,384 ÷ 7 can be rounded off to 500.

Find two estimates that the quotient is between.
Question 9.
22 ÷ 3
Answer: The Quotient of 22 ÷ 3 is between 7 and 8.

Explanation:
Use 21. , so 21 ÷ 3 = 7.
Use 24.  so 24 ÷ 3 = 8 .
22 is between 21 and 24.
So, the quotient of 22 ÷ 3 is between 7 and 8.

Question 10.
165 ÷ 9
Answer: The Quotient of 165 ÷ 9 is between 10 and 20

Explanation:
Use 90. 9 ÷ 9 = 1, so 90 ÷ 9 = 10.
Use 180. 18 ÷ 9 = 2, so 180 ÷ 9 = 20 .
165 is between 90 and 180.
So, the quotient of 165 ÷ 9 is between 10 and 20

Question 11.
2,387 ÷ 5
Answer: The Quotient of 2,387 ÷ 5 is between 400 and 500.

Explanation:
Use 2,000. 20 ÷ 5 = 4, so 2,000 ÷ 5 = 400.
Use 2,500. 25 ÷ 5 = 5, so 2,500 ÷ 5 = 500 .
2,387 is between 2,000 and 2,500.
So, the quotient of 2,387 ÷ 5 is between 400 and 500.

Question 12.
3,813 ÷ 4
Answer: The Quotient of 3,813 ÷ 4 is between 900 and 1,000

Explanation:
Use 3,600. 36 ÷ 4 = 9, so 3,600 ÷ 4 = 900.
Use 4,000. 40 ÷ 4 = 10, so 4,000 ÷ 4 = 1,000 .
3,813 is between 3,600 and 4,000.
So, the quotient of 2,387 ÷ 5 is between 400 and 500.

DIG DEEPER!
Estimate to compare.
Question 13.
Big Ideas Math Answers 4th Grade Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.2 4
Answer: 3 is greater than 2

Explanation;
Given Expressions are 26 ÷ 9 and 2
Let 26 be rounded to 27.
Now,
27 ÷ 9 = 3
Hence, from the above values,
We can conclude that 3 is greater than 2

Question 14.
Big Ideas Math Answers 4th Grade Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.2 5
Answer: 50 is equal to 50.

Explanation;
Given Expressions are 142 ÷ 3 and 50
Let 142 be rounded to 150.
Now,
150 ÷ 3 = 15 tens ÷ 3
= 5 tens
= 50
Hence, from the above values,
We can conclude that 50 is equal to 50.

Question 15.
Big Ideas Math Answers 4th Grade Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.2 6
Answer: 80 is less than 90

Explanation;
Given Expressions are 645 ÷ 8 and 816 ÷ 9
Let 645 be rounded to 640.
Now,
640 ÷ 8 = 64 tens ÷ 8
= 8 tens
= 80
Let 816 be rounded to 810.
Now,
810 ÷ 9 = 81 tens ÷ 9
= 9 tens
= 90
Hence, from the above results,
We can conclude that 80 is less than 90

Question 16.
Big Ideas Math Answers 4th Grade Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.2 7
Answer: 200 is equal to 200

Explanation;
Given Expressions are 1,200 ÷ 6 and 800 ÷ 4
Now,
1,200 ÷ 6 = 12 hundreds ÷ 6
= 2 hundred
= 200
Now,
800 ÷ 4 = 80 tens ÷ 4
= 20 tens
= 200
Hence, from the above results,
We can conclude that 200 is equal to 200

Question 17.
A pizza shop owner has 2,532 coupons in pamphlets of 4 coupons each. He wants to determine whether he has enough pamphlets to give one to each of his first 600 customers. Can he use an estimate, or is an exact answer required? Explain.
Big Ideas Math Answers 4th Grade Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.2 8

Answer: He has to use an Estimate.

Explanation:
Given that a pizza owner has 2,532 coupons in pamphlets of 4 coupons each and he wants to give one to each of his first 600 customers.
So, we have to first find the total number of coupons he has.
But, there are 2,532 coupons in pamphlets for 4 coupons each. The coupons are in such a way that we can cot divide all the coupons equally.
So, we have to find the estimate of 2,532
Let 2,532 be rounded to 2,400 ( Since the coupons are to be divided into 4 coupons each)
Hence,
2,400 ÷ 4 = 24 hundreds ÷ 4
= 6 hundred
= 600
From this,
We can conclude that the pizza owner has to use an Estimate.

YOU BE THE TEACHER
Your friend finds a quotient. Is his answer reasonable? Estimate to check.
Question 18.
Big Ideas Math Answers 4th Grade Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.2 9
Answer: His answer is reasonable

Explanation:
Given Expression is 273 ÷ 3
Let 273 be rounded to 270
Now,
270 ÷ 3 = 27 tens ÷ 3
= 9 tens
= 90
Since there is not much difference between the Estimate quotient and the quotient of your friend, his answer is reasonable.

Question 19.
Big Ideas Math Answers 4th Grade Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.2 10
Answer: His answer is not reasonable.

Explanation:
Given Expression is 4,290 ÷ 6
Let 4,290 be rounded to 4,200
Now,
4,200 ÷ 6 = 42 hundreds ÷ 6
= 7 hundred
= 700
Since the Estimate quotient and the quotient of your friend has large difference, the answer of your friend is not reasonable.

Think and Grow: Modeling Real Life

Example
Mount Nantai is 2,486 meters above sea level. It is about 8 times as many meters above sea level as the Taal Volcano. About how many meters above sea level is the Taal Volcano?

Big Ideas Math Answers 4th Grade Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.2 11

Mount Nantai is about 8 times as many meters above sea level as the Taal Volcano, so estimate 2,486 ÷ 8.
Think: What number close to 2,486 is easily divided by 8?
Try 2,400. 24 ÷ 8 = 3, so 2,400 ÷ 8 = 300.
Try 3,200. 32 ÷ 8 = 4, so 3,200 ÷ 8 = 400.
Choose 2,400 because 2,486 is closer to 2,400.
So, the Taal Volcano is about 400 meters above sea level.

Show and Grow

Question 20.
There are about 3,785 milliliters in 1 gallon. There are 4 times as many milliliters in 1 gallon as there are in 1 quart. About how many milliliters are in 1 quart?
Answer: 900 milliliters

Explanation:
Given that there are about 3,785 milliliters in 1 gallon and there are 4 times as many milliliters in 1 gallon as there are in 1 quart.
So,
The number of milliliters in 1 gallon = the number of milliliters in 1 quant
So, we have to find 3,785 ÷ 4
Now,
Let 3,785 be rounded to 3,600
So,
3,600 ÷ 4 = 36 hundreds ÷ 4
= 9 hundred
= 900
Hence, from the above,
We can conclude that the number of milliliters in 1 quant is: 900 milliliters.

Question 21.
A teenager works at an amusement park for 3 months and earns $2,178. She earns the same amount each month. About how much money does she earn each month?

Answer: $700

Explanation:
Given that a teenager works at an amusement park for 3 months ad earns $2,178 and she earns the same amount each month.
So, the amount of money she earns each month = $2,178 ÷ 3
Now,
Let 2,178 be rounded to 2,100 (Since the amount of money is the same each month)
Now,
2,100 ÷ 3 = 21 hundreds ÷ 3
= 7 hundred
=700
Hence, from the above,
We can conclude that she earns $700 each month.

Question 22.
An animal shelter has a bin filled with 456 pounds of dog food. There are 4 large dogs at the shelter who each eat 2 pounds of dog food each day. For about how many days can the dogs eat from the bin of food?

Big Ideas Math Answers 4th Grade Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.2 12

Answer: The dogs eat food from the bin of food for about 60 days.

Explanation:
Given that an animal shelter has a bin filled with 456 pounds of dog food.
It is also given that there are 4 large dogs at the shelter who each eat 2 pounds of dog food each day.
So,
The total amount of food dogs eat = 4 × 2 = 8 pounds
So,
The number of days the total amount of food the dogs eat = Total amount of food ÷ Total amount of food dogs eat per each day
= 456 ÷ 8
Let,
456 be rounded to 480
Now,
480 ÷ 8 = 48 tens ÷ 8
= 6 tens
= 60
Hence, from the above,
We can conclude that the dogs complete the total food in about 60 days.

Estimate Quotients Homework & Practice 5.2

Estimate the quotient.

Question 1.
33 ÷ 4.
Answer: 8

Explanation;
Let 33 be rounded off to 32.
So, now we have to find 32 ÷ 4
Now,
32 ÷ 4 = 8
Hence, from the above,
We can conclude that 33 ÷ 4 can be rounded off to 8.

Question 2.
527 ÷ 9
Answer: 60

Explanation;
Let 527 be rounded off to 540.
So, now we have to find 540 ÷ 9
Now,
540 ÷ 9 = 54 tens ÷ 9
= 6 tens
= 60
Hence, from the above,
We can conclude that 527 ÷ 9 can be rounded off to 60.

Find two estimates that the quotient is between.

Question 3.
308 ÷ 7
Answer: The Quotient of 308 ÷ 7 is between 40 and 50

Explanation:
Use 280. 28 ÷ 7 = 4, so 280 ÷ 7 = 40.
Use 350. 35 ÷ 7 = 5, so 350 ÷ 7 = 50 .
308 is between 280 and 350.
So, the quotient of 308÷ 7 is between 40 and 50.

Question 4.
3,421 ÷ 6
Answer: The Quotient of 3,421 ÷ 6 is between 500 and 600

Explanation:
Use 3,000. 30 ÷ 6 = 5, so 3,000 ÷ 6 = 500.
Use 3,600. 36 ÷ 6 = 6, so 3,600 ÷ 6 = 600 .
3,421 is between 3,000 and 3,600.
So, the quotient of 3,421 ÷ 6 is between 500 and 600.

DIG DEEPER!
Estimate to compare
Question 5.
Big Ideas Math Answers 4th Grade Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.2 13
Answer: 30 is less than 40

Explanation;
Given Expressions are 97 ÷ 3 and 40
Let 97 be rounded to 90.
Now,
90 ÷ 3 = 30
Hence, from the above values,
We can conclude that 30 is greater than 40

Question 6.
Big Ideas Math Answers 4th Grade Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.2 14
Answer: 80 is greater than 20

Explanation;
Given Expressions are 425 ÷ 5 and 182 ÷7
Let 425 be rounded to 400.
Now,
400 ÷ 5 = 40 tens ÷ 5
= 8 tens
= 80
Let 182 be rounded to 140.
Now,
140 ÷ 7 =14 tens ÷ 7
= 2 tens
= 20
Hence, from the above results,
We can conclude that 80 is greater than 20

Question 7.
Three friends want to share 261 tickets equally. They want to determine whether they can each have at least 87 tickets. Can they use an estimate, or is an exact answer required? Explain.
Big Ideas Math Answers 4th Grade Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.2 15

Answer: The three friends don’t have to use an Estimate and they needed an exact answer.

Explanation:
Given that 3 friends want to share 261 tickets equally and they want to determine whether they each have at least 87 tickets.
By using Estimate:
Let 261 be rounded to 270.
So, 270 tickets have to be shared among 3 friends.
Now,
270 ÷ 3 = 27 tens ÷ 3
= 9 tens
= 90
So, each friend have 90 tickets.
Now,
The given actual value is 261 tickets.
So,
261 ÷ 3 = ( 240 + 21 ) ÷ 3 ( Divide 261 into multiples of 3 so that all the partitioned numbers can be divided exactly)
= ( 240 ÷ 3) + ( 21 ÷ 3)
= 80 + 7
= 87 tickets.
Hence, from the above,
We can conclude that the 3 friends required an exact number.

Question 8.
Reasoning
Explain how to find a better estimate for 462 ÷ 5 than the one shown.

Round 462 to 500. Estimate 500 ÷ 5. 500 ÷ 5 = 100, so 462 ÷ 5 is about 100.

Answer: The better estimate to find 462 ÷ 5 is to round off 462 to 450.

Explanation:
Given Expression is 462 ÷ 5
Let 462 be rounded to 450
So,
450 ÷ 5 = 45 tens ÷ 5
= 9 tens
= 90
Now,
Let 462 be rounded to 500.
So,
500 ÷ 5 = 5 hundreds ÷ 5
= 1 hundred
=100
But, 462 is near to 450 when compared to 450.
So,
We can conclude that
462 ÷ 5 = 90

Question 9.
Modeling Real Life
A machine that makes toy spinners is in operation for 8 hours each day. The machine makes 7,829 toy spinners in 1 day. About how many toy spinners does the machine make each hour?

Answer: About 1,000 toy spinners the machine make each hour.

Explanation:
Given that a machine that makes toy spinners is in operation for 8 hours each day and the machine makes 7,829 toy spinners in 1 day.
So, to find the number of toy spinners the machine make each hour, we have to find the value of 7,829 ÷ 8
Now,
Let 7,829 be rounded to 8,000
So, 8,000 ÷ 8 = 80 hundreds ÷ 8
= 10 hundred
=1,000
So, from the above
We can conclude that about 1,000 toy spinners the machine make each hour.

Question 10.
Modeling Real Life
A little penguin has 10,235 feathers. The penguin has about 3 times as many feathers as a blue jay. About how many feathers does the blue jay have?

Big Ideas Math Answers 4th Grade Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.2 16
Answer: The blue jay have about 3,412 feathers.

Explanation;
Given that a penguin has 10,235 feathers and it has about 3 times as many feathers as a blue jay.
So,
The feathers of a blue jay = 10,235 ÷ 3
Now,
Let 10,235 be rounded to 10,236
So,
10,236 ÷ 3 = ( 9,000 + 1,236) ÷ 3
=( 9,000 ÷ 3 )+ ( 1,236 ÷ 3)
= 3,000 + 412
= 3,412 feathers
Hence, from the above,
We can conclude that there are about 3,412 feathers in a blue jay.

Review & Refresh

Write an equation for the comparison sentence.

Question 11.
15 is 9 more than 6.

Answer:

Question 12.
56 is 7 times as many as 8.

Answer: 7 ×8 = 56

Explanation:
Given that 56 is 7 times as many as 8
That means 8 + 8 + 8+ 8 + 8 +8 + 8 = 56
= 7 ×8 = 56

Lesson 5.3 Understand Division and Remainder

Explore and Grow

Use base ten blocks to determine whether 14 can be divided equally among 2, 3, 4, or 5 groups. Draw and describe your models.
Big Ideas Math Answers Grade 4 Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.3 1
Big Ideas Math Answers Grade 4 Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.3 2

Answer:

Structure
Explain why the units that are leftover cannot be put into a group.

Answer: The leftover units of the above group can not be put into a group because it is clearly mentioned that the blocks will have to be divided into equal parts.

Think and Grow: Find and Interpret Remainder

Sometimes you cannot divide a number evenly and there is an amount left over.
Big Ideas Math Answers Grade 4 Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.3 3
The amount left over is called the remainder. Use an R to represent the remainder.

 

Show and Grow

Use a model to find the quotient and the remainder.

Question 1.
19 ÷ 6 = _____ R _____
Big Ideas Math Answers Grade 4 Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.3 5

Answer: 6 R 1

Explanation:
Number of Units in each group = 6
Number of units leftover = 1
So,
19 ÷ 6 = 6 R 1
Where R is the Remainder (or) the number of units leftover

Question 2.
34 ÷ 5 = _____ R ______
Big Ideas Math Answers Grade 4 Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.3 6
Answer: 6 R 4

Explanation:
Number of Units in each group = 6
Number of units leftover = 4
So,
34 ÷ 5 = 6 R 4
Where R is the Remainder (or) the number of units leftover

Question 3.
26 ÷ 3 = _____ R _____
Answer: 8 R 2

Explanation:
Divide 26 into 3 equal parts.
So, we will get
Number of Units in each group = 8
Number of units leftover = 2
Hence,
26 ÷ 3 = 8 R 2
Where R is the Remainder (or) the number of units leftover

Question 4.
20 ÷ 7 = ______ R _____
Answer: 14 R 6

Explanation:
Divide 20 into 7 equal parts.
So, we will get
Number of Units in each group = 2
Number of units leftover = 6
Hence,
20 ÷ 7 = 14 R 6
Where R is the Remainder (or) the number of units leftover

Apply and Grow: Practice

Use a model to find the quotient and the remainder.
Question 5.
13 ÷ 2 = _____ R _____
Answer: 6 R 1

Explanation:
Divide 13 into 2 equal parts.
So, we will get
Number of Units in each group = 6
Number of units leftover = 1
Hence,
13 ÷ 2 = 6 R 1
Where R is the Remainder (or) the number of units leftover

Question 6.
25 ÷ 9 = ____ R ____
Answer: 2 R 7

Explanation:
Divide 25 into 9 equal parts.
So, we will get
Number of Units in each group = 2
Number of units leftover = 7
Hence,
25 ÷ 9 = 2 R 7
Where R is the Remainder (or) the number of units leftover

Question 7.
28 ÷ 8 = _______ R _____
Answer: 3 R 4

Explanation:
Divide 28 into 8 equal parts.
So, we will get
Number of Units in each group = 3
Number of units leftover = 4
Hence,
28 ÷ 8 = 3 R 4
Where R is the Remainder (or) the number of units leftover

Question 8.
15 ÷ 4 = _____ R _____
Answer: 3 R 3

Explanation:
Divide 15 into 4 equal parts.
So, we will get
Number of Units in each group = 3
Number of units leftover = 3
Hence,
15 ÷ 4 = 3 R 3
Where R is the Remainder (or) the number of units leftover

Question 9.
29 ÷ 6 = _____ R ______
Answer: 4 R 5

Explanation:
Divide 29 into 6 equal parts.
So, we will get
Number of Units in each group = 4
Number of units leftover = 5
Hence,
29 ÷ 6 = 4 R 5
Where R is the Remainder (or) the number of units leftover

Question 10.
11 ÷ 5 = ______ R ______
Answer: 2 R 1

Explanation:
Divide 11 into 5 equal parts.
So, we will get
Number of Units in each group = 2
Number of units leftover = 1
Hence,
11 ÷ 5 = 2 R 1
Where R is the Remainder (or) the number of units leftover

Question 11.
Descartes has 23 cat treats to divide equally among 4 friends. How many treats does he give each friend? How many treats are left over?
Answer:
The number of treats he gives to each friend = 5
The number of treats leftover = 3

Explanation:
Given that Descartes has 23 cat treats to divide equally among 4 friends.
So,
We have to find 23 ÷ 4 to find the number of treats he gives to each friend and the number of treats leftover
Now,
Divide 23 into 4 equal parts.
So, we will get
Number of Units in each group = 5
Number of units leftover = 3
So,
23 ÷ 4 = 5 R 3
Where R is the Remainder (or) the number of units leftover
Hence, from the above,
We can conclude that,
The number of treats he gives to each friend = 5
The number of treats leftover = 3

Question 12.
You have 26 markers. How many groups of 3 markers can you make? How many markers are left over?
Answer:

The number of markers can you make = 8
The number of markers leftover = 2

Explanation:
Given that you have 26 markers and you have to make a group of 3 markers each.
So, we have to find 26 ÷ 3 so that we can find the number of markers that you make and the number of markers leftover.
Now,
Divide 26 into 3 equal parts.
So, we will get
Number of Units in each group = 8
Number of units leftover = 2
So,
26 ÷ 3 = 8 R 2
Where R is the Remainder (or) the number of units leftover
Hence, from the above,
We can conclude that,
The number of markers can you make = 8
The number of markers leftover = 2

Question 13.
Structure
Write a division equation represented by the model.
Big Ideas Math Answers Grade 4 Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.3 7
Answer: 19 ÷ 3 = 5 R 4

Explanation:
Given model is
Big Ideas Math Answers Grade 4 Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.3 7
So, from the above model, we can see that
The number of groups = 3
The number of units in each group = 5
The number of units leftover = 4
So, to find the total number of units,
Total number of units = (The number of groups × The number of units in each group ) + ( The number of units leftover)
= (5 × 3) + 4 = 15 + 4 = 19
Hence, from the above, the division equation represented by the model is:
19 ÷ 3 = 5 R 4

Question 14.
YOU BE THE TEACHER
Is Newton correct? Draw a model to support your answer.

Big Ideas Math Answers Grade 4 Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.3 8

Answer: Yes, Newton is correct.

Explanation:
The given  division equation is:
30 ÷ 4
From the division equation, we can see that
The number of units in each group = 6
The number of units leftover = 6
So, we can write
30 ÷ 4 = 6 R 6
Newton has also given the same division equation as we got above.
Hence, Newton is correct.

Think and Grow: Modeling Real Life
Example
A water taxi transports passengers to an island. The taxi holds no more than 8 passengers at a time. There are 53 people in line to ride the water taxi.

Big Ideas Math Answers Grade 4 Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.3 9
• How many trips to the island are full?
• How many trips to the island are needed?
• How many passengers are on the last trip?
Use a model to find 53 ÷ 8.
Big Ideas Math Answers Grade 4 Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.3 10
The quotient is 6. The remainder is 5.
Interpret the quotient and the remainder.

How many trips to the island are full?
“6”  is the number of trips that have 8 passengers.
So, 6 trips to the island are full.

How many trips to the island are needed?
6 trips are full and 1 trip is not full.
So, 7 trips are needed.

How many passengers are on the last trip?
“5”  is the number of passengers that are on the last trip.
So, 5 passengers are on their last trip.

Show and Grow

Question 15.
Tours of a crayon factory have no more than 9 guests. There are 87 guests in line to tour the factory.
• How many tours are full?
• How many tours are needed?
• How many guests are on the last tour?
Big Ideas Math Answers Grade 4 Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.3 11

Answer:
a) 9 Tours are full.
b) 10 Tours are needed.
c) 6 guests are on the last tour.

Explanation:
Given that there are 87 guests in line to tour a crayon factory It is also given that there are no more than 9 guests in the tours of a crayon factory.
We have to observe that to make all the guests full on all the trips without leftovers, we will need 10 trips. i.e.., 90 guests.
So, now we have to find the number of guests in each tour of a crayon factory by finding the quotient and remainder of 87 ÷ 9.
Now,
87 ÷ 9
From this, we can see
The number of guests in each tour that are full = 9
The number of guests leftover on the last trip = 6
Hence, from the above,
We can conclude that
a) 9 Tours are full.
b) 10 Tours are needed.
c) 6 guests are on the last tour.

Understand Division and Remainder Homework & Practice 5.3

Use a model to find the quotient and the remainder.
Question 1.
25 ÷ 7 = _____ R _____
Answer: 3 R 4

Explanation:
Divide 25 into 7 equal parts.
So, we will get
Number of Units in each group = 3
Number of units leftover = 4
Hence,
25 ÷ 7 = 3 R 4
Where R is the Remainder (or) the number of units leftover

Question 2.
19 ÷ 2 = ______ R ______
Answer: 9 R 1

Explanation:
Divide 19 into 2 equal parts.
So, we will get
Number of Units in each group = 9
Number of units leftover = 1
Hence,
19 ÷ 2 = 9 R 1
Where R is the Remainder (or) the number of units leftover

Question 3.
27 ÷ 6 = _____ R ______
Answer: 4 R 3

Explanation:
Divide 27 into 6 equal parts.
So, we will get
Number of Units in each group = 4
Number of units leftover = 3
Hence,
27 ÷ 6 = 4 R 3
Where R is the Remainder (or) the number of units leftover

Question 4.
26 ÷ 4 = ______ R ______
Answer: 6 R 2

Explanation:
Divide 26 into 4 equal parts.
So, we will get
Number of Units in each group = 6
Number of units leftover = 2
Hence,
26 ÷ 4 = 6 R 2
Where R is the Remainder (or) the number of units leftover

Use a model to find the quotient and the remainder.
Question 5.
29 ÷ 8 = _____ R _____
Answer: 3 R 5

Explanation:
Divide 29 into 8 equal parts.
So, we will get
Number of Units in each group = 3
Number of units leftover = 5
Hence,
29 ÷ 8 = 3 R 5
Where R is the Remainder (or) the number of units leftover

Question 6.
11 ÷ 2 = _____ R ______
Answer: 5 R 1

Explanation:
Divide 11 into 2 equal parts.
So, we will get
Number of Units in each group = 5
Number of units leftover = 1
Hence,
11 ÷ 2 = 5 R 1
Where R is the Remainder (or) the number of units leftover

Question 7.
DIG DEEPER!
A number divided by 4 has a remainder. What numbers might the remainder be? Explain.
Answer:

Question 8.
Modeling Real Life
Tours of a space center can have no more than 7 guests. There are 31 guests in line to tour the space center.
• How many tours are full?
• How many tours are needed?
• How many guests are on the last tour?
Answer:
a) 4 Tours are full.
b) 5 Tours are needed.
c) 3 guests are on the last tour.

Explanation:
Given that there are 31 guests in line to tour the space center. It is also given that there are no more than 7 guests in the tours of a space center.
We have to observe that to make all the guests full on all the trips without leftovers, we will need 5 trips. i.e.., 35 guests.
So, now we have to find the number of guests in each tour of a crayon factory by finding the quotient and remainder of 31 ÷ 7.
Now,
31 ÷ 7
From this, we can see
The number of guests in each tour that are full = 4
The number of guests leftover on the last trip = 3
Hence, from the above,
We can conclude that
a) 4 Tours are full.
b) 5 Tours are needed.
c) 3 guests are on the last tour.

Question 9.
Modeling Real Life
You need 3 googly eyes to make one monster puppet. You have 28 googly eyes. How many monster puppets can you make?
Big Ideas Math Answers Grade 4 Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.3 12
Answer: You can make 9 monster puppets.

Explanation:
Given that you need 3  googly eyes to make 1 monster puppet.
It is also given that you have 28 googly eyes.
So, to find the number of monster puppets that you can make with the given googly eyes, we have to find the quotient and remainder of 28 ÷ 3.
The quotient of 28 ÷ 3.is the number of monster puppets that you can make with the help of given googly eyes.
Now,
28 ÷ 3
From this, we can find
The number of units in each group = 9
The number of units leftover = 1
Now, we have to observe that “The number of units in each group” is the “Quotient”
Hence, from the above,
We can conclude that You can make 9 monster puppets with 1 leftover.

Question 10.
Modeling Real Life
Forty-one students attend tryouts for a debate league. Each team can have 6 students. How many students will not be on a team?
Answer: 5 students will not be on a team

Explanation:
Given that 41 students attend tryouts for a debate league and each team can have 6 students.
So, to find the number of students that can not form a team, we have to find the quotient and remainder of 41 ÷6
Now,
41 ÷ 6
From this, we can observe that
The number of students in each team = 6
The number of students that can not form a team = 5
Hence, from the above,
We can conclude that 5 students will not be on a team.

Question 11.
Modeling Real Life
A book has 37 pages. You read 7 pages each day. How many days will it take you to finish the book?
Answer: About 5 days

Explanation:
Given that a book has 37 pages and you read 7 pages each day.
To find the number of days it will take to finish the book, we have to find the quotient of 37 ÷ 7.
Now,
37 ÷ 7
From this,
The number of groups that can be filled = 5
The number of groups that are leftover = 2
Hence, from the above,
We can conclude that the book will be finished in about 5 days.

Review & Refresh

Estimate the sum or difference.
Question 12.
50,917 – 23,846 = _____
Answer: 27,071

Question 13.
499,042 + 181,765 = ______
Answer: 680,807

5.4 Use Partial Quotients

Explore and Grow

Use the area models to find 3 × 12 and 36 ÷ 3.

Big Ideas Math Solutions Grade 4 Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.4 1
Answer:

Reasoning
How does the Distributive Property relate to each of the area models? Explain.

Answer: The “Distributive Property for Partial Quotients” works in a similar way to ” The Distributive Property of Partial Products”

Explanation:
Take the above 2 Area models as examples.
In the 1st Area model, we have to find 3 × 12.
Now, by using the Distributive Property of Partial Products,
3 × 12 = 3 × (10 + 2)
= ( 3 ×10 ) + ( 3 × 2 )
= 30 + 6
= 36
Hence, 3 × 12 = 36
In the 2ndArea model, we have to find 36 ÷ 3
Now, by using the Distributive Property for Partial Quotients,
36 ÷ 3 = ( 30 + 6 ) ÷ 3
= ( 30 ÷ 3 ) + ( 6 ÷ 3 )
= 10 + 2
= 12
Hence, 36 ÷ 3 = 12

Note:  When the quotient has to divide into Partial quotients, the partial quotients has to be the multiples of the number that has to  divide the quotient.

Think and Grow: Use Partial Quotients to Divide

To divide using partial quotients, subtract a multiple of the divisor that is less than the dividend. Continue to subtract multiples until the remainder is less than the divisor. The factors that are multiplied by the divisor are called partial quotients. Their sum is the quotient.
Example
Use an area model and partial quotients to find 235 ÷ 5.

Show and Grow

Use an area model and partial quotients to divide.
Question 1.
60 ÷ 4 = _____
Big Ideas Math Solutions Grade 4 Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.4 3

Answer:

Explanation:
By using the Distributive Property for Partial Quotients,
60 ÷ 4 = ( 40 + 20 ) ÷ 4
= ( 40 ÷ 4 ) + ( 20 ÷ 4 )
= 10 + 5
= 15
Hence, 60 ÷ 4 = 15

Question 2.
192 ÷ 3 = _____
Big Ideas Math Solutions Grade 4 Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.4 4

Answer:

Explanation:
By using the Distributive Property for Partial Quotients,
192 ÷ 3 = ( 180 + 12 ) ÷ 3
= ( 180 ÷ 3 ) + ( 12 ÷ 3 )
= 60 + 4
= 64
Hence, 192 ÷ 3 = 64

Apply and Grow: Practice

Question 3.
Use an area model and partial quotients to find 264 ÷ 8.

Big Ideas Math Solutions Grade 4 Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.4 5

Answer:

Explanation:
By using the Distributive Property for Partial Quotients,
264 ÷ 8 = ( 240 + 24 ) ÷ 8
= ( 240 ÷ 8 ) + ( 24 ÷ 8 )
= 30 + 3
= 33
Hence, 264 ÷ 8 = 33

Use partial quotients to divide
Question 4.
\(\sqrt [ 4 ]{ 96 } \)

Answer: 

Explanation:
By using the Distributive Property for Partial Quotients,
96 ÷ 4 = ( 80 + 16 ) ÷ 4
= ( 80 ÷ 4 ) + ( 16 ÷ 4 )
= 20 + 4
= 24
Hence, 96 ÷ 4 = 24

Question 5.
\(\sqrt [ 9 ]{ 405 } \)
Answer:

Explanation:
By using the Distributive Property for Partial Quotients,
405 ÷ 9 = ( 360 + 45 ) ÷ 9
= ( 360 ÷ 9 ) + ( 45 ÷ 9 )
= 40 + 5
= 45
Hence, 405 ÷ 9 = 45

Question 6.
\(\sqrt [ 6 ]{ 378 } \)
Answer:

Explanation:
By using the Distributive Property for Partial Quotients,
378 ÷ 6 = ( 360 + 18 ) ÷ 6
= ( 360 ÷ 6 ) + ( 18 ÷ 6 )
= 60 + 3
= 63
Hence, 378 ÷ 6 = 63

Question 7.
\(\sqrt [ 7 ]{ 84 } \)
Answer:

Explanation:
By using the Distributive Property for Partial Quotients,
84 ÷ 7 = ( 70 + 14 ) ÷ 7
= ( 70 ÷ 7 ) + ( 14 ÷ 7 )
= 10 +2
= 12
Hence, 84 ÷ 7 = 12

Question 8.
\(\sqrt [ 5 ]{ 735 } \)
Answer:

Explanation:
By using the Distributive Property for Partial Quotients,
735 ÷ 5 = ( 700 + 35 ) ÷ 5
= ( 700 ÷ 5 ) + ( 35 ÷ 5 )
= 140 + 7
= 147
Hence, 735 ÷ 5 = 147

Question 9.
Structure
Find the missing numbers.
Big Ideas Math Solutions Grade 4 Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.4 6
Answer: The missing numbers are: 60, 20, and 3.

Explanation:

By using the Distributive Property for Partial Quotients,
332 ÷ 4 = ( 240 + 80 + 12 ) ÷ 4
= ( 240 ÷ 4 ) + ( 80 ÷ 4 ) + ( 12 ÷ 4 )
= 60 + 20 + 3
= 83
Hence, 332 ÷ 4 = 63

Think and Grow: Modeling Real Life

Example
There are 8 students on each tug-of-war team. How many tug-of-war teams are there?
Big Ideas Math Solutions Grade 4 Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.4 7
Use an area model and partial quotients to find 128 ÷ 8.

 

Show and Grow

Use the table above.
Question 10.
There are 5 students on each relay race team. How many relay race teams are there?
Answer: 18 relay race teams

Explanation:
The given number of students in relay race = 90
Given that there are 5 students on each relay race team.
SO, to find the number of relay race teams, we have to find the value of 90 ÷ 5.

Hence, from the above,
We can conclude that there are 18 relay race teams.

Question 11.
DIG DEEPER!
There are 6 students on beach volleyball team. There are 4 fewer students on each water balloon toss team than each volleyball team. How many of each team are there?

Answer: There are 78 teams of water balloon toss team.

Explanation:
Given that,
The total number of students in the volleyball team = 96
The total number of students in the water balloon toss team = 156
It is also given that,
There are 6 students on each volleyball team and there are 4 fewer students on each water balloon toss team.
So,
The number of students on  water balloon toss team = 6 – 4 = 2 students.
Hence, to find the number of students on each water balloon toss team, we have to find the value of 156 ÷ 2 by using the Distributive Property of Partial quotients.
Using the Distributive Property of Partial quotients,
156 ÷ 2 = ( 140 + 16 ) ÷ 2
= ( 140 ÷ 2 ) + ( 16 ÷ 2 )
= 70 + 8
= 78
Hence, from the above,
We can conclude that there are 78 students on each water balloon toss team.

Question 12.
Twenty-seven students were absent on the day of sign-ups. They all decide to play kickball. There are 9 students on each kickball team. How many kickball teams are there?

Answer:  There are 3 kickball teams

Explanation:
Given that there are 27 students who were absent on the day of sign-ups and they all decide to play kickball. It is also given that there are 9 students on each kickball team.
So, to find the number of kickball teams, we have to find the value of 27 ÷ 9
Now, by using the Distributive Property of partial quotients,
27 ÷ 9 = ( 18 + 9 ) ÷ 9
= ( 18 ÷ 9 ) + ( 9 ÷ 9 )
= 2 + 1
= 3
Hence, from the above,
We can conclude that there are 3 kickball teams.

Use Partial Quotients Homework & Practice 5.4

Question 1.
Use an area model and partial quotients to find 345 ÷ 5.

Big Ideas Math Solutions Grade 4 Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.4 9

Answer:

Explanation:
Using the Distributive Property of Partial quotients,
345 ÷ 5 = ( 300 + 45 ) ÷ 5
= ( 300 ÷ 5 ) + ( 45 ÷ 5 )
= 60 + 9
= 69
Hence, 345 ÷ 5 = 63

Use partial quotients to divide.
Question 2.
\(\sqrt [ 6 ]{ 90 } \)
Answer: 15

Explanation:
Using the Distributive Property of Partial quotients,
90 ÷ 6 = ( 60 + 30 ) ÷ 6
= ( 60 ÷ 6 ) + ( 30 ÷ 6 )
= 10 + 5
= 15
Hence, 90 ÷ 6 = 15

Question 3.
\(\sqrt [ 3 ]{ 48 } \)
Answer: 16

Explanation:
Using the Distributive Property of Partial quotients,
48 ÷ 3 = ( 18 + 30 ) ÷ 3
= ( 30 ÷ 3 ) + ( 18 ÷ 3 )
= 10 + 6
= 16
Hence, 48 ÷ 3 = 16

Question 4.
\(\sqrt [ 8 ]{ 200 } \)
Answer: 25

Explanation:
Using the Distributive Property of Partial quotients,
200 ÷ 8 = ( 160 + 40 ) ÷ 8
= ( 160 ÷ 8 ) + ( 40 ÷ 8 )
= 20 + 5
= 25
Hence, 200 ÷ 8 = 25

Use partial quotients to divide.
Question 5.
\(\sqrt [ 4 ]{ 56 } \)
Answer: 14

Explanation:
Using the Distributive Property of Partial quotients,
56 ÷ 4 = ( 40 + 16 ) ÷ 4
= ( 40 ÷ 4 ) + ( 16 ÷ 4 )
= 10 + 4
= 14
Hence, 56 ÷ 4 = 14

Question 6.
\(\sqrt [ 7 ]{ 511 } \)
Answer: 73

Explanation:
Using the Distributive Property of Partial quotients,
511 ÷ 7 = ( 490 + 21 ) ÷ 7
= ( 490 ÷ 7 ) + ( 21 ÷ 7 )
= 70 + 3
= 73
Hence, 511 ÷ 7 = 73

Question 7.
\(\sqrt [ 9 ]{ 423 } \)
Answer: 47

Explanation;
Using the Distributive Property of Partial quotients,
423 ÷ 9 = ( 360 + 63 ) ÷ 9
= ( 360 ÷ 9 ) + ( 63 ÷ 9 )
= 40 + 7
= 47
Hence, 423 ÷ 9 = 47

Question 8.
YOU BE THE TEACHER
Descartes finds 952 ÷ 8. Is he correct? Explain.

Big Ideas Math Solutions Grade 4 Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.4 10

Answer: Descartes is correct.

Explanation:

By using the Distributive Property for Partial Quotients,
952 ÷ 8 = ( 800 + 80 + 72 ) ÷ 8
= ( 800 ÷ 8 ) + ( 80 ÷ 8 ) + ( 72 ÷ 8 )
= 100 + 10 + 9
= 119
Hence, 952 ÷ 8 = 119

Question 9.
Writing
Explain how you can solve a division problem in more than one-way using partial quotients.
Answer: We can solve partial quotients in more than one-way by dividing the partial quotients in more than 2 multiples.

Explanation:
Take an example of 952 ÷ 8.

From the above, we can see that the partial quotients are divided in to 3 quotients which are the multiples of 8.

Question 10.
Modeling Real Life
Each shelter animal gets 3 toys. How many shelter animals are there?

Big Ideas Math Solutions Grade 4 Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.4 11

Answer: There are 56 shelter animals.

Explanation:
From the given table,
The number of toys donated = 168
It is also given that each shelter animal gets 3 toys.
So,
The number of shelter animals = The number of toys donated ÷ The number of toys that each shelter animal gets
= 168 ÷ 3
By using the Distributive property of quotients,
168 ÷ 3 = ( 150 + 18 ) ÷ 3
= ( 150 ÷ 3 ) + ( 18 ÷ 3 )
= 50 + 6
= 56
Hence, from the above,
We can conclude that there are 56 shelter animals.

Review & Refresh

Find the product.
Question 11.
40 × 70 = _____
Answer: 2,800

Explanation:
Using the Place-value method,
40 × 70 = 40 × 7 tens
= 4 tens × 7 tens
= 28 × 1 ten × 1 ten
= 28 × 10 × 10
= 2,800
Hence, 40 × 70 = 2,800

Question 12.
30 × 58 = ______
Answer: 1740

Explanation:
Using the Distributive Property of partial products,
30 × 58 = 30 × ( 50 + 8 )
= ( 30 × 50 ) + ( 30 × 8 )
= 1,500 + 240
= 1,740
Hence, 30 × 58 = 1,740

Question 13.
62 × 90 = ______
Answer: 5580

Explanation:
Using the place-value method,
62 × 90 = 62 × 9 tens
= 558 tens
= 558 × 10
= 5,580
Hence,62 × 90 = 5,580

Lesson 5.5 Use Partial Quotients with a Remainder

Explore and Grow

Use an area model to find 125 ÷ 5.

Big Ideas Math Answer Key Grade 4 Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.5 1

Answer:

Explanation:
Using the Distributive Property of partial quotients,
125 ÷ 5 = ( 100 + 25 ) ÷ 5
= ( 100 ÷ 5 ) + ( 25 ÷ 5 )
= 20 + 5
= 25
Hence, 125 ÷ 5 = 25

Can you use an area model to find 128 ÷ 5? Explain your reasoning.
Big Ideas Math Answer Key Grade 4 Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.5 2

Answer: We can’t use an area model to find 128 ÷ 5 because 128 can’t be divided by 5 as 128 leaves a remainder.

Explanation:
Given division Expression is 128 ÷ 5
From this Expression, we can say that 128 can’t be divided by 5
When 128 is divided by 5, it leaves a remainder of 3.
So,
128 ÷ 5
From this,
The number of units in each group = 25
The number of units that are leftover = 3
Hence,
128 ÷ 5 = 25 R 3

Construct Arguments
Explain to your partner how your model shows that 5 does not divide evenly into 128.

Answer:


                               


Like the above arrangement, do the remaining 20 times.
Hence, we arranged 125 models in to a group and there are 3 models that are not fit in the group.
In this way, 128 will not be divided by 5 evenly.

Think and Grow: Practice

Example
Use partial quotients to find 2,918 ÷ 4.

So, 2,918 ÷ 4 = 2,916 R 2.

Show and Grow

Use partial quotients to divide.
Question 1.
82 ÷ 3 = _____
Big Ideas Math Answer Key Grade 4 Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.5 5
Answer: 27 R 1

Explanation:

Using Distributive property of partial quotients,
82 ÷ 3 = ( 60 + 21 ) ÷ 3
= ( 60 ÷ 3 ) + ( 21 ÷ 3)
= 20 + 7
= 27 R 1
Hence, 82 ÷ 3 = 27 R 1

Question 2.
754 ÷ 9 = _____
Big Ideas Math Answer Key Grade 4 Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.5 6
Answer: 83 R 7

Explanation:

Using Distributive property of partial quotients,
754 ÷ 9 = ( 720 + 27 ) ÷ 9
= ( 720 ÷ 9 ) + ( 27 ÷ 9 )
= 80 + 3
= 83 R 7
Hence, 754 ÷ 9 = 83 R 7

Question 3.
\(\sqrt [ 8 ]{ 460 } \)
Answer:  57 R 4

Explanation:
Using Distributive property of partial quotients,
460 ÷ 8 = ( 400 + 56 ) ÷ 8
= ( 400 ÷ 8 ) + ( 56 ÷ 8 )
= 50 + 7
= 57 R 4
Hence, 460 ÷ 8 = 57 R 4

Question 4.
\(\sqrt [ 5 ]{ 3,242 } \)
Answer: 648 R 2

Explanation:
Using Distributive property of partial quotients,
3,242 ÷ 5 = ( 3,000 + 200 + 40 ) ÷ 5
= ( 3,000 ÷ 5 ) + ( 200 ÷ 5 ) + ( 40 ÷ 5 )
= 600 + 40 + 8
= 648 R 2
Hence, 3,242 ÷ 5 = 648 R 2

Question 5.
\(\sqrt [ 6 ]{ 5,850 } \)
Answer: 975 R 0

Explanation:
Using Distributive property of partial quotients,
5,850 ÷ 6 = ( 5,400 + 420 + 30 ) ÷ 6
= ( 5,400 ÷ 6 ) + ( 420 ÷ 6 ) + ( 30 ÷ 6 )
= 900 + 70 + 5
Hence, 5,850 ÷ 6 = 975 R 0

Apply and Grow: Practice

Use partial quotients to divide.
Question 6.
\(\sqrt [ 5 ]{ 63 } \)
Answer: 12 R 3

Explanation:
Using Distributive property of partial quotients,
63 ÷ 5 = ( 50 + 10 ) ÷ 5
= ( 50 ÷ 5 ) + ( 10 ÷ 5 )
= 10 + 2
= 12 R 3
Hence, 63 ÷ 5 = 12 R 3

Question 7.
\(\sqrt [ 7 ]{ 401 } \)
Answer: 57 R 2

Explanation:
Using Distributive property of partial quotients,
401 ÷ 7 = ( 350 + 49 ) ÷ 7
= ( 350 ÷ 7 ) + ( 49 ÷ 7 )
= 50 + 7
= 57 R 2
Hence, 401 ÷ 7 = 57 R 2

Question 8.
\(\sqrt [ 4 ]{ 5,237 } \)
Answer: 1,309 R 1

Explanation:
Using Distributive property of partial quotients,
5,237 ÷ 4 = ( 5,200 + 36 ) ÷ 4
= ( 5,200 ÷ 4 ) + ( 36 ÷ 4 )
= 1,300 + 9
= 1,309 R 1
Hence, 5,237 ÷ 4 = 1,309 R 1

Question 9.
\(\sqrt [ 9 ]{ 256 } \)
Answer: 28 R 4

Explanation:
Using Distributive property of partial quotients,
256 ÷ 9 = ( 180 + 72 ) ÷ 9
= ( 180 ÷ 9 ) + ( 72 ÷ 9 )
= 20 + 8
= 28 R 4
Hence, 256 ÷ 9 = 28 R 4

Question 10.
\(\sqrt [ 8 ]{ 945 } \)
Answer: 118 R 1

Explanation:
Using Distributive property of partial quotients,
945 ÷ 8 = ( 880 + 64 ) ÷ 8
= ( 880 ÷ 8 ) + ( 64 ÷ 8 )
= 110 + 8
= 118 R 1
Hence, 945 ÷ 8 = 118 R 1

Question 11.
\(\sqrt [ 2 ]{ 7,043 } \)
Answer: 3521 R 1

Explanation:
Using Distributive property of partial quotients,
7,043 ÷ 2 = ( 7,000 + 42 ) ÷ 2
= ( 7,000 ÷ 2 ) + ( 42 ÷ 2 )
= 3,500 + 21
= 3,521 R 1
Hence, 7,043 ÷ 2 = 3,521 R 1

Question 12.
The third, fourth, and fifth grades make 146 science projects for a fair. Did each grade make the same number of projects? Explain.

Big Ideas Math Answer Key Grade 4 Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.5 7

Answer:  No, each grade did not make the same number of projects.

Explanation:
It is given that the third, fourth, and fifth grades make 146 science projects for a fair.
From this,
The total number of grades = 3
Now, to find whether each grade makes the same number of projects or not, we can find out by knowing the value of 146 ÷ 3.
Now,
Using Distributive property of partial quotients,
146 ÷ 3 = ( 120 + 24 ) ÷ 3
= ( 120 ÷ 3 ) + ( 24 ÷ 3 )
= 40 + 8
= 48 R 2
Hence, 146 ÷ 3 = 48 R 2

Question 13.
Structure
Newton found 315 ÷ 6. Explain how the steps would be different if he had used 50 as the first partial quotient?
Big Ideas Math Answer Key Grade 4 Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.5 8
Answer:

By changing the first partial quotient to 50, the number of partial quotients reduced to 2, and hence the calculation of the division becomes easy.

Think and Grow: Modeling Real Life

Example
There are 1,862 people attending a mud run. Each wave of runners can have8 people. How many waves of runners are needed?
Big Ideas Math Answer Key Grade 4 Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.5 9
Use partial quotients to find 1,862 ÷ 8.

1,862 ÷ 8 = 232 R 6
Interpret the quotient and the remainder.
The quotient is 232. This means that 232 waves of runners will have 8 people.
The remainder is 6. This means that 1 wave of runners will have 6 people.
So, 1864  waves of runners are needed.

Show and Grow

Question 14.
A juice factory has 768 fluid ounces of juice for guests to sample. A worker pours the juice into 5-fluid ounce cups. How many cups does the worker fill?
Answer: The number of cups that the worker fill = 153 cups

Explanation:
Given that a juice factory has 78 fluid ounces of juice for guests to sample and a worker pours the juice into 5- fluid ounce cups.
To find the number of cups that the worker fill, we have to find the quotient of 768 ÷ 5
Now,
768 ÷ 5 = ( 700 + 60 + 5 ) ÷ 5
= ( 700 ÷ 5 ) + ( 60 ÷ 5 ) + ( 5 ÷ 5 )
= 140 + 12 + 1
= 153 R 3
Hence,
The number of cups that the worker fill = 153 cups

Question 15.
A toy company designs 214 collectible figures. The company releases 6 of the figures each month. How many months will it take the company to release all of the collectible figures? How many years will it take?
Big Ideas Math Answer Key Grade 4 Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.5 11
Answer:
The number of months that the company released all the toys = About 35 years
The number of years that the company released all the toys = About 3 years

Explanation:
Given that a toy company designs 214 collectible figures and the company releases 6 of the figures each month.
Now,
The number of months that the company released all toys = 214 ÷ 6
Now,
214 ÷ 6 = ( 180 + 30 ) ÷ 6
= ( 180 ÷ 6 ) + ( 30 ÷ 6 )
= 30 + 5
= 35 R 4
We know that,
1 year = 12 months
It is given that 6 figures released each month.
So, for 1 year, 72 toys will be released.
Now,
The number of years that the company released all the toys = 214 ÷ 72
Now,
214 ÷ 72 = 2 R 70
Hence, from the above,
We can conclude that
The number of months that the company released all the toys = About 35 years
The number of years that the company released all the toys = About 3 years

Use Partial Quotients with a Remainder Homework & Practice 5.5

Use partial quotients to divide.
Question 1.
\(\sqrt [ 4 ]{ 67 } \)
Answer: 16 R 3

Explanation:
Using Distributive property of partial quotients,
67 ÷ 4 = ( 60 + 4 ) ÷ 4
= ( 60 ÷ 4 ) + ( 4 ÷ 4 )
= 15 + 1
= 16 R 3
Hence, 67 ÷ 4 = 16 R 3

Question 2.
\(\sqrt [ 2 ]{ 715 } \)
Answer: 357 R 1

Explanation:
Using Distributive property of partial quotients,
715 ÷ 2 = ( 700 + 14 ) ÷ 2
= ( 700 ÷ 2 ) + ( 14 ÷ 2 )
= 350 + 7
= 357 R 1
Hence, 715 ÷ 2 = 357 R 1

Question 3.
\(\sqrt [ 5 ]{ 1,308 } \)
Answer: 261 R 3

Explanation:
Using Distributive property of partial quotients,
1,308 ÷ 5 = ( 1,000 + 300 + 5 ) ÷ 5
= ( 1,000 ÷ 5 ) + ( 300 ÷ 5 ) + ( 5 ÷ 5 )
= 200 + 60 + 1
= 261 R 3
Hence, 1,308 ÷ 5 = 261 R 3

Question 4.
\(\sqrt [ 3 ]{ 516 } \)
Answer: 172 R 0

Explanation:
Using Distributive property of partial quotients,
516 ÷ 3 = ( 510 + 6 ) ÷ 3
= ( 510 ÷ 3 ) + ( 6 ÷ 3 )
= 170 + 2
= 172 R 0
Hence, 516 ÷ 3 = 172 R 0

Question 5.
\(\sqrt [ 9 ]{ 2,497 } \)
Answer: 277 R 4

Explanation:
Using Distributive property of partial quotients,
2,497 ÷ 9 = ( 1,800 + 630 + 36 + 27 )  ÷ 9
= ( 1,800 ÷ 9 ) + ( 630 ÷ 9 ) + ( 36 ÷ 9 ) + ( 27 ÷ 9 )
= 200 + 70 + 4 + 3
= 277 R 4
Hence, 2,497 ÷ 9 = 277 R 4

Question 6.
\(\sqrt [ 6 ]{ 831 } \)
Answer: 138 R 3

Explanation:
Using Distributive property of partial quotients,
831 ÷ 6 = ( 780 + 48 ) ÷ 6
= (780 ÷ 6 ) + ( 48 ÷ 6 )
= 130 + 8
= 138 R 3
Hence, 831 ÷ 6 = 138 R 3

Use partial quotients to divide.
Question 7.
\(\sqrt [ 9 ]{ 476 } \)
Answer: 52 R 8

Explanation:
Using Distributive property of partial quotients,
476 ÷ 9 = ( 450 + 18 ) ÷ 9
= ( 450 ÷ 9 ) + ( 18 ÷ 9 )
= 50 + 2
= 52 R 8
Hence, 476 ÷ 9 = 52 R 8

Question 8.
\(\sqrt [ 7 ]{ 2,254 } \)
Answer: 322 R 0

Explanation:
Using Distributive property of partial quotients,
2,254 ÷ 7 = ( 2,100 + 140 + 14 ) ÷ 7
= ( 2,100 ÷ 7 ) + ( 140 ÷ 7 ) + ( 14 ÷ 7 )
= 300 + 20 + 2
= 322 R 0
Hence, 2,254 ÷ 7 = 322 R 0

Question 9.
\(\sqrt [ 4 ]{ 3,018 } \)
Answer: 754 R 2

Explanation:
Using Distributive property of partial quotients,
3,018 ÷ 4 = ( 2,800 + 200 + 16 ) ÷ 4
= ( 2,800 ÷ 4 ) + ( 200 ÷ 4 ) + ( 16 ÷ 4 )
= 700 + 50 + 4
= 754 R 2
Hence, 3,018 ÷ 4 = 754 R 2

Question 10.
Reasoning
Show how to use the least number of partial quotients to find 3,526 ÷ 4

Answer:
By using the Distributive property of partial quotients,
3,526 ÷ 4 = ( 3,520 + 4) ÷ 4
= ( 3,520 ÷ 4 ) + ( 4 ÷ 4 )
= 880 + 1
= 881 R 2
Hence, 3,526 ÷ 4 = 881 R 2
Hence, from the above,
We can conclude that the least number of partial quotients are “2” in 3,526 ÷ 4

Question 11.
Modeling Real Life
A gardening center has 1,582 pots to fill. Each bag of soil can fill 4 pots. How many bags of soil are needed?
Big Ideas Math Answer Key Grade 4 Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.5 12

Answer: The number of bags of soil needed = 395 R 2

Explanation:
Given that a gardening center has 1,582 pots to fill and each bag of soil can fill 4 pots.
So, the number of bags of soil needed = Number of pots ÷ Number of pots that each bag of soil fill
= 1,582 ÷ 4
By using the Distributive property of quotients,
1,582 ÷ 4 = ( 1,200 + 360 + 20 ) ÷ 4
= ( 1,200 ÷ 4 ) + ( 360 ÷ 4 ) + ( 20 ÷ 4 )
= 300 + 90 + 5
= 395 R 2
Hence, from the above,
We can conclude that the number of bags of soil needed are: 395 R 2

Question 12.
DIG DEEPER!
You have 178 photos. You put 3 photos on each page of an album. Your friend has 354 photos. She puts 6 photos on each page of an album. Who uses more pages? Explain.
Answer: The one who has 178 photos uses more pages.

Explanation:
Given that you 178 photos and you put 3 photos on each page of an album.
Hence, The number of pages used by you = The number of photos ÷ The number of photos on each page of an album
= 178 ÷ 3
By using the Distributive property of quotients,
178 ÷ 3 = ( 150 + 27 ) ÷ 3
= ( 150 ÷ 3 ) + ( 27 ÷ 3 )
= 50 + 9
= 59 R 1 pages
It is also given that your friend has 354 photos and she puts 6 photos on each page of an album.
Hence, The number of pages used by your friend = The number of photos ÷ The number of photos on each page of an album
= 354 ÷ 6
By using the Distributie property of quotients,
354 ÷ 6 = ( 300 + 48 + 6 ) ÷ 6
= ( 300 ÷ 6 ) + ( 48 ÷ 6 ) + ( 6 ÷ 6 )
= 50 + 8 + 1
= 59 pages
Hence, from the above,
we can conclude that you have more pages wh0 have 60 pages.

Review & Refresh

Question 13.
An Olympic swimmer wants to eat 10,000 calories each day. He eats 3,142 calories at breakfast and 3,269 calories at lunch. How many more calories must the swimmer eat to reach his goal?
Answer: The swimmer must eat 3,589 calories to reach his goal.

Explanation:
Given that an Olympic swimmer wants to eat 10,000 calories each day.
It is also given that he eats 3,142 calories at breakfast and 3,269 calories at lunch.
So,
The total calories he eat = 3,142 + 3,269 = 6,411 calories
Hence,
The number of calories he wants to eat more to reach his goal = Total number of calories – The number of calories he eats
= 10,000 – 6,411
= 3,589 calories
Hence, from the above,
We can conclude that the swimmer must eat 3,589 calories to reach his goal.

Lesson 5.6 Divide Two-Digit Numbers by One-Digit Numbers

Explore and Grow

Use a model to find each quotient. Draw each model.
Big Ideas Math Answers 4th Grade Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.6 1
84 ÷ 4

85 ÷ 5
Answer:
84 ÷ 4 = 21
85 ÷ 5 = 17

Explanation:
Let the given Expressions be marked as A) and B)
So,
A) 84 ÷4  B) 85 ÷ 5
Now,
By using the Distributive property of partial quotients,
84 ÷ 4 = ( 80 + 4 ) ÷ 4
= ( 80 ÷ 4 ) + ( 4 ÷ 4 )
= 20 + 1
= 21
Hence, 84 ÷ 4 = 21
By using the Distributive property of partial quotients,
85 ÷ 5 = ( 80 + 5 0 ÷ 5
= ( 80 ÷ 5 ) + ( 5 ÷ 5 )
= 16 + 1
= 17
Hence, 85 ÷ 5 = 17

Construct Arguments
Explain to your partner how your methods for finding the quotients above are the same. Then explain how they are different.
Answer: Let you and your partner are finding the value of 70 ÷ 5.
You want to use the Partial quotients method and your partner wants to use the area model method.
Even though the methods of your’s and your partner are different, the answer will be the same.
Now,
By using the Distributive Property of partial quotients,
70 ÷ 5 = ( 65 + 5) ÷ 5
= ( 65 ÷ 5 ) + ( 5 ÷ 5 )
= 13 + 1
= 14
So, 70 ÷ 5 = 14
Now,
By using the Area Model,
70 ÷ 5
From this,
The number of units that are grouped together = 14
The number of units that are leftover = 0
So, 70 ÷ 5 = 14 R 0

Think and Grow: Use Regrouping to Divide

Show and Grow

Divide. Then check your answer.
Question 1.
Big Ideas Math Answers 4th Grade Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.6 3
Answer: 96 ÷ 6 = 16

Explanation:

By using the partial quotients method,
96 ÷ 6 = ( 90 + 6 ) ÷ 6
= ( 90 ÷ 6 ) + ( 6 ÷ 6 )
= 15 + 1
= 16
Hence, 96 ÷ 6 = 16

Question 2.
Big Ideas Math Answers 4th Grade Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.6 4
Answer: 88 ÷ 2 = 44

Explanation:

By using the partial quotients method,
88 ÷ 2 = ( 80 + 8 ) ÷ 2
= ( 80 ÷ 2 ) + ( 8 ÷ 2 )
= 40 + 4
= 44
So, 88 ÷ 2 = 44

Question 3.
Big Ideas Math Answers 4th Grade Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.6 5
Answer: 74 ÷ 5 = 14 R 4

Explanation:

By using the partial quotients method,
74 ÷ 5 = ( 65 + 5 ) ÷ 5
= ( 65 ÷ 5 ) + ( 5 ÷ 5 )
= 13 + 1
= 14
So, 74 ÷ 5 = 14 R 4

Apply and Grow: Practice

Divide. Then check your answer.
Question 4.
\(\sqrt [ 5 ]{ 60 } \)
Answer: 60 ÷ 5 = 12

Explanation:
By using the partial quotients method,
60 ÷ 5 = ( 55 + 5 ) ÷ 5
= ( 55 ÷ 5 ) + ( 5 ÷ 5 )
= 11 + 1
= 12
Hence, 60 ÷ 5 = 12

Question 5.
\(\sqrt [ 6 ]{ 70 } \)
Answer: 70 ÷ 6 = 11 R 4

Explanation:
70 ÷ 6 = ( 60 + 6 ) ÷ 6
= ( 60 ÷ 6 ) + ( 6 ÷ 6 )
= 10 + 1
= 11
Hence, 70 ÷ 6 = 11 R 4

Question 6.
\(\sqrt [ 8 ]{ 90 } \)
Answer: 90 ÷ 8 = 11 R 2

Explanation:
By using the partial quotients method,
90 ÷ 8 = ( 80 + 8 ) ÷ 8
= ( 80 ÷ 8 ) + ( 8 ÷ 8 )
= 10 + 1
= 11
Hence, 90 ÷ 8 = 11 R 2

Question 7.
\(\sqrt [ 3 ]{ 93 } \)
Answer: 93 ÷ 3 = 31 R 0

Explanation:
By using the partial quotients method,
93 ÷ 3 = ( 90 + 3 ) ÷ 3
= ( 90 ÷ 3 ) + ( 3 ÷ 3 )
= 30 + 1
= 31
Hence, 93 ÷ 3 = 31 R 0

Question 8.
\(\sqrt [ 2 ]{ 45 } \)
Answer: 45 ÷ 2 = 22 R 1

Explanation:
By using the partila quotients method,
45 ÷ 2 = ( 40 + 4 ) ÷ 2
= ( 40 ÷ 2 ) + ( 4 ÷ 2 )
= 20 + 2
= 22
Hence, 45 ÷ 2 = 22 R 1

Question 9.
\(\sqrt [ 3 ]{ 64 } \)
Answer: 64 ÷ 3 = 21 R 1

Explanation:
By using the partial quotients method,
64 ÷ 3 = ( 60 + 3 ) ÷ 3
= ( 60 ÷ 3 ) + ( 3 ÷ 3 )
= 20 + 1
= 21
Hence, 64 ÷ 3 = 21 R 1

Question 10.
\(\sqrt [ 6 ]{ 42} \)
Answer: 42 ÷ 6 = 7

Explanation:
By using the partial quotients method,
42 ÷ 6 = ( 36 + 6 ) ÷ 6
= ( 36 ÷ 6 ) + ( 6 ÷ 6 )
= 6 + 1
= 7
Hence, 42 ÷ 6 = 7

Question 11.
\(\sqrt [ 8 ]{ 36 } \)
Answer: 36 ÷ 8 = 4 R 4

Explanantion:
By using the partial quotients method,
36 ÷ 8 = ( 24 + 8 ) ÷ 8
= ( 24 ÷ 8 ) + ( 8 ÷ 8 )
= 3 + 1
= 4 R 4
Hence, 36 ÷ 8 = 4 R 4

Question 12.
\(\sqrt [ 7 ]{ 50 } \)
Answer: 50 ÷ 7 = 7 R 1

Explanation:
By using the partial quotients method,
50 ÷ 7 = ( 42 + 7 ) ÷ 7
= ( 42 ÷ 7 ) + ( 7 ÷ 7 )
= 6 + 1
= 7
Hence, 50 ÷ 7 = 7 R 1

Question 13.
Writing
Explain how you can use estimation to check the reasonableness of your answer when dividing a two-digit number by a one-digit number.
Answer:  We can check the estimation to check the reasonableness of the answer when dividing a two-digit number by a one-digit number by 2 methods. They are:
A) The partial quotients method  B) Regrouping  C) Area model method

Question 14.
Structure
Find the missing numbers.
Big Ideas Math Answers 4th Grade Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.6 6
Answer: The missing numbers are: 4, 2, 5, 2, and 2 ( The order of numbers is from top to bottom)

Explanation:

By using the partial quotients method,
72 ÷ 5 = ( 65 + 5 ) ÷ 5
= ( 65 ÷ 5 ) + ( 5 ÷ 5 )
= 13 + 1
= 14
Hence, 72 ÷ 5 = 14 R 2

Think and Grow: Modeling Real Life

Example
A house cat has 64 muscles in its ears. It has the same number of muscles in each ear. How many muscles does the house cat have in each ear?
Big Ideas Math Answers 4th Grade Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.6 7
The house cat has 2 ears, so find 64 ÷ 2.
Think: 64 is 6 tens and 4 ones.
Divide the tens.

Show and Grow

Question 15.
You earn 5 cents for each plastic bottle you recycle. You recycle some bottles and earn 75 cents. How many bottles did you recycle?
Answer:  75 ÷ 5 = 15 bottles

Explanation:
Given,
The cost of each plastic bottle you recycle = 5 cents
The total money earned by recycling plastic bottles = 75 cents
So,
The number of plastic bottles = The total money earned by recycling plastic bottles ÷ The cost of each plastic bottle you recycle
=75 ÷ 5
Now,
By using the partial quotients method,
75 ÷ 5 = ( 70 + 5 ) ÷ 5
= ( 70 ÷ 5 ) + ( 5 ÷ 5 )
= 14 + 1
= 15
Hence, from the above,
We can conclude that there are 15 plastic bottles that are recycled.

Question 16.
DIG DEEPER!
A cross-country runner must run 80 miles in 1 week. He wants to run about the same number of miles each day. How many miles should he run each day? How can you interpret the remainder?
Answer: 80 ÷ 7 = 11 R 3

Explanation:

By using the partial quotients method,
80 ÷ 7 = ( 70 + 7 ) ÷ 7
= ( 70 ÷ 7 ) + ( 7 ÷ 7 )
= 10 + 1
= 11
Hence, 80 ÷ 7 = 11 R 3

Question 17.
Admission to a go-kart park costs a total of $78 for 3 adults and 3 children. The price is the same for all ages. What is the cost of admission for each person?
Big Ideas Math Answers 4th Grade Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.6 9
Answer: 78 ÷ 6 = $13

Explanation:
Given that admission to a go-kart, park costs a total of $78 for 3 adults and 3 children and it is also given that the price is the same for all ages.
Hence, the price for children and adults are the same.
Hence, total number of people = 3 adults + 3 children = 6 people
To find the cost of admission for each person, we have to find the value of 78 ÷ 6
Now,
By using the partial products method,
78 ÷ 6 = ( 60 + 18 ) ÷ 6
= ( 60 ÷ 6 ) + ( 18 ÷ 6 )
= 10 + 3
= 13
Hence, from the above,
We can conclude that the cost of admission for each person is: $13.

Divide Two-Digit Numbers by One-Digit Numbers Homework & Practice 5.6

Divide. Then check your answer.
Question 1.
Big Ideas Math Answers 4th Grade Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.6 10
Answer: 85 ÷ 5 = 17

Explanation:

By using the partial quotients method,
85 ÷ 5 = ( 80 + 5 ) ÷ 5
= ( 80 ÷ 5 ) + ( 5 ÷ 5 )
= 16 + 1
= 17
Hence, 85 ÷ 5 = 17

Question 2.
Big Ideas Math Answers 4th Grade Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.6 11
Answer: 63 ÷ 3 = 21

Explanation:

By using the partial quotients method,
63 ÷ 3 = ( 60 + 3 ) ÷ 3
= ( 60 ÷ 3 ) + ( 3 ÷ 3 )
=20 + 1
= 21
Hence, 63 ÷ 3 = 21

Question 3.
Big Ideas Math Answers 4th Grade Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.6 12
Answer: 94 ÷ 7 =13 R 3

Explanation:

By using the partial quotients method,
94 ÷7 = ( 84 + 7 ) ÷ 7
= ( 84 ÷ 7 ) + ( 7 ÷ 7 )
= 12 + 1
= 13 R 3
Hence, 94 ÷ 7 = 13 R 3

Divide. Then check your answer.
Question 4.
\(\sqrt [ 6 ]{ 74 } \)
Answer: 74 ÷ 6 = 12 R 2

Explanation:
By using the partial products method,
74 ÷ 6 = ( 60 + 12 ) ÷ 6
= ( 60 ÷ 6 ) + ( 12 ÷ 6 )
= 10 + 2
= 12 R 2
Hence, 74 ÷ 6 = 12 R 2

Question 5.
\(\sqrt [ 8 ]{ 92 } \)
Answer: 92 ÷ 8 = 11 R 4

Explanation:
By using the partial products method,
92 ÷ 8 = ( 80 + 8 ) ÷ 8
= ( 80 ÷ 8 ) + ( 8 ÷ 8 )
= 10 + 1
= 11 R 4
Hence, 92 ÷ 8 = 11 R 4

Question 6.
\(\sqrt [ 3 ]{ 50 } \)
Answer: 50 ÷ 3 = 16 R 2

Explanation:
By using the partial quotients method,
50 ÷ 3 = ( 42 + 6 ) ÷ 3
= ( 42 ÷ 3 ) + ( 6 ÷ 3 )
= 14 + 2
= 16 R 2
Hence, 50 ÷ 3 = 16 R 2

Question 7.
\(\sqrt [ 2 ]{ 83 } \)
Answer: 83 ÷ 2 = 41 R 1

Explanation:
By using the partial quotients method,
83 ÷ 2 = ( 80 + 2 ) ÷ 2
= ( 80 ÷ 2 ) + ( 2 ÷ 2 )
= 40 + 1
= 41 R 1
Hence, 83 ÷ 2 = 41 R 1

Question 8.
\(\sqrt [ 9 ]{ 72 } \)
Answer: 72 ÷ 9 = 8 R 0

Explanation:
By using the partial quotients method,
72 ÷ 9 = ( 63 + 9 ) ÷ 9
= ( 63 ÷ 9 ) + ( 9 ÷ 9 )
= 7 + 1
= 8 R 0
Hence, 72 ÷ 9 = 8 R 0

Question 9.
\(\sqrt [ 7 ]{ 65 } \)
Answer: 65 ÷ 7 = 9 R 2

Explanation:
By using the artial products method,
65 ÷ 7 = ( 56 + 7 ) ÷ 7
= ( 56 ÷ 7 ) + ( 7 ÷ 7 )
= 8 + 1
= 9 R 2
Hence, 65 ÷ 7 = 9 R 2

Question 10.
Which One Doesn’t Belong?
Which problem does not require regrouping not to solve?
Big Ideas Math Answers 4th Grade Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.6 13
Answer: Let the given Expressions be named as A), B), C), and D)
Hence,
A) 36 ÷ 2
B) 55 ÷ 2
C) 47 ÷ 2
D) 92 ÷ 2
From the above Expressions, we can conclude that Expression A) does not require Regrouping to solve.

Explanation:
Let the given Expressions be named as A), B), C), and D)
Hence,
A) 36 ÷ 2
B) 55 ÷ 2
C) 47 ÷ 2
D) 92 ÷ 2
A) 36 ÷ 2 = ( 3 tens + 6 ones ) ÷ 2
= ( 3 tens ÷ 2 ) + ( 6 ones ÷ 2 )
= 15 + 3
= 18
Here, as there are no tens left to regroup, there is no need for Regrouping again.
B) 55 ÷ 2 = ( 5 tens + 5 ones ) ÷ 2
= ( 5 tens ÷ 2 ) + ( 5 ones ÷ 2 )
It can’t be solved further and there is need for us for further Regrouping.
Hence, the remaining two will also be the same.
Hence, from the above,
We can conclude that Expression A) does not require Regrouping.

Question 11.
Modeling Real Life
A team of 6 students finishes an obstacle course in 66 minutes. Each student spends the same number of minutes on the course. How many minutes is each student on the course?

Big Ideas Math Answers 4th Grade Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.6 14

Answer: Each student takes 11 minutes to finish an obstacle course.

Explanation:
Given that a team of 6 students finishes an obstacle course in 66 minutes and it is also given that each student spends the same number of minutes on the course.
So,
The number of minutes each student takes on the course = The number of total minutes taken to complete an obstacle course ÷ The number of members on each team
= 66 ÷ 6
Now,
By using the partial quotients method,
66 ÷ 6 = ( 60 + 6 ) ÷ 6
= ( 60 ÷ 6 ) + ( 6 ÷ 6 )
= 10 + 1
= 11 minutes
Hence, from the above,
We can conclude that each student takes 11 minutes to finish an obstacle race.

Question 12.
DIG DEEPER!
You want to make 40 origami animals in 3 days. You want to make about the same number of animals each day. How many animals should you make each day? How can you interpret the remainder?

Big Ideas Math Answers 4th Grade Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.6 15

Answer: You should make 13 animals each day.

Explanation:

Given that you want to make 40 origami animals in 3 days and you want to make about the same number of animals each day.
So,
The number of animals you want to make on each day = Total number of animals ÷ 3
= 40 ÷ 3
Now,
By using the partial quotients method,
40 ÷ 3 = ( 30 + 9 ) ÷ 3
= ( 30 ÷ 3 ) + ( 9 ÷ 3 )
= 10 + 3
= 13 R 1
Hence, from the above,
We can conclude that you have to make 13 animals each day with one remaining at last.

Review & Refresh

Estimate the product.
Question 13.
32 × 67
Answer:  32 × 67 = 2,144

Explanation:
By using the partial products method,
32 × 6 = ( 30 + 2 ) × ( 60 + 7 )
= ( 30 × 60 ) + ( 2 × 60 ) + ( 30 × 7 ) + ( 2 × 7 )
= 1,800 + 120 + 210 + 14
= 2,144
Hence, 32 × 67 = 2,144

Question 14.
24 × 51
Answer: 24 × 51 = 1,224

Explanation:
By using the partial products method,
24 × 51 = ( 20 + 4 ) × ( 50 + 1 )
= ( 20 × 50 ) + ( 20 × 1 ) + ( 4 × 50 ) + ( 4 × 1 )
= 1,000 + 20 + 200 + 4
= 1,244
Hence, 24 × 51 = 1,244

Question 15.
96 × 75
Answer: 96 × 75 = 7,200

Explanation:
By using the partial products method,
96 × 75 = ( 90 + 6 ) × ( 70 + 5 )
= ( 90 × 70 ) + ( 90 × 5 ) + ( 6 × 70 ) + ( 6 × 5 )
= 6,300 + 450 + 420 + 30
= 7,200
Hence, 96 × 75 = 7,200

Lesson 5.7 Divide Multi-Digit Numbers by One-Digit Numbers

Explore and Grow

Use a model to divide. Draw each model.
348 ÷ 3
Big Ideas Math Answers Grade 4 Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.7 1
148 ÷ 3
Answer:
348 ÷ 3 = 116
148 ÷ 3 = 49 R 1

Explanation:

By using the partial quotients method,
348 ÷ 3 = ( 330+ 18 ) ÷ 3
= ( 330 ÷ 3 ) + ( 18 ÷ 3 )
= 110 + 6
= 116
Hence, 348 ÷ 3 = 116
148 ÷ 3 = ( 120 + 27 ) ÷ 3
= ( 120 ÷ 3 ) + ( 27 ÷ 3 )
= 40 + 9
= 49 R 1
Hence, 148 ÷ 3 = 49 R 1

Reasoning
Explain why the quotient of 148 ÷ 3 does not have a digit in the hundreds place.
Answer:
When 148 is divided by 3, the hundreds place in 148 won’t be divided by 3. This is the reason why the quotient of 148 ÷ 3 does not have the hundreds place.

Think and Grow: Practice

Example
Find 907 ÷ 5.
Estimate: 1,000 ÷ 5 = 200 R 0
Big Ideas Math Answers Grade 4 Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.7 2
So, 907 ÷ 5 = 181 R 2.
Check: Because 181 R 2 is close to the estimate, the answer is reasonable.

Show and Grow

Divide. Then check your answer.
Question 1.
Big Ideas Math Answers Grade 4 Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.7 3
Answer: 531 ÷ 4 = 132 R 3

Explanation:

By using the partial quotients method,
531 ÷ 4 = ( 480 +48 ) ÷ 4
= ( 480 ÷ 4 ) + ( 48 ÷ 4 )
= 120 + 12
= 132 R 3
Hence, 531 ÷ 4 = 132 R 3

Question 2.
Big Ideas Math Answers Grade 4 Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.7 4
Answer: 7,180 ÷ 5 = 1,436

Explanation:

By using the partial quotients method,
7,180 ÷ 5 = ( 7,000 + 150 + 30 ) ÷ 5
= ( 7,000 ÷ 5 ) + ( 150 ÷ 5 ) + ( 30 ÷ 5 )
= 1,400 + 10 + 30 + 6
= 1,436
Hence, 7,180 ÷ 5 = 1,436

Question 3.
Big Ideas Math Answers Grade 4 Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.7 5
Answer: 8,385 ÷ 7 = 1,197 R 6

Explanation:

By using the partial quotients method,
8,385 ÷ 7 = ( 7,700 + 630 + 49 ) ÷ 7
= ( 7,700 ÷ 7 ) + ( 630 ÷ 7 ) + ( 49 ÷ 7 )
= 1,100 + 90 + 7
=1,197 R 6
Hence, 8,385 ÷ 7 = 1,197 R 6

Apply and Grow: Practice

Divide. Then check your answer.
Question 4.
\(\sqrt [ 5 ]{ 6,381 } \)
Answer: 6,381 ÷ 5 = 1,276 R 1

Explanation:
By using the partial quotients method,
6,381 ÷ 5 = ( 6,000 + 350 + 30 ) ÷ 5
= ( 6,000 ÷ 5 ) + ( 350 ÷ 5 ) + ( 30 ÷ 5 )
=1,200 + 70 + 6
=1,276 R 1
Hence, 6,381 ÷ 5 = 1,276 R 1

Question 5.
\(\sqrt [ 3 ]{ 4,605 } \)
Answer: 4,605 ÷ 3 = 1,535 R 0

Explanation:
4,605 ÷ 3 = ( 4,500 + 105 ) ÷ 3
= ( 4,500 ÷ 3 ) + ( 105 ÷ 3 )
= 1,500 + 35
= 1,535 R 0
Hence, 4,605 ÷ 3 = 1,535 R 0

Question 6.
\(\sqrt [ 6 ]{ 820 } \)
Answer: 820 ÷ 6 = 136 R 4

Explanation:
By using the partial quotients method,
820 ÷ 6 = ( 720 + 90 + 6 ) ÷ 6
= ( 720 ÷ 6 ) + ( 90 ÷ 6 ) + ( 6 ÷ 6 )
= 120 + 15 + 1
= 136 R 4
Hence, 820 ÷ 6 = 136 R 4

Question 7.
\(\sqrt [ 6 ]{ 7,039 } \)
Answer: 7,039 ÷ 6 = 1,173 R 1

Explanation:
By using the partial quotients method,
7,039 ÷ 6 = ( 6,600 + 420 + 18 ) ÷ 6
= ( 6,600 ÷ 6 ) + ( 420 ÷ 6 ) + ( 18 ÷ 6 )
= 1,100 + 70 + 3
= 1,173 R 1
Hence,
7,039 ÷ 6 = 1,173 R 1

Question 8.
\(\sqrt [ 4 ]{ 855 } \)
Answer: 855 ÷ 4 = 213 R 3

Explanation:
By using the partial quotients method,
855 ÷ 4 = ( 800 + 52 ) ÷ 4
= ( 800 ÷ 4 ) + ( 52 ÷ 4 )
= 200 + 13
= 213 R 3
Hence, 855 ÷ 4 = 213 R 3

Question 9.
\(\sqrt [ 2 ]{ 367 } \)
Answer: 367 ÷ 2 = 183 R 1

Explanation:
By using the partial quotients method,
367 ÷ 2 = ( 360 + 6 ) ÷ 2
= ( 360 ÷ 2 ) + ( 6 ÷ 2 )
= 180 + 3
= 183 R 1
Hence, 367 ÷ 2 = 183 R 1

Question 10.
\(\sqrt [ 8 ]{ 9,692 } \)
Answer: 9,692 ÷ 8 = 1,211 R 4

Explanation:
By using the partial quotients method,
9,692 ÷ 8 = ( 9,600 + 88 ) ÷ 8
= ( 9,600 ÷ 8 ) + ( 88 ÷ 8 )
= 1,200 + 11
= 1,211 R 4
Hence, 9,692 ÷ 8 = 1,211 R 4

Question 11.
\(\sqrt [ 7 ]{ 8,345 } \)
Answer: 8,345 ÷ 7 = 1,192 R 1

Explanation:
By using the partial quotients method,
8,345 ÷ 7 = ( 7,700 +630 +14 ) ÷ 7
= ( 7,700 ÷ 7 ) + ( 630 ÷ 7 ) + ( 14 ÷ 7 )
= 1,100 + 90 + 2
=1,192 R 1
Hence, 8,345 ÷ 7 = 1,192 R 1

Question 12.
\(\sqrt [ 7 ]{ 971 } \)
Answer: 971 ÷ 7 = 138 R 5

Explanation:
By using the partial quotients method,
971 ÷ 7 = ( 910 + 56 ) ÷ 7
= ( 910 ÷ 7 ) + ( 56 ÷ 7 )
= 130 + 8
= 138 R 5
Hence, 971 ÷ 7 = 138 R 5

Question 13.
There are 8,274 people at an air show. The people are divided into equally 6 sections. How many people are in each section?

Big Ideas Math Answers Grade 4 Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.7 6

Answer: There are 1,379 people in each section.

Explanation:
Given that there are 8,274 people at an air show and the people are divided equally into 6 sections.
So,
The number of people in each section = Total number of people ÷ Number of sections
= 8,274 ÷ 6
Now,
By using the partial quotients method,
8,274 ÷ 6 = ( 7,800 + 420 + 54 ) ÷ 6
= ( 7,800 ÷ 6 ) + ( 420 ÷ 6 ) + ( 54 ÷ 6 )
= 1,300 + 70 + 9
= 1,379
Hence, from the above,
We can conclude that there are 1,379 people in each section.

Question 14.
YOU BE THE TEACHER
Newton finds 120 ÷ 5. Is he correct? Explain.

Big Ideas Math Answers Grade 4 Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.7 7

Answer: Newton is not correct.

Explanation:
According to Newton,
For 120 ÷ 5,
Quotient = 114 and Remainder = 0
But,
By using the partial quotients method,
120 ÷ 5 = ( 100 + 20 ) ÷ 5
= ( 100 ÷ 5 ) + ( 20 ÷ 5 )
= 20 + 4 )
= 24
Hence,
According to the partial quotients method,
For 120 ÷ 5,
Quotient = 24 and Remainder = 0
Hence, from the above,
We can conclude that Newton is not correct.

Think and Grow: Modeling Real Life

Example
There are 1,014 toy car tires at a factory. Each car needs 4 tires. How many toy cars can the factory workers make with the tires?
Each car needs 4 tires, so find 1,014 ÷ 4.
1 thousand cannot be shared among 4 groups without regrouping. So, regroup 1 thousand as 10 hundreds.

Interpret the quotient and the remainder.
The quotient is 253. The factory workers can make 253 toy cars.
The remainder is 2. There are 2 tires left over.

Show and Grow

Question 15.
A principal orders 750 tablets. The distributor can fit 8 tablets in each box. How many boxes are needed to ship all of the tablets?
Answer:
The number of boxes needed to ship all of the tablets= 93 boxes with 6 leftovers = 93 R 6

Explanation:
Given that a principal orders 750 tablets and the distributor can fit 8 tablets in each box.
So,
The number of boxes needed to ship all of the tablets = The number of tablets ordered ÷ The number of tablets that fits into each box
= 750 ÷ 8
Now,
By using the partial quotients method,
750 ÷ 8 = ( 720 + 24 ) ÷ 8
= ( 720 ÷ 8 ) + ( 24 ÷ 8 )
= 90 + 3
= 93 R 6
Hence, from the above,
We can conclude that
The number of boxes needed to ship all the tablets = 93 boxes with 6 leftovers

Question 16.
An athlete’s heart rate after a 5-mile run is 171 beats per minute, which is 3 times as fast as her resting heart rate. What is the athlete’s resting heart rate?
Big Ideas Math Answers Grade 4 Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.7 9
Answer: The athlete’s resting heart rate is 57 beats per minute.

Explanation:
Given that an athlete’s heart rate after a 5-mile run is 171 beats per minute. It is also given that 171 beats per minute which are 3 times as fast as her resting heart rate.
So,
The athlete’s heart rating rate = Athlete’s heart rate after a 5-mile run ÷ 3
= 171 ÷ 3
Now,
By using the partial quotients method,
171 ÷ 3 = ( 150 + 21 ) ÷ 3
= ( 150 ÷ 3 ) + ( 21 ÷ 3 )
= 50 + 7
= 57 beats per minute.
Hence, from the above,
We can conclude that the athlete’s resting heart rate is 57 beats per minute.

Question 17.
A car costs $5,749. The taxes and fees for the car cost an additional $496. A customer uses a 5-year  interest-free loan to buy the car. How much money will the customer pay for the car each year?
Answer:  The customer will pay $1,249 for the car each year

Explanation:
Given that a car costs $5,749 and the taxes and fees for the car cost an additional $496.
So,
The total cost of the car = 5,749 + 496 = $6,245
It is also given that the customer uses a 5-year interest – free loan to buy the car.
So,
The money that the customer will pay each year = The total cost of the car ÷ 5
= 6,245 ÷ 5
Now,
By using the partial quotients method,
6,245 ÷ 5 = ( 6,000 +  200 + 45 ) ÷ 5
= ( 6,000 ÷ 5 ) + ( 200 ÷ 5 ) + ( 45 ÷ 5 )
= 1,200 +  40 + 9
= $1,249
Hence, from the above,
We can conclude that
The money that the customer pay each year = $1,249

Divide Multi-Digit Numbers by One-Digit Numbers Homework & Practice 5.7

Divide. Then check your answer.
Question 1.
Big Ideas Math Answers Grade 4 Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.7 10
Answer: 473 ÷ 9 = 52 R 5

Explanation:

By using the partial quotients method,
473 ÷ 9 = ( 450 ÷ 9) + ( 18 ÷ 9)
= 50 + 2
= 52 R 5
Hence, 473 ÷ 9 = 52 R 5

Question 2.
Big Ideas Math Answers Grade 4 Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.7 11
Answer: 3,904 ÷ 2 = 1,952

Explanation:

By using the partial quotients method,
3,904 ÷ 2 = ( 3,000 + 900 + 4 ) ÷ 2
= ( 3,000 ÷ 2 ) + ( 900 ÷ 2 ) + ( 4 ÷ 2 )
= 1,500 + 450 +2
= 1,902
Hence, 3,904 ÷ 2 = 1,902

Question 3.
Big Ideas Math Answers Grade 4 Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.7 12
Answer: 2,531 ÷6 = 421 R 5

Explanation:

By using the partial quotients method,
2,531 ÷ 6 = ( 2,400 + 120 + 6 ) ÷ 6
= ( 2,400 ÷ 6 ) + ( 120 ÷ 6 ) + ( 6 ÷ 6 )
= 400 + 20 + 1
= 421 R 5
Hence, 2,531 ÷ 6 = 421 R 5

Question 4.
\(\sqrt [ 5 ]{ 8,271 } \)
Answer: 8,271 ÷ 5 = 1,654 R 1

Explanation:
By using the partial quotients method,
8,271 ÷ 5 = ( 8,000 + 250 + 20 ) ÷ 5
= ( 8,000 ÷ 5 ) + ( 250 ÷ 5 ) + (20 ÷ 5 )
= 1,600 + 50 + 4
= 1,654 R 1
Hence, 8,271 ÷ 5 = 1,654 R 1

Question 5.
\(\sqrt [ 7 ]{ 952 } \)
Answer: 952 ÷ 7 = 136

Explanation:
By using the partial quotients method,
952 ÷ 7 = ( 910 + 42 ) ÷ 7
= (910 ÷ 7 ) + ( 42 ÷ 7 )
= 130 + 6
= 136
Hence, 952 ÷ 7 = 136

Question 6.
\(\sqrt [ 8 ]{ 9,107 } \)
Answer: 9,107 ÷ 8 = 1,138 R 3

Explanation:
9,107 ÷ 8 = ( 8,800 + 240 + 56 +8 ) ÷ 8
= ( 8,800 ÷ 8 ) + ( 240 ÷ 8 ) + ( 56 ÷ 8 ) + ( 8 ÷ 8 )
= 1,100 + 30 + 7 + 1
= 1,138 R 3
Hence, 9,107 ÷ 8 = 1,138 R 3

Divide. Then check your answer.
Question 7.
\(\sqrt [ 3 ]{ 509 } \)
Answer: 509 ÷ 3 = 169 R 2

Explanation:
By using the partial quotients method,
509 ÷ 3 = ( 480 + 27 ) ÷ 3
= ( 480 ÷ 3 ) + ( 27 ÷ 3 )
= 160 + 9
= 169 R 2
Hence, 509 ÷ 3 = 169 R 2

Question 8.
\(\sqrt [ 4 ]{ 7,150 } \)
Answer: 7,150 ÷ 4 = 1,787 R 2

Explanation:
7,150 ÷ 4 = ( 6,800 + 320 + 28 ) ÷ 4
= ( 6,800 ÷ 4 ) + ( 320 ÷ 4 ) + ( 28 ÷ 4 )
= 1,700 + 80 + 7
= 1,787 R 2
Hence, 7,150 ÷ 4 = 1,787 R 2

Question 9.
\(\sqrt [ 2 ]{ 5,547 } \)
Answer: 5,547 ÷ 2 = 2,773 R 1

Explanation:
5,547 ÷ 2 = ( 5,000 + 500 +46 ) ÷ 2
= ( 5,000 ÷ 2 ) + ( 500 ÷ 2 ) + ( 46 ÷ 2 )
= 2,500 + 250 + 23
= 2,273 R 1
Hence, 5,547 ÷ 2 = 2,273 R 1

Question 10.
\(\sqrt [ 3 ]{ 756 } \)
Answer: 756 ÷ 3 = 252 R 0

Explanation:
756 ÷ 3 = ( 720 +36 ) ÷ 3
= ( 720 ÷ 3 ) + ( 36 ÷ 3 )
= 240 + 12
= 252
Hence, 756 ÷ 3 = 252 R 0

Question 11.
\(\sqrt [ 4 ]{ 6,871 } \)
Answer: 6,871 ÷ 4 = 1,717 R 3

Explanation:
6,871 ÷ 4 = ( 6,000 + 800 + 64 + 4 ) ÷ 4
= ( 6,000 ÷ 4 ) + ( 800 ÷ 4 ) + ( 64 ÷ 4 ) + ( 4 ÷ 4 )
= 1,500 + 200 + 16 +1
= 1,717 R 3
Hence, 6,871 ÷ 4 = 1,717 R 3

Question 12.
\(\sqrt [ 5 ]{ 108 } \)
Answer: 108 ÷ 5 = 21 R 3

Explanation:
108 ÷ 5 = ( 100 +5 ) ÷ 5
= ( 100 ÷ 5 ) + ( 5 ÷ 5 )
= 20 + 1
= 21 R 3
Hence, 108 ÷ 5 = 21 R 3

Question 13.
A deli sells 203 pretzel roll sandwiches. Each bag has 7 pretzel rolls. How many bags of pretzel rolls are used?
Big Ideas Math Answers Grade 4 Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.7 13

Answer: The number of pretzel rolls is used = 29 rolls

Explanation:
Given that a deli sells 203 pretzel roll sandwiches and each bag has 7 pretzel rolls.
So,
The number of pretzel rolls used = Total number of pretzel roll sandwiches ÷ 7
= 203 ÷ 7
Now,
By using the partial quotients method,
203 ÷ 7 = ( 140 + 63 ) ÷ 7
= ( 140 ÷ 7 ) + ( 63 ÷ 7 )
= 20 + 9
= 29 bags
Hence, from the above,
We can conclude that there are 29 bags of pretzel rolls of sandwiches

Question 14.
DIG DEEPER!
Which expression does have a 4-digit quotient? Explain not how you know without solving.
6,197 ÷ 2
3, 261 ÷ 5
5,240 ÷ 4
9,045 ÷ 8
Answer:
Let the Expressions be named as A), B), C), and D)
So,
A) 6,197 ÷ 2
B) 3, 261 ÷ 5
C) 5,240 ÷ 4
D) 9,045 ÷ 8
From the above,
Expression C) does have a 4- digit quotient without leaving a remainder.

Explanation;
Let the Expressions be named as A), B), C), and D)
So,
A) 6,197 ÷ 2
B) 3, 261 ÷ 5
C) 5,240 ÷ 4
D) 9,045 ÷ 8
A) 6,197 ÷ 2
By using the partial quotients method,
6,197 ÷ 2 = ( 6,000 + 196 ) ÷ 2
= ( 6,000 + 196 ) ÷ 2
= 3,000 + 98
=3,098 R 1
Hence, 6,197 ÷ 2 = 3,098 R 1
B) 3,261 ÷ 5 = ( 3,000 + 250 + 10 ) ÷ 5
= ( 3,000 ÷ 5 ) + ( 250 ÷ 5 ) + ( 10 ÷ 5 )
= 600 + 50 + 2
= 652 R 1
Hence, 3,261 ÷ 5 = 652 R 1
C) 5,240 ÷ 4 = ( 5,200 + 40 ) ÷ 4
= ( 5,200 ÷ 4 ) + ( 40 ÷ 4 )
= 1,300 + 10
= 1,310 R 0
Hence, 5,240 ÷ 4 = 1,310
D) 9,045 ÷ 8 = ( 9,000 + 40 ) ÷ 8
= ( 9,000 ÷ 8 ) + ( 40 ÷ 8 )
= 1,125 + 5
= 1,130 R 5
Hence, from all the 4 Expressions,
We can conclude that the Expression C)  has 4-digit quotient having remainder 0

Question 15.
Modeling Real Life
There are 505 pieces of chalk donated for a community event. The chalk is bundled into groups of 7 pieces. How many bundles can be made?
Big Ideas Math Answers Grade 4 Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.7 14

Answer: The number of bundles that can be made = 72 R 1

Explanation:
Given that there are 505 pieces of chalk donated for a community event and the chalk is bundled into groups of 7 pieces.
So,
The number of bundles that can be made = 505 ÷ 7
Now,
By using the partial quotients method,
505 ÷ 7 = ( 490 + 14 ) ÷ 7
= ( 490 ÷ 7 ) + ( 14 ÷ 7 )
= 70 + 2
= 72 R 1
Hence, from the above,
We can conclude that
The number of bundles that can be made = 72 bundles with 1 leftover

Review & Refresh

Find the product. Check whether your answer is reasonable.
Question 16.
Estimate: _____
61 × 3 = _____
Answer: 61 × 3 =183

Explanation:
By using theDistributive Property,
61 × 3 = ( 60 + 1 ) × 3
= ( 60 × 3 ) + ( 1 × 3 )
= 180 + 3
= 183
So, 61 × 3 = 183
Estimate:
Let 61 be rounded to 60.
So, by using the place-value method,
60 × 3 = 6 tens × 3
= 18 tens
= 18 × 10
= 180
So, 60 × 3 = 180
Hence, from the above,
We can conclude that the actual answer is near to the Estimate. So, the answer is reasonable.

Question 17.
Estimate: ______
54 × 9 = ______
Answer: 54 × 9 = 486

Explanation:
By using the Distributive Property,
54 × 9 = ( 50 + 4 ) × 9
= ( 50 × 9 ) + ( 4 × 9 )
= 450 + 36
= 486
So, 54 × 9 = 486
Estimate:
Let 54 be rounded to 55.
Let 9 be rounded to 10
So, by using the place-value method,
10 × 55 = 1 ten × 55
= 55 tens
= 55 × 10
= 550
So, 55 × 10 = 550
Hence, from the above,
We can conclude that the actual answer is not near to the Estimate. So, the answer is not reasonable.

Question 18.
Estimate: _______
82 × 7 = ______
Answer: 82 × 7 = 574

Explanation:
By using the Distributive Property,
82 × 7 = ( 80 + 2 ) × 7
= ( 80 × 7 ) + ( 2 × 7 )
= 560 + 14
= 574
So, 82 × 7 = 574
Estimate:
Let 82 be rounded to 80.
Let 7 be rounded to 5.
So, by using the place-value method,
80 × 5 = 8 tens × 5
= 40 tens
= 40 × 10
= 400
So, 80 × 5 = 400
Hence, from the above,
We can conclude that the actual answer is not near to the Estimate. So, the answer is not reasonable.

Lesson 5.8 Divide by One-Digit Numbers

Explore and Grow

Use a model to find each quotient. Draw each model.
312 ÷ 3
Big Ideas Math Solutions Grade 4 Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.8 1
312 ÷ 4
Answer:
312 ÷ 3 = 104
312 ÷ 4 = 78

Explanation:
           
By using the partial quotients method,
312 ÷ 3 = ( 300 + 12 )÷ 3
= ( 300 ÷ 3 ) + ( 12 ÷ 3 )
= 100 + 4
= 104
Hence, 312 ÷ 3 = 104
By using the partial quotients method,
312 ÷ 4 = ( 280 + 20 + 12 ) ÷ 4
= ( 280 ÷ 4 ) + ( 20 ÷ 4 ) + ( 12 ÷ 4 )
= 70 + 5 + 3
= 78
Hence, 312 ÷ 4 = 78

Structure
Compare your models for each quotient. What is the same? What is different? What do you think this means when using regrouping to divide?
Answer: The Divisor for both the models is the same and the quotient is different for both Expressions.

Think and Grow: Divide by One-Digit Numbers

Example
Find 4,829 ÷ 8.
4 thousand cannot be shared among 8 groups without regrouping. So, regroup 4 thousands as 40 hundred and combine with 8 hundred.

Show and Grow

Divide. Then check your answer.
Question 1.
Big Ideas Math Solutions Grade 4 Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.8 3
Answer: 756 ÷ 7 = 108

Explanation:

By using the partial quotients method,
756 ÷ 7 = ( 700 + 56 ) ÷ 7
= ( 700 ÷ 7 ) + ( 56 v 7 )
= 100 + 8
= 108
Hence, 756 ÷ 7 = 108

Question 2.
Big Ideas Math Solutions Grade 4 Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.8 4
Answer: 364 ÷ 6 = 6 R 4

Explanation:

By using the partial quotients method,
364 ÷ 6 = ( 300 + 60 ) ÷ 6
= ( 300 ÷ 6 ) + ( 60 ÷ 6 )
= 50 + 10
= 60 R 4
Hence, 364 ÷ 6 = 60 R 4

Question 3.
Big Ideas Math Solutions Grade 4 Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.8 5
Answer: 3,190 ÷ 3 = 1,063 R 1

Explanation:

By using the partial quotients method,
3,190 ÷ 3 = ( 3,000 + 180 + 9 ) ÷ 3
= ( 3,000 ÷ 3 ) + ( 180 ÷ 3 ) + ( 9 ÷ 3 )
= 1,000 + 60 + 3
= 1,063 R 1
Hence, 3,190 ÷ 3 = 1,063 R 1

Apply and Grow: Practice

Divide. Then check your answer.
Question 4.
\(\sqrt [ 2 ]{ 81 } \)
Answer: 81 ÷ 2 = 40 R 1

Explanation:
By using the partial quotients method,
81 ÷ 2 = ( 60 + 20 ) ÷ 2
= ( 60 ÷ 2 ) + ( 20 ÷ 2 )
= 30 + 10
= 40 R 1
Hence, 81 ÷ 2 = 40 R 1

Question 5.
\(\sqrt [ 4 ]{ 428 } \)
Answer: 428 ÷ 4 = 107

Explanation:
By using the partial quotients method,
428 ÷ 4 = ( 400 + 28 ) ÷ 4
= ( 400 ÷ 4 ) + ( 28 ÷ 4 )
= 100 + 7
= 107
Hence, 428 ÷ 4 = 107

Question 6.
\(\sqrt [ 6 ]{ 842 } \)
Answer: 842 ÷ 6 = 140 R 2

Explanation:
By using the partial quotients method,
842 ÷ 6 = ( 780 + 60 ) ÷ 6
= ( 780 ÷ 6 ) + ( 60 ÷ 6 )
= 130 + 10
= 140 R 2
Hence, 842 ÷ 6 = 140 R 2

Question 7.
\(\sqrt [ 3 ]{ 2,724 } \)
Answer: 2,724 ÷ 3 = 908

Explanation:
By using partial quotients method,
2,724 ÷ 3 = ( 2,700 + 24 ) ÷ 3
= ( 2,700 ÷ 3 ) + ( 24 ÷ 3 )
= 900 + 8
= 908
Hence, 2,724 ÷ 3 = 908

Question 8.
\(\sqrt [ 9 ]{ 635 } \)
Answer: 635 ÷ 9 = 70 R 5

Explanation:
By using partial products method,
635 ÷ 9 = ( 540 + 90 ) ÷ 9
= ( 540 ÷ 9 ) + ( 90 ÷ 9 )
= 60 + 10
= 70 R 5
Hence, 635 ÷ 9 = 70 R 5

Question 9.
\(\sqrt [ 6 ]{ 1,442 } \)
Answer: 1,442 ÷ 6 = 240 R 2

Explanation:
By using partial quotients method,
1,442 ÷ 6 = ( 1,200 +240 ) ÷ 6
= ( 1,200 ÷ 6 ) + ( 240 ÷ 6 )
= 200 +40
= 240 R 2
Hence, 1,442 ÷ 6 = 240 R 2

Question 10.
\(\sqrt [ 6 ]{ 303 } \)
Answer: 303 ÷ 6 = 50 R 3

Explanation:
By using partial quotients method,
303 ÷ 6 = ( 240 + 60 ) ÷ 6
= ( 240 ÷ 6 ) + ( 60 ÷ 6 )
= 40 + 10
= 50 R 3
Hence, 303 ÷ 6 = 50 R 3

Question 11.
\(\sqrt [ 5 ]{ 2,530 } \)
Answer: 2,530 ÷ 5 = 506

Explanation:
By using partial quotients method,
2,530 ÷ 5 = ( 2,500 + 30 ) ÷ 5
= ( 2,500 ÷ 5 ) + ( 30 ÷ 5 )
= 500 + 6
= 506
Hence, 2,530 ÷ 5 = 506

Question 12.
\(\sqrt [ 8 ]{ 7,209 } \)
Answer: 7,209 ÷ 8 = 900 R 9

Explanation:
By using partial quotients method,
7,209 ÷ 8 = ( 6,400 + 800 ) ÷ 8
= ( 6,400 ÷ 8 ) + ( 800 ÷ 8 )
= 800 + 100
= 900 R 9
Hence, 7,209 ÷ 8 = 900 R  9

Question 13.
The 5 developers of a phone app earn a profit of $4,535 this month. They divide the profit equally. How much money does each developer get?
Big Ideas Math Solutions Grade 4 Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.8 6
Answer: Each developer gets the money of $907

Explanation:
Given that the 5 developers of a phone app earn a profit of $4,535 this month and it is also given that they divide the profit equally.
So,
The money each developer get = The total profit earned by 5 developers ÷ 5
= $4,535 ÷ 5
Now,
By using the partial quotients method,
4,535 ÷ 5 = ( 4,500 + 35 ) ÷ 5
= ( 4,500 ÷ 5 ) + ( 35 ÷ 5 )
= 900 + 7
= 907
Hence, from the above,
We can conclude that the money got by each developer is $907.

Question 14.
Newton finds 817 ÷ 4. Is he correct? Explain.
Big Ideas Math Solutions Grade 4 Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.8 7
Answer: Newton is not correct.

Explanation;

By using the partial quotients method,
817 ÷ 4 = ( 800 + 16 ) ÷ 4
= ( 800 ÷ 4 ) + ( 16 ÷ 4 )
= 200 + 4
= 204 R 1
According to Newton,
817 ÷ 4 = 24 R 1
But according to the partial quotients method,
817 ÷ 4 = 204 R 1
Hence, from the above,
We can conclude that Newton is not correct.

Think and Grow: Modeling Real Life

Example
Seven players are placed on each basketball team. Remaining basketball players are added to the teams, so some of the teams have 8 players. How many basketball teams have 7 players? 8 players?
There are 1,839 players signed up for basketball, so find 1,839 ÷ 7.
1 thousand be shared cannot among 7 groups without regrouping. So, regroup 1 thousand as 10 hundreds and combine with 8 hundreds.

Interpret the quotient and the remainder.
The quotient is262. So, there are 262 basketball teams in all.
The remainder is 5. So, 5 basketball teams have 8 players.
Subtract to find how many teams have 7 players. 262 −5 = 257
So, 257 basketball teams have 7 players and 5 have 8 players.

Show and Grow

Use the table above.
Question 15.
Nine players are placed on each ball hockey team. Remaining players are added to the teams, so some of the teams have 10 players. How many ball hockey teams have9 players? 10 players?
Answer:
Given that there are 9 players on each ball hockey team.
So, the quotient represents the ball hockey teams who have 9 players and the remainder represents the ball hockey teams who have 10 players.
Now,
By using the partial quotients method,
952 ÷ 9 = ( 900 + 45 ) ÷ 9
= ( 900 ÷ 9 ) + ( 45 ÷ 9 )
= 100 + 5
= 105 R 7
The quotient is 105. So, there are 105 ball hockey teams in all.
The remainder is 7. So, 7 ball hockey teams have 10 players.
Subtract to find the number of teams has 9 players. 105 −7 = 98
So, 98 basketball teams have 9 players and 7 have 10 players.

Question 16.
Eighty-four players who signed up to play soccer decide not to play. Eight players are placed on each soccer team. How many soccer teams are there?
Answer: There are 299 soccer teams

Explanation:
Given that 84 players who signed up to play soccer decide not to play and it is also given that 8 players are placed on each soccer team.
It is also given that,
The total number of players who play soccer = 2,476 players
So,
The number of players who played soccer = 2,476 – 84 = 2,392 players
Now,
The number of soccer teams = 2,392 ÷ 8
Now,
By using the partial quotients method,
2,392 ÷ 8 = ( 1,600 + 720 + 72 ) ÷ 8
= ( 1,600  ÷ 8 ) + ( 720 ÷ 8 ) + ( 72 ÷ 8 )
= 200 + 90 + 9
= 299
Hence, from the above,
we can conclude that there are 299 soccer teams.

Divide by One-Digit Numbers Homework & Practice 5.8

Divide. Then check your answer.
Question 1.
Big Ideas Math Solutions Grade 4 Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.8 9
Answer: 832 ÷ 8 = 104

Explanation:

By using the partial quotients method,
832 ÷ 8 = ( 800 + 32 ) ÷ 8
= ( 800 ÷ 8 ) + ( 32 ÷ 8 )
= 100 + 4
= 104
Hence, 832 ÷ 8 = 104

Question 2.
Big Ideas Math Solutions Grade 4 Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.8 10
Answer: 215 ÷ 7 = 30 R 5

Explanation:

By using the partial quotients method,
215 ÷ 7 = ( 140 + 70 ) ÷7
( 140 ÷ 7 ) + ( 70 ÷ 7 )
= 20 + 10
= 30 R 5
Hence, 215 ÷7 = 30 R 5

Question 3.
Big Ideas Math Solutions Grade 4 Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.8 11
Answer: 5,078 ÷ 5 = 1,015 R 3

Explanation:

By using the partial quotients method,
5,078 ÷ 5 = ( 5,000 + 75 ) ÷ 5
= ( 5,000 ÷5 ) + ( 75 ÷ 5 )
= 1,000 + 15
= 1,015 R 3
Hence, 5,078 ÷ 5 = 1,015 R 3

Question 4.
\(\sqrt [ 7 ]{ 94 } \)
Answer: 94 ÷ 7 = 13 R 3

Explanation:
By using the partial quotients method,
94 ÷ 7 = ( 84 + 7 ) ÷ 7
= ( 84 ÷ 7 ) + ( 7 ÷ 7 )
= 12 + 1
= 13 R 1
Hence, 94 ÷ 7 = 13 R 1

Question 5.
\(\sqrt [ 6 ]{ 731 } \)
Answer: 731 ÷ 6 = 121 R 5

Explanation:
By using the partial quotients method,
731 ÷ 6 = ( 720 + 6 ) ÷ 6
= ( 720 ÷ 6 ) + ( 6 ÷ 6 )
= 120 + 1
= 121 R 5
Hence, 731 ÷ 6 = 121 R 5

Question 6.
\(\sqrt [ 4 ]{ 6,514 } \)
Answer: 6,514 ÷ 4 = 1,628 R 2

Explanation:
By using the partial quotients method,
6,514 ÷ 4 = ( 6,000 + 480 + 32 ) ÷ 4
= ( 6,000 ÷ 4 ) + ( 480 ÷ 4 ) + ( 32 ÷ 4 )
= 1,500 + 120 + 8
= 1,628 R 2
Hence, 6,514 ÷ 4 = 1,628 R 2

Question 7.
\(\sqrt [ 3 ]{ 62 } \)
Answer: 62 ÷ 3 = 20 R 2
Explanation:
By using the partial quotients method,
62 ÷3 = ( 33 + 27 ) ÷ 3
= ( 33 ÷ 3 ) + ( 27 ÷ 3 )
= 11 + 9
= 20 R 2
Hence, 62 ÷ 3 = 20 R 2

Question 8.
\(\sqrt [ 5 ]{ 548 } \)
Answer: 548 ÷ 5 = 109 R 3

Explanation:
By using the partial quotients method,
548 ÷ 5 = ( 500 + 45 ) ÷ 5
= ( 500 ÷ 5 ) + ( 45 ÷ 5 )
= 100 + 9
= 109 R 3
Hence, 548 ÷ 5 = 109 R 3

Question 9.
\(\sqrt [ 2 ]{ 4,136 } \)
Answer: 4,136 ÷ 2 = 2,068

Explanation:
By using the partial quotients method,
4,136 ÷ 2 = ( 4,000 + 100 + 36 ) ÷ 2
= ( 4,000 ÷ 2 ) + ( 100 ÷ 2 ) + ( 36 ÷ 2 )
= 2,000 + 50 + 18
= 2,068
Hence, 4,136 ÷ 2 = 2,068

Divide. Then check your answer.
Question 10.
\(\sqrt [ 7 ]{ 214 } \)
Answer: 214 ÷ 7 = 30 R 4

Explanation:
By using the partial quotients method,
214 ÷ 7 = ( 140 + 70 ) ÷ 7
= ( 140 ÷ 7 ) + ( 70 ÷ 7 )
= 20 + 10
= 30 R 4
Hence, 214 ÷ 7 = 30 R 4

Question 11.
\(\sqrt [ 4 ]{ 321 } \)
Answer: 321 ÷ 4 = 80 R 1

Explanation:
By usng the partial quotients method,
321 ÷ 4 = ( 280 + 40 ) ÷ 4
= ( 280 ÷ 4 ) + ( 40 ÷ 4 )
= 70 + 10
= 80 R 1
Hence, 321 ÷ 4 = 80 R 1

Question 12.
\(\sqrt [ 6 ]{ 5,162 } \)
Answer: 5,162 ÷6 = 860 R 2

Explanation:
By using the partial quotients method,
5,162 ÷ 6 = ( 4,800 + 300 + 60 ) ÷ 6
= ( 4,800 ÷ 6 ) + ( 300 ÷ 6 ) + ( 60 ÷ 6 )
= 800 + 50 + 10
= 860 R 2
Hence, 5,162 ÷ 6 = 860 R 2

Question 13.
\(\sqrt [ 2 ]{ 7,301 } \)
Answer: 7,301 ÷ 2 = 3,650 R 1

Explanation:
By using the partial quotients method,
7,301 ÷ 2 = ( 7,000 + 300 ) ÷ 2
= ( 7,000 ÷ 2 ) + ( 300 ÷ 2 )
= 3,500 + 150
= 3,650 R 1
Hence, 7,301 ÷ 2 = 3,650 R 1

Question 14.
\(\sqrt [ 5 ]{ 603 } \)
Answer: 603 ÷ 5 = 120 R 3

Explanation:
By using the partial products method,
603 ÷ 5 = ( 550 + 50 ) ÷ 5
= ( 550 ÷ 5 ) + ( 50 ÷ 5 )
= 110 + 10
= 120 R 3
Hence, 603 ÷ 5 = 120 R 3

Question 15.
\(\sqrt [ 3 ]{ 6,082 } \)
Answer: 6,082 ÷ 3 = 2,027 R 1

Explanation:
By using the partial quotients method,
6,082 ÷ 3 = ( 6,000 + 81 ) ÷ 3
= ( 6,000 ÷ 3 ) + ( 81 ÷ 3 )
= 2,000 + 27
= 2,027 R 1
Hence, 6,082 ÷ 3 = 2,027 R 1

Question 16.
There are 450 pounds of grapes for a grape stomping contest. They are divided equally into 5 barrels. How many pounds of grapes are in each barrel?

Big Ideas Math Solutions Grade 4 Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.8 12

Answer: There are 90 pounds of grapes in each barrel.

Explanation:
Given that there are 450 pounds of grapes for a grape stomping contest and they are divided equally into 5 barrels.
So,
The number of pounds of grapes in each barrel = Total number of pounds of groups ÷ 5
= 450 ÷ 5
Now,
By using the partial quotients method,
450 ÷ 5 = ( 400 + 50 ) ÷ 5
= ( 400 ÷ 5 ) + ( 50 ÷ 5 )
= 80 + 10
= 90 pounds
Hence, from the above,
We can conclude that there are 90 pounds of grapes in each barrel.

Question 17.
DIG DEEPER!
How could you change the dividend in Exercise 11 so that there would be no remainder? Explain.
Answer: The dividend given in Exercise 11 is 321 and leaves the remainder 1 when the dividend is divided by 4.
So, to make the dividend a number in such a way that it is divisible by 4 and leaves the remainder 0
By observing the given number 321, we can say that it is close to 320 which is divisible by 4 and leaves the remainder 0.
Hence, the number which we have to change so that the remainder will be 0 is: 320

Question 18.
Modeling Real Life
Five actresses are placed on each team. Remaining actresses are added to the teams, so some of the teams have 6 actresses. How many teams have 5 actresses? 6 actresses?

Big Ideas Math Solutions Grade 4 Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.8 13

Answer:
The number of teams that have 5 actresses = 40
The number of teams that have 6 actresses = 2

Explanation:
Given that actresses are placed on each team and the remaining actresses are added to the teams.
So, from the table, we can see that
The total number of actresses = 202
Now, to find the number of teams that have 5 actresses, we have to find the value of 202 ÷ 5
The remainder of 202 ÷ 5 gives the number of teams that have 6 actresses
Now,
By using the partial quotients method,
202 ÷ 5 = ( 150 + 50 ) ÷ 5
= ( 150 ÷ 5 ) + ( 50 ÷ 5 )
= 30 + 10
= 40 R 2
Hence, from the above,
We can conclude that
The number of teams that have 5 actresses = 40 teams
The number of teams that have 6 actresses = 2 teams

Review & Refresh

Write the value of the underlined digit.
Question 19.
86,109
Answer: The place- value of 6 in 86,109 is: 6,000

Explanation:
We know that the position of a given digit in any number depends on the place-value of that number.
So,
The place-value of 6 in 86,109 is: 6,000

Question 20.
15,327
Answer: The place-value of 1 in 15,327 is: 10,000

Explanation:
We know that the position of a given digit in any number depends on the place-value of that number.
So,
The place-value of 1 in 15,327 is: 10,000

Question 21.
914,263
Answer: The place-value of 9 in  914,263 is: 900,000

Explanation:
We know that the position of a given digit in any number depends on the place-value of that number.
So,
The place-value of 9 in  914,263 is: 900,000

Question 22.
284,505
Answer: The place-value of 0 in 284,505 is: 0 ( Since 0 × 10 = 0 )

Explanation:
We know that the position of a given digit in any number depends on the place-value of that number.
So,
The place-value of 0 in 284,505 is: 0 ( Since 0 × 10 = 0 )

Lesson 5.9 Problem Solving: Division

Explore & Grow

Make a plan to solve the problem.
A fruit vendor has 352 green apples and 424 red apples. The vendor uses all of the apples to make fruit baskets. He puts 8 apples in each basket. How many fruit baskets does the vendor make?
Big Ideas Math Answer Key Grade 4 Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.9 1

Answer: The fruit vendor makes 97 fruit baskets

Explanation:
Given that a fruit vendor has 352 green apples and 424 red apples.
So,
Total number of apples = Number of red apples + Number of green apples
= 352 + 424
= 776 apples.
It is also given that the vendor uses all of the apples to make fruit baskets and he puts 8 apples in each basket.
Hence,
The number of fruit baskets that the vendor makes = Total number of apples ÷ Number of apples the vendor puts in each basket
= 776 ÷ 8
Now,
By using the partial quotients method,
776 ÷ 8 = ( 720 + 56 ) ÷ 8
= ( 720 ÷ 8 ) + ( 56 ÷ 8 )
= 90 + 7
= 97 baskets
Hence, from the above,
We can conclude that there are 97 fruit baskets that the vendor made.

Make Sense of Problems
The vendor decides that each basket should have 8 of the same colored apples. Does this change your plan to solve the problem? Will this change the answer? Explain.
Answer: Yes, you have to change your plan to solve the problem but this way will not change the answer as we will get the same answer as above.

Explanation:
Given that the vendor decides that each basket should have 8 of the same colored apples.
So,
The number of fruit baskets that have red apples = 424 ÷ 8
Now,
By using the partial quotients method,
424 ÷ 8 = ( 400 + 24 ) ÷ 8
= ( 400 ÷ 8 ) + ( 24 ÷ 8 )
= 50 + 3
= 53 fruit baskets ( for every 8 red apples)
Now,
The number of fruit baskets that have green apples = 352 ÷ 8
Now,
By using the partial quotients method,
352 ÷ 8 = ( 320 + 32 ) ÷ 8
= ( 320 ÷ 8 ) + ( 32 ÷ 8 )
= 40 + 4
= 44 fruit baskets ( for every 8 green apples)
So,
Total number of fruit baskets that a vendor make = Number of fruit baskets that have red apples + Number of fruit baskets that have green apples
= 44 + 53 = 97 fruit baskets
Hence, from the above,
We can conclude that we change the plan of solving the problem but the answer remains the same.

Think and Grow: Problem Solving: Division

Example
The speed of sound in water is 1,484 meters per second. Sound travels 112 more than 4 times as many meters per second in water as it does in air. What is the speed of sound in air?
Big Ideas Math Answer Key Grade 4 Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.9 2
Understand the Problem
What do you know?
• The speed of sound in water is sound 1,484 meters per second.
• Sound travels 112 more than 4 times as many meters per second in water as it does in air.

What do you need to find?
• You need to find the speed in the air.

Make a Plan
How will you solve it?
• Subtract 112 from 1,484 to find 4 times the speed of sound in the air.
• Then divide the difference by 4 to find the speed of sound in the air.

Solve

So,
The speed of sound in air is 343 meters per second.

Show and Grow
Question 1.
Explain how you can check whether your answer above is reasonable.
Answer:

Apply and Grow: Practice

Understand the problem. What do you know? What do you need to find? Explain.
Question 2.
A surf shop owner divides 635 stickers evenly among all of her surfboards. Each surfboard has 3 tiki stickers and 2 turtle stickers. How many surfboards does she have?
Answer: She has 127 surfboards

Explanation:
Given that a surf shop owner divides 635 stickers evenly among all of her surfboards and each surfboard has 3 tiki stickers and 2 turtle stickers.
So,
The total number of stickers on each surfboard = Number of tiki stickers + Number of turtle stickers
= 3 + 2
= 5 stickers
So,
Total number of surfboards = Total number of stickers ÷ The number of stickers on each surfboard
= 635 ÷ 5
By using the partial quotients method,
635 ÷ 5 = ( 600 + 35 ) ÷ 5
= ( 600 ÷ 5 ) + ( 35 ÷ 5 )
= 120 + 7
= 127 surfboards
Hence, from the above,
We can conclude that there are 127 surfboards.

Question 3.
There are 1,008 projects in a science fair. The projects are divided into equally 9 rooms. Each room has 8 equal rows of projects. How many projects are in each row?
Big Ideas Math Answer Key Grade 4 Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.9 4
Answer: There are 14 projects in each row.

Explanation:
Given that there are 1,008 projects in a science fair and the projects are divided into equally 9 rooms. It is also given that each room has 8 equal rows of projects.
So,
Total number of rooms = 9 × 8 =72 rooms
Now,
Number of projects in each row = Total number of projects ÷ Total number of rooms
= 1,008 ÷ 72
Now,
By using the partial quotients method,
1,008 ÷ 72 = ( 720 + 288 )÷ 72
= ( 720 ÷ 72 ) + ( 288 ÷ 72 )
= 10 + 4
= 14 projects in each row
Hence, from the above,
We can conclude that there are 14 projects in each row.

Understand the problem. Then make a plan. How will you solve it? Explain.
Question 4.
Of 78 students who work on a mural, 22 students design it, and the rest of the students paint it. The painters are divided equally among 4 areas of the mural. How many painters are assigned to each area?
Answer: 14 painters are assigned to each area.

Explanation:
Given that there are 78 students who work on a mural and 22 students designed the mural and the remaining students painted it.
So,
The students who painted the mural = Total number of students – The students who designed the mural
= 78 – 22
= 56 students
It is also given that the painters are divided equally among 4 areas of the mural.
So,
The number of painters divided = The number of students who painted the mural ÷ 4
= 56 ÷ 4
Now,
By using the partial quotients method,
56 ÷ 4 = ( 40 + 16 ) ÷ 4
= ( 40 ÷ 4 ) + ( 16 ÷ 4 )
= 10 + 4
= 14 students
Hence, from the above,
we can conclude that there are 14 students are assigned to each area.

Question 5.
The Winter Olympics occur twice every 8 years. How many times will the Winter Olympics occur in 200 years?
Answer:  50 times

Explanation:
Given that the Winter Olympics occur twice every 8 years.
So, the Winter Olympics occur every 4 years. ( 8 ÷ 2 = 4 )
So,
The number of times Winter Olympics occur in 200 years = 200 ÷ 4
Now,
By using the partial quotients method,
200 ÷ 4 = ( 160 + 40 ) ÷ 4
= ( 160 ÷ 4 ) + ( 40 ÷ 4 )
= 40 + 10
= 50 times
Hence, from the above,
We can conclude that the Winter Olympics occur 50 times in 200 years.

Question 6.
A party planner wants to put 12 balloons at each of 15 tables. The balloons come in packages of 8. How many packages of balloons must the party planner buy?
Answer: The party planner buy 176 packages of balloons with 4 leftover balloons

Explanation:
Given that a part planner wants to put 12 balloons at each of 15 tables.
So,
Total number of balloons = 15 × 12
Now,
By using the partial products method,
12 × 15 = ( 10 + 2 ) × ( 10 + 5 )
= ( 10 × 10 ) + ( 10 × 5 ) + ( 2 × 10 ) + ( 2 × 5 )
= 100 + 50 + 20 +10
= 180 balloons
It is also given that the balloons come in packages of 8.
So,
The number of balloons that each package contain = Total number of balloons ÷ 8
= 180 ÷ 8
Now,
By using the partial quotients method,
180 ÷ 8 = ( 160 + 16 ) ÷ 8
= ( 160 ÷ 8 ) + ( 16 ÷ 8 )
= 20 + 2
= 22 R 4
Hence, from the above,
We can conclude that the number of balloons that each package contain is: 22 with 4 leftovers

Question 7.
An art teacher has 8 boxes of craft sticks. Each box has 235 sticks. The students use the sticks to make as many hexagons as possible. How many sticks are used?
Big Ideas Math Answer Key Grade 4 Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.9 5

Answer: The number of sticks used is: 313 with 2 leftover

Explanation:
Given that an art teacher has 8 boxes of craft sticks and each box has 235 sticks.
So,
Total number of sticks = 235 × 8
Now,
By using the partial products method,
235 × 8 = ( 200 + 35 ) × 8
= ( 200 × 8 ) + ( 35 × 8 )
= 1,600 + 280
= 1,880 sticks
It is also given the sticks are used to make hexagons.
We know that,
The number of sides of Hexagon = 6
So,
The number of sticks used to make hexagons = 1,880 ÷ 6
Now,
By using the partial quotients method,
1,880 ÷ 6 = ( 1,800 + 60 + 18 ) ÷ 6
= ( 1,800 ÷ 6 ) + ( 60 ÷ 6 ) + ( 18 ÷ 6 )
= 300 + 10 + 3
= 313 R 2
Hence, from the above,
We can conclude that the number of sticks used for making hexagons is: 313 with 2 leftover

Think and Grow: Modeling Real Life

Example
A book enthusiast has $200 to buy an e-reader and e-books. He uses a $20 off coupon and buys the e-reader shown. Each e-book costs $6. How many e-books can the book enthusiast buy?
Think: What do you know? What do you need to find? How will you solve it?
Big Ideas Math Answer Key Grade 4 Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.9 6
Step 1: How much money does the book enthusiast pay for the e-reader?
Subtract $20 from $119.

Step 2: Subtract to find how much money he has left to spend on e-books.
$200 – _____ = d
d is the unknown difference.

Step 3: Use to find the number of e-books the book enthusiast can buy.

So, the book enthusiast can buy 16 e-books.

Show and Grow

Question 8.
You run 17 laps around a track. Newton runs 5 times as many laps as you. Descartes runs 35 more laps than Newton. Eight laps around the track are equal to 1 mile. How many miles does Descartes run?
Big Ideas Math Answer Key Grade 4 Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.9 10
Answer: Descartes runs 15 miles

Explanation:
Given that you run 17 laps around a track and Newton runs 5 times as many laps as you.
So,
The number of laps covered by Newton = 17 × 5 = 85 laps
It is also given that Descartes runs 35 more laps than Newton.
So,
The number of laps covered by Descartes = 35 + Number of laps covered by Newton
= 35 + 85
= 120 laps
Given that,
8 laps = 1 mile
So,
The number of miles covered by Descartes = The number of laps covered by Descartes ÷ 8
= 120 ÷ 8
Now,
By using the partial quotients method,
120 ÷ 8 = ( 80 + 40 ) ÷ 8
= ( 80 ÷ 8 ) + ( 40 ÷ 8 )
= 10 + 5
= 15 miles
Hence, from the above,
We can conclude that Descartes covers 15 miles.

Problem Solving: Division Homework & Practice 5.9

Understand the problem. Then make a plan. How will you solve it? Explain.
Question 1.
You borrow a 235-page book from the library. You read 190 pages. You have 3 days left until you have to return the book. You want to read the same number of pages each day to finish the book. How many pages should you read each day?

Answer:  The number of pages you should read each day is: 15 pages

Explanation:
Given that you borrow a 235-page book and you read 190 pages. It is also given that you have only 3 days left until you have to return the book.
So,
The number of remaining pages = 235 – 190 = 45 pages
So,
The number of pages you should read each day = 45 ÷ 3
Now,
By using the partial quotients method,
45 ÷ 3 = ( 30 + 15 ) ÷ 3
= ( 30 ÷ 3 ) + ( 15 ÷ 3 )
= 10 + 5
= 15
Hence, from the above,
We can conclude that the number of pages you should read each day is: 15 pages

Question 2.
There are 24 fourth-graders and 38 fifth graders traveling to a math competition. If 8 students can fit into each van, how many vans are needed?
Answer: The number of vans needed is: 7 R 6

Explanation:
Given that there are 24 fourth-graders and 38 fifth-graders traveling to a math competition.
So,
The total number of students who are traveling to a math competition = Number of fourth-graders + Number of fifth-graders
= 24 + 38
=  62 students
It is also given that 8 students can fit into each van.
So,
The number of vans needed = Total number of students ÷ Number of students fit into each van
= 62 ÷ 8
Now,
By using the partial quotients method,
62 ÷ 8 = ( 40 + 16 ) ÷ 8
= ( 40 ÷ 8 ) + ( 16 ÷ 8 )
= 5 + 2
= 7 vans with 6 students leftover
= 7 R 6
Hence, from the above,
We can conclude that the number of vans needed is: 7 vans with 6 students leftover

Question 3.
Your class has 3 bags of buttons to make riding horses for a relay race. Each bag has 54 buttons. What is the greatest number of horses your class can make?
Big Ideas Math Answer Key Grade 4 Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.9 11
Answer: The greatest number of horses your class can make = 40 horses

Explanation:
Given that your class has 3 bags of buttons to make riding horses for a relay race and each bag contains 54 buttons
So,
The total number of buttons present = 54 × 3
Now,
By using the Distributive Property of multiplication,
54 × 3 = ( 50 + 4 ) × 3
( 50 × 3 ) + ( 4 × 3 )
= 150 + 12
= 162 buttons
It is also given that each horse needs 4 buttons.
So,
The greatest number of horses you can make = Total number of buttons ÷ 4
= 162 ÷ 4
NOw,
By using the partial quotients method,
162 ÷ 4 = ( 120 + 40 ) ÷ 4
= ( 120 ÷ 4 ) + ( 40 ÷ 4 )
= 30 + 10
= 40 horses with 2 buttons leftover
= 40 R 2
Hence, from the above,
We can conclude that the greatest number of horses you can make = 40 horses

Question 4.
Factory workers make 2,597 small, 2,597 medium, and 2,597 large plush toys. The workers pack the toys into boxes with 4 toys in each box. How many toys are left over?
Answer: The number of toys leftover are: 3 toys

Explanation:
Given that factory workers make 2,597 small, 2,597 medium, and 2,597 large plush toys.
So,
Total number of toys that factory workers can made = 2,597 + 2,597 + 2,597
= 7,791 toys
It is also given that the workers pack the toys into boxes with 4 toys in each box.
So,
The number of toys in each box = Total number of toys that the factory workers can make ÷ 4
= 7,791 ÷ 4
Now,
By using the partial quotients method,
7,791 ÷ 4 = ( 7,200 + 520 + 68 ) ÷ 4
= ( 7,200 ÷ 4 ) + ( 520 ÷ 4 ) + ( 68 ÷ 4 )
= 1,200 + 130 + 17
=1,347 R 3
Hence, from the above,
We can conclude that there are 3 leftovers after all the toys packed into  group of 4.

Question 5.
Writing
Write and solve a two-step word problem that can be solved using division.
Answer:

Question 6.
Modeling Real Life
You exercise for 300 minutes this week. Outside of jogging, you divide your exercising time equally among 3 other activities. How many minutes do you spend on each of your other 3 activities?
Big Ideas Math Answer Key Grade 4 Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.9 12
Answer: You spend 100 minutes on each of your other 3 activities.

Explanation:
Given that you exercise for 300 minutes this week.
It is also given that outside of jogging, you divide your exercising time equally among 3 other activities.
So,
The number of minutes you spend on each of your 3 other activities = Total minutes of exercising ÷ Total number of activities outside jogging
= 300 ÷ 3
Now,
By using the partial quotients method,
300 ÷ 3 = ( 270 + 30 ) ÷ 3
= ( 270 ÷ 3 ) + ( 30 ÷ 3 )
= 90 + 10
= 100 minutes.
Hence, from the above,
we can conclude that you spent 100 minutes on each of the 3 other activities outside of jogging.

Question 7.
Modeling Real Life
Drones are used to help protect orangutans and their habitats. A drone takes a picture every 2 seconds. How many pictures does the drone take in 30 minutes?
Big Ideas Math Answer Key Grade 4 Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 5.9 13

Answer: The drone makes 900 pictures in 30 minutes.

Explanation:
Given that the drones are used to help protect orangutans and their habitats
It is also given that a drone takes a picture every 2 seconds.
We know that,
1 minute = 60 seconds
So,
30 minutes = 30 × 60 seconds
Now,
By using the place-value method,
30 × 60 = 3 tens × 60
= 3 tens × 6 tens
= 18 × 10 × 10
= 1,800 seconds
So,
The number of pictures does the drone takes in 30 minutes = 1,800 ÷ 2
Now,
By using the partial quotients method,
1,800 ÷ 2 = ( 1,000 +800 ) ÷ 2
= ( 1,000 ÷ 2 ) + ( 800 ÷ 2 )
= 500 + 400
= 900 pictures
Hence, from the above,
we can conclude that the drone takes 900 pictures in 30 minutes.

Review & Refresh

Find the product. Check whether your answer is reasonable.
Question 8.
Estimate: _____
41 × 22 = _____
Answer: 41 × 22 = 902

Explanation:
By using the partial products method,
41 × 22 = ( 40 + 1 ) × ( 20 + 2 )
=( 40 × 20 ) + ( 1 × 20 ) + ( 40 × 2 ) + ( 1 × 2 )
= 800 + 20 + 80 + 2
= 902
So, 41 × 22 = 902
Estimate:
Let 41 be rounded to 40.
Let 22 be rounded to 20.
So, by using the place-value method,
40 × 20 = 4 tens × 20
= 80 tens
= 80 × 10
= 800
So, 40 × 20 = 400
Hence, from the above,
We can conclude that the actual answer is not near to the Estimate. So, the answer is not reasonable.

Question 9.
Estimate: ______
87 × 19 = ______

Answer: 87 × 19 = 1,653

Explanation:
By using the partial products method,
87 × 19 = ( 80 + 17 ) × ( 10 + 9 )
=( 80 × 10 ) + ( 80 × 9 ) + ( 17 × 10 ) + ( 17 × 9 )
= 800 + 720 + 170 + 153
= 1,653
So, 87 × 19 = 1,653
Estimate:
Let 87 be rounded to 85.
Let 19 be rounded to 20.
So, by using the place-value method,
85 × 20 = 2 tens × 85
= 170 tens
= 170 × 10
= 1,700
So, 85 × 20 = 1,700
Hence, from the above,
We can conclude that the actual answer is near to the Estimate. So, the answer is reasonable.

Question 10.
Estimate: _____
36 × 59 = ______
Answer: 36 × 59 = 2,124

Explanation:
By using the partial products method,
36 × 59 = ( 30 + 6 ) × ( 50 + 9 )
=( 30 × 50 ) + ( 30 × 9 ) + ( 6 × 50 ) + ( 6 × 9 )
= 1,500 + 270 + 300 + 54
= 2,124
So, 36 × 59 = 2,124
Estimate:
Let 36 be rounded to 35.
Let 59 be rounded to 60.
So, by using the place-value method,
35 × 60 = 6 tens × 35
= 210 tens
= 210 × 10
= 2,100
So, 35 × 60 = 2,100
Hence, from the above,
We can conclude that the actual answer is near to the Estimate. So, the answer is reasonable.

Divide Multi-Digit Numbers by One-Digit Numbers Performance Task

The students in fourth grade go on a field trip to a planetarium.
Question 1.
The teachers have $760 to buy all of the tickets for the teachers and students. They receive less than $6 in change.

a. Each ticket costs $6. How many tickets do the teachers buy?
b. Exactly how much money is left over?
Big Ideas Math Answers 4th Grade Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 1
c. There are 6 groups on the field trip. Each group has 1 teacher. There is an equal number of students in each group. How many students are in each group?
d. Two groups can be in the planetarium for each show. The planetarium has 7 rows of seats with 8 seats in each row. How many seats are empty during each show?

Answer:
a) The teachers buy 126 tickets.
b) The money that is leftover: $4
c) The number of students in each group: 20
d) The number of seats empty during each show is: 0

Explanation:
Given that the teachers have $760 to buy all of the tickets for the teachers and students and they received less than $6 change.
a) Given that each ticket cost $6
So,
The number of tickets that the teachers can buy = 760 ÷ 6
Now,
By using the partial quotients method,
760 ÷ 6 = ( 600 + 120 + 36 ) ÷ 6
= ( 600 ÷ 6 ) + ( 120 ÷ 6 ) + ( 36 ÷ 6 )
= 100 + 20 + 6
= 126 tickets with 4 leftover as change=
= 126 R 4
b) From the above,
The money that is leftover = $4
c) Given that there are 6 groups on the field trip and each group has 1 teacher.
It is also given that there is an equal number of students in each group.
From the number of tickets, we can conclude that there are 126 students.
So,
The number of groups of students = 126 ÷ 6
Now,
By using the partial quotients method,
126 ÷ 6 = ( 60 + 60 ) ÷ 6
= ( 60 ÷ 6 ) + ( 60 ÷ 6 )
= 10 + 10
= 20 R 6
Hence, from the above,
we can conclude that there are 20 student groups.
d) Given that there are two groups in the planetarium for each show and the planetarium has 7 rows of seats with 8 seats in each row.
So,
The total number of seats = 7 × 8 = 56 seats
So,
The number of seats that are empty during each show = 56 ÷ 2 = 28
Hence, there are no empty seats left in each show.

Question 2.
The groups will be at the planetarium from 11:00 A.M. until 2:30 P.M. During that time they will rotate through 7 events: the planetarium show, 5 activities, and lunch. The planetarium show lasts 45 minutes. Each activity lasts 22 minutes. Students have 5 minutes between each event. How long does each group have to eat lunch?
Big Ideas Math Answers 4th Grade Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 2

Answer: Each group has 10 minutes to eat lunch.

Explanation:
Given that,
The groups will be at the planetarium from 11:00 A.M. until 2:30 P.M.
The number of events = 7
The number of activities = 5
The time of planetarium show = 45 minutes
The time each activity lasts = 22 minutes
So,
The time at which 5 activities lasts = 22 × 5 = 110 minutes
The time for which students have time between each event = 5 minutes
Now,
The total time that the groups will be at the planetarium = From 11:00 A.M. until 2:30 P.M.
= 3 hours 30 minutes
= 210 minutes
The time taken by the groups for the activities and events = 45 + 110 =155 minutes
So,
The remaining time = 210 – 155 = 55 minutes
The time is taken by all the students between the events = 5 × 7 = 35 minutes
So, the remaining time = 55 – 35 = 20 minutes
Given that there are 2 groups in the above Exercise.
So, The time is taken for lunch by each group = 20 ÷ 2 = 10 minutes.
Hence, from the above,
we can conclude that there are 10 minutes for each group to eat lunch.

Question 3.
You learn that the distance around Mars is about twice the distance around the moon. The distance around Mars is 13,263 miles. To find the distance around the moon, do you think an estimate or an exact answer is needed? Explain.

Answer: We need an Estimate.

Explanation:
Given that the distance around Mars is about twice the distance around the moon and the distance around Mars is 13,263 miles.
So,
The distance of Moon = The distance around Mars ÷ 2
= 13,263 ÷ 2
If we find the value of 13,263 ÷ 2, we will get the answer with decimals.
So, we have to take an Estimate to calculate the distance around the Moon.
Estimate:
let 13,263 be rounded to 13,262
Now,
By using the partial quotients method,
13,262 ÷ 2 = ( 12,000 + 1,000 + 260 + 2 ) ÷ 2
= ( 12,000 ÷ 2 ) + ( 1,000 ÷ 2 ) + ( 260 ÷ 2 ) + ( 2 ÷ 2 )
= 6,000 + 500 + 130 +1
= 6,631 miles.
Hence, from the above,
We can conclude that the estimated distance around the Moon = 6,631 miles

Divide Multi-Digit Numbers by One-Digit Numbers Activity

Division Dots
Directions:
1. Players take turns connecting two dots, each using a different color.
2. On your turn, connect two dots, vertically or horizontally. If you close a square around a division problem, find and write the quotient and the remainder. If you do not close a square, your turn is over.
3. Continue playing until all division problems are solved.
4. The player with the most completed squares wins!
Big Ideas Math Answers 4th Grade Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers 3
Answer:

Divide Multi-Digit Numbers by One-Digit Numbers Chapter Practice

5.1 Divide Tens, Hundreds, and Thousands

Find the quotient.
Question 1.
90 ÷ 9 = _____
Answer: The quotient is: 10

Explanation:
The given Expression is:
90 ÷ 9 =9 tens ÷ 9
= 1 ten
=10
So, 90 ÷ 9 = 10

Note: The “Quotient” of an Expression is defined as a number that divides the Dividend and the quotient should be an Integer.

Question 2.
560 ÷ 7 = _____
Answer: The quotient is: 80

Explanation:
The given Expression is:
560 ÷ 7 = 56 tens ÷ 7
= 8 tens
=80
So, 560 ÷ 7 = 80

Note: The “Quotient” of an Expression is defined as a number that divides the Dividend and the quotient should be an Integer.

Question 3.
2,700 ÷ 9 = _____
Answer: The quotient is: 300

Explanation:
The given Expression is:
2,700 ÷ 9 = 27 hundreds ÷ 9
= 3 hundred
=300
So, 2,700 ÷ 9 = 300

Note: The “Quotient” of an Expression is defined as a number that divides the Dividend and the quotient should be an Integer.

Question 4.
240 ÷ 4 = ______
Answer: The quotient is: 60

Explanation:
The given Expression is:
240 ÷ 4 = 24 tens ÷ 4
= 6 tens
=60
So, 240 ÷ 4 = 60

Note: The “Quotient” of an Expression is defined as a number that divides the Dividend and the quotient should be an Integer.

Question 5.
4,500 ÷ 5 = _____
Answer: The quotient is: 900

Explanation:
The given Expression is:
4,500 ÷ 5 = 45 hundreds ÷ 5
= 9 hundred
=900
So, 4,500 ÷ 5 = 900

Note: The “Quotient” of an Expression is defined as a number that divides the Dividend and the quotient should be an Integer.

Question 6.
60 ÷ 6 = _____
Answer: The quotient is: 10

Explanation:
The given Expression is:
60 ÷ 6 = 6 tens ÷ 6
= 1 ten
=10
So, 60 ÷ 6 = 10

Note: The “Quotient” of an Expression is defined as a number that divides the Dividend and the quotient should be an Integer.

Question 7.
1,600 ÷ 8 = _____
Answer: The quotient is: 200

Explanation:
The given Expression is:
1,600 ÷ 8 = 16 hundreds ÷ 8
= 2 hundred
=200
So, 1,600 ÷ 8 = 200

Note: The “Quotient” of an Expression is defined as a number that divides the Dividend and the quotient should be an Integer.

Question 8.
30 ÷ 3 = ______
Answer: The quotient is: 10

Explanation:
The given Expression is:
30 ÷ 3 = 3 tens ÷ 3
= 1 ten
=10
So, 30 ÷ 3 = 10

Note: The “Quotient” of an Expression is defined as a number that divides the Dividend and the quotient should be an Integer.

Question 9.
540 ÷ 9 = _____
Answer: The quotient is: 60

Explanation:
The given Expression is:
540 ÷ 9 = 54 tens ÷ 9
= 6 tens
=60
So, 540 ÷ 9 = 60

Note: The “Quotient” of an Expression is defined as a number that divides the Dividend and the quotient should be an Integer.

Find the missing number.
Question 10.
720 ÷ _____ = 80
Answer: The missing number is: 9

Explanation:
Let the missing number be X.
The given Expression is:
720 ÷ X = 80
So, X can be calculated by
X= 720 ÷ 80 = 9

Question 11.
_____ ÷ 7 = 10
Answer: 70

Explanation:
Let the missing number be X.
The given Expression is:
X ÷ 7 = 10
So, X can be calculated by
X= 7 × 10 = 70

Question 12.
1,800 ÷ _____ = 600
Answer: The missing number is: 3

Explanation:
Let the missing number be X.
The given Expression is:
1,800 ÷ X = 600
So, X can be calculated by
X= 1,800 ÷ 600 = 3

5.2 Estimate Quotients

Question 13.
47 ÷ 7
Answer: The estimated quotient is: 7

Explanation;
Let 47 be rounded off to 49.
So, now we have to find 49 ÷ 7
Now,
49 ÷ 7 = ( 42 + 7 ) ÷ 7
= ( 42 ÷ 7 ) + ( 7 ÷ 7 )
= 6 + 1
= 7
Hence, from the above,
We can conclude that 47 ÷ 7 can be rounded off to 7.

Question 14.
593 ÷ 6
Answer: The estimated quotient is: 99

Explanation;
Let 593 be rounded off to 594.
So, now we have to find 594 ÷ 6
Now,
594 ÷ 6 = ( 540 + 54 ) ÷ 6
= ( 540 ÷ 6 ) + ( 54 ÷ 6 )
= 90 + 19
= 99
Hence, from the above,
We can conclude that 593 ÷ 6 can be rounded off to 99

Find two estimates that the quotient is between.
Question 15.
261 ÷ 8
Answer: The quotient of 261 ÷ 8 is between 30 and 40.

Explanation:
Use 240. 24 ÷ 8 = 3, so 240 ÷ 8 = 30.
Use 320. 32 ÷ 8 = 4, so 320 ÷ 8 = 40 .
261 is between 240 and 320.
So, the quotient of 261 ÷ 8 is between 30 and 40.

Question 16.
7,012 ÷ 9
Answer: The quotient of 7,012 ÷ 9 is between 700 and 800.

Explanation:
Use 6,300. 63 ÷ 9 = 7, so 6,300 ÷ 9 = 700.
Use 7,200. 72 ÷ 9 = 8, so 7,200 ÷ 9 = 800 .
7,012 is between 6,300 and 7,200.
So, the quotient of 7,012 ÷ 9 is between 700 and 800.

Question 17.
Reasoning
Explain how to find a better estimate for 2,589 ÷ 6 than the one shown.
Round 2,589 to 3,000. Estimate 3,000 ÷ 6.
3,000 ÷ 6 = 500, so 2,589 ÷ 6 is about 500.
Answer: The better Estimate to find 2,589 ÷ 6 is to round off 2,589 to 2,580

Explanation:
Given Expression is 2,589 ÷ 6
Let 2,589 be rounded to 3,000
So,
3,000 ÷ 6 = 300 tens ÷ 6
= 50 tens
= 500
Now,
Let 2,589 be rounded to 2,580.
So,
2,580 ÷ 6 = 258 tens ÷ 6
= 43 tens
=430
But, 2,589 is near to 2,580 when compared to 3,000.
So,
We can conclude that
2,580 ÷ 6 = 43

5.3 Understand Division and Remainder

Use a model to find the quotient and the remainder.
Question 18.
14 ÷ 4 = _____ R _____
Answer: 3 R 2

Explanation:
Divide 14 into 4 equal parts.
So, we will get
Number of Units in each group = 3
Number of units leftover = 2
Hence,
26 ÷ 3 = 3 R 2
Where R is the Remainder (or) the number of units leftover

Question 19.
28 ÷ 6 = _____ R ______
Answer: 4 R 4

Explanation:
Divide 28 into 6 equal parts.
So, we will get
Number of Units in each group = 4
Number of units leftover = 4
Hence,
28 ÷ 6 = 4 R 4
Where R is the Remainder (or) the number of units leftover

Question 20.
18 ÷ 7 = _____ R ______
Answer: 2 R 4

Explanation:
Divide 18 into 7 equal parts.
So, we will get
Number of Units in each group = 2
Number of units leftover = 4
Hence,
18 ÷ 7 = 2 R 4
Where R is the Remainder (or) the number of units leftover

Question 21.
23 ÷ 3 = _____ R _______
Answer: 7 R 2

Explanation:
Divide 23 into 3 equal parts.
So, we will get
Number of Units in each group = 7
Number of units leftover = 2
Hence,
23 ÷ 3 = 7 R 2
Where R is the Remainder (or) the number of units leftover

Question 22.
Modeling Real Life
Tours of a factory can have no more than 9 guests. There are 76 guests in line to tour the factory.
• How many tours are full?
• How many tours are needed?
• How many guests are on the last tour?
Big Ideas Math Answers 4th Grade Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers chp 22
Answer:

a) The number of tours that are full = 8 tours
b) The number of tours needed = 9 tours
c) The number of guests on the last tour = 4 guests

Explanation:
Given that there are 76 guests in line to tour a  factory It is also given that there are no more than 9 guests on the tours of a  factory.
We have to observe that to make all the guests full on all the trips without leftovers, we will need 9 trips. i.e.., 81 guests.
So, now we have to find the number of guests in each tour of a crayon factory by finding the quotient and remainder of 76 ÷ 9.
Now,
76 ÷ 9
From this, we can see
The number of guests in each tour that are full = 8
The number of guests leftover on the last trip = 4
Hence, from the above,
We can conclude that
a) 8 Tours are full.
b) 9 Tours are needed.
c) 4 guests are on the last tour.

5.4 Use Partial Quotients

Use partial quotients to divide.
Question 23.
\(\sqrt [ 8 ]{ 504 } \)
Answer: 504 ÷ 8 = 63

Explanation:
By using the partial quotients method,
504 ÷ 8 = ( 480 + 24 ) ÷ 8
( 480 ÷ 8 ) + ( 24 ÷ 8 )
= 60 + 3
= 63
Hence, 504 ÷ 8 = 63

Question 24.
\(\sqrt [ 4 ]{ 52 } \)
Answer: 52 ÷ 4 = 13

Explanation:
By using the partial quotients method,
52 ÷ 4 = ( 40 + 12 ) ÷ 4
= ( 40 ÷ 4 ) + ( 12 ÷ 4 )
= 10 + 3
= 13
Hence, 52 ÷ 4 = 13

Question 25.
\(\sqrt [ 7 ]{ 119 } \)
Answer: 119 ÷ 7 = 17

Explanation:
By using the partial quotients method,
119 ÷ 7 = ( 105 + 14 ) ÷ 7
= ( 105 ÷ 7 ) + ( 14 ÷ 7 )
= 15 + 2
= 17
Hence, 119 ÷ 7 = 17

5.5 Use Partial Quotients with an Remainder

Use partial quotients to divide.
Question 26.
\(\sqrt [ 5 ]{ 82 } \)
Answer: 82 ÷ 5 = 16 R 2

Explanation:
By using the partial quotients method,
82 ÷ 5 = ( 75 + 5 ) ÷ 5
= ( 75 ÷ 5 ) + ( 5 ÷ 5 )
= 15 + 1
= 16 R 2
Hence, 82 ÷ 5 = 16 R 2

Question 27.
\(\sqrt [ 8 ]{ 759 } \)
Answer: 759 ÷ 8 = 94 R 7

Explanation:
By using the partial quotients method,
759 ÷ 8 = ( 720 + 32 ) ÷ 8
= ( 720 ÷ 8 ) + ( 32 ÷ 8 )
= 90 + 4
= 94 R 7
Hence, 759 ÷ 8 = 94 R 7

Question 28.
\(\sqrt [ 3]{ 5,468 } \)
Answer: 5,468 ÷ 3 = 1,822 R 2

Explanation:
By using the partial quotients method,
5,468 ÷ 3 = ( 5,400 + 66 ) ÷ 3
= ( 5,400 ÷ 3 ) + ( 66 ÷ 3 )
= 1,800 + 22
= 1,822 R 2
Hence, 5,468 ÷ 3 = 1,822 R 2

5.6 Divide Two-Digit Numbers by One-Digit Numbers

Divide. Then check your answer.
Question 29.
\(\sqrt [ 3 ]{ 58 } \)
Answer: 58 ÷ 3 = 19 R 1

Explanation:
By using the partial quotients method,
58 ÷ 3 = ( 48 + 9 ) ÷ 3
= ( 48 ÷ 3 ) + ( 9 ÷ 3 )
= 16 + 3
= 19 R 1
Hence, 58 ÷ 3 = 19 R 1

Question 30.
\(\sqrt [ 4 ]{ 90 } \)
Answer: 90 ÷ 4 = 22 R 2

Explanation:
By using the partial quotients method,
90 ÷ 4 = ( 80 + 8 ) ÷ 4
= ( 80 ÷ 4 ) + ( 8 ÷ 4 )
= 20 + 2
= 22 R 2
Hence, 90 ÷ 4 = 22 R 2

Question 31.
\(\sqrt [ 2 ]{ 67 } \)
Answer: 67 ÷ 2 = 33 R 1

Explanation:
By using the partial quotients method,
67 ÷ 2 = ( 60 + 6 ) ÷ 2
= ( 60 ÷ 2 ) + ( 6 ÷ 2 )
= 30 + 3
= 33 R 1
Hence, 67 ÷ 2 = 33 R 1

5.7 Divide Multi-Digit Numbers by One-Digit Numbers

Divide. Then check your answer.
Question 32.
\(\sqrt [ 5 ]{ 865 } \)
Answer: 865 ÷ 5 = 173

Explanation:
By using the partial quotients method,
865 ÷ 5 = ( 800 + 65 )÷ 5
= ( 800 ÷ 5 ) + ( 65 ÷ 5 )
=  160 + 13
= 173
Hnece, 865 ÷ 5 = 173

Question 33.
\(\sqrt [ 2 ]{ 7,532 } \)
Answer: 7,532 ÷ 2 = 3,766

Explanation:
By using the partial quotients method,
7,532 ÷ 2 = ( 7,000 + 500 + 3 ) ÷ 2
= ( 7,000 ÷ 2 ) + ( 500 ÷ 2 )  + ( 32 ÷ 2 )
= 3,500 + 250 + 16
= 3,766
Hence, 7,532 ÷ 2 = 3,766

Question 34.
\(\sqrt [ 4 ]{ 507 } \)
Answer: 507 ÷ 4 = 126 R 3

Explanation:
By using the partial quotients method,
507 ÷ 4 = ( 480 + 24 ) ÷ 4
= ( 480 ÷ 4 ) + ( 24 ÷ 4 )
= 120 + 6
= 126 R 3
Hence, 507 ÷ 4 = 126 R 3

Question 35.
\(\sqrt [ 6 ]{ 9,127 } \)
Answer: 9,127 ÷ 6 = 1,521 R 1

Explanation:
By using the partial quotients method,
9,127 ÷ 6 = ( 9,000 + 126 ) ÷ 6
= ( 9,000 ÷ 6 ) + ( 126 ÷ 6 )
= 1,500 + 21
= 1,521 R 1

Question 36.
\(\sqrt [ 8 ]{ 253 } \)
Answer: 253 ÷ 8 = 31 R 5

Explanation:
By using the partial quotients method,
253 ÷ 8 = ( 240 + 8 ) ÷ 8
= ( 240 ÷ 8 ) + ( 8 v 8 )
= 30 + 1
= 31 R 5
Hence,253 ÷ 8 = 31 R 5

Question 37.
\(\sqrt [ 6 ]{ 429 } \)
Answer: 429 ÷ 6 = 71 R 3

Explanation:
By using the partial quotients,
429 ÷ 6 = ( 420 + 6 ) ÷ 6
= ( 420 ÷ 6 ) + ( 6 ÷ 6 )
= 70 + 1
= 71 R 3
Hence, 429 ÷ 6 = 71 R 3

5.8 Divide by One-Digit Numbers

Divide. Then check your answer.
Question 38.
\(\sqrt [ 3 ]{ 91 } \)
Answer: 91 ÷ 3 = 30 R 1

Explanation:
By using the partial quotients method,
91 ÷ 3 = ( 60 + 30 ) ÷ 3
= ( 60 ÷ 3 ) + ( 30 ÷ 3 )
= 20 + 10
= 30 R 1
Hence, 91 ÷ 3 = 30 R 1

Question 39.
\(\sqrt [ 7 ]{ 914 } \)
Answer: 914 ÷ 7 = 130 R 4

Explanation:
By using the partial quotients method,
914 ÷ 7 = ( 840 + 70 ) ÷ 7
= ( 840 ÷ 7 ) + ( 70 ÷ 7 )
= 120 + 10
= 130 R 4
Hence, 914 ÷ 7 = 130 R 4

Question 40.
\(\sqrt [ 2 ]{ 6,075 } \)
Answer: 6,075 ÷ 2 = 3,037 R 1

Explanation:
By using the partial quotients method,
6,075 ÷ 2 = ( 6,000 + 74 ) ÷ 2
= ( 6,000 ÷ 2 ) + ( 74 ÷ 2 )
= 3,000 + 37
= 3,037 R 1
Hence, 6,075 ÷ 2 = 3,037 R 1

5.9 Problem Solving: Division

Question 41.
A young snake sheds its skin every 2 weeks. How many times will the snake shed its skin in 3 years?
Big Ideas Math Answers 4th Grade Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers chp 41.1
Big Ideas Math Answers 4th Grade Chapter 5 Divide Multi-Digit Numbers by One-Digit Numbers chp 41.2
Answer: A young snake sheds its skin 78 times in 3 years.

Explanation:
Given that a young snake sheds its skin every 2 weeks
We know that,
1 year = 52 weeks
So,
3 years = 3 × 52 = 156 weeks
So,
The number of times the young snake sheds its skin in 3 years = 156 ÷ 2
Now,
By using the partial quotients method,
156 ÷ 2 = ( 100 + 56 ) ÷ 2
= ( 100 ÷ 2 ) + ( 56 ÷ 2 )
= 50 + 28
= 78 times
Hence, from the above,
We can conclude that the young snake will shed its skin 78 times in 3 years

Conclusion:

After your preparation please test your knowledge by solving the problems provided at the end of the chapter. You can also prepare the questions on your own and solve the problems if you learn the concepts in depth. Prepare well and secure highest marks in the exams. All the Best!!!

Big Ideas Math Answers Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison

Big Ideas Math Answers Grade 4 Chapter 7

Other than marks it is important for the students to understand the in-depth concepts in math. So, make your students learn the real-time examples along with learning the concepts. We have given questions with real-time problems so that it will be easy to understand the concept deeply. Refer to our Big Ideas Math Answers Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison for a better understanding of the concept. Keeping the growth of the students in mind our team has prepared the Big Ideas Math Answers Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison topic-wise.

Big Ideas 4th Grade Math Book Chapter 7 Understand Fraction Equivalence and Comparison Answer Key

Start your preparation with the Big Ideas 4th Grade Math Book Answer Key. Know your strengths and weaknesses and concentrate on the weak concepts to become a math expert. Find out different problems with a detailed explanation here. We have included problems on every topic individually. therefore, preparing with our Big Ideas Math Book 4th Grade Solution Key Chapter 7 Understand Fraction Equivalence and Comparison is easy and quick. Click on the below-given links and begin practicing the problems.

Lesson: 1 Model Equivalent Fractions

Lesson: 2 Generate Equivalent Fractions by Multiplying

Lesson: 3 Generate Equivalent Fractions by Dividing

Lesson: 4 Compare Fractions Using Benchmarks

Lesson: 5 Compare Fractions

Performance Task

Lesson 7.1 Model Equivalent Fractions

Explore and Grow

Use the model to write fractions that are the same size as \(\frac{1}{2}\).
Big Ideas Math Answer Key Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.1 1

Answer: Using the model, the fractions that have the same size as \(\frac{1}{2}\) are:
\(\frac{2}{4}\), \(\frac{3}{6}\), \(\frac{4}{8}\), \(\frac{6}{12}\)

Explanation:
Given model is:
Big Ideas Math Answer Key Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.1 1
In the above table, the whole 1 is divided in to different fractions like \(\frac{1}{2}\),
\(\frac{1}{3}\), \(\frac{1}{4}\), \(\frac{1}{5}\), \(\frac{1}{6}\),
\(\frac{1}{8}\), \(\frac{1}{10}\), and \(\frac{1}{12}\)
So,
From this table, the fractions that give us the same size as of \(\frac{1}{2}\) are:
\(\frac{2}{4}\), \(\frac{3}{6}\), \(\frac{4}{8}\), \(\frac{6}{12}\)

Hence, from the above,
We can conclude that the fractions that are having the same size as \(\frac{1}{2}\) are:
\(\frac{2}{4}\), \(\frac{3}{6}\), \(\frac{4}{8}\), \(\frac{6}{12}\)

Reasoning
Can you write a fraction with a denominator of 12 that is to the same size as \(\frac{2}{3}\) ? Explain.

Answer: The fraction with the denominator 12 that is to the same size as \(\frac{2}{3}\) is:
\(\frac{8}{12}\)

Explanation:
The given model is:
Big Ideas Math Answer Key Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.1 1
From the above table, the fractions having denominator 12 have only 1 row.
But, we want the same value of \(\frac{2}{3}\) by having the denominator 12
So, from the table,
When we add all \(\frac{1}{12}\), we will get the value \(\frac{8}{12}\)
So,
The value of \(\frac{2}{3}\) is equal to the \(\frac{8}{12}\)
Hence, from the above,
We can conclude that the fraction with denominator 12 that is having the same size as \(\frac{2}{3}\) is:
\(\frac{8}{12}\)

Think and Grow: Model Equivalent Fractions

Two or more numbers that have the same value are. Two or equivalent more fractions that name the same part of a whole are equivalent. Equivalent fractions name the same point on a number line. fractions.
Example
Use models to find equivalent fractions for \(\frac{2}{3}\) .
Big Ideas Math Answer Key Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.1 2
One Way:
Draw models that show the same whole divided into different numbers of parts.

Show and Grow

Use the model to find an equivalent fraction for \(\frac{2}{5}\) .
Answer:
Big Ideas Math Answer Key Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.1 4
Answer: The equivalent fraction for \(\frac{2}{5}\) is: \(\frac{4}{10}\)

Explanation:
The given model is:
Big Ideas Math Answer Key Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.1 4
From the above model, we can say the fraction is: \(\frac{2}{5}\)
Now, the model for the equivalent fraction of \(\frac{2}{5}\) is:

From the above model, we can say that the fraction is: \(\frac{4}{10}\)
Hence, from the above,
We can conclude that the equivalent of \(\frac{2}{5}\) is: \(\frac{4}{10}\)

Question 2.
Use the number line to find an1equivalent fraction for \(\frac{1}{6}\) .
Answer:
Big Ideas Math Answer Key Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.1 5.1
The equivalent fraction for \(\frac{1}{6}\) is: \(\frac{2}{12}\)

Explanation:
The given number line is:
Big Ideas Math Answer Key Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.1 5.1
From the given line, we can say that every 2 lines represent \(\frac{1}{6}\) multiple.
So,
Each line of the given number line represents \(\frac{1}{12}\) multiple
So,
The number line with \(\frac{1}{12}\) will be like:

So, from the above number line,
We can say that the equivalent fraction of \(\frac{1}{6}\) is: \(\frac{1}{12}\)

Hence, from the above,
We can conclude that the equivalent fraction of \(\frac{1}{6}\) is: \(\frac{1}{12}\)

Apply and Grow: Practice

Use the model to find an equivalent fraction.
Question 3.
Big Ideas Math Answer Key Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.1 5
Answer: The equivalent fraction of \(\frac{3}{6}\) is: \(\frac{1}{2}\)

Explanation:
The given model is:
Big Ideas Math Answer Key Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.1 5
From the above model, we can say that the fraction is: \(\frac{3}{6}\)
So,
The model for the equivalent fraction is:

From the above, we can see that the model represents \(\frac{1}{2}\)
Note: \(\frac{3}{6}\) is equal to \(\frac{1}{2}\) because when we divide \(\frac{3}{6}\) with 3, then we can get the answer.
Hence, from the above,
We can conclude that \(\frac{3}{6}\) = \(\frac{1}{2}\)

Question 4.
Big Ideas Math Answer Key Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.1 6
Answer: The  equivalent fraction of \(\frac{1}{5}\) is: \(\frac{2}{10}\)

Explanation:
The given model is:
Big Ideas Math Answer Key Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.1 6
From the above model, we can say that the fraction is: \(\frac{1}{5}\)
So,
The model for the equivalent fraction is:

So,
The equivalent fraction from the above model is: \(\frac{2}{10}\)
So,
When \(\frac{2}{10}\) is divided by 2, we will get the equivalent fraction
Hence, from the above,
We can say that \(\frac{1}{5}\) = \(\frac{2}{10}\)

Question 5.
Big Ideas Math Answer Key Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.1 7
Answer: The equivalent fraction of \(\frac{4}{5}\) is: \(\frac{8}{10}\)

Explanation:
The given model is:
Big Ideas Math Answer Key Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.1 7
From the above model, we can say the fraction is \(\frac{4}{5}\)
So,
The model for the equivalent fraction is:

From the above model, we can say that the equivalent fraction is: \(\frac{8}{10}\)
So,
When \(\frac{8}{10}\) is divided by 2, we will get the equivalent fraction.
Hence, from the above,
We can conclude that \(\frac{4}{5}\) = \(\frac{8}{10}\)

Question 6.
Big Ideas Math Answer Key Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.1 8
Answer: The equivalent fraction of \(\frac{1}{2}\) is: \(\frac{5}{10}\)

Explanation:
The given model is:
Big Ideas Math Answer Key Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.1 8
From the above model, we can say that the fraction is: \(\frac{1}{2}\)
So,
The model for the equivalent fraction is:

So, the model that represents the equivalent fraction is: \(\frac{5}{10}\)
So,
When \(\frac{5}{10}\) is divided by 5, we will get the equivalent fraction
Hence, from the above,
We can conclude that \(\frac{1}{2}\) = \(\frac{5}{10}\)

Use a number line to find an equivalent fraction.
Question 7.
Big Ideas Math Answer Key Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.1 9
Answer: The equivalent fraction of \(\frac{3}{4}\) is: \(\frac{6}{8}\)

Explanation:
The given number line is:
Big Ideas Math Answer Key Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.1 9
From the above-given number line,
Every 2 lines represent the value.i.e., \(\frac{1}{4}\)
So,
The value of each line represents \(\frac{1}{8}\)
So,
The model for the equivalent fraction is:

So,
From the equivalent number line,
We can say that \(\frac{3}{4}\) is equivalent to \(\frac{6}{8}\)
Hence, from the above,
We can conclude that \(\frac{3}{4}\) = \(\frac{6}{8}\)

Question 8.
Big Ideas Math Answer Key Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.1 10
Answer: The equivalent fraction of \(\frac{1}{3}\) is: \(\frac{3}{9}\)

Explanation:
The given number line is:
Big Ideas Math Answer Key Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.1 10
From the above number line,
The given fraction is: \(\frac{1}{3}\)
From the above number line,
Every 2 lines represent a value.
So,
if we divide the lines into 3 lines between the values, then we will get each line value with the denominator 9
So,
The number line for the equivalent fraction is:

So, from the above number line,
We can say that the equivalent fraction of \(\frac{1}{3}\) is: \(\frac{3}{9}\)
Hence, from the above,
We can conclude that \(\frac{1}{3}\) = \(\frac{3}{9}\)

Question 9.
Open-Ended
Write two equivalent fractions to describe the portion of the eggs that are white
Big Ideas Math Answer Key Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.1 11
Answer:
The two equivalent fractions to describe the portion of eggs that are white are:
\(\frac{6}{12}\) and \(\frac{1}{2}\)

Explanation:
The given model is:
Big Ideas Math Answer Key Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.1 11
From the above model,
We can say that,
The number of colored eggs is: 6
The number of white eggs is: 6
So,
The total number of eggs are: 12
So,
The fraction form for the white eggs is = \(\frac{Number of white eggs}{Total number of eggs}\)
= \(\frac{6}{12}\)
Equivalent form for the white eggs:
Consider the colored eggs as 1 group and the white eggs as 1 group
So,
The number of White eggs is: 1
The number of colored eggs is: 1
So,
The total number of eggs are: 2
So,
The fraction that the eggs are white = \(\frac{Number of white eggs}{Total number of eggs}\)
= \(\frac{1}{2}\)
Hence, from the above,
We can say that
The two equivalent fractions to describe the portion of eggs that are white are:
\(\frac{6}{12}\) and \(\frac{1}{2}\)

Question 10.
YOU BE THE TEACHER
Your friend says the models show equivalent fractions. Is your friend correct? Explain.
Big Ideas Math Answer Key Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.1 12
Answer: Yes, your friend is correct

Explanation:
Given models are:
Big Ideas Math Answer Key Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.1 12
Let the two models be named as 1st model and 2nd model.
So,
From the 1st model,
The total number of parts are: 4
The colored part is: 1
So,
The fraction form is: \(\frac{The colored part}{The total number of parts}\) = \(\frac{1}{4}\)
Now,
From the 2nd model,
The total number of parts are: 4
The colored part is: 1
So,
The fraction form is: \(\frac{The colored part}{The total number of parts}\) = \(\frac{1}{4}\)
Hence, from the above,
We can conclude that the 2 fractions are equal and so, your friend is correct.

Think and Grow: Modeling Real Life

Example
You and your friend make braided paper bookmarks. Yours is \(\frac{2}{3}\) foot long. Your friend’s is \(\frac{7}{12}\) foot long. Are the bookmarks the same length?
Big Ideas Math Answer Key Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.1 13
Determine whether the fractions are equivalent. Plot the fractions on the same number line. Then compare.
Big Ideas Math Answer Key Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.1 14
From the above number line,
we can observe that \(\frac{2}{3}\) is after the \(\frac{7}{12}\)
So,
\(\frac{2}{3}\) is greater than \(\frac{7}{12}\)
So,
The bookmarks do not have the same length.

Show and Grow

Question 11.
The lasagna pans are the same size. Are the amounts of lasagna left in the pans equal?
Big Ideas Math Answer Key Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.1 15
Answer: No, the amount of lasagna left in the pans are not equal

Explanation:
Given lasagna pans are:
Big Ideas Math Answer Key Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.1 15
Let the two lasagna pans be named as 1st model and 2nd model
From the first model,
The total number of parts are: 12
The number of parts that are occupied is: 4
The number of parts that are left is: 8
From the 2nd model,
The total number of parts are: 6
The number of parts that are occupied is: 2
The number of parts that are left is: 4
Hence, from the above,
We can conclude that the amount of lasagna left in the pans is not equal.

Question 12.
DIG DEEPER!
You run 3 laps around an outdoor track, where 4 laps are equal to 1 mile. Your friend runs 6 laps around the indoor track shown. Do you and your friend run the same distance? Explain.
Big Ideas Math Answer Key Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.1 16
Answer: Yes, you and your friend run the same distance.

Explanation:
Given that you run 3 laps around the outdoor track whereas for you, 4 laps is equal to 1 mile.
So,
The total distance covered by you = \(\frac{The number of laps that you run}{The number of laps that is equal to 1 mile}\)= \(\frac{3}{4}\)
It is also given that your friend runs 6 laps around the indoor track whereas, for your friend, 8 laps is equal to 1 mile.
So,
The total distance covered by your friend = \(\frac{The number of laps that your friend run}{The number of laps that is equal to 1 mile}\) = \(\frac{6}{8}\)
When \(\frac{3}{4}\) is multiplied and divide by 2 , we get \(\frac{6}{8}\)
So,
\(\frac{3}{4}\) = \(\frac{6}{8}\)
Hence, from the above,
We can conclude that you and your friend runs the same distance

Model Equivalent Fractions Homework & Practice 7.1

Use the model to find an equivalent fraction.
Question 1.
Big Ideas Math Answer Key Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.1 17
Answer: The equivalent fraction of \(\frac{3}{5}\) is: \(\frac{6}{10}\)

Explanation:
The given model is:
Big Ideas Math Answer Key Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.1 17
From the above model,
We can say that the fraction is: \(\frac{3}{5}\)
So,
The model for the Equivalent fraction is:

From the above model, we can say that the equivalent fraction is: \(\frac{6}{10}\)
So,
When we divide \(\frac{6}{10}\) by 2, we will get the equivalent value.
Hence, from the above,
We can conclude that \(\frac{3}{5}\) = \(\frac{6}{10}\)

Question 2.
Big Ideas Math Answer Key Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.1 18
Answer: The equivalent fraction of \(\frac{1}{4}\) is: \(\frac{2}{8}\)

Explanation:
The given model is:
Big Ideas Math Answer Key Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.1 18
From the above model, we can say that the fraction is: \(\frac{1}{4}\)
So,
The model for the equivalent fraction is:

From the above model,
So, we can say that the equivalent fraction is: \(\frac{2}{8}\)
So,
When \(\frac{1}{4}\) is multiplied and divided by 2, we can get the equivalent value.
Hence, from the above,
We can conclude that \(\frac{1}{4}\) = \(\frac{2}{8}\)

Use a number line to find an equivalent fraction.
Question 3.
Big Ideas Math Answer Key Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.1 19
Answer: The equivalent fraction of \(\frac{4}{6}\) is: \(\frac{8}{12}\)

Explanation:
The given Number line is:
Big Ideas Math Answer Key Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.1 19
From the given number line,
we can see that every 2 lines represent a value.i.e., \(\frac{4}{6}\)
So, when there will be 2 lines between 2 values, then each line will become \(\frac{1}{12}\)
So,
The model for the equivalent fraction is:

So, from the equivalent number line,
We can say that \(\frac{4}{6}\) is equal to \(\frac{8}{12}\)
Hence, from the above,
We can conclude that \(\frac{4}{6}\) = \(\frac{8}{12}\)

Question 4.
Big Ideas Math Answer Key Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.1 20
Answer: The equivalent fraction of \(\frac{2}{3}\) is: \(\frac{4}{6}\)

Explanation:
The given number line is:
Big Ideas Math Answer Key Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.1 20
From the above number line,
we can say that 2 lines represent a value. i.e., \(\frac{2}{3}\)
So,
The value of each line when divided into 2 parts represent: \(\frac{1}{6}\)
So,
The equivalent number line is:

So, from the above equivalent number line,
We can see that \(\frac{2}{3}\) is equal to \(\frac{4}{6}\)
Hence, from the above,
We can conclude that \(\frac{2}{3}\) = \(\frac{4}{6}\)

Find the equivalent fraction.
Question 5.
Big Ideas Math Answer Key Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.1 21
Answer: The equivalent fraction of \(\frac{1}{6}\) is: \(\frac{2}{12}\)

Explanation:
When we multiply and divide \(\frac{1}{6}\) with 2, we will get the equivalent value of \(\frac{1}{6}\)
So,

Hence, from the above,
We can conclude that \(\frac{1}{6}\) = \(\frac{2}{12}\)

Question 6.
Big Ideas Math Answer Key Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.1 22
Answer: The equivalent fraction of \(\frac{2}{5}\) is: \(\frac{4}{10}\)

Explanation:
When we multiply and divide \(\frac{2}{5}\) with 2, we will get the equivalent value of \(\frac{2}{5}\)
So,

Hence, from the above,
We can conclude that \(\frac{2}{5}\) = \(\frac{4}{10}\)

Question 7.
Big Ideas Math Answer Key Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.1 23
Answer: The equivalent fraction of \(\frac{1}{4}\) is: \(\frac{2}{8}\)

Explanation:
When we multiply and divide \(\frac{1}{4}\) with 2, we will get the equivalent value of \(\frac{1}{4}\)
So,

Hence, from the above,
We can conclude that \(\frac{1}{4}\) = \(\frac{2}{8}\)

Question 8.
Big Ideas Math Answer Key Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.1 24
Answer: The equivalent fraction of \(\frac{9}{12}\) is: \(\frac{3}{4}\)

Explanation:
When we divide the \(\frac{9}{12}\) by 3, we will get the equivalent value of \(\frac{9}{12}\)
So,

Hence, from the above,
We can conclude that \(\frac{9}{12}\) = \(\frac{3}{4}\)

Question 9.
Which One Doesn’tBelong?
Which model does not belong with the other three? Explain
Big Ideas Math Answer Key Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.1 25
Answer:
Let the given figures be named as A), B), C) and D)
So,
from the fraction values, figure C) does not belong with the other three because the values of the denominators of A), B), D) are all the multiples of 3

Explanation:
There are 4 given figures.
Let it be named as A), B), C) and D)
Big Ideas Math Answer Key Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.1 25
From A),
The total number of parts are: 3
The colored part is: 1
So,
The fraction form is: \(\frac{The colored part}{The total number of parts}\) = \(\frac{1}{3}\)
From B),
The total number of parts are: 6
The colored part is: 3
So,
The fraction form is: \(\frac{The colored part}{The total number of parts}\) = \(\frac{3}{6}\)
From C),
The total number of parts are: 5
The colored part is: 2
So,
The fraction form is: \(\frac{The colored part}{The total number of parts}\) = \(\frac{2}{5}\)
From D),
The total number of parts are: 12
The colored part is: 4
So,
The fraction form is: \(\frac{The colored part}{The total number of parts}\) = \(\frac{4}{12}\)
Hence, from all the figures,
We can conclude that figure C) does not belong with the other three because the values of the denominators of A), B), D) are all the multiples of 3

Question 10.
Modeling Real Life
Your crayon is \(\frac{1}{6}\) foot long. Your friend’s crayon is \(\frac{3}{12}\) foot long. Are the crayons the same length?
Big Ideas Math Answer Key Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.1 26
Answer: No, the crayons are not of the same length because your crayon is shorter than the crayon of your friend.

Explanation:
Given that,
The length of your crayon is: \(\frac{1}{6}\) foot
The length of your friend’s crayon is: \(\frac{3}{12}\) foot
So,
When we compare \(\frac{1}{6}\) foot and \(\frac{3}{12}\) foot,
Now,
Multiply and divide \(\frac{1}{6}\) with 2, we will get \(\frac{2}{12}\) foot
So,
we can sayt that \(\frac{1}{6}\) foot is less than \(\frac{3}{12}\) foot
Hence, from the above,
We can conclude that the length of the crayons are not equal.

Review & Refresh

Find the product
Question 11.
5 × 437 = _______
Answer: 5 × 437 = 2,185

Explanation:
By using the partial products method,
5 × 437 = 5 × ( 400 + 30 + 7 )
= ( 5 × 400 ) + ( 5 × 30 ) + ( 5 × 7 )
= 2,000 + 150 + 35
= 2,185
Hence, 5 × 437 = 2,185

Question 12.
6,982 × 9 = _______
Answer: 6,982 × 9 = 62,838

Explanation:
By using the partial products method,
6,982 × 9 = ( 6,000 + 900 + 80 + 2 ) × 9
= ( 6,000 × 9 ) + ( 900 × 9 ) + ( 80 × 9 ) + ( 2 × 9 )
= 54,000 + 8,100 + 720 + 18
= 62,838
Hence, 6,982 × 9 = 62,838

Question 13.
8 × 708 = _______
Answer: 8 × 708 = 5,664

Explanation:
By using the partial products method,
8 × 708 = 8 × ( 700 + 8 )
= ( 8 × 700 ) + ( 8 × 8 )
= 5,600 + 64
= 5,664
Hence, 8 × 708 = 5,664

Lesson 7.2 Generate Equivalent Fractions by Multiplying

Explore and Grow

Shade the second model in each pair to show an equivalent fraction.Then write the fraction.

Describe the relationship between each pair of numerators and each pair of denominators.
Answer: From the three models, we can say that the equivalent fractions of the three models obtained by multiplying and dividing \(\frac{1}{2}\) with 2, 3 and 4
Hence,
In the first model,
\(\frac{1}{2}\) = \(\frac{2}{4}\)
In the second model,
\(\frac{1}{2}\) = \(\frac{3}{6}\)
In the third model,
\(\frac{1}{2}\) = \(\frac{4}{8}\)

Structure
How can you use multiplication to write equivalent fractions? Explain. Then use your method to find another fraction that is equivalent to \(\frac{1}{2}\).
Answer: The equivalent fraction of \(\frac{1}{2}\) is: \(\frac{3}{6}\)
So,
\(\frac{1}{2}\) = \(\frac{3}{6}\)

Explanation:
We can multiply the given fraction by another fraction that have the same numerator and denominator to write equivalent fractions.
For example,
To write the equivalent fractions of \(\frac{1}{2}\), we can multiply so many fractions with the same numerator and denominator like \(\frac{2}{2}\), \(\frac{3}{3}\) etc.
Hence,
The equivalent fraction of \(\frac{1}{2}\) is: \(\frac{3}{6}\) or \(\frac{2}{4}\) or \(\frac{4}{8}\) or \(\frac{5}{10}\)

Think and Grow: Multiply to find Equivalent Fractions

You can find an equivalent fraction by multiplying the numerator and the denominator by the same number.
Big Ideas Math Answers 4th Grade Chapter 7 Understand Fraction Equivalence and Comparison 7.2 2
Example
Find an equivalent fraction for \(\frac{3}{5}\).

Show and Grow

Find an equivalent fraction.
Question 1.
Big Ideas Math Answers 4th Grade Chapter 7 Understand Fraction Equivalence and Comparison 7.2 4
Answer: The equivalent fraction of \(\frac{5}{6}\) is: \(\frac{10}{12}\)

Explanation:
When we multiply the fraction with the fraction having the same number as the numerator as the denominator, we will get the equivalent fraction.
So,
\(\frac{5}{6}\) is multiplied with \(\frac{2}{2}\)
Hence,

Question 2.
Big Ideas Math Answers 4th Grade Chapter 7 Understand Fraction Equivalence and Comparison 7.2 5
Answer: The equivalent fraction of \(\frac{8}{5}\) is: \(\frac{16}{10}\)

Explanation:
When we multiply the fraction with the fraction having the same number as the numerator as the denominator, we will get the equivalent fraction.
So,
\(\frac{8}{5}\) is multiplied with \(\frac{2}{2}\)
Hence,

Find the equivalent fraction.
Question 3.
Big Ideas Math Answers 4th Grade Chapter 7 Understand Fraction Equivalence and Comparison 7.2 6
Answer: The equivalent fraction of \(\frac{1}{2}\) is: \(\frac{4}{8}\)

Explanation:
The given model is:
Big Ideas Math Answers 4th Grade Chapter 7 Understand Fraction Equivalence and Comparison 7.2 6
The two models are the original model and its equivalent model.
So,
According to the models,
\(\frac{1}{2}\) = \(\frac{4}{8}\)
When we multiply the fraction with the fraction having the same number as the numerator as the denominator, we will get the equivalent fraction.
So according to the Equivalet model,
\(\frac{1}{2}\) is multiplied with \(\frac{4}{4}\)
Hence,

Question 4.
Big Ideas Math Answers 4th Grade Chapter 7 Understand Fraction Equivalence and Comparison 7.2 7
Answer: The equivalent fraction is: \(\frac{2}{3}\) is: \(\frac{4}{6}\)

Explanation:
The given model is:
Big Ideas Math Answers 4th Grade Chapter 7 Understand Fraction Equivalence and Comparison 7.2 7
The two models are the original model and the equivalent model.
So,
According to the models,
\(\frac{2}{3}\) = \(\frac{4}{6}\)
When we multiply the fraction with the fraction having the same number as the numerator as the denominator, we will get the equivalent fraction.
So according to the Equivalent model,
\(\frac{1}{2}\) is multiplied with \(\frac{2}{2}\)
Hence,

Apply and Grow: Practice

Find the equivalent fraction.
Question 5.
Big Ideas Math Answers 4th Grade Chapter 7 Understand Fraction Equivalence and Comparison 7.2 8
Answer: The equivalent model of \(\frac{3}{4}\) is: \(\frac{6}{8}\)

Explanation:
The given model is:
Big Ideas Math Answers 4th Grade Chapter 7 Understand Fraction Equivalence and Comparison 7.2 8
The two models are the original model and the equivalent model.
So,
According to the models,
\(\frac{3}{4}\) = \(\frac{6}{8}\)
When we multiply the fraction with the fraction having the same number as the numerator as the denominator, we will get the equivalent fraction.
So according to the Equivalent model,
\(\frac{3}{4}\) is multiplied with \(\frac{2}{2}\)
Hence,

Question 6.
Big Ideas Math Answers 4th Grade Chapter 7 Understand Fraction Equivalence and Comparison 7.2 9
Answer: The equivalent fraction of \(\frac{1}{3}\) is: \(\frac{4}{12}\)

Explanation:
The given model is:
Big Ideas Math Answers 4th Grade Chapter 7 Understand Fraction Equivalence and Comparison 7.2 9
The two models are the original model and the equivalent model.
So,
According to the models,
\(\frac{1}{3}\) = \(\frac{4}{12}\)
When we multiply the fraction with the fraction having the same number as the numerator as the denominator, we will get the equivalent fraction.
So according to the Equivalent model,
\(\frac{1}{3}\) is multiplied with \(\frac{4}{4}\)
Hence,

Question 7.
Big Ideas Math Answers 4th Grade Chapter 7 Understand Fraction Equivalence and Comparison 7.2 10
Answer: The equivalent fraction of \(\frac{9}{6}\) is: \(\frac{18}{12}\)

Explanation:
When we multiply the fraction with the fraction having the same number as the numerator as the denominator, we will get the equivalent fraction.
So,
\(\frac{9}{6}\) is multiplied with \(\frac{2}{2}\)
Hence,

Question 8.
Big Ideas Math Answers 4th Grade Chapter 7 Understand Fraction Equivalence and Comparison 7.2 11
Answer: The equivalent fraction of \(\frac{7}{5}\) is: \(\frac{140}{100}\)

Explanation:
When we multiply the fraction with the fraction having the same number as the numerator as the denominator, we will get the equivalent fraction.
So,
\(\frac{7}{5}\) is multiplied with \(\frac{20}{20}\)
Hence,

Find an equivalent fraction.
Question 9.
\(\frac{7}{6}\)
Answer: The equivalent fraction of \(\frac{7}{6}\) is: \(\frac{14}{12}\)

Explanation:
When we multiply the fraction with the fraction having the same number as the numerator as the denominator, we will get the equivalent fraction.
So,
\(\frac{7}{6}\) is multiplied with \(\frac{2}{2}\)
Hence,
\(\frac{7}{6}\) = \(\frac{14}{12}\)

Question 10.
\(\frac{10}{10}\)
Answer: The equivalent fraction of \(\frac{10}{10}\) is: \(\frac{20}{20}\)

Explanation:
When we multiply the fraction with the fraction having the same number as the numerator as the denominator, we will get the equivalent fraction.
So,
\(\frac{10}{10}\) is multiplied with \(\frac{2}{2}\)
Hence,
\(\frac{10}{10}\) = \(\frac{20}{20}\)

Question 11.
\(\frac{2}{4}\)
Answer: The equivalent fraction of \(\frac{2}{4}\) is: \(\frac{4}{8}\)

Explanation:
When we multiply the fraction with the fraction having the same number as the numerator as the denominator, we will get the equivalent fraction.
So,
\(\frac{2}{4}\) is multiplied with \(\frac{2}{2}\)
Hence,
\(\frac{2}{4}\) = \(\frac{4}{8}\)

Find two equivalent fractions.
Question 12.
\(\frac{5}{5}\)
Answer:
The two equivalent fractions of \(\frac{5}{5}\) are: \(\frac{10}{10}\) and \(\frac{15}{15}\)

Explanation:
When we multiply the fraction with the fraction having the same number as the numerator as the denominator, we will get the equivalent fraction.
So,
\(\frac{5}{5}\) is multiplied with \(\frac{2}{2}\) and \(\frac{3}{3}\)
Hence,
\(\frac{5}{5}\) = \(\frac{10}{10}\) and \(\frac{15}{15}\)

Question 13.
\(\frac{4}{3}\)
Answer:
The two equivalent fractions of \(\frac{4}{3}\) are: \(\frac{8}{6}\) and \(\frac{12}{9}\)

Explanation:
When we multiply the fraction with the fraction having the same number as the numerator as the denominator, we will get the equivalent fraction.
So,
\(\frac{4}{3}\) is multiplied with \(\frac{2}{2}\) and \(\frac{3}{3}\)
Hence,
\(\frac{4}{3}\) = \(\frac{8}{6}\) and \(\frac{12}{9}\)

Question 14.
\(\frac{1}{10}\)
Answer:
The two equivalent fractions of \(\frac{1}{10}\) are: \(\frac{2}{20}\) and \(\frac{3}{30}\)

Explanation:
When we multiply the fraction with the fraction having the same number as the numerator as the denominator, we will get the equivalent fraction.
So,
\(\frac{1}{10}\) is multiplied with \(\frac{2}{2}\) and \(\frac{3}{3}\)
Hence,
\(\frac{1}{10}\) = \(\frac{2}{20}\) and \(\frac{3}{30}\)

Question 15.
Writing
Explain how \(\frac{1}{4}\) and \(\frac{2}{8}\) are equivalent using multiplication. Use models to support your answer.
Answer: The equivalent fraction of \(\frac{1}{4}\) is: \(\frac{2}{8}\)

Explanation:
For the \(\frac{1}{4}\) and \(\frac{2}{8}\), the models are:

The two models are the original model and the equivalent model.
So,
According to the original model,
The total number of parts are: 4
The colored part is: 1
According to the Equivalent model,
The total number of parts are: 8
The colored part is: 2
Now,
When we multiply the fraction with the fraction having the same number as the numerator as the denominator, we will get the equivalent fraction.
So,
When we multiply \(\frac{1}{4}\) with \(\frac{2}{2}\), we will get the \(\frac{2}{8}\) which is the equivalent value of \(\frac{1}{4}\)
Hence,
The equivalent value of \(\frac{1}{4}\) is: \(\frac{2}{8}\)

DIG DEEPER!
Write true or false for the statement. If false, explain why.
Question 16.
Big Ideas Math Answers 4th Grade Chapter 7 Understand Fraction Equivalence and Comparison 7.2 12
Answer: \(\frac{4}{2}\) = \(\frac{24}{12}\)

Explanation:
Given fractions are: \(\frac{4}{2}\) and \(\frac{24}{12}\)
When we multiply the fraction with the fraction having the same number as the numerator as the denominator, we will get the equivalent fraction.
So,
When \(\frac{4}{2}\) is multiplied with \(\frac{6}{6}\), we will get  \(\frac{24}{12}\) which is the equivalent value of \(\frac{4}{2}\)
Hence,
\(\frac{4}{2}\) = \(\frac{24}{12}\)

Question 17.
Big Ideas Math Answers 4th Grade Chapter 7 Understand Fraction Equivalence and Comparison 7.2 13
Answer: \(\frac{3}{5}\) is not equal to \(\frac{6}{100}\)

Explanation:
Given fractions are: \(\frac{3}{5}\) and \(\frac{6}{100}\)
When we multiply the fraction with the fraction having the same number as the numerator as the denominator, we will get the equivalent fraction.
So,
When \(\frac{3}{5}\) is multiplied with \(\frac{2}{2}\), we will get  \(\frac{6}{10}\) which is not the equivalent value of \(\frac{3}{5}\)
Hence,
\(\frac{3}{5}\) is not equal to \(\frac{6}{100}\)

Think and Grow: Modeling Real Life

Example
A recipe calls for \(\frac{3}{4}\) cup of oats. You only have a \(\frac{1}{8}\) cup measuring cup. What fraction of a cup of oats, in eighths,do you need? Use multiplication to write an equivalent fraction for \(\frac{3}{4}\) in eighths.
Big Ideas Math Answers 4th Grade Chapter 7 Understand Fraction Equivalence and Comparison 7.2 14

Show and Grow

Question 18.
You need \(\frac{1}{2}\) cup of water for a science experiment. You only have a \(\frac{1}{4}\) cup measuring cup. What fraction of a cup of water, in fourths, do you need?
Answer: You need \(\frac{2}{4}\), in fourths, of a cup of water

Explanation:
Given that you need \(\frac{1}{2}\) cup of water for a science experiment and you only have a \(\frac{1}{4}\) cup measuring cup.
So, in terms of \(\frac{1}{4}\), \(\frac{1}{2}\) can be written as:
\(\frac{1}{2}\) = \(\frac{1 × 2}{2 × 2}\)
= \(\frac{2}{4}\)
Hence, from the above,
We can conclude that we need \(\frac{2}{4}\) fraction of a cup of water, in fourths

Question 19.
A pedestrian needs to walk \(\frac{4}{5}\) mile to meet her goal. The path is marked at every tenth of a mile. What fraction of a mile, in tenths, should she walk?
Answer: She should walk \(\frac{8}{10}\), in tenths of a mile

Explanation:
Given that a pedestrian needs to walk \(\frac{4}{5}\) mile to meet her goal.
It is also given that the path is marked at every tenth of a mile.
So, in terms of \(\frac{1}{10}\), \(\frac{4}{5}\) can be written as:
\(\frac{4}{5}\) = \(\frac{4 × 2}{5 × 2}\)
= \(\frac{8}{10}\)
Hence, from the above,
We can conclude that she should walk \(\frac{8}{10}\), in tenths of a mile

Question 20.
DIG DEEPER!
You put together \(\frac{7}{10}\) of a puzzle. The puzzle has pieces. What fraction of the puzzle, in hundredths, is not put together? Explain.
Big Ideas Math Answers 4th Grade Chapter 7 Understand Fraction Equivalence and Comparison 7.2 16
Answer: The fraction of the puzzle not put together, in hundredths is: \(\frac{30}{100}\)

Explanation:
Given that you put together \(\frac{7}{10}\) of a puzzle.
So, from this, the part of the puzzle that is not put together is: \(\frac{3}{10}\)
We have to find the fraction of the puzzle that is not put together, in hundredths
So, in terms of \(\frac{1}{100}\), \(\frac{3}{10}\) can be written as:
\(\frac{3}{10}\) = \(\frac{3 × 10}{10 × 10}\)
= \(\frac{30}{100}\)
Hence, from the above,
We can conclude that the fraction of the puzzle not put together, in hundredths is: \(\frac{30}{100}\)

Question 21.
You have \(\frac{3}{5}\) of a dollar in coins. What fraction of a dollar, in hundredths, do you have? Write one possible combination of coins that you have.
Big Ideas Math Answers 4th Grade Chapter 7 Understand Fraction Equivalence and Comparison 7.2 17
Answer: The fraction of a dollar in hundredths is: \(\frac{60}{100}\)

Explanation:
Given that you have \(\frac{3}{5}\) of a dollar in coins.
So, in terms of \(\frac{1}{100}\), \(\frac{3}{5}\) can be written as:
\(\frac{3}{5}\) = \(\frac{3 × 20}{5 × 20}\)
= \(\frac{60}{100}\)
Hence, from the above,
We can conclude that the fraction of a dollar in hundredths is: \(\frac{60}{100}\)

Generate Equivalent Fractions by Multiplying Homework & Practice 7.2

Find an equivalent fraction.
Question 1.
Big Ideas Math Answers 4th Grade Chapter 7 Understand Fraction Equivalence and Comparison 7.2 18
Answer: The equivalent fraction of \(\frac{1}{5}\) is: \(\frac{2}{10}\)

Explanation:
When we multiply the fraction with the fraction having the same number as the numerator as the denominator, we will get the equivalent fraction.
So,
\(\frac{1}{5}\) is multiplied with \(\frac{2}{2}\)
Hence,

Question 2.
Big Ideas Math Answers 4th Grade Chapter 7 Understand Fraction Equivalence and Comparison 7.2 19
Answer: The equivalent fraction of \(\frac{11}{6}\) is: \(\frac{22}{12}\)

Explanation:
When we multiply the fraction with the fraction having the same number as the numerator as the denominator, we will get the equivalent fraction.
So,
\(\frac{11}{6}\) is multiplied with \(\frac{2}{2}\)
Hence,

Find an equivalent fraction.
Question 3.
Big Ideas Math Answers 4th Grade Chapter 7 Understand Fraction Equivalence and Comparison 7.2 20
Answer: The equivalent fraction of \(\frac{4}{6}\) is: \(\frac{8}{12}\)

Explanation:
The given model is:
Big Ideas Math Answers 4th Grade Chapter 7 Understand Fraction Equivalence and Comparison 7.2 20
The two models are the original model and its equivalent model.
So,
According to the models,
\(\frac{4}{6}\) = \(\frac{8}{12}\)
When we multiply the fraction with the fraction having the same number as the numerator as the denominator, we will get the equivalent fraction.
So according to the Equivalent model,
\(\frac{4}{6}\) is multiplied with \(\frac{2}{2}\)
Hence,

Question 4.
Big Ideas Math Answers 4th Grade Chapter 7 Understand Fraction Equivalence and Comparison 7.2 21
Answer: The equivalent fraction of \(\frac{2}{5}\) is: \(\frac{4}{10}\)

Explanation:
The given model is:
Big Ideas Math Answers 4th Grade Chapter 7 Understand Fraction Equivalence and Comparison 7.2 21
The two models are the original model and its equivalent model.
So,
According to the models,
\(\frac{2}{5}\) = \(\frac{4}{10}\)
When we multiply the fraction with the fraction having the same number as the numerator as the denominator, we will get the equivalent fraction.
So according to the Equivalent model,
\(\frac{2}{5}\) is multiplied with \(\frac{2}{2}\)
Hence,

Question 5.
Big Ideas Math Answers 4th Grade Chapter 7 Understand Fraction Equivalence and Comparison 7.2 22
Answer: The equivalent fraction of \(\frac{3}{3}\) is: \(\frac{6}{6}\)

Explanation:
When we multiply the fraction with the fraction having the same number as the numerator as the denominator, we will get the equivalent fraction.
So,
\(\frac{3}{3}\) is multiplied with \(\frac{2}{2}\)
Hence,

Question 6.
Big Ideas Math Answers 4th Grade Chapter 7 Understand Fraction Equivalence and Comparison 7.2 23
Answer: The equivalent fraction of \(\frac{7}{10}\) is: \(\frac{70}{100}\)

Explanation:
Explanation:
When we multiply the fraction with the fraction having the same number as the numerator as the denominator, we will get the equivalent fraction.
So,
\(\frac{7}{10}\) is multiplied with \(\frac{10}{10}\)
Hence,

Find an equivalent fraction.
Question 7.
\(\frac{5}{3}\)
Answer: The equivalent value of \(\frac{5}{3}\) is: \(\frac{10}{6}\)

Explanation:
When we multiply the fraction with the fraction having the same number as the numerator as the denominator, we will get the equivalent fraction.
So,
\(\frac{5}{3}\) is multiplied with \(\frac{2}{2}\)
Hence,
\(\frac{5}{3}\) = \(\frac{10}{6}\)

Question 8.
\(\frac{4}{4}\)
Answer: The equivalent value of \(\frac{4}{4}\) is: \(\frac{8}{8}\)

Explanation:
When we multiply the fraction with the fraction having the same number as the numerator as the denominator, we will get the equivalent fraction.
So,
\(\frac{4}{4}\) is multiplied with \(\frac{2}{2}\)
Hence,
\(\frac{4}{4}\) = \(\frac{8}{8}\)

Question 9.
\(\frac{5}{10}\)
Answer: The equivalent value of \(\frac{5}{10}\) is: \(\frac{10}{20}\)

Explanation:
When we multiply the fraction with the fraction having the same number as the numerator as the denominator, we will get the equivalent fraction.
So,
\(\frac{5}{10}\) is multiplied with \(\frac{2}{2}\)
Hence,
\(\frac{5}{10}\) = \(\frac{10}{20}\)

Find two equivalent fractions.
Question 10.
\(\frac{3}{2}\)
Answer:
The two equivalent fractions of \(\frac{3}{2}\) are: \(\frac{6}{4}\) and \(\frac{9}{6}\)

Explanation:
When we multiply the fraction with the fraction having the same number as the numerator as the denominator, we will get the equivalent fraction.
So,
\(\frac{3}{2}\) is multiplied with \(\frac{2}{2}\) and \(\frac{3}{3}\)
Hence,
\(\frac{3}{2}\) = \(\frac{6}{4}\) and \(\frac{9}{6}\)

Question 11.
\(\frac{4}{10}\)
Answer:
The two equivalent fractions of \(\frac{4}{10}\) are: \(\frac{8}{20}\) and \(\frac{12}{30}\)

Explanation:
When we multiply the fraction with the fraction having the same number as the numerator as the denominator, we will get the equivalent fraction.
So,
\(\frac{4}{10}\) is multiplied with \(\frac{2}{2}\) and \(\frac{3}{3}\)
Hence,
\(\frac{4}{10}\) = \(\frac{8}{20}\) and \(\frac{12}{30}\)

Question 12.
\(\frac{10}{5}\)
Answer:
The two equivalent fractions of \(\frac{10}{5}\) are: \(\frac{20}{10}\) and \(\frac{30}{15}\)

Explanation:
When we multiply the fraction with the fraction having the same number as the numerator as the denominator, we will get the equivalent fraction.
So,
\(\frac{10}{5}\) is multiplied with \(\frac{2}{2}\) and \(\frac{3}{3}\)
Hence,
\(\frac{10}{5}\) = \(\frac{20}{10}\) and \(\frac{30}{15}\)

Question 13.
DIG DEEPER!
What is Descartes’s fraction?
Big Ideas Math Answers 4th Grade Chapter 7 Understand Fraction Equivalence and Comparison 7.2 24
Answer:

Question 14.
YOU BE THE TEACHER
Your friend says she can write a fraction equivalent to \(\frac{3}{4}\) that has a denominator of 10 and 4 and a whole number in the numerator. Is your friend correct? Explain.
Answer: Your friend is not correct

Explanation:
Given that your friend says she can write a fraction equivalent to \(\frac{3}{4}\) that has a denominator of 10 and 4 and a whole number in the numerator.
When we multiply the fraction with the fraction having the same number as the numerator as the denominator, we will get the equivalent fraction.
So,
\(\frac{3}{4}\) will be multiplied with \(\frac{10}{25}\) to get the value of 10 in the denominator
\(\frac{3}{4}\) will be multiplied \(\frac{1}{1}\) to get the value of 4 in the denomintor.
So,
\(\frac{3}{4}\) = \(\frac{30}{100}\) But it is said this fraction has to ahve the denominator of 10 but we get the denominator of 100
Hence, from the above,
we can conclude that your friend is not correct

Question 15.
Modeling Real Life
A recipe calls for 1 teaspoon of cinnamon. You only have a \(\frac{1}{2}\) teaspoon measuring spoon. What fraction of a teaspoon of cinnamon, in halves, do you need?
Big Ideas Math Answers 4th Grade Chapter 7 Understand Fraction Equivalence and Comparison 7.2 25
Answer: You need \(\frac{1}{2}\) of a fraction of teaspoon, in halves

Explanation:
Given that a recipe calls for 1 teaspoon of cinnamon and you only have a \(\frac{1}{2}\) teaspoon measuring spoon.
So,
In terms of \(\frac{1}{2}\), we can write \(\frac{1}{2}\) as:
\(\frac{1}{2}\) = \(\frac{1× 1}{2 × 1}\)
Hence, from the above,
We can conclude that you need \(\frac{1}{2}\) of a fraction of teaspoon, in halves

Question 16.
A couple lives in Florida for \(\frac{1}{3}\) of the year. Each year has 12 months. What fraction of a year,  in twelfths, does the couple live in Florida?
Answer: The couple lives in Florida for \(\frac{4}{12}\) of a year, in twelfths.

Explanation:
Given that a couple lives in Florida for \(\frac{1}{3}\) of the year.
It is also given that each year has 12 months
So,
In terms of \(\frac{1}{12}\), we can write \(\frac{1}{3}\) as:
\(\frac{1}{3}\) = \(\frac{1× 4}{3 × 4}\)
Hence, from the above,
We can conclude that the couple lives in Florida for \(\frac{4}{12}\) of a year, in twelfths.

Review & Refresh

Divide. Then check your answer.
Question 17.
7√891
Answer: 891 ÷ 7 =127 R 2

Explanation:
By using the partial quotients method,
891 ÷ 7 = ( 840 + 49 ) ÷ 7
= ( 840 ÷ 7 ) + ( 49 ÷ 7 )
= 120 + 7
= 127 R 2
Hence, 891 ÷ 7 = 127 R 2

Question 18.
3√2,395
Answer: 2,395 ÷ 3 =798 R 1

Explanation:
By using the partial quotients method,
2,395 ÷ 3 = ( 2,100 + 180 + 39 + 72 + 3 ) ÷ 3
= ( 2,100 ÷ 3 ) + ( 180 ÷ 3 ) + ( 39 ÷ 3 ) + ( 72 ÷ 3 ) + ( 3 ÷ 3 )
=700 + 60 + 13 + 24 + 1
= 798 R 1
Hence, 2,395 ÷ 3 = 798 R 1

Question 19.
6√627
Answer: 627 ÷ 6 = 104 R 3

Explanation:
By using the partial quotients method,
627 ÷ 6 = ( 600 + 24 ) ÷ 6
= ( 600 ÷ 6 ) + ( 24 ÷ 6 )
= 100 + 4
= 104 R 3
Hence, 627 ÷ 6 = 104 R 3

Lesson 7.3 Generate Equivalent Fractions by Dividing

Explore and Grow

Shade the second model in each pair to show an equivalent fraction.Then write the fraction.

Describe the relationship between each pair of numerators and each pair of denominators.
Answer: The division of the fractions depends on the divisibility rules.
So,
When the numerator and denominator are even numbers, then they can only be divided by even numbers.
When the numerator and denominator are the multiples of 3, then they can only be divided by 3.
So, like these, we will divide the numerator and the denominator

Structure
How can you use division to write equivalent fractions? Explain and then use your method to find a fraction that is equivalent to \(\frac{6}{10}\).
Answer: The equivalent fraction of \(\frac{6}{10}\) is: \(\frac{3}{5}\)

Explanation:
The division of the fractions depends on the divisibility rules of the numbers.
Now,
Divisibility rule of 2: If the one’s digit ends with 0, 2, 4, 6, 8, then that number is divisible by 2
So,
\(\frac{6}{10}\) has the numbers 6 and 10 which has the one’s digit of 6 and 0.
So,
We can divide the \(\frac{6}{10}\) by 2
So,
We will get ,
\(\frac{6}{10}=\frac{6 \div 2}{10 \div 2}=\frac{3}{5}\)
Hence, from the above,
We can conclude that from the above method, the equivalent of \(\frac{6}{10}\) is: \(\frac{3}{5}\)

Think and Grow: Divide to Find Equivalent Fractions

A factor that is shared bytwo or more given numbers is a common factor. You can find an equivalent fraction by dividing the numerator and the denominator by a common factor.
\(\frac{2}{4}=\frac{2 \div 2}{4 \div 2}=\frac{1}{2}\)
Example
Find an equivalent fraction for \(\frac{8}{12}\) .
Find the common factors of 8 and 12.

Show and Grow

Find an equivalent fraction.
Question 1.
Big Ideas Math Answers Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.3 3
Answer: The equivalent fraction of \(\frac{3}{6}\) is: \(\frac{1}{2}\)

Explanation:
The given fraction is \(\frac{3}{6}\)
From the above fraction, the numerator and denominator are: 3 and 6
3 and 6 are the multiples of 3.
So,
We have to divide the \(\frac{3}{6}\) with 3
So,
\(\frac{3}{6}=\frac{3 \div 3}{6 \div 3}=\frac{1}{2}\)
Hence,

Question 2.
Big Ideas Math Answers Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.3 4
Answer: The equivalent fraction of \(\frac{20}{8}\) is: \(\frac{10}{4}\)

Explanation:
The given fraction is \(\frac{20}{8}\)
From the above fraction, the numerator and denominator are: 20 and 8
20 and 8 are the multiples of 2.( SInce the one’s digit are 0 and 8)
So,
We have to divide the \(\frac{20}{8}\) with 2
So,
\(\frac{20}{8}=\frac{20 \div 2}{8 \div 2}=\frac{10}{4}\)
Hence,

Find the equivalent fraction.
Question 3.
Big Ideas Math Answers Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.3 5
Answer: The equivalent fraction of \(\frac{4}{10}\) is: \(\frac{2}{5}\)

Explanation:
The given fraction is \(\frac{4}{10}\)
From the above fraction, the numerator and denominator are: 4 and 10
4 and 10 are the multiples of 2.( SInce the one’s digit are 0 and 4)
So,
We have to divide the \(\frac{4}{10}\) with 2
So,
\(\frac{4}{10}=\frac{4 \div 2}{10 \div 2}=\frac{2}{5}\)
Hence,

Question 4.
Big Ideas Math Answers Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.3 6
Answer: The equivalent fraction of \(\frac{90}{100} is: [latex]\frac{9}{10}

Explanation:
The given fraction is [latex]\frac{90}{100}\)
From the above fraction, the numerator and denominator are: 90 and 100
90 and 100 are the multiples of 10.
So,
We have to divide the \(\frac{90}{100}\) with 10
So,
\(\frac{90}{100}=\frac{90 \div 10}{100 \div 10}=\frac{9}{10}\)
Hence,

Question 5.
Big Ideas Math Answers Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.3 7
Answer: The equivalent fraction of \(\frac{14}{4}\) is: \(\frac{7}{2}\)

Explanation:
The given fraction is \(\frac{14}{4}\)
From the above fraction, the numerator and denominator are: 14 and 4
14 and 4 are the multiples of 2. ( Since, the one’s digit is 4 )
So,
We have to divide the \(\frac{14}{4}\) with 2
So,
\(\frac{14}{4}=\frac{14 \div 2}{4 \div 2}=\frac{7}{2}\)
Hence,

Apply any Grow: Practice

Find the equivalent fraction.
Question 6.
Big Ideas Math Answers Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.3 8
Answer: The equivalent fraction of \(\frac{2}{6}\) is: \(\frac{1}{3}\)

Explanation:
The given fraction is \(\frac{2}{6}\)
From the above fraction, the numerator and denominator are: 2 and 6
2 and 6 are the multiples of 2. ( Since, the one’s digit is 2 and 6 )
So,
We have to divide the \(\frac{2}{6}\) with 2
So,
\(\frac{2}{6}=\frac{2 \div 2}{6 \div 2}=\frac{1}{3}\)
Hence,

Question 7.
Big Ideas Math Answers Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.3 9
Answer: The equivalent fraction of \(\frac{16}{12}\) is: \(\frac{4}{3}\)

Explanation:
The given fraction is \(\frac{16}{12}\)
From the above fraction, the numerator and denominator are: 16 and 12
16 and 12 are the multiples of 4.
So,
We have to divide the \(\frac{16}{12}\) with 4
So,
\(\frac{16}{12}=\frac{16 \div 4}{12 \div 4}=\frac{4}{3}\)
Hence,

Question 8.
Big Ideas Math Answers Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.3 10
Answer: The equivalent fraction of \(\frac{80}{100}\) is: \(\frac{8}{10}\)

Explanation:
The given fraction is \(\frac{80}{100}\)
From the above fraction, the numerator and denominator are: 80 and 100
80 and 100 are the multiples of 10.
So,
We have to divide the \(\frac{80}{100}\) with 10
So,
\(\frac{80}{100}=\frac{80 \div 10}{100 \div 10}=\frac{8}{10}\)
Hence,

Question 9.
Big Ideas Math Answers Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.3 11
Answer: The equivalent fraction of \(\frac{8}{8}\) is: \(\frac{1}{1}\)

Explanation:
The given fraction is \(\frac{8}{8}\)
From the above fraction, the numerator and denominator are: 8 and 8
8 and 8 are the multiples of 8.
So,
We have to divide the \(\frac{8}{8}\) with 8
So,
\(\frac{8}{8}=\frac{8 \div 8}{8 \div 8}=\frac{1}{1}\)
Hence,

Question 10.
Big Ideas Math Answers Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.3 12
Answer: The equivalent fraction of \(\frac{2}{4}\) is: \(\frac{1}{2}\)

Explanation:
The given fraction is \(\frac{2}{4}\)
From the above fraction, the numerator and denominator are: 2 and 4
2 and 4 are the multiples of 2.
So,
We have to divide the \(\frac{2}{4}\) with 2
So,
\(\frac{2}{4}=\frac{2 \div 2}{4 \div 2}=\frac{1}{2}\)
Hence,

Question 11.
Big Ideas Math Answers Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.3 13
Answer: The equivalent fraction of \(\frac{30}{6}\) is: \(\frac{10}{2}\)

Explanation:
The given fraction is \(\frac{30}{6}\)
From the above fraction, the numerator and denominator are: 30 and 6
30 and 6 are the multiples of 3.  ( Since the sum of the digits of the numerator and the denominator are the multiples of 3 )
So,
We have to divide the \(\frac{30}{6}\) with 3
So,
\(\frac{30}{6}=\frac{30 \div 3}{6 \div 3}=\frac{10}{2}\)
Hence,

Find an equivalent fraction for the point on the number line.
Question 12.
Big Ideas Math Answers Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.3 14
Answer: The equivalent fraction of \(\frac{6}{8}\) is: \(\frac{3}{4}\)

Explanation:
The Given number line is:
Big Ideas Math Answers Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.3 14
In the given line, the point on the marked line is at \(\frac{6}{8}\)
So,
The numerator and denominator of \(\frac{6}{8}\) are: 6 and 8 which are the multiples of 2
So,
\(\frac{6}{8}\) is divided by 2
So,
\(\frac{6}{8}=\frac{6 \div 2}{8 \div 2}=\frac{3}{4}\)
Hence,
The equivalent fraction of \(\frac{6}{8}\) is: \(\frac{3}{4}\)

Question 13.
Big Ideas Math Answers Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.3 15
Answer: The equivalent fraction of \(\frac{4}{12}\) is: \(\frac{1}{3}\)

Explanation:
The given number line is:
Big Ideas Math Answers Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.3 15
In the given line, the point on the marked line is at \(\frac{4}{12}\)
So,
The numerator and denominator of \(\frac{4}{12}\) are: 4 and 12 which are the multiples of 2
So,
\(\frac{4}{12}\) is divided by 2
So,
\(\frac{4}{12}=\frac{4 \div 2}{12 \div 2}=\frac{2}{6}\)
Hence,
The equivalent fraction of \(\frac{4}{12}\) is: \(\frac{2}{6}\)

Find an equivalent fraction.
Question 14.
\(\frac{3}{12}\)
Answer: The equivalent fraction of \(\frac{3}{12}\) is: \(\frac{1}{4}\)

Explanation:
The given fraction is \(\frac{3}{12}\)
From the above fraction, the numerator and denominator are: 3 and 12
3 and 12 are the multiples of 3.  ( Since the sum of the digits of the numerator and the denominator are the multiples of 3 )
So,
We have to divide the \(\frac{3}{12}\) with 3
So,
\(\frac{3}{12}=\frac{3 \div 3}{12 \div 3}=\frac{1}{4}\)
Hence,
The equivalent fraction of \(\frac{3}{12}\) is: \(\frac{1}{4}\)

Question 15.
\(\frac{18}{6}\)
Answer: The equivalent fraction of \(\frac{18}{6}\) is: \(\frac{6}{2}\)

Explanation:
The given fraction is \(\frac{18}{6}\)
From the above fraction, the numerator and denominator are: 18 and 6
18 and 6 are the multiples of 3.  ( Since the sum of the digits of the numerator and the denominator are the multiples of 3 )
So,
We have to divide the \(\frac{18}{6}\) with 3
So,
\(\frac{18}{6}=\frac{18 \div 3}{6 \div 3}=\frac{6}{2}\)
Hence,
The equivalent fraction of \(\frac{18}{6}\) is: \(\frac{3}{1}\)

Find two equivalent fractions.
Question 16.
\(\frac{20}{10}\)
Answer: The two equivalent fractions of \(\frac{20}{10}\) is: \(\frac{2}{1}\) and \(\frac{10}{5}\)

Explanation:
The given fraction is \(\frac{20}{10}\)
From the above fraction, the numerator and denominator are: 20 and 10
20 and 10 are the multiples of 10 and 2
So,
We have to divide the \(\frac{20}{10}\) with 10 and 2
So,
\(\frac{20}{10}=\frac{20 \div 10}{10 \div 10}=\frac{2}{1}\)
\(\frac{20}{10}=\frac{20 \div 2}{10 \div 2}=\frac{10}{5}\)
Hence,
The two equivalent fractions of \(\frac{20}{10}\) is: \(\frac{2}{1}\) and \(\frac{10}{5}\)

Question 17.
\(\frac{75}{100}\)
Answer: The two equivalent fractions of \(\frac{75}{100}\) is: \(\frac{15}{20}\) and \(\frac{3}{4}\)

Explanation:
The given fraction is \(\frac{75}{100}\)
From the above fraction, the numerator and denominator are: 75 and 100
75 and 100 are the multiples of 5 and 25 ( Since the one’s digit is 5 and 0 ).
So,
We have to divide the \(\frac{75}{100}\) with 5 and 25
So,
\(\frac{75}{100}=\frac{75 \div 5}{100 \div 5}=\frac{15}{20}\)
\(\frac{75}{100}=\frac{75 \div 25}{100 \div 25}=\frac{3}{4}\)
Hence,
The two equivalent fractions of \(\frac{75}{100}\) is: \(\frac{15}{20}\) and \(\frac{3}{4}\)

Question 18.
Reasoning
Your friend begins to divide the numerator and denominator of \(\frac{12}{6}\) by 4 and then gets stuck. Explain why your friend gets stuck.
Answer: The numerator and denominator of \(\frac{12}{6}\) is 12 and 6
when we find out the sum of the digits of the numerator and denominator, we find out that they are the multiples of 3.
So, we have to divide the \(\frac{12}{6}\) by 3 instead of 4
So, your friend get struck

Explanation:
The given fraction is: \(\frac{12}{6}\)
From the above fraction, the numerator and the denominator are: 12 and 6
When we find the sum of the digits of the numerator and the denominator, we can find out that the sum of the digits is the multiples of 3.
So,
We have to divide \(\frac{12}{6}\) by 3 instead of 4
Hence, from the above,
We can conclude that your friend is struck

Question 19.
DIG DEEPER!
Can you write an equivalent fraction with a lesser numerator and denominator when the numerator and denominator of a fraction are both odd numbers? Explain.
Answer:
Let the fraction with both the numerator and denominator odd numbers is: \(\frac{15}{9}\)
So,
The equivalent fraction of \(\frac{15}{9}\) is: \(\frac{5}{3}\)

Explanation:
Given that you have to write an equivalent fraction with a lesser numerator and denominator when the numerator and denominator of a fraction are both odd numbers.
Now,
Let the fraction with numerator and denominator odd numbers are: \(\frac{15}{9}\)
From the above fraction, the numerator and denominator are: 15 and 9
15 and 9 are the multiples of 3 ( Since the sum of the digits is a multiple of 3 ).
So,
We have to divide the \(\frac{15}{9}\) with 3
So,
\(\frac{15}{9}=\frac{15 \div 3}{9 \div 3}=\frac{5}{3}\)
Hence,
The equivalent fraction of \(\frac{15}{9}\) is: \(\frac{5}{3}\)

Think and Grow: Modeling Real Life

Example
The Lechtal High Trail is a 100-kilometer hiking trail in Austria. A hiker has completed 70 kilometers of the trail. What fraction of the trail, in tenths, has the hiker completed?
Big Ideas Math Answers Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.3 16
Use the distances to write the fraction of the trail the hiker has completed.

Show and Grow

Question 20.
A puzzle cube has 54 stickers. Nine of the stickers are orange. A cube has 6 faces. What fraction of the stickers, in sixths, are orange?
Big Ideas Math Answers Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.3 18
Answer: The fraction of the stickers, in sixths, that are orange is: \(\frac{1}{6}\)

Explanation:
Given that a puzzle has 54 stickers and each cube has 6 faces.
It is also given that 9 of the stickers are orange.
So,
The fraction of the stickers that are in orange = The number of stickers that are in orange ÷ The total number of stickers
= 9 ÷ 54
=\(\frac{1}{6}\)
Hence, from the above,
We can conclude that the fraction of the stickers, in sixths, that are orange are: \(\frac{1}{6}\)

Question 21.
There are 28 students in a class. Seven of the students pack their lunch. What fraction of the students, in fourths, pack their lunch?
Answer: The fraction of the students, in fourths that pack their lunch, is: \(\frac{1}{4}\)

Explanation:
It is given that there are 28 students in a class and 7 of the students pack their lunch.
So,
The fraction of students, in fourths, that pack their lunch = The number of students that pack their lunch ÷ The total number of students
= 7 ÷ 28
= \(\frac{1}{4}\)
Hence, from the above,
We can conclude that the fraction of students, in fourths that pack their lunch, are: \(\frac{1}{4}\)

Question 22.
DIG DEEPER!
There are 45 apps on a tablet. Nine of the apps are games. What fraction of the apps, in fifths, are not games? Explain.
Answer: The fraction of the apps, in fifths that are not games, is: \(\frac{4}{5}\)

Explanation:
Given that there are 45 apps on a tablet and nine of the apps are games.
So,
The fraction, in fifths that are games = The number of apps that are games ÷ The total number of games
= 9 ÷ 45
= \(\frac{1}{5}\)
Now,
The fraction of the apps, in fifths that are not games = 1- ( \(\frac{1}{5}\) )
= ( 5 – 1 ) ÷ 5
= \(\frac{4}{5}\)
Hence, from the above,
We can conclude that the fraction of apps, in fifths that are not games, is: \(\frac{4}{5}\)

Generate Equivalent Fractions by Dividing Homework & Practice 7.3

Find an equivalent fraction.
Question 1.
Big Ideas Math Answers Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.3 19
Answer: The equivalent fraction of \(\frac{8}{10}\) is: \(\frac{4}{5}\)

Explanation:
The given fraction is \(\frac{8}{10}\)
From the above fraction, the numerator and denominator are: 8 and 10
8 and 10 are the multiples of 2. ( Since, the one’s digit is 8 and 0 )
So,
We have to divide the \(\frac{8}{10}\) with 2
So,
\(\frac{8}{10}=\frac{8 \div 2}{10 \div 2}=\frac{4}{5}\)
Hence,

Question 2.
Big Ideas Math Answers Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.3 20
Answer: The equivalent fraction of \(\frac{24}{6}\) is: \(\frac{8}{2}\)

Explanation:
The given fraction is \(\frac{24}{6}\)
From the above fraction, the numerator and denominator are: 24 and 6
24 and 6 are the multiples of 3. ( Since, the sum of the digits is the multiples of 3 )
So,
We have to divide the \(\frac{24}{6}\) with 3
So,
\(\frac{24}{6}=\frac{24 \div 3}{6 \div 3}=\frac{8}{2}\)
Hence,

Find the equivalent fraction.
Question 3.
Big Ideas Math Answers Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.3 21
Answer: The equivalent fraction of \(\frac{4}{6}\) is: \(\frac{2}{3}\)

Explanation:
The given fraction is \(\frac{4}{6}\)
From the above fraction, the numerator and denominator are: 4 and 6
4 and 6 are the multiples of 2. ( Since, the one’s digits are 4 and 6 )
So,
We have to divide the \(\frac{4}{6}\) with 2
So,
\(\frac{4}{6}=\frac{4 \div 2}{6 \div 2}=\frac{2}{3}\)
Hence,

Question 4.
Big Ideas Math Answers Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.3 22
Answer: The equivalent fraction of \(\frac{25}{100}\) is: \(\frac{5}{20}\)

Explanation:
The given fraction is \(\frac{25}{100}\)
From the above fraction, the numerator and denominator are: 25 and 100
25 and 100 are the multiples of 5. ( Since, the one’s digits are 5 and 0 )
So,
We have to divide the \(\frac{25}{100}\) with 5
So,
\(\frac{25}{100}=\frac{25 \div 5}{100 \div 5}=\frac{5}{20}\)
Hence,

Find an equivalent fraction for the point on the number line.
Question 5.
Big Ideas Math Answers Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.3 23
Answer: The equivalent fraction of \(\frac{9}{12}\) is: \(\frac{3}{4}\)

Explanation:
The Given number line is:
Big Ideas Math Answers Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.3 23
In the given line, the point on the marked line is at \(\frac{9}{12}\)
So,
The numerator and denominator of \(\frac{9}{12}\) are: 9 and 12 which are the multiples of 3
So,
\(\frac{9}{12}\) is divided by 3
So,
\(\frac{9}{12}=\frac{9 \div 3}{12 \div 3}=\frac{3}{4}\)
Hence,
The equivalent fraction of \(\frac{9}{12}\) is: \(\frac{3}{4}\)

Question 6.
Big Ideas Math Answers Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.3 24
Answer: The equivalent fraction of \(\frac{6}{10}\) is: \(\frac{3}{5}\)

Explanation:
The Given number line is:
Big Ideas Math Answers Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.3 24
In the given line, the point on the marked line is at \(\frac{6}{10}\)
So,
The numerator and denominator of \(\frac{6}{10}\) are: 6 and 10 which are the multiples of 2
So,
\(\frac{6}{10}\) is divided by 2
So,
\(\frac{6}{10}=\frac{6 \div 2}{10 \div 2}=\frac{3}{5}\)
Hence,
The equivalent fraction of \(\frac{6}{10}\) is: \(\frac{3}{5}\)

Find an equivalent fraction.
Question 7.
\(\frac{3}{6}\)
Answer: The equivalent fraction of \(\frac{3}{6}\) is: \(\frac{1}{2}\)

Explanation:
The given fraction is \(\frac{3}{6}\)
From the above fraction, the numerator and denominator are: 3 and 6
3 and 6 are the multiples of 3.  ( Since the  the numerator and the denominator are the multiples of 3 )
So,
We have to divide the \(\frac{3}{6}\) with 3
So,
\(\frac{3}{6}=\frac{3 \div 3}{6 \div 3}=\frac{1}{2}\)
Hence,
The equivalent fraction of \(\frac{3}{6}\) is: \(\frac{1}{2}\)

Question 8.
\(\frac{8}{4}\)
Answer: The equivalent fraction of \(\frac{8}{4}\) is: \(\frac{4}{2}\)

Explanation:
The given fraction is \(\frac{8}{4}\)
From the above fraction, the numerator and denominator are: 8 and 4
8 and 4 are the multiples of 2.  ( Since the one’s digit is 8 and 4 )
So,
We have to divide the \(\frac{8}{4}\) with 2
So,
\(\frac{8}{4}=\frac{8 \div 2}{4 \div 2}=\frac{4}{2}\)
Hence,
The equivalent fraction of \(\frac{8}{4}\) is: \(\frac{4}{2}\)

Question 9.
\(\frac{15}{5}\)
Answer: The equivalent fraction of \(\frac{15}{5}\) is: \(\frac{3}{1}\)

Explanation:
The given fraction is \(\frac{15}{5}\)
From the above fraction, the numerator and denominator are: 15 and 5
15 and 5 are the multiples of 5.  ( Since the last digits are 5 )
So,
We have to divide the \(\frac{15}{5}\) with 5
So,
\(\frac{15}{5}=\frac{15 \div 5}{5 \div 5}=\frac{3}{1}\)
Hence,
The equivalent fraction of \(\frac{15}{5}\) is: \(\frac{3}{1}\)

Find two equivalent fractions.
Question 10.
\(\frac{4}{100}\)
Answer: The two equivalent fractions of \(\frac{4}{100}\) is: \(\frac{1}{25}\) and \(\frac{2}{50}\)

Explanation:
The given fraction is \(\frac{4}{100}\)
From the above fraction, the numerator and denominator are: 4 and 100
4 and 100 are the multiples of 2 and 4
So,
We have to divide the \(\frac{4}{100}\) with 4 and 2
So,
\(\frac{4}{100}=\frac{4 \div 2}{100 \div 2}=\frac{2}{50}\)
\(\frac{4}{100}=\frac{4 \div 4}{100 \div 4}=\frac{1}{25}\)
Hence,
The two equivalent fractions of \(\frac{4}{100}\) is: \(\frac{2}{50}\) and \(\frac{1}{25}\)

Question 11.
\(\frac{6}{6}\)
Answer: The two equivalent fractions of \(\frac{6}{6}\) is: \(\frac{1}{1}\) and \(\frac{2}{2}\)

Explanation:
The given fraction is \(\frac{6}{6}\)
From the above fraction, the numerator and denominator are: 6 and 6
6 and 6 are the multiples of 3 and 6
So,
We have to divide the \(\frac{6}{6}\) with 6 and 2
So,
\(\frac{6}{6}=\frac{6 \div 2}{6 \div 2}=\frac{3}{3}\)
\(\frac{6}{6}=\frac{6 \div 6}{6 \div 6}=\frac{1}{1}\)
Hence,
The two equivalent fractions of \(\frac{6}{6}\) is: \(\frac{1}{1}\) and \(\frac{3}{3}\)

Question 12.
\(\frac{24}{8}\)
Answer: The two equivalent fractions of \(\frac{24}{8}\) is: \(\frac{6}{2}\) and \(\frac{12}{4}\)

Explanation:
The given fraction is \(\frac{24}{8}\)
From the above fraction, the numerator and denominator are: 24 and 8
24 and 8 are the multiples of 4 and 2
So,
We have to divide the \(\frac{24}{8}\) with 4 and 2
So,
\(\frac{24}{8}=\frac{24 \div 4}{8 \div 4}=\frac{6}{2}\)
\(\frac{24}{8}=\frac{24 \div 2}{2 \div 2}=\frac{12}{4}\)
Hence,
The two equivalent fractions of \(\frac{24}{8}\) is: \(\frac{6}{2}\) and \(\frac{12}{4}\)

Question 13.
Writing
Explain how to find an equivalent fraction using division.
Answer:
Let the fraction be \(\frac{a}{b}\)
Let the equivalent fraction of \(\frac{a}{b}\) is: \(\frac{x}{y}\)
Let the number that divides \(\frac{a}{b}\) be p.
So,
Now, by using division,
\(\frac{a}{b}=\frac{a \div p}{b \div p}=\frac{x}{y}\)
Hence, like the above,
We can find th eequivalent fraction by using division.

Question 14.
Patterns
Describe and complete the pattern.
Big Ideas Math Answers Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.3 25
Answer: The remaining pattern is: \(\frac{8}{200}\) and \(\frac{4}{100}\)

Explanation:
The given pattern is:
Big Ideas Math Answers Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.3 25
From the above pattern, we can say that the given pattern is divided by 2.
Now,
\(\frac{64}{1,600}=\frac{64 \div 2}{1,600 \div 2}=\frac{32}{800}\)
So, like the above,
The remaining two patterns will be:
\(\frac{16}{400}=\frac{16 \div 2}{400 \div 2}=\frac{8}{200}\)
\(\frac{8}{200}=\frac{8 \div 2}{200 \div 2}=\frac{4}{100}\)
Hence, from the above,
We can conclude the pattern will be:

Question 15.
Modeling Real Life
A book shows100 hieroglyphic symbols. You have learned the meanings of 30 of them. What fraction of the symbols’ meanings, in tenths, have you learned?
Big Ideas Math Answers Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.3 26
Answer: The fraction of the symbols’ meaning, in tenths you have learned, is: \(\frac{3}{10}\)

Explanation:
Given that a book shows 100 hieroglyphic symbols
It is also given that you have learned the meanings of 30 of them.
So,
The fraction of the symbols’ meaning, in tenths, you have read = The number of meanings of the symbols’ you have learned ÷ The total number of symbols
= 30 ÷ 100
= \(\frac{3}{10}\)
Hence, from the above,
We can conclude that the fraction of the symbols’ meaning, in tenths, you have learned is: \(\frac{3}{10}\)

Question 16.
DIG DEEPER!
There are 54 players in a beach volleyball club. Nine of the players cannot attend a game night. The coach needs to make even teams with the players that are there. What fraction of the players, in sixths, are at the game night?
Big Ideas Math Answers Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.3 27
Answer: The fraction of the players, in sixths, are at the game night is: \(\frac{5}{6}\)

Explanation:
Given that there are 54 players in a beach volleyball club.
It is also given that nine of the players can not attend a game night and also the coach needs to make even teams with the players that are there.
So,
The fraction of the players, in sixths, that are not in the game night = \(\frac{The number of players that are not in the game night}{The total number of players}\)
= \(\frac{9}{54}\)
So,
we have to divide \(\frac{9}{54}\) with 9 ( Since 9 and 54 are the multiples of 9 )
So,
\(\frac{9}{54}=\frac{9 \div 9}{54 \div 9}=\frac{1}{6}\)
So,
The fraction of players, in sixths, that attended a game night = 1 – \(\frac{1}{6}\)
= \(\frac{5}{6}\)
Hence, from the above,
We can conclude that the fraction of the players, in sixths, that attended a game night is: \(\frac{5}{6}\)

Review & Refresh

Find the product. Check whether your answer is reasonable.
Question 17.
71 × 63 = _______
Answer: 71 × 63 = 4,473

Explanation:
Let 71 be rounded to 70
Let 63 be rounded to 65
So,
By using the place-value method,
70 × 65 = 7 tens × 65
= 455 tens
= 4,550
Hence, from the above 2 values 4,473 and 4,550,
We can conclude that the answer is not reasonable.

Question 18.
24 × 98 = _______
Answer: 24 × 98 = 2,352

Explanation:
Let 24 be rounded to 25
Let 98 be rounded to 100
So,
By using the place-value method,
25 × 100 = 10 tens × 25
= 250 tens
= 2,500
Hence, from the above 2 values 2,352 and 2,500,
We can conclude that the answer is not reasonable

Question 19.
85 × 27 = _______
Answer: 85 × 27 = 2,295

Explanation:
Let 27 be rounded to 25
So,
By using the partial products method,
85 × 25 = ( 80 + 5 ) × ( 20 + 5 )
= ( 80 × 20 ) + ( 80 × 5 ) + ( 5 × 20 ) + ( 5 × 5 )
= 1,600 + 400 + 100 + 25
= 2,125
Hence, from the above 2 values 2,295 and 2,125,
We can conclude that the answer is not reasonable

Lesson 7.4 Compare Fractions Using Benchmarks

Explore and Grow

Big Ideas Math Solutions Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.4 1
Answer: 

By using the above models,
5 is greater than half of the value of 8 i.e., 4
So,
We can conclude that \(\frac{5}{8}\) is greater than \(\frac{2}{5}\)

Structure
How does the numerator of a fraction compare tothe11denominator when the fraction is less than \(\frac{1}{2}\)? greater than \(\frac{1}{2}\) ? equal to \(\frac{1}{2}\)? Explain.
Answer:

Think and Grow: Compare Fractions Using Benchmarks

A benchmark is a commonly used number that you can use to compare other numbers. You can use the benchmarks \(\frac{1}{2}\) and 1 to help you compare fractions.
Big Ideas Math Solutions Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.4 2
Example
Use fraction strips to compare \(\frac{7}{10}\) and \(\frac{3}{8}\).

Show and Grow

Compare. Use a model to help
Question 1.
Big Ideas Math Solutions Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.4 4
Answer: \(\frac{5}{12}\) is less than \(\frac{3}{5}\)

Explanation:

From the above model,
5 is less than half of the value of 12 i.e., 6
So,
We can conclude that \(\frac{5}{12}\) is less than \(\frac{3}{5}\)

Question 2.
Big Ideas Math Solutions Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.4 5
Answer: \(\frac{3}{4}\) is equal to \(\frac{6}{8}\) which is the equivalent fraction of \(\frac{3}{4}\)

Explanation:

From the above model,
When we simplify \(\frac{6}{8}\) i.e., divide it by 2, we will get \(\frac{3}{4}\)
So,
We can conclude that \(\frac{3}{4}\) is equal to \(\frac{6}{8}\)

Question 3.
Big Ideas Math Solutions Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.4 6
Answer: \(\frac{6}{5}\) is greater than \(\frac{9}{10}\)

Explanation:

From the above model,
6 is greater than half of the value of 5.
So,
We can conclude that \(\frac{6}{5}\) is greater than \(\frac{9}{10}\)

Apply and Grow: Practice

Compare. Use a model to help
Question 4.
Big Ideas Math Solutions Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.4 7
Answer: \(\frac{4}{12}\) is less than \(\frac{7}{10}\)

Explanation:
Given fractions are: \(\frac{4}{12}\) and \(\frac{7}{10}\)
Now,
From \(\frac{4}{12}\),
we can see that 4 is less than half of the value of 12 ie., 6
Hence,
We can conclude that \(\frac{4}{12}\) is less than \(\frac{7}{10}\)

Question 5.
Big Ideas Math Solutions Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.4 8
Answer: \(\frac{1}{2}\) is equal to \(\frac{3}{6}\) which is the equivalent fraction of \(\frac{3}{6}\)

Explanation:
Given fractions are: \(\frac{1}{2}\) and \(\frac{3}{6}\)
Now,
When we simplify \(\frac{3}{6}\), by divideing with 3, we will get \(\frac{1}{2}\)
So,
We can conclude that \(\frac{1}{2}\) isequal to \(\frac{3}{6}\) which is the equivalent fraction of
\(\frac{3}{6}\)

Question 6.
Big Ideas Math Solutions Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.4 9
Answer: \(\frac{2}{10}\) is less than \(\frac{5}{6}\)

Explanation:
Given fractions are: \(\frac{2}{10}\) and \(\frac{5}{6}\)
Now,
From  \(\frac{2}{10}\),
We can see that 2 is less than half of the value of 10 i.e., 5
So,
We can conclude that \(\frac{2}{10}\) is less than \(\frac{5}{6}\)

Question 7.
Big Ideas Math Solutions Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.4 10
Answer: \(\frac{5}{5}\) is equal to \(\frac{12}{12}\)

Explanation:
Given fractions are: \(\frac{5}{5}\) and \(\frac{12}{12}\)
Now,
From \(\frac{5}{5}\),
We can see that the numerator and denominator are equal, so we will get the same result in both the numerator and the denominator
So,
We can conclude that \(\frac{5}{5}\) is equal to \(\frac{12}{12}\)

Question 8.
Big Ideas Math Solutions Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.4 11
Answer: \(\frac{4}{2}\) is greater than \(\frac{7}{10}\)

Explanation:
Given fractions are: \(\frac{4}{2}\) and \(\frac{7}{10}\)
Now,
From \(\frac{4}{2}\),
We can see that 4 is greater than half of the value of 2.
So,
We can conclude that \(\frac{4}{2}\) is greater than \(\frac{7}{10}\)

Question 9.
Big Ideas Math Solutions Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.4 12
Answer: \(\frac{4}{6}\) is greater than \(\frac{1}{3}\)

Explanation:
Given fractions are: \(\frac{4}{6}\) and \(\frac{1}{3}\)
When we simplify \(\frac{4}{6}\), we will get \(\frac{2}{3}\)
Now,
From \(\frac{2}{3}\),
We can see that 2 is greater than half of the value of 3.
So,
We can conclude that \(\frac{4}{6}\) is greater than \(\frac{1}{3}\)

Question 10.
Big Ideas Math Solutions Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.4 13
Answer: \(\frac{5}{4}\) is greater than \(\frac{3}{8}\)

Explanation:
Given fractions are: \(\frac{5}{4}\) and \(\frac{3}{8}\)
Now,
From \(\frac{5}{4}\),
We can say that 5 is greater than half of the value of 4 i.e., 2
So,
We can conclude that \(\frac{5}{4}\) is greater than \(\frac{3}{8}\)

Question 11.
Big Ideas Math Solutions Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.4 14
Answer: \(\frac{6}{12}\) is less than \(\frac{4}{5}\)

Explanation:
Given fractions are: \(\frac{6}{12}\) and \(\frac{4}{5}\)
When we simplify \(\frac{6}{12}\), we will get \(\frac{1}{2}\)
Now,
From \(\frac{4}{5}\),
We can say that 4 is greater than half of the value of 5
So,
We can conclude that \(\frac{6}{12}\) is less than \(\frac{4}{5}\)

Question 12.
Big Ideas Math Solutions Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.4 15
Answer: \(\frac{3}{2}\) is greater than \(\frac{80}{100}\)

Explanation:
The given fractions are: \(\frac{3}{2}\) and \(\frac{80}{100}\)
When we simplify \(\frac{80}{100}\) by dividing with 20, we will get \(\frac{4}{5}\)
Now,
From \(\frac{3}{2}\),
We can say that 3 is greater than half of the value of 2 i.e., 1
So,
We can conclude that \(\frac{3}{2}\) is greater than \(\frac{80}{100}\)

Question 13.
A black bear hibernates for \(\frac{7}{12}\) of 1 year. A bat hibernates for \(\frac{1}{4}\) of 1 year. Which animal hibernates longer? How do you know?
Big Ideas Math Solutions Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.4 16

Answer: The black bear hibernates for a longer time than the bat

Explanation:
It is given that a black bear hibernates for \(\frac{7}{12}\) of 1 year and a bat hibernates for \(\frac{1}{4}\) of 1 year.
So,
The time that a black bear hibernates is: \(\frac{7}{12}\) of 1 year
The time that a bat hibernates is: \(\frac{1}{4}\) of 1 year
So,
To make the total time 1 year, we have to multiply  \(\frac{1}{4}\) with \(\frac{3}{3}\)
So,
\(\frac{1}{4}\) = \(\frac{3}{12}\)
Hence,
When we compare \(\frac{7}{12}\) and \(\frac{3}{12}\),
We can conclude that a black bear hibernates for a long time.

Question 14.
Writing
Explain how you can tell whether a fraction is greater than, less than, or equal to1, just by looking at the numerator and the denominator
Answer: We will compare the numerator with half of the value of the denominator. In this way, we will determine the relationship between the fractions.
Now,
Let the 2 fractions be \(\frac{a}{b}\) and \(\frac{x}{y}\)
So,
Compare the first fraction’s a and b
When a > ( b / 2 ), \(\frac{a}{b}\) > \(\frac{x}{y}\)
When a < ( b / 2 ), \(\frac{a}{b}\) < \(\frac{x}{y}\)
When a = ( b / 2 ), \(\frac{a}{b}\) = \(\frac{x}{y}\)
Hence, from the above,
We can conclude that we can compare the two fractions just by looking at the numerator and the denominator

Question 15.
DIG DEEPER.!
You and your friend pack lunch. You eat \(\frac{2}{6}\) of your lunch. Your friend eats \(\frac{3}{4}\) of his lunch. Can you tell who ate more? Explain.
Answer: Your friend ate more lunch than you

Explanation:
Given that you and your friend pack lunch.
It is also given that you eat \(\frac{2}{6}\) of your lunch and your friend eats \(\frac{3}{4}\) of his lunch.
So,
The amount of lunch you ate = \(\frac{2}{6}\)
By simplifying \(\frac{2}{6}\) i.e., dividing by 2, we will get \(\frac{1}{3}\)
So,
The amount of lunch you ate = \(\frac{1}{3}\)
Now,
The amount of lunch your friend ate = \(\frac{3}{4}\)
So, when we compare \(\frac{1}{3}\) and \(\frac{3}{4}\),
Now,
From \(\frac{1}{3}\),
1 is less than half of the value of the 3
From \(\frac{3}{4}\),
3 is greater than half of the value of 4
Hence, from the above,
We can conclude that your friend ate more than you

Think and Grow: Modeling Real Life

Example
You have \(\frac{3}{5}\) of a bottle of blue paint and \(\frac{7}{8}\) of a bottle of yellow paint. Do you have enough of each paint color to make the recipe? Explain.
Big Ideas Math Solutions Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.4 17
Compare each fraction of a paint bottle to the benchmark \(\frac{1}{2}\).

So,
You have enough of each paint color to make the recipe.

Show and Grow

Question 16.
You have \(\frac{3}{8}\) tablespoon of baking soda, \(\frac{2}{3}\) tablespoon of contact lens solution, and \(\frac{5}{3}\) cups of glue. Do you have enough of each ingredient to make the recipe? Explain.
Big Ideas Math Solutions Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.4 19

Answer: You have enough ingredients of contact lens solution and glue.

Explanation:
It is given that you have \(\frac{3}{8}\) tablespoon of baking soda, \(\frac{2}{3}\) tablespoon of contact lens solution, and \(\frac{5}{3}\) cups of glue.
But it is given that
Big Ideas Math Solutions Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.4 19
So, when we compare the two values,
For the tablespoon of baking soda,
\(\frac{3}{8}\) is less than \(\frac{3}{4}\)
For the tablespoon of contact lens solution,
\(\frac{2}{3}\) is less than \(\frac{3}{2}\)
For the tablespoon of glue,
\(\frac{5}{3}\) is greater than \(\frac{6}{8}\)
Hence, from the above,
We can conclude that you have enough ingredients to make the recipe.

Question 17.
DIG DEEPER!
You and your friend are making posters for a science fair. The posters are the same size. Your poster has 8 equal parts and you are halfway done. Your friend’s poster has 12 equal parts. Your friend has completed \(\frac{9}{12}\) of her poster. Who has a greater amount of posters left to complete?

Answer: You have a greater amount of posters left to complete

Explanation:
Given that you and your friend are making posters for a science fair and the posters are the same size.
Given that,
Your poster has 8 equal parts and you are halfway done.
So,
The representation of your work is: \(\frac{4}{8}\)
The simple form of \(\frac{4}{8}\) is: \(\frac{1}{2}\)
Now,
It is also given that your friend’s poster has 12 equal parts and you have completed \(\frac{9}{12}\) of her poster.
So,
The representation of your friend’s work is: \(\frac{9}{12}\)
The simplified form of \(\frac{9}{12}\) is: \(\frac{3}{4}\)
So,
The amount of work left by you is: \(\frac{1}{2}\)
The amount of work left by your friend is: \(\frac{1}{4}\)
Hence, from the above,
We can conclude that you have a greater amount of posters left to complete.

Compare Fractions Using Benchmarks Homework & Practice 7.4

Compare. Use a model to help.
Question 1.
Big Ideas Math Solutions Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.4 20
Answer: \(\frac{8}{12}\) is less than \(\frac{1}{4}\)

Explanation:
The given fractions are: \(\frac{8}{12}\) and \(\frac{8}{6}\)
Now,
The simple form of \(\frac{8}{12}\) is: \(\frac{2}{3}\)
The simple form of \(\frac{8}{6}\) is: \(\frac{4}{3}\)
So,
When we compare \(\frac{2}{3}\) and \(\frac{4}{3}\),
We can say that \(\frac{2}{3}\) is less than \(\frac{4}{3}\)

Question 2.
Big Ideas Math Solutions Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.4 21
Answer: \(\frac{9}{10}\) is greater than \(\frac{1}{3}\)

Explanation:
The given fractions are: \(\frac{9}{10}\) and \(\frac{1}{3}\)
For \(\frac{9}{10}\) and \(\frac{1}{3}\)
When we compare 9 with half of the value of 10 i.e., 5, and when we compare 1 with half of the value of 3,
We will conclude that \(\frac{9}{10}\) is greater than \(\frac{1}{3}\)

Question 3.
Big Ideas Math Solutions Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.4 22
Answer: \(\frac{5}{2}\) is greater than \(\frac{7}{8}\)

Explanation:
The given fractions are: \(\frac{5}{2}\) and \(\frac{7}{8}\)
For \(\frac{5}{2}\) and \(\frac{7}{8}\)
When we compare 5 with half of the value of 2 i.e., 1, and when we compare 7 with half of the value of 8,
We will conclude that \(\frac{5}{2}\) is greater than \(\frac{7}{8}\)

Question 4.
Big Ideas Math Solutions Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.4 23
Answer: \(\frac{1}{4}\) is less than \(\frac{7}{12}\)

Explanation:
The given fractions are: \(\frac{1}{4}\) and \(\frac{7}{12}\)
For \(\frac{1}{4}\) and \(\frac{7}{12}\)
When we compare 1 with half of the value of 4 i.e., 2, and when we compare 7 with half of the value of 6,
We will conclude that \(\frac{1}{4}\) is less than \(\frac{7}{1}\)

Question 5.
Big Ideas Math Solutions Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.4 24
Answer: \(\frac{2}{2}\) is less than \(\frac{10}{8}\)

Explanation:
The given fractions are: \(\frac{2}{2}\) and \(\frac{10}{8}\)
For \(\frac{2}{2}\) and \(\frac{10}{8}\)
When we compare 2 with the value of 2, and when we compare 10 with half of the value of 8,
We will conclude that \(\frac{2}{2}\) is less than \(\frac{10}{8}\)

Question 6.
Big Ideas Math Solutions Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.4 25
Answer: \(\frac{60}{100}\) is equal to \(\frac{3}{5}\)

Explanation:
The given two fractions are: \(\frac{60}{100}\) and \(\frac{3}{5}\)
When we simplify (i.e., divide by 20) \(\frac{60}{100}\), we will get \(\frac{3}{5}\)
So,
When we compare the two fractions,
We will conclude that \(\frac{60}{100}\) is equal to \(\frac{3}{5}\)

Question 7.
Big Ideas Math Solutions Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.4 26
Answer: \(\frac{4}{12}\) is equal to \(\frac{2}{6}\)

Explanation:
The two given fractions are: \(\frac{4}{12}\) and \(\frac{2}{6}\)
When \(\frac{4}{12}\) is divided by 4, we will get \(\frac{1}{3}\)
When \(\frac{2}{6}\) is divided by 2, we will get \(\frac{1}{3}\)
So, when we compare the two fractions,
We will conclude that the two fractions are equal.

Question 8.
Big Ideas Math Solutions Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.4 27
Answer: \(\frac{4}{6}\) is greater than \(\frac{5}{100}\)

Explanation:
The given two fractions are: \(\frac{4}{6}\) and \(\frac{5}{100}\)
When \(\frac{4}{6}\) is divided by 2, we will get \(\frac{2}{3}\)
When \(\frac{5}{100}\) is divided by 5, we will get \(\frac{1}{20}\)
So,
When we compare 2 with half of the value of 3,
We will conclude that \(\frac{4}{6}\) is greater than \(\frac{5}{100}\)

Question 9.
Big Ideas Math Solutions Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.4 28
Answer: \(\frac{8}{10}\) is less than \(\frac{9}{1}\)

Explanation:
The given two fractions are: \(\frac{8}{10}\) and \(\frac{9}{1}\)
When we cross multiply the two fractions, i.e., 8 × 1 and 9 × 10,
We will get the values of 8 and 90,
So, from the above,
We will conclude that \(\frac{8}{10}\) is less than \(\frac{9}{1}\)

Question 10.
In a litter of kittens, \(\frac{3}{4}\) are white, and \(\frac{2}{8}\) are gray. Are there more white or more gray kittens?
Big Ideas Math Solutions Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.4 29
Answer: There are more white kittens than grey kittens

Explanation:
Given that,
The number of white kittens are: \(\frac{3}{4}\)
The number of gray kittens are: \(\frac{2}{8}\)
Simplify \(\frac{2}{8}\) by dividing it with 2
So,
\(\frac{2}{8}\) = \(\frac{1}{4}\)
So,
The number of white kittens are: \(\frac{1}{4}\)
The number of gray kittens are: \(\frac{3}{4}\)
So,
When we compare the number of white and gray kittens,
We will conclude that white kittens are more than gray kittens

Open-Ended
Complete the statement.
Question 11.
Big Ideas Math Solutions Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.4 30
Answer: The missing fraction is: \(\frac{1}{3}\)

Explanation:
The given fraction is: \(\frac{5}{6}\)
Let the missing fraction be: \(\frac{1}{3}\)
So,
When we compare the 2 fractions by cross multiplying the two fractions,
We will conclude that

Question 12.
Big Ideas Math Solutions Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.4 31
Answer: The missing fraction is:\(\frac{5}{8}\)

Explanation:
The given fraction is: \(\frac{7}{8}\)
Let the missing fraction be: \(\frac{5}{8}\)
So,
When we compare the two fractions by cross multiplying the two fractions,
We will conclude that

Question 13.
Big Ideas Math Solutions Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.4 32
Answer: The missing fraction is:  \(\frac{2}{5}\)

Explanation:
The given fraction is: \(\frac{12}{5}\)
Let the missing fraction be: \(\frac{2}{5}\)
So,
When we compare the two fractions by cross multiplication,
We will conclude that

Question 14.
Number Sense
Which statements are true
Big Ideas Math Solutions Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.4 33

Answer:
Let the given statements be named as A), B), ) and D)
So,
From these four statements, A), B) statements are true

Explanation:
Let the four statements be named as: A), B), C) and D)
Now, the given statements are:
Big Ideas Math Solutions Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.4 33
Cross multiply each statement
So,
From A), we will get 48 >15
From B), we will get 100 >48
From C), we will get 12 < 9
From D), we will get 8 < 8
Hence, from the above,
We can conclude that A) and B) statements are true

Question 15.
Modeling Real Life
You have \(\frac{1}{3}\) cup of oranges and \(\frac{5}{4}\) cups of strawberries. Do you have enough of each ingredient to make the smoothie? Explain.
Big Ideas Math Solutions Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.4 34

Answer: We have only enough strawberries ingredient

Explanation:
The given recipe is:
Big Ideas Math Solutions Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.4 34
It is also given that you have \(\frac{1}{3}\) cup of oranges and \(\frac{5}{4}\) cups of strawberries.
But, from the recipe,

It is given that \(\frac{3}{2}\) cup of oranges and \(\frac{7}{8}\) cups of strawberries.
So,
When we compare \(\frac{1}{3}\) and \(\frac{3}{2}\) cup of oranges, we can say that we have less amount of oranges to make the recipe.
When we compare \(\frac{5}{4}\) and \(\frac{7}{8}\) cups of strawberries, we can say that we have enough amount of strawberries to make the recipe.
Hence, from the above,
We can conclude that we have only enough strawberry ingredients to make the recipe.

Question 16.
DIG DEEPER!
Newton and Descartes are picking apples at a farm. Newton’s bag of 4apples weighs \(\frac{4}{5}\) pound. Descartes’s bag weighs \(\frac{3}{2}\) pounds. How much money will Newton and Descartes each pay for their bag of apples?
Big Ideas Math Solutions Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.4 35
Answer: The amount of money will Newton and Descartes each pay for their bag of apples is: $301

Explanation:
It is given that Newton and Descartes are picking apples at a farm.
It is also given that Newton’s bag of 4 apples weighs \(\frac{4}{5}\) pound.
Big Ideas Math Solutions Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.4 35
So,
\(\frac{4}{5}\) is in between \(\frac{1}{2}\) pound and 1 pound
So,
As per the figure, the pice of Newton’s bag is: $75
So,
The amount paid by Newton = 4 × 75 = $300
It is also given that Descartes’s bag weighs \(\frac{3}{2}\) pounds.
So,
\(\frac{3}{2}\) is greater than 1 pound
So,
The amount paid by Descartes is: $1
Hence, from the above,
The amount paid by Newton and Descartes = 300 + 1 = $301

Review & Refresh

Find the factor pairs for the number.
Question 17.
12
Answer: The factor pairs of 12 are: 1 and 12, 2 and 6, 3 and 4

Explanation:
Factors are the numbers that divide the original number completely.
Hence,
The factor pairs of 12 are: 1 × 12, 2 × 6, 3 × 4

Question 18.
50
Answer: The factor pairs of 50 are: 1 and 50, 2 and 25, 5 and 10

Explanation:
Factors are the numbers that divide the original number completely.
Hence,
The factor pairs of 50 are: 1 × 50, 2 × 25, 5 × 10

Question 19.
17
Answer: The factor pairs of 17 are: 1 and 17, 17 and 1

Explanation:
Factors are the numbers that divide the original number completely.
hence,
The factor pairs of 17 are: 1 × 17, 17 × 1

Lesson 7.5 Compare Fractions

Explore and Grow

Shade each pair of models to compare \(\frac{1}{3}\) and \(\frac{5}{12}\). Explain how each pair of models helps you compare the fractions differently.
Big Ideas Math Answer Key Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.5 1
Answer:

From the first pattern,
The first fraction is:
\(\frac{1}{3}\) = \(\frac{4}{12}\)
The second fraction is: \(\frac{5}{12}\)
By comparing the two fractions’ numerators,
We can conclude that \(\frac{4}{12}\) is less than \(\frac{5}{12}\)
From the second pattern,
The first fraction is:
\(\frac{1}{3}\) = \(\frac{5}{15}\)
The second fraction is: \(\frac{5}{12}\)
By comparing the two fractions’ denominators,
We can conclude that \(\frac{5}{15}\) is greater than \(\frac{5}{12}\)

Reasoning
How can you use equivalent fractions to compare fractions with different numerators and different denominators?
Answer:
We can use the equivalent fractions to compare the different numerators and different denominators by making either both the fractions’ numerators or both the fractions’ denominators.

Think and Grow: Compare Fractions

Example
Compare \(\frac{3}{5}\) and \(\frac{9}{10}\).
One Way:
Use a like denominator. Find an equivalent fraction for \(\frac{3}{5}\) that has a denominator of 10.
Multiply the numerator and the denominator of \(\frac{3}{5}\) by 2

Another Way:
Use a like numerator. Find an equivalent fraction for \(\frac{3}{5}\) that has a numerator of 10.
Multiply the numerator and the denominator of \(\frac{3}{5}\) by 3

Show and Grow

Compare. Use a model to help.
Question 1.
Big Ideas Math Answer Key Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.5 4
Answer: \(\frac{7}{8}\) is greater than \(\frac{3}{4}\)

Explanation:
The given fractions are: \(\frac{7}{8}\) and \(\frac{3}{4}\)
So,
Let make the denominator of the two fractions equal so that we can compare the numerators
To make the denominators equal, multiply the second fraction with 2
So,
\(\frac{3}{4}\) = \(\frac{6}{8}\)
So, when we compare the 2 fractions,
We can conclude that \(\frac{7}{8}\) is greater than \(\frac{3}{4}\)

Question 2.
Big Ideas Math Answer Key Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.5 5
Answer: \(\frac{4}{6}\) is equal to \(\frac{2}{3}\)

Explanation:
The given fractions are: \(\frac{4}{6}\) and \(\frac{2}{3}\)
So,
Let make the denominator of the two fractions equal so that we can compare the numerators
To make the denominators equal, multiply the second fraction with 2
So,
\(\frac{2}{3}\) = \(\frac{4}{6}\)
So, when we compare the 2 fractions,
We can conclude that \(\frac{4}{6}\) is equal to \(\frac{2}{3}\)

Question 3.
Big Ideas Math Answer Key Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.5 6
Answer: \(\frac{4}{3}\) is greater than \(\frac{5}{4}\)

Explanation:
The given fractions are: \(\frac{4}{3}\) and \(\frac{5}{4}\)
So,
Let make the denominator of the two fractions equal so that we can compare the numerators
To make the denominators equal, multiply the first fraction with 4 and the second fraction with 3
So,
\(\frac{4}{3}\) = \(\frac{16}{12}\)
\(\frac{5}{4}\) = \(\frac{15}{12}\)
So, when we compare the 2 fractions,
We can conclude that \(\frac{4}{3}\) is greater than \(\frac{5}{4}\)

Apply and Grow: Practice

Compare. Use a model to help
Question 4.
Big Ideas Math Answer Key Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.5 7
Answer: \(\frac{56}{12}\) is greater than \(\frac{1}{3}\)

Explanation:
The given fractions are: \(\frac{6}{12}\) and \(\frac{1}{3}\)
So,
Let make the denominator of the two fractions equal so that we can compare the numerators
To make the denominators equal, multiply the second fraction with 4
So,
\(\frac{1}{3}\) = \(\frac{4}{12}\)
So, when we compare the 2 fractions,
We can conclude that \(\frac{6}{12}\) is greater than \(\frac{1}{3}\)

Question 5.
Big Ideas Math Answer Key Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.5 8
Answer: \(\frac{8}{10}\) is equal to \(\frac{4}{5}\)

Explanation:
The given fractions are: \(\frac{8}{10}\) and \(\frac{4}{5}\)
So,
Let make the denominator of the two fractions equal so that we can compare the numerstors
To make the denominators equal, multiply the second fraction with 2
So,
\(\frac{4}{5}\) = \(\frac{8}{10}\)
So, when we compare the 2 fractions,
We can conclude that \(\frac{8}{10}\) is equal to \(\frac{4}{5}\)

Question 6.
Big Ideas Math Answer Key Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.5 9
Answer: \(\frac{3}{10}\) is greater than \(\frac{1}{4}\)

Explanation:
The given two fractions are: \(\frac{3}{10}\) and \(\frac{1}{4}\)
So,
Let make the denominators equal so that we can compare the numerators
So,
To make the denominators equal, multiply the first fraction with 4 and the second fraction with 10
So,
\(\frac{3}{10}\) = \(\frac{12}{40}\)
\(\frac{1}{4}\) = \(\frac{10}{40}\)
So, when we compare the 2 fractions,
We can conclude that \(\frac{3}{10}\) is greater than \(\frac{1}{4}\)

Question 7.
Big Ideas Math Answer Key Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.5 10
Answer: \(\frac{3}{8}\) is less than \(\frac{2}{5}\)

Explanation:
The given two fractions are: \(\frac{3}{8}\) and \(\frac{2}{5}\)
So,
Let make the denominators equal so that we can compare the numerators
So,
To make the denominators equal, multiply the first fraction with 5 and the second fraction with 8
So,
\(\frac{3}{8}\) = \(\frac{15}{40}\)
\(\frac{2}{5}\) = \(\frac{16}{40}\)
So, when we compare the 2 fractions,
We can conclude that \(\frac{3}{8}\) is less than \(\frac{2}{5}\)

Question 8.
Big Ideas Math Answer Key Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.5 11
Answer: \(\frac{9}{6}\) is equal to \(\frac{15}{10}\)

Explanation:
The given two fractions are: \(\frac{9}{6}\) and \(\frac{15}{10}\)
So,
Let make the denominators equal so that we can compare the numerators
So,
To make the denominators equal, multiply the first fraction with 10 and the second fraction with 6
So,
\(\frac{9}{6}\) = \(\frac{90}{60}\)
\(\frac{15}{10}\) = \(\frac{90}{60}\)
So, when we compare the 2 fractions,
We can conclude that \(\frac{9}{6}\) is equal to \(\frac{15}{10}\)

Question 9.
Big Ideas Math Answer Key Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.5 12
Answer: \(\frac{7}{10}\) is less than \(\frac{9}{12}\)

Explanation:
The given two fractions are: \(\frac{7}{10}\) and \(\frac{9}{12}\)
So,
Let make the denominators equal so that we can compare the numerators
So,
To make the denominators equal, multiply the first fraction with 12 and the second fraction with 10
So,
\(\frac{7}{10}\) = \(\frac{84}{120}\)
\(\frac{9}{12}\) = \(\frac{90}{120}\)
So, when we compare the 2 fractions,
We can conclude that \(\frac{7}{10}\) is less than \(\frac{9}{12}\)

Question 10.
Writing
Explain why writing fractions with like denominators or like numerators is helpful when comparing them.
Answer: If we write the fractions with the like numerators and like denominators, then the comparison will be easy when we do the cross multiplication of the fractions.

Explanation:
\(\frac{1}{2}\) is less than \(\frac{3}{4}\)
So, to make the like denominators, multiply the first fraction with 2
So, \(\frac{1}{2}\) = \(\frac{2}{4}\)
Since the denominators are equal, compare the numerstors and we will get the conclusion \(\frac{1}{2}\) is less than \(\frac{3}{4}\)

Question 11.
DIG DEEPER!
Use the fractions and symbols to make two true statements.
Big Ideas Math Answer Key Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.5 13
Answer:
The two statements are:
\(\frac{9}{12}\) < \(\frac{5}{6}\)
\(\frac{7}{8}\) > \(\frac{2}{3}\)

Explanation:
Given pattern is:
Big Ideas Math Answer Key Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.5 13
Given fractions from the above are:
\(\frac{9}{12}\), \(\frac{5}{6}\), \(\frac{7}{8}\) and \(\frac{2}{3}\)
Now, first take the 2 fractions for comparison.
The 2 fractions are: \(\frac{9}{12}\), \(\frac{5}{6}\)
So, to make the denominators of these 2 fractions equal, multiply the second fraction with 2
So,
\(\frac{5}{6}\) = \(\frac{10}{12}\)
Hence,
The first statement is:
\(\frac{9}{12}\) < \(\frac{5}{6}\)
Now, take the remaining 2 fractions.
The remaining 2 fractions are: \(\frac{7}{8}\) and \(\frac{2}{3}\)
So, to make the denominators of these 2 fractions equal, multiply the first fraction with 3 and the second fraction with 8
So,
\(\frac{7}{8}\) = \(\frac{21}{24}\)
\(\frac{2}{3}\) \(\frac{16}{24}\)
Hence,
The second statement is:
\(\frac{7}{8}\) > \(\frac{2}{3}\)

Think and Grow: Modeling Real Life

Example
You try to use a \(\frac{13}{16}\)-inch socket to tighten a bolt, but it is too big. Should you try a \(\frac{3}{4}\) -inch socket or a \(\frac{7}{8}\)-inch socket next?
Big Ideas Math Answer Key Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.5 14
Write each fraction using a like denominator.
Because 4 and 8 are both factors of 16, use 16 as the denominator.

Show and Grow

Question 12.
You drill a hole using a \(\frac{5}{16}\) -inch drill bit. The hole is too small. Which drill bit should you use to enlarge the hole?
Big Ideas Math Answer Key Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.5 16
Answer: You should use \(\frac{3}{8}\)– inch drill bit to enlarge the hole.

Explanation:
It is given that you drill a hole using a \(\frac{5}{16}\) and by using this, the hole is too small.
To enlarge a drill bit, it is also given 2 types of drill bits. They are:
A) \(\frac{3}{8}\) B) \(\frac{1}{4}\)
Now, to find the type of drill bit which is used to enlarge the hole, compare \(\frac{5}{16}\) with the given 2 drill bits
Now,
A) \(\frac{3}{8}\) and \(\frac{5}{16}\)
So, to make the denominators equal, \(\frac{3}{8}\) with 2.
So,
\(\frac{3}{8}\)= \(\frac{6}{16}\)
Hence,
\(\frac{6}{16}\) > \(\frac{5}{16}\)
Now,
B) \(\frac{5}{16}\) and \(\frac{1}{4}\)
So, to make the denominators equal, multiply \(\frac{1}{4}\) with 4
So,
\(\frac{1}{4}\) = \(\frac{4}{16}\)
Hence,
\(\frac{5}{16}\) < \(\frac{4}{16}\)
Hence, from the above two,
We can conclude that we can use \(\frac{3}{8}\) – inch bit drill to enlarge a hole which is made by \(\frac{5}{16}\) – inch bit drill

Question 13.
DIG DEEPER!
Order the animals from lightest to heaviest.
Big Ideas Math Answer Key Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.5 17
Answer: The order of animals from the lightest to heaviest is: Elk < Grizzly Bear < Mouse

Explanation:
The given table is:
Big Ideas Math Answer Key Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.5 17
So,
The weights of the animals from the above table are:
Mouse: \(\frac{4}{5}\)
Elk: \(\frac{1}{3}\)
Grizzly bear: \(\frac{3}{8}\)
To find the weights of all the 3 animals, we have to make the denominators of the three equal.
So, to make the denominators equal, multiply the three fractions by 120.
So, we will get
The weight of Mouse is:
\(\frac{4}{5}\) = \(\frac{96}{120}\)
The weight of Elk is:
\(\frac{1}{3}\) = \(\frac{40}{120}\)
The weight of Grizzly bear is:
\(\frac{3}{8}\) = \(\frac{45}{120}\)
Hence, from the above,
We can say that the order of the weights of the animals from the lightest to heaviest is: Elk < Grizzly Bear < Mouse

Compare Fractions Homework & Practice 7.5

Compare. Use model to help.
Question 1.
Big Ideas Math Answer Key Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.5 18
Answer: \(\frac{3}{10}\) is greater than \(\frac{1}{5}\)

Explanation:
The given two fractions are: \(\frac{3}{10}\) and \(\frac{1}{5}\)
So,
Let make the denominators equal so that we can compare the numerators
So,
To make the denominators equal, multiply the second fraction with 2
So,
\(\frac{1}{5}\) = \(\frac{2}{10}\)
So, when we compare the 2 fractions,
We can conclude that \(\frac{3}{10}\) is greater than \(\frac{1}{5}\)

Question 2.
Big Ideas Math Answer Key Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.5 19
Answer: \(\frac{4}{5}\) is greater than \(\frac{2}{3}\)

Explanation:
The given two fractions are: \(\frac{4}{5}\) and \(\frac{2}{3}\)
So,
Let make the denominators equal so that we can compare the numerators
So,
To make the denominators equal, multiply the first fraction with 3 and the second fraction with 5
So,
\(\frac{4}{5}\) = \(\frac{12}{15}\)
\(\frac{2}{3}\) = \(\frac{10}{15}\)
So, when we compare the 2 fractions,
We can conclude that \(\frac{4}{5}\) is greater than \(\frac{2}{3}\)

Question 3.
Big Ideas Math Answer Key Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.5 20
Answer: \(\frac{5}{8}\) is less than \(\frac{2}{1}\)

Explanation:
The given two fractions are: \(\frac{5}{8}\) and \(\frac{2}{1}\)
So,
Let make the denominators equal so that we can compare the numerators
So,
To make the denominators equal, multiply the second fraction with 8
So,
\(\frac{2}{1}\) = \(\frac{16}{8}\)
So, when we compare the 2 fractions,
We can conclude that \(\frac{5}{8}\) is less than \(\frac{2}{1}\)

Compare. Use a model to help.
Question 4.
Big Ideas Math Answer Key Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.5 21
Answer: \(\frac{9}{10}\) is less than \(\frac{97}{100}\)

Explanation:
The given two fractions are: \(\frac{9}{10}\) and \(\frac{97}{100}\)
So,
Let make the denominators equal so that we can compare the numerators
So,
To make the denominators equal, multiply the first fraction with 10
So,
\(\frac{9}{10}\) = \(\frac{90}{100}\)
So, when we compare the 2 fractions,
We can conclude that \(\frac{9}{10}\) is less than \(\frac{97}{100}\)

Question 5.
Big Ideas Math Answer Key Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.5 22
Answer: \(\frac{3}{8}\) is greater than \(\frac{2}{6}\)

Explanation:
The given two fractions are: \(\frac{3}{8}\) and \(\frac{2}{6}\)
So,
Let make the denominators equal so that we can compare the numerators
So,
To make the denominators equal, multiply the first fraction with 3 and the second fraction with 4
So,
\(\frac{3}{8}\) = \(\frac{9}{24}\)
\(\frac{2}{6}\) = \(\frac{8}{24}\)
So, when we compare the 2 fractions,
We can conclude that \(\frac{3}{8}\) is greater than \(\frac{2}{6}\)

Question 6.
Big Ideas Math Answer Key Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.5 23
Answer: \(\frac{1}{3}\) is equal to \(\frac{4}{12}\)

Explanation:
The given two fractions are: \(\frac{1}{3}\) and \(\frac{4}{12}\)
So,
Let make the denominators equal so that we can compare the numerators
So,
To make the denominators equal, multiply the first fraction with 4
So,
\(\frac{1}{3}\) = \(\frac{4}{12}\)
So, when we compare the 2 fractions,
We can conclude that \(\frac{1}{3}\) is equal to \(\frac{4}{12}\)

Question 7.
Big Ideas Math Answer Key Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.5 24
Answer: \(\frac{7}{2}\) is greater than \(\frac{6}{5}\)

Explanation:
The given two fractions are: \(\frac{7}{2}\) and \(\frac{6}{5}\)
So,
Let make the denominators equal so that we can compare the numerators
So,
To make the denominators equal, multiply the first fraction with 5 and the second fraction with 2
So,
\(\frac{7}{2}\) = \(\frac{35}{10}\)
\(\frac{6}{5}\) = \(\frac{12}{10}\)
So, when we compare the 2 fractions,
We can conclude that \(\frac{7}{2}\) is less than \(\frac{6}{5}\)

Question 8.
Big Ideas Math Answer Key Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.5 25
Answer: \(\frac{1}{10}\) is less than \(\frac{2}{12}\)

Explanation:
The given two fractions are: \(\frac{1}{10}\) and \(\frac{2}{12}\)
So,
Let make the denominators equal so that we can compare the numerators
So,
To make the denominators equal, multiply the first fraction with 12 and the second fraction with 10
So,
\(\frac{1}{10}\) = \(\frac{12}{120}\)
\(\frac{2}{12}\) = \(\frac{20}{120}\)
So, when we compare the 2 fractions,
We can conclude that \(\frac{1}{10}\) is less than \(\frac{2}{12}\)

Question 9.
Big Ideas Math Answer Key Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.5 26
Answer: \(\frac{3}{4}\) is greater than \(\frac{4}{6}\)

Explanation:
The given two fractions are: \(\frac{3}{4}\) and \(\frac{4}{6}\)
So,
Let make the denominators equal so that we can compare the numerators
So,
To make the denominators equal, multiply the first fraction with 3 and the second fraction with 2
So,
\(\frac{3}{4}\) = \(\frac{9}{12}\)
\(\frac{4}{6}\) = \(\frac{8}{12}\)
So, when we compare the 2 fractions,
We can conclude that \(\frac{3}{4}\) is greater than \(\frac{4}{6}\)

Question 10.
Structure
Compare \(\frac{3}{8}\) and \(\frac{1}{4}\) two different ways.

Answer: \(\frac{3}{8}\) is greater than \(\frac{1}{4}\)

Explanation:
The given two fractions are: \(\frac{3}{8}\) and \(\frac{1}{4}\)
First way of Comparison:
Let make the denominators equal so that we can compare the numerators
So,
To make the denominators equal, multiply the second fraction with 2
So,
\(\frac{1}{4}\) = \(\frac{2}{8}\)
So, when we compare the 2 fractions,
We can conclude that \(\frac{3}{8}\) is greater than \(\frac{1}{4}\)
Second Way of Comparison:
The given two fractions are: \(\frac{3}{8}\) and \(\frac{1}{4}\)
So,
Let make the numerators equal so that we can compare the denominators
So,
To make the numerators equal, multiply the second fraction with 3
So,
\(\frac{1}{4}\) = \(\frac{3}{4}\)
So, when we compare the 2 fractions,
We can conclude that \(\frac{3}{8}\) is greater than \(\frac{1}{4}\)

Question 11.
Modeling Real Life
A sailor is making a ship in a bottle. The last thing he needs to do is seal the bottle with a cork stopper. He tries a \(\frac{3}{4}\) -inch cork stopper, but it is too small. Should he try a \(\frac{1}{2}\)-inch cork stopper or a \(\frac{4}{5}\)-inch cork stopper next? Explain.
Big Ideas Math Answer Key Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.5 27
Answer: He had to use \(\frac{4}{5}\) -inch cork stopper

Explanation:
It is given that a sailor is making a ship in a bottle and he needs to seal the bottle after making the ship.
It is also given that he tried a \(\frac{3}{4}\) -inch cork stopper which is small
So, to make the big cork stopper so that it can close the bottle, there are 2 choices given.
The two choices are:
A) \(\frac{1}{2}\) B) \(\frac{4}{5}\)
Now,
Compare \(\frac{1}{2}\) with \(\frac{3}{4}\)
Now,
Let make the denominators equal so that we can compare the numerators
So,
To make the denominators equal, multiply the second fraction with 2
So,
\(\frac{1}{2}\) = \(\frac{2}{4}\)
So, when we compare the 2 fractions,
We can conclude that \(\frac{1}{2}\) is less than \(\frac{3}{4}\)
Now,
Compare \(\frac{4}{5}\) with \(\frac{3}{4}\)
Now,
Let make the denominators equal so that we can compare the numerators
So,
To make the denominators equal, multiply the  first fraction with 4 and the second fraction with 5
So,
\(\frac{4}{5}\) = \(\frac{16}{20}\)
\(\frac{3}{4}\) = \(\frac{15}{20}\)
So, when we compare the 2 fractions,
We can conclude that \(\frac{4}{5}\) is greater than \(\frac{3}{4}\)
Hence, from the above 2 comparisons,
We can conclude that we can use \(\frac{4}{5}\) – inch cork stopper.

Question 12.
DIG DEEPER!
Order the lengths of hair donated from greatest to least.
Big Ideas Math Answer Key Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.5 28

Answer: The order of lengths of hair donated from the greatest to the least is: C > B > A

Explanation:
The given table is:
Big Ideas Math Answer Key Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.5 28
From the above table,
The length of the hair donated by student A is: \(\frac{3}{4}\)
The length of the hair donated by student B is: \(\frac{11}{12}\)
The length of the hair donated by student C is: \(\frac{5}{6}\)
Now,
Let make the denominators equal so that we can compare the numerators
So,
To make the denominators equal, multiply the first fraction with 3, the second fraction with 1 and the third fraction with 2
So,
\(\frac{3}{4}\) = \(\frac{9}{12}\)
\(\frac{5}{6}\) = \(\frac{10}{12}\)
\(\frac{11}{12}\) = \(\frac{11}{12}\)
So, when we compare the 3 fractions,
We can conclude that the order of the lengths of hair donated from highest to least is: C > B > A

Review & Refresh

Question 13.
Extend the pattern of shapes by repeating the rule “triangle, pentagon, octagon.”What is the 48th shape in the pattern?
Big Ideas Math Answer Key Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.5 29
Answer: The 48th shape in the given pattern is: Octagon

Explanation:
The given pattern is:
Big Ideas Math Answer Key Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison 7.5 29
From the above pattern,
The given rule is: Triangle, pentagon, octagon
So,
The total number of figures in the given pattern = 3
So,
The 48th figure in the given pattern = 48 ÷ 3
Now,
By using the partial quotients method,
48 ÷ 3 = ( 30 + 18 ) ÷ 3
= ( 30 ÷ 3 ) + ( 18 ÷ 3 )
= 10 + 6
= 16 R 0
Hence, from the above,
We can conclude that the 48th figure in the given pattern is: Octagon

Understand Fraction Equivalence and Comparison Performance Task

Question 1.
a. Your art teacher wants you to complete the design below. Half of the squares are colored black. Complete the table. Then use the table to finish the design.

b.Which two colors cover the same portion of the design? Explain.

Answer: Green and Blue colors cover the same portion of the design
Explanation:
The teacher has given a board that has a total of 100 squares and according to the table, we have to complete the design of the squares.
The table is:

Hence, from the above table,
We can conclude that Blue and Yellow colors cover the same portion of the design.

Question 2.
Your teacher displays 30 designs in a rectangular array on the wall. Show two different ways your teacher can arrange the designs.

Answer: The  different ways that your teacher can arrange the designs are: 1 × 30, 2 × 15, 3 × 5, 5 × 3, 15 × 2, 30 × 1

Explanation:
It is given that your teacher displays 30 designs in a rectangular array on the wall.
So, we can arrange the 30 designs in different ways by using the factor pairs of 30
Now,
The factor pairs of 30 are: 1 × 30, 2 × 15, 3 × 5, 5 × 3, 15 × 2, 30 × 1
Hence, from the above,
We can conclude that the different ways to arrange the 30 designs in a rectangular array is: 1 × 30, 2 × 15, 3 × 5, 5 × 3, 15 × 2, 30 × 1

Understand Fraction Equivalence and Comparison Activity

Fraction Boss
Directions:
1. Divide the Fraction Boss Cards equally between both players.
2. Each player flips a Fraction Boss Card.
3. Players compare their fractions. The player with the greater fraction takes both cards.
4. If the fractions are equal, each player flips another card. Players compare their fractions. The player with the greater fraction takes all four cards.
5. The player with the most cards at the end of the round wins!
Big Ideas Math Answers 4th Grade Chapter 7 Understand Fraction Equivalence and Comparison 2

Understand Fraction Equivalence and Comparison Chapter Practice

7.1 Model Equivalent Fractions

Question 1.
Use the model to find an equivalent fraction for \(\frac{3}{4}\).
Big Ideas Math Answers 4th Grade Chapter 7 Understand Fraction Equivalence and Comparison chp 1
Answer: The equivalent fraction of \(\frac{3}{4}\) is: \(\frac{6}{8}\)

Explanation:
The given model is:
Big Ideas Math Answers 4th Grade Chapter 7 Understand Fraction Equivalence and Comparison chp 1
From the above model,
We can say that the fraction is: \(\frac{3}{4}\)
So,
The model for the Equivalent fraction is:

From the above model, we can say that the equivalent fraction is: \(\frac{6}{8}\)
So,
When we divide \(\frac{6}{8}\) by 2, we will get the equivalent value.
Hence, from the above,
We can conclude that \(\frac{3}{4}\) = \(\frac{6}{8}\)

Question 2.
Use the number line to find an equivalent fraction for \(\frac{2}{5}\).
Big Ideas Math Answers 4th Grade Chapter 7 Understand Fraction Equivalence and Comparison chp 2
Answer: The equivalent fraction of \(\frac{2}{5}\) is: \(\frac{4}{10}\)

Explanation:
The given number line is:
Big Ideas Math Answers 4th Grade Chapter 7 Understand Fraction Equivalence and Comparison chp 2
From the above number line,
we can say that 2 lines represent a value. i.e., \(\frac{2}{5}\)
So,
The value of each line when divided into 2 parts represent: \(\frac{1}{10}\)
So,
The equivalent number line is:

So, from the above equivalent number line,
We can see that \(\frac{2}{5}\) is equal to \(\frac{4}{10}\)
Hence, from the above,
We can conclude that \(\frac{2}{5}\) = \(\frac{4}{10}\)

Question 3.
Open-Ended
Write two equivalent fractions to describe the portion of the apples that are red.
Big Ideas Math Answers 4th Grade Chapter 7 Understand Fraction Equivalence and Comparison chp 3
Answer:
The two equivalent fractions to describe the portion of apples that are red is:
\(\frac{4}{8}\) and \(\frac{1}{2}\)

Explanation:
The given model is:
Big Ideas Math Answers 4th Grade Chapter 7 Understand Fraction Equivalence and Comparison chp 3
From the above model,
We can say that,
The number of colored apples is: 4
The number of red apples is: 4
So,
The total number of apples are: 8
So,
The fraction form for the red apples is = \(\frac{Number of red apples}{Total number of apples}\)
= \(\frac{4}{8}\)
Equivalent form for the red apples:
Consider the colored apples as 1 group and the red apples as 1 group
So,
The number of red apples is: 1
The number of colored apples is: 1
So,
The total number of apples are: 2
So,
The fraction that the apples are red = \(\frac{Number of red apples}{Total number of apples}\)
= \(\frac{1}{2}\)
Hence, from the above,
We can say that
The two equivalent fractions to describe the portion of apples that are red are:
\(\frac{4}{8}\) and \(\frac{1}{2}\)

7.2 Generate Equivalent Fractions by Multiplying

Find the equivalent fraction.
Question 4.
Big Ideas Math Answers 4th Grade Chapter 7 Understand Fraction Equivalence and Comparison chp 4
Answer: The equivalent fraction of \(\frac{3}{4}\) is: \(\frac{9}{12}\)

Explanation:
The given fraction is \(\frac{3}{4}\)
From the above fraction, the numerator and denominator are: 3 and 4
It is given that the denominator value is 12.
So, to make the value 12 from 4, we have to multiply the fraction with 3
So,
\(\frac{3}{4} is multiplied with [latex]\frac{3}{3}.
Hence,

Question 5.
Big Ideas Math Answers 4th Grade Chapter 7 Understand Fraction Equivalence and Comparison chp 5
Answer: The equivalent fraction of [latex]\frac{1}{2}\) is: \(\frac{5}{10}\)

Explanation:
The given fraction is \(\frac{1}{2}\)
From the above fraction, the numerator and denominator are: 1 and 2
It is given that the numerator value is 5.
So, to make the value 5 from 1, we have to multiply the fraction with 5
So,
\(\frac{1}{2} is multiplied with [latex]\frac{5}{5}.
Hence,

Question 6.
Big Ideas Math Answers 4th Grade Chapter 7 Understand Fraction Equivalence and Comparison chp 6
Answer: The equivalent fraction of [latex]\frac{8}{5}\) is: \(\frac{160}{100}\)

Explanation:
The given fraction is \(\frac{8}{5}\)
From the above fraction, the numerator and denominator are: 8 and 5
It is given that the denominator value is 100.
So, to make the value 100 from 5, we have to multiply the fraction with 20
So,
\(\frac{8}{5} is multiplied with [latex]\frac{20}{20}.
Hence,

Find an equivalent fraction.
Question 7.
[latex]\frac{1}{6}\)
Answer: The equivalent fraction of \(\frac{1}{6}\) is: \(\frac{2}{12}\)

Explanation:
When we multiply the fraction with the fraction having the same number as the numerator as the denominator, we will get the equivalent fraction.
So,
\(\frac{1}{6}\) is multiplied with \(\frac{2}{2}\)
Hence,
\(\frac{1}{6}\) = \(\frac{2}{12}\)

Question 8.
\(\frac{5}{5}\)
Answer: The equivalent fraction of \(\frac{5}{5}\) is: \(\frac{10}{10}\)

Explanation:
When we multiply the fraction with the fraction having the same number as the numerator as the denominator, we will get the equivalent fraction.
So,
\(\frac{5}{5}\) is multiplied with \(\frac{2}{2}\)
Hence,
\(\frac{5}{5}\) = \(\frac{10}{10}\)

Question 9.
\(\frac{1}{4}\)
Answer: The equivalent fraction of \(\frac{1}{4}\) is: \(\frac{2}{8}\)

Explanation:
When we multiply the fraction with the fraction having the same number as the numerator as the denominator, we will get the equivalent fraction.
So,
\(\frac{1}{4}\) is multiplied with \(\frac{2}{2}\)
Hence,
\(\frac{1}{4}\) = \(\frac{2}{8}\)

Find two equivalent fractions.
Question 10.
\(\frac{3}{2}\)
Answer:
The two equivalent fractions of \(\frac{3}{2}\) is: \(\frac{6}{4}\) and \(\frac{9}{6}\)

Explanation:
When we multiply the fraction with the fraction having the same number as the numerator as the denominator, we will get the equivalent fraction.
So,
\(\frac{3}{2}\) is multiplied with \(\frac{2}{2}\) and \(\frac{3}{3}\)
Hence,
\(\frac{3}{2}\) = \(\frac{6}{4}\) and \(\frac{9}{6}\)

Question 11.
\(\frac{1}{3}\)
Answer:
The two equivalent fractions of \(\frac{1}{3}\) is: \(\frac{2}{6}\) and \(\frac{3}{9}\)

Explanation:
When we multiply the fraction with the fraction having the same number as the numerator as the denominator, we will get the equivalent fraction.
So,
\(\frac{1}{3}\) is multiplied with \(\frac{2}{2}\) and \(\frac{3}{3}\)
Hence,
\(\frac{1}{3}\) = \(\frac{2}{6}\) and \(\frac{3}{9}\)

Question 12.
\(\frac{4}{5}\)
Answer:
The two equivalent fractions of \(\frac{4}{5}\) is: \(\frac{8}{10}\) and \(\frac{12}{15}\)

Explanation:
When we multiply the fraction with the fraction having the same number as the numerator as the denominator, we will get the equivalent fraction.
So,
\(\frac{4}{5}\) is multiplied with \(\frac{2}{2}\) and \(\frac{3}{3}\)
Hence,
\(\frac{4}{5}\) = \(\frac{8}{10}\) and \(\frac{12}{15}\)

7.3 Generate Equivalent Fractions by Dividing

Find the equivalent fraction.
Question 13.
Big Ideas Math Answers 4th Grade Chapter 7 Understand Fraction Equivalence and Comparison chp 13
Answer: The equivalent fraction of \(\frac{3}{12}\) is: \(\frac{1}{4}\)

Explanation:
The given fraction is: \(\frac{3}{12}\)
From the above fraction, the numerator and denominator are: 3 and 12
3 and 12 are the multiples of 3.
So,
\(\frac{3}{12}=\frac{3 \div 3}{12 \div 3}=\frac{1}{4}\)
Hence,

Question 14.
Big Ideas Math Answers 4th Grade Chapter 7 Understand Fraction Equivalence and Comparison chp 14
Answer: The equivalent fraction of \(\frac{18}{100}\) is: \(\frac{9}{50}\)

Explanation:
The given fraction is: \(\frac{18}{100}\)
From the above fraction, the numerator and denominator are: 18 and 100
18 and 100 are the multiples of 2.
So,
\(\frac{18}{100}=\frac{18 \div 2}{100 \div 2}=\frac{9}{50}\)
Hence,

Question 15.
Big Ideas Math Answers 4th Grade Chapter 7 Understand Fraction Equivalence and Comparison chp 15
Answer: The equivalent fraction of \(\frac{20}{10}\) is: \(\frac{4}{2}\)

Explanation:
The given fraction is: \(\frac{20}{10}\)
From the above fraction, the numerator and denominator are: 20 and 10
20 and 10 are the multiples of 5. ( 20 and 10 are also multiples of 2 and 10 but in the denominator, 2i s given which can only be obtained by divideing 10 with 5 )
So,
\(\frac{20}{10}=\frac{20 \div 5}{10 \div 5}=\frac{4}{2}\)
Hence,

Find an equivalent fraction.
Question 16.
\(\frac{4}{6}\)
Answer: The equivalent fraction of \(\frac{4}{6}\) is: \(\frac{2}{3}\)

Explanation:
The given fraction is: \(\frac{4}{6}\)
From the above fraction, the numerator and denominator are: 4 and 6
4 and 6 are the multiples of 2. ( 4 and 6 are multiples of 2 )
So,
\(\frac{4}{6}=\frac{4 \div 2}{6 \div 2}=\frac{2}{3}\)
Hence,
The equivalent fraction of \(\frac{4}{6}\) is: \(\frac{2}{3}\)

Question 17.
\(\frac{16}{4}\)
Answer: The equivalent fraction of \(\frac{16}{4}\) is: \(\frac{8}{2}\)

Explanation:
The given fraction is: \(\frac{16}{4}\)
From the above fraction, the numerator and denominator are: 16 and 4
16 and 4 are the multiples of 2.
So,
\(\frac{16}{4}=\frac{16 \div 2}{4 \div 2}=\frac{8}{2}\)
Hence,
The equivalent fraction of \(\frac{16}{4}\) is: \(\frac{8}{2}\)

Question 18.
\(\frac{20}{8}\)
Answer: The equivalent fraction of \(\frac{20}{8}\) is: \(\frac{10}{4}\)

Explanation:
The given fraction is: \(\frac{20}{8}\)
From the above fraction, the numerator and denominator are: 20 and 8
20 and 8 are the multiples of 2.
So,
\(\frac{20}{8}=\frac{20 \div 2}{8 \div 2}=\frac{10}{4}\)
Hence,
The equivalent fraction of \(\frac{20}{8}\) is: \(\frac{10}{4}\)

Find two equivalent fractions.
Question 19.
\(\frac{80}{100}\)
Answer: The two equivalent fractions of \(\frac{80}{100}\) is: \(\frac{4}{5}\) and \(\frac{20}{25}\)

Explanation:
The given fraction is: \(\frac{80}{100}\)
From the above fraction, the numerator and denominator are: 80 and 100
80 and 100 are the multiples of 20 and 4
So,
We have to divide \(\frac{80}{100}\) with 20and 4
So,
\(\frac{80}{100}=\frac{80 \div 20}{100 \div 20}=\frac{4}{5}\)
\(\frac{80}{100}=\frac{80 \div 4}{100 \div 4}=\frac{20}{25}\)
Hence,
The two equivalent fractions of \(\frac{80}{100}\) is: \(\frac{4}{5}\) and \(\frac{20}{25}\)

Question 20.
\(\frac{6}{12}\)
Answer: The two equivalent fractions of \(\frac{6}{12}\) is: \(\frac{2}{4}\) and \(\frac{1}{2}\)

Explanation:
The given fraction is: \(\frac{6}{12}\)
From the above fraction, the numerator and denominator are: 6 and 12
6 and 12 are the multiples of 3 and 6
So,
We have to divide \(\frac{6}{12}\) with 3 and 6
So,
\(\frac{6}{12}=\frac{6 \div 3}{12 \div 3}=\frac{2}{4}\)
\(\frac{6}{12}=\frac{6 \div 6}{12 \div 6}=\frac{1}{2}\)
Hence,
The two equivalent fractions of \(\frac{6}{12}\) is: \(\frac{2}{4}\) and \(\frac{1}{2}\)

Question 21.
\(\frac{40}{4}\)
Answer: The two equivalent fractions of \(\frac{40}{4}\) is: \(\frac{20}{2}\) and \(\frac{10}{1}\)

Explanation:
The given fraction is: \(\frac{40}{4}\)
From the above fraction, the numerator and denominator are: 40 and 4
40 and 4 are the multiples of 2 and 4
So,
We have to divide \(\frac{40}{4}\) with 2 and 4
So,
\(\frac{40}{4}=\frac{40 \div 2}{4 \div 2}=\frac{20}{2}\)
\(\frac{40}{4}=\frac{40 \div 4}{4 \div 4}=\frac{10}{1}\)
Hence,
The two equivalent fractions of \(\frac{40}{4}\) is: \(\frac{20}{2}\) and \(\frac{10}{1}\)

Question 22.
Modeling Real Life
You have 90$. What fraction of a dollar, in tenths, do you have?
Answer: The fraction of a dollar, in tenths, is: \(\frac{1}{10}\)

Explanation:
It is given that you have 90$
So,
The fraction of a dollar, in tenths that pack their lunch = The number of dollars ÷ The total number of dollars
= 90 ÷ 100
= \(\frac{9}{10}\)
Hence, from the above,
We can conclude that the number of dollars, in tenths, is: \(\frac{9}{10}\)

7.4 Compare Fractions Using Benchmarks

Compare. Use a model to help.
Question 23.
Big Ideas Math Answers 4th Grade Chapter 7 Understand Fraction Equivalence and Comparison chp 23
Answer: \(\frac{7}{8}\) is greater than \(\frac{2}{5}\)

Explanation:
The given fractions are: \(\frac{7}{8}\) and \(\frac{2}{5}\)
For the two fractions, the models are:

So,
Let make the denominators equal so that we can compare the numerators
So,
To make the denominators equal, multiply the first fraction with 5 and the second fraction with 8
So,
\(\frac{7}{8}\) = \(\frac{35}{40}\)
\(\frac{2}{5}\) = \(\frac{16}{40}\)
So, when we compare the 2 fractions,
We can conclude that \(\frac{7}{8}\) is greater than \(\frac{2}{5}\)

Question 24.
Big Ideas Math Answers 4th Grade Chapter 7 Understand Fraction Equivalence and Comparison chp 24
Answer: \(\frac{6}{10}\) is less than \(\frac{4}{3}\)

Explanation:
The given fractions are: \(\frac{6}{10}\) and \(\frac{4}{3}\)
For the two fractions, the models are:

So,
Let make the denominators equal so that we can compare the numerators
So,
To make the denominators equal, multiply the first fraction with 3 and the second fraction with 10
So,
\(\frac{6}{10}\) = \(\frac{18}{30}\)
\(\frac{4}{3}\) = \(\frac{40}{30}\)
So, when we compare the 2 fractions,
We can conclude that \(\frac{6}{10}\) is less than \(\frac{4}{3}\)

Question 25.
Big Ideas Math Answers 4th Grade Chapter 7 Understand Fraction Equivalence and Comparison chp 25
Answer: \(\frac{1}{6}\) is equal to \(\frac{2}{12}\)

Explanation:
The given fractions are: \(\frac{1}{6}\) and \(\frac{2}{12}\)
For the two fractions, the models are:

So,
Let make the denominators equal so that we can compare the numerators
So,
To make the denominators equal, multiply the first fraction with 2
So,
\(\frac{1}{6}\) = \(\frac{2}{12}\)
So, when we compare the 2 fractions,
We can conclude that \(\frac{1}{6}\) is equal to \(\frac{2}{12}\)

7.5 Compare Fractions

Compare. Use a model to help.
Question 26.
Big Ideas Math Answers 4th Grade Chapter 7 Understand Fraction Equivalence and Comparison chp 26
Answer: \(\frac{1}{4}\) is equal to \(\frac{3}{12}\)

Explanation:
The given two fractions are: \(\frac{1}{4}\) and \(\frac{3}{12}\)
So,
Let make the denominators equal so that we can compare the numerators
So,
To make the denominators equal, multiply the first fraction with 3
So,
\(\frac{1}{4}\) = \(\frac{3}{12}\)
So, when we compare the 2 fractions,
We can conclude that \(\frac{1}{4}\) is equal to \(\frac{3}{12}\)

Question 27.
Big Ideas Math Answers 4th Grade Chapter 7 Understand Fraction Equivalence and Comparison chp 27
Answer: \(\frac{2}{3}\) is greater than \(\frac{6}{10}\)

Explanation:
The given two fractions are: \(\frac{2}{3}\) and \(\frac{6}{10}\)
So,
Let make the denominators equal so that we can compare the numerators
So,
To make the denominators equal, multiply the first fraction with 10 and the second fraction with 3
So,
\(\frac{2}{3}\) = \(\frac{20}{30}\)
\(\frac{6}{10}\) = \(\frac{18}{30}\)
So, when we compare the 2 fractions,
We can conclude that \(\frac{2}{3}\) is greater than \(\frac{6}{10}\)

Question 28.
Big Ideas Math Answers 4th Grade Chapter 7 Understand Fraction Equivalence and Comparison chp 28
Answer: \(\frac{7}{8}\) is greater than \(\frac{5}{6}\)

Explanation:
The given two fractions are: \(\frac{7}{8}\) and \(\frac{5}{6}\)
So,
Let make the denominators equal so that we can compare the numerators
So,
To make the denominators equal, multiply the first fraction with 6 and the second fraction with 8
So,
\(\frac{7}{8}\) = \(\frac{42}{48}\)
\(\frac{5}{6}\) = \(\frac{40}{48}\)
So, when we compare the 2 fractions,
We can conclude that \(\frac{7}{8}\) is greater than \(\frac{5}{6}\)

Understand Fraction Equivalence and Comparison of Cumulative Practice

Question 1.
Which number is a common factor of 12, 16, and 40?
A. 5
B. 8
C. 3
D. 4
Answer: D) 4

Explanation:
Given numbers are: 12, 16 and 40
The factors are the numbers that divide the original number completely.
So,
The factors of 12 are: 1, 2, 3, 4, 6, 12
The factors of 16 are: 1, 2, 4, 8, 16
The factors of 40 are: 1, 2, 4, 5, 8, 10, 20, 40
So,
The common factors of 12, 16, and 20 are: 1, 2, 4

Question 2.
Which statements describe the difference between 77,986 and 21,403?
Big Ideas Math Answers Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison cp 2

Answer: The given numbers are: 77,986 and 21,403
So,
The difference between 77,986 and 21,403 is: 56,583
The below statements describe the difference between 77,986 and 21,403

Question 3.
Which product is between 5,050 and 5,100?
A. 652 × 8
B. 566 × 9
C. 1,023 × 5
D. 1,257 × 4

Answer: B) 566 × 9

Explanation:
Let the given product Expressions be named as  A), B), C) and D)
So,
The given products are:
A) 652 × 8   B) 566 × 9  C) 1,023 × 5  D) 1,257 × 4
Now,
By using the partial products method,
A. 652 × 8 = 5,216
B. 566 × 9 = 5,094
C. 1,023 × 5 = 5,115
D. 1,257 × 4 = 5,028
So,
From the above values,
The product that is between 5,050 and 5,100 is: 566 × 9

Question 4.
Use the advertisement to answer the question.
Big Ideas Math Answers Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison cp 4
What is the greatest number of miles that the car could have been driven?
A. 84,699
B. 84,750
C. 84,749
D. 84,650
Answer: The greatest number of miles that the car should have been driven is: 84,699 miles

Explanation:
Given that the car should have been driven approximate 84,700 miles ( rounded to the nearest 100 )
The given options are: A) 84,699 B) 84,750 C) 84,749 D) 84,650
The greatest number that is close to 84,700 is: 84, 699
Hence, from the above,
We can conclude that the greatest number of miles that the car should have been driven is: 84,699 miles

Question 5.
Which multiplication expressions could be represented by the area model?

Answer: The multiplication Expression represented by the area model is: 84 × 34

Explanation:
The given area model is:

So, from the area model,
The factor pairs of 2,400 are: 80 × 30, 60 × 40
The factor pairs of 120 are: 4 × 30, 6 × 20
From the above two factors, we can say that
The area model can be formed from 80, 4 and 30
Hence,
The product for the given area model is: 84 × 34

Question 6.
Your neighbor buys yogurt in packages of 4. If your neighbor only buys complete packages, how many yogurts could he have bought?
Big Ideas Math Answers Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison cp 6
A. 2
B. 20
C. 34
D. 18
Answer: The complete packages of yogurts that he could have brought is: 20

Explanation:
It is given that your neighbor bought yogurt in packages of 4
So, for complete packages, the number of packages that he takes should not have any extra yogurt i.e., remainder yogurt when packing completely in the packs of 4.
So,
From the above options,
We have only 20 yogurts that can be packed completely in packs of 4.

Question 7.
Which fraction makes the statement true?
Big Ideas Math Answers Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison cp 7
Answer: \(\frac{1}{4}\) makes the statement true.

Explanation:
Given fraction is: \(\frac{6}{8}\)
So,
We have to compare \(\frac{6}{8}\) with the given options to satisfy the relation the option < \(\frac{6}{8}\)
Now,
A) The two fractions are:  \(\frac{1}{4}\) and \(\frac{6}{8}\)
So, to make the denominators equal, multiply the first fraction with 2
So,
\(\frac{1}{4}\) = \(\frac{2}{8}\)
Hence,
\(\frac{1}{4}\) < \(\frac{6}{8}\)

Question 8.
Which division equation is represented by the counters?
Big Ideas Math Answers Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison cp 8
Answer: D) 26 ÷ 6 = 4 R 2

Explanation:
Given counters are:

From the above model,
The total number of counters are: 6
The number of parts that each counter has: 4
The number of parts that has leftover is: 2
So,
The total number of parts in 6 counters = 4 + 4 + 4 + 4 + 4 + 4 = 24
The total number of parts = The number of parts in 6 counters + The number of parts leftover = 24 + 2 = 26
Hence, from the above, the division Expression can be represented as:
26 ÷ 6 = 4 R 2

Question 9.
Which statements are true?
Big Ideas Math Answers Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison cp 9
Answer: The below statements are true:

Prime numbers:
The numbers that have exactly 2 factors 1 and itself are ” Prime numbers”
Composite numbers:
The numbers that have more than 2 factors are “Composite numbers”

Question 10.
What is the quotient of 3,258 and 3?
Big Ideas Math Answers Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison cp 10
Answer: The quotient of 3,258 and 3 is: 1,086

Explanation:
By using the partial quotients method,
3,258 ÷ 3 = ( 3,000 + 240 + 18 ) ÷ 3
= ( 3,000 ÷ 3 ) + ( 240 ÷ 3 ) + ( 18 ÷ 3 )
= 1,000 + 80 + 6
= 1,086
Hence, from the above,
The quotient of 3,258 and 3 is: 1,086

Question 11.
Which model shows an equivalent fraction for the fraction shown by the model below?
Big Ideas Math Answers Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison cp 11
Answer: The equivalent fraction for the given model in shown in Option C)

Explanation:
From the given model,
The fraction is:\(\frac{3}{6}\)
From\(\frac{3}{6}\),
The number of colored parts is: 3
The total number of parts are: 6
So,
As 3 and 6 are the multiples of 3, divide \(\frac{3}{6}\) with 3
So, we get
\(\frac{3}{6}\) = \(\frac{1}{2}\)
So,
The equivalent fraction of \(\frac{3}{6}\) is \(\frac{1}{2}\) which is shown in the model of Option C)

Question 12.
What is the product of 68 and 45?
A. 8,420
B. 3,060
C. 612
D. 27,360
Answer: 68 × 45 = 3,060

Explanation:
By using the partial products method,
68 × 45 = ( 60 + 8 ) × ( 40 + 5 )
= ( 60 × 40 ) + ( 8 × 5 ) + ( 60 × 5 ) + ( 8 × 40 )
= 2,400 + 40 + 300 + 320
= 3,060
Hence,
68 × 45 = 3,060 which is Option B)

Question 13.
Which pattern uses the same rule as the pattern below?
2, 10, 18, 26, 34, 42
A. 15, 23, 31, 39, 47, 55
B. 5, 25, 125, 625, 3, 125, 15, 625,
C. 70, 62, 54, 46, 38, 30
D. 25, 34, 43, 52, 61, 70
Answer: Option A) follows the same pattern as the given pattern

Explanation:
Given pattern is: 2, 10, 18, 26, 34, 42
From the above pattern,
The rule is: Add 8
So,
By checking the options,
We can conclude that option A) satisfies the same rule as in the given pattern

Question 14.
Which statements are true?
Big Ideas Math Answers Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison cp 12
Answer: The statements that are true are:

Explanation:
Compare the given fractions by making either both the denominators or the numerators equal.

Question 15.
Part A At a summer camp, 67 students are in a line to rent kayaks. Each kayak can hold 4 people. How many kayaks will be full?
Think
Solve
Explain
Big Ideas Math Answers Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison cp 15
Part B How many kayaks will be used?
Part C How many students will be in the last kayak? Explain.

Answer:
Part A: 16 kayaks will be full
Part B: 17 kayaks will be used
Part C: 3 students will be in the last kayak

Explanation:
Given that there are 67 students standing in a line to rent a kayak and each kayak holds 4 people.
So,
The number of kayaks is: 67 ÷ 4
Now,
67 ÷ 4 = ( 60 + 4 ) ÷ 4
= ( 60 ÷ 4 ) + ( 4 ÷ 4 )
= 15 + 1
= 16 R 3
Hence, from the above
The number of kayaks that are full is: 16
The number of kayaks that will be used is: 17
The number of students that will be in the last kayak is: 3

Question 16.
How many zeros will the product of 80 and 50 have?
A. 1
B. 2
C. 3
D. 4
Answer: The product of 80 and 50 will have 3 zeroes

Explanation:
By using the place-value method,
80 × 50 = 8 tens × 50
= 400 tens
= 4,000
Hence,
The number of zeroes in 80 × 50 is: 3

Question 17.
There are 57 electronic books checked out of a library. There are 8 times as many printed books checked out as electronic books. How many total books are checked out of the library?
A. 513
B. 4,113
C. 122
D. 456
Answer: A) The total number of books that are checked out of the library: 513

Explanation:
It is given that there are 57 electronic books checked out of a library.
It is also given that there are 8 times as many printed books checked out as electronic books
So,
The total number of printed books = 57 × 8 = 456
Hence,
The total number of books = The number of electronic books + The number of printed books
= 57 + 456
= 513 books

Question 18.
Which number is equal to 100,000 + 5,000 + 80 + 4?
A. 1,584
B. 105084
C. 15,084
D. 105,840
Answer: B) 1,05,084

Explanation:
The given Expanded form is: 100,000 + 0 + 5,000 + 0 + 80 + 4
Hence,
The standard form is: 105,084

Understand Fraction Equivalence and Comparison STEAM Performance Task

Question 1.
Sea level is the average level of the oceans on Earth. The global sea level is rising about \(\frac{1}{8}\) inch each year.
a. If this pattern continues, how much will the sea level rise in 80 years?
b. You read from another source that the sea level is1rising about \(\frac{1}{2}\) inch every 4 years. Did your source use the same fact as above? Explain.
Big Ideas Math Answers Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison spt 1
c. In 10 years, will the sea level rise more or less than an inch? Explain.
Big Ideas Math Answers Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison spt 2
d. How many years will it take the sea level to rise about 1 foot?
e. Use the Internet or some other resource to learn about rising sea levels. Write one interesting fact that you learn.
Answer:
a) The amount of sea-level rise in 80 years is: 10 inches
b) Yes, your source does not use the same fact as above
c) In 10 years, the sea level will be an inch.
d) The amount of years that the sea-level will rise to about 1 foot is: 96 years
e) The sea-level will be more in the coastal areas

Explanation:
It is given that the global sea-level is rising about \(\frac{1}{8}\)  inch every year
So,
The amount of sea-level that will rise in 80 years = 80 × \(\frac{1}{8}\) = 10 inches
It is also given that another Internet source given that the sea-level is rising about \(\frac{1}{4}\) inches every 4 years
So,
The amount of sea-level that will rise every year = \(\frac{1}{4}\) × \(\frac{1}{4}\) = \(\frac{1}{16}\)
So,
Your source does not use the same source as above
We know that,
1 inch = 12 foot
But from above,
We can see that the amount of sea-level that will rise in 80 years in only 10 inch
So,
The amount os sea-level will be less than an inch in 10 year
The number of years that will take for the sea-level that will rise about 1 foot = 96 years

Question 2.
The gravitational pull of the moon affects the high and low tides of the oceans on Earth. Cities along the coasts use tide tables each day. Use the tide table to answer the questions.
Big Ideas Math Answers Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison spt 3
a. When swimming, why is it important to understand tide tables?
Answer: When swimming, the tide tables is important because we can’t swim when there will be high tides. Otherwise, there is a chance, that we will be suffocated by water and die

b. What are the common factors of the water heights?
Answer: The common factors of water heights are: 1, 2, 4
c. Make a picture graph of the water heights.
Big Ideas Math Answers Grade 4 Chapter 7 Understand Fraction Equivalence and Comparison spt 4
d. What pattern do you notice about the water heights? Explain.
Answer: All water heights are the multiples of 4

Conclusion:
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