Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120

Big Ideas Math Answers Grade 1 Chapter 6

Download Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120 pdf for free from here. The solutions for all the questions are given in a simple manner by subject experts. The Chapter Count and Write Numbers to 120 includes Count to 120 by Ones, Count to 120 by Tens, Count and Write Numbers to 120, etc. This chapter discusses the methods to count the numbers to 120. Go through the below sections to find topic-wise links.

Big Ideas Math Book 1st Grade Answer Key Chapter 6 Count and Write Numbers to 120

To make your preparation perfect you have to follow the best guide. Big Ideas Math Book 1st Grade Answer Key Chapter 6 Count and Write Numbers to 120 is the best study material which consists of clear-cut explanations for all the problems. So, access the links and start your preparation for the exams.

Vocabulary

Lesson: 1 Count to 120 by Ones

Lesson: 2 Count to 120 by Tens

Lesson: 3 Compose Numbers 11 to 19

Lesson: 4 Tens

Lesson: 5 Tens and Ones

Lesson: 6 Make Quick Sketches

Lesson: 7 Understand Place Value

Lesson: 8 Write Numbers in Different Ways

Lesson: 9 Count and Write Numbers to 120

Chapter: 6 – Count and Write Numbers to 120

Count and Write Numbers to 120 Vocabulary

Organize It

Review Words:
column
decade numbers
hundred chart
row

Use the review words to complete the graphic organizer.

Big Ideas Math Answer Key Grade 1 Chapter 6 Count and Write Numbers to 120 1
Answer:
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120-1

Explanation:
The given graphic organizer title is called as hundred chart,
and numbers 41 to 50 are called as row, numbers 4,14,24
to 84, 94 are called as column and numbers 10,20,30 to 90,100
are called decade numbers.

Define It

Use your vocabulary cards to identify the words.

Big Ideas Math Answer Key Grade 1 Chapter 6 Count and Write Numbers to 120 2

Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120-2

Explanation:
Given in the picture to identify words 2 – Two and 3 – Three.

Lesson 6.1 Count to 120 by Ones

Explore and Grow

Point to each number as you count to 120. Color the first two rows and the last two rows. How are the rows the same? How are they different?
Big Ideas Math Answer Key Grade 1 Chapter 6 Count and Write Numbers to 120 3
Answer:
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120-3

Explanation:
As we move from 1,2,3 count to 120  have pointed with an arrow for each number. Colored with green first two rows from 1 to 10, 11 to 20 and the last two rows from 101 to 110, 111 to 120.
The rows are same – as we move from number 1 to 120 from first number to second number the number is increased by 1 as 1, 1 + 1 = 2, 2 + 1 = 3 and so on 119 + 1 = 120 means the next number is addition of 1 plus the previous number.
The rows are different – as we see the numbers are not the same they differ its 1,2 ,3 to 119, 120 from 1 to 120.

Show and Grow

Count by ones to write the missing numbers.

Question 1.
82, _____83_____, ____84____, _____85_____, ____86______, ____87______
Answer:
The missing numbers after 82 are  83, 84, 85, 86 ,87

Explanation:
Given the starting number as 82 we count by ones and
write the next missing numbers as  82 + 1 = 83,
83 + 1 = 84, 84 + 1 = 85 , 85 +1 = 86, 86 + 1 = 87,
now the missing numbers after 82  are 83, 84, 85, 86 ,87.

Question 2.
103, ____104______, ____105______, ___106_______, ____107______, ____108______
Answer:
The missing numbers after 103 are 104,105,106,107,108.

Explanation:
Given the starting number as 103 we count by ones and
write the next missing numbers as 103 + 1  = 104,
104 + 1 =105, 105 + 1 = 106, 106 + 1 = 107,107 + 1 = 108,
now the missing numbers after 103 are 104,105,106,107,108.

Apply and Grow: Practice

Count by ones to write the missing numbers.

Question 3.
56, ____57______, ____58____, ____59______, ____60______, ____61______
Answer:
The missing numbers after 56 are 57,58,59,60,61

Explanation:
Given the starting number as 56 we count by ones and
write the next missing numbers as 56 + 1 = 57, 57 + 1 = 58,
58 + 1 = 59, 59 + 1 = 60, 60 + 1 = 61,
now the missing numbers after 56 are 57,58,59,60,61.

Question 4.
98, _____99_____, ____100______, ____101______, ____102______, ____103______
Answer:
The missing numbers after 98 are 99,100,101,102,103

Explanation:
Given the starting number as 98 we count by ones and
write the next missing numbers as 98 + 1 = 99,99 + 1 = 100,
100 + 1 = 101, 101 + 1 = 102, 102 + 1 = 103,
So the missing numbers after 98 are 98,99,100,101,102,103.

Question 5.
115, ____116______, ____117______, ____118______, ____119______, ___120_______
Answer:
The missing numbers after 115 are 116,117,118,119,120

Explanation:
Given the starting number as 115 we count by ones and
write the next missing numbers as 115 + 1 = 116,
116 + 1 = 117, 117 + 1 = 118, 118 + 1 = 119, 119 + 1 = 120,
So the missing numbers after 115 are 116,117,118,119,120.

Question 6.
___40_______, ____41______, 42, ____43______, ____44______, ____45______
Answer:
The missing numbers before and after 42 are 40,41,43,44,45

Explanation:
Given number is 42 and 2 numbers before 42 are
count 1 before 42 is 42- 1= 41 and count 1 before 41 is
41 – 1 = 40 and count 1 after 42 is 42 + 1 = 43, 43 + 1 = 44,
44 + 1 = 45, So the missing numbers before and after 42 are 40,41,43,44,45.

Write the missing numbers in the chart.

Question 7.
Big Ideas Math Answer Key Grade 1 Chapter 6 Count and Write Numbers to 120 4
Answer:
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120-4
The missing numbers in the chart are 33,35,42,44.

Explanation:
The missing number after 32 is 32 + 1 = 33,
the number after 34 is 34 + 1 = 35,
the number before 43 is 43 – 1 = 42 and
number after 43 is 43 + 1 = 44, So the missing
numbers in the chart are 33,35,42,44.

Question 8.
Big Ideas Math Answer Key Grade 1 Chapter 6 Count and Write Numbers to 120 5
Answer:
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120-5
The missing numbers in the chart are 81,83,84,92,94.

Explanation:
The missing number before 82 is 82 – 1 = 81,
the number after 82 is 82 + 1 = 83,now number
after 83 is 83 + 1 = 84.
The number after 91 is 91 + 1 = 92,
the number after 93 is 93 + 1 = 94, therefore the
missing numbers in the chart are 81,83,84,92,94.

Question 9.
MP Structure
Structure Write a number between 95 and 105. Then count by ones to write the next 7 numbers.
____96___, _____97____, _____98_____, ____99____, ___ 100___, ____101___, ___102__, ___103___
Answer:
Let the number be 96 which is between 95 and 105 and
the next 7 numbers after 96 are 97,98,99,100,101,102,103.

Explanation:
Given to write a number between 95 and 105,
Let us take the number as 96, now we will write the
next 7 numbers are 96 + 1 = 97,97+ 1 = 98, 98 + 1 =99,
99 + 1 = 100,100 + 1 = 101, 101 + 1 = 102, 102 + 1 = 103,
therefore the next 7 numbers after 96 are 97,98,99,100,101,102,103.

Think and Grow: Modeling Real Life

You hove 108 bouncy balls. You want 112. How many more bouncy bails do you need?
Draw more bails to show 112:
Big Ideas Math Answer Key Grade 1 Chapter 6 Count and Write Numbers to 120 6

______4______ more bouncy balls
Answer:
4 more bouncy  balls are needed,
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120-5

Explanation:
Given I have 108 bouncy balls and I want 112,
So I need 108 + 1 = 109, 109 + 1 = 110, 110 + 1 = 111,
111+ 1 = 112 or 112 – 108 = 4, so we need 4 more bouncy  balls.

Show and Grow

Question 10.
You have 66 rocks. You want 75. How many more rocks do you need?
Draw more rocks to show 75:
Big Ideas Math Answer Key Grade 1 Chapter 6 Count and Write Numbers to 120 7

______9_______ more rocks
Answer:
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120-6
I need 9 more rocks.

Explanation:
Given I have 66 rocks and I want 75,
So more rocks I needed are 75 – 66 = 9.

Count to 120 by Ones Practice 6.1

Count by ones to write the missing numbers.

Question 1.
57, _____58_____, _____59_____, ____60______, ____61______, ____62______
Answer:
The missing numbers after 57 are 58,59,60,61,62.

Explanation:
Given the starting number as 57 we count by ones and
write the next missing numbers as 57 + 1 = 58,
58 + 1 = 59, 59 + 1 = 60, 60 + 1 = 61, 61 + 1 = 62,
So the missing numbers after 57 are 58,59,60,61,62.

Question 2.
109, ____110___, ____111___, ____112___, ____113___, __114_____
Answer:
The missing numbers after 109 are 110,111,112,113,114

Explanation:
Given the starting number as 109 we count by ones and
write the next missing numbers as 109 + 1 = 110,
110 + 1 =111,111 + 1 = 112, 112 + 1 = 113, 113 + 1 = 114,
So the missing numbers after 109 are 110,111,112,113,114.

Question 3.
40, ____41___, ____42___, ___43___, ___44___, ___45___
Answer:
The missing numbers after 40 are 41,42,43,44,45

Explanation:
Given the starting number as 40 we count by ones and
write the next missing numbers as 40 + 1 = 41,
41 + 1 = 42,42 + 1 = 43, 43 + 1 = 44, 44 + 1 = 45,
So the missing numbers after 40 are 40,41,42,43,44,45.

Question 4.
_____97_____, ____98______, ____99______, 100, ____101______, ____102______
Answer:
The missing numbers before and after 100 are 97,98,99,101,102

Explanation:
Given number is 100 and 3 numbers before 100 are
count 1 before 100 is 100- 1= 99, count 1
before 99 is 99 – 1 = 98 and count 1 before 98 is
98 – 1 = 97, numbers after 100 are 100 + 1 = 101 and
101 + 1 = 102,therefore the missing numbers
before and after 100 are 97,98,99,101,102.

Write the missing numbers in the chart.

Question 5.
Big Ideas Math Answer Key Grade 1 Chapter 6 Count and Write Numbers to 120 8
Answer:
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120-7

The missing numbers in the chart are 21,23,30,32.

Explanation:
The missing number after 20 is 20 + 1 = 21,
the number after 22 is 22 + 1 = 23,
the number before 31 is 31 – 1 = 30 and
number before 33 is 33 – 1 = 32, So the missing
numbers in the chart are 21,23,30,32.

Question 6.
Big Ideas Math Answer Key Grade 1 Chapter 6 Count and Write Numbers to 120 9
Answer:
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120-8

The missing numbers in the chart are 102,104,105,113,115.

Explanation:
The missing number before 103 is 103 – 1 = 102,
the number after 103 is 103 + 1 = 104,
the number after 104 is 104 + 1 = 105,
the number after 112 is 112 + 1 = 113 and
the number after 114 is 114 + 1 = 115
So the missing numbers in the chart are 102,104,105,113,115.

Question 7.
MP Structure
Write a number between 85 and 95. Then count by ones to write the next 7 numbers.

__87__, _88___, _89___, __90___, _91__, _92_, _93_, _94__
Answer:
Let the number be 87 which is between 85 and 95 and
the next 7 numbers after 87 are 88,89,90,91,92,93,94.

Explanation:
Given to write a number between 85 and 95,
Let us take the number as 87, now we will write the
next 7 numbers are 87 + 1 = 88,88+ 1 = 89, 89 + 1 =90,
90 + 1 = 91,91 + 1 = 92, 92 + 1 = 93, 93 + 1 = 94,
therefore the next 7 numbers after 87 are 88,89,90,91,92,93,94.

Question 8.
Modeling Real Life
There are 110 tokens. You want 119. How many more tokens do you need?
Big Ideas Math Answer Key Grade 1 Chapter 6 Count and Write Numbers to 120 10

____9_____ more tokens
Answer:
I need 9 more tokens.

Explanation :
Given there are 110 tokens. and I want 119,
so I needed 119 – 110 = 9 more tokens.

Review & Refresh

Question 9.
? – 6 = 4
Big Ideas Math Answer Key Grade 1 Chapter 6 Count and Write Numbers to 120 11
Think 6 + 4 = ___10_____ .
So, ______10__ – 6 = 4.
Answer:
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120-9
6 + 4 = 10
so, 10 – 6 = 4
Explanation:
Given ___ – 6 = 4, Let us take the missing number as X,
therefore X – 6 = 4, X = 4 + 6 =10,the equation is 10 – 6 = 4.

Lesson 6.2 Count to 120 by Tens

Explore and Grow

Count to 10. Circle the number. Count 10 more. Circle the number. Continue until you reach 120.

Big Ideas Math Answers 1st Grade 1 Chapter 6 Count and Write Numbers to 120 12
Answer:
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120-10

Explanation:

Start counting from 1 to 10 circled, Next from 11 to 20 counted
circled, counted till 120 and circled.

Show and Grow

Count by tens to write the missing numbers.

Question 1.
70, __80___, __90__, __100__, _110__, __120__
Answer:
The missing numbers after 70 are 80,90,100,110,120

Explanation:
Started from 70 counting by 10’s so 70 + 10 = 80,
80 + 10 = 90, 90 + 10 = 100, 100 + 10 = 110, 110 + 10 = 120,
So the missing numbers after 70 are 80,90,100,110,120

Question 2.
31, __41___, __51__, _61__, __71__, _81__
Answer:
The missing numbers after 31 are 41,51,61,71,81

Explanation:
Started from 31 counting by 10’s so 31 + 10 = 41,
41 + 10 = 51, 51 + 10 =61, 61 + 10 = 71 ,71 +10 = 81,
So the missing numbers after 31 are 41,51,61,71,81.

Apply and Grow: Practice

Count by tens to write the missing numbers

Question 3.
62, __72__, __82__, _92__, _102__, _112___
Answer:
The missing numbers after 62 are 72,82,92,102,112

Explanation:
Started from 62 counting by 10’s are 62 + 10 = 72,
72 + 10 = 82, 82 + 10 =92, 92 + 10 = 102 ,102 +10 = 112,
So the missing numbers after 62 are 72,82,92,102,112

Question 4.
43, __53__, _63__, __73__, __83__, _93__
Answer:
The missing numbers after 43 are 53,63,73,83,93

Explanation:
Started from 43 counting by 10’s are 43 + 10 = 53,
53 + 10 = 63, 63 + 10 =73, 73 + 10 = 83 ,83 +10 = 93,
So the missing numbers after 43 are 53,63,73,83,93.

Question 5.
___10__, __20___, 30, __40__, __50___, __60__
Answer:
The missing numbers before and after 30 are 10,20,40,50,60

Explanation:
Started from 30 counting by 10’s numbers before and after 30 are
before 30 – 10 = 20, 20 – 10 = 10 and after 30 are 30 + 10 = 40,
40 + 10 = 50, 50 + 10 = 60, therefore numbers before and after 30
are 10,20,40,50,60.

Question 6.
Write the missing numbers from the chart. Then count on by tens to write the next three numbers.
Big Ideas Math Answers 1st Grade 1 Chapter 6 Count and Write Numbers to 120 13
Answer:
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120-11
The missing numbers in the chart are 71,81 and
count on 10’s the next three numbers after 81 are 91,101,111.

Explanation:
Given numbers in the chart, the number missing before 72
is 72 – 1 = 71 and number missing before 82 in the chart is
82 – 1 = 81,now counting by 10’s next three numbers after 81 are
81 + 10 = 91,91 + 10 = 101,101 + 10 = 111.
Therefore the missing numbers in the chart are 71,81 and
count on 10’s the next three numbers after 81 are 91,101,111.

Question 7.
YOU BE THE TEACHER
Your friend counts by tens starting with 27. Is your friend correct? Show how you know.
27, 37, 47, 67, 77, 87
Answer:
No, Friend is incorrect. As 57 is missing.

Explanation:
Given friend counts by tens starting with 27 and
gets 27, 37, 47, 67, 77, 87 but after 47 we should
get 47 + 10 =  57 as 57 is missing friend is incorrect.

Think and Grow: Modeling Real Life

You have 50 points. On your next turn, you knock over 6 cans. How many points do you have now?
Big Ideas Math Answers 1st Grade 1 Chapter 6 Count and Write Numbers to 120 14
Write the numbers:

____110______ points
Answer:
I have 110 points now.

Explanation:
Given I have 50 points on my next turn I knock
over 6 cans and 10 points for each knocked over can
means I got 6 X 10 = 60 points, So in total I have
50 + 60 = 110 points now.

Show and Grow

Question 8.
You have 21 points. On your next turn, 3 beanbags land in the circle. How many points do you have now?
Big Ideas Math Answers 1st Grade 1 Chapter 6 Count and Write Numbers to 120 15
Write the numbers:

_____51_____ points
Answer:
I have 51 points now.

Explanation:
Given I have 21 points on my next turn 3 beanbags
land in the circle 10 points for each beanbag
means I got 3 X 10 = 30 points, So in total I have
21 + 30 = 51 points now.

Count to 120 by Tens Practice 6.2

Count by tens to write the missing numbers.

Question 1.
69, __79_, _89__, _99_, _109_, _119__
Answer:
The missing numbers after 69 are 79,89,99,109,119

Explanation:
Started from 69 counting by 10’s numbers after 69 are
69 + 10 = 79, 79 + 10 = 89, 89 + 10 = 99, 99 + 10 = 109,
109 + 10 = 119, therefore the missing numbers after 69
are 79,89,99,109,119.

Question 2.
41, __51__, _61___, _71__, _81__, _91__
Answer:
The missing numbers after 41 are 51,61,71,81,91

Explanation:
Started from 41 counting by 10’s numbers after 41 are
41 + 10 = 51,51 + 10 = 61, 61 + 10 = 71, 71 + 10 = 81,
81 + 10 =91, So the missing numbers after 41 are 51,61,71,81,91.

Question 3.
16, __26__, _36__, _46__, _56__, __66__
Answer:
The missing numbers after 16 are 26,36,46,56,66

Explanation:
Started from 16 counting by 10’s numbers after 16 are
16 + 10 = 26, 26 + 10 = 36,36 + 10 = 46, 46 + 10 = 56,
56 + 10 = 66, So the missing numbers after 16 are 26,36,46,56,66.

Question 4.
_64__, __74__, __84__, 94, __104_, __114__
Answer:
The missing numbers before 94 are 84,74 ,64 and
after 94 are 104, 114.

Explanation:
Started from 94 counting by 10’s numbers before and after 94 are
before 94 – 10 = 84, 84 – 10 = 74 , 74 – 10 = 64 and after are
94 + 10 = 104,104 + 10 = 114, So the missing numbers before
94 are 84,74 ,64 and after 94 are 104, 114.

Question 5.
Write the missing numbers from the chart. Then count on by tens to write the next three numbers.
Big Ideas Math Answers 1st Grade 1 Chapter 6 Count and Write Numbers to 120 16
Answer:
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120-12

The missing numbers in the chart are 6,16 and
count on 10’s the next three numbers after 16 are 26,36,46.

Explanation:
Given numbers in the chart, the number missing after 5
is 5 + 1  = 6 and number missing after 15 in the chart is
15 + 1 = 16,now counting by 10’s next three numbers after 16 are
16 + 10 = 26,26 + 10 = 36,36 + 10 = 46.
Therefore the missing numbers in the chart are 6,16 and
count on 10’s the next three numbers after 16 are 26,36,46.

Question 6.
DIG DEEPER!
You count to 50. You only count 5 numbers. Did you count by ones or by tens? Show how you know.
Answer:
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120-13
I count by tens.

Explanation:
Given I count to 50 and count only 5 numbers , If I count by ones
I will get 50 numbers like 1,1+1=2, 2+ 1=3,…till I reach 49+ 1 = 50
as shown in the above picture, If I count by tens I get 5 numbers,
Like 10,10 + 10 = 20,20 + 30 = 40, 40 + 10 = 50, as I count 5 numbers
So I count by tens.

Question 7.
Modeling Real Life
You have 30 points. On your next turn, 4 balls stick to the wall. How many points do you have now?
Big Ideas Math Answers 1st Grade 1 Chapter 6 Count and Write Numbers to 120 17
____70_____ points
Answer:
I have 70 points now.

Explanation:
Given I have 30 points. On my next turn,
4 balls stick to the wall and each ball we have 10 points,
So 4 X 10 = 40 points means in total I have 30 + 40 = 70 points now.

Review & Refresh

Question 8.
3 + 1 = ___4_____
Answer:
3 + 1 = 4

Explanation:
We are going to add 1 to 3 we get 4.

Question 9.
5 – 1 = ____4_____
Answer:
5 – 1 = 4

Explanation:
We will subtract 1 from 5 we get 4.

Lesson 6.3 Compose Numbers 11 to 19

Explore and Grow

Color to show 13 and 17. What is the same about the numbers? What is different?
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120 18
Answer:
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120-14
Both are same- 13, 17 are prime numbers.
Both 13 and 17 are different as
both are not same numbers, they both are not equal.

Explanation:
Taken numbers from 1 to 20 in the chart,
colored numbers 13 and 17 with green.
Both are the same- 13 and 17 are prime numbers-
these are the numbers, which are only divisible by 1 or
the number itself as 13 and 17 are divisible by 1 and only
by them they both are prime.
Both are different-As 13 and 17 are equal they differ.

Show and Grow

Question 1.
Circle 10 feathers. Complete the sentence.
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120 19
____One____ ten and ___six____ ones is __sixteen____ .
Answer:

One ten and six ones is sixteen.
1 ten and 6 ones is 16.

Explanation:
Circled 10 feathers, after counting we have total
16 feathers we write the sentence as
one ten and six ones is sixteen means 10 + 6 = 16.
1 ten and 6 ones is 16.

Apply and Grow: Practice

Circle 10 feathers. Complete the sentence.

Question 2.
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120 20
___One_____ ten and __five___ ones is ____fifteen_____ .
Answer:
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120-16
One ten and five ones is fifteen.
1 ten and 5 ones is 15.

Explanation:
Circled 10 fishes, after counting we have total
15 fishes we write the sentence as
one ten and five ones is fifteen means 10 + 5 = 15.
1 ten and 5 ones is 15.

Question 3.
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120 21
____One____ ten and ___three__ ones is ____thirteen_____ .

Answer:
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120-17
One ten and three ones is thirteen.
1 ten and 3 ones is 13.

Explanation:
Circled 10 sun flowers, after counting we have total
13 sunflowers we write the sentence as
one ten and three ones is thirteen means 10 + 3 = 13.
1 ten and 3 ones is 13.

Question 4.
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120 22
___One_ ten and __seven__ ones is ___seventeen___ .
Answer:
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120-18

One ten and seven ones is seventeen.
1 ten and 7 ones is 17.

Explanation:
Circled 10 bees, after counting we have total
17 bees we write the sentence as
one ten and seven ones is seventeen means 10 + 7 = 17.
1 ten and 7 ones is 17.

Question 5.
MP Number Sense
Color to show the number. Complete the sentence.
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120 23
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120 24
___one___ ten and __six___ ones is __sixteen____ .
Answer:
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120-19
One ten and six ones is sixteen.
1 ten and 6 ones is 16.

Explanation:
Colored the shown number 16 with blue,
we write the sentence as
one ten and six ones is sixteen means 10 + 6 = 16.
1 ten and 6 ones is 16.

Think and Grow: Modeling Real Life

You have 15 footballs. A bag can hold 10. You fill a bag. How many footballs are not in the bag?
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120 25
Draw a picture:

Write the missing numbers: __one_ ten and __five_ ones is __fifteen__ footballs
Answer:
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120-19
One ten and five ones is fifteen footballs.
1 ten and 5 ones is 15.
5 footballs are not there in the bag,
Zero ten and five ones or 0 ten and 5 ones is five or 5 footballs are missing.

Explanation:
Given I have 15 footballs means one ten and five ones is
fifteen or 1 ten and 5 ones is 15 and a bag can hold 10 footballs,
In the picture there are total 15 footballs in that I
have taken 10 in the bag so footballs which are not there
in the bag are 15 – 10 = 5, So missing number of footballs
in the bag are zero ten and five ones is five or 0 ten and 5 ones is 5.

Show and Grow

Question 6.
Your teacher has 18 calculators. A case can hold 10. Your teacher fills a case. How many calculators are not in the case?
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120 26
Draw a picture:

Write the missing numbers: __one__ ten and __eights ones _ is eighteen_ calculators
Answer:
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120-20
One ten and eight ones is eighteen calculators.
1 ten and 8 ones is 18.
8 calculators are not there in the case,
Zero ten and eight ones is eight or
0 ten and 8 ones is 8 calculators are missing,

Explanation:
Given teacher ha 18 calculators means one ten and eight ones is
eighteen or 1 ten and 8 ones is 18 and a case can hold 10,
In the picture there are total 18 calculators  in that I
have taken 10 in the case so number of calculators which
are not there in the case are 18 – 10 = 8, So missing calculators
in the case are zero ten and eight ones is eight or 0 ten and 8 ones is 8.

Compose Numbers 11 to 19 Practice 6.3

Circle 10 objects. Complete the sentence.

Question 1.
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120 27
__one___ ten and ___one___ ones is eleven_ .
Answer:
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120-21

One ten and one ones is eleven.
1 ten and 1 ones is 11.

Explanation:
Circled 10 balls, after counting we have total
11 balls we write the sentence as
one ten and one ones is eleven means 10 + 1 = 11.
1 ten and 1 ones is 11.

Question 2.
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120 28
__one____ ten and __nine____ ones is _nineteen_____ .
Answer:
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120-22

One ten and nine ones is nineteen.
1 ten and 9 ones is 19.

Explanation:
Circled 10 fishes, after counting we have total
19 fishes we write the sentence as
one ten and nine ones is nineteen means 10 + 9 = 19.
1 ten and 9 ones is 19.

Question 3.
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120 29
___one___ ten and __three____ ones __thirteen____ .
Answer:
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120-23

One ten and three ones is thirteen.
1 ten and 3 ones is 13.

Explanation:
Circled 10 honeybees, after counting we have total
13 we write the sentence as
one ten and three ones is thirteen means 10 + 3 = 13.
1 ten and 3 ones is 13.

Question 4.
MP Number Sense
Color to show the number. Complete the sentence.
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120 30
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120 31
__one____ ten and _eight___ ones is eighteen__ .
Answer:
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120-24

One ten and eight ones is eighteen.
1 ten and 8 ones is 18.

Explanation:
Colored the shown number 18 with yellow,
we write the sentence as
one ten and eight ones is eighteen means 10 + 8 = 18.
1 ten and 8 ones is 18.

Question 5.
MP Number Sense Match.
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120 32
Answer:
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120-25

Explanation:
Matched 1 ten and 3 ones with 13, 1 ten and 8 ones with 12
and 12 ones with 12 in the picture above.

Question 6.
Modeling Real Life
You have 16 books. A backpack can hold 10. You fill a backpack. How many books are not in the backpack?
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120 33

_____6_____ books
Answer:
There are 6 books which are not in the backpack.

Explanation:
Given I have 16 books means and a backpack
can hold 10 books, out of 16 books I
have taken 10 in the backpack so books which are not there
in the backpack are 16 – 10 = 6 books.

Review & Refresh

Question 7.
10 + 10 = ___20_____
Answer:
10 + 10 = 20

Explanation:
We are going to add 10 with 10 we get 20.

Question 8.
10 + 10 = ___20______
Answer:
10 + 10 = 20

Explanation:
We are going to add 10 with 10 we get 20.

Lesson 6.4 Tens

Explore and Grow

Circle groups of 10. Write the number of groups.

Big Ideas Math Solutions Grade 1 Chapter 6 Count and Write Numbers to 120 34
Answer:

How many counters are there in all?
____20______ counters
Answer:
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120-26
There are 20 counters in all.

Explanation:
In the given picture circled groups of 10, There are 2 groups,
and counted there are total 20 counters in all.

Show and Grow

Circle groups of 10. Complete the sentence.

Question 1.
Big Ideas Math Solutions Grade 1 Chapter 6 Count and Write Numbers to 120 35

____6__ ten and _0__ ones is 60______ .
Answer:
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120-27
6 groups, 6 ten and 0 ones is 60.

Explanation:
In the given picture circled groups of 10, There are 6 groups,
and counted there are total 60 in all. 6 ten and 0 ones is 60.

Question 2.
Big Ideas Math Solutions Grade 1 Chapter 6 Count and Write Numbers to 120 36

__5____ ten and ___0___ ones is 50_ .
Answer:
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120-28
5 groups, 5 ten and 0 ones is 50.

Explanation:
In the given picture circled groups of 10, There are 5 groups,
and counted there are total 50 in all. 5 ten and 0 ones is 50.

Apply and Grow: Practice

Circle groups of 10. Complete the sentence.

Question 3.
Big Ideas Math Solutions Grade 1 Chapter 6 Count and Write Numbers to 120 37

___7___ ten and ___0___ ones is 70______ .
Answer:
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120-29
7 groups, 7 ten and 0 ones is 70.

Explanation:
In the given picture circled groups of 10, There are 7 groups,
and counted there are total 70 in all. 7 ten and 0 ones is 70.

Question 4.
Big Ideas Math Solutions Grade 1 Chapter 6 Count and Write Numbers to 120 38

___9___ ten and ___0___ ones is _90__ .
Answer:
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120-30
9 groups, 9 ten and 0 ones is 90.

Explanation:
In the given picture circled groups of 10, There are 9 groups,
and counted there are total 90 in all. 9 ten and 0 ones is 90.

Question 5.
Big Ideas Math Solutions Grade 1 Chapter 6 Count and Write Numbers to 120 39

___1___ ten and ___0___ ones is __10____ .
Answer:
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120-31
1 group, 1 ten and 0 ones is 10.

Explanation:
In the given picture circled groups of 10, There are 1 group,
and counted there are total 10 in all. 1 ten and 0 ones is 10.

Question 6.
MP Number Sense
You have 4 groups of 10 linking cubes. How many linking cubes do you have?

_____40_______ linking cubes
Answer:
I have 40 linking cubes.

Explanation:
Given I have 4 groups of 10 linking cubes, So I have
4 X 10 = 40 linking cubes.

Think and Grow: Modeling Real Life

You read 10 books every month. You 40 books. How many months does it take?
Big Ideas Math Solutions Grade 1 Chapter 6 Count and Write Numbers to 120 40
Draw a picture:
Write the missing numbers: ___4___ tens and ___0___ ones __4____ months

Answer:
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120-32

4 tens and 0 ones is 40 books.
It will take 4 months.

Explanation:
Given I read 10 books every month and read 40 books,
4 tens and 0 ones is 40 books.
So we will divide 40 by 10 we get 4, therefore it will take 4 months,
Show and Grow

Question 7.
There are 10 dog bones in each box. You need 20 bones. How many boxes do you need?
Big Ideas Math Solutions Grade 1 Chapter 6 Count and Write Numbers to 120 41
Draw a picture:

Write the missing numbers: ___2___ tens and ___0___ ones __2____ boxes
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-33

2 tens and 0 ones is 20 boxes.
We need 2 boxes.

Explanation:
Given there are 10 dog bones in each box and need 20 bones.
2 tens and 0 ones is 20 bones.
So we will divide 20 by 10 we get 2, therefore we need 2 boxes.

Tens Practice 6.4

Circle groups of 10. Complete the sentence.

Question 1.
Big Ideas Math Solutions Grade 1 Chapter 6 Count and Write Numbers to 120 42

___8___ tens and __0____ ones is 80__ .
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-34

8 groups, 1 ten and 0 ones is 10.

Explanation:
In the given picture circled groups of 10, There are 8 groups,
and counted there are total 80 in all. 8 ten and 0 ones is 80.

Question 2.
Big Ideas Math Solutions Grade 1 Chapter 6 Count and Write Numbers to 120 43

___6___ tens and __0____ ones is __60____ .
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-35

6 groups, 6 ten and 0 ones is 60.

Explanation:
In the given picture circled groups of 10, There are 6 groups,
and counted there are total 60 in all. 6 ten and 0 ones is 60.

Question 3.
Big Ideas Math Solutions Grade 1 Chapter 6 Count and Write Numbers to 120 44

___3___ tens and __0____ ones _is 30_____ .
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-36

3 groups, 3 ten and 0 ones is 30.

Explanation:
In the given picture circled groups of 10, There are 3 groups,
and counted there are total 30 in all. 3 ten and 0 ones is 30.

Circle groups of 10. Complete the sentence.

Question 4.
Big Ideas Math Solutions Grade 1 Chapter 6 Count and Write Numbers to 120 45

___2___ tens and __0____ ones is __20____ .
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-37

2 groups, 2 ten and 0 ones is 20.

Explanation:
In the given picture circled groups of 10, There are 2 groups,
and counted there are total 20 in all. 2 ten and 0 ones is 20.

Question 5.
MP Number Sense
You have 7 groups of 10 linking cubes. How many linking cubes do you have?

_____70______ linking cubes
Answer:

I have 70 linking cubes.

Explanation:
Given I have 7 groups of 10 linking cubes, So I have
7 X 10 = 70 linking cubes.

Question 6.
Modeling Real Life
You swim 10 laps at every practice. You want to swim 50 laps. How many practices will it take?

_____5______ Practices
Answer:

5 tens and 0 ones is 50 laps
It will take 5 practices.

Explanation:
Given I swim 10 laps at every practice, I want to swim 50 laps
5 tens and 0 ones is 50 laps.
So I will divide 50 by 10 I get 5, therefore I will take 5 practices.

Review & Refresh

Question 7.
4 + 3 + 4 = ____11________
Answer:
4 + 3 + 4 = 11

Explanation:
First we add 4 and 3 then add 4 we get 11.

Question 8.
1 + 5 + 9 = _____15______
Answer:
1 + 5 + 9 = 15

Explanation:
First we add 1 and 5 then add 9 we get 15.

Question 9.
2 + 2 + 1 = _____5______
Answer:
2 + 2 + 1 = 5

Explanation:
First we add 2 and 2 then add 1 we get 5.

Question 10.
7 + 3 + 6 = _____16______
Answer:
7 + 3 + 6 = 16

Explanation:
First we add 7 and 3 then add 6 we get 16.

Lesson 6.5 Tens and Ones

Explore and Grow

Model 2 tens and 3 ones. Write the number.

Big Ideas Math Answer Key Grade 1 Chapter 6 Count and Write Numbers to 120 46
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-38
2 tens and 3 ones is 23.

Explanation:
Model shown in the picture 2 tens and 3 ones,
So placed 2 ten blocks in tens place and 3 blocks in ones place,
making 2 tens and 3 ones ,therefore the number is 23.

Show and Grow

Question 1.
Big Ideas Math Answer Key Grade 1 Chapter 6 Count and Write Numbers to 120 47

__2___ tens and ___1___ ones is ___21___ .
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-39

2 tens and 1 ones is 21

Explanation:
Given in the picture there are 2 tens block in tens place
and 1 block in ones make 2 tens and 1 ones is 21.

Apply and Grow: Practice

Question 2.
Big Ideas Math Answer Key Grade 1 Chapter 6 Count and Write Numbers to 120 48

___3___ tens and __5____ ones is ___35___ .
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-40
3 tens and 5 ones is 35

Explanation:
Given in the picture there are 3 ten blocks in tens place
and 5 blocks in ones make 3 tens and 5 ones is 35.

Question 3.
Big Ideas Math Answer Key Grade 1 Chapter 6 Count and Write Numbers to 120 49

___6___ tens and ___6___ ones is ___6___ .
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-41
6 tens and 6 ones is 66

Explanation:
Given in the picture there are 6 ten blocks in tens place
and 6 blocks in ones make 6 tens and 6 ones is 66.

Question 4.
Big Ideas Math Answer Key Grade 1 Chapter 6 Count and Write Numbers to 120 50

__8____ tens and __9____ ones is __89____ .
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-42
8 tens and 9 ones is 89

Explanation:
Given in the picture there are 8 ten blocks in tens place
and 9 blocks in ones make 8 tens and 9 ones is 89.

Question 5.
YOU BE THE TEACHER
You have 92 linking cubes. Your friend says that there are 2 tens and 9 ones.
Is your friend correct? Show how you know.
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-43
No, friend is incorrect.

Explanation:
Given I have 92 linking cubes, friend says that there are 2 tens and 9 ones,
but 92 means as shown in picture we get 9 tens and 2 ones,
therefore friend is in correct.

Think and Grow: Modeling Real Life

Your teacher has 2 packages of dice and 3 extra dice. Each package has 10 dice. How many dice are there in all?
Big Ideas Math Answer Key Grade 1 Chapter 6 Count and Write Numbers to 120 51
Draw a picture:

Write the missing numbers: ___2___ tens and __3____ ones _is__23___ dice
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-44
2 tens and 3 ones is 23 dice are there in all.

Explanation:
Given my teacher has 2 packages of dice and 3 extra dice.
Each package has 10 dice, 2 packages means 2 X 10 = 20 dice
and 3 extra dice making in all 2 tens and 3 ones is 23 dice as shown in picture.

Show and Grow

Question 6.
You have 3 boxes of colored pencils and 4 extra colored pencils. Each box has 10 pencils. How many colored pencils are there in all?
Big Ideas Math Answer Key Grade 1 Chapter 6 Count and Write Numbers to 120 52
Draw a picture:

Write the missing numbers: ___3___ tens and __4____ ones _is_34__ colored pencils
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-45
3 tens and 4 ones is 34 colored pencils.

Explanation:
Given I have 3 boxes of colored pencils and 4 extra colored pencils.
Each box has 10 pencils, So 3 boxes have 3 X 10 = 30 pencils and 4 extra
makes 30 + 4 = 34, 3 tens and 4 ones is 34 colored pencils are there in all.

Tens and Ones Practice 6.5

Question 1.
Big Ideas Math Answer Key Grade 1 Chapter 6 Count and Write Numbers to 120 53

__7____ tens and ___9___ ones is __79____ .
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-46
7 tens and 9 ones is 79.

Explanation:
Given in the picture there are 7 ten blocks in tens place
and 9 blocks in ones make 7 tens and 9 ones is 79.

Question 2.
Big Ideas Math Answer Key Grade 1 Chapter 6 Count and Write Numbers to 120 54

__8____ tens and ___1___ ones _is 81_____ .
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-47
8 tens and 1 ones is 81.

Explanation:
Given in the picture there are 8 ten blocks in tens place
and 1 blocks in ones make 8 tens and 1 ones is 81.

Question 3.
Big Ideas Math Answer Key Grade 1 Chapter 6 Count and Write Numbers to 120 55

___2___ tens and __4___ ones is 24_ .
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-48
2 tens and 4 ones is 24.

Explanation:
Given in the picture there are 2 ten blocks in tens place
and 4 blocks in ones make 2 tens and 4 ones is 24.

Question 4.
YOU BE THE TEACHER
You have 17 linking cubes. Your friend says that there is 1 ten and 7 ones. Is your friend correct? Show how you know.
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-49

Yes, friend is correct.

Explanation:
Given I have 17 linking cubes, friend says that there is 1 ten and 7 ones,
means as shown in picture we get 1 ten and 7 ones, similar to what friend says,
therefore friend is  correct.

Question 5.
Modeling Real Life
You have 5 bags of apples and 1 extra apple. Each bag has 10 apples. How many apples are there in all?
Big Ideas Math Answer Key Grade 1 Chapter 6 Count and Write Numbers to 120 56

_____51______ apples
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-50
5 tens and 1 ones is 51 apples.

Explanation:
Given I have 5 bags of apples and 1 extra apple.
Each bag has 10 apples, So 5 bags have 5 X 10 = 50 apples and 1 extra apple
makes 50 + 1 = 51, 5 tens and 1 ones is 51 apples are there in all.

Review & Refresh

Question 6.
_____4_____ + 6 = 10
Answer:
4 + 6 = 10

Explanation:
Given ____ + 6 = 10, let us take the missing
number as X , X + 6 = 10, X = 10 – 6 = 4 making
the equation as 4 + 6 = 10.

Question 7.
_____6______ + 2 = 8
Answer:
6 + 2 = 8

Explanation:
Given ____ + 2 = 8, let us take the missing
number as X , X + 2 = 8, X = 8 – 2 = 6 making
the equation as 6 + 2 = 8.

Lesson 6.6 Make Quick Sketches

Explore and Grow

Model the number 2.
Big Ideas Math Answers 1st Grade 1 Chapter 6 Count and Write Numbers to 120 57
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-51

2 ones is 2.

Explanation:
Model shown in the picture are 2 ones,
So 2 blocks in ones place, as shown in picture
making 2 ones ,therefore the number is 2.

Show and Grow

Make a quick sketch. Complete the sentence.

Question 1.
72
Big Ideas Math Answers 1st Grade 1 Chapter 6 Count and Write Numbers to 120 58
__72____ is ____7____ tens and __2____ ones.
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-52

72 is 7 tens and 2 ones.

Explanation:
The picture has 7 ten blocks in tens place
and 2 blocks in ones makes 72 as 7 tens and 2 ones.

Question 2.
36
Big Ideas Math Answers 1st Grade 1 Chapter 6 Count and Write Numbers to 120 59
__36____ is ___3__ tens and __6_ ones.
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-53

36 is 3 tens and 6 ones.

Explanation:
The picture has 3 ten blocks in tens place
and 6 blocks in ones makes 36 as 3 tens and 6 ones.

Apply and Grow: Practice

Make a quick sketch. Complete the sentence.

Question 3.
45
Big Ideas Math Answers 1st Grade 1 Chapter 6 Count and Write Numbers to 120 60
___45___ is ___4_____ tens and ___5___ ones.
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-54

45 is 4 tens and 5 ones.

Explanation:
The picture has 4 ten blocks in tens place
and 5 blocks in ones makes 45 as 4 tens and 5 ones.

Question 4.
87
Big Ideas Math Answers 1st Grade 1 Chapter 6 Count and Write Numbers to 120 60
___87___ is ___8___ tens and ___7___ ones.
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-55

87 is 8 tens and 7 ones.

Explanation:
The picture has 8 ten blocks in tens place
and 7 blocks in ones makes 87 as 8 tens and 7 ones.

Question 5.
64
Big Ideas Math Answers 1st Grade 1 Chapter 6 Count and Write Numbers to 120 60
__64____ is ___6____ tens and ___4___ ones.
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-56

64 is 6 tens and 4 ones.

Explanation:
The picture has 6 ten blocks in tens place
and 4 blocks in ones makes 64 as 6 tens and 4 ones.

Question 6.
DIG DEEPER!
Which sketch shows 54?
Big Ideas Math Answers 1st Grade 1 Chapter 6 Count and Write Numbers to 120 61
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-57
First sketch shows 54.

Explanation:
Given 2 sketches in that first counts 54 and second counts
45, So first sketch shows 54, So selected first one for 54.

Think and Grow: Modeling Real Life

You need 58 plates for a party. You have 51. How many more plates do you need?
Big Ideas Math Answers 1st Grade 1 Chapter 6 Count and Write Numbers to 120 62
Complete the model:
Big Ideas Math Answers 1st Grade 1 Chapter 6 Count and Write Numbers to 120 63

____7_______ more plates
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-58
I need 7 more plates.

Explanation:
Given I need 58 plates for a party and I have 51,
therefore more plates needed are 58 – 51 = 7 plates
as shown in the picture above.

Show and Grow

Question 7.
You need 80 tickets for a prize. You have 73. How many more tickets do you need?
Big Ideas Math Answers 1st Grade 1 Chapter 6 Count and Write Numbers to 120 64
Complete the model:
Big Ideas Math Answers 1st Grade 1 Chapter 6 Count and Write Numbers to 120 65

_____7______ more tickets
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-59
I need 7 more tickets.

Explanation:
Given I need 80 tickets for a prize and I have 73,
therefore more tickets needed are 80 – 73 = 7 plates
as shown in the picture above.

Make Quick Sketches Practice 6.6

Make a quick sketch. Complete the sentence.

Question 1.
27
Big Ideas Math Answers 1st Grade 1 Chapter 6 Count and Write Numbers to 120 66
__27____ is ____2____ tens and ___7___ ones.
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-60

27 is 2 tens and 7 ones.

Explanation:
The picture has 2 ten blocks in tens place
and 7 blocks in ones makes 27 as 2 tens and 7 ones.

Question 2.
61
Big Ideas Math Answers 1st Grade 1 Chapter 6 Count and Write Numbers to 120 66
__61__ is ___6__ tens and ___1___ ones.
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-61

61 is 6 tens and 1 one.

Explanation:
The picture has 6 ten blocks in tens place
and 1 blocks in one makes 61 as 6 tens and 1 one.

Question 3.
92
Big Ideas Math Answers 1st Grade 1 Chapter 6 Count and Write Numbers to 120 66
___92___ is ___9_____ tens and ___2___ ones.
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-62

92 is 9 tens and 2 ones.

Explanation:
The picture has 9 ten blocks in tens place
and 2 blocks in ones makes 92 as 9 tens and 2 ones.

Question 4.
DIG DEEPER
Which sketch shows 87?
Big Ideas Math Answers 1st Grade 1 Chapter 6 Count and Write Numbers to 120 67
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-63

First sketch shows 87.

Explanation:
Given 2 sketches in that first counts 87 and second counts
78, So first sketch shows 87, So selected first one for 87.

Question 5.
Modeling Real Life
You need 55 beads to make a necklace. You have 48. How many more beads do you need?
Big Ideas Math Answers 1st Grade 1 Chapter 6 Count and Write Numbers to 120 68

____7______ more beads
Answer:
I need 7 more beads.

Explanation:
Given I need 55 beads to make necklace and I have 48,
therefore more beads needed are 55 – 48 = 7 beads.

Review & Refresh

Question 6.
Color the shapes that have only 4 sides.
Big Ideas Math Answers 1st Grade 1 Chapter 6 Count and Write Numbers to 120 69
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-64

Explanation:
Given some shapes and to color that have 4 sides so selected shapes
that has 4 sides and  have colored as shown above in the picture.

Lesson 6.7 Understand Place Value

Explore and Grow

Newton has 2 rods. Make a quick sketch.
Write the number.
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120 70
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-65
Newton has 2 rods or 20 cubes

Explanation:
Given Newton has 2 rods as shown in the picture means 20 cubes.

Descartes has 2 cubes. Make a quick sketch.
Write the number.
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120 71
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-66
Descartes has 2 cubes

Explanation:
Given Descartes has 2 cubes as shown in picture 2 cubes.

How are the models alike? How are they different?
Answer:
Both use cubes so alike,
Both differ in number so different.

Explanation:
Models are alike means both use cubes.
They are different Newton has 20 cubes and Descartes have 2 cubes,
20 and 2 are not the same and are not equal so they are different.

Show and Grow

Question 1.
Make a quick sketch. Complete the sentences.
64
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120 72
_____6_____ tens is ____60______ .
____4______ ones is ____4______ .
_____6_____ tens and ____4______ ones is ___64_______.
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-67
6 tens is 60,
4 ones is 4, 6 tens and 4 ones is 64.

Explanation:
As shown in the picture above first we take 6 tens cubes as 60
and 4 ones cubes as 4, making 6 tens and 4 ones is 64.

Apply and Grow: Practice

Make a quick sketch. Complete the sentences.

Question 2.
72
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120 72
_____7_____ tens is ____70______ .
_____2_____ ones is ____2______ .
______7____ tens and ____2______ ones is ___72__.
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-68
7 tens is 70,
2 ones is 2, 7 tens and 2 ones is 72.

Explanation:
As shown in the picture above first we take 7 tens cubes as 70
and 2 ones cubes as 2, making 7 tens and 2 ones is 72.

Question 3.
98
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120 72
__9____ tens is ___90____ .
___8____ ones is ___8_____ .
___9____ tens and ____8____ ones is ___98___.
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-69

9 tens is 90,
8 ones is 8, 9 tens and 8 ones is 98.

Explanation:
As shown in the picture above first we take 9 tens cubes as 90
and 8 ones cubes as 8, making 9 tens and 8 ones is 98.

Question 4.
57
____5______ tens is ___50_______ .
_____7_____ ones is ____7______ .
______5____ tens and _____7_____ ones is ____57______.
Answer:
5 tens is 50,
7 ones is 7, 5 tens and 7 ones is 57.

Explanation:
First we take 5 tens cubes as 50
and 7 ones cubes as 7, making 5 tens and 7 ones is 57.

Think and Grow: Modeling Real Life

You have 94 charms to make bracelets. There are lo charms on each bracelet. How many bracelets can you make?
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120 73
Model:

Write the missing numbers: ___9___ tens and ___4___ ones ___9___ bracelets
Answer:
9 tens and 4 ones is 94,
I make 9 bracelets.

Explanation:
Given I have 94 charms to make bracelets. There are lo charms
on each bracelet. So I divide 94 with 10 I get 9 as whole
and 4 as remainder so I cannot make with 4 charms,
I consider only whole so I take 9,therefore I can make 9 bracelets
or as I have 9 tens and 4 ones I take only tens so 9 bracelets I can make.

Show and Grow

Question 5.
You have 67 seeds. You plant 10 seeds in a row. How many rows can you plant?
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120 74
Model:

Write the missing numbers: ___6___ tens and __7____ ones __6____ rows
Answer:
6 tens and 7 ones is 67,
I can plant 6 rows.

Explanation:
Given I have 67 seeds. I can plant 10 seeds in a row
So I divide 67 with 10 I get 6 as whole
and 7 as remainder so I cannot make with 7 seeds a row ,
I consider only whole so I take 6,therefore I can make 6 rows only,
or as I have 6 tens and 7 ones I take only tens so 6 rows I can make.

Understand Place Value Practice 6.7

Make a quick sketch. Complete the sentences.

Question 1.
81
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120 75
_____8_____ tens is ___80_______ .
_____1_____ ones is ____1______ .
_____8_____ tens and ____1______ ones is ____81______.
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-70

8 tens is 80,
1 one is 1, 8 tens and 1 one is 81.

Explanation:
As shown in the picture above first we take 8 tens cubes as 80
and 1 one cubes as 1, making 8 tens and 1 one is 81.

Question 2.
53
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120 75
_____5_____ tens is _____50_____ .
______3____ ones is ____3______ .
______5____ tens and _____3____ ones is ___53_____.
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-71

5 tens is 50,
3 ones is 3, 5 tens and 3 ones is 53.

Explanation:
As shown in the picture above first we take 5 tens cubes as 50
and 3 ones cubes as 3, making 5 tens and 3 ones is 53.

Question 3.
49
___4_______ tens is ____40______ .
_____9_____ ones is ____9______ .
______4____ tens and ____9__ ones is __49__.
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-72

4 tens is 40,
9 ones is 9, 4 tens and 9 ones is 49.

Explanation:
As shown in the picture above first we take 4 tens cubes as 40
and 9 ones cubes as 9, making 4 tens and 9 ones is 49.

Question 4.
76
____7______ tens is ____70______ .
______6____ ones is ____6______ .
____7______ tens and ____6______ ones is ____76______.
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-73

7 tens is 70,
6 ones is 6, 7 tens and 6 ones is 76.

Explanation:
As shown in the picture above first we take 7 tens cubes as 70
and 6 ones cubes as 6, making 7 tens and 6 ones is 76.

Question 5.
Modeling Real Life
You have 77 crayons. A box can hold 10 crayons. How many boxes can you fill?
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120 76

_____7____ boxes
Answer:
7 tens and 7 ones is 77,
I can fill 7 boxes.

Explanation:
Given I have 77 crayons, A box can hold 10 crayons,
So I divide 77 with 10 I get 7 as whole
and 7 as remainder I consider only whole so I take 7,
therefore I need 7 boxes to fill or as I have 7 tens and 7 ones to
fill I take only tens so 7 boxes I can fill.

Review & Refresh

Is the equation true or false?

Question 6.
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120 77
Answer:
4 + 9 ≠  2 + 3 + 5
Not equal, So false

Equation:
First we see 4 and 9 as 4 + 9 is 13 and
if we add 2,3 and 5 we get 10 ,
therefore 4 + 9 ≠ 2 + 3 +5, So false.

Question 7.
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120 78
Answer:
5 + 3 = 4 + 4
Is equal, So true

Explanation:
First we see 5 + 3 as we add 5 and 3 we get 8 and
4 + 4 is 8 both sides are equal, So equations are true.

Lesson 6.8 Write Numbers in Different Ways

Explore and Grow

Model 27 two ways.
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120 79
Big Ideas Math Solutions Grade 1 Chapter 6 Count and Write Numbers to 120 80
_____2___ tens and __7_____ ones is ___27_____ .
_____1___ ten and ___17____ ones is ___27_____ .
Answer:
one way is
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-74
2 tens and 7 ones is 27.
Other way is
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-75
1 ten and 17 ones is 27.

Explanation:
One way is we add 2 tens and 7 ones equal to 27,
20 + 7 = 27 and other way is 1 ten and 17 ones is 27,
10 + 17 = 27.

Show and Grow

Question 1.
Model 25 two ways.
Big Ideas Math Solutions Grade 1 Chapter 6 Count and Write Numbers to 120 81
____2____ tens and ____5___ ones is 25.
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-76
2 tens and 5 ones is 25.

Explanation:
we add 20 and 5 we get 25, So 2 tens and 5 ones is 25.

Big Ideas Math Solutions Grade 1 Chapter 6 Count and Write Numbers to 120 82
____1____ tens and __15_____ ones is 25.
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-77
1 ten and 15 ones is 25

Explanation:
we add 10 and 25 we get 25, So 1 ten and 15 ones is 25.

Apply and Grow: Practice

Question 2.
Model 52 two ways.
Big Ideas Math Solutions Grade 1 Chapter 6 Count and Write Numbers to 120 82
____5____ tens and ___2____ ones is 52.
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-78
5 tens and 2 ones is 52.

Explanation:
We add 50 and 2 we get 52, So 5 tens and 2 ones is 52.

Big Ideas Math Solutions Grade 1 Chapter 6 Count and Write Numbers to 120 82
______4__ tens and ____12___ ones is 52.
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-79
4 tens and 12 ones is 52.

Explanation:
First we take 40 and add to 12 we get 52,
40 + 12 = 52, therefore 4 tens and 12 ones is 52.

Question 3.
Model 14 two ways.
Big Ideas Math Solutions Grade 1 Chapter 6 Count and Write Numbers to 120 82
___1_____ tens and ___4____ ones is 14.
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-80
1 ten and 4 ones is 14

Explanation:
we add 10 + 4 = 14,
So 1 ten and 4 ones is 14.

Big Ideas Math Solutions Grade 1 Chapter 6 Count and Write Numbers to 120 82
___0_____ tens and ___14____ ones is 14.
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-81
14 ones is 14

Explanation:
we add 0 + 14 = 14,
0 tens and 14 ones is 14.

Question 4.
DIG DEEPER!
Circle all of the ways that show 39.
3 + 9          2 tens and 9 ones      9 tens and 3 ones
10 + 29      3 tens and 19 ones    39 ones
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-82
Circled the ways that show 39 are 10 + 29 and 39 ones

Explanation:
As shown in picture circled all of the ways that show 39,
3 + 9 = 12 ≠ 39 so not circled,
2 tens and 9 ones is 29 ≠ 39 so not circled,
9 tens and 3 ones is 93 ≠ 39 so not circled,
10 + 29 = 39 = 39 so circled it,
3 tens and 19 ones is 30 + 19 is 49 ≠ 39 so not circled,
39 ones is 39 = 39 so circled it.

Think and Grow: Modeling Real Life

The models show how many seashells you and your friend have. Does your friend have the same number of seashells as you?
Big Ideas Math Solutions Grade 1 Chapter 6 Count and Write Numbers to 120 83
Big Ideas Math Solutions Grade 1 Chapter 6 Count and Write Numbers to 120 84
Circle: Yes         No

Show how you know:
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-83
No friend does not have the same number of
seashells as much as I have.

Explanation:
After counting I have total 25 seashells and friend have 30,
Both do not have same or equal number of seashells ,
So circled No as shown in the picture above.

Show and Grow

Question 5.
The models show how many erasers you and your friend have. Does your friend have the same number of erasers as you?
Big Ideas Math Solutions Grade 1 Chapter 6 Count and Write Numbers to 120 85
Big Ideas Math Solutions Grade 1 Chapter 6 Count and Write Numbers to 120 86
Circle: Yes          No

Show how you know:
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-84
Yes, I and my friend have the same number of erasers.

Explanation:
After counting I have total 40 erasers and friend also have 40,
Both have the same or equal number of erasers,
So circled Yes as shown in the picture above.

Write Numbers in Different Ways Practice 6.8

Question 1.
Model 49 two ways.
Big Ideas Math Solutions Grade 1 Chapter 6 Count and Write Numbers to 120 87
__4______ tens and
___9____ ones
is 49.
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-84
4 tens and 9 ones is 49.

Explanation:
Adding 40 to 9 is 40 + 9 = 49,
so 4 tens and 9 ones is 49.
Big Ideas Math Solutions Grade 1 Chapter 6 Count and Write Numbers to 120 87
___30_____ tens and
____19____ ones
is 49.
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-85

30 tens and 19 ones is 49.

Explanation:
We add 30 and 19 we get 49,
So 30 tens and 19 ones make 49.

Question 2.
DIG DEEPER!
Circle all of the ways that show 45.
40 + 5              45 tens and 0 ones       4 tens and 5 ones
20 + 15            2 tens and 25 ones       54 ones
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-86

Circled the ways that show 45 are 40 + 5,
4 tens and 5 ones and 2 tens and 25 ones.

Explanation:
As shown in picture circled all of the ways that show 45,
40 + 5 = 45 = 45 so circled it,
45 tens and 0 ones is 450 ≠ 45 so not circled,
4 tens and 5 ones is 45 = 45 so circled it,
20 + 15 = 35 ≠ 45 so not circled,
2 tens and 25 ones is 20 + 25 is 45 = 45  so circled it,
54 ones is 54 ≠ 45 so not circled.

Question 3.
Modeling Real Life
The models show the number of toy cars you and your friend have. Does your friend have the same number of toy cars as you?
Big Ideas Math Solutions Grade 1 Chapter 6 Count and Write Numbers to 120 88
Big Ideas Math Solutions Grade 1 Chapter 6 Count and Write Numbers to 120 89
Circle: Yes        No

Show how you know:
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-87
Yes, friend have the same number of toy cars as I have.

Explanation:
After counting I have total 34 toy cars and friend also have 34 ,
Both have same or equal number of toy cars,
So circled Yes as shown in the picture above.

Review & Refresh

Circle the heavier object.

Question 4.
Big Ideas Math Solutions Grade 1 Chapter 6 Count and Write Numbers to 120 90
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-88
Circled ball.

Explanation:
Given ball and balloon as ball is
heavier object than balloon I circled it.

Question 5.
Big Ideas Math Solutions Grade 1 Chapter 6 Count and Write Numbers to 120 91
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-89
Circled 4 bananas

Explanation:
Given 4 bananas and 1 banana as 4 bananas are
heavier object than 1 banana I circled it.

Lesson 6.9 Count and Write Numbers to 120

Explore and Grow

How many balls are there? How did you count?
Big Ideas Math Answer Key Grade 1 Chapter 6 Count and Write Numbers to 120 92
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-89
Number of balls are 106. Counted one after another.

Explanation:
There are total 106 balls, we start pointing and counting
from first ball as one add next ball and so on counting
till we reach the last ball ,we get total 106 balls.

Show and Grow

Question 1.
Big Ideas Math Answer Key Grade 1 Chapter 6 Count and Write Numbers to 120 93
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-90
10 tens and 7 ones is 107.

Explanation:
Given in the picture if we count there are
100 + 7 = 107 , 10 tens and 7 ones is 107.

Question 2.
Big Ideas Math Answer Key Grade 1 Chapter 6 Count and Write Numbers to 120 94
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-91
11 tens and 5 ones is 115.

Explanation:
Given in the picture if we count there are
110 + 5 = 107 , 11 tens and 5 ones is 115.

Apply and Grow: Practice

Question 3.
Big Ideas Math Answer Key Grade 1 Chapter 6 Count and Write Numbers to 120 95
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-92
6 tens and 4 ones is 64.

Explanation:
Given in the picture if we count there are
60 + 4 = 64 , 6 tens and 4 ones is 64.

Question 4.
Big Ideas Math Answer Key Grade 1 Chapter 6 Count and Write Numbers to 120 96
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-93
9 tens and 8 ones is 98.

Explanation:
Given in the picture if we count there are
90 + 8 = 98 , 9 tens and 8 ones is 98.

Question 5.
Big Ideas Math Answer Key Grade 1 Chapter 6 Count and Write Numbers to 120 97
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-94
10 tens and 1 one is 101.

Explanation:
Given in the picture if we count there are
100 + 1 = 101 , 10 tens and 1 one is 101.

Question 6.
Big Ideas Math Answer Key Grade 1 Chapter 6 Count and Write Numbers to 120 98
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-95
10 tens and 3 ones is 103.

Explanation:
Given in the picture if we count there are
100 + 3 = 103 , 10 tens and 3 ones is 103.

Question 7.
Big Ideas Math Answer Key Grade 1 Chapter 6 Count and Write Numbers to 120 99
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-96
11 tens and 0 ones is 110.

Explanation:
Given in the picture if we count there are
110 + 0 = 110 , 11 tens and 0 ones is 110.

Question 8
DIG DEEPER!
What number is equal to lo tens and 8 ones? Show how you know.
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-97
The number equal to 10 tens and 8 ones is 108

Explanation:
Given to find number equal to 10 tens and 8 ones,
as shown in figure taken 10 tens and 8 ones ,
the number is equal to 100 + 8 = 108.

Think and Grow: Modeling Real Life

Your teacher has 12 bags of balloons. Each bag has 10 balloons. How many balloons are there in all?
Big Ideas Math Answer Key Grade 1 Chapter 6 Count and Write Numbers to 120 100
Model:

____120_______ balloons
Answer:
There are 120 balloons are there in all.

Explanation:
Given my teacher has 12 bags of balloons and
each bag has 10 balloons so in all there are
12 X 10 = 120 balloons.

Show and Grow

Question 9.
A dentist has 10 boxes of toothbrushes and 9 extra toothbrushes. Each box has 10 toothbrushes. How many toothbrushes are there in all?
Big Ideas Math Answer Key Grade 1 Chapter 6 Count and Write Numbers to 120 101
Model:

___109________ toothbrushes
Answer:

10 tens and 9 ones is 109.
There are 109 tooth brushes are there in all.

Explanation:
Given a dentist has 10 boxes of toothbrushes and
9 extra toothbrushes, each box has 10 toothbrushes
means 10 x 10 = 100 and 9 extra makes 100 + 9 = 109 or
10 tens and 9 ones is 109 tooth brushes are there in all.

Count and Write Numbers to 120 Practice 6.9

Question 1.
Big Ideas Math Answer Key Grade 1 Chapter 6 Count and Write Numbers to 120 102
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-98

7 tens and 9 ones is 79.

Explanation:
Given in the picture if we count there are
70 + 9 = 79 , 7 tens and 9 ones is 79.

Question 2.
Big Ideas Math Answer Key Grade 1 Chapter 6 Count and Write Numbers to 120 103
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-99

11 tens and 5 ones is 115.

Explanation:
Given in the picture if we count there are
110 + 5 = 115 , 11 tens and 5 ones is 115.

Question 3.
Big Ideas Math Answer Key Grade 1 Chapter 6 Count and Write Numbers to 120 104
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-100

12 tens and 0 ones is 120.

Explanation:
Given in the picture if we count there are
120 + 0 = 120 , 12 tens and 0 ones is 120.

Question 4.
Big Ideas Math Answer Key Grade 1 Chapter 6 Count and Write Numbers to 120 105
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-101

9 tens and 2 ones is 92.

Explanation:
Given in the picture if we count there are
90 + 2 = 92 , 9 tens and 2 ones is 92.

Question 5.
DIG DEEPER!
What number is equal to 11 tens and 2 ones? Show how you know.
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-102
The number equal to 11 tens and 2 ones is 112.

Explanation:
We have 11 tens and 2 ones as shown in picture
11 tens is 110 and 2 ones is 2 so 110 + 2= 112.

Question 6.
Modeling Real Life
You have 4 packs of baseball cards and 6 packs of football cards. Each pack has 10 cards. How many cards do you have in all?
Big Ideas Math Answer Key Grade 1 Chapter 6 Count and Write Numbers to 120 106
_____100_______ cards
Answer:
I have in total 100 cards in all.

Explanation:
Given I have 4 packs of baseball cards and
6 packs of football cards, Each pack has 10 cards,
therefore 4 packs of baseball means 4 X 10 = 40 cards,
6 packs of football cards have 6 X 10 = 60 cards, In total
I have 40 + 60 =100 cards in all.

Review & Refresh

Make a 10 to add.

Question 7.
Big Ideas Math Answer Key Grade 1 Chapter 6 Count and Write Numbers to 120 107
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-103

Sum is 6 + 4 + 1 = 11  or  10 + 1 = 11,
So 6 + 5 = 11

Explanation:
First we make sum 10 , For 6 to make 10 we need 10-6=4,
as we have to  add 4 to 6 now  we will minus 4 from 5 we get 5 – 4 = 1,
Now we write the total sum as 6 + 4 + 1 = 11  or
10 + 1 = 11 or 6 + 5 = 11.

Question 8.
Big Ideas Math Answer Key Grade 1 Chapter 6 Count and Write Numbers to 120 108
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-104

Sum is 7 + 3 + 5 = 15  or  10 + 5 = 15,
So 7 + 8 = 15

Explanation:
First we make sum 10 , For 7 to make 10 we need 10-7=3,
as we have to  add 3 to 7 now  we will minus 3 from 8 we get 8 – 3 = 5,
Now we write the total sum as 7 + 3 + 5 = 15  or
10 + 5 = 15 or 7 + 8 = 15.

Count and Write Numbers to 120 Performance Task

Question 1.
Your class sells candles for a fundraiser. You earn 10 dollars for every large candle you sell and 1 dollar for every small candle.

a. You sell 6 large candles. How much money do you raise?
Big Ideas Math Answers 1st Grade 1 Chapter 6 Count and Write Numbers to 120 109
Answer:
I raised 60 dollars.

Explanation:
Given my class sells candles for a fundraiser.
I earn 10 dollars for every large candle I sell ,
I sold 6 large candles means 6 X 10 = 60 dollars,
So I raised 60 dollars.

b. You want to raise 72 dollars. How much more money do you need to raise?

_____12_____ more dollars
Answer:
I need 12 more dollars to raise 72 dollars.

Explanation:
I know I raised 60 dollars to raise 72 dollars
I need more is 72 – 60 = 12 dollars.

c. You also sell 12 small candles. Do you reach your goal?
Big Ideas Math Answers 1st Grade 1 Chapter 6 Count and Write Numbers to 120 110
Yes          No
Answer:
Yes, I reached my goal.

Explanation:
Given I earn1 dollar for every small candle I sell,
I sold 12 small candles means 12 X 1= 12 dollars,
In total I sold 6 large candles and 12 small candles
I raised total 60 + 12 = 72 dollars which is similar to
which I want to raise, So yes I reached my goal.

Question 2.
Your friend wants to raise 54 dollars. What are two ways your friend can sell large and small candles to reach her goal?
_____5____ large candles and ____4______ small candles
_____4_____ large candles and ___14_______ small candles
Answer:
5 large candles and 4 small candles,
4  large candles and 14 small candles.

Explanation:
I know if I sell 1 large candle I earn 10 dollars
and if I sell 1 small candles I earn 1 dollar to raise
54 dollars my friend can sell 5 tens and 4 ones is 54,
5 X 10 + 4 X 1 = 50 + 4 = 54,
So 5 large candles and 4 small candles makes 54 dollars
She can even sell 4 tens and 14 ones is 54,
4 X 10 + 14 X 1 = 40 + 14 = 54,
So 4  large candles and 14 small candles makes 54 dollars.

Count and Write Numbers to 120 Chapter Practice

Count to 120 by Ones Homework & Practice 6.1

Question 1.
Count by ones to write the missing numbers.
99, __100___, ___101___, __102___, __103___, ___104___
Answer:
The missing numbers after 99 are 100,101,102,103,104

Explanation:
Started from 99 counting by 1’s numbers after 99 are
99 + 1 = 100, 100 + 1 = 101,101 + 1 = 102, 102 + 1 = 103,
104 + 1 = 105, So the missing numbers after 99 are 100,101,102,103,104.

write the missing numbers.

Question 2.
Big Ideas Math Answers 1st Grade 1 Chapter 6 Count and Write Numbers to 120 111
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-105

The missing numbers in the chart are 69,71,78,80.

Explanation:
The missing number after 68 is 68 + 1 = 69,
the number after 70 is 70 + 1 = 71,
the number before 79 is 79 – 1 = 78 and
number before 81 is 81 – 1 = 80, So the missing
numbers in the chart are 69,71,78 and 80.

Question 3.
Big Ideas Math Answers 1st Grade 1 Chapter 6 Count and Write Numbers to 120 112
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-106

The missing numbers in the chart are 101,103,110,112,113

Explanation:
The missing number after 100 is 100 + 1 = 101,
the number after 102 is 102 + 1 = 103,
the number before 111 is 111 – 1 = 110,
the number after 111 is 111 + 1 = 112,
the number after 112 is 112 + 1 = 113,
So the missing numbers in the chart are 101,103,110,112,113.

Count to 120 by Tens Homework & Practice 6.2

Question 4.
Write the missing numbers from the chart. Then count on by tens to write the next two numbers.
Big Ideas Math Answers 1st Grade 1 Chapter 6 Count and Write Numbers to 120 113
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-107

The missing numbers in the chart are 84,94 and
count on 10’s the next two numbers after 94 are 104,114.

Explanation:
Given numbers in the chart, the number missing before 85
is 85 – 1 = 84 and number missing after 93 and before 94 in the chart is
93 + 1= 94 or 95 – 1 = 94,now counting by 10’s next
two numbers after 94 are 94 + 10 = 104,104 + 10 = 114,
Therefore the missing numbers in the chart are 84 and 94
count on 10’s the next two numbers after 94 are 104,114.

Question 5.
YOU BE THE TEACHER
Your friend counts by tens starting with 53. Is your friend correct? Show how you know.
53, 63, 73, 83, 103
Answer:
No friend is Incorrect.

Explanation:
Given friend counts by tens starting with 53 after 53 its is
53 + 10 = 63 then 63 + 10 = 73, 73 + 10 = 83 and after 83
it is 83 + 10 = 93 not 103 therefore friend is incorrect.

Compose Numbers 11 to 19 Homework & Practice 6.3

Question 6.
Circle 10 ducks. Complete the sentence.
Big Ideas Math Answers 1st Grade 1 Chapter 6 Count and Write Numbers to 120 114
____1______ ten and ____2______ ones is ___12____ .
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-108

1 ten and 2 ones is 12.

Explanation:
Circled 10 ducks, after counting we have total
12 ducks we write the sentence as
one ten and 2 ones is twelve means 10 + 2 = 12.
1 ten and 2 ones is 12.

Question 7.
Modeling Real Life
You have 19 tennis balls. A bag can hold 10. You fill a bag. How many tennis balls are not in the bag?

_____9_______ tennis balls
Answer:

9 tennis balls are not there in the bag.

Explanation:
Given I have 19 tennis balls and a bag can hold 10 and
I have filled a bag so balls that are not there in the bag are
19 – 10 = 9 , So 9 tennis balls are not there in the bag.

Tens Homework & Practice 6.4

Question 8.
Circle groups of 10. Complete the sentence.
Big Ideas Math Answers 1st Grade 1 Chapter 6 Count and Write Numbers to 120 115

______5____ tens and ___0___ ones is ___50______ .
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-109

fife tens and zero ones is fifty.
5 tens and 0 ones is 50.

Explanation:
Circled 5 tens ,five tens and zero ones is fifty or
5 tens and 0 ones is 50.

Tens and Ones Homework & Practice 6.5

Question 9.
Big Ideas Math Answers 1st Grade 1 Chapter 6 Count and Write Numbers to 120 116
____9_____ tens and ____9_____ ones is ___99_____ .
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-110

9 tens and 9 ones is 99

Explanation:
Given in the picture  there are 9 ten blocks in tens place
and 9 blocks in ones make 9 tens and 9 ones is 99.

Question 10.
Modeling Real Life
You have 6 boxes of plastic cups and 3 extra cups. Each box has 10 cups. How many cups are there in all?

_____63_____ plastic cups
Answer:
In all there are 63 plastic cups.

Explanation:
Given I have 6 boxes of plastic cups and 3 extra cups and
each box has 10 cups means 6 X 10 = 60 and 3 extra cups make
60 + 3 = 63 or 6 tens and 3 ones is  63 Plastic cups.

Make Quick Sketches Homework & Practice 6.6

Make a quick sketch. Complete the sentence.

Question 11.
17

____17____ is ____1___ ten and ____7____ ones.
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-111

17 is 1 tens and 7 ones.

Explanation:
Given in the picture  there are 1 ten blocks in tens place
and 7 blocks in ones make 17 as 1 tens and 7 ones.

Question 12.
84

___84____ is ____8___ tens and ___4__ ones.
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-112

84 is 8 tens and 4 ones is 84

Explanation:
Given in the picture  there are 8 ten blocks in tens place
and 4 blocks in ones make 84 as 8  tens and 4 ones.

Understand Place Value Homework & Practice 6.7

Question 13.
39
______3____ tens is ____30______ .
______9____ ones is ____9______ .
_____3____ tens and ____9______ ones is ____39______ .
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-113

3 tens is 30,
9 ones is 9,
So 3 tens and 9 ones is 39.

Explanation:
As shown in the picture above first we take 3 tens cubes as 30
and 9 ones cubes as 9, making 3 tens and 9 ones is 39.

Write Numbers in Different Ways Homework & Practice 6.8

Question 14.
Model 59 two ways.
Big Ideas Math Answers 1st Grade 1 Chapter 6 Count and Write Numbers to 120 117
_____5_____ tens and _____9_____ ones is 59.
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-114

5 tens and 9 ones is 59

Explanation:
we add 5 X 10 = 50 and 9 X 1 = 9
we get 50 + 9 = 59,
So 5 tens and 9 ones is 59.

Big Ideas Math Answers 1st Grade 1 Chapter 6 Count and Write Numbers to 120 117
_____2_____ tens and ____39______ ones is 59.
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-115
2 tens and 39 ones is 59

Explanation:
we add 2  X 10 = 20 and 39 X 1 = 39
we get  20 + 39 = 59,
So 2 tens and 39 ones is 59.

Count and Write Numbers to 120 Homework & Practice 6.9

Question 15.
Big Ideas Math Answers 1st Grade 1 Chapter 6 Count and Write Numbers to 120 118
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-116

11 tens and 4 ones is 114

Explanation:
As shown in the picture there are 11 ten blocks in tens place
and 4 blocks in ones make 11 tens and 4 ones is 114.

Question 16.
Big Ideas Math Answers 1st Grade 1 Chapter 6 Count and Write Numbers to 120 119
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-117

10 tens and 8 ones is 108

Explanation:
As shown in the picture there are 10 ten blocks in tens place
and 8 blocks in ones make 10 tens and 8 ones is 108.

Conclusion:

Start solving the problems in the performance task and chapter test which are at the end of the chapter and check the solutions from here. Hope the information about Big Ideas Math Book Grade 1 Answer Key Chapter 6 Count and Write Numbers to 120 is beneficial for you to overcome the difficulties in maths. Check out the links and start solving all the questions.

Big Ideas Math Answers Grade 5 Chapter 4 Multiply Whole Numbers

Big Ideas Math Answers Grade 5 Chapter 4 Multiply Whole Numbers

Get the free pdf of Big Ideas Math Answers Grade 5 Chapter 4 Multiply Whole Numbers. Every problem is explained with the solution and explanation on Big Ideas Multiply Whole Numbers Grade 5 Chapter 4 Math Answers. Our highly trained experts are provided the concepts in a simple way where every student can understand the solving methods. Also, various methods are included to solve the math problems. Therefore, practice all problems and score good marks in the exam. Every explanation is mainly explained with images, graphs, tables, and, etc. So, students will not face any problems to solve the basic questions too difficult questions.

Big Ideas 5th Grade Chapter 4 Multiply Whole NumbersMath Book Answer Key

Save your time by finding every concept and its related problems individually. Quickly go through the below links and check out all the topics with problems. Mainly, concentrate on concepts to prepare and solve the questions on your own. If you understand the concept, it becomes easier to solve difficult problems easily. Download Big Ideas Math Book 5th Grade Answers Chapter 4 Multiply Whole Numbers pdf and start your practice now. We have also included exercises for the better practice of students.

Lesson: 1 Multiplication Patterns

Lesson: 2 Estimate Products

Lesson: 3 Multiply by One-Digit Numbers

Lesson: 4 Multiply by Two-Digit Numbers

Lesson: 5 Multiply by Multi-Digit Whole Numbers

Performance Task

Lesson 4.1 Multiplication Patterns

Explore and Grow

Complete the table. Find the value of each expression.
Big Ideas Math Answer Key Grade 5 Chapter 4 Multiply Whole Numbers 4.1 1
What patterns do you notice?
Answer:

Power Expression Value
8 x 101 8 x 10 80
8 x 102 8 x 100 800
8 x 103 8 x 1,000 8,000
8 x 104 8 x 10,000 80,000

Explanation:

In each product, the number of zeros after 8 is the same as the exponent.

Patterns
Explain how to multiply any number by a power of 10.
Answer:
By multiplying the given number with the 10 as many times as to its the given power value.
Explanation :
Example  4 x 102  we take as  4 x 10 x 10 means 10 is multiplied twice
means as the power of 10 is 2 so we  multiply the number 4 with 10 twice,
if it 4 x 103 we take it as 4 x 10 x 10 x 10 means we take 10 by 3 times as power is 3

Think and Grow: Use Patterns to Find Products

Example
Find 46 × 103.
One Way:
Use powers and look for a pattern
Big Ideas Math Answer Key Grade 5 Chapter 4 Multiply Whole Numbers 4.1 2

46,000

Another Way:
Use place value.
46 × 103 = 46 × 1,000
= 46 thousands
= ___46,000___
Big Ideas Math Answer Key Grade 5 Chapter 4 Multiply Whole Numbers 4.1 3
So, 46 × 103.
Notice the pattern: In each product, the number of zeros after 46 is the same as the exponent.

Example
Find 40 × 600.
Use a multiplication fact and look for a pattern.
Big Ideas Math Answer Key Grade 5 Chapter 4 Multiply Whole Numbers 4.1 4
So, 40 × 600 = __24,000____.
Notice the pattern: In each product, the number of zeros after 24 is equal to the number of zeros in the factors.

Show and Grow

Find the product.
Question 1.
78 × 104
Answer:
The product of 78 × 10= 78 x 10,000 = 780,000
Explanation: As 78 is multiplied by 104  we get the result as 780,000

Question 2.
20 × 400 = ______
Answer:
The product of 20 x 400 = 2x 10 x 4 x 100 = 2 x 4 x 10 x 100 = 8,000
Explanation:
As 20 X 400 has 3 zeros we have 20 x 400 as 8,000

Apply and Grow: Practice

Find the product
Question 3.
11 × 103 = _______
Answer:
The product of 11 × 103 = 11 x 1,000=11,000
Explanation :
The result of  11 × 103  is 11,000

Question 4.
40 × 70 = _______
Answer:
The product of 40 x 70 = 4 x 10 x 7 x 10 =4 x 7 x 10 x 10 = 2,800
Explanation :
When 40 is multiplied by  70 the result is 2,800

Question 5.
300 × 90 = ________
Answer:
The product of 300 x 90 = 3 x 100 x 9 x 10 = 3 x 9 x 100 x 10 = 27,000
Explanation :
When 300 is multiplied by  90 the result is 27,000

Question 6.
104 × 60 = _______
Answer:
The product of 104 × 60 = 10 x 10 x 10 x 10 x 60 = 6,00,000
Explanation :
When 104  is multiplied by 60 the result is 6,00,000

Question 7.
206 × 102 = _______
Answer:
The product of 206 × 102 = 206 x 10 x 10 = 20,600
Explanation :
When 206 is multiplied by 102 the result is 20,600

Question 8.
15 × 200 = _________
Answer:
The product of 15 x 200 = 15 x 2 x 10 x 10 = 3,000
Explanation :
When 15 is multiplied by 200 the result is 3,000

Find the missing factor.
Question 9.
30 × ______ = 15,000
Answer:
30 x ____= 15,000,15,000/30 = 500
Explanation:
So when 15,000 is divided by 30 we get 500 the missing factor is 500

Question 10.
_____ × 103 = 54,000
Answer:
_____ × 103 = 54,000
the missing factor is 54
Explanation :
When 54,000 is divided by 10or 1000
we get 54,000 x 1/ 1000= 54

Question 11.
500 × ______ =40,000
Answer:
500 × ______ =40,000
the missing factor is 8,000
Explanation:
so when 40,000 is dived by 500 its is equal to
40,000 x 1/500=8,000

Question 12.
A heart pumps about 2,000 gallons of blood each day. How many gallons of blood does the heart pump in 7 days?
Answer:
Given that heart pumps about 2,000 gallons of blood each day and for 7 days it is
2,000 x 7 = 14,000 gallons
Explanation :
So we multiply 2000 by 7 for 7 days and the result is 14,000 gallons

Question 13.
Number Sense
Which expressions are equal to 24,000?
Big Ideas Math Answer Key Grade 5 Chapter 4 Multiply Whole Numbers 4.1 5
Answer:
240 x 100= 24 x 10 x 10 x 10  = 24,000 so it is equal
30 x 800 = 3 x 10 x 8 x 10 x 10 = 24,000 so this is also equal
400 x 600= 4 x 10 x 10 x 6 x 10 x 10 =24,000 so this also equal
and 24 x 1,000 = 24 x 10 x 10 x 10 = 24,000 which is also equal
so all the given expressions are equal to 24,000
Explanation :
When Big Ideas Math Answer Key Grade 5 Chapter 4 Multiply Whole Numbers 4.1 5

these numbers are multiplied all results are equal to 24,000

Question 14.
Logic
When you multiply any whole number by 100, what do you know about the digits in the product?
Answer:
When we multiply any whole number by 100, it becomes the digit with two zeros in the right , number x 100= number x 102 =number,00
or we get the whole number with additional 2 zeros in the right
Explanation:
For example if the number is 5 it becomes as 5 x 100 = 5 x 102= 500, 2 zeros after 5

Question 15.
DIG DEEPER!
Show how to find 20 × 103 using the Associative Property of Multiplication.
Answer:
The Associative Property of Multiplication of 20 × 103  is
(20 x 10) x 10 x 10 = 20 x ( 10 x 10 ) x 10
(200) x 10 x 10 = 20 x (100) x 10
20,000 = 20,000
Explanation:
Both sides are equal by using the Associative Property of Multiplication Law.

Think and Grow: Modeling Real Life

Example
A black bear consumes 12,000 calories each day for 30 days. A grizzly bear consumes 400,000 calories in 30 days. Which bear consumes more calories in 30 days?
Multiply 12,000 and 30 to find the number of calories the black bear consumes in 30 days. Use a multiplication fact and a pattern.
12 × 3 = ____36____ Multiplication fact
The multiplication fact is 12 × 3 = 36. The factors 12,000 and 30 have a total of _____4_ zeros. So, the product has __4___ zeros after 36.
12,000 × 30 = ___3,60,000______ calories
The black bear consumes 3,60,000 calories in 30 days. Compare the numbers of calories the bears consume in 30 days.
A grizzly bear consumes 4,00,000 x 30 = 120,00,000
So  grizzly bear consumes more calories than black bear
The ___grizzly ___ bear consumes more calories in 30 days.

Show and Grow

Question 16.
The attendance at a theme park is about 300,000 guests each month. The attendance at a water park is 2,280,000 guests in 1 year. Which park has a greater attendance in 1 year?
Answer:
The theme park has a greater attendance in 1 year.
Explanation:
The attendance at theme park is about 3,00,000 guests in one month so in a year it is 3,00,000 x 12 = 36,00,000 guests.
and given water park has 22,80,000 guests= 228 x 104  in 1 year
so as compared between the two parks 36,00,000 guests = 360 x104  is greater than 22,80,000 guests= 228 x 10 means the theme park has a greater attendance in 1 year.

Question 17.
Your heart beats 120,000 times each day. In 1 week, a professional athlete’s heart beats 607,000 times. How many more times does your heart beat in 1 week compared to the athlete?
Answer:
Our heart beats 233,000 times more compared to the athlete in 1 week
Explanation :
Each day our heart beats 120,000 times each day
so in 1 week it is 120,000 x 7 = 840,000 times our heart beats  in 1 week given  athlete’s heart beats in 1 week is 607,000 times so compared to both it is
840,000 – 607,000 = 233,000 times
So our heart beats 233,000 times more compared to the athlete in 1 week

Question 18.
A whale’s brain weighs 103 times as much as a platypus’s brain. The whale’s brain weighs 9,000 grams. How many grams does the platypus’s brain weigh?
Big Ideas Math Answer Key Grade 5 Chapter 4 Multiply Whole Numbers 4.1 6
Answer:
platypus’s brain weighs 9 grams
Explanation:
Given whale’ brain weighs 103 times  as much as a platypus’s brain and also given that whale’s brain weighs 9,000 grams
so platypus’s brain weighs 9,000 x 1/ 103 times = 9,000x 1/1000 = 9 grams.

Multiplication Patterns Homework & Practice 4.1

Find the product
Question 1.
37 × 104
Answer:
37 × 10= 370,000
Explanation:
The product of 37 × 10= 37 x 10 x 10 x 10 x 10= 370,000

Question 2.
400 × 90 = ______
Answer:
400 × 90 = 36,000
Explanation:
The product of 400 × 90 = 4 x 10 x 10 x 9 x 10 = 4 x 9 x 103 = 36,000

Question 3.
100 × 900 = _____

Answer:
100 x 900 = 90,000
Explanation:
The product of 100 x 900 = 10 x 10 x 9 x 10 x 10 = 9 x 10= 90,000

Question 4.
800 × 700 = _____
Answer:
800 x 700 = 560,000
Explanation:
The product of 800 x 700 = 8 x 10 x 10 x 7 x 10 x 10 = 8 x 7 x 10=560,000

Question 5.
867 × 103 = ______
Answer:
867 × 103 =867,000
Explanation:
The product of 867 × 103 = 867 x 10 x 10 x 10 = 867,000

Question 6.
12 × 6,000 = _______
Answer:
12 x 6,000 = 72,000
Explanation:
The product of 12 x 6,000 = 12 x 6 x 10 x 10 x 10= 12 x 6 x 103 = 72,000

Find the missing factor.
Question 7.
____ × 102 = 3,000
Answer:
30 x 102 = 3,000
Explanation:
The missing factor of  ____ × 102 = 3,000 is 3,000 x 1/100= 3 x 103 / 10=30

Question 8.
______ × 500 = 450,000
Answer:
900 x 500 = 450,000
Explanation:
The missing factor of ______ × 500 = 450,000 is 450,000 x 1/500= 450,000/500=
45 x 104  / 5 x 10= 9 x 10= 900

Question 9.
80 × _____ = 640,000
Answer:
80 x 8000 = 640,000
Explanation:
The missing factor of 80 × _____ = 640,000 is 640,000 x 1/80 = 64 x 104  / 8 x 10=
8 x 10=8,000

Question 10.
Reasoning
Do 50 × 8,000 and 50 × 9,000 have the same number of zeros in their products? Explain.
Answer:
Yes, 50 × 8,000 and 50 × 9,000 have the same number of zeros in their products
Explanation:
Why because 50 x 8,000 = 5 x 10 x 8 x 10 x 10 x 10 = 5 x 8 x 104
and 50 x 9,000 = 5 x 10 x 9 x 10 x 10 x 10 = 5 x 9 x 104
so when compared between 5 x 8 x 104  and 5 x 9 x 10we have 4 zeros in both the products.

Question 11.
DIG DEEPER!
The product of a number and twice that number is 180,000. Write the multiplication equation.
Answer:
2 n2 = 18 x 104
Explanation:
let the number be n and its twice is 2 n is 180,000
n x 2n = 18 x 10 x 10 x 10 x 10
2 n2 = 18 x (10 x 10) x (10 x 10)
so the multiplication equation is 2 n2 = 18 x 10

Question 12.
Modeling Real Life
A hummingbird flutters its wings 80 times in 1 second. A bumblebee flutters its wings 7,800 times in 1 minute. Which animal flutters its wings more times in 1 minute?
Big Ideas Math Answer Key Grade 5 Chapter 4 Multiply Whole Numbers 4.1 7
Answer:
bumblebee flutters more

Explanation:
Given humming bird flutters its wings 80 times in 1 second equals to 8 x 10,
bumblebee flutters its wings 7,800 times in 1 second equals to 78 x 102
as 78 x 10> 8 x 10 means bumblebee flutters its wings more than humming bird in 1 minute.

Question 13.
DIG DEEPER!
In 1 day, the blood in your body travels a total of 12,000 miles. This is about 4 times the distance across the United States from coast to coast. About how many miles is the distance across the United States?
Answer:
The distance across the United States is 3,000 miles.
Explanation:
Given in 1 day the blood in our body travels a total of 12,000 miles which is equal to 4 times the distance across the United States from coast to coast.
so the number of miles in the distance across the United States is
12,000 miles / 4=3,000 miles.

Review & Refresh

Classify the triangle by its sides.
Question 14.
Big Ideas Math Answer Key Grade 5 Chapter 4 Multiply Whole Numbers 4.1 8
Answer:
Equilateral Triangle
Explanation:
In the  given figure it is showing all the 3 sides are equal so we call  the triangle as Equilateral Triangle.

Question 15.
Big Ideas Math Answer Key Grade 5 Chapter 4 Multiply Whole Numbers 4.1 9
Answer:
Isosceles Triangle
Explanation:
In the above figure it is showing only 2 sides are equal so we call the given triangle as Isosceles Triangle.

Question 16.
Big Ideas Math Answer Key Grade 5 Chapter 4 Multiply Whole Numbers 4.1 10
Answer:
Scalene Triangle
Explanation:
In the above figure it is showing no side is equal so we call the given triangle as
Scalene Triangle.

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 5th Grade Chapter 4 Multiply Whole Numbers 4.2 1
Big Ideas Math Answers 5th Grade Chapter 4 Multiply Whole Numbers 4.2 2
Compare your answers with a partner’s. Did you choose the same expressions?
Answer:
29 x 26
30 x 25
Explanation:
so for 29 x 26 i am choosing the expression 30 x 25 as 29 is near to 30 and 26 is near to 26 i have selected the expression 30 x 25
21 x 24
20 x 25
Explanation :
for 21 x 24 i am choosing the expression 20 x 25 as 21 is near to 20 and 24 is near to 25 i have selected the expression 20 x 25
32 x 26
30 x 25
Explanation :
for 32 x 26 i am choosing the expression 30 x 25 as in the given expressions only 32 is near to 30 and 26 is near to 25 i have selected the expression 30 x 25
28 x 24
30 x 25
Explanation :
for 28 x 24 also i am choosing the expression 30 x 25 why because 28 is near to 30 and 24 is near to 25 so i am selecting the expression  30 x 25
Yes i have selected the same expression 30 x 25  for all the three products 29 x 26, 32 x 26,28 x 24 because for all the given three products the nearest expression we are getting is 30 x 25

Construct Arguments
You estimate the product of 23 and 26 using the expression 25 × 30. Without multiplying, determine whether the estimate is greater than or less than the actual product. Explain.
Answer:
Given the product 23 x 26 as 25 x 30
So the result of given estimate is greater than the actual value
Explanation:
Because as 25 is greater than 23 and 30 is greater than 26 so obviously the expression 25 x 30 has greater value than that of the product of 23 X 26

Think and Grow: Estimate Products

An overestimate is greater than the actual value, and an underestimate is less than the actual value. When you estimate a product, your answer will be an overestimate when both numbers are greater than the original factors. Your answer will be an underestimate when both numbers are less than the original factors.

Example
Use rounding to estimate 204 × 61. Explain whether your estimate is an overestimate or an underestimate.
Big Ideas Math Answers 5th Grade Chapter 4 Multiply Whole Numbers 4.2 3

12 x 1,000 = 12,000
So, 204 × 61 is about __12,444___. This is an __underestimate______ because about both rounded numbers are ___200___, __60____ the original factors.

Example
Use compatible numbers to estimate 23 × 194. Explain whether your estimate is an overestimate or an underestimate.
Choose compatible numbers. Then multiply.
Big Ideas Math Answers 5th Grade Chapter 4 Multiply Whole Numbers 4.2 4

50 x100 = 5,000
So, 23 × 194 is about __4,462____. This is an __overestimate_____ because both compatible numbers are __25____ __200____ the original factors.

Show and Grow

Estimate the product. Explain whether your estimate is an overestimate or an underestimate.
Question 1.
387 × 29
Answer:
387 x 29 =11,223
400  X 30 = 12,000 is overestimate
Explanation :
As our estimate after  rounding each factor becomes 387 as 400 nearest hundred  and 29 as 30 to nearest ten so it is 400 x 30 = 12 x 1,000=12,000 as 387 x 29 < 400 x 30 as the estimate result is greater than the given product results our estimate is an overestimate

Question 2.
52 × 913
Answer:
52 x 913 = 47,476
50 x 900 = 45,000 is underestimate

Explanation:
As our estimate after rounding each factor becomes 52 as 50 to nearest ten and 913 as 900 nearest hundred so it is 50 X 900 = 45,000
as 52 x 913 > 50 x 900 as the estimate result is less than the given product results
our estimate is an underestimate.

Apply and Grow: Practice

Estimate the product. Explain whether your estimate is an overestimate or an underestimate.
Question 3.
45 × 98
Answer:

45 x 98 = 4,410
50 x 100 = 5,000 is overestimate
Explanation:
As our estimate after rounding each factor becomes 45 as 50 to nearest ten and 98 as 100 to nearest hundred so it is 50 X 100 = 5,000
as 45 x 98 < 50 x 100 as the estimate result is greater than the given product results our estimate is an overestimate

Question 4.
21 × 404
Answer:
21 x 404 = 8,484
20 x 400 = 8,000 is underestimate

Explanation:
As our estimate after rounding each factor becomes 21 as 20 nearest ten and 404 as 400 to nearest hundred so it is 20 X 400 = 8,000
as 21 x 404 > 20 x 400 as the estimate result is less than the given product results our estimate is an underestimate.

Question 5.
394 × 285
Answer:
394 x 285 = 112,290
400 x 300 = 120,000 is overestimate
Explanation:
our estimate after rounding each factor becomes 394  as 400 to nearest hundred and 285 as 300 to nearest hundred so it is 400 X 300 = 120,000
as 394 x 285 < 400 x 300 as the estimate result is greater than the given product results our estimate is an overestimate

Question 6.
596 × 488
Answer:
596 x 488 = 290,848
600 x 500 = 300,000 is overestimate

Explanation:
As our estimate after rounding each factor becomes 596  as 600 to nearest hundred and 488 as 500 to nearest hundred so it is 600 X 500 = 300,000
as 596 x 488 < 600 x 500 as the estimate result is greater than the given product results our estimate is an overestimate

Question 7.
194 × 46
Answer:
194 x 46=8,924
200 x 50 = 10,000 overestimate

Explanation:
As our estimate after rounding each factor becomes 194  as 200 to nearest hundred and 46 as 50 to nearest ten so it is 200 X 50 = 10,000
as 194 x 46 < 600 x 500 as the estimate result is greater than the given product results our estimate is an overestimate

Question 8.
22 × 221
Answer:
22 x 221 =4,862
20 x 200 = 4,000 underestimate

Explanation :
As our estimate after rounding each factor becomes 22  as 20 to nearest ten and 221 as 200 to nearest hundred so it is 20 X 200 = 4,000
as 22 x 221 > 20 x 200 as the estimate result is less than the given product results our estimate is an underestimate.

Question 9.
Number Sense
You estimate 27 × 408 by rounding each number to the nearest ten. Your friend estimates the product using the compatible numbers 25 and 400. Without multiplying, determine which answer is an overestimate and which answer is an underestimate. Explain.
Answer:
27 x 408
30 x 410 Mine is overestimate and
27 x 408
25 x 400 Friend is underestimate
Explanation:
Given my estimate as 27 x 408 by rounding the number to the nearest ten it becomes 27 as 30 and 408 as 410 as 27 is less than 30 and 408 is also less than 410 so the product 27 x 408 is less than 30 x 410
27 x 408
30 x 410
as 27 x 408 < 30 x 410 mine is overestimate.
Now 27 x 408
as     25 x 400
my friend estimates the product using the compatible numbers 25 and 400 as 25 is less than 27 and 400 is less than 408 obviously the product results 25 X 400 <27 x 408 so my friend results are underestimate.
therefore mine is overestimate and friends is underestimate

Question 10.
Writing

Explain two different methods to estimate 49 × 305. Which do you prefer?
Answer:
I prefer 50 x 300
Explanation:
49 x 305 by rounding each number to the nearest ten it is 50 x 310 ,
49 x 305
50 x 310
49 x 305=14,945 and 50 x 310 = 15,500
49 as 50 and 305 as 310 product so 49 is less than 50 and 305 is also less than 310 so the product 49 x 305 is less than 50 x 310
49 x 305 <  50 x 310 which is overestimate
now 49 x 305
50 x 300
49 x 305 compatible numbers are 50 and 300 the results 49 x 305 = 14,945 and 50 x 300 = 15,000 means 49 x 305 < 50 x 300 which is also overestimate,
as both the results are overestimate but i prefer 50,300  compatible numbers only as these number results are very near to the given product 49 x 305 as compared to the rounding the numbers to nearest ten.
so i prefer 50 x 300
Question 11.
DIG DEEPER!
You estimate 9 × 26 using the compatible numbers 10 and 25. Is your estimate an overestimate or an underestimate? Explain.
Answer:
My estimate is underestimate.
Explanation:
My estimate is given as 9 x 26 = 234
and compatible numbers result is  10 x 25 = 250,
as 9 x26 < 10 x 25 as my estimate results is less than the compatible numbers result so my estimate is underestimate

Think and Grow: Modeling Real Life

Example
Earth travels about 1,118 miles in 1 minute. About how far does Earth travel in 1 hour?
Think: What do you know? What do you need to find? How will you solve?
Big Ideas Math Answers 5th Grade Chapter 4 Multiply Whole Numbers 4.2 5
There are 60 minutes in 1 hour, so multiply the distance by 60.
60 × 1,118 = ?
Use rounding to estimate the product.
Big Ideas Math Answers 5th Grade Chapter 4 Multiply Whole Numbers 4.2 6
So, Earth travels about _60,000_____ miles in 1 hour.

Show and Grow

Question 12.
Old Faithful erupts about 19 times in 1 day. Each eruption sprays about 6,050 gallons of water. About how many gallons of water does Old Faithful spray in 1 day?
Big Ideas Math Answers 5th Grade Chapter 4 Multiply Whole Numbers 4.2 7
Answer:
Old faithful spray in 1 day19 x 6,000= 114,000 gallons of water
Explanation :
Given Old Faithful erupts about 19 times in 1 day and each eruption sprays about 6,050 gallons of water, so Old faithful sprays about 19 x 6,050=114,950 gallons of water in 1 day.
Now using rounding to estimate the product it becomes 19 as 20 and 6,050 as 6,000
so Old faithful sprays about 19 x 6,000=114,000 gallons of water in 1 day.

Question 13.
Workers place a foam finger on every seat in a stadium for an upcoming football game. There are 29 sections in the stadium and about 245 seats in each section. About how many foam fingers are needed? Is it better to have an overestimate or an underestimate? Explain.
Big Ideas Math Answers 5th Grade Chapter 4 Multiply Whole Numbers 4.2 8
Answer:
Needed are 7,105 number of foam fingers and It is better to have overestimate
Explanation:
Given there are total 29 sections in the stadium
and 245 seats in each section.
The number of foam fingers needed would be 29 x 245 = 7,105 foam fingers
So we need 7,105 number of foam fingers
So now by using rounding 29 becomes 30 and 245 as 250 so it becomes
30 x 250=7,500 foam fingers.
Now on comparing 29 x 245 < 30 x 350 so the results is overestimate.
So it is better to have overestimate only

Question 14.
DIG DEEPER!
One acre of land is equal to the area of a rectangular piece of land that is 22 yards wide and 220 yards long. Central Park in New York City is 843 acres. What is the approximate area of Central Park in square yards?
Answer:
Central Park in New York City has 22 x 220 x 843 square yards
Explanation:
Given one acre of land is equal to 22 yards wide and 220 yards long
means one acre is 22 x 220 square yards
So Central Park in New York City has 22 x 220 x 843 square yards.

Estimate Products Homework & Practice 4.2

Use rounding to estimate the product. Explain whether your estimate is an overestimate or an underestimate.
Question 1.
85 × 96
Answer:
After rounding 90x 100= 9,000, overestimate
Explanation:
85 x 96=8,160  after rounding it becomes 90 x 100 =9,000
as 85 is less than 90 and 96 is less than 100 so it is overestimate

Question 2.
41 × 21
Answer:
After rounding 40 x 20=800, underestimate
Explanation:
41 x 21 = 861 after rounding it becomes 40 x 20 = 800
as 41 is greater than 40 and 21 is also greater than 20 so it becomes underestimate

Use compatible numbers to estimate the product. Explain whether your estimate is an overestimate or an underestimate.
Question 3.
56 × 106
Answer:
56 x 106 = 5,936 and 55 x 100 = 5,500
as 5,936 > 5,500 ,So my estimate is overestimate
Explanation:
For 56 x 106 compatible numbers are 55 and 100 and its product is 5,500,
56 is greater than 55 and 106 is greater than 100,so the product of given numbers is greater than the product of compatible numbers, so my estimate is overestimate

Question 4.
23 × 597
Answer:
23 x 597 = 13,731 and 25 x 600 = 15,000
as 13,731 < 15,000, So my estimate is underestimate
Explanation:
For 23 x 597 compatible numbers are 25 and 600
56 is greater than 55 and 106 is greater than 100,so the product of given numbers is lesser than the products of compatible numbers, so my estimate is underestimate

Estimate the product
Question 5.
203 × 85
Answer:
203 x 85 = 17,255 and 200 x 90 = 18,000
as 17,255 < 18,000 So my estimate is underestimate
Explanation:
For 203 x 85 compatible numbers are 200 and 90
203 is greater than 200 and 85 is lesser than 90,But the product of given numbers is lesser than the product of compatible numbers so my estimate is underestimate
Question 6.
67 × 405
Answer:
67 x 405 = 27,135 and 70 x 400 = 28,000
as 27,135 < 28,000 So my estimate is underestimate
Explanation:
For 67 x 405 compatible numbers are 70 and 400
As 67 is less than 70 and 405 is greater than 400,but the product of given numbers is lesser than the product of compatible numbers, so my estimate is underestimate

Question 7.
Number Sense
Estimate 426 × 2,045. Is your answer an overestimate or an underestimate? Explain.
Answer:
426 x 2,045=871,170 and 430 x 2,000= 860,000
as 871,170 > 860,000 so mine is overestimate
Explanation:
For 426 x 2,045 compatible numbers are 430 and 2000
As 426 is less than  and 430 and 2,045 is greater than 2,000 but the product of given numbers is greater than the product of compatible numbers, so my estimate is overestimate
Question 8.
Open-Ended
Write 2 three-digit numbers. Then estimate their product by rounding so that the answer is an underestimate.
Answer:
188 x 478 = 89,864
200 x 500 = 100,000
as 89,864 < 100,000 , so 188 x 478 is underestimate
Explanation:
To make the product as an underestimate let us take the 2 three-digit numbers as 188 and 478 and their compatible numbers 200 and 500,
As 188 is lesser than 200 and 478 is also less than 500 obviously the product becomes lesser than the product of compatible numbers and their product becomes underestimate.
Question 9.
Reasoning
Without multiplying, determine which estimate is closer to the actual product. Explain.
Big Ideas Math Answers 5th Grade Chapter 4 Multiply Whole Numbers 4.2 9
Answer:
The closer product to 21 x 372  is 20 x 400
Explanation :
If we compare 21 x 372 with 20 x 400 we see 20 is less than 21 which is very close and 372 is near to 400 and the result is much closer than when compared to the result of product of 25 x 400 as 25 is greater than to 21 and even 400 is greater than 372, so obviously the product is much far, So the closer product is 20 x 400

Question 10.
Modeling Real Life
Your cousin sells jars of jam at a farmers’ market. He sells 48 jars in 1 day for $12 each. About how much money does he earn in sales in 1 day?
Answer:
48 x $12 = $ 576 in 1 day
Explanation:
Given that cousin sells 48 jars in 1 day, each jar cost $12 , so the total cost per 1 day is product of number of jars X cost of 1 jar =48 x $12 = $ 576 in 1 day
So my cousin sells $ 576 in 1 day
Question 11.
DIG DEEPER!
SupersaurusA weighed 45 tons. About how many more pounds did an Argentinosaurus weigh than a Supersaurus?
Big Ideas Math Answers 5th Grade Chapter 4 Multiply Whole Numbers 4.2 10
Answer:
Argentinosaurus weighs 72,000 pounds more than a Supersaurus
Explanation:
As we know 1 ton = 2000 pounds, given Supersaurus weighed 45 tons = 45 x 2000=90,000 pounds and Argentinosaurus weighs 81 tons = 81 x 2000= 162,000 pounds, So 162,000 – 90,000= 72,000 means Argentinosaurus weigh 72,000 more than a Supersaurus

Review & Refresh

Use the figure.
Big Ideas Math Answers 5th Grade Chapter 4 Multiply Whole Numbers 4.2 11
Question 12.
Name a pair of lines that appear to be parallel.
Answer:
( A,G ) and ( C,J) appear as parallel
Explanation:
In geometry, parallel lines can be defined as two lines in the same plane that are at equal distance from each other and never meet so ( A,G ) and ( C,J) appear as parallel
Question 13.
Name two lines that are perpendicular.
Answer:
( D, F ) and ( B, H ) are perpendicular
Explanation:
In geometry,
Two distinct lines intersecting each other at 90° or a right angle are called perpendicular lines
Question 14.
Name two intersecting lines.
Answer:
(D,F) and (C,J) are two intersecting lines
Explanation:
As the line DF are CJ are intersecting at point F so DF,CJ are called as intersecting lines

Lesson 4.3 Multiply by One-Digit Numbers

Explore and Grow

Use the area model and partial products to find 1,985 × 4.
Big Ideas Math Answers Grade 5 Chapter 4 Multiply Whole Numbers 4.3 1
Answer:
1,985 x 4 = 7,940
Reasonable 7,940  is close to 8,000
Explanation :
Step 1: Multiply the ones. Regroup
4 x 5 ones = 20 ones, Regroup  20 ones as 2 tens and 0 ones
2
1985
x    4
      0
Step 2 : Multiply the tens. Regroup
4 x 8 tens  = 32 tens,  32 tens + 2 tens = 34 tens, Regroup 34 tens as
3 hundreds and 4 tens
3
1985
x    4 
40
Step 3 : Multiply the hundreds. Regroup hundreds
4 x 9 hundreds = 36 hundreds, 36 hundreds + 3 hundreds = 39 hundreds
Regroup 39 hundreds as 3 thousands and 9 hundred
3
1985
x    4 
940
Step 4 : Multiply the thousands. Add the regrouped thousands
4 x 1 thousands = 4 thousands,4 thousands +3 thousands=7 thousands
3
1985
x    4
7,940
1,985 x 4 = 7,940         Reasonable 7,940  is close to 8,000

Structure
Explain how you can use an area model and partial products to find 2,083 × 3.
Answer:
2,083 x 3 = 6,249
Reasonable 6,249  is close to 6,000 or 6,300
Explanation:

Step 1: Multiply the ones.
3 x 3 ones = 9 ones,
2083
x    3
     9
Step 2 : Multiply the tens. Regroup
3 x 8 tens  = 24 tens, Regroup 24 tens as 2 hundreds and 4 tens
2
2083
x    3
   49
Step 3 : Multiply the hundreds.
3 x 0 hundreds = 0 hundred, 0 hundred + 2 hundreds = 2 hundreds
2083
x    3
  249
Step 4 : Multiply the thousands.
3 x 2  thousands = 6 thousands
2083
x    3
6,249
2,083 x 3 = 6,249         Reasonable 6,249  is close to 6,000 or 6,300

Think and Grow: Multiply by One-Digit a Number

Example
Find 8,902 × 4. Estimate ___35,608___
Big Ideas Math Answers Grade 5 Chapter 4 Multiply Whole Numbers 4.3 2

So 8,902 x 4 = 35,608.                  Reasonable 35,608 is close to 35,600

Show and Grow

Find the product. Check whether your answer is reasonable.
Question 1.
Big Ideas Math Answers Grade 5 Chapter 4 Multiply Whole Numbers 4.3 3
Big Ideas Math Answers Grade 5 Chapter 4 Multiply Whole Numbers 4.3 4
Answer:
1. 7,346 x 2= 14,692               Reasonable 14,692 and  is close to 15,000
2. 12,053 x 3 = 36,159            Reasonable 36,159 and  is close to 36,000
Explanation:
1.7,346 x 2=14,692
Step 1: Multiply the ones. Regroup
2 x 6 ones = 12 ones, Regroup 12 ones as 1 tens and 2 ones
1
7,346
x    2
     2
Step 2 : Multiply the tens.
2 x 4 tens  = 8 tens,  8 tens + 1 tens = 9 tens
7,346
x    2
    92
Step 3 : Multiply the hundreds.
2 x 3 hundreds = 6 hundreds
7,346
x     2
   692
Step 4 : Multiply the thousands.
2 x 7 thousands = 14 thousands
7,346
x     2
14,692
So 7,346 x 2 = 14,692        Reasonable 14,692 and  is close to 15,000

2. 12,053 x 3 =36,159
Step 1: Multiply the ones.
3 x 3 ones = 9 ones
12,053
x      3
        9
Step 2 : Multiply the tens. Regroup
3 x 5 tens  = 15 tens, Regroup 15 tens as 1 hundred and 5 tens
1
12,053
x       3
       59
Step 3 : Multiply the hundreds. Regroup
3 x 0 hundreds = 0 hundreds, Regroup 0 hundred +1 hundred = 1 hundred
12,053
x      3
159
Step 4 : Multiply the thousands.
3 x 2 thousands = 6 thousands
12,053
x      3
  6,159
Step 5  : Multiply the ten thousands.
3 x 1 ten thousands = 3 ten thousands
12,053
x      3
36,159
12,053 x 3 = 36,159              Reasonable 36,159 is close to 36,000

Apply and Grow: Practice

Find the product. Check whether your answer is reasonable.
Question 3.
Big Ideas Math Answers Grade 5 Chapter 4 Multiply Whole Numbers 4.3 5
Answer:
503 x 5 = 2,515           Reasonable as 2,515 is close to 2,500
Explanation:
Step 1: Multiply the ones.
5 x 3 ones = 15 ones, Regroup 15 ones as 1 tens and 5 ones
1
503
x  5
    5
Step 2 : Multiply the tens. Regroup
5 x 0 tens  = 0 tens,  0 tens + 1 tens = 1 tens
503
x  5
  15
Step 3 : Multiply the hundreds.
5 x 5 hundreds = 25 hundreds
503
x  5
2,515
503 x 5 = 2,515        Reasonable 2,515 is close to 2,500

Question 4.
Big Ideas Math Answers Grade 5 Chapter 4 Multiply Whole Numbers 4.3 6
Answer:
5,295 x 9 = 47,655             Reasonable 47,655 is close to 48,000
Explanation:
Step 1: Multiply the ones. Regroup
9 x 5 ones = 45 ones, Regroup 45 ones as 4 tens and 5 ones
4
5,295
x     9
      5
Step 2 : Multiply the tens. Regroup
9 x 9 tens  = 81 tens, Regroup 81 tens+ 4 tens = 85 tens as 8 hundred and 5 tens
8
5,295
x     9
55
Step 3 : Multiply the hundreds. Regroup
9 x 2 hundreds = 18 hundreds, Regroup 18 hundred +8 hundred =
26 hundreds as 2 thousand and 6 hundreds
2
5,295
x     9
655
Step 4 : Multiply the thousands.
9 x 5 thousands = 45 thousands, Regroup 45 thousands+2 thousand=47 thousand
2
5,295
x     9 
47,655
5,295 x 9  =47,655          Reasonable 47,655 is close to 48,000

Question 5.
Big Ideas Big Ideas Math Answers Grade 5 Chapter 4 Multiply Whole Numbers 4.3 7
Answer:
87,294 x 3 = 2,61,882        Reasonable 2,61,888 is close to 2,60,000
Explanation:
Step 1: Multiply the ones. Regroup
3 x 4 ones = 12 ones, Regroup 12 ones as 1 ten and 2 ones
1
87,294
x      3
        2
Step 2 : Multiply the tens. Regroup
3 x 9 tens  = 27 tens, Regroup  27 tens + 1 ten = 28 tens as 2 hundred and 8 tens
2
87,294
x       3
82
Step 3 : Multiply the hundreds. Regroup
3 x 2 hundreds = 6 hundreds, Regroup 6 hundred +2 hundred = 8 hundreds
87,294
x      3
882
Step 4 : Multiply the thousands. Regroup
3 x 7 thousands = 21 thousands, Regroup as 2 ten thousands and 1 thousand
2
87,294
x      3
1882
Step 5  : Multiply the ten thousands.
3 x 8 ten thousands = 24 ten thousands,24 ten thousands+ 2 ten thousands
=26 ten thousands, regroup as 2 hundred thousand and 6 ten thousand
2
87,294
x      3
2,61,882
87,294 x 3 =2,61,882        Reasonable 2,61,888 is close to 2,60,000

Question 6.
399 × 2 = _____
Answer:
399 x 2 = 798        Reasonable 798 is close to 800
Explanation:
Step 1: Multiply the ones.
2 x 9 ones = 18 ones, Regroup 1 ones as 1 tens and 8 ones
1
399
x    2
      8
Step 2 : Multiply the tens. Regroup
2 x 9 tens  = 18 tens,  18 tens + 1 tens = 19 tens, Regroup 19 tens as 1 hundred  and 9 tens
1
399
x   2                            
   98
Step 3 : Multiply the hundreds.
2 x 3 hundreds = 6 hundreds, Regroup 6 hundreds+1 hundreds= 7 hundreds
399
x   2 
798
So 399 x 2 = 798       Reasonable 798 is close to 800

Question 7.
33,184 × 2 = _____
Answer:
33,184 x 2 =66,368  Reasonable 66,368 is close to 66,000
Explanation:
Step 1: Multiply the ones.
2 x 4 ones =8 ones
33,184
x      2
        8
Step 2 : Multiply the tens. Regroup
2 x 8 tens  = 16 tens, Regroup 16 tens as 1 hundred  and 6 tens
1
33,184
x      2
68
Step 3 : Multiply the hundreds. Regroup
2 x 1 hundreds = 2 hundreds, Regroup 2 hundreds +1 hundred =  3 hundreds
33,184
x      2
368
Step 4 : Multiply the thousands.
2 x 3 thousands = 6 thousands
33,184
x      2
  6368
Step 5  : Multiply the ten thousands.
2 x 3 ten thousands = 6 ten thousands
33184
x      2
66,368
66,368  x 2 = 66,368        Reasonable 66,368 is close to 66,000

Question 8.
60,759 × 4 = _______
Answer:
60,759 x 4 = 2,43,036        Reasonable 2,43,036 is close to 2,40,000
Explanation:
Step 1: Multiply the ones.
4 x 9 ones =36 ones, Regroup 36 ones as 3 tens and 6 ones
3
60759
x      4
        6
Step 2 : Multiply the tens. Regroup
4 x 5 tens  = 20 tens, Regroup 20 tens + 3 tens = 23 tens as 2 hundred and 3 tens
2
60759
x      4
36
Step 3 : Multiply the hundreds. Regroup
4 x 7 hundreds = 28 hundreds, Regroup 28 hundred +2 hundred =30 hundreds as 3 thousand and 0 hundreds
3
60759
x      4
036
Step 4 : Multiply the thousands.
4 x 0 thousands = 0 thousands , Regroup 0 thousands + 3 thousands=3 thousands
60759
x     4
3036
Step 5  : Multiply the ten thousands.
4 x 6 ten thousands =24 ten thousands
60759
x      4
243068
60,759 x 4 = 2,43,036        Reasonable 2,43,036 is close to 2,40,000

Question 9.
A multiplex has 9 different movie theaters. Each theater has 455 seats. How many seats are in the multiplex?
Answer:
9 x 455 = 4,095 seats
Explanation:
Given a multiplex has 9 different movie theaters, and each theater has 455 seats , so total number of seats in the multiplex is 9 x 455=4,095
there are total 4,095 seats available.

Question 10.
An interstellar object travels 85,700 miles in 1 hour. How far does the object travel in 5 hours?
Answer:
85,770 x 5 = 428,850 miles
Explanation:
Given interstellar object travels 85,700 miles in 1 hour, in 5 hours the interstellar object will travel 85,770 x 5 = 428,850 miles

Question 11.
Without multiplying, determine how much greater 4,395 × 7 is than 4,395 × 6. Explain.
Answer:
it is more by 4,395
Explanation:
Comparing the values both the sides 4,395 is same and as 7 is one time more than 6 we multiply the value by one more time,

so the value of 4,395 x 7 to 4,395 x 6 is one time greater by 4,395

Question 12.
DIG DEEPER!
Find the missing digits.
Big Ideas Math Answers Grade 5 Chapter 4 Multiply Whole Numbers 4.3 8
Answer:
40,716 ÷ 3 = 13572 , Missing digits are 1 , 5, 2
Explanation:
To get the missing digits we divide 40,716 by 3
3)40716(13572
3              3x 1=3
4-3 =1 , we take next digit 0 so it becomes10
——
10          3 x 3=9
09         10- 9=1 , we take next digit 7 becomes 17
——
17            3x 5 = 15
15           17-15=2, we take next digit 1 becomes 21
——
21           3 x 7 = 21
21           21-21 = 0,then we take next digit becomes 6
——–
06          3 x 2 =6
6            6-6=0 and we ends as next  no number
——
0

If 40,716 is divided  by 3 we will get results as 13572  and missing digits are 1 , 5, 2

Think and Grow: Modeling Real Life

Example
A Cuvier’s beaked whale can dive 1,324 feet deeper than 4 times the depth a beluga whale can dive. How deep can a Cuvier’s beaked whale dive?
Big Ideas Math Answers Grade 5 Chapter 4 Multiply Whole Numbers 4.3 9
Write and solve an equation to find the depth.
Big Ideas Math Answers Grade 5 Chapter 4 Multiply Whole Numbers 4.3 10
Let d represent the diving depth of a Cuvier’s beaked whale.
d = (4 × 2,123) + 1,324
= __8,492___ + 1,324
= __9,816___
A Cuvier’s beaked whale can dive   9,816 feet.

Show and Grow

Question 13.
The height of Chicago’s Willis Tower is 469 feet more than 2 times the height of the Gateway Arch. How tall is Willis Tower?
Big Ideas Math Answers Grade 5 Chapter 4 Multiply Whole Numbers 4.3 11
Answer:
The height of Chicago’s Willis Tower is 1,729 feet
Explanation:
Willis Tower = ( 2 X height of the Gateway Arch) +469 feet
Given the height of the Gateway Arch in  St.Louis as 630 feet
Willis Tower = ( 2 X 630 feet ) + 469 feet
Willis Tower = 1,260 + 469 = 1,729 feet

Question 14.
DIG DEEPER!
The Amazon River is 350 miles shorter than 3 times the length of the Colorado River. How much longer is the Amazon River than the Mississippi River?
Big Ideas Math Answers Grade 5 Chapter 4 Multiply Whole Numbers 4.3 12
Answer:
Amazon River is 1,660 miles longer than the Mississippi River
Explanation:
Given Amazon River is = ( 3 x Colorado River )- 350 miles
Amazon River = (3 x 1,450)-350 miles
Amazon River = 4,350 – 350 = 4,000 miles
Now comparing Amazon River with Mississippi River = Mississippi River – Amazon River = 4,000-2,340=1,660 miles
So Amazon River is 1,660 miles longer than the Mississippi River

Multiply by One-Digit Numbers Homework & Practice 4.3

Find the product. Check whether your answer is reasonable.
Question 1.
Big Ideas Math Answers Grade 5 Chapter 4 Multiply Whole Numbers 4.3 13
Answer:
814 x 2=1,628    Reasonable 1,624 is close to 1,600 or 2,000
Explanation:
Step 1: Multiply the ones.
2 x 4 ones = 8 ones.
814
x   2
     8
Step 2 : Multiply the tens.
2 x 1 tens  = 2 tens,
814
x   2                            
  28
Step 3 : Multiply the hundreds.
2 x 8 hundreds = 16 hundreds
814
x  2
1,628
814 x 2=1,628      Reasonable 1,624 is close to 2,000 or 1,600

Question 2.
Big Ideas Math Answers Grade 5 Chapter 4 Multiply Whole Numbers 4.3 14
Answer:
206 x 4 = 824             Reasonable 824 is close to 1,000
Step 1: Multiply the ones.
4 x 6 ones = 24 ones. Regroup 24 ones as 2 tens and 4 ones
2
206
x   4
     4
Step 2 : Multiply the tens. Regroup
4 x 0 tens  = 0 tens, Regroup 0 tens + 2 tens = 2 tens
206
x   4                            
   24
Step 3 : Multiply the hundreds.
4 x 2 hundreds = 8 hundreds
206
x  4
824
206 x 4=824      Reasonable 824 is close to 1,000

Question 3.
Big Ideas Math Answers Grade 5 Chapter 4 Multiply Whole Numbers 4.3 15
Answer:
2,425 x 6 = 14,550             Reasonable 14,550 is close to 14,500
Explanation:
Step 1: Multiply the ones. Regroup
6 x 5 ones = 30 ones, Regroup 30 ones as 3 tens and 0 ones
3
2,425
x     6
      0
Step 2 : Multiply the tens. Regroup
6 x 2 tens  = 12 tens, Regroup 12 tens+ 3 tens = 15 tens as 1 hundred  and 5 tens
1
2425
x    6
50
Step 3 : Multiply the hundreds. Regroup
6 x 4 hundreds = 24 hundreds, Regroup 24 hundred +1 hundred = 25 hundreds as 2 thousand and 5 hundreds
2
2425
x    6
550
Step 4 : Multiply the thousands.
6 x 2 thousands = 12 thousands, Regroup 12 thousands+2 thousands as
14 thousand
2425
x     6
14,550
2,425 x 6 = 14,550        Reasonable 14,550 is close to 14,500

Question 4.
Big Ideas Math Answers Grade 5 Chapter 4 Multiply Whole Numbers 4.3 16
Answer:
5,149 x 9 = 46,341      Reasonable 46,341 is close to 46,000
Explanation:
Step 1: Multiply the ones. Regroup
9 x 9 ones = 81 ones, Regroup 81 ones as 8 tens and 1 ones
8
5149
x     9
      1
Step 2 : Multiply the tens. Regroup
9 x 4 tens  = 36 tens, Regroup 36 tens+ 8 tens =44 tens as 4 hundred  and 4 tens
4
5149
x    9
41
Step 3 : Multiply the hundreds. Regroup
9 x 1 hundreds = 9 hundreds, Regroup 9 hundred +4 hundred =13 hundreds as
1 thousand and 3 hundreds
1
5149
x    9
341
Step 4 : Multiply the thousands.
9 x 5 thousands = 45 thousands, Regroup 45 thousands+1 thousand as 46 thousand
5149
x    9
46,341
5,149 x 9 = 46,341      Reasonable 46,341 is close to 46,000

Question 5.
Big Ideas Math Answers Grade 5 Chapter 4 Multiply Whole Numbers 4.3 17
Answer:
17,823 x 3 = 53,469        Reasonable 53,469 is close to 50,000 or 53,000
Explanation:
Step 1: Multiply the ones.
3 x 3 ones = 9 ones
17,823
x      3
        9
Step 2 : Multiply the tens.
3 x 2 tens  = 6 tens
17,823
x       3
69
Step 3 : Multiply the hundreds. Regroup
3 x 8 hundreds = 24 hundreds, Regroup as 2 thousands and 4 hundreds
2
17823
x      3
    469
Step 4 : Multiply the thousands. Regroup
3 x 7 thousands = 21 thousands, Regroup 21 thousands+ 2 thousands= 23 thousands as 2 ten thousands and 3 thousands
2
17823
x      3
3,469
Step 5  : Multiply the ten thousands.
3 x 1 ten thousands = 3 ten thousands ,Regroup 3 ten thousands + 2 ten thousands = 5 ten thousands
17823
x      3
53,469
17823 x 3 = 53,469      Reasonable 53,469 is close to 50,000 or 53,000

Question 6.
Big Ideas Math Answers Grade 5 Chapter 4 Multiply Whole Numbers 4.3 18
Answer:
61,837 x 8 = 494,696        Reasonable 494,696 is close to 500,000
Explanation:
Step 1: Multiply the ones.
8 x 7 ones =56 ones, Regroup 56 ones as 5 tens and 6 ones
5
61,837
x      8
        6
Step 2 : Multiply the tens. Regroup
8 x 3 tens  = 24 tens, Regroup 24 tens + 5 tens = 29 tens as 2 hundred and 9 tens
2
61,837
x      8
96
Step 3 : Multiply the hundreds. Regroup
8 x 8 hundreds = 64 hundreds, Regroup 64 hundred +2 hundred =66  hundreds as 6 thousand and 6 hundreds
6
61837
x      8
696
Step 4 : Multiply the thousands. Regroup
8 x 1 thousands = 8 thousands , Regroup 8 thousands + 6 thousands=
14 thousands as 1 ten thousand and 4 thousands
1
61837
x     8
4,696
Step 5  : Multiply the ten thousands. Regroup
8 x 6 ten thousands =48 ten thousands, Regroup 48 ten thousands +1 ten thousand = 49 ten thousands
61837
x      8
494,696
61,837 x 8 = 494,696        Reasonable 494,696 is close to 500,000

Find the product. Check whether your answer is reasonable.
Question 7.
4,257 × 5 = _______
Answer:
4,257 x 5 = 21,285                Reasonable 21,285 is close to 20,000
Explanation:
Step 1: Multiply the ones. Regroup
5 x 7 ones = 35 ones, Regroup  35 ones as 3 tens and 5 ones
3
4257
x    5
     5
Step 2 : Multiply the tens. Regroup
5 x 5 tens  = 25 tens, Regroup 25 tens+ 3 tens =28 tens as 2 hundreds and 8 tens
2
4257
x    5
85
Step 3 : Multiply the hundreds. Regroup
5 x 2 hundreds = 10 hundreds, Regroup 10 hundreds + 2 hundreds = 12 hundreds as 1 thousand and 2 hundreds
1
4257
x    5
285
Step 4 : Multiply the thousands.
5 x 4 thousands = 20 thousands, Regroup 20 thousands + 1 thousand =21 thousands
4257
x    5
21,285
4,257 x 5 = 21,285                Reasonable 21,285 is close to 20,000

Question 8.
6,702 × 6 = _____
Answer:
6,702 x 6 = 40,212                    Reasonable 40,212 is close to 40,000
Explanation:
Step 1: Multiply the ones. Regroup
6 x 2 ones = 12 ones, Regroup  12 ones as 1 ten and 2 ones
1
6702
x    6
     2
Step 2 : Multiply the tens. Regroup
6 x 0 tens  = 0 tens, Regroup  0 tens+ 1 tens =1 tens
6702
x    6
12
Step 3 : Multiply the hundreds. Regroup
6 x 7 hundreds = 42 hundreds, Regroup 42 hundreds as 4 thousands and 2 hundreds
4
6702
x    6
212
Step 4 : Multiply the thousands.
6 x 6 thousands = 36 thousands, Regroup 36 thousands+ 4 thousand = 40 thousands
6702
x    6
40,212
6,702 x 6 = 40,212                Reasonable 40,212 is close to 40,000
Question 9.
28,956 × 7 = ______
Answer:
28,956 x 7 = 2,02,692                 Reasonable 2,02,692 is close to 2,00,000
Explanation:
Step 1: Multiply the ones.
7 x 6 ones =42 ones, Regroup 42 ones as 4 tens and 2 ones
4
28,956
x       7
        2
Step 2 : Multiply the tens. Regroup
7 x 5 tens  = 35 tens, Regroup 35 tens + 4 tens = 39 tens as 3 hundreds and 9 tens
3
28956
x      7
92
Step 3 : Multiply the hundreds. Regroup
7 x 9 hundreds = 63 hundreds, Regroup 63 hundreds +3 hundred =
66  hundreds as 6 thousand and 6 hundreds
6
28956
x      7
692
Step 4 : Multiply the thousands. Regroup
7 x 8 thousands = 56 thousands , Regroup 56 thousands + 6 thousands=
62 thousands as 6 ten thousand and 2 thousands
6
28956
x      7
2692
Step 5  : Multiply the ten thousands. Regroup
7 x 2 ten thousands = 14  ten thousands, Regroup 14 ten thousands +
6 ten thousands = 20 ten thousands
28956
x      7
20,2692
28,956 x 7 = 2,02,692                 Reasonable 2,02,692 is close to 2,00,000

Question 10.
A parking garage has 8 levels. There are 194 parking spots on each level. How many parking spots does the garage have?
Answer:
194 x 8 =1,552 parking spots
Explanation:
Given a parking garage has 8 levels and each level has 194 parking spots so total number of parking spots are 194 is multiplied by 8 ,194 x 8 =1,552 , there are total
1,552 parking spots available in the garage.
Question 11.
YOU BE THE TEACHER
Newton finds 16,041 × 8. Is Newton correct? Explain.
Big Ideas Math Answers Grade 5 Chapter 4 Multiply Whole Numbers 4.3 19
Answer:
Yes, Newton is correct
Explanation:
Step 1: Multiply the ones.
8 x 1 ones = 8 ones
16,041
x       8
        8
Step 2 : Multiply the tens. Regroup
8 x 4 tens  = 32 tens, Regroup 3 hundreds and 2 tens
3
16,041
x      8
28
Step 3 : Multiply the hundreds. Regroup
8 x 0 hundreds = 0 hundreds, Regroup 0 hundreds +3 hundreds = 3 hundreds
16,041
x      8
328
Step 4 : Multiply the thousands. Regroup
8 x 6 thousands = 48 thousands , Regroup 48 thousands as 4 ten thousand and 8 thousands
4
16,041
x      8 
8328
Step 5  : Multiply the ten thousands. Regroup
8 x 1 ten thousands = 8 ten thousands, Regroup 8 ten thousands + 4 ten thousands = 12 ten thousands
16,041
x      8
128,328
As the value of 16,041 x 8 = 128,328, Yes Newton is correct
Question 12.
DIG DEEPER!
Write a different multiplication problem that has the same product as 726 × 5.
Answer:
726 x 5 =3,630
605 x 6 =3,630
Explanation:
we divide the product 3630 by 6


we get Quotient as 605 means  when we multiply 605 x 6 we get the  same product result 3,630 similarly when 726 x 5 is multiplied,
So 726 x 5 = 605 x 6

Question 13.
Modeling Real Life
The land area of Florida is 8,056 square miles more than 5 times the land area of Vermont. What is the land area of Florida?
Big Ideas Math Answers Grade 5 Chapter 4 Multiply Whole Numbers 4.3 20
Answer:
The land area of Florida is 49,496 square miles
Explanation:
Given land area of Florida is 8,056 square miles more than 5 times the land area of Vermont and land area of Vermont is 9,216 square miles
land area of Florida = ( 5 x land area of Vermont ) + 9,216 square miles
land area of Florida = ( 5 x 8056 ) + 9,216 square miles
land area of Florida =40280+9,216 = 49,496 square miles
So the land area of Florida is 49,496 square miles.

Question 14.
DIG DEEPER!
A solar farm has 6 rectangular arrays of solar panels. Each array has 105 rows with 8 panels in each row. How many solar panels are on the solar farm?
Answer:
6 x 105 x 8 = 5,040 panels
Explanation:
Given a solar farm has 6 rectangular arrays of solar panels and each array has 105 rows with 8 panels in each row so there are 105 x 8 = 840 panels and we have total 6 rectangular arrays
So total number of solar panels available are 6 x 840 =5,040 panels

Review & Refresh

Multiply.
Question 15.
2 × \(\frac{1}{4}\)
Answer:
2 x 1/4=1/2 or 1 by 2
Explanation:
When 2 is multiplied by fraction of 1,4 we get 2 x 1/4 as 1/2 or 1 by 2

Question 16.
3 × \(\frac{2}{6}\)
Answer:
3 x 2/6 = 6/6 =1
Explanation:
When 3 is multiplied by fraction of 2,6 we get 3 x 2/6 as 6/6 equals to 1

Question 17.
1 × \(\frac{7}{10}\)
Answer:
1 x 7/10 = 7/10 or 0.7
Explanation:
When 1 is multiplied by fraction of 7,10 we get 1 x 7/10 as 7/10 or 0.7

Lesson 4.4 Multiply by Two-Digit Numbers

Explore and Grow

Use the area model and partial products to find 28 × 13.
Big Ideas Math Solutions Grade 5 Chapter 4 Multiply Whole Numbers 4.4 1
Answer:
28 x 13 = 364            Reasonable 364 is close to 400
Explanation:
Step 1: Multiply 28 by 3 ones. Regroup
3 ones x 28 = 84 ones
2
28
x 13
  84
Step 2: Multiply 28 by 1 tens or 10, Regroup
1 tens x 28 =28 tens =280 ones
28
x 13
  84
280
Step 3 : Add the partial products.
28
x   13
    84
+280
364
28 x 13 =364            Reasonable 364 is close to 400

Repeated Reasoning
Explain how you can use an area model and partial products to find 128 × 13.
Answer:
128 x 13 =1,664           Reasonable 1,664 is close to 1,600
Explanation:

Step 1: Multiply 128 by 3 ones. Regroup
3 ones x 128 = 384 ones
2
128
x  13
 384
Step 2: Multiply 128 by 1 tens or 10, Regroup
1 tens x 128 =128 tens =1280 ones
128
    x 13
    384
1280
Step 3 : Add the partial products.
128
x     13
      384
+ 1280
1664
128 x 13 =1,664            Reasonable 1,664 is close to 1,600

Think and Grow: Multiply by a Two-Digit Number

Example
Find 312 × 82. Estimate _______
Big Ideas Math Solutions Grade 5 Chapter 4 Multiply Whole Numbers 4.4 2

So 312 x 82= 25,584                  Reasonable 25,584 is close to 25,500
Show and Grow

Find the product. Check whether your answer is reasonable.
Question 1.
Big Ideas Math Solutions Grade 5 Chapter 4 Multiply Whole Numbers 4.4 3
Answer:
184 x 23 = 4,232                      Reasonable 4,232 is close to 4,000

Explanation:
Step 1: Multiply 184 by 3 ones. Regroup
3 ones x 184 = 552 ones
21
184
x 23
552
Step 2: Multiply 184 by 2 tens or 20, Regroup
2 tens x 184 =368 tens =3680 ones
1
184
  x 23
  552
3680
Step 3 : Add the partial products.
184
x   23
     552
+3680
4232
184 x 23 =4,232            Reasonable 4,232 is close to 4,000

Question 2.
Big Ideas Math Solutions Grade 5 Chapter 4 Multiply Whole Numbers 4.4 4
Answer:
817 x 49 = 40,033           Reasonable 40,033 is close to 40,000
Explanation:
Step 1: Multiply 817 by 9 ones. Regroup
9 ones x 817 = 7353 ones
1 6
817
x 49
7353
Step 2: Multiply 817 by 4 tens or 40, Regroup
4 tens x 817 =3,268 tens =32680 ones
2
817
  x 49
  7353
32680
Step 3 : Add the partial products.
817
x   49
     7353
+32680
40033
817 x 49 =40,033            Reasonable 40,033 is close to 40,000

Question 3.
Big Ideas Math Solutions Grade 5 Chapter 4 Multiply Whole Numbers 4.4 5
Answer:
5046 x 91 = 4,59,186          Reasonable 4,59,186 is close to 4,60,000

Explanation:
Step 1: Multiply 5046 by 1 ones. Regroup
1 ones x  = 5046 ones
5046
x  91
5046
Step 2: Multiply 5046 by 9 tens or 90, Regroup
9 tens x 5046 =45,414 tens =45,4140 ones
45
5046
  x   91
    5046
454140
Step 3 : Add the partial products.
5046
x      91
        5046
+ 454140
4,59,186
5046 x 91 = 4,59,186        Reasonable 4,59,186 is close to 4,60,000

Apply and Grow: Practice

Find the product. Check whether your answer is reasonable.
Question 4.
Big Ideas Math Solutions Grade 5 Chapter 4 Multiply Whole Numbers 4.4 6
Answer:
96 x 24 = 2,304               Reasonable 2,304 is close to 2,300

Explanation:
Step 1: Multiply 96 by 4 ones. Regroup
4 ones x 96 =  ones
2
96
x 24
384
Step 2: Multiply 96 by 2 tens or 20, Regroup
2 tens x 96 =192 tens =1920 ones
1
96
 x 24
  384
1920
Step 3 : Add the partial products.
96
x   24
     384
+1920
2,304
96 x 24 = 2,304        Reasonable 2,304 is close to 2,300

Question 5.
Big Ideas Math Solutions Grade 5 Chapter 4 Multiply Whole Numbers 4.4 7
Answer:
108 x 19 = 2,052             Reasonable 2,052 is close to 2,000

Explanation:
Step 1: Multiply 108 by 9 ones. Regroup
9 ones x 108 = 972 ones
7
108
x 19
972
Step 2: Multiply 108 by 1 tens or 10, Regroup
1 tens x 108 =108 tens =1080 ones
108
  x 19
  972
1080
Step 3 : Add the partial products.
108
x  19
     972
+1080
2,052
108 x 19 = 2,052             Reasonable 2,052 is close to 2,000

Question 6.
Big Ideas Math Solutions Grade 5 Chapter 4 Multiply Whole Numbers 4.4 8
Answer:
6420 x 75 = 4,81,500           Reasonable 4,81,500 is close to 5,00,000
Explanation:
Step 1: Multiply 6420 by 5 ones. Regroup
5 ones x 6420 = 32100 ones
21
6420
x  75
32,100
Step 2: Multiply 6420 by 7 tens or 70, Regroup
7 tens x 6420 =44,940 tens =449400 ones
6420
  x 75
  32100
449400
Step 3 : Add the partial products.
6420
x   75
     32100
+449400
481500
6420 x 75 = 4,81,500           Reasonable 4,81,500 is close to 5,00,000

Question 7.
802 × 41 = ______
Answer:
802 x 41 = 32,882                  Reasonable 32,882 is close to 33,000

Explanation:
Step 1: Multiply 802 by 1 ones. Regroup
1 ones x 802 = 802 ones
802
x 41
802
Step 2: Multiply 802 by 4 tens or 40, Regroup
4 tens x 802 =3208 tens =32080 ones
802
  x 41
  802
32080
Step 3 : Add the partial products.
802
x       41
       802
+32080
32,882
802 x 41 = 32,882          Reasonable 32,882 is close to 33,000

Question 8.
577 × 86 = _____
Answer:
577 x 86 = 49,622                 Reasonable 49,622 is close to 50,000
Explanation:
Step 1: Multiply 577 by 6 ones. Regroup
6 ones x 577 =3,462 ones
4 4
577
x 86
3462
Step 2: Multiply 577 by 8 tens or 80, Regroup
8 tens x 577 =4616 tens =46160 ones
577
  x 86
  3462
46160
Step 3 : Add the partial products.
577
x      86
    3462
+46160
49622
577 x 86 = 49,622                 Reasonable 49,622 is close to 50,000

Question 9.
9,513 × 67 = _____
Answer:
9513 x 67 = 6,37,371             Reasonable  6,37,371   is close to 6,50,000
Explanation:
Step 1: Multiply 9513 by 7 ones. Regroup
7 ones x 9513 =66,591 ones
2
9513
x 67
66,591
Step 2: Multiply 9513 by 6 tens or 60, Regroup
6 tens x 9513 =57078 tens = 570780 ones
9513
  x 67
  66591
570780
Step 3 : Add the partial products.
9513
x      67
    66591
+570780
637371
9513 x 67 = 6,37,371             Reasonable  6,37,371   is close to 6,50,000

Question 10.
Is 6,364 a reasonable product for 706 × 45? Explain your reasoning.
Answer:
No 6,344 is not a reasonable product for 706 x 45 as 706 x 45 = 31,770
Explanation:
When we multiply 706 x 45 we get result as 31,770
as 6,364  is not at all close to 31,770, So 6,344 is not reasonable product for 706 x 45

Question 11.
DIG DEEPER!
Which two-digit number when multiplied by itself has a product of 625?
Answer:
25, because 25 x 25 = 625
Explanation:
Given that a two-digit number when multiplied by itself has a product of 625
lets take the two digit number as X and X x X = 625
X2=625 so X = square root of 625 = √625= √25 x 25, X = 25
So 25 when multiplied by itself we get 625

Think and Grow: Modeling Real Life

Example
A blood bank receives 58 donors in 1 day. Each person donates 1 pint, or about 473 milliliters of blood. About how many milliliters of blood are donated that day?
Big Ideas Math Solutions Grade 5 Chapter 4 Multiply Whole Numbers 4.4 9
Multiply the number of people by the amount of blood each person donates to find the total amount of blood donated.
Multiply 473 and 58.
Big Ideas Math Solutions Grade 5 Chapter 4 Multiply Whole Numbers 4.4 10
So, about _27,434_ milliliters of blood are donated that day.

Show and Grow

Question 12.
A store sells 15 drones. How much money does the store collect?
Big Ideas Math Solutions Grade 5 Chapter 4 Multiply Whole Numbers 4.4 11
Answer:
248 x 15 = $3,720
Explanation:
Given the cost of drone as $248 and the store sells 15 drones, total money collected in the store is $248 x 15=$3,720
Step 1: Multiply 248 by 5 ones. Regroup
5 ones x 248 = 1240 ones
4
248
x15
1240
Step 2: Multiply 248 by 1 tens or 10, Regroup
1 tens x 248 =248 tens = 2480 ones
248
  x 15
 1240
2480
Step 3 : Add the partial products.
248
x  15
  1240
+2480
3720
The store collected total amount is $3,720

Question 13.
What is the area of the soccer field?
Big Ideas Math Solutions Grade 5 Chapter 4 Multiply Whole Numbers 4.4 12
Answer:
The area of the soccer field is 120 x 80 =9,600 square yards
Explanation:
Step 1: Multiply 120 by 0 ones.
0 ones x 120 =120 ones
120
x 80
000
Step 2: Multiply 120 by 8 tens or 80, Regroup
8 tens x 120 =960 tens =  9600 ones
120
x 80
 000
9600
Step 3 : Add the partial products.
120
x 80
   000
+9600
9600
The total area of the soccer field is 120 x 80 =9,600 square yards

Question 14.
The Cassini spacecraft orbited Saturn for 13 years and 77 days. Three of those years were leap years. For how many days did the Cassini spacecraft orbit Saturn? Justify your answer.
Answer:
Cassini spacecraft orbited Saturn for total 4,825 days
Explanation:
Given the Cassini spacecraft orbited Saturn for 13 years and 77 days and three of those years were leap, First we need to convert years to days and add to 77 days, we know  each year has 365 days so first we multiply 365 with 13= 365 x 13 = 4,745 days,
365
x 13
4,745
And we know leap year has 366 days and in 13 years we have 3 leap years so we will add 3 more days to 4,745+3=4,748 days or
adding 3 days as it is saying in 13 years leap years 3 years are leap to 4,745 days=4,745+3=4,748 days
3 days as 3 are leap years
4745
+   3
4,748
Now adding 4,748 and 77 days
4,748
+   77
4,825
Cassini spacecraft orbited Saturn for total 4,825 days

Multiply by Two-Digit Numbers Homework & Practice 4.4

Find the product. Check whether your answer is reasonable.
Question 1.
Big Ideas Math Solutions Grade 5 Chapter 4 Multiply Whole Numbers 4.4 13
Answer:
107 x 11 =1,177     reasonable 1,177 is close to 1,000 or 12,00
Explanation:
Step 1: Multiply 107 by 1 ones.
1 ones x 107 =107 ones
107
x 11
107
Step 2: Multiply 107 by 1 tens or 10, Regroup
1 tens x 107 =107 tens =  1070 ones
107
x 11
 107
1070
Step 3 : Add the partial products.
107
x 11
    107
+1070
1,177
107 x 11 =1,177      reasonable 1,177 is close to 1,000 or 12,00

Question 2.
Big Ideas Math Solutions Grade 5 Chapter 4 Multiply Whole Numbers 4.4 14
Answer:
72 x 13 = 936       reasonable 936 is close to 1,000
Explanation:
Step 1: Multiply 72 by 3 ones.
3 ones x 72 =216 ones
72
x 13
216
Step 2: Multiply 72 by 1 tens or 10, Regroup
1 tens x 72 =72 tens =  720 ones
72
x 13
 216
 720
Step 3 : Add the partial products.
72
x 13
   216
+720
936
72 x 13 =936      reasonable 936 is close to 1,000

Question 3.
Big Ideas Math Solutions Grade 5 Chapter 4 Multiply Whole Numbers 4.4 15
Answer:
466 x 27 = 12,582       reasonable  12,582 is close to 13000 or 12,600
Explanation:
Step 1: Multiply 466 by 7 ones.
7 ones x 466 = 3262 ones
44
466
x 27
3262
Step 2: Multiply 466 by 2 tens or 20, Regroup
2 tens x 466 =932 tens = 9320 ones
466
x 27
 3262
 9320
Step 3 : Add the partial products.
466
x 27
  3262
+9320
12,582
466 x 27 = 12,582    reasonable   12,582 is close to 13000 or 12,600

Find the product. Check whether your answer is reasonable.
Question 4.
Big Ideas Math Solutions Grade 5 Chapter 4 Multiply Whole Numbers 4.4 16
Answer:
83 x 57 = 4,731             reasonable 4731 is close to 5,000
Explanation:
Step 1: Multiply 83 by 7 ones.
7 ones x 83 = 581 ones
2
83
x 57
581
Step 2: Multiply 83 by 5 tens or 50, Regroup
5 tens x 83 =415 tens =  4150 ones
83
x 57
  518
4150
Step 3 : Add the partial products.
83
x 57
     581
+4150
4731
83 x 57 = 4,731          reasonable 4731 is close to 5,000

Question 5.
Big Ideas Math Solutions Grade 5 Chapter 4 Multiply Whole Numbers 4.4 17
Answer:
893 x 44 = 39,292    reasonable 39,292 is close to 40,000
Explanation:
Step 1: Multiply 893 by 4 ones.
4 ones x 893 = 3,572 ones
3 1
893
x 44
3572
Step 2: Multiply 893 by 4 tens or 40, Regroup
4 tens x 893 =3572 tens = 35720 ones
893
x 44
  3572
35720
Step 3 : Add the partial products.
893
x 44
     3572
+35720
39292
893 x 44 = 39,292    reasonable 39,292 is close to 40,000

Question 6.
Big Ideas Math Solutions Grade 5 Chapter 4 Multiply Whole Numbers 4.4 18
Answer:
2079 x 82 = 170,478   reasonable 170,478 is close to 170,000
Explanation:
Step 1: Multiply 2079 by 2 ones.
2 ones x 2079 = 4,158 ones
11
2079
x  82
4158
Step 2: Multiply 2079 by 8 tens or 80, Regroup
8 tens x 2079 = 16,632 tens = 16,6320 ones
2079
x 82
    4158
16,6320
Step 3 : Add the partial products.
2079
x      82
       4158
+16,6320
170478
2079 x 82 = 170,478   reasonable 170,478 is close to 170,000

Question 7.
6,082 × 25 = ______
Answer:
6,082 x 25 = 152,050    reasonable 152,050 is close to 150,000
Explanation:
Step 1: Multiply  by 5 ones.
5 ones x 6082 = 30,410 ones
41
6082
x   25
30410
Step 2: Multiply 6082 by 2 tens or 20, Regroup
2 tens x 6082 = 12164 tens = 121640 ones
6082
x  25
  30410
12,1640
Step 3 : Add the partial products.
6082
x      25
     30410
+121640
15,2050
6,082 x 25 = 152,050    reasonable 152,050 is close to 150,000

Question 8.
158 × 96 = ______
Answer:
158 x 96 = 15,168         reasonable 15168 is close to 15,000
Explanation:
Step 1: Multiply 158 by 96 ones.
6 ones x 158 = 948 ones
34
158
x 96
948
Step 2: Multiply 158 by 9 tens or 90, Regroup
9 tens x 158 =1422 tens = 14220 ones
57
158
x 96
    948
14220
Step 3 : Add the partial products.
158
x 96
      948
+14220
15,168
158 x 96 = 15,168         reasonable 15168 is close to 15,000

Question 9.
98 × 74 = ______
Answer:
98 x 74 = 7252             reasonable 7252 is close to 7000
Explanation:
Step 1: Multiply 98 by 4 ones.
4 ones x 98 = 392 ones
3
98
x74
392
Step 2: Multiply 98 by 7 tens or 70, Regroup
7 tens x 98  = 686 tens = 6860 ones
98
x74
392
6860
Step 3 : Add the partial products.
98
x 74
     392
+6860
7252
98 x 74 = 7252             Reasonable 7252 is close to 7000

Question 10.
Writing
Why is one partial product always greater than the other partial product when multiplying by a two-digit number?
Answer:
The first partial product cannot have a multiplicand greater than 9.
The second partial product cannot have a multiplicand less than 10.
So one partial product is always greater than the other partial product when multiplying by a two-digit number
Explanation:
Lets multiply 315 with 26 a two digit number(26)
The multiplicand 26 can be decomposed to (20 + 6). The first partial product is the product of 315 x 6= 1890. The second partial product is the product of 315 x 20 = 6300.
Here the first partial product cannot have a multiplicand greater than 9  which is 6 and the second partial product cannot have a multiplicand less than 10 which is 20,therefore  one partial product always greater than the other partial product when multiplying by a two-digit number.

Question 11.
DIG DEEPER!
Use the numbers 2, 4, 6, and 8 once to form the greatest product.
Big Ideas Math Solutions Grade 5 Chapter 4 Multiply Whole Numbers 4.4 19
Answer:
The greatest product is 806 x 42 = 33,852
Explanation:
Given numbers are 2,4,6,8 , first greatest number among given numbers are  8 and next is 6 then 4 and then 2 we have the pattern  given as 3 digit number multiplied by  2 digit number, so first 3 great digits numbers are 8,6,0 in these the biggest is 8 then 6 but the middle digit is already given as 0 so the first three digits are 806,
Now the left two digits are 2,4 now to make the greatest 2 digit number we take from the combination of 2 and 4 ,in these the greatest number is 4 first then 2 making it as 42 as second greatest two digit number,
therefore the product is 806 X 42 =33,852 making it as the greatest product.
806
x    42
33,852

Question 12.
Modeling Real Life
A store sells 79 virtual reality headsets. How much money does the store collect?
Big Ideas Math Solutions Grade 5 Chapter 4 Multiply Whole Numbers 4.4 20
Answer:
$125 X 79 = $9,875
Explanation:
Given that the virtual reality headset cost as $ 79 and the store sells 79 virtual reality headsets, Total money collected by the store is $125 x 79=$9,875
125
x 79
9,875

Question 13.
Modeling Real Life
Your friend needs a trumpet for 12 months. She can rent a trumpet for $42 each month and pay a yearly fee of $25, or she can buy a trumpet for $550. Should she buy or rent the trumpet? Explain.
Answer:
Friend should take the trumpet for rent because it is costing less than buying it.
Rent for 12 months is $529
Buying costs to $550.
Explanation:
Friend needs a trumpet for 12 months and rent of trumpet is $42 for each month, so for 12 months it costs as $42 x 12=  $504 and also has to pay  a yearly fee of $25 means total cost for entire 12 months is $504+$25= $529.
Given that if she buys its cost to $550, now comparing between rent and buy as $529<$550 ,as rent is less than buying she has to take it for rent so that she can save $21
Review & Refresh

Use a place value chart to answer the question.
Question 14.
4,000 is 10 times as great as what number?
Answer:
4,000 is 10 times as great as 400
Explanation:
Lets us take the number as X,
Given 4,000 is 10 times as great as X means X  x 10= 4,000
the value of X equals to 4000 x 1/10= 4000/10=400
X=400 therefore 4,000 is 10 times as great as 400

Question 15.
50 is \(\frac{1}{10}\) of what number?
Answer:
50 is \(\frac{1}{10}\) of 500
Explanation:
Let the number be X and given 50 = 1/10 x X, so X = 50 x 10 = 500

Lesson 4.5 Multiply by Multi-Digit Whole Numbers

Explore and Grow

Use any strategy to find each product.
Big Ideas Math Answer Key Grade 5 Chapter 4 Multiply Whole Numbers 4.5 1.1
Answer:
425 x 2 = 850  , 425 x 12 =5,100  , 425 x 112 = 47,600
Explanation:
1) 425 x 2 =850, multiply 425 by 2 ones or 2
1
425
x  2
850
2) 425 x 12 = 51,00
Explanation:
Step 1 : Multiply 425 by 2 ones or 2
1
425
x 12
850
Step 2 : Multiply 425 by 1 tens or 10.
425
x 12
  850
4250
Step 3: Add the partial products.
425
x 12
    850
+4250
  5100
3) 425 x 112 = 47,600
Explanation:
Step 1 : Multiply 425 by 2 ones or 2
1
425
x112
850
Step 2 :Multiply 425 by 1 tens or 10.
425
x 112
  850
4250
Step 3 : Multiply 425 by 1 hundred or 100. Regroup
425
x 112
    850
4250
42500
Step 4: Add the partial products.
425
x 112
        850
4250
+ 42500
47,600 

Structure
How can you use the first product to find the second product? How can you use the second product to find the third product?
Answer:
Yes we can use the first product to find the second product and use the second product to find the third product.
Explanation:
For example let the numbers be 324 x 125 = 40,500
First product result will be product of 324 is first multiplied by last digit 5 at ones place as 324 x 5=1620
Second product result will be the first product results in addition to multiple of 324 with the second digit which is at tens place as 324 x 20 and add to first product results, so here we have used the first product result to find the second product results
324
x25
1620
6480
8100
Third product result will be the second product results and we add it to the multiple of product of third digit which is at hundreds place as  324 x 100 and add to second product results, so here we have used the second product result to find the third product results
324
x125
  8100
32400
40,500

Think and Grow: Multiply Multi-Digit Whole Numbers

Example
Find 2,043 × 132. Estimate __2,69,676_____
Big Ideas Math Answer Key Grade 5 Chapter 4 Multiply Whole Numbers 4.5 1

2,043 x 132 = 2,69,676    reasonable 2,69,676 is close to 2,70,000

Show and Grow

Find the product. Check whether your answer is reasonable.
Question 1.
Big Ideas Math Answer Key Grade 5 Chapter 4 Multiply Whole Numbers 4.5 2
Answer:
318 x 523 =166,314            reasonable 1,66,314 is close to 1,70,000
Explanation:
Step 1 : Multiply 318 by 3 ones or 3
2
318
x 523
  954
Step 2 :Multiply 318 by 2 tens or 20.
1
318
x 523
  954
6360
Step 3 : Multiply 318 by 5 hundred or 500. Regroup
318
x 523
      954
6360
159000
Step 4: Add the partial products.
318
x     523
        954
6360
+159000
166314
So 318 x 523 =166,314            reasonable 1,66,314 is close to 1,70,000

Question 2.
Big Ideas Math Answer Key Grade 5 Chapter 4 Multiply Whole Numbers 4.5 3
Answer:
7291 x 308 = 2,245,628  reasonable 2,245,628 is close to 2,300,000
Explanation:
Step 1 : Multiply 7291 by 8 ones or 8
27
7291
x 308
58328
Step 2 :Multiply 7219 by 0 tens .
7291
x 308
58328
00000
Step 3 : Multiply 7219 by 3 hundred or 300. Regroup
2
7291
x 308
   58328
00000
2187300
Step 4: Add the partial products.
7291
x 308__
     58328
00000
+2187300
2245628
7291 x 308 = 2,245,628     Reasonable 2,245,628 is close to 2,300,000

Apply and Grow: Practice

Find the product. Check whether your answer is reasonable.
Question 3.
Big Ideas Math Answer Key Grade 5 Chapter 4 Multiply Whole Numbers 4.5 4
Answer:
521 x 317 = 1,65,157    Reasonable 165,157  is close to 1,700,00
Explanation:
Step 1 : Multiply 521 by 7 ones or 7
1
521
x 317
3647
Step 2 :Multiply 521 by 1 tens or 10 .
521
x 317
3647
5210
Step 3 : Multiply  521 by 3 hundred or 300. Regroup
521
x 317
    3647
5210
156300
Step 4: Add the partial products.
 521
x 317
       3647
5210
+156300
165157
521 x 317 = 1,65,157    Reasonable 165,157  is close to 1,700,00

Question 4.
Big Ideas Math Answer Key Grade 5 Chapter 4 Multiply Whole Numbers 4.5 5
Answer:
631 x 574 = 3,62,194  Reasonable 3,62,194 is close to 3,60,000
Explanation:
Step 1 : Multiply 631 by 4 ones or 4
631
x 574
2524
Step 2 :Multiply 631 by 7 tens or 70 .
631
x 574
 2524
44170
Step 3 : Multiply 631 by 5 hundred or 500. Regroup
631
x 574
    2524
44170
315500
Step 4: Add the partial products.
631
x 574
     2524
44170
+315500
362194
        
631 x 574 = 3,62,194          Reasonable 3,62,194 is close to 3,60,000  

Question 5.
Big Ideas Math Answer Key Grade 5 Chapter 4 Multiply Whole Numbers 4.5 6
Answer:
2496 x 358 =8,93,568       Reasonable 8,93,568 is close to 9,00,000
Explanation:
Step 1 : Multiply 2496 by 8 ones or 8
374
2496
x 358
19,968
Step 2 :Multiply 2496 by 5 tens or 50 .
243
2496
x 358
 19968
124800
Step 3 : Multiply 2496 by 3 hundred or 300. Regroup
2496
x 358
 19968
124800
748800
Step 4: Add the partial products.
2496
x 358
  19968
124800
748800   
893568
2496 x 358 =8,93,568      Reasonable 8,93,568 is close to 9,00,000
Question 6.
155 × 956 = ______
Answer:
155 x 956 = 1,48,180      Reasonable 1,48,180 is close to 1,50,000
Explanation:
Step 1 : Multiply 155 by 6 ones or 6
33
155
x 956
  930
Step 2 :Multiply 155 by 5 tens or 50 .
22
155
x 956
  930
7750
Step 3 : Multiply 155 by 9 hundred or 900. Regroup
155
x 956
      930
7750
139500
Step 4: Add the partial products.
155
x 956
      930
7750
139500
148180
155 x 956 = 1,48,180      Reasonable 1,48,180 is close to 1,50,000

Question 7.
748 × 239 = ______
Answer:
748 x 239 = 1,78,772     Reasonable  1,78,772 is close to or 1,80,000 or  2,00,000
Explanation:
Step 1 : Multiply 748 by 9 ones or 9
47
748
x239
 6732
Step 2 :Multiply 748 by 3 tens or 30 .
12
748
x 239
  6732
22440
Step 3 : Multiply 748 by 2 hundred or 200. Regroup
1
748
x 239
   6732
22440
149600
Step 4: Add the partial products.
748
x 239___
       6732
22440
+149600
178772
748 x 239 = 1,78,772     Reasonable  1,78,772 is close to or 1,80,000 or  2,00,000

Question 8.
1,907 × 218 = ______
Answer:
1907 x 218 = 415,726  Reasonable 4,15,726 is close to 4,20,000
Explanation:
Step 1 : Multiply 1907 by 8 ones or 8
7 5
1907
x 218
15256
Step 2 :Multiply 1907 by 1 tens or 10 .
1907
x 218
 15256
19070
Step 3 : Multiply  1907 by 2 hundred or 200. Regroup
1  1
1907
x 218
  15256
19070
381400
Step 4 :Add the partial products.
1907
x 218
     15256
19070
+381400
415726
1907 x 218 = 415,726  Reasonable 4,15,726 is close to 4,20,000

Question 9.
The delta of the Nile River is about 150 miles wide. How many feet wide is the delta?
Big Ideas Math Answer Key Grade 5 Chapter 4 Multiply Whole Numbers 4.5 7
Answer:
5280 x 150 = 7,92,000 feet
Explanation:
Given the delta of the Nile River is about 150 miles wide and 1 mile is equal to 5,280 feet so total feet of Nile River is product of 5280 and 150 , 5280 x 150 = 7,92,000 feet

Question 10.
Reasoning
When you multiply a three-digit number by a four-digit number, what is the greatest number of digits the product can have? Explain.
Answer:
999 x 9999 = 9,989,001 the product will have 7 greatest number of digits.
Explanation:
We know 9 is the greatest digit, so the greatest three-digit number is 999 and the greatest four-digit number is 9999 when 999 x 9999 we get the results as 999 x 9999 = 9,989,001 which consists of total 7 digits in it, therefore when we multiply a three-digit number by a four-digit number, we get 7 greatest number of digits in the results.

Question 11.
DIG DEEPER!
Find the missing digit so that both products are the same.
Big Ideas Math Answer Key Grade 5 Chapter 4 Multiply Whole Numbers 4.5 8
Answer:
The missing digit is 0, as 850 x 150= 1,27,500 and 375 x 340 = 1,27,500 are same
Explanation:
To find the missing digit first we multiply 850 X 150 we get 1,27,500 and  it is mentioned that the products are same means
850 x 150= 1,27,500 = 375 x 34___. So to find the missing digit from 34___, we divide 1,27,500 by 375 and the result is 340. So obviously the missing  digit is 0.

Think and Grow: Modeling Real Life

Example
The employees at a baseball bat factory need to make 350,000 bats in 6 months. They make 2,750 bats each day and work 127 days in the 6 months. Do the employees make enough bats?
Big Ideas Math Answer Key Grade 5 Chapter 4 Multiply Whole Numbers 4.5 9
Multiply the number of bats produced each day by the number of days.
Big Ideas Math Answer Key Grade 5 Chapter 4 Multiply Whole Numbers 4.5 10

Compare the number of bats produced to the number of bats needed.
2750 x 127 = 3,49,250 bats as 3,49,250 < 3,50,000
The employees does not___ make enough bats.
Explanation:
Each day the employees make 2750 bats and has worked 127 days in 6 months , so in 6 months the employees have made
2750 x 127 =3,49,250 bats, given that the employees at a baseball bat factory need to make 350,000 bats in 6 months, now comparing  we are getting difference as the employees made to the required number of bats.
We are getting employees made is less when compared to the needed number of bats as 3,49,250 < 3,50,000. Both are not matching so employees have not made enough bats.

Show and Grow

Question 12.
The drivers at a warehouse need to deliver 40,000 packages each day. The warehouse has 128 trucks. Each truck has 350 packages. The drivers deliver all of the packages on the trucks. Do the warehouse drivers deliver enough packages?
Answer:
350 x 128= 44,800 and required is 40,000 as compared it is less,
Yes, the warehouse drivers can deliver enough packages each day.
Explanation:
Each truck has 350 packages and the ware house has 128 trucks, so the total number of packages in a day are 350 x 128 = 44,800 packages. Now comparing the drivers need only 40,000 packages each day as  44,800 is greater than 40,000, 44,800>40,000.Easily the drivers can deliver the packages each day.

Question 13.
A school buys 5 tablets and 5 laptops. Each tablet costs $379. Each laptop costs $449. How much more does the school pay for laptops than it pays for tablets?
Answer:
Tablets costs = $379 x 5=  $1,895
Laptop costs = $449 x 5 = $2,245, Laptop to Tablets = $ 2,245-$1,895=$350, School pays $350 more.
Explanation:
School buys 5 tablets and 5 laptops and each tablet costs $379 and laptop costs $449,the total amount for tablets costs = $379 x 5=  $1,895 and for laptop costs = $449 x 5 = $2,245,Now comparing the costs of laptops to tablets $2,245 to $1895 we are getting a difference of  $350 among them. So the school pays $379
Question 14.
DIG DEEPER!
A koala sleeps 18 hours each day. How many minutes does the koala sleep in 1 year?
Big Ideas Math Answer Key Grade 5 Chapter 4 Multiply Whole Numbers 4.5 11
Answer:
The koala sleeps 3,94,200 minutes in 1 year.
Explanation:
We know that in 1 hour equals to 60 minutes and in 1 year we have total 365 days a koala sleep 18 hours each day means
18 x 60 =1,080 minutes in each day, so in 1 year koala sleeps 1080 minutes x 365= 3,94,200 minutes in 1 year.

Multiply by Multi-Digit Whole Numbers Homework & Practice 4.5

Find the product. Check whether your answer is reasonable.
Question 1.
Big Ideas Math Answer Key Grade 5 Chapter 4 Multiply Whole Numbers 4.5 12
Answer:
102 x 104 =10,608     Reasonable 10,608  is close to 10,000
Explanation:
Step 1 : Multiply 102 by 4 ones or 4
102
x 104
  408
Step 2 : Multiply 102 by 0 tens .
102
x 104
408
000
Step 3 : Multiply 102 by 1 hundred or 100. Regroup
102
x 104
   408
000
10200
Step 4 : Add the partial products.
102
x 104
       408
000
+ 10200
10608
102 x 104 =10,608     Reasonable 10,608 is close to 10,000

Question 2.
Big Ideas Math Answer Key Grade 5 Chapter 4 Multiply Whole Numbers 4.5 13
Answer:
185 x 123 = 22,755  Reasonable 22,755 is close to 20,000
Explanation:
Step 1 : Multiply 18 by 3 ones or 3
21
185
x 123
  555
Step 2 : Multiply 185 by 2 tens or 20
11
185
x 123
  555
3700
Step 3 : Multiply 185 by 1 hundred or 100. Regroup
185
x 123
  555
3700
18500
Step 4 : Add the partial products.
185
x 123
       555
3700
+ 18500
22755

185 x 123 = 22,755  Reasonable 22,755 is close to 20,000
Question 3.
Big Ideas Math Answer Key Grade 5 Chapter 4 Multiply Whole Numbers 4.5 14
Answer:
3410 x 251 = 855,910    Reasonable 855,910 is close to 8,60,000
Explanation:
Step 1 : Multiply 3410 by 1 ones or 1
3410
x 251
3410
Step 2 : Multiply 3410 by 5 tens or 50
2
3410
x 251
    3410
170500
Step 3 : Multiply 3410 by 2 hundred or 200. Regroup
3410
x 251
    3410
170500
682000
Step 4 : Add the partial products.
3410
x 251
3410
170500
+682000
855910
3410 x 251 = 855,910    Reasonable 855,910 is close to 8,60,000

Find the product. Check whether your answer is reasonable.
Question 4.
Big Ideas Math Answer Key Grade 5 Chapter 4 Multiply Whole Numbers 4.5 15
Answer:
954 x 176 = 1,67,904      Reasonable 1,67,904 is close to 1,68,000 or 1,70,000
Explanation:
Step 1 : Multiply 954 by 6 ones or 6
32
954
x 176
5724
Step 2 : Multiply 954 by 7 tens or 70
32
954
x 176
   5724
66780
Step 3 : Multiply 954 by 1 hundred or 100. Regroup
954
x 176
   5724
66780
95400
Step 4 : Add the partial products.
954
x 176
    5724
66780
+95400
167904

954 x 176 = 1,67,904      Reasonable 1,67,904 is close to 1,68,000 or 1,70,000

Question 5.
Big Ideas Math Answer Key Grade 5 Chapter 4 Multiply Whole Numbers 4.5 16
Answer:
818 x 524 = 428,632    Reasonable 4,28,632 is close to 4,30,000 or 4,00,000
Explanation:
Step 1 : Multiply 818 by 4 ones or 4
3
818
x 524
3272
Step 2 : Multiply 818 by 2 tens or 20
1
818
x  524
   3272
16360
Step 3 : Multiply 818 by  5 hundred or 500. Regroup
4
818
x  524
   3272
16360
409000
Step 4 : Add the partial products.
818
x  524
       3272
16360
+ 409000
428632
818 x 524 = 428,632    Reasonable 4,28,632 is close to 4,30,000 or 4,00,000

Question 6.
Big Ideas Math Answer Key Grade 5 Chapter 4 Multiply Whole Numbers 4.5 17
Answer:
6301 x 472 =2,974,072   Reasonable 2,974,072 is close to 3,000,000
Explanation:
Step 1 : Multiply 6301 by 2 ones or 2
6301
x 472
12602
Step 2 :  Multiply 6301 by 7 tens or 70
2
6301
x 472
   12602
441070
Step 3 : Multiply 6301 by  4 hundred or 400. Regroup
6301
x 472
     12602
441070
2520400
Step 4 : Add the partial products.
6301
x 472
      12602
441070
+2520400
2974072
6301 x 472 =2,974,072   Reasonable 2,974,072 is close to 3,000,000

Question 7.
999 × 186 = ______
Answer:
999 x 186 = 185,814     Reasonable 185,814 is close to 2,00,000
Explanation:
Step 1 : Multiply 999 by 6 ones or 6
55
999
x 186
5994
Step 2 :  Multiply 999 by 8 tens or 80
77
999
x 186
  5994
79920
Step 3 : Multiply 999 by  1 hundred or 100. Regroup
999
x 186
    5994
79920
99900
Step 4 : Add the partial products.
999
x 186
    5994
79920
+99900
185814

999 x 186 = 185,814     Reasonable 185,814 is close to 2,00,000
Question 8.
2,445 × 356 = ______
Answer:
2445 x 356 = 870420     Reasonable 870420 is close to 8,70,000
Explanation:
Step 1 : Multiply 2445 by 6 ones or 6
223
2445
x 356
14670
Step 2 :  Multiply 2445 by 5 tens or 50
222
2445
x  356
  14670
122250
Step 3 : Multiply 2445 by  3 hundred or 300. Regroup
111
2445
x 356
  14670
122250
733500
Step 4 : Add the partial products.
2445
x 356
     14670
122250
+733500
  870420
2445 x 356 = 870420     Reasonable 870420 is close to 8,70,000

Question 9.
1,564 × 389 = ______
Answer:
1564 x 389 = 608396     Reasonable 6,08,396 is close to 6,00,000
Step 1 : Multiply 1564 by 9 ones or 9
553
1564
x 389
14076
Step 2 :  Multiply 1564 by 8 tens or 80
453
1564
x 389
  14076
125120
Step 3 : Multiply 1564 by  3 hundred or 300. Regroup
111
1564
x 389
  14076
125120
469200
Step 4 : Add the partial products.
1564
x 389
    14076
125120
+469200
608396 

1564 x 389 = 608396     Reasonable 6,08,396 is close to 6,00,000

Question 10.
Reasoning
Your friend says the product of 4,164 and 137 is 70,460. How will finding an estimate help your friend realize the answer is not reasonable?
Answer:
4164 X 137 = 570,468, 5,70,468 is not equal to 70,460.
Explanation:
Friend says product of 4,164 and 137 is 70,460 , Now finding the estimate the product of 4,164 x 137 = 5,70,468 but friend says it is 70,460. So after finding friend realize the answer is not reasonable and not even close.

Question 11.
YOU BE THE TEACHER
Your friend says that when multiplying 300 by 126, she can multiply 3 × 126 and write two zeros after the product. Is your friend correct? Explain.
Answer:
Yes, Friend is correct
Explanation:
When first we multiply 300 x 126 we get the product as 37,800 and when we multiply 3 x 126 we get result as 378 and if we write two zeros after the product it becomes 37800 which is similar to the result of 300 x 126, so friend is correct.
We can write 300 x 126 as 3 x 102 x 126=   (3 x 126 ) x 102 , We know that if any whole number is multiplied by 10 with powers  we get the product whole number with the  given number of power value additional number of  zeros in the right.
as (3 x 126 ) x 102   has 2 as power to 10 we write the result with 2 zeros after the product which is equal to 300 x 126,So friend is correct.

Question 12.
Modeling Real Life
Your friend starts a video channel and wants to have 50,000 subscribers by the end of 1 year. She gets 140 new subscribers each day for 365 days. Does she meet her goal?
Answer:
140 x 365 = 51,100 as 51,100 > 50,000, Yes friend meets her goal.
Explanation:
As friend starts a video channel and wants to have 50,000 subscribers by the end of 1 year, and each day friend gets 140 new subscribers for 365 days , So total number of subscribers for 365 days or 1 year is 140 x 365= 51,100.
As 51,100 is greater than 50,000 means friend can easily meets her goal.

Review & Refresh

Evaluate the expression.
Question 13.
(5 + 8) × 4
Answer:
(5+8) x 4 = 13 x 4 = 52
Explanation:
Expressions inside parentheses are evaluated first from left to right, i.e (5+8) and then we proceed with multiplication.
So first we add the values inside the parentheses or values of bracket (5+8) x 4 ,5+8=13 and multiply by 4= 13 x 4 = 52,
(5 + 8) × 4 = 52
Question 14.
16 + (9 ÷ 3)
Answer:
16 + (9 ÷ 3) = 16 + 3 = 19
Explanation:
First we take expressions inside parentheses (9 ÷ 3) and evaluate then we proceed with addition,
When  9 ÷ 3 we get  3 , Now we go with addition so 16 + 3 = 19.
16 + (9 ÷ 3) = 19

Question 15.
7 × (4 + 6)
Answer:
7 x (4 +6) = 7 x 10 = 70
Explanation:
First we take expressions inside parentheses ( 4 + 6)=10 and proceed with multiplication, 7 x 10 = 70,
7 x (4 +6) = 70

Multiply Whole Numbers Performance Task

The Grand Coulee Dam is located on the Columbia River in Washington. The dam is used to control flood waters, provide irrigation, and generate power.
Big Ideas Math Answers 5th Grade Chapter 4 Multiply Whole Numbers 1
Question 1.
The dam has 12 pumps that each transfer the same amount of water through pipes from Lake Roosevelt to Banks Lake. Together, the pumps can transfer about 12,000 gallons of water each second.
a. About how many gallons of water can each pump transfer in 1 second?
b. About how many gallons of water can each pump transfer in 1 hour? Explain.
Answer:
a. 12,000 ÷ 12 = 1,000 gallons of water can each pump transfer in 1 second.
b. 1000 x 3,600 seconds = 3,600,000 gallons of water can each pump transfer in 1 hour.
Explanation :
a. Given a dam has 12 pumps that each transfer the same amount of water through pipes from Lake Roosevelt to Banks Lake, and together, the pumps can transfer about 12,000 gallons of water each second.
12 pumps = 12,000 gallons of water each second, Each pump transfer in 1 second is equal to 12,000 divided by 12= 12,000 ÷ 12 =
1,000 gallons of water can each pump transfer in 1 second.
b. Now number of gallons of water can each pump transfer in 1 hour, we know in 1 hour we have 60 minutes and 1 minute has 60 seconds so making it as 3,600 seconds in an hour.1 hour equals to 3,600 seconds. If in 1 second each pump can transfer 1000 gallons of water then in hour it will transfer 1000 x 3,600= 3,600,000 gallons of water can each pump transfer in 1 hour.
Question 2.
The dam generates power Capacity of Power at the Grand Coulee Dam using 33 generators located in 4 different powerhouses.
Big Ideas Math Answers 5th Grade Chapter 4 Multiply Whole Numbers 2
a. What is the total Power for the Grand Coulee Dam?
b. Only about one-third of the capacity of power for each powerhouse is actually generated each year. About how many megawatts does the Third Powerhouse actually generate in 1 year?
c. There are 106 watts in 1 megawatt. Without calculating, how can you find about how many watts the Third Powerhouse actually generates in 1 year?
Answer:
a. The total Power for the Grand Coulee Dam is 6,701 megawatts.
b. The Third Powerhouse actually generates only 1/3 in 1 year is 1800 x 1/3 = 1800/3=1800 ÷ 1/3 = 600 megawatts
c. The number of watts the Third Powerhouse actually generates in 1 year =600 x 106 watts.
Explanation:
a. First we calculate the total capacity of the power house in each power house with the capacity of power from each generator with the number of generators as per the powerhouse, Total Capacity of the power house = capacity of powerhouse from each generator multiply by number of generators  separately. Then adding all capacities we get the total Power for the Grand Coulee Dam as shown in the below table, So total Power for the Grand Coulee Dam is 6,701 megawatts.

b. We have Third Powerhouse has total capacity as 1,800 megawatts as calculated in that only about one-third of the capacity of power for each powerhouse is actually generated each year means total capacity in third powerhouse multiply by 1/3=  1800 x 1/3 = 1800/3=1800 ÷ 1/3 = 600 megawatts.
Third Powerhouse actually generates 600 megawatts in 1 year.
c. Given 1 mega watt equals to 106 watts , We have Third Powerhouse generates 600 megawatts in 1 year, so to convert megawatts  into watts we multiply it by 106 watts =600 x 106 watts, Third Powerhouse actually generates 600 x 106 watts  in 1 year.

Multiply Whole Numbers Activity

Multiplication Adventure
Directions:
1. Players take turns rolling a die. Players solve problems on their boards to race the explorers to their destinations.
2. On your turn, solve the next multiplication problem in the row of your roll.
3. The first player to get an explorer to a destination wins!
Big Ideas Math Answers 5th Grade Chapter 4 Multiply Whole Numbers 3
Answer:

Multiply Whole Numbers Chapter Practice

4.1 Multiplication Patterns

Find the product.
Question 1.
72 × 103 = ______
Answer:
72 × 103 = 72,000
Explanation :
72 × 103 =  72 x 10 x 10 x 10 =72,000 , we multiply 72 with 10 three times as 10 power is 3.

Question 2.
30 × 900 = ______
Answer:
30 × 900 =27,000
Explanation:
When we multiply 30 by 900 we get 27,000 or 30 × 900 = 3 x 10 x 9 x 10 x 10 = 3 x 9 x 103= 27 x 103= 27,000
Question 3.
40 × 500 = ______
Answer:
40 × 500 = 20,000
Explanation:
When we multiply 40 by 500 we get 20,000 or 40 × 500 = 4 x 10 x 5 x 10 x 10 = 4 x 5 x 103= 20 x 103=20,000

Question 4.
60 × 800 = ______
Answer:
60 x 800 = 48,000
Explanation:
When we multiply 60 by 800 we get 48,000 or 60 × 800 = 6 x 10 x 8 x 10 x 10 = 6 x 8 x 103= 48 x 103=48,000

Question 5.
20 × 90 = ______
Answer:
20 x 90 = 1800
Explanation:
When we multiply 20 by 90 we get 1800 or 20 × 90 = 2 x 10 x 9 x 10 = 2 x 9 x 102= 18 x 102=1800

Question 6.
102 × 41 = _____
Answer:
102 × 41 =4100
Explanation:
When we multiply  102 × 41 =10 x 10 x 41 = 100 x 41 = 41,00

Find the missing factor.
Question 7.
_____ × 103 = 26,000
Answer:
The missing factor is 26
Explanation:
To get the missing factor, let us take it as X ,So  X x 103 = 26,000, therefore X = 26,000 / 103 =26,000/1000=26

Question 8.
600 × ____ = 24,000
Answer:
The missing factor is 40

Explanation:
To get the missing factor, let us take it as X , So 600 x X = 24,000, therefore X = 24,000/600 = 40

Question 9.
5,000 × _____ = 250,000
Answer:
The missing factor is 50

Explanation:
To get the missing factor, let us take it as X , So 5,000 x X = 250,000, therefore X = 250,000/5000 = 50

Question 10.
Logic
The product of a number and twice that number is 320,000. Write the multiplication equation.
Answer:
Let the number be X the multiplication equation is X x 2X =320,000
Explanation:
Let us take the number as X given the product of a number and twice that number equals to 320,000 so the equation becomes as x X x 2X = 320,000 means X multiplied by 2X result is 320,000.

4.2 Estimate Products

Estimate the product. Explain whether your estimate is an overestimate or an underestimate.
Question 11.
44 × 81
Answer:
44 x 81 = 3,564
50 x 80 = 4,000 is overestimate
Explanation :
As our estimate after rounding each factor becomes as 44 as 50 to nearest ten and 81 as 80 to nearest ten as 44 x 81 < 50 x 80 as the estimate result is greater than the given product results our estimate is an overestimate

Question 12.
29 × 67
Answer:
29 x 67 = 1,943
30 x 70 = 2,100 is overestimate
Explanation :
As our estimate after rounding each factor becomes 29 as 30  to nearest ten and 67 as 70 to nearest ten as 29 x 67 < 30 x 70 as the estimate result is greater than the given product results our estimate is an overestimate.

Question 13.
504 × 302
Answer:
504 x 302 = 152,208
500 x 300 = 150,000 is underestimate
Explanation :
As our estimate after rounding each factor becomes 504 as 500 to nearest hundred and 302 as 300 to nearest hundred as 504 x 302  > 500 x 300 as the estimate result is lesser than the given product results our estimate is an underestimate.

Question 14.
797 × 29
Answer:
792 x 29 = 22,968
800 x 30 = 24,000 is overestimate
Explanation :
As our estimate after rounding each factor becomes 792 as 800 to nearest hundred and 29 as 30 to nearest ten as 792 x 29 < 800 x 30 as the estimate result is greater than the given product results our estimate is an overestimate.

4.3 Multiply by One-Digit Numbers

Find the product. Check whether your answer is reasonable.
Question 15.
Big Ideas Math Answers 5th Grade Chapter 4 Multiply Whole Numbers chp 15
Answer:
663 x 5 = 3,315     Reasonable 3,315 is close to 3,300
Explanation:
Step 1: Multiply the ones. Regroup
5 x 3 ones = 15 ones, Regroup  15 ones as 1 tens and 5 ones
1
663
x   5
     5
Step 2 : Multiply the tens. Regroup
5 x 6 tens  = 30 tens,  30 tens + 1 tens = 31 tens, Regroup 31 tens as
3 hundreds and 1 tens
3
663
x    5 
15
Step 3 : Multiply the hundreds. Regroup hundreds
5 x 6 hundreds = 30 hundreds, 30 hundreds + 3 hundreds = 33 hundreds
Regroup 33 hundreds as 3 thousands and 3 hundred
663
x   5  
3315
663 x 5 = 3,315     Reasonable 3,315 is close to 3,300

Question 16.
Big Ideas Math Answers 5th Grade Chapter 4 Multiply Whole Numbers chp 16
Answer:
1973 x 7 = 13,811      Reasonable 13,811  is close to 14,000
Explanation:
Step 1: Multiply the ones. Regroup
7 x 3 ones = 21 ones, Regroup  21 ones as 2 tens and 1 ones
2
1973
x    7
     1
Step 2 : Multiply the tens. Regroup
7 x 7 tens  = 49 tens,  49 tens + 2 tens = 51 tens, Regroup 51 tens as
5 hundreds and 1 tens
5
1973
x    7 
11
Step 3 : Multiply the hundreds. Regroup hundreds
7 x 9 hundreds = 63 hundreds, 63 hundreds + 5 hundreds = 68 hundreds
Regroup 68 hundreds as 6 thousands and 8 hundreds
6
1973
x   7  
811
Step 4 : Multiply the thousands. Add the regrouped thousands
7 x 1 thousands = 7 thousands,7 thousands +6 thousands=13 thousands
1973
x    7
13811
1973 x 7 = 13,811      Reasonable 13,811  is close to 14,000

Question 17.
Big Ideas Math Answers 5th Grade Chapter 4 Multiply Whole Numbers chp 17
Answer:
75,028 x 6 = 450,168        Reasonable  4,50,168 is close to 4,50,000
Explanation:
Step 1: Multiply the ones. Regroup
6 x 8 ones = 48 ones, Regroup  48 ones as 4 tens and 8 ones
4
75028
x      6
       8
Step 2 : Multiply the tens. Regroup
6 x 2 tens  = 12 tens,  12 tens + 4 tens = 16 tens, Regroup 16 tens as
1 hundreds and 6 tens
1
75028
x      6 
68
Step 3 : Multiply the hundreds. Regroup hundreds
6 x 0 hundreds = 0 hundreds, 0 hundreds + 1 hundreds = 1 hundred
75028
x     6  
168
Step 4 : Multiply the thousands. Regroup
6 x 5 thousands = 30 thousands, Regroup 30 thousands as 3 ten thousands and 0 thousands
3
75028
x      6
   0168
Step 5  : Multiply the ten thousands.
6 x 7 ten thousands = 42 ten thousands,42 ten thousands+ 3 ten thousands
=45 ten thousands, regroup as 4 hundred thousand and 5 ten thousand
75028
x      6
450168
75,028 x 6 = 4,50,168     Reasonable 4,50,168 is close to 4,50,000

Question 18.
Modeling Real Life
The Great Western Trail is 355 miles longer than 4 times the length of the Iditarod Trail. The Iditarod Trail is 1,025 miles long. How long is the Great Western Trail?
Answer:
The Great Western Trail = 355 + ( 4 x 1,025)= 355 + 4,100 = 4,455 miles
Explanation :
Given that The Great Western Trail is 355 miles longer than 4 times the length of the Iditarod Trail,
The Great Western Trail= 355 + ( 4 x the Iditarod Trail),
and the Iditarod Trail is 1,025 miles ,
The Great Western Trail = 355 + ( 4 x 1025) = 355 + 4100 = 4,455 miles, The Great Western Trail is 4,455 miles long.
4.4 Multiply by Two-Digit Numbers

Find the product. Check whether your answer is reasonable.
Question 19.
Big Ideas Math Answers 5th Grade Chapter 4 Multiply Whole Numbers chp 19
Answer:
45 x 35 = 1575            Reasonable 1575 is close to 1500
Explanation:
Step 1: Multiply 45 by 5 ones. Regroup
5 ones x 45 = 225 ones
2
45
x 35
225
Step 2: Multiply 45 by 3 tens or 30, Regroup
3 tens x 45 =135 tens =1350 ones
45
x 35
  225
1350
Step 3 : Add the partial products.
45
x   35
     225
+1350
1575
45 x 35 = 1575            Reasonable 1575 is close to 1500

Question 20.
Big Ideas Math Answers 5th Grade Chapter 4 Multiply Whole Numbers chp 20
Answer:
206 x 74 = 15,244         Reasonable 15,244 is close to 15,000
Explanation:
Step 1: Multiply 206 by 4 ones. Regroup
4 ones x 206 = 824 ones
2
206
x 74
824
Step 2: Multiply 206 by 7 tens or by 70, Regroup
7 tens x 206 =1442 tens =14420 ones
206
x 74
    824
14420
Step 3 : Add the partial products.
206
x        74
        824
+ 14420
   15244
206 x 74 = 15,244         Reasonable 15,244 is close to 15,000

Question 21.
Big Ideas Math Answers 5th Grade Chapter 4 Multiply Whole Numbers chp 21
Answer:
5,082 x 69 =  350,658                Reasonable 3,50,658 is close to 3,50,000
Explanation:
Step 1: Multiply 5082 by 9 ones. Regroup
9 ones x 5082 = 45,738 ones
71
5082
x  69
45,738
Step 2: Multiply 5082 by 6 tens or 60, Regroup
6 tens x 5082 = 30492 tens =304920 ones
5082
x  69
  45738
304920
Step 3 : Add the partial products.
5082
x    69
    45738
+304920
350658
5,082 x 69 =  350,658                Reasonable 3,50,658 is close to 3,50,000

4.5 Multiply by Multi-Digit Whole Numbers

Find the product. Check whether your answer is reasonable.
Question 22.
Big Ideas Math Answers 5th Grade Chapter 4 Multiply Whole Numbers chp 22
Answer:
612 x 143 = 87,516      Reasonable 87,516 is close to 90,000
Explanation:
Step 1 : Multiply 612 by 3 ones or 3
612
x 143
1836
Step 2 :Multiply 612 by 4 tens or 40.
612
x 143
 1836
24480
Step 3 : Multiply 612 by 1 hundred or 100. Regroup
612
x 143
  1836
24480
61200
Step 4: Add the partial products.
 612
x 143
     1836
   24480
+61200
87516
612 x 143 = 87,516      Reasonable 87,516 is close to 90,000

Question 23.
Big Ideas Math Answers 5th Grade Chapter 4 Multiply Whole Numbers chp 23
Answer:
899 x 475 = 427,025    Reasonable 4,27,025 is close to 4,30,000
Explanation:
Step 1 : Multiply 899 by 5 ones or 5
44
899
x 475
 4495
Step 2 :Multiply 899 by 7 tens or 70.
66
899
x 475
 4495
62930
Step 3 : Multiply 899 by 4 hundred or 400. Regroup
33
899
x 475
 4495
62930
359600
Step 4: Add the partial products.
899
x 475
      4495
62930
+359600
427025
899 x 475 = 427,025    Reasonable 4,27,025 is close to 4,30,000

Question 24.
Big Ideas Math Answers 5th Grade Chapter 4 Multiply Whole Numbers chp 24
Answer:
6182 x 794 = 4,908,508   Reasonable 4,908,508  is close to 5,000,000
Explanation:
Step 1 : Multiply 6182 by 4 ones or 4
3
6182
x 794
24728
Step 2 :Multiply 6182 by 9 tens or 90.
71
6182
x 794
  24728
556380
Step 3 : Multiply 6182 by 7 hundred or 700. Regroup
51
6182
x 794
  24728
556380
4327400
Step 4: Add the partial products.
6182
x 794
    24728
556380
4327400
4908508
6182 x 794 = 4,908,508   Reasonable 4,908,508  is close to 5,000,000

Conclusion:

We wish the Big Ideas Math Grade 5 Answer Key Chapter 4 Multiply Whole Numbers is beneficial for you all. There are different methods to solve the problems in multiply whole numbers. Prepare the questions on your own and solve the problems. Students are notified to improve their math skills by referring to the Big Ideas Math Answers Grade 5 Chapter 4 Multiply Whole Numbers.

Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals

Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals

Clearing the math exams without proper guidance and preparation is really tough. Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals preparatory material is prepared by highly experienced subject experts. After great research and easy-to-understand concepts, the material is prepared which helps the candidates in all-over preparation. While practicing the various questions, Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals helps you. You can find all the questions and cross-check them with the solutions present in the book.

Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals

Regular practice of all questions present in Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals will indulge confidence in you. To attempt the exam with confidence and to score better marks, practice all the questions given in that book. All the topics in Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals are prepared in a certain way that all the candidates can understand them easily. Follow all the material given in here and click on the links below to kickstart your preparation and to score the expected marks.

Fractions and Decimals

Lesson: 1 Multiplying Fractions

Lesson: 2 Dividing Fractions

Lesson: 3 Dividing Mixed Numbers

Lesson: 4 Adding and Subtracting Decimals

Lesson: 5 Multiplying Decimals

Lesson: 6 Dividing Whole Numbers

Lesson: 7 Dividing Decimals

Chapter: 2 – Fractions and Decimals

Fractions and Decimals STEAM VIDEO/Performance

STEAM Video

Space is Big

An astronomical unit (AU) is the average distance between Earth and the Sun, about 93 million miles. Why do astronomers use astronomical units to measure distances in space? In what different ways can you compare the distances between objects and the locations of objects using the four mathematical operations?
Answer:
The space beyond Earth is so incredibly vast that units of measure
which are convenient for us in our everyday lives can become GIGANTIC.
Distances between the planets, and especially between the stars,
can become so big when expressed in miles and kilometers that they’re unwieldy.
So for cosmic distances, we switch to whole other types of units:
astronomical units, light years and parsecs.
Astronomical units, abbreviated AU, are a useful unit of measure
within our solar system. One AU is the distance from the Sun to Earth’s orbit,
which is about 93 million miles (150 million kilometers).

Watch the STEAM Video “Space is Big.” Then answer the following questions.
1. You know the distances between the Sun and each planet.
How can you find the minimum and maximum distances between
two planets as they rotate around the Sun?
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 1
Given the distances between the sun and each planet we
can find the minimum and maximum approximately distances
between two planets around the sun, we have each
planet distance now we will see which is farthest and
which planet is near to sun, then subtract the larger one
from smaller and tell the distances from them,
Example the distance from sun to mars is 5 AU and
sun to mercury is 2 AU,
then the distance from mercury to mars is 5 – 2 = 3 AU.

2. The table shows the distances of three celestial bodies from Earth.
It takes about three days to travel from Earth to the Moon.
How can you estimate the amount of time it would take to travel from
Earth to the Sun or to Venus?

Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 2
Earth to the Sun ≈ 1190 days ,
Earth to Venus ≈ 330 days

Explanation:
Speed = Distance ÷ Time ,
We have distance from Earth to Moon is 0.00256 and
time is 3 days = 3 X 24 = 72 hours
now speed is = 0.00256 ÷ 72 = 0.0000355 AU per hour,
now we have distance from Earth to Sun is 1 AU and speed as
0.000035 per hour, Time taken is distance by speed =
1 ÷ 0.000035 = 28571.428 hours, now we divide by 24 to get days as
28571.428 ÷ 24 = 1190.4 days ≈ 1190 days to travel from Earth to Sun,
now we have distance from Earth to Venus as 0.277 Au and speed as
0.000035 per hour, Time taken is 0.277 ÷ 0.000035 = 7914.28 hours,
now we divide by 24 to get number of days as 7914.28 ÷ 24 = 329.7 ≈
330 days to travel from Earth to Venus.

Performance Task

Space Explorers

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 6 Chapter 2 Fractions and Decimals 3
You will use a table that shows the average distances between the Sun and each planet in our solar system to find several distances in space. Then you will use the speed of the Orion spacecraft to answer questions about time and distance.
Is it realistic for a manned spacecraft to travel to each planet in our solar system? Explain why or why not.

Fractions and Decimals Getting Ready for Chapter 2

Chapter Exploration

Work with a partner. The area model represents the multiplication of two fractions. Copy and complete the statement.

Question 1.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 4

Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals-1
\(\frac{2}{3}\) X \(\frac{3}{4}\) = \(\frac{6}{12}\)

Explanation:
Step I: We multiply the numerators as 2 X 3 = 6
Step II: We multiply the denominators as 3 X 4 =12
Step III: We write the fraction in the simplest form as
\(\frac{6}{12}\),
So \(\frac{2}{3}\) X \(\frac{3}{4}\) = \(\frac{6}{12}\)

Question 2.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 5

Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals-2
\(\frac{1}{2}\) X \(\frac{2}{3}\) = \(\frac{2}{6}\)

Explanation:
First part is \(\frac{1}{2}\) and second part is \(\frac{2}{3}\)
Step I: We multiply the numerators as 1 X 2 = 2
Step II: We multiply the denominators as 2 X 3 = 6
Step III: We write the fraction in the simplest form as
\(\frac{2}{6}\),
So \(\frac{1}{2}\) X \(\frac{2}{3}\) = \(\frac{2}{6}\)

Question 3.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 6

Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals-3
\(\frac{2}{5}\) X \(\frac{2}{3}\) = \(\frac{4}{15}\)

Explanation:
First part is \(\frac{2}{5}\) and second part is \(\frac{2}{3}\)
Step I: We multiply the numerators as 2 X 2 = 4
Step II: We multiply the denominators as 5 X 3 = 15
Step III: We write the fraction in the simplest form as
\(\frac{4}{15}\),
So \(\frac{2}{5}\) X \(\frac{2}{3}\) = \(\frac{4}{15}\)

Question 4.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 7

Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals-4
\(\frac{3}{4}\) X \(\frac{1}{4}\) = \(\frac{3}{16}\)

Explanation:
First part is \(\frac{3}{4}\) and second part is \(\frac{1}{4}\)
Step I: We multiply the numerators as 3 X 1 = 3
Step II: We multiply the denominators as 4 X 4 = 16
Step III: We write the fraction in the simplest form as \(\frac{3}{16}\),
So \(\frac{3}{4}\) X \(\frac{1}{4}\) = \(\frac{3}{16}\)

Work with a partner. Use an area model to find the product.

Question 5.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 8

Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals-5
So \(\frac{1}{2}\) X \(\frac{1}{3}\) = \(\frac{1}{6}\)

Explanation:
We draw an area model to find the product
Step 1: We are given with \(\frac{1}{2}\) so
we take shade 1 part out of 2 ,
Step 2: We have \(\frac{1}{3}\) now we shade
1 part out of 3
Step 4 : We multiply the numerators as 1 X 1 = 1
Step 5: We multiply the denominators as 2 X 3 = 6
Step 6 : The purple area came through overlapping which is the product as
1 part out of 6 and write the fraction
in the simplest form as \(\frac{1}{6}\).
So \(\frac{1}{2}\) X \(\frac{1}{3}\) = \(\frac{1}{6}\)

Question 6.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 9

Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals-6

So \(\frac{4}{5}\) X \(\frac{1}{4}\) = \(\frac{4}{20}\)

Explanation:
We draw an area model to find the product
Step 1: We are given with \(\frac{4}{5}\) so
we take shade 4 parts out of 5 ,
Step 2: We have \(\frac{1}{4}\) now we shade
1 part out of 4
Step 4 : We multiply the numerators as 4 X 1 = 4
Step 5: We multiply the denominators as 5 X 4 = 20
Step 6 : The purple area came through overlapping which is the product as
4 parts out of 20 and write the fraction
in the simplest form as \(\frac{4}{20}\).
So \(\frac{4}{5}\) X \(\frac{1}{4}\) = \(\frac{4}{20}\).

Question 7.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 10

Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals-7
So \(\frac{1}{6}\) X \(\frac{3}{4}\) = \(\frac{3}{24}\)

Explanation:
We draw an area model to find the product
Step 1: We are given with \(\frac{1}{6}\) so
we take shade 1 part out of 6 ,
Step 2: We have \(\frac{3}{4}\) now we shade
3 parts out of 4
Step 4 : We multiply the numerators as 1 X 3 = 3
Step 5: We multiply the denominators as 6 X 4 = 24
Step 6 : The purple area came through overlapping which is the product as
3 parts out of 24 and write the fraction
in the simplest form as \(\frac{3}{24}\).
So \(\frac{1}{6}\) X \(\frac{3}{4}\) = \(\frac{3}{24}\).

Question 8.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 11
Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals-8

So \(\frac{3}{5}\) X \(\frac{1}{4}\) = \(\frac{3}{20}\)

Explanation:
We draw an area model to find the product
Step 1: We are given with \(\frac{3}{5}\) so
we take shade 3 parts out of 5 ,
Step 2: We have \(\frac{1}{4}\) now we shade
1 part out of 4
Step 4 : We multiply the numerators as 3 X 1 = 3
Step 5: We multiply the denominators as 5 X 4 = 20
Step 6 : The purple area came through overlapping which is the product as
3 parts out of 20 and write the fraction
in the simplest form as \(\frac{3}{20}\).
So \(\frac{3}{5}\) X \(\frac{1}{4}\) = \(\frac{3}{20}\).

Question 9.
MODELING REAL LIFE
You have a recipe that serves 6 people. The recipe uses three-fourths of a cup of milk.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 12
a. How can you use the recipe to serve more people? How much milk would you need? Give 2 examples.
b. How can you use the recipe to serve fewer people? How much milk would you need? Give 2 examples.
a. Example 1: To serve 8 people we need
\(\frac{8}{6}\) recipe + \(\frac{24}{4}\) milk
Example 2 : To serve 10 people we need
\(\frac{10}{6}\) recipe + \(\frac{30}{4}\) milk.
b. Example 1: To serve 5 people we need
\(\frac{5}{6}\) recipe + \(\frac{15}{4}\) milk
Example 2 : To serve 4 people we need
\(\frac{4}{6}\) recipe + \(\frac{12}{4}\) milk.

Explanation:
Given 1 recipe serves 6 people and uses three-fourths of a cup of milk,
a. We can use the recipe to serve more people and milk would we need are
6 persons = 1 recipe + \(\frac{3}{4}\) milk
1 person = \(\frac{1}{6}\) recipe + \(\frac{3}{4}\) milk
Example 1 : 8 persons = \(\frac{8}{6}\) recipe + \(\frac{24}{4}\) milk
Example 2 : 10 persons = \(\frac{10}{6}\) recipe + \(\frac{30}{4}\) milk

b. We can use the recipe to serve fewer people and milk would we need are
6 persons = 1 recipe + \(\frac{3}{4}\) milk
1 person = \(\frac{1}{6}\) recipe + \(\frac{3}{4}\) milk
5 persons = \(\frac{5}{6}\) recipe + \(\frac{15}{4}\) milk
4 persons = \(\frac{4}{6}\) recipe + \(\frac{12}{4}\) milk.

Vocabulary

The following vocabulary terms are defined in this chapter. Think about what each term might mean and record your thoughts.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 13
Reciprocals : In mathematics, the reciprocal, also known as multiplicative inverse,
is the inverse of a number x. denoted as 1/x or x-1.
This means that the product of a number x and its reciprocal yields 1.
The reciprocal of a number is simply the number that has been flipped or
inverted upside-down. This entails transposing a number such that
the numerator and denominator are placed at the bottom and top respectively.
To find the reciprocal of a whole number, just convert it into a fraction in
which the original number is the denominator and the numerator is 1.
The reciprocal of 2/3 is 3/2.
The product of 2/3 and its reciprocal 3/2 is 1.
2/3 x 3/2 = 1.

Lesson 2.1 Multiplying Fractions

EXPLORATION 1
Using Models to Solve a Problem
Work with a partner. A bottle of water is \(\frac{1}{2}\) full. You drink \(\frac{2}{3}\) of the water. Use one of the models to find the portion of the bottle of water that you drink. Explain your steps.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 14

Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals-9

We use area model to give the portion of the bottle of water
that I drink is \(\frac{1}{3}\) of the bottle.

Explanation:
Given a bottle of water is \(\frac{1}{2}\) full.
I drink \(\frac{2}{3}\) of the water. So the portion of the bottle of water
that I drink is \(\frac{1}{2}\) X \(\frac{2}{3}\) = \(\frac{2}{6}\) =
\(\frac{1}{3}\) of the bottle.
We draw an area model to find the product
Step 1: We are given with \(\frac{1}{2}\) so
we take shade 1 part out of 2,
Step 2: We have \(\frac{2}{3}\) now we shade
2 parts out of 3
Step 4 : We multiply the numerators as 1 X 2 = 2
Step 5: We multiply the denominators as 2 X 3 = 6
Step 6 : The purple area came through overlapping which is the product as
2 parts out of 6 or 1 part out of 3 and write the fraction
in the simplest form as \(\frac{2}{6}\) or \(\frac{1}{3}\)
So the portion of the bottle of water that I drink is
\(\frac{2}{6}\) = \(\frac{1}{3}\) of the bottle.

EXPLORATION 2
Work with a partner. A park has a playground that is \(\frac{3}{4}\) of its width and \(\frac{4}{5}\) of its length.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 15
a. Use a model to find the portion of the park that is covered by the playground. Explain your steps.
b. How can you find the solution of part(a) without using a model?
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 16
a.
Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals-10
The portion of the park that is covered by the playground is
\(\frac{12}{20}\) = \(\frac{3}{5}\)

b. Without using a model the portion of the park that is covered by the playground is \(\frac{12}{20}\) = \(\frac{3}{5}\).

Explanation:
a. Used area model to find the portion of the park that is covered by the playground.
Given a park has a playground that is \(\frac{3}{4}\) of its width and
\(\frac{4}{5}\) of its length. So the portion of the park
that is covered by the playground is \(\frac{3}{4}\) X \(\frac{4}{5}\)
we explain this by an area model as
Step 1: We are given with \(\frac{3}{4}\) so we take shade 3 parts out of 4,
Step 2: We have \(\frac{4}{5}\) now we shade 4 parts out of 5
Step 4 : We multiply the numerators as 3 X 4 = 12
Step 5: We multiply the denominators as 4 X 5 = 20
Step 6 : The purple area came through overlapping which is the product as
12 parts out of 20 or 3 part out of 5 and write the fraction
in the simplest form as \(\frac{12}{20}\) or \(\frac{3}{5}\)
The portion of the park that is covered by the playground is
\(\frac{12}{20}\) = \(\frac{3}{5}\).

b. Without using a model we have first part is \(\frac{3}{4}\) and
second part is \(\frac{4}{5}\)
Step I: We multiply the numerators as 3 X 4 = 12
Step II: We multiply the denominators as 4 X 5 = 20
Step III: We write the fraction in the simplest form as
\(\frac{12}{20}\) or \(\frac{3}{5}\),
The portion of the park that is covered by the playground is \(\frac{12}{20}\) =
\(\frac{3}{5}\).

2.1 Lesson

Try It
Multiply.

Question 1.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 17

\(\frac{1}{3}\) X \(\frac{1}{5}\) = \(\frac{1}{15}\)

Explanation:
Step I: We multiply the numerators as 1 X 1 = 1
Step II: We multiply the denominators as 3 X 5 =15
Step III: We write the fraction in the simplest form as
\(\frac{1}{15}\),
So \(\frac{1}{3}\) X \(\frac{1}{5}\) = \(\frac{1}{15}\).

Question 2.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 18
\(\frac{2}{3}\) X \(\frac{3}{4}\) = \(\frac{6}{12}\)
= \(\frac{2}{4}\) = \(\frac{1}{2}\)

Explanation:
Step I: We multiply the numerators as 2 X 3 = 6
Step II: We multiply the denominators as 3 X 4 =12
Step III: We write the fraction in the simplest form as \(\frac{6}{12}\),
So \(\frac{2}{3}\) X \(\frac{3}{4}\) = \(\frac{6}{12}\)=
= \(\frac{2}{4}\) as both can go in 2 we get latex]\frac{1}{2}[/latex].

Question 3.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 19
So \(\frac{1}{2}\) X \(\frac{5}{6}\) = \(\frac{5}{12}\)

Explanation:
Step I: We multiply the numerators as 1 X 5 = 5
Step II: We multiply the denominators as 2 X 6 =12
Step III: We write the fraction in the simplest form as \(\frac{5}{12}\),
So \(\frac{1}{2}\) X \(\frac{5}{6}\) = \(\frac{5}{12}\).

Key Idea
Multiplying Fractions
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 20

Try It

Multiply. Write the answer in simplest form.

Question 4.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 21

\(\frac{3}{7}\) X \(\frac{2}{3}\) = \(\frac{6}{21}\) = \(\frac{2}{7}\)

Explanation:
Step I: We multiply the numerators as 3 X 2 = 6
Step II: We multiply the denominators as 7 X 3 =21
Step III: We write the fraction in the simplest form as \(\frac{6}{21}\),
So \(\frac{3}{7}\) X \(\frac{2}{3}\) = \(\frac{6}{21}\) =
further simplified as both go in 3 we get \(\frac{2}{7}\).

Question 5.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 22
\(\frac{4}{9}\) X \(\frac{3}{10}\) = \(\frac{12}{90}\) =
\(\frac{2}{15}\)

Explanation:
Step I: We multiply the numerators as 4 X 3 = 12
Step II: We multiply the denominators as 9 X 10 =90
Step III: We write the fraction in the simplest form as \(\frac{12}{90}\),
So \(\frac{4}{9}\) X \(\frac{3}{10}\) = \(\frac{12}{90}\)
further can be simplified as both can be divided by 6 we get 2 X 6 = 12 and
15 x 6 = 90, (2, 15) So \(\frac{12}{90}\) = \(\frac{2}{15}\).

Question 6.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 23
\(\frac{6}{5}\) X \(\frac{5}{8}\) = \(\frac{30}{40}\) =\(\frac{3}{4}\)

Explanation:
Step I: We multiply the numerators as 6 X 5 = 30
Step II: We multiply the denominators as 5 X 8 =40
Step III: We write the fraction in the simplest form as \(\frac{30}{40}\),
So \(\frac{6}{5}\) X \(\frac{5}{8}\) = \(\frac{30}{40}\)
further can be simplified as both can be divided by 10 we get 3 X 10 = 30 and
4 x 10 = 40,(3,4) So \(\frac{30}{40}\) = \(\frac{3}{4}\).

Try It

Question 7.
WHAT IF?
You use \(\frac{1}{4}\) of the flour to make the dough.
How much of the entire bag do you use to make the dough?

\(\frac{1}{4}\) of the entire bag we do use to make the dough

Explanation:
Given we use \(\frac{1}{4}\) of the flour to make the dough,
We take entire bag as 1 which has flour , So we use
\(\frac{1}{4}\) out of 1 means \(\frac{1}{4}\) X 1 = \(\frac{1}{4}\)
of the entire bag we do use to make the dough.

Key Idea
Multiplying Mixed Numbers
Write each mixed number as an improper fraction. Then multiply as you would with fractions.

Try It
Multiply. Write the answer in simplest form.

Question 8.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 24
\(\frac{1}{3}\) X \(\frac{7}{6}\) = \(\frac{7}{18}\)

Explanation:
Given \(\frac{1}{3}\) X 1 \(\frac{1}{6}\) so
first we write mixed number 1 \(\frac{1}{6}\)  as 1 X 6 + 1 by 6 = \(\frac{7}{6}\) now
we multiply \(\frac{1}{3}\)  X  \(\frac{7}{6}\),
Step I: We multiply the numerators as 1 X 7 = 7
Step II: We multiply the denominators as 3 X 6 =18
Step III: We write the fraction in the simplest form as \(\frac{7}{18}\),
So \(\frac{1}{3}\) X \(\frac{7}{6}\) = \(\frac{7}{18}\).

Question 9.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 25
\(\frac{7}{2}\) X \(\frac{4}{9}\) = \(\frac{28}{18}\) = \(\frac{14}{9}\) = 1 \(\frac{5}{9}\)

Explanation:
Given 3 \(\frac{1}{2}\) X \(\frac{4}{9}\) so
first we write mixed number 3 \(\frac{1}{2}\) as 3 X 2 + 1 by 2 = \(\frac{7}{2}\) now we multiply \(\frac{7}{2}\)  X  \(\frac{4}{9}\),
Step I: We multiply the numerators as 7 X 4 = 28
Step II: We multiply the denominators as 2 X 9 =18
Step III: We write the fraction in the simplest form as \(\frac{28}{18}\),
we can further simplify as both goes in 2, 14 X 2 = 28 and 9 X 2 = 18, (14,9)
So 3 \(\frac{1}{2}\) X \(\frac{4}{9}\) = \(\frac{28}{18}\)  = \(\frac{14}{9}\). As numerator is greater than denominator we write in mixed fraction also as (1 X 9 + 5 by 9 ), 1\(\frac{5}{9}\).

Question 10.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 26
4 \(\frac{2}{3}\) X \(\frac{3}{4}\) = \(\frac{42}{12}\)  = \(\frac{7}{2}\) = 3 \(\frac{1}{2}\).

Explanation:
Given 4 \(\frac{2}{3}\) X \(\frac{3}{4}\) so
first we write mixed number 4 \(\frac{2}{3}\) as 4 X 3 + 2 by 3 = \(\frac{14}{3}\) now we multiply \(\frac{14}{3}\)  X  \(\frac{3}{4}\),
Step I: We multiply the numerators as 14 X 3 = 42
Step II: We multiply the denominators as 3 X 4 =12
Step III: We write the fraction in the simplest form as \(\frac{42}{12}\),
we can further simplify as both goes in 6, 6 X 7 = 42 and 6 X 2 = 12, (7,2)
So 4 \(\frac{2}{3}\) X \(\frac{3}{4}\) = \(\frac{42}{12}\)  = \(\frac{7}{2}\). As numerator is greater than denominator we write in
mixed fraction also as (3 X 2 + 1 by 2 ), 3 \(\frac{1}{2}\).

Try It
Multiply. Write the answer in simplest form.

Question 11.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 27
1 \(\frac{7}{8}\) X 2 \(\frac{2}{5}\) = \(\frac{180}{40}\) = \(\frac{9}{2}\) = 4 \(\frac{1}{2}\).

Explanation :
1 \(\frac{7}{8}\) X 2 \(\frac{2}{5}\) , We write mixed fractions
1 \(\frac{7}{8}\) as 1 X 8 + 7 by 8 = \(\frac{15}{8}\) and 2 \(\frac{2}{5}\) as 2 X 5 + 2 by 5 = \(\frac{12}{5}\) Now we multiply
\(\frac{15}{8}\) X \(\frac{12}{5}\),
Step I: We multiply the numerators as 15 X 12 = 180
Step II: We multiply the denominators as 8 X 5 =40
Step III: We write the fraction in the simplest form as \(\frac{180}{40}\),
we can further simplify as both goes in 20, 20 X 9 = 180 and 20 X 2 = 40, (9,2),
1 \(\frac{7}{8}\) X 2 \(\frac{2}{5}\) = \(\frac{180}{40}\) = \(\frac{9}{2}\) , As numerator is greater than denominator we write in
mixed fraction also as (4 X 2 + 1 by 2 ), 4 \(\frac{1}{2}\).

Question 12.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 28
5 \(\frac{5}{7}\) X 2 \(\frac{1}{10}\) = \(\frac{840}{70}\) = 12

Explanation :
5 \(\frac{5}{7}\) X 2 \(\frac{1}{10}\) , We write mixed fractions
5 \(\frac{5}{7}\) as 5 X 7 + 5 by 7 = \(\frac{40}{7}\) and 2 \(\frac{1}{10}\) as 2 X 10 + 1 by 10 = \(\frac{21}{10}\) Now we multiply
\(\frac{40}{7}\) X \(\frac{21}{10}\),
Step I: We multiply the numerators as 40 X 21 = 840
Step II: We multiply the denominators as 7 X 10 =70
Step III: We write the fraction in the simplest form as \(\frac{840}{70}\),
we can further simplify as both goes in 70, 70 X 12 = 840 and 70 X 1 = 70, (12,1), therefore
5 \(\frac{5}{7}\) X 2 \(\frac{1}{10}\) = \(\frac{840}{70}\) = 12 .

Question 13.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 29
2 \(\frac{1}{3}\) X 7 \(\frac{2}{3}\) = \(\frac{161}{9}\) = 17 \(\frac{8}{9}\)

Explanation:
2 \(\frac{1}{3}\) X 7 \(\frac{2}{3}\) , We write mixed fractions
2 \(\frac{1}{3}\) as 2 X 3 + 1 by 3 = \(\frac{7}{3}\) and 7 \(\frac{2}{3}\) as 7 X 3 + 2 by 3 = \(\frac{23}{3}\) Now we multiply
\(\frac{7}{3}\) X \(\frac{23}{3}\),
Step I: We multiply the numerators as 7 X 23 = 161
Step II: We multiply the denominators as 3 X 3 =9
Step III: We write the fraction in the simplest form as \(\frac{161}{9}\),
therefore 2 \(\frac{1}{3}\) X 7 \(\frac{2}{3}\) = \(\frac{161}{9}\),
As numerator is greater than denominator we write in
mixed fraction also as (17 X 9 + 8 by 9 ), 17 \(\frac{8}{9}\).

Self-Assessment for Concepts & Skills

Solve each exercise. Then rate your understanding of the success criteria in your journal.

MULTIPLYING FRACTIONS AND MIXED NUMBERS
Multiply. Write the answer in simplest form.

Question 14.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 30
\(\frac{1}{8}\) X \(\frac{1}{6}\) = \(\frac{1}{48}\)

Explanation:
Step I: We multiply the numerators as 1 X 1 = 1
Step II: We multiply the denominators as 8 X 6 =48
Step III: We write the fraction in the simplest form as \(\frac{1}{48}\),
So \(\frac{1}{8}\) X \(\frac{1}{6}\) = \(\frac{1}{48}\).

Question 15.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 31

\(\frac{3}{8}\) X \(\frac{2}{3}\) = \(\frac{6}{24}\) = \(\frac{1}{4}\)

Explanation:
Step I: We multiply the numerators as 3 X 2 = 6
Step II: We multiply the denominators as 8 X 3 =24
Step III: We write the fraction in the simplest form as \(\frac{6}{24}\),
we can further simplify as both goes in 6, 6 X 1 = 6 and 6 X 4 = 24, (1,4),
So \(\frac{3}{8}\) X \(\frac{2}{3}\) = \(\frac{6}{24}\) = \(\frac{1}{4}\).

Question 16.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 32

2 \(\frac{1}{6}\) X 4 \(\frac{2}{5}\) = \(\frac{286}{30}\) = \(\frac{143}{15}\) = 9 \(\frac{8}{15}\)

Explanation:
2 \(\frac{1}{6}\) X 4 \(\frac{2}{5}\) , We write mixed fractions
2 \(\frac{1}{6}\) as 2 X 6 + 1 by 6 = \(\frac{13}{6}\) and 4 \(\frac{2}{5}\) as 4 X 5 + 2 by 5 = \(\frac{22}{5}\) Now we multiply
\(\frac{13}{6}\) X \(\frac{22}{5}\),
Step I: We multiply the numerators as 13 X 22 = 286
Step II: We multiply the denominators as 6 X 5 =30
Step III: We write the fraction in the simplest form as \(\frac{286}{30}\),
we can further simplify as both goes in 2, 2 X 143 = 286 and 2 X 15 = 30, (143,15),
therefore 2 \(\frac{1}{6}\) X 4 \(\frac{2}{5}\) = \(\frac{286}{30}\) =
\(\frac{143}{15}\). As numerator is greater than denominator we write in
mixed fraction also as (9 X 15 + 8 by 15 ), 9 \(\frac{8}{15}\).

Question 17.
REASONING
What is the missing denominator?
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 33
The missing denominator is 4

Explanation:
Let us take the missing denominator as x now we have
denominators as 7 X x = 28, so x = \(\frac{28}{7}\) = 4.

Question 18.
USING TOOLS
Write a multiplication problem involving fractions that is represented by the model. Explain your reasoning.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 34

The multiplication problem involving fractions that is represented by the model
\(\frac{4}{5}\) X \(\frac{1}{3}\) = \(\frac{4}{15}\)

Explanation:
Given the area model as shown in figure we take the first fraction part
as 4 out of 5 as yellow area represents 4 parts out of 5 and the second fraction
part is 1 out of 3 with blue now the products of fraction is
\(\frac{4}{5}\) X \(\frac{1}{3}\) =
Step I: We multiply the numerators as 4 X 1 = 4
Step II: We multiply the denominators as 5 X 3 = 15
Step III: We write the fraction in the simplest form as \(\frac{4}{15}\),
which is the green area came through overlapping which is the product as
4 parts out of 15, Therefore the multiplication problem involving fractions
that is represented by the model is \(\frac{4}{5}\) X \(\frac{1}{3}\) = \(\frac{4}{15}\).

Question 19.
USING TOOLS
Use the number line to find Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 35 Explain your reasoning.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 35.1
Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals-11
\(\frac{3}{4}\) X \(\frac{1}{2}\) = \(\frac{3}{8}\)

Explanation:
We write 1/2 as 4 by 8 on the number line,
now we take 3 parts of 4 from 4/8 we get results as
\(\frac{3}{8}\) on the number line or we take 3 times \(\frac{1}{8}\) on
number line we get \(\frac{3}{8}\).
Therefore \(\frac{3}{4}\) X \(\frac{1}{2}\) = \(\frac{3}{8}\).

Self-Assessment for Problem Solving

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 20.
You spend \(\frac{5}{12}\) of a day at an amusement park. You spend \(\frac{2}{5}\) of that time riding waterslides. How many hours do you spend riding waterslides? Draw a model to show why your answer makes sense.
I spent 4 hours for riding waterslides
Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals-11

Explanation:
Given I spend \(\frac{5}{12}\) of a day at an amusement park means
out of 24 hours so hours spent at the park is \(\frac{5}{12}\) X 24 = 10 hours.
Now in 10 hours we spend \(\frac{2}{5}\) of that time riding waterslides means
number of hours spent riding waterslides is 10 X \(\frac{2}{5}\) = 4 hours.
So I spent 4 hours for riding waterslides.
We take area model as shown in the picture first part as 5 out of 12 and second part as 24
we get 120 by 12 as 10 hours and from 10 hours we spent 2 out of 5 we get 4 hours.

Question 21.
A venue is preparing for a concert on the floor shown. The width of the red carpet is \(\frac{1}{6}\) of the width of the floor. What is the area of the red carpet?
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 36
The area of the red carpet is 1057 square feet

Explanation:
Given the width of floor is 63 feet, So width of the red carpet is \(\frac{1}{6}\) X 63 =
\(\frac{63}{6}\) we can simplify as both goes in 3we get \(\frac{21}{2}\),
Now we write length of floor 100 \(\frac{2}{3}\) as 100 X 3 + 2 by 3 as \(\frac{302}{3}\), Now we has width and length of the carpet so area of the carpet is
\(\frac{302}{3}\) X \(\frac{21}{2}\) = \(\frac{6342}{6}\)
both goes in 6 so we get 1057 square feet.

Question 22.
You travel 9\(\frac{3}{8}\) miles from your house to a shopping mall. You travel \(\frac{2}{3}\) of that distance on an interstate. The only road construction you encounter is on the first \(\frac{2}{5}\) of the interstate. On how many miles of your trip do you encounter construction?

I encounter \(\frac{5}{2}\) or 2 \(\frac{1}{2}\) miles of construction of my trip.

Explanation:
Given I travel 9\(\frac{3}{8}\) we write mixed fraction as fraction as
9 X 8 + 3 by 8 = \(\frac{75}{8}\) now I travel \(\frac{2}{3}\) of that distance on an interstate,
So the distance on the interstate is \(\frac{2}{3}\)  X \(\frac{75}{8}\) = \(\frac{150}{24}\) both goes in 6 we get \(\frac{25}{4}\) miles.
Now in \(\frac{25}{4}\) the only construction I encounter is \(\frac{2}{5}\) of
\(\frac{25}{4}\),
So \(\frac{2}{5}\) X \(\frac{25}{4}\) =
\(\frac{50}{20}\) = \(\frac{5}{2}\) miles,
As numerator is greater than denominator we write in mixed fraction
also as (2 X 2 + 1 by 2 ), 2 \(\frac{1}{2}\) miles.

Multiplying Fractions Homework & Practice 2.1

Review & Refresh

Find the LCM of the numbers.

Question 1.
8, 10
The LCM of 8, 10 is 40

Explanation:
The LCM is the smallest positive number that all of the numbers divide into evenly.
One way is to list the multiples of each number, then choose the common multiples,
then the least one.
Multiples of 8:{8,16,24,32,40,48,56,64,72,80,}
Multiples of 10:{10,20,30,40,50,60,70,80,90,}
Common Multiples are {40,80,..}
therefore LCM(8,10)=40.

Question 2.
5, 7
The LCM of 5, 7 is 35

Explanation:
The LCM is the smallest positive number that all of the numbers divide into evenly.
One way is to list the multiples of each number, then choose the common multiples,
then the least one.
Multiples of 5:{5,10,15,20,25,30,35,40,45,50,}
Multiples of 7:{7,14,21,28,35,42,49,56,}
Common Multiples  {35,..}
therefore LCM(5,7)=35.

Question 3.
2, 5, 7
The LCM of 2,5,7 is 70

Explanation:
The LCM is the smallest positive number that all of the numbers divide into evenly.
1. List the prime factors of each number.
2. Multiply each factor the greatest number of times it occurs in either number.
Since 2 has no factors besides  and 2,
2 is a prime number,
Since  has no factors besides 1 and 7.
7 is a prime number,
Since 5 has no factors besides 1 and 5.
5 is a prime number,
The LCM of 2,5,7 is the result of multiplying all prime factors
the greatest number of times they occur in either number 2 X 5 X 7 = 70,
So the LCM of 2,5,7 is 70.

Question 4.
6, 7, 10
The LCM of 6, 7, 10 is 210

Explanation:
Common division of 6 ,7, 10 is
Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals-12

∴ So the LCM of the given numbers is 2 X 5 X 3 X 7 = 210.

Divide. Use a diagram to justify your answer.

Question 5.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 36.1
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 36.1= 6 X 2 = 12
Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals-13
Explanation:
Given 6 divides \(\frac{5}{2}\)  we get 6 X 2 = 12,
we take 6 parts in that we are dividing by \(\frac{1}{2}\) we
are getting 12 parts of a whole.

Question 6.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 37
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 37= 8 X 4 = 32
Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals-14
Explanation:
Given 8 divides \(\frac{1}{4}\)  we get 8 X 4 = 32,
we take 8 parts in that we are dividing by \(\frac{1}{4}\) we
are getting 32 parts of a whole.

Question 7.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 38
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 38= 12
Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals-15
Explanation:
Given 4 divides \(\frac{1}{3}\)  we get 4 X 3 = 12,
we take 4 parts in that we are dividing by \(\frac{1}{3}\) we
are getting 12 parts of a whole.

Question 8.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 39
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 39= 4 X 5 = 20
Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals-16
Explanation:
Given 4 divides \(\frac{1}{5}\)  we get 4 X 5 = 20,
we take 4 parts in that we are dividing by \(\frac{1}{5}\) we
are getting 20 parts of a whole.

Write the product as a power.

Question 9.
10 × 10 × 10

10 X 10 X 10 =103

Explanation:
Given expression as 10 X 10 X 10 as 10 is multiplied by 3 times we write it as
power of 10 as 103.

Question 10.
5 × 5 × 5 × 5

5 × 5 × 5 × 5 = 54

Explanation:
Given expression as 5 X 5 X 5 X 5 as 5 is multiplied by 4 times we write it as
power of 5 as 54.

Question 11.
How many inches are in 5\(\frac{1}{2}\) yards?
A. 15\(\frac{1}{2}\)
B. 16\(\frac{1}{2}\)
C. 66
D. 198

D, 5\(\frac{1}{2}\) yards = 198

Explanation:
Given 5\(\frac{1}{2}\) yards first we write mixed fractions into
fractions as 5 X 2 + 1 by 2 as \(\frac{11}{2}\) yards as we know 1 yard is equal to
36 inches so \(\frac{11}{2}\) = \(\frac{11}{2}\)  X 36 = \(\frac{396}{2}\) = 198 , So it matches with D bit.

Concepts, Skills, & Problem Solving
CHOOSE TOOLS A bottle of water is \(\frac{2}{3}\) full. You drink the given portion of the water. Use a model to find the portion of the bottle of water that you drink. (See Exploration 1, p. 45.)

Question 12.
\(\frac{1}{2}\)
Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals-17

We use area model to give the portion of the bottle of water
that I drink is \(\frac{1}{3}\) of the bottle.

Explanation:
Given a bottle of water is \(\frac{2}{3}\) full.
I drink \(\frac{1}{2}\) of the water. So the portion of the bottle of water
that I drink is \(\frac{2}{3}\) X \(\frac{1}{2}\) = \(\frac{2}{6}\) =
\(\frac{1}{3}\) of the bottle.
We draw an area model to find the product
Step 1: We are given with \(\frac{2}{3}\) so
we take shade 2 part out of 3,
Step 2: We have \(\frac{1}{2}\) now we shade
1 part out of 2
Step 4 : We multiply the numerators as 2 X 1 = 2
Step 5: We multiply the denominators as 3 X 2 = 6
Step 6 : The purple area came through overlapping which is the product as
2 parts out of 6 or 1 part out of 3 and write the fraction
in the simplest form as \(\frac{2}{6}\) or \(\frac{1}{3}\)
So the portion of the bottle of water that I drink is \(\frac{2}{6}\) =
\(\frac{1}{3}\) of the bottle.

Question 13.
\(\frac{1}{4}\)
Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals-18

We use area model to give the portion of the bottle of water
that I drink is \(\frac{1}{6}\) of the bottle.

Explanation:
Given a bottle of water is \(\frac{2}{3}\) full.
I drink \(\frac{1}{4}\) of the water. So the portion of the bottle of water
that I drink is \(\frac{2}{3}\) X \(\frac{1}{4}\) = \(\frac{2}{12}\) =
\(\frac{1}{6}\) of the bottle.
We draw an area model to find the product
Step 1: We are given with \(\frac{2}{3}\) so
we take shade 2 part out of 3,
Step 2: We have \(\frac{1}{4}\) now we shade
1 part out of 4
Step 4 : We multiply the numerators as 2 X 1 = 2
Step 5: We multiply the denominators as 3 X 4 = 12
Step 6 : The purple area came through overlapping which is the product as
2 parts out of 12 or 1 part out of 6 and write the fraction
in the simplest form as \(\frac{2}{12}\) or \(\frac{1}{6}\)
So the portion of the bottle of water that I drink is \(\frac{2}{12}\) =
\(\frac{1}{6}\) of the bottle.

Question 14.
\(\frac{3}{4}\)
Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals-19

We use area model to give the portion of the bottle of water
that I drink is \(\frac{1}{2}\) of the bottle.

Explanation:
Given a bottle of water is \(\frac{2}{3}\) full.
I drink \(\frac{3}{4}\) of the water. So the portion of the bottle of water
that I drink is \(\frac{2}{3}\) X \(\frac{3}{4}\) = \(\frac{6}{12}\) =
\(\frac{1}{2}\) of the bottle.
We draw an area model to find the product
Step 1: We are given with \(\frac{2}{3}\) so
we take shade 2 part out of 3,
Step 2: We have \(\frac{3}{4}\) now we shade
3 parts out of 4
Step 4 : We multiply the numerators as 2 X 3 = 6
Step 5: We multiply the denominators as 3 X 4 = 12
Step 6 : The purple area came through overlapping which is the product as
6 parts out of 12 or 1 part out of 2 and write the fraction
in the simplest form as \(\frac{6}{12}\) or \(\frac{1}{2}\)
So the portion of the bottle of water that I drink is \(\frac{6}{12}\) =
\(\frac{1}{2}\) of the bottle.

MULTIPLYING FRACTIONS
Multiply. Write the answer in simplest form.

Question 15.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 40
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 40= \(\frac{2}{21}\)

\(\frac{1}{7}\) X \(\frac{2}{3}\) = \(\frac{2}{21}\)

Explanation:
Given expression as \(\frac{1}{7}\) X \(\frac{2}{3}\)
Step I: We multiply the numerators as 1 X 2 = 2
Step II: We multiply the denominators as 7 X 3 = 21
Step III: We write the fraction in the simplest form as \(\frac{2}{21}\),
So \(\frac{1}{7}\) X \(\frac{2}{3}\) = \(\frac{2}{21}\).

Question 16.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 41
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 41= \(\frac{5}{16}\)

\(\frac{5}{8}\) X \(\frac{1}{2}\) = \(\frac{5}{16}\)

Explanation:
Given expression as \(\frac{5}{8}\) X \(\frac{1}{2}\)
Step I: We multiply the numerators as 5 X 1 = 5
Step II: We multiply the denominators as 8 X 2 = 16
Step III: We write the fraction in the simplest form as \(\frac{5}{16}\),
So \(\frac{5}{8}\) X \(\frac{1}{2}\) = \(\frac{5}{16}\).

Question 17.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 42
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 42= \(\frac{1}{10}\)

\(\frac{1}{4}\) X \(\frac{2}{5}\) = \(\frac{2}{20}\) = \(\frac{1}{10}\)

Explanation:
Given expression as \(\frac{1}{4}\) X \(\frac{2}{5}\)
Step I: We multiply the numerators as 1 X 2 = 2
Step II: We multiply the denominators as 4 X 5 = 20
Step III: We write the fraction in the simplest form as \(\frac{2}{20}\),
we can further simplify as both goes in 2, 2 X 1 = 2 and 2 X 10 = 20, (1,10),
So \(\frac{1}{4}\) X \(\frac{2}{5}\) = \(\frac{2}{20}\) = \(\frac{1}{10}\)

Question 18.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 43
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 43= \(\frac{3}{28}\)

\(\frac{3}{7}\) X \(\frac{1}{4}\) = \(\frac{3}{28}\)

Explanation:
Given expression as \(\frac{3}{7}\) X \(\frac{1}{4}\)
Step I: We multiply the numerators as 3 X 1 = 3
Step II: We multiply the denominators as 7 X 4 = 28
Step III: We write the fraction in the simplest form as \(\frac{3}{28}\),
So \(\frac{3}{7}\) X \(\frac{1}{4}\) = \(\frac{3}{28}\).

Question 19.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 44
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 44= \(\frac{8}{21}\)

\(\frac{2}{3}\) X \(\frac{4}{7}\) = \(\frac{8}{21}\)

Explanation:
Given expression as \(\frac{2}{3}\) X \(\frac{4}{7}\)
Step I: We multiply the numerators as 3 X 1 = 3
Step II: We multiply the denominators as 7 X 4 = 28
Step III: We write the fraction in the simplest form as \(\frac{3}{28}\),
So \(\frac{2}{3}\) X \(\frac{4}{7}\) = \(\frac{8}{21}\).

Question 20.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 45
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 45 = \(\frac{5}{8}\)

\(\frac{5}{7}\) X \(\frac{7}{8}\) = \(\frac{35}{56}\) = \(\frac{5}{8}\)

Explanation:

Given expression as \(\frac{5}{7}\) X \(\frac{7}{8}\)
Step I: We multiply the numerators as 5 X 7 = 35
Step II: We multiply the denominators as 7 X 8 = 56
Step III: We write the fraction in the simplest form as \(\frac{35}{56}\),
we can further simplify as both goes in 7, 7 X 5 = 35 and 7 X 8 = 56, (5,8),
So \(\frac{5}{7}\) X \(\frac{7}{8}\) = \(\frac{35}{56}\) = \(\frac{5}{8}\).

Question 21.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 46

Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 46= \(\frac{1}{24}\)

\(\frac{3}{8}\) X \(\frac{1}{9}\) = \(\frac{3}{72}\) = \(\frac{1}{24}\)

Explanation:
Given expression as \(\frac{3}{8}\) X \(\frac{1}{9}\)
Step I: We multiply the numerators as 3 X 1 = 3
Step II: We multiply the denominators as 8 X 9 = 72
Step III: We write the fraction in the simplest form as \(\frac{3}{72}\),
we can further simplify as both goes in 3, 3 X 1 = 3 and 3 X 24 = 72, (1,24),
So \(\frac{3}{8}\) X \(\frac{1}{9}\) = \(\frac{3}{72}\) = \(\frac{1}{24}\).

Question 22.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 47
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 47= \(\frac{1}{3}\)

\(\frac{5}{6}\) X \(\frac{2}{5}\) = \(\frac{10}{30}\) = \(\frac{1}{3}\)

Explanation:

Given expression as \(\frac{5}{6}\) X \(\frac{2}{5}\)
Step I: We multiply the numerators as 5 X 2 = 10
Step II: We multiply the denominators as 6 X 5 = 30
Step III: We write the fraction in the simplest form as \(\frac{10}{30}\),
we can further simplify as both goes in 10, 10 X 1 = 10 and 10 X 3 = 30, (1,3),
So \(\frac{5}{6}\) X \(\frac{2}{5}\) = \(\frac{10}{30}\) = \(\frac{1}{3}\).

Question 23.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 48
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 48= \(\frac{25}{6}\) = 4 \(\frac{1}{6}\)

\(\frac{5}{12}\) X 10 = \(\frac{50}{12}\) = \(\frac{25}{6}\) =
4 \(\frac{1}{6}\)

Explanation:

Given expression as \(\frac{5}{12}\) X 10
Step I: We multiply the numerators as 5 X 10 = 50
Step II: Denominators will be same as 12
Step III: We write the fraction in the simplest form as \(\frac{50}{12}\),
we can further simplify as both goes in 2, 2 X 25 = 50 and 2 X 6 = 12, (25,6),
\(\frac{25}{6}\) . As numerator is greater than denominator we write in mixed fraction also as ( 4 X 6 + 1 by 6 ), 4 \(\frac{1}{6}\). So \(\frac{5}{12}\) X 10 = \(\frac{50}{12}\) = \(\frac{25}{6}\) = 4 \(\frac{1}{6}\).

Question 24.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 49
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 49= \(\frac{21}{4}\) = 5\(\frac{1}{4}\)

Explanation:
Given expression as 6 X \(\frac{7}{8}\)
Step I: We multiply the numerators as 6 X 7 = 42
Step II: Denominators will be same as 8
Step III: We write the fraction in the simplest form as \(\frac{42}{8}\),
we can further simplify as both goes in 2, 2 X 21 = 42 and 2 X 4 = 8, (21,4),
\(\frac{21}{4}\). So 6 X \(\frac{7}{8}\)  = \(\frac{42}{8}\) = \(\frac{21}{4}\) as numerator is greater we write as (5 X 4 + 1 by 4), 5\(\frac{1}{4}\).

Question 25.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 50
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 50= \(\frac{2}{5}\)

\(\frac{3}{4}\) X \(\frac{8}{15}\) = \(\frac{24}{60}\) = \(\frac{2}{5}\)

Explanation:
Given expression as \(\frac{3}{4}\) X \(\frac{8}{15}\)
Step I: We multiply the numerators as 3 X 8 = 24
Step II: We multiply the denominators as 4 X 15 = 60
Step III: We write the fraction in the simplest form as \(\frac{24}{60}\),
we can further simplify as both goes in 12, 12 X 2 = 24 and 12 X 5 = 60, (2,5),
So \(\frac{3}{4}\) X \(\frac{8}{15}\) = \(\frac{24}{60}\) = \(\frac{2}{5}\).

Question 26.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 51
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 51= \(\frac{16}{45}\)

\(\frac{4}{9}\) X \(\frac{4}{5}\) = \(\frac{16}{45}\)

Explanation:

Given expression as \(\frac{4}{9}\) X \(\frac{4}{5}\)
Step I: We multiply the numerators as 4 X 4 = 16
Step II: We multiply the denominators as 9 X 5 = 45
Step III: We write the fraction in the simplest form as \(\frac{16}{45}\),
So \(\frac{4}{9}\) X \(\frac{4}{5}\) = \(\frac{16}{45}\).

Question 27.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 52
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 52= \(\frac{9}{49}\)

\(\frac{3}{7}\) X \(\frac{3}{7}\) = \(\frac{9}{49}\)

Explanation:
given expression as \(\frac{3}{7}\) X \(\frac{3}{7}\)
Step I: We multiply the numerators as 3 X 3 = 9
Step II: We multiply the denominators as 7 X 7 = 49
Step III: We write the fraction in the simplest form as \(\frac{9}{49}\),
So \(\frac{3}{7}\) X \(\frac{3}{7}\) = \(\frac{9}{49}\).

Question 28.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 53
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 53= \(\frac{5}{27}\)

\(\frac{5}{6}\) X \(\frac{2}{9}\) = \(\frac{10}{54}\) = \(\frac{5}{27}\)

Explanation:

given expression as \(\frac{5}{6}\) X \(\frac{2}{9}\)
Step I: We multiply the numerators as 5 X 2 = 10
Step II: We multiply the denominators as 6 X 9 = 54
Step III: We write the fraction in the simplest form as \(\frac{10}{54}\),
we can further simplify as both goes in 2, 2 X 5 = 10 and 2 X 27 = 54, (5,27),
So \(\frac{5}{6}\) X \(\frac{2}{9}\) = \(\frac{10}{54}\) = \(\frac{5}{27}\).

Question 29.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 54
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 54= \(\frac{13}{21}\)

\(\frac{13}{18}\) X \(\frac{6}{7}\) = \(\frac{78}{126}\) = \(\frac{13}{21}\)

Explanation:

Given expression as \(\frac{13}{18}\) X \(\frac{6}{7}\)
Step I: We multiply the numerators as 13 X 6 = 78
Step II: We multiply the denominators as 18 X 7 = 126
Step III: We write the fraction in the simplest form as \(\frac{78}{126}\),
we can further simplify as both goes in 6, 6 X 13 = 78 and 6 X 21 = 126, (13,21),
So \(\frac{13}{18}\) X \(\frac{6}{7}\) = \(\frac{78}{126}\) = \(\frac{13}{21}\).

Question 30.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 55
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 55= \(\frac{49}{30}\) = 1 \(\frac{19}{30}\)

\(\frac{7}{9}\) X \(\frac{21}{10}\) = \(\frac{147}{90}\) = \(\frac{49}{30}\) = 1 \(\frac{19}{30}\)

Explanation:
Given expression as \(\frac{7}{9}\) X \(\frac{21}{10}\)
Step I: We multiply the numerators as 7 X 21 = 147
Step II: We multiply the denominators as 9 X 10 = 90
Step III: We write the fraction in the simplest form as \(\frac{147}{90}\),
we can further simplify as both goes in 3, 3 X 49 = 147 and 3 X 30 = 90, (147,90),
\(\frac{49}{30}\). As numerator is greater than denominator we write in mixed fraction also as (1 X 30 + 19 by 30 ), 1 \(\frac{19}{30}\). Therefore \(\frac{7}{9}\) X \(\frac{21}{10}\) = \(\frac{147}{90}\) = \(\frac{49}{30}\) =
1 \(\frac{19}{30}\).

Question 31.
MODELING REAL LIFE
In an aquarium, \(\frac{2}{5}\) of the fish are surgeonfish. Of these, \(\frac{3}{4}\) are yellow tangs. What portion of all fish in the aquarium are yellow tangs?
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 56
\(\frac{3}{10}\) portion of all fish in the aquarium are yellow tangs

Explanation:
Given in an aquarium, \(\frac{2}{5}\) of the fish are surgeonfish. Of these, \(\frac{3}{4}\) are yellow tangs. So the portion of all fish in the aquarium yellow tangs are
\(\frac{2}{5}\) X \(\frac{3}{4}\),
Step I: We multiply the numerators as 2 X 3 = 6
Step II: We multiply the denominators as 5 X 4 = 20
Step III: We write the fraction in the simplest form as \(\frac{6}{20}\),
we can further simplify as both goes in 2, 2 X 3 = 6 and 2 X 10 = 20, (3,10),\(\frac{3}{10}\).
So \(\frac{2}{5}\) X \(\frac{3}{4}\) = \(\frac{3}{10}\), therefore \(\frac{3}{10}\) portion of all fish in the aquarium are yellow tangs.

Question 32.
MODELING REAL LIFE
You exercise for \(\frac{3}{4}\) of an hour. You jump rope for \(\frac{1}{3}\) of that time. What portion of the hour do you spend jumping rope?
\(\frac{1}{4}\) of the hour I do spend jumping rope

Explanation:
Given I exercise for \(\frac{3}{4}\) of an hour and I jump rope for \(\frac{1}{3}\) of that time.
So portion of the hour I do spend jumping rope is
\(\frac{3}{4}\) X \(\frac{1}{3}\),
Step I: We multiply the numerators as 3 X 1 = 3
Step II: We multiply the denominators as 4 X 3 = 12
Step III: We write the fraction in the simplest form as \(\frac{3}{12}\),
we can further simplify as both goes in 3, 3 X 1 = 3 and 3 X 4 = 12, (1,4),
\(\frac{1}{4}\). So \(\frac{3}{4}\) X \(\frac{1}{3}\) = \(\frac{1}{4}\), Therefore \(\frac{1}{4}\) of the hour I do spend jumping rope.

REASONING
Without finding the product, copy and complete the statement using <, >, or =. Explain your reasoning.

Question 33.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 57
> as \(\frac{9}{10}\) < 1

Explanation:
As both sides we have \(\frac{4}{7}\) both gets cancelled and we get 1 in left side
and \(\frac{9}{10}\) in right side as we know 1 is greater than \(\frac{9}{10}\)
so \(\frac{4}{7}\) > \(\frac{9}{10}\) X \(\frac{4}{7}\).

Question 34.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 58
> as \(\frac{22}{15}\) > 1

Explanation:
As both sides we have \(\frac{5}{8}\) both gets cancelled and we get 1 in right side
and \(\frac{22}{15}\) in left side as we know 1 is less than \(\frac{22}{15}\)
as numerator is greater than denominator so \(\frac{5}{8}\) X \(\frac{22}{15}\) >  \(\frac{5}{8}\).

Question 35.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 59
= as \(\frac{7}{7}\) = 1
Explanation:
As both sides we have \(\frac{5}{6}\) both gets cancelled and we get 1 in left side
and \(\frac{7}{7}\) = 1 in right side as both are equal to 1, so \(\frac{5}{6}\) = \(\frac{5}{6}\) X \(\frac{7}{7}\).

MULTIPLYING FRACTIONS AND MIXED NUMBERS
Multiply. Write the answer in simplest form.

Question 36.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 60
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 60= \(\frac{8}{9}\)
1\(\frac{1}{3}\) X \(\frac{2}{3}\) = \(\frac{8}{9}\)

Explanation:
Given 1 \(\frac{1}{3}\) X \(\frac{2}{3}\) so
first we write mixed number 1 \(\frac{1}{3}\) as 1 X 3 + 1 by 3 = \(\frac{4}{3}\) now we multiply \(\frac{4}{3}\)  X  \(\frac{2}{3}\),
Step I: We multiply the numerators as 4 X 2 = 8
Step II: We multiply the denominators as 3 X 3 = 9
Step III: We write the fraction in the simplest form as \(\frac{8}{9}\),
So 1\(\frac{1}{3}\) X \(\frac{2}{3}\) = \(\frac{8}{9}\).

Question 37.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 61
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 61= 2
6\(\frac{2}{3}\) X \(\frac{3}{10}\) = 2

Explanation:
Given 6\(\frac{2}{3}\) X \(\frac{3}{10}\) so
first we write mixed number 6\(\frac{2}{3}\) as 6 X 3 + 2 by 3 = \(\frac{20}{3}\) now we multiply \(\frac{20}{3}\)  X  \(\frac{3}{10}\),
Step I: We multiply the numerators as 20 X 3 = 60
Step II: We multiply the denominators as 3 X 10 = 30
Step III: We write the fraction in the simplest form as \(\frac{60}{30}\),
we can further simplify as both goes in 30, 30 X 2 = 60 and 30 X 10 = 30, (2,1),
So 6\(\frac{2}{3}\) X \(\frac{3}{10}\) = 2.

Question 38.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 62
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 62= 2
2\(\frac{1}{2}\) X \(\frac{4}{5}\) = 2

Explanation:
Given 2\(\frac{1}{2}\) X \(\frac{4}{5}\) so
first we write mixed number 6\(\frac{2}{3}\) as 2 X 2 + 1 by 2 = \(\frac{5}{2}\) now we multiply \(\frac{5}{2}\)  X  \(\frac{4}{5}\),
Step I: We multiply the numerators as 5 X 4 = 20
Step II: We multiply the denominators as 2 X 5 = 10
Step III: We write the fraction in the simplest form as \(\frac{20}{10}\),
we can further simplify as both goes in 10, 10 X 2 = 20 and 10 X 1 = 10, (2,1),
So 2\(\frac{1}{2}\) X \(\frac{4}{5}\) = 2.

Question 39.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 63
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 63= 2

\(\frac{3}{5}\) X 3\(\frac{1}{3}\) = \(\frac{30}{15}\) = 2.

Explanation:
Given \(\frac{3}{5}\) X 3\(\frac{1}{3}\) so
first we write mixed number 3\(\frac{1}{3}\) as 3 X 3 + 1 by 3 = \(\frac{10}{3}\) now we multiply \(\frac{3}{5}\)  X  \(\frac{10}{3}\),
Step I: We multiply the numerators as 3 X 10 = 30
Step II: We multiply the denominators as 5 X 3 = 15
Step III: We write the fraction in the simplest form as \(\frac{30}{15}\),
we can further simplify as both goes in 15, 15 X 2 = 30 and 15 X 1 = 15, (2,1),
So \(\frac{3}{5}\) X 3\(\frac{1}{3}\) = \(\frac{30}{15}\) = 2.

Question 40.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 64
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 64 = 5

7\(\frac{1}{2}\) X \(\frac{2}{3}\) = \(\frac{30}{6}\) = 5

Explanation:
Given 7\(\frac{1}{2}\) X \(\frac{2}{3}\) so
first we write mixed number 7\(\frac{1}{2}\) as 7 X 2 + 1 by 2 = \(\frac{15}{2}\) now we multiply \(\frac{15}{2}\)  X  \(\frac{2}{3}\),
Step I: We multiply the numerators as 15 X 2 = 30
Step II: We multiply the denominators as 2 X 3 = 6
Step III: We write the fraction in the simplest form as \(\frac{30}{6}\),
we can further simplify as both goes in 6, 6 X 5 = 30 and 6 X 1 = 6, (5,1),
So 7\(\frac{1}{2}\) X \(\frac{2}{3}\) = \(\frac{30}{6}\) = 5.

Question 41.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 65
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 65= 2

\(\frac{5}{9}\) X 3\(\frac{3}{5}\) = \(\frac{90}{45}\) = 2.

Explanation:
Given \(\frac{5}{9}\) X 3\(\frac{3}{5}\) so
first we write mixed number 3\(\frac{3}{5}\) as 3 X 5 + 3 by 5 = \(\frac{18}{5}\) now we multiply \(\frac{5}{9}\)  X  \(\frac{18}{5}\),
Step I: We multiply the numerators as 5 X 18 = 90
Step II: We multiply the denominators as 9 X 5 = 45
Step III: We write the fraction in the simplest form as \(\frac{90}{45}\),
we can further simplify as both goes in 45, 45 X 2 = 90 and 45 X 1 = 45, (2,1),
So \(\frac{5}{9}\) X 3\(\frac{3}{5}\) = \(\frac{90}{45}\) = 2.

Question 42.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 66
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 66= 1

\(\frac{3}{4}\) X 1\(\frac{1}{3}\) = \(\frac{12}{12}\) = 1

Explanation:
Given \(\frac{3}{4}\) X 1\(\frac{1}{3}\) so
first we write mixed number 1\(\frac{1}{3}\) as 1 X 3 + 1 by 3 = \(\frac{4}{3}\) now we multiply \(\frac{3}{4}\)  X  \(\frac{4}{3}\),
Step I: We multiply the numerators as 3 X 4 = 12
Step II: We multiply the denominators as 4 X 3 = 12
Step III: We write the fraction in the simplest form as \(\frac{12}{12}\),
we can further simplify as both goes in 12, 12 X 1 = 12 and 12 X 1 = 12, (1,1),
So \(\frac{3}{4}\) X 1\(\frac{1}{3}\) = \(\frac{12}{12}\) = 1.

Question 43.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 67
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 67= \(\frac{3}{2}\) = 1\(\frac{1}{2}\)

3\(\frac{3}{4}\) X \(\frac{2}{5}\) = \(\frac{3}{2}\) = 1\(\frac{1}{2}\)

Explanation:
Given 3\(\frac{3}{4}\) X \(\frac{2}{5}\) so
first we write mixed number 3\(\frac{3}{4}\) as 3 X 4 + 3 by 4 = \(\frac{15}{4}\) now we multiply \(\frac{15}{2}\)  X  \(\frac{2}{5}\),
Step I: We multiply the numerators as 15 X 2 = 30
Step II: We multiply the denominators as 4 X 5 = 20
Step III: We write the fraction in the simplest form as \(\frac{30}{20}\),
we can further simplify as both goes in 10, 10 X 3 = 30 and 10 X 2 = 20, (3,2),
So 3\(\frac{3}{4}\) X \(\frac{2}{5}\) = \(\frac{3}{2}\).
As numerator is greater than denominator we write in mixed fraction
also as (1 X 2 + 1 by 2), 1 \(\frac{1}{2}\).

Question 44.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 68
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 68= \(\frac{7}{2}\) = 3\(\frac{1}{2}\)

4 \(\frac{3}{8}\) X \(\frac{4}{5}\) = \(\frac{7}{2}\) = 3\(\frac{1}{2}\)

Explanation:
Given 4\(\frac{3}{8}\) X \(\frac{4}{5}\) so
first we write mixed number 4\(\frac{3}{8}\) as 8 X 4 + 3 by 8 = \(\frac{35}{8}\) now we multiply \(\frac{35}{8}\)  X  \(\frac{4}{5}\),
Step I: We multiply the numerators as 35 X 4 = 140
Step II: We multiply the denominators as 8 X 5 = 40
Step III: We write the fraction in the simplest form as \(\frac{140}{40}\),
we can further simplify as both goes in 20, 20 X 7 = 140 and 20 X 2 = 40, (7,2),
So 4\(\frac{3}{8}\) X \(\frac{4}{5}\) = \(\frac{7}{2}\).
As numerator is greater than denominator we write in mixed fraction
also as (3 X 2 + 1 by 2), 3\(\frac{1}{2}\).

Question 45.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 69
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 69= \(\frac{17}{14}\) = 1\(\frac{3}{14}\)

\(\frac{3}{7}\) X 2\(\frac{5}{6}\) = \(\frac{17}{14}\) = 1\(\frac{3}{14}\)

Explanation:
Given \(\frac{3}{7}\) X 2\(\frac{5}{6}\) so
first we write mixed number 2\(\frac{5}{6}\) as 2 X 6 + 5 by 6 = \(\frac{17}{6}\) now we multiply \(\frac{3}{7}\)  X  \(\frac{17}{6}\),
Step I: We multiply the numerators as 17 X 3 = 51
Step II: We multiply the denominators as 7 X 6 = 42
Step III: We write the fraction in the simplest form as \(\frac{51}{42}\),
we can further simplify as both goes in 3, 3 X 17 = 51 and 3 X 14 = 12, (17,14),
So \(\frac{3}{7}\) X 2\(\frac{5}{6}\) = \(\frac{17}{14}\),
As numerator is greater than denominator we write in mixed fraction
also as (1 X 14 + 3 by 14), 1\(\frac{3}{14}\).

Question 46.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 70
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 70= \(\frac{117}{5}\) = 23\(\frac{2}{5}\)

1\(\frac{3}{10}\) X 18 = \(\frac{117}{5}\) = 23\(\frac{2}{5}\)

Explanation:
Given 1\(\frac{3}{10}\) X 18 so
first we write mixed number 1\(\frac{3}{10}\) as 1 X 10  + 3 by 10 = \(\frac{13}{10}\) now we multiply \(\frac{13}{10}\)  X  18,
Step I: We multiply the numerators as 13 X 18 = 234
Step II: Denominators will be same as 10
Step III: We write the fraction in the simplest form as \(\frac{234}{18}\),
we can further simplify as both goes in 2, 2 X 117 = 234 and 2 X 5 = 10, (117,5)
So 1\(\frac{3}{10}\) X 18 = \(\frac{117}{5}\).
As numerator is greater than denominator we write in mixed fraction
also as (23 X 5 + 2 by 5), 23\(\frac{2}{5}\).

Question 47.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 71
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 71= \(\frac{110}{3}\) = 36\(\frac{2}{3}\)

15 X 2\(\frac{4}{9}\) = \(\frac{110}{3}\) = 36\(\frac{2}{3}\)

Explanation:
Given 15 X 2\(\frac{5}{6}\) so first we write mixed number
2\(\frac{4}{9}\) as 2 X 9 + 4 by 9 = \(\frac{22}{9}\) now we multiply 15 X  \(\frac{22}{9}\),
Step I: We multiply the numerators as 15 X 22 = 330
Step II: Denominator will be same as  9
Step III: We write the fraction in the simplest form as \(\frac{330}{9}\),
we can further simplify as both goes in 3, 3 X 110 = 330 and 3 X 3 = 9, (110,3),
So 15 X 2\(\frac{4}{9}\) = \(\frac{110}{3}\),
As numerator is greater than denominator we write in mixed fraction
also as (36 X 3 + 2 by 3), 36\(\frac{2}{3}\).

Question 48.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 72
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 72= \(\frac{63}{8}\) = 7\(\frac{7}{8}\)

Explanation:
Given 1\(\frac{1}{6}\) X 6\(\frac{3}{4}\) so
first we write mixed numbers 1\(\frac{1}{6}\) as 6 X 1 + 1 by 6 = \(\frac{7}{6}\) and 6\(\frac{3}{4}\) as 4 X 6 + 3 by 4 = \(\frac{27}{4}\)
now we multiply \(\frac{7}{6}\)  X  \(\frac{27}{4}\),
Step I: We multiply the numerators as 7 X 27 = 189
Step II: We multiply the denominators as 6 X 4 = 24
Step III: We write the fraction in the simplest form as \(\frac{189}{24}\),
we can further simplify as both goes in 3, 3 X 63 = 189 and 3 X 8 = 24, (63,8),
So 1\(\frac{1}{6}\) X 6\(\frac{3}{4}\) = \(\frac{63}{8}\).
As numerator is greater than denominator we write in mixed fraction
also as (7 X 8 + 7 by 8 ), 7\(\frac{7}{8}\).

Question 49.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 73
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 73= \(\frac{58}{9}\) = 6\(\frac{4}{9}\)

Explanation:
Given 2\(\frac{5}{12}\) X 2\(\frac{2}{3}\) so
first we write mixed numbers 2\(\frac{5}{12}\) as 12 X 2 + 5 by 12 = \(\frac{29}{12}\) and 2\(\frac{2}{3}\) as 3 X 2 + 2 by 3 = \(\frac{8}{3}\)
now we multiply \(\frac{29}{12}\)  X  \(\frac{8}{3}\),
Step I: We multiply the numerators as 29 X 8 = 232
Step II: We multiply the denominators as 12 X 3 = 36
Step III: We write the fraction in the simplest form as \(\frac{232}{36}\),
we can further simplify as both goes in 4, 4 X 58  = 232 and 4 X 9 = 36, (58,9),
So 2\(\frac{5}{12}\) X 2\(\frac{2}{3}\) = \(\frac{58}{9}\).
As numerator is greater than denominator we write in mixed fraction
also as (6 X 9 + 4 by 9), 6\(\frac{4}{9}\).

Question 50.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 74
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 74 = \(\frac{125}{7}\) = 17\(\frac{6}{7}\)

Explanation:
Given 5\(\frac{5}{7}\) X 3\(\frac{1}{8}\) so
first we write mixed numbers 5\(\frac{5}{7}\) as 7 X 5 + 5 by 7 = \(\frac{40}{7}\) and 3\(\frac{1}{8}\) as 8 X 3 + 1 by 8 = \(\frac{25}{8}\)
now we multiply \(\frac{40}{7}\)  X  \(\frac{25}{8}\),
Step I: We multiply the numerators as 40 X 25 = 1000
Step II: We multiply the denominators as 7 X 8 = 56
Step III: We write the fraction in the simplest form as \(\frac{1000}{56}\),
we can further simplify as both goes in 8, 8 X 125 = 1000 and 8 X 7 = 56, (125,7),
So 5\(\frac{5}{7}\) X 3\(\frac{1}{8}\) = \(\frac{125}{7}\).
As numerator is greater than denominator we write in mixed fraction
also as (17 X 7 + 6 by 7), 17\(\frac{6}{7}\).

Question 51.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 75
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 75= \(\frac{91}{8}\) = 11\(\frac{3}{8}\)

Explanation:
Given 2\(\frac{4}{5}\) X 4\(\frac{1}{16}\) so
first we write mixed numbers 2\(\frac{4}{5}\) as 2 X 5 + 4 by 5 = \(\frac{14}{5}\) and 4\(\frac{1}{16}\) as 4 X 16 + 1 by 16 = \(\frac{65}{16}\)
now we multiply \(\frac{14}{5}\)  X  \(\frac{65}{16}\),
Step I: We multiply the numerators as 14 X 65 = 910
Step II: We multiply the denominators as 5 X 16 = 80
Step III: We write the fraction in the simplest form as \(\frac{910}{80}\),
we can further simplify as both goes in 10, 10 X 91 = 910 and 10 X 8 = 80, (91,8),
So 2\(\frac{4}{5}\) X 4\(\frac{1}{16}\) = \(\frac{91}{8}\).
As numerator is greater than denominator we write in mixed fraction
also as (11 X 8 + 3 by 8 ), 11\(\frac{3}{8}\).

YOU BE THE TEACHER
Your friend finds the product. Is your friend correct? Explain your reasoning.

Question 52.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 76
No friend is incorrect.
4 X 3\(\frac{7}{10}\) = 14\(\frac{8}{10}\) ≠ 12\(\frac{7}{10}\)

Explanation:
4 X 3\(\frac{7}{10}\) so first we write mixed numbers 3\(\frac{7}{10}\)
as 3 X 10 + 7 by 10 = \(\frac{37}{10}\)
now we multiply 4 X \(\frac{37}{10}\)
Step I: We multiply the numerators as 4 X 37 = 148
Step II: Denominator will be same as 10
Step III: We write the fraction in the simplest form as \(\frac{148}{10}\),
So 4 X 3\(\frac{7}{10}\) = \(\frac{148}{10}\),As numerator is greater than denominator we write in mixed fraction also as (14 X 10 + 8 by 10), 14\(\frac{8}{10}\). As friend says 4 X 3\(\frac{7}{10}\)  =12\(\frac{7}{10}\) which is incorrect
because 4 X 3\(\frac{7}{10}\)  = 14\(\frac{8}{10}\). No friend is incorrect as
4 X 3\(\frac{7}{10}\) = 14\(\frac{8}{10}\) ≠ 12\(\frac{7}{10}\)

Question 53.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 77

No friend is incorrect.
2\(\frac{1}{2}\) X 7\(\frac{4}{5}\) = 19\(\frac{1}{2}\) ≠ 14\(\frac{2}{5}\)

Explanation:
Given 2\(\frac{1}{2}\) X 7\(\frac{4}{5}\) so
first we write mixed numbers 2\(\frac{1}{2}\) as 2 X 2 + 1 by 2 = \(\frac{5}{2}\) and 7\(\frac{4}{5}\) as 7 X 5 + 4 by 5 = \(\frac{39}{5}\)
now we multiply \(\frac{5}{2}\)  X  \(\frac{39}{5}\),
Step I: We multiply the numerators as 5 X 39 = 195
Step II: We multiply the denominators as 2 X 5 = 10
Step III: We write the fraction in the simplest form as \(\frac{195}{10}\),
we can further simplify as both goes in 5, 5 X 39 = 195 and 5 X 2 = 10, (39,2),
So 2\(\frac{1}{2}\) X 7\(\frac{4}{5}\) = \(\frac{39}{2}\).
As numerator is greater than denominator we write in mixed fraction
also as (19 X 2 + 1 by 2), 19\(\frac{1}{2}\).
As friend says  2\(\frac{1}{2}\) X 7\(\frac{4}{5}\) =14\(\frac{2}{5}\) which is incorrect because 2\(\frac{1}{2}\) X 7\(\frac{4}{5}\) = 19\(\frac{1}{2}\). No friend is incorrect as 2\(\frac{1}{2}\) X 7\(\frac{4}{5}\) = 19\(\frac{1}{2}\) ≠ 14\(\frac{2}{5}\).

Question 54.
MODELING REAL LIFE
A vitamin C tablet contains \(\frac{1}{4}\) of a gram of vitamin C. You take 1\(\frac{1}{2}\) tablets every day. How many grams of vitamin C do you take every day?
\(\frac{3}{8}\) grams of vitamin C I do take every day.

Explanation:
Given a vitamin C tablet contains \(\frac{1}{4}\) of a gram of vitamin C.
I take 1\(\frac{1}{2}\) tablets every day. So number of grams of vitamin C
I do take every day is 1\(\frac{1}{2}\) X \(\frac{1}{4}\) so
first we write mixed number 1\(\frac{1}{2}\) as 1 X 2  + 1 by 2 =
\(\frac{3}{2}\) now we multiply \(\frac{3}{2}\)  X  \(\frac{1}{4}\)
Step I: We multiply the numerators as 3 X 1 = 3
Step II: We multiply the denominators as 2 X 4 = 8
Step III: We write the fraction in the simplest form as \(\frac{3}{8}\),
So 1\(\frac{1}{2}\) X \(\frac{1}{4}\) = \(\frac{3}{8}\).
Therefore \(\frac{3}{8}\) grams of vitamin C I do take every day.

Question 55.
PROBLEM SOLVING
You make a banner for a football rally.
a. What is the area of the banner?
b. You add a \(\frac{1}{4}\)-foot border on each side. What is the area of the new banner?
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 78

a. Area of the banner is 7 square feet
b. The area of new banner is 10\(\frac{15}{16}\) square feet

Explanation:
a. Given width of banner as 4\(\frac{2}{3}\) ft and height as 1\(\frac{1}{2}\) feet,Therefore the area of banner is 4\(\frac{2}{3}\) X 1\(\frac{1}{2}\) so
first we write mixed numbers 4\(\frac{2}{3}\) as 4 X 3 + 2 by 3 = \(\frac{14}{3}\) and 1\(\frac{1}{2}\) as 1 X 2 + 1 by 2 = \(\frac{3}{2}\)
now we multiply \(\frac{14}{3}\)  X  \(\frac{3}{2}\),
Step I: We multiply the numerators as 14 X 3 = 42
Step II: We multiply the denominators as 3 X 2 = 6
Step III: We write the fraction in the simplest form as \(\frac{42}{6}\),
we can further simplify as both goes in 6, 6 X 7 = 42 and 6 X 1 = 6, (7,1),
4\(\frac{2}{3}\) X 1\(\frac{1}{2}\) = 7 square feet.
b. Now we add a \(\frac{1}{4}\)-foot border on each side now the width changes to
\(\frac{14}{3}\) X \(\frac{1}{4}\) =\(\frac{14 x 1}{3 X 4}\)= \(\frac{14}{12}\) on simplification both goes in 2 we get \(\frac{7}{6}\),
the new width is \(\frac{14}{3}\) + \(\frac{7}{6}\) we get 14 x 2 + 7 by 6 = \(\frac{35}{6}\) and new height changes to \(\frac{3}{2}\) X \(\frac{1}{4}\) = \(\frac{3}{8}\) now the new height becomes \(\frac{3}{2}\) + \(\frac{3}{8}\) we get (3 X 4 + 3 by 8) = \(\frac{15}{8}\), Now the area of the new banner is \(\frac{35}{6}\) X \(\frac{15}{8}\)
Step I: We multiply the numerators as 35 X 15 = 525
Step II: We multiply the denominators as 6 X 8 = 48
Step III: We write the fraction in the simplest form as \(\frac{525}{48}\),
we can further simplify as both goes in 3, 3 X 175 = 525 and 3 X 16 = 48, (175,16),
\(\frac{175}{16}\), As numerator is greater than denominator we write in mixed fraction also as (10 X 16 + 15 by 16), 10\(\frac{15}{16}\) square feet.

MULTIPLYING FRACTIONS AND MIXED NUMBERS
Multiply. Write the answer in simplest form.

Question 56.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 79
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 79= \(\frac{2}{15}\)

Explanation:
Step I: We multiply the numerators as 1 X 3 X 4 = 12
Step II: We multiply the denominators as 2 X 5 X 9 = 90
Step III: We write the fraction in the simplest form as \(\frac{12}{90}\),
we can further simplify as both goes in 6, 6 X 2 = 12 and 6 X 15 = 90, (2,15),
So \(\frac{1}{2}\) X \(\frac{3}{5}\) X \(\frac{4}{9}\) = \(\frac{2}{15}\).

Question 57.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 80
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 80= \(\frac{25}{12}\) = 2\(\frac{1}{12}\)
Explanation:
Given expression as \(\frac{4}{7}\) X 4\(\frac{3}{8}\) X \(\frac{5}{6}\),
first we write mixed numbers 4\(\frac{3}{8}\) as 4 X 8 + 3 by 8 = \(\frac{35}{8}\) now we multiply \(\frac{4}{7}\)  X  \(\frac{35}{8}\) X \(\frac{5}{6}\)
Step I: We multiply the numerators as 4 X 35 X 5 = 700
Step II: We multiply the denominators as 7 X 8 X 6 = 336
Step III: We write the fraction in the simplest form as \(\frac{700}{336}\),
we can further simplify as both goes in 28, 28 X 25 = 700 and 28 X 12 = 336, (25,12),
So \(\frac{4}{7}\) X 4\(\frac{3}{8}\) X \(\frac{5}{6}\) = \(\frac{25}{12}\), As numerator is greater than denominator we write in mixed fraction
also as (2 X12 + 1 by 12), 2\(\frac{1}{12}\).

Question 58.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 81
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 81= \(\frac{132}{5}\) = 26\(\frac{2}{5}\).

Explanation:
Given expression as 1\(\frac{1}{15}\) X 5\(\frac{2}{5}\) X 4\(\frac{7}{12}\),first we write mixed numbers 1\(\frac{1}{15}\) as 1 X 15 + 1 by 15 = \(\frac{16}{15}\), 5\(\frac{2}{5}\) as 5 X 5 + 2 by 5 = \(\frac{27}{5}\),
4\(\frac{7}{12}\) as 4 X 12 + 7 by 12 = 55 by 12 = \(\frac{55}{12}\)
now we multiply \(\frac{16}{15}\) X \(\frac{27}{5}\) X \(\frac{55}{12}\)
Step I: We multiply the numerators as 16 X 27 X 55 = 23760
Step II: We multiply the denominators as 15 X 5 X 12 = 900
Step III: We write the fraction in the simplest form as \(\frac{23760}{900}\),
we can further simplify as both goes in 180, 180 X 132 = 23760 and 180 X 5 = 900, (132,5),
So 1\(\frac{1}{15}\) X 5\(\frac{2}{5}\) X 4\(\frac{7}{12}\) =
\(\frac{132}{5}\). As numerator is greater than denominator we write in
mixed fraction also as (26 X 5 + 2 by 5), 26\(\frac{2}{5}\).

Question 59.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 82
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 82 = \(\frac{27}{125}\)
Explanation:
Given expression as (\(\frac{3}{5}\))3 we write as  \(\frac{3}{5}\) X \(\frac{3}{5}\) X \(\frac{3}{5}\) now
Step I: We multiply the numerators as 3 X 3 X 3 = 27
Step II: We multiply the denominators as 5 X 5 X 5 = 125
Step III: We write the fraction in the simplest form as \(\frac{27}{125}\).
therefore (\(\frac{3}{5}\))3 = (\(\frac{27}{125}\)).

Question 60.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 83
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 83= \(\frac{9}{25}\)
Explanation :
Now we write the expression as (\(\frac{4}{5}\))2 X (\(\frac{3}{4}\))2 =
\(\frac{4}{5}\) X \(\frac{4}{5}\) X \(\frac{3}{4}\) X \(\frac{3}{4}\) now Step I: We multiply the numerators as 4 X 4 X 3 X 3 = 144
Step II: We multiply the denominators as 5 X 5 X 4 X 4 = 400
Step III: We write the fraction in the simplest form as \(\frac{144}{400}\)
we can further simplify as both goes in 16, 16 X 9 = 144 and 16 X 25 = 400, (9,25),
therefore (\(\frac{4}{5}\))2 X (\(\frac{3}{4}\))2 = \(\frac{9}{25}\).

Question 61.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 84
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 84= \(\frac{121}{144}\)
Explanation:
First we write mixed fraction into fraction as 1\(\frac{1}{10}\) = 1 X 10 +1 by 10 =
\(\frac{11}{10}\), Now we write (\(\frac{5}{6}\))2 X (\(\frac{11}{10}\))2 = \(\frac{5}{6}\) X \(\frac{5}{6}\) X \(\frac{11}{10}\) X \(\frac{11}{10}\) now Step I: We multiply the numerators as 5 X 5 X 11 X 11 = 3025
Step II: We multiply the denominators as 6 X 6 X 10 X 10 = 3600
Step III: We write the fraction in the simplest form as \(\frac{3025}{3600}\),
we can further simplify as both goes in 25, 25 X 121 = 3025 and 25 X 144 = 3600, (121,144),
therefore (\(\frac{5}{6}\))2 X (1\(\frac{1}{10}\))2 = \(\frac{121}{144}\).

Question 62.
OPEN-ENDED
Find a fraction that, when multiplied by \(\frac{1}{2}\), is less than \(\frac{1}{4}\).
\(\frac{1}{3}\)

Explanation:
Let us take on fraction as \(\frac{1}{3}\) which when multiplied by \(\frac{1}{2}\) we get numerator as 1 X 1 = 1 and denominator as 3 X 2 = 6 as \(\frac{1}{6}\) < \(\frac{1}{4}\), So we take \(\frac{1}{3}\) as
\(\frac{1}{3}\) X \(\frac{1}{2}\) < \(\frac{1}{4}\).

Question 63.
LOGIC
You are in a bike race. When you get to the first checkpoint, you are \(\frac{2}{5}\) of the distance to the second checkpoint. When you get to the second checkpoint, you are \(\frac{1}{4}\) of the distance to the finish. What is the distance from the start to the first checkpoint?
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 85
The distance from the start to the first checkpoint is 4 miles

Explanation:
Given total distance is 40 miles and first check point is \(\frac{2}{5}\)
of the distance to the second checkpoint and second checkpoint is
\(\frac{1}{4}\) of the distance to the finish, So second checkpoint is \(\frac{1}{4}\) X 40 as both goes in 4 we get 10 miles, At 10 miles we have second checkpoint,
now the distance from the start to the first checkpoint is \(\frac{2}{5}\) X Second checkpoint \(\frac{2}{5}\) X 10,
Step 1 : We multiply the numerators as 2 X 10
Step II: Denominator is 5,
Step III: We write the fraction in the simplest form as \(\frac{20}{5}\)
we can further simplify as both goes in 5, 5 X 4 = 20 and 5 X 1 = 5, (4,1) = 4 miles,
therefore the distance from the start to the first checkpoint is 4 miles.

Question 64.
NUMBER SENSE
Is the product of two positive mixed numbers ever less than 1? Explain.

No, the product of two positive mixed numbers never ever be less than 1,

Explanation:
We know a mixed must be greater than 1 and two numbers
that are greater than one that are multiplied together end up
being greater that either number by itself. So the product of two
positive mixed numbers never ever be less than 1.

Question 65.
REASONING
You plan to add a fountain to your garden.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 86
a. Draw a diagram of the fountain in the garden. Label the dimensions.
b. Describe two methods for finding the area of the garden that surrounds the fountain.
c. Find the area. Which method did you use, and why?
a.
b. 1. Subtract area of the fountain from the total area of the garden or
2. We use rectangles to find the area of each piece of the garden and add these areas.
c. Area of the garden is 44\(\frac{3}{8}\) square feet, Used
Subtract method because of fewer calculations.

Explanation:
a. We have taken measurements of fountain and drawn the fountain in the
garden as shown with labels of width of fountain as 3\(\frac{1}{3}\) and
height of the fountain as 5\(\frac{1}{4}\).
b. To find the area of the garden that surrounds the fountain first we
subtract area of the fountain from the total area of the garden or
We use rectangles to find the area of each piece of the garden and add these areas.
c. We use subtraction method first we calculate area of garden as
9\(\frac{1}{6}\) X 6\(\frac{3}{4}\)
First we write mixed fraction into fraction as 9\(\frac{1}{6}\) = 9 X 6 + 1 by 6 =
\(\frac{55}{6}\), 6\(\frac{3}{4}\) = 6 X 4 + 3 by 4 = \(\frac{27}{4}\) Now we multiply \(\frac{55}{6}\) X \(\frac{27}{4}\)
Step I: We multiply the numerators as 55 X 27 = 1485
Step II: We multiply the denominators as 6 X 4 = 24
Step III: We write the fraction in the simplest form as \(\frac{1485}{24}\),
we can further simplify as both goes in 3, 3 X 495 = 1485 and 3 X 8 = 24, (495,8),
9\(\frac{1}{6}\) X 6\(\frac{3}{4}\) = \(\frac{495}{8}\).

Now we calculate area of fountain as 5\(\frac{1}{4}\) X 3 \(\frac{1}{3}\),
First we write mixed fraction into fraction as 5\(\frac{1}{4}\) = 5 X 4  + 1 by 4 =
\(\frac{21}{4}\) and 3 \(\frac{1}{3}\) = 3 X 3 + 1 by 3 = \(\frac{10}{3}\) ,Step I: We multiply the numerators as 21 X 10 = 210
Step II: We multiply the denominators as 4 X 3 = 12
Step III: We write the fraction in the simplest form as \(\frac{210}{12}\),
we can further simplify as both goes in 6, 6 X 35 = 210 and 6 X 2 = 12, (35,2),
5\(\frac{1}{4}\) X 3\(\frac{1}{3}\) = \(\frac{35}{2}\).
Therefore area of garden is total area of garden – area of fountain so
\(\frac{495}{8}\) – \(\frac{35}{2}\) = (495 – 35 X 4) by 8 = \(\frac{355}{8}\) ,As numerator is greater than denominator we write in
mixed fraction also as (44 X 8 + 3 by 8), 44\(\frac{3}{8}\) square feet,
Therefore area of the garden is 44\(\frac{3}{8}\) square feet,
Here we have used subtraction method because we can do fewer calculations.

Question 66.
PROBLEM SOLVING
The cooking time for a ham is \(\frac{2}{5}\) of an hour for each pound. What time should you start cooking a ham that weighs 12\(\frac{3}{4}\) pounds so that it is done at 4:45 P.M.?
We should start cooking at 11:39 am so that it is done at 4:45pm

Explanation:
Given the cooking time for a ham is \(\frac{2}{5}\) of an hour for each pound,
The minutes required is \(\frac{2}{5}\)  X 60 = 24 minutes,
so 24 minutes for each pound, Now we multiply minutes by pounds
First we write 12\(\frac{3}{4}\) as 12 X 4 + 3 by 4 is \(\frac{51}{4}\) X 24
Step I: We multiply the numerators as 51 X 24 =1224
Step II: Denominators is same 4
Step III: We write the fraction in the simplest form as \(\frac{1224}{4}\), we get
306 as both goes in 4, Now we convert 306 minutes into hours as
\(\frac{306}{60}\) = 5.1 and .1 in an hour is 6 minutes, Therefore
we cook the ham for 5 hours and 6 minutes. As it is done at 4:45 pm means
16 hours,45 minutes – 5 hours 6 minutes = 11 : 39 am,
So we should start cooking at 11:39 am.

Question 67.
PRECISION
Complete the Four Square for \(\frac{7}{8}\) × \(\frac{1}{3}\).
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 87
Answer is \(\frac{7}{24}\)
Meaning is \(\frac{7}{8}\) of \(\frac{1}{3}\)
Model is
Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals-21
Application : The path around a park is \(\frac{1}{3}\) mile long,
I walk \(\frac{7}{8}\) of the path, How far did I walk?

Explanation:
First we write answer as \(\frac{7}{8}\) × \(\frac{1}{3}\) =
Step I: We multiply the numerators as 7 X 1 = 7
Step II: We multiply denominators as 8 X 3 = 24
Step III: We write the fraction in the simplest form as \(\frac{7}{24}\),
So \(\frac{7}{8}\) × \(\frac{1}{3}\) = \(\frac{7}{24}\)
The meaning is \(\frac{7}{8}\) of \(\frac{1}{3}\) and Model is
as shown above the product is purple color overlapping 7 out of 24,
Applying the question as The path around a park is \(\frac{1}{3}\) mile long,
I walk \(\frac{7}{8}\) of the path, How far did I walk?

Question 68.
DIG DEEPER!
You ask 150 people about their pets. The results show that \(\frac{9}{25}\) of the people own a dog. Of the people who own a dog, \(\frac{1}{6}\) of them also own a cat.
a. What portion of the people own a dog and a cat?
b. How many people own a dog but not a cat? Explain.

a. 54 people, 36% portion of the people own a dog and a cat.
b. 45 people own a dog but not a cat.

Explanation:
Given I ask 150 people about their pets, in that \(\frac{9}{25}\) of the people own a dog means 150 X \(\frac{9}{25}\) = \(\frac{1350}{25}\) = 54 people own a dog. So portion of the people own a dog and cat in 150 are 54 X 100 by 150 = 36%.
Now Of the people who own a dog, \(\frac{1}{6}\) of them also own a cat,
Number of them also own a cat are 54 X \(\frac{1}{6}\) = \(\frac{54}{6}\)= 9
So 9 people owns a cat. We have 54 people owns dog and cat ,
so only own a dog but not cat are 54 – 9 = 45 people own a dog but not a cat.

Question 69.
NUMBER SENSE
Use each of the numbers from 1 to 9 exactly once to create three mixed numbers with the greatest possible product. Then use each of the numbers exactly once to create three mixed numbers with the least possible product. Find each product. Explain your reasoning. The fraction portion of each mixed number should be proper.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 88
Greatest possible product is
9\(\frac{1}{2}\) X 8\(\frac{3}{4}\) X 7\(\frac{5}{6}\) = 651\(\frac{7}{48}\)
Least possible product is
1\(\frac{4}{9}\) X 2\(\frac{5}{8}\) X 3\(\frac{6}{7}\) = 14\(\frac{5}{8}\)

Explanation:
First we use the greatest digits for the whole numbers ,then we use
remaining digits to form the greatest(or least) fractional parts later
use the least(or greatest) digits for the whole numbers ,
to calculate greatest possible product as
9\(\frac{1}{2}\) X 8\(\frac{3}{4}\) X 7\(\frac{5}{6}\),
we write mixed fraction as 9\(\frac{1}{2}\) = 9 X 2 + 1 by 2 = \(\frac{19}{2}\),
8\(\frac{3}{4}\) = 8 X 4 + 3 by 4 = \(\frac{35}{4}\), 7\(\frac{5}{6}\) =
7 X 6 + 5 by 6 = \(\frac{47}{6}\),Now we multiply \(\frac{19}{2}\) X \(\frac{35}{4}\) X \(\frac{47}{6}\),
Step I: We multiply the numerators as 19 X 35 X 47 = 31255
Step II: We multiply denominators as 2 X 4 X 6 = 48
Step III: We write the fraction in the simplest form as \(\frac{31255}{48}\),
As numerator is greater than denominator we write in
mixed fraction also as (651 X 48 + 7 by 48), 651\(\frac{7}{48}\),

Now calculate least possible product as
1\(\frac{4}{9}\) X 2\(\frac{5}{8}\) X 3\(\frac{6}{7}\),
we write mixed fraction as 1\(\frac{4}{9}\) = 1 X 9 + 4 by 9 = \(\frac{13}{9}\),
2\(\frac{5}{8}\) = 2 X 8 + 5 by 8 = \(\frac{21}{8}\), 3\(\frac{6}{7}\) =
3 X 7 + 6 by 7 = \(\frac{27}{7}\),Now we multiply \(\frac{13}{9}\) X \(\frac{21}{8}\) X \(\frac{27}{7}\)
Step I: We multiply the numerators as 13 X 21 X 27 = 7371
Step II: we multiply denominators as 9 X 8 X 7 = 504
Step III: We write the fraction in the simplest form as \(\frac{7371}{504}\),
we can further simplify as both goes in 63, 63 X 117 = 7371 and 63 X 8 = 504, (117,8) =
\(\frac{117}{8}\),As numerator is greater than denominator we write in
mixed fraction also as (14 X 8 + 5 by 8), 14\(\frac{5}{8}\).

Therefore greatest possible product is 9\(\frac{1}{2}\) X 8\(\frac{3}{4}\) X 7\(\frac{5}{6}\) = 651\(\frac{7}{48}\),least possible product is 1\(\frac{4}{9}\) X 2\(\frac{5}{8}\) X 3\(\frac{6}{7}\) = 14\(\frac{5}{8}\).

Lesson 2.2 Dividing Fractions

EXPLORATION 1

Dividing by Fractions
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 89
Work with a partner. Answer each question using a model.
a. How many two-thirds are in four?
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 90
b. How many three-fourths are in three?
c. How many two-fifths are in four-fifths?
d. How many two-thirds are in three?
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 91
e. How many one-thirds are in five-sixths?
a.
Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals-22
Six, two-thirds are there in four,
b.
Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals-23

Four, three-fourths are there in three,
c.
Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals-24
Two, two-fifths are in four-fifths,
d.
Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals-25
4\(\frac{1}{2}\), two-thirds are there in three,
e.
Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals-26
2\(\frac{1}{2}\), one-thirds are in five-sixths

Explanation:
There are
a. Given to find Two-Thirds are there in four is four divides \(\frac{2}{3}\)
we write reciprocal of the fraction \(\frac{2}{3}\) as \(\frac{3}{2}\)=
4  X \(\frac{3}{2}\) = \(\frac{12}{2}\) =6,
Six, Two-Thirds are there in four as shown in the model above,
b. Given to find three-fourths are there in three is 3 divides  \(\frac{3}{4}\)
we write reciprocal of the fraction \(\frac{3}{4}\) as \(\frac{4}{3}\)=
3 X \(\frac{4}{3}\)= \(\frac{12}{3}\) = 4
Four, three-fourths are there in three shown in the model above,
c. Given to find two-fifths are in four-fifths so four-fifths divides two -fifths,first we write
\(\frac{2}{5}\) as reciprocal \(\frac{5}{2}\) now multiply as
\(\frac{4}{5}\) X \(\frac{5}{2}\) = \(\frac{4 X 5}{5 X 2}\) = 2
Two, two-fifths are in four-fifths shown in the model above,
d. Given to find Two-Thirds are there in three is three divides \(\frac{2}{3}\)
we write reciprocal of the fraction \(\frac{2}{3}\) as \(\frac{3}{2}\)=
3  X \(\frac{3}{2}\) = \(\frac{9}{2}\) as numerator is greater we write as
4 X 2 + 1 by 2 = 4\(\frac{1}{2}\)  4\(\frac{1}{2}\), two-thirds are there in three shown in the model above,
e. Given to find one-thirds are in five-sixths first we write
\(\frac{1}{3}\) as reciprocal \(\frac{3}{1}\) now multiply as
\(\frac{5}{6}\) X \(\frac{3}{1}\) = \(\frac{5 X 3}{6 X 1}\) = \(\frac{5}{2}\)] as numerator is greater we write as
(2 X 2 + 1 by 2) = 2\(\frac{1}{2}\),So 2\(\frac{1}{2}\), one-thirds are in five-sixths as shown in the model above.

EXPLORATION 2

Finding a Pattern
Work with a partner. The table shows the division expressions from Exploration 1. Complete each multiplication expression so that it has the same value as the division expression above it. What can you conclude about dividing by fractions?
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 92
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 93
Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals-27
Explanation:
Dividing a fraction by another fraction is the same as multiplying the fraction
by the reciprocal (inverse) of the other. We get the reciprocal of a fraction
by interchanging its numerator and denominator.
Yes the pattern found can be applied to division by a whole number as
Step 1: The whole number is converted into the fraction by
applying the denominator value as 1
Step 2: Take the reciprocal of the number
Step 3: Now, multiply the fractional value by a given fraction
Step 4: Simplify the given expression

Example: Divide \(\frac{6}{5}\) by 10
Step 1: Convert 10 into a fraction: \(\frac{10}{1}\)
Step 2: Take reciprocal: \(\frac{1}{10}\)
Step 3: Multiply \(\frac{6}{5}\) and \(\frac{1}{10}\)
we get \(\frac{6}{50}\) on simplification as both goes in 2, 2 X 3 = 6 and 2 x 25 = 50,(3,25), So we get \(\frac{3}{25}\).

2.2 Lesson

Two numbers whose product is 1 are reciprocals, or multiplicative inverses. To write the reciprocal of a number, first write the number as a fraction. Then invert the fraction. So, the reciprocal of a fraction \(\frac{a}{b}\) is \(\frac{b}{a}\), where a ≠ 0 and b ≠ 0.

Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 94

Try It

Question 1.
\(\frac{3}{4}\)
\(\frac{3}{4}\) the reciprocals is  \(\frac{4}{3}\)

Explanation:
We write the reciprocals of \(\frac{3}{4}\) as the number is already in fraction
we write as \(\frac{4}{3}\)

Question 2.
5
5 the reciprocals is \(\frac{1}{5}\)
Explanation:
We write the reciprocals of 5 as the number is whole number we convert
into the fraction by applying the denominator value as 1 and take
the reciprocal of the number as \(\frac{1}{5}\).

Question 3.
\(\frac{7}{2}\)
\(\frac{7}{2}\) the reciprocals is  \(\frac{2}{7}\)

Explanation:
We write the reciprocals of \(\frac{7}{2}\) as the number is already in fraction
we write as \(\frac{2}{7}\).

Question 4.
\(\frac{4}{9}\)
\(\frac{4}{9}\) the reciprocals is  \(\frac{9}{4}\)

Explanation:
We write the reciprocals of \(\frac{4}{9}\) as the number is already in fraction
we write as \(\frac{9}{4}\).

Key Idea
Dividing Fractions
Words
To divide a number by a fraction, multiply the number by the reciprocal of the fraction.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 95

Divide. Write the answer in simplest form. Use a model to justify your answer.

Question 5.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 96
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 96= 4
Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals-28

Explanation:
Given \(\frac{1}{2}\) ÷ \(\frac{1}{8}\) we write reciprocal of the fraction \(\frac{1}{8}\) as \(\frac{8}{1}\) and multiply as \(\frac{1}{2}\) X 8 =\(\frac{8}{2}\) = 4. As shown in the model we take half of 8 we get 4.

Question 6.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 97
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 97= 1\(\frac{1}{3}\)
Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals-29

Explanation:
Given \(\frac{2}{5}\) ÷ \(\frac{3}{10}\) we write reciprocal of the fraction \(\frac{3}{10}\) as \(\frac{10}{3}\) and multiply as \(\frac{2}{5}\) X
\(\frac{10}{3}\)=\(\frac{2 X 10 }{5 X 3}\) = \(\frac{4}{3}\) = 1\(\frac{1}{3}\) and we have shown in the area model above.

Question 7.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 98
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 98=  \(\frac{1}{2}\)
Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals-30
Explanation:
Given \(\frac{3}{8}\) ÷ \(\frac{3}{4}\) we write reciprocal of the fraction \(\frac{3}{4}\) as \(\frac{4}{3}\) and multiply as \(\frac{3}{8}\) X
\(\frac{4}{3}\) = \(\frac{3 X 4 }{8 X 3}\) = \(\frac{1}{2}\)
and we have shown in the area model as above.
Question 8.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 99
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 99= \(\frac{4}{9}\)
Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals-31
Explanation:
Given \(\frac{2}{7}\) ÷ \(\frac{9}{14}\) we write reciprocal of the fraction \(\frac{9}{14}\) as \(\frac{14}{9}\) and multiply as \(\frac{2}{7}\) X
\(\frac{14}{9}\) = \(\frac{2 X 14 }{7 X 9}\) = \(\frac{4}{9}\)
and we have shown in the area model above.

Try It

Divide. Write the answer in simplest form.

Question 9.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 100
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 100= \(\frac{1}{9}\)

Explanation:
\(\frac{1}{3}\)÷ 3 as the number 3 is whole number we convert
into the fraction by applying the denominator value as 1 and take
the reciprocal of the number as \(\frac{1}{3}\) and multiply as \(\frac{1}{3}\) X \(\frac{1}{3}\) Step I: We multiply the numerators as 1 X 1 = 1
Step II: We multiply denominators as 3 X 3 = 9
Step III: We write the fraction in the simplest form as \(\frac{1}{9}\),
So \(\frac{1}{3}\)÷ 3 = \(\frac{1}{9}\).

Question 10.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 101
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 101= \(\frac{1}{15}\)

Explanation:
\(\frac{2}{3}\)÷ 10 as the number 10 is whole number we convert
into the fraction by applying the denominator value as 1 and take
the reciprocal of the number as \(\frac{1}{10}\) and multiply as \(\frac{2}{3}\) X \(\frac{1}{10}\) Step I: We multiply the numerators as 2 X 1 = 2
Step II: We multiply denominators as 3 X 10 = 30
Step III: We write the fraction in the simplest form as \(\frac{2}{30}\),
we can further simplify as both goes in 2, 2 X 1 = 2 and 2 X 15 = 30, (1,15) =
So \(\frac{2}{3}\)÷ 10 = \(\frac{1}{15}\).

Question 11.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 102
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 102= \(\frac{5}{32}\)

Explanation:
\(\frac{5}{8}\)÷ 4 as the number 4 is whole number we convert
into the fraction by applying the denominator value as 1 and take
the reciprocal of the number as \(\frac{1}{4}\) and multiply as \(\frac{5}{8}\) X \(\frac{1}{4}\) Step I: We multiply the numerators as 5 X 1 = 5
Step II: We multiply denominators as 8 X 4 = 32
Step III: We write the fraction in the simplest form as \(\frac{5}{32}\),
So \(\frac{5}{8}\)÷ 4 = \(\frac{5}{32}\).

Question 12.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 103
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 103= \(\frac{3}{14}\)
Explanation:
\(\frac{6}{7}\)÷ 4 as the number 4 is whole number we convert
into the fraction by applying the denominator value as 1 and take
the reciprocal of the number as \(\frac{1}{4}\) and multiply as \(\frac{6}{7}\) X \(\frac{1}{4}\) Step I: We multiply the numerators as 6 X 1 = 6
Step II: We multiply denominators as 7 X 4 = 28
Step III: We write the fraction in the simplest form as \(\frac{6}{28}\),
we can further simplify as both goes in 2, 2 X 3 = 6 and 2 X 14 = 28, (3,14) =
So \(\frac{6}{7}\)÷ 4 = \(\frac{3}{14}\).

Self-Assessment for Concepts & Skills

Solve each exercise. Then rate your understanding of the success criteria in your journal.

DIVIDING FRACTIONS
Divide. Write the answer in simplest form. Draw a model to justify your answer.

Question 13.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 104
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 104=\(\frac{12}{15}\)= \(\frac{4}{5}\)
Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals-32

Explanation:
Given \(\frac{2}{3}\) ÷ \(\frac{5}{6}\) we write reciprocal of the fraction \(\frac{5}{6}\) as \(\frac{6}{5}\) and multiply as \(\frac{2}{3}\) X
\(\frac{6}{5}\) = \(\frac{2 X 6}{3 X 5}\) = \(\frac{12}{15}\)
we can further simplify as both goes in 3, 3 X 4 = 12 and 3 X 5 = 15, (4,5) = \(\frac{4}{5}\).

Question 14.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 105
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 105= \(\frac{2}{7}\)
Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals-34

Explanation:
\(\frac{6}{7}\)÷ 3 as the number 3 is whole number we convert
into the fraction by applying the denominator value as 1 and take
the reciprocal of the number as \(\frac{1}{3}\) and multiply as \(\frac{6}{7}\) X \(\frac{1}{3}\) Step I: We multiply the numerators as 6 X 1 = 6
Step II: We multiply denominators as 7 X 3 = 21
Step III: We write the fraction in the simplest form as \(\frac{6}{21}\),
we can further simplify as both goes in 3, 3 X 2 = 6 and 3 X 7 = 21, (2,7),
So \(\frac{6}{7}\)÷ 3 = \(\frac{2}{7}\).

Question 15.
WHICH ONE DOESN’T BELONG?
Which of the following does not belong with the other three? Explain your reasoning.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 106
\(\frac{3}{2}\) X \(\frac{4}{5}\) does not belong with the other three

Explanation:
a. \(\frac{2}{3}\) ÷ \(\frac{4}{5}\) we write reciprocal of the fraction \(\frac{4}{5}\) as \(\frac{5}{4}\) and multiply as \(\frac{2}{3}\) X
\(\frac{5}{4}\) = \(\frac{2 X 5}{3 X 4}\)  = \(\frac{10}{12}\)
we can further simplify as both goes in 2, 2 X 5 = 10 and 2 X 6 = 12, (5,6),
\(\frac{2}{3}\) ÷ \(\frac{4}{5}\) = \(\frac{5}{6}\).
b. \(\frac{3}{2}\) X \(\frac{4}{5}\) = \(\frac{3 X 4}{2 X 5}\)  =\(\frac{12}{10}\) we can further simplify as both goes in 2, 2 X 6 = 12 and 2 X 5 = 10, (6,5), \(\frac{6}{5}\).
c. \(\frac{5}{4}\) X \(\frac{2}{3}\) = \(\frac{5 X 2}{4 X 3}\) =
\(\frac{10}{12}\) we can further simplify as both goes in 2, 2 X 5 = 10 and 2 X 6 = 12, (5,6),\(\frac{5}{4}\) X \(\frac{2}{3}\) = \(\frac{5}{6}\).
d. \(\frac{5}{4}\) ÷ \(\frac{3}{2}\) we write reciprocal of the fraction \(\frac{3}{2}\) as \(\frac{2}{3}\) and multiply as \(\frac{5}{4}\) X
\(\frac{2}{3}\) = \(\frac{5 X 2}{4 X 3}\)  = \(\frac{10}{12}\)
we can further simplify as both goes in 2, 2 X 5 = 10 and 2 X 6 = 12, (5,6),
\(\frac{5}{4}\) ÷ \(\frac{3}{2}\) = \(\frac{5}{6}\).
As \(\frac{2}{3}\) ÷ \(\frac{4}{5}\), \(\frac{5}{4}\) X \(\frac{2}{3}\), \(\frac{5}{4}\) ÷ \(\frac{3}{2}\) = \(\frac{5}{6}\) only
\(\frac{3}{2}\) X \(\frac{4}{5}\) = \(\frac{6}{5}\),
so \(\frac{3}{2}\) X \(\frac{4}{5}\) does not belong with the other three.

MATCHING
Match the expression with its value.

Question 16.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 107
\(\frac{2}{5}\) ÷ \(\frac{8}{15}\) =\(\frac{3}{4}\) matches with B

Explanation:
Given expressions as \(\frac{2}{5}\) ÷ \(\frac{8}{15}\) we write reciprocal of the fraction \(\frac{8}{15}\) as \(\frac{15}{8}\) and multiply as \(\frac{2}{5}\) X \(\frac{15}{8}\) = \(\frac{2 X 15}{5 X 8}\) = \(\frac{30}{40}\),we can further simplify as both goes in 10, 10 X 3 = 30 and 10 X 4 = 40, (3,4),
\(\frac{3}{4}\) therefore \(\frac{2}{5}\) ÷ \(\frac{8}{15}\) =\(\frac{3}{4}\) matches with B.

Question 17.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 108
\(\frac{8}{15}\) ÷ \(\frac{2}{5}\) = 1\(\frac{1}{3}\) matches with D

Explanation:
Given expressions as \(\frac{8}{15}\) ÷ \(\frac{2}{5}\) we write reciprocal of the fraction \(\frac{2}{5}\) as \(\frac{5}{2}\) and multiply as \(\frac{8}{15}\) X \(\frac{5}{2}\) = \(\frac{8 X 5}{15 X 2}\) = \(\frac{40}{30}\),we can further simplify as both goes in 10, 10 X 4 = 40 and 10 X 3 = 30, (4,3),
\(\frac{4}{3}\) As numerator is greater than denominator we write in
mixed fraction also as (1 X 3 + 1 by 3), 1\(\frac{1}{3}\), therefore \(\frac{8}{15}\) ÷ \(\frac{2}{5}\) = 1\(\frac{1}{3}\) matches with D.

Question 18.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 109
\(\frac{2}{15}\) ÷ \(\frac{8}{5}\) =\(\frac{1}{12}\) matches with A

Explanation:
Given expressions as \(\frac{2}{15}\) ÷ \(\frac{8}{5}\) we write reciprocal of the fraction \(\frac{8}{5}\) as \(\frac{5}{8}\) and multiply as \(\frac{2}{15}\) X \(\frac{5}{8}\) = \(\frac{2 X 5}{15 X 8}\) = \(\frac{10}{120}\),we can further simplify as both goes in 10, 10 X 1 = 10 and 10 X 12= 120, (1,12),
\(\frac{1}{12}\) ,therefore \(\frac{2}{15}\) ÷ \(\frac{8}{5}\) =\(\frac{1}{12}\) matches with A.

Question 19.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 110
\(\frac{8}{5}\) ÷ \(\frac{2}{15}\) = 12 matches with C

Explanation:
Given expressions as \(\frac{8}{5}\) ÷ \(\frac{2}{15}\) we write reciprocal of the fraction \(\frac{2}{15}\) as \(\frac{15}{2}\) and multiply as \(\frac{8}{5}\) X \(\frac{15}{2}\) = \(\frac{8 X 15}{5 X 2}\) = \(\frac{120}{10}\),we can further simplify as both goes in 10, 10 X 12 = 120 and 10 X 1 = 10, (12,1),= 12,
So \(\frac{8}{5}\) ÷ \(\frac{2}{15}\) = 12 matches with C.

Self-Assessment for Problem Solving

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 20.
You have 5 cups of rice to make bibimbap, a popular Korean meal. The recipe calls for \(\frac{4}{5}\) cup of rice per serving. How many full servings of bibimbap can you make? How much rice is left over?
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 111

6 full servings of bibimbap I can make, \(\frac{1}{4}\) cup of rice is left over

Explanation:
Given I have 5 cups of rice to make bibimbap, a popular Korean meal and
the recipe calls for \(\frac{4}{5}\) cup of rice per serving so we divide 5 cups with
\(\frac{4}{5}\) cup of rice per serving as 5 ÷ \(\frac{4}{5}\) we write reciprocal of the fraction \(\frac{4}{5}\) as \(\frac{5}{4}\) and multiply as
5 X \(\frac{5}{4}\), Step I: We multiply the numerators as 5 X 5 = 25
Step II: Denominator is same 4
Step III: We write the fraction in the simplest form as \(\frac{25}{4}\),
As numerator is greater than denominator we write in
mixed fraction also as ( 6 X 4 + 1 by 4), 6\(\frac{1}{4}\) we got whole as 6, therefore
6 full servings of bibimbap I can make and \(\frac{1}{4}\) cup of rice is left over.

Question 21.
A band earns \(\frac{2}{3}\) of their profit from selling concert tickets and \(\frac{1}{5}\) of their profit from selling merchandise. The band earns a profit of $1500 from selling concert tickets. How much profit does the band earn from selling merchandise?

$450 profit does the band earn from selling merchandise

Explanation :
Let us take the amount be x and a band earns \(\frac{2}{3}\) of their profit from selling concert tickets and makes a profit of $1500 from selling concert tickets,
So x X \(\frac{2}{3}\) = $1500, on simplification 2x = 1500 X 3 = 4500,
therefore x = \(\frac{4500}{2}\) = $2250 and \(\frac{1}{5}\) of their profit from selling merchandise means $2250 X \(\frac{1}{5}\) = \(\frac{2250}{5}\)=
$450.

Dividing Fractions Homework & Practice 2.2

Review & Refresh

Multiply. Write the answer in simplest form.

Question 1.
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 112
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 112= \(\frac{21}{40}\)
Explanation:
Given expression as \(\frac{7}{10}\) X \(\frac{3}{4}\),
Step I: We multiply the numerators as 7 X 3 = 21
Step II: We multiply denominators as 10 X 4 = 40
Step III: We write the fraction in the simplest form as \(\frac{21}{40}\),
So \(\frac{7}{10}\) X \(\frac{3}{4}\) = \(\frac{21}{40}\).

Question 2.
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 113
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 113= 1\(\frac{17}{18}\)
Explanation:
Given \(\frac{5}{6}\) X 2\(\frac{1}{3}\) so
first we write mixed number 2\(\frac{1}{3}\)  as 2 X 3 + 1 by 3 =
\(\frac{7}{3}\) now
we multiply \(\frac{5}{6}\)  X  \(\frac{7}{3}\),
Step I: We multiply the numerators as 5 X 7 = 35
Step II: We multiply the denominators as 6 X 3 =18
Step III: We write the fraction in the simplest form as \(\frac{35}{18}\),
As numerator is greater than denominator we write in
mixed fraction also as ( 1 X 18 + 17 by 18 ), 1\(\frac{17}{18}\)
So \(\frac{5}{6}\) X 2\(\frac{1}{3}\) = 1\(\frac{17}{18}\).

Question 3.
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 114
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 114= \(\frac{1}{6}\).
Explanation:
Given expression as \(\frac{4}{9}\) X \(\frac{3}{8}\),
Step I: We multiply the numerators as 4 X 3 = 12
Step II: We multiply denominators as 9 X 8 = 72
Step III: We write the fraction in the simplest form as \(\frac{12}{72}\),
we can further simplify as both goes in 12, 12 X 1 = 12 and 12 X 6 = 72, (1,6),
So \(\frac{4}{9}\) X \(\frac{3}{8}\) = \(\frac{1}{6}\).

Question 4.
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 115
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 115= 16

Explanation :
2\(\frac{2}{5}\) X  6\(\frac{2}{3}\), We write mixed fractions
2\(\frac{2}{5}\) as 2 X 5 +2 by 5 = \(\frac{12}{5}\) and
6\(\frac{2}{3}\) as 6 X 3 + 2 by 3 = \(\frac{20}{3}\) Now we multiply
\(\frac{12}{5}\) X \(\frac{20}{3}\),
Step I: We multiply the numerators as 12 X 20 = 240
Step II: We multiply the denominators as 5 X 3 = 15
Step III: We write the fraction in the simplest form as \(\frac{240}{15}\),
we can further simplify as both goes in 15, 15 X 16 = 240 and 15 X 1 = 15, (16,1),
therefore 2\(\frac{2}{5}\) X  6\(\frac{2}{3}\) = 16.

Match the expression with its value.

Question 5.
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 116
3 + 2 X 42 = 35 matches with B

Explanation:
Given expression as 3 + 2 X 42 = 3 + ( 2 X 4 X 4) = 3 + 32 = 35 which matches with B.

Question 6.
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 117
(3 + 2) X 42 = 80 matches with D

Explanation:
Given expression as (3 + 2) X 42 = 5 X (4 X 4) = 5 X 16 = 80 which matches with D.

Question 7.
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 118

2 + 3 X 42 = 50 matches with C

Explanation:
Given expression as 2 + 3 X 42 = 2 + (3 X 4 X 4) = 2 + 48 = 50 which matches with C.

Question 8.
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 119

42 + 2 X 3 = 22 matches with A

Explanation:
Given expression as 42 + 2 X 3 = 16 + 6 = 22 which matches with A.

Find the area of the rectangle.

Question 9.
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 120
The area of rectangle is 14 square feet

Explanation:
Given width as 4 ft and height as 3.5 ft, we can write 3.5 as \(\frac{35}{10}\),
the area of rectangle is 4 X \(\frac{35}{10}\) = \(\frac{140}{10}\)
as both goes in 10 we get \(\frac{140}{10}\) =14 square feet.

Question 10.
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 121
The area of rectangle is \(\frac{1}{6}\) square feet

Explanation:
Given width as \(\frac{2}{3}\) ft and height as \(\frac{1}{4}\)
the area of rectangle is \(\frac{2}{3}\)  X \(\frac{1}{4}\)
Step I: We multiply the numerators as 2 X 1 = 2
Step II: We multiply the denominators as 3 X 4 = 12
Step III: We write the fraction in the simplest form as \(\frac{2}{12}\),
we can further simplify as both goes in 2, 2 X 1 = 2 and 2 X 6 = 12, (1,6),
\(\frac{1}{6}\), therefore the area of rectangle is \(\frac{1}{6}\) square feet.

Question 11.
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 122
The area of rectangle is \(\frac{15}{16}\) square yards

Explanation:
Given width as 1\(\frac{1}{2}\) ft and height as \(\frac{5}{8}\)
We write mixed fractions 1\(\frac{1}{2}\) as 1 X 2 +1 by 2 =
\(\frac{3}{2}\) the area of rectangle is \(\frac{3}{2}\)  X \(\frac{5}{8}\)
Step I: We multiply the numerators as 3 X 5 = 15
Step II: We multiply the denominators as 2 X 8 = 16
Step III: We write the fraction in the simplest form as \(\frac{15}{16}\),
therefore the area of rectangle is \(\frac{15}{16}\) square yards.

Concepts, Skills, & Problem Solving
CHOOSE TOOLS
Answer the question using a model. (See Exploration 1, p. 53.)

Question 12.
How many three-fifths are in three?
Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals-43
Five, three-fifths are in three

Explanation:
Given to find three-fifths are in three is three divides \(\frac{3}{5}\)
we write reciprocal of the fraction \(\frac{3}{5}\) as \(\frac{5}{3}\)=
3  X \(\frac{5}{3}\) = \(\frac{15}{3}\) = 5.

Question 13.
How many two-ninths are in eight-ninths?
Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals-44
Four, two-ninths are in eight-ninths

Explanation:
Given to find two-ninths are in eight-ninths means \(\frac{8}{9}\) ÷
\(\frac{2}{9}\) we write reciprocal of the fraction \(\frac{2}{9}\)
as \(\frac{9}{2}\)= \(\frac{8}{9}\) X \(\frac{9}{2}\) =
\(\frac{8 X 9}{9 X 2}\)= \(\frac{8}{2}\) as both goes in 2
as 2 X 4 = 8, 2 X 1= 2,(4,1), \(\frac{8}{2}\) = 4.

Question 14.
How many three-fourths are in seven-eighths?
Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals-45
\(\frac{7}{6}\) or 1\(\frac{1}{6}\), three-fourths are in seven-eighths

Explanation:
Given to find three-fourths are in seven-eighths means \(\frac{7}{8}\) ÷
\(\frac{3}{4}\) we write reciprocal of the fraction \(\frac{3}{4}\) as \(\frac{4}{3}\)= \(\frac{7}{8}\) X \(\frac{4}{3}\) =
\(\frac{7 X 4}{8 X 3}\)= \(\frac{28}{24}\) as both goes in 4
as 4 X 7 = 28, 4 X 6= 24,(7,6), \(\frac{7}{6}\) as numerator is
greater we write as ( 1 X 6 + 1 by 6) = 1\(\frac{1}{6}\).

WRITING RECIPROCALS
Write the reciprocal of the number.

Question 15.
8
8 the reciprocal is \(\frac{1}{8}\)

Explanation:
We write the reciprocal of a number, first write the number as a fraction.
Then invert the fraction, so reciprocal of 8 is \(\frac{8}{1}\) is \(\frac{1}{8}\).

Question 16.
\(\frac{6}{7}\)
\(\frac{6}{7}\) = \(\frac{7}{6}\)

Explanation:
Already its in fraction so the reciprocal of \(\frac{6}{7}\) is inverting the fraction
\(\frac{7}{6}\).

Question 17.
\(\frac{2}{5}\)
\(\frac{2}{5}\) = \(\frac{5}{2}\)

Explanation:
Already its in fraction so the reciprocal of \(\frac{2}{5}\) is inverting the fraction
\(\frac{5}{2}\).

Question 18.
\(\frac{11}{8}\)
\(\frac{11}{8}\) = \(\frac{8}{11}\)

Explanation:
Already its in fraction so the reciprocal of \(\frac{11}{8}\) is inverting the fraction
\(\frac{8}{11}\).

DIVIDING FRACTIONS
Divide. Write the answer in simplest form.

Question 19.
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 123
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 123= \(\frac{2}{3}\)
Explanation:
Given expressions as \(\frac{1}{3}\) ÷ \(\frac{1}{2}\)
we write reciprocal of the fraction \(\frac{1}{2}\) as \(\frac{2}{1}\) and multiply as \(\frac{1}{3}\) X \(\frac{2}{1}\) =
\(\frac{1 X 2}{3 X 1}\) = \(\frac{2}{3}\),
therefore \(\frac{1}{3}\) ÷ \(\frac{1}{2}\) = \(\frac{2}{3}\).

Question 20.
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 124
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 124= \(\frac{1}{2}\)
Explanation:
Given expressions as \(\frac{1}{8}\) ÷ \(\frac{1}{4}\)
we write reciprocal of the fraction \(\frac{1}{4}\) as
\(\frac{4}{1}\) and multiply as \(\frac{1}{8}\) X
\(\frac{4}{1}\) = \(\frac{1 X 4}{8 X 1}\) = \(\frac{4}{8}\),
we can further simplify as both goes in 4, 4 X 1 = 4 and 4 X 2 = 8, (1,2),
therefore \(\frac{1}{8}\) ÷ \(\frac{1}{4}\) = \(\frac{1}{2}\).

Question 21.
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 125
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 125= \(\frac{1}{7}\)
Explanation:
Given expressions as \(\frac{2}{7}\) ÷ 2, we write reciprocal of 2 as
\(\frac{1}{2}\) and multiply as \(\frac{2}{7}\) X \(\frac{1}{2}\) = \(\frac{2 X 1}{7 X 2}\) = \(\frac{2}{14}\),we can further
simplify as both goes in 2, 2 X 1 = 2 and 2 X 7 = 14, (1,7), \(\frac{1}{7}\),
therefore \(\frac{2}{7}\) ÷ 2 = \(\frac{1}{7}\).

Question 22.
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 126
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 126= \(\frac{2}{5}\)
Explanation:
Given expressions as \(\frac{6}{5}\) ÷ 3, we write reciprocal of 3 as
\(\frac{1}{3}\) and multiply as \(\frac{6}{5}\) X
\(\frac{1}{3}\) = \(\frac{6 X 1}{5 X 3}\) = \(\frac{6}{15}\),
we can further simplify as both goes in 3, 3 X 2 = 6 and 3 X 5 = 15, (2,5),
\(\frac{2}{5}\),therefore \(\frac{6}{5}\) ÷ 3 = \(\frac{2}{5}\).

Question 23.
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 127

Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 127= \(\frac{3}{2}\) = 1\(\frac{1}{2}\)
Explanation:
Given expressions as \(\frac{2}{3}\) ÷ \(\frac{4}{9}\),
we write reciprocal of the fraction \(\frac{4}{9}\) as \(\frac{9}{4}\)
and multiply as \(\frac{2}{3}\) X \(\frac{9}{4}\) = \(\frac{2 X 9}{3 X 4}\) = \(\frac{18}{12}\),we can further simplify as both goes in 6,
6 X 3 = 18 and 6 X 2 = 12, (3,2),
therefore \(\frac{2}{3}\) ÷ \(\frac{4}{9}\) = \(\frac{3}{2}\) as numerator is greater than denominator we can write in mixed
fraction as (1 X 2 + 1 by 2) = 1\(\frac{1}{2}\).

Question 24.
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 128
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 128= \(\frac{35}{12}\) = 2\(\frac{11}{12}\)
Explanation:
Given expressions as \(\frac{5}{6}\) ÷ \(\frac{2}{7}\),
we write reciprocal of the fraction \(\frac{2}{7}\) as \(\frac{7}{2}\)
and multiply as \(\frac{5}{6}\) X \(\frac{7}{2}\) = \(\frac{5 X 7}{6 X 2}\) = \(\frac{35}{12}\),as numerator is greater than
denominator we can write in mixed fraction as (2 X 12 + 11) by 12 =2\(\frac{11}{12}\).Therefore \(\frac{5}{6}\) ÷ \(\frac{2}{7}\) = 2\(\frac{11}{12}\).

Question 25.
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 129
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 129= 16
Explanation:
Given expressions as 12 ÷ \(\frac{3}{4}\),
we write reciprocal of the fraction \(\frac{3}{4}\) as \(\frac{4}{3}\)
and multiply as 12 X \(\frac{4}{3}\) = \(\frac{12 X 4}{1 X 3}\) =
\(\frac{48}{3}\),we can further simplify as both goes in 3,
3 X 16 = 48 and 3 X 1 = 3, (16,1),therefore 12 ÷ \(\frac{3}{4}\) = 16.

Question 26.
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 130
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 130= 20
Explanation:
Given expressions as 8 ÷ \(\frac{2}{5}\), we write reciprocal of the fraction \(\frac{2}{5}\) as \(\frac{5}{2}\) and multiply as
8 X \(\frac{5}{2}\) = \(\frac{8 X5}{1 X 2}\) =
\(\frac{40}{2}\),we can further simplify as both goes in 2,
2 X 20 = 40 and 2 X 1 = 2, (20,1),therefore 8 ÷ \(\frac{2}{5}\) = 20.

Question 27.
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 131
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 131= \(\frac{1}{14}\)
Explanation:
Given expressions as \(\frac{3}{7}\) ÷ 6, we write reciprocal of 6
as \(\frac{1}{6}\) and multiply as \(\frac{3}{7}\) X
\(\frac{1}{6}\) = \(\frac{3 X 1}{7 X 6}\) = \(\frac{3}{42}\),
we can further simplify as both goes in 3, 3 X 1 = 3 and 3 X 14 = 42, (1,14),
therefore \(\frac{3}{7}\) ÷ 6 = \(\frac{1}{14}\).

Question 28.
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 132
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 132= \(\frac{3}{25}\)
Explanation:
Given expressions as \(\frac{12}{25}\) ÷ 4, we write reciprocal of 4
as \(\frac{1}{4}\) and multiply as \(\frac{12}{25}\) X \(\frac{1}{4}\) = \(\frac{12 X 1}{25 X 4}\) = \(\frac{12}{100}\),
we can further simplify as both goes in 4, 4 X 3 = 12 and 4 X 25 = 100, (3,25),
therefore \(\frac{12}{25}\) ÷ 4 = \(\frac{3}{25}\).

Question 29.
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 133
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 133= \(\frac{1}{3}\)
Explanation:
Given expressions as \(\frac{2}{9}\) ÷ \(\frac{2}{3}\),
we write reciprocal of the fraction \(\frac{2}{3}\) as \(\frac{3}{2}\)
and multiply as \(\frac{2}{9}\) X \(\frac{3}{2}\) = \(\frac{2 X 3}{9 X 2}\) = \(\frac{6}{18}\),we can further simplify as both goes in 6,
6 X 1 = 6 and 6 X 3 = 18, (1,3)= \(\frac{1}{3}\).
Therefore \(\frac{2}{9}\) ÷ \(\frac{2}{3}\) = \(\frac{1}{3}\).

Question 30.
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 134
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 134= \(\frac{2}{3}\)
Explanation:
Given expressions as \(\frac{8}{15}\) ÷ \(\frac{4}{5}\),
we write reciprocal of the fraction \(\frac{4}{5}\) as \(\frac{5}{4}\)
and multiply as \(\frac{8}{15}\) X \(\frac{5}{4}\) =
\(\frac{8 X 5}{15 X 4}\) = \(\frac{40}{60}\),
we can further simplify as both goes in 20, 20 X 2 = 40 and
20 X 3 = 60, (2,3)=\(\frac{2}{3}\).Therefore \(\frac{8}{15}\) ÷
\(\frac{4}{5}\) = \(\frac{2}{3}\).

Question 31.
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 135
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 135= 3
Explanation:
Given expressions as \(\frac{1}{3}\) ÷ \(\frac{1}{9}\),
we write reciprocal of the fraction \(\frac{1}{9}\) as \(\frac{9}{1}\)
and multiply as \(\frac{1}{3}\) X \(\frac{9}{1}\) =
\(\frac{1 X 9}{3 X 1}\) = \(\frac{9}{3}\),
we can further simplify as both goes in 3, 3 X 3 = 9 and 3 X 1 = 3, (3,1)= 3.
Therefore \(\frac{1}{3}\) ÷ \(\frac{1}{9}\) = 3.

Question 32.
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 136
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 136= \(\frac{28}{15}\) = 1\(\frac{13}{15}\)
Explanation:
Given expressions as \(\frac{7}{10}\) ÷ \(\frac{3}{8}\),
we write reciprocal of the fraction \(\frac{3}{8}\) as \(\frac{8}{3}\)
and multiply as \(\frac{7}{10}\) X \(\frac{8}{3}\) =
\(\frac{7 X 8}{10 X 3}\) = \(\frac{56}{30}\),
we can further simplify as both goes in 2, 2 X 28 = 56 and 2 X 15 = 30,
(28,15)=\(\frac{28}{15}\),as numerator is greater than
denominator we can write in mixed fraction as (1 X 15 + 13 by 15) =
1\(\frac{13}{15}\).Therefore \(\frac{7}{10}\) ÷
\(\frac{3}{8}\) = 1\(\frac{13}{15}\).

Question 33.
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 137
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 137= \(\frac{2}{27}\)
Explanation:
Given expressions as \(\frac{14}{27}\) ÷ 7,
we write reciprocal of 7 as \(\frac{1}{7}\) and multiply as
\(\frac{14}{27}\) X \(\frac{1}{7}\) = \(\frac{14 X 1}{27 X 7}\) = \(\frac{14}{189}\),we can further simplify as both goes in 7,
7 X 2 = 14 and 7 X 27 = 189, (2,27), \(\frac{2}{27}\),therefore
\(\frac{14}{27}\) ÷ 7 = \(\frac{2}{27}\).

Question 34.
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 138
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 138= \(\frac{1}{24}\)
Explanation:
Given expressions as \(\frac{5}{8}\) ÷ 15, we write reciprocal of 15
as \(\frac{1}{15}\) and multiply as \(\frac{5}{8}\) X
\(\frac{1}{15}\) = \(\frac{5 X 1}{8 X 15}\) = \(\frac{5}{120}\),
we can further simplify as both goes in 5, 5 X 1 = 5 and 5 X 24 = 120, (1,24),
\(\frac{1}{24}\),therefore \(\frac{5}{8}\) ÷ 15
= \(\frac{1}{24}\).

Question 35.
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 139
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 139= \(\frac{27}{28}\)
Explanation:
Given expressions as \(\frac{27}{32}\) ÷ \(\frac{7}{8}\),
we write reciprocal of the fraction \(\frac{7}{8}\) as \(\frac{8}{7}\)
and multiply as \(\frac{27}{32}\) X \(\frac{8}{7}\) =
\(\frac{27 X 8}{32 X 7}\) = \(\frac{216}{224}\),
we can further simplify as both goes in 8, 8 X 27 = 216 and 8 X 28 = 224,
(27,28)=\(\frac{27}{28}\).Therefore \(\frac{27}{32}\) ÷
\(\frac{7}{8}\) = \(\frac{27}{28}\).

Question 36.
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 140
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 140= \(\frac{26}{75}\)
Explanation:
Given expressions as \(\frac{4}{15}\) ÷ \(\frac{10}{13}\),
we write reciprocal of the fraction \(\frac{10}{13}\) as
\(\frac{13}{10}\) and multiply as \(\frac{4}{15}\) X
\(\frac{13}{10}\) = \(\frac{4 X 13}{15 X 10}\) =
\(\frac{52}{150}\),we can further simplify as both goes in 2,
2 X 26 = 52 and 2 X 75 = 150, (26,75)=\(\frac{26}{75}\).
Therefore \(\frac{4}{15}\) ÷ \(\frac{10}{13}\) = \(\frac{26}{75}\).

Question 37.
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 141
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 141= \(\frac{81}{4}\) = 20\(\frac{1}{4}\)
Explanation:
Given expressions as 9 ÷ \(\frac{4}{9}\), we write reciprocal of
the fraction \(\frac{4}{9}\) as \(\frac{9}{4}\)
and multiply as 9 X \(\frac{9}{4}\) = \(\frac{9 X 9}{1 X 4}\) =
\(\frac{81}{4}\),as numerator is greater than denominator
we can write in mixed fraction as (20 X 4 + 1 by 4) = 20\(\frac{1}{4}\),
therefore 9 ÷ \(\frac{4}{9}\) = 20\(\frac{1}{4}\).

Question 38.
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 142
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 142= 24
Explanation:
Given expressions as 10 ÷ \(\frac{5}{12}\), we write reciprocal
of the fraction \(\frac{5}{12}\) as \(\frac{12}{5}\) and
multiply as 10 X \(\frac{12}{5}\) = \(\frac{10 X 12}{1 X 5}\) =
\(\frac{120}{5}\),we can further simplify as both goes in 5,
5 X 24 = 120 and 5 X 1 = 5, (24,1)= 24.Therefore 10 ÷ \(\frac{5}{12}\) = 24.

YOU BE THE TEACHER
Your friend finds the quotient. Is your friend correct? Explain your reasoning.

Question 39.
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 143

Yes, friend is correct as \(\frac{4}{7}\) ÷ \(\frac{13}{28}\) =
1\(\frac{3}{13}\).

Explanation:
Given expression as \(\frac{4}{7}\) ÷ \(\frac{13}{28}\),
we write reciprocal of the fraction \(\frac{13}{28}\) as
\(\frac{28}{13}\) and multiply as \(\frac{4}{7}\) X
\(\frac{28}{13}\) = \(\frac{4 X 28}{7 X 13}\) = \(\frac{112}{91}\),
we can further simplify as both goes in 7, 7 X 16 = 112 and 7 X 13 = 91,
(16,13) = \(\frac{16}{13}\), as numerator is greater than denominator
we can write in mixed fraction as (1 X 13 + 3 by 13) = 1\(\frac{3}{13}\).
Therefore \(\frac{4}{7}\) ÷ \(\frac{13}{28}\) = 1\(\frac{3}{13}\).
So friend is correct.

Question 40.
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 144
No, Friend is incorrect as \(\frac{2}{5}\) ÷ \(\frac{8}{9}\) =
\(\frac{9}{20}\) ≠ 2\(\frac{2}{9}\)

Explanation:
Given expression as \(\frac{2}{5}\) ÷ \(\frac{8}{9}\),
we write reciprocal of the fraction \(\frac{8}{9}\) as
\(\frac{9}{8}\) and multiply as \(\frac{2}{5}\) X
\(\frac{9}{8}\) = \(\frac{2 X 9}{5 X 8}\) = \(\frac{18}{40}\),
we can further simplify as both goes in 2, 2 X 9 = 18 and 2 X 20 = 40,
(9,20) = \(\frac{9}{20}\).Therefore \(\frac{2}{5}\) ÷
\(\frac{8}{9}\) = \(\frac{9}{20}\).
No, Friend is incorrect as \(\frac{2}{5}\) ÷ \(\frac{8}{9}\) =
\(\frac{9}{20}\) ≠ 2\(\frac{2}{9}\).

Question 41.
REASONING
You have \(\frac{3}{5}\) of an apple pie. You divide the remaining pie into 5 equal slices. What portion of the original pie is each slice?

\(\frac{3}{25}\) of the original pie is each slice

Explanation:
Given I have \(\frac{3}{5}\) of an apple pie and divide
the remaining pie into 5 equal slices. So the portion of the original pie
in each slice is \(\frac{3}{5}\) ÷  5 we write reciprocal of 5
as \(\frac{1}{5}\) and multiply as \(\frac{3}{5}\) X
\(\frac{1}{5}\) = \(\frac{3 X 1}{5 X 5}\) = \(\frac{3}{25}\),
therefore \(\frac{3}{25}\) of the original pie is each slice.

Question 42.
PROBLEM SOLVING
How many times longer is the baby alligator than the baby gecko?
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 145
5\(\frac{5}{8}\) times longer is the baby alligator than the baby gecko

Explanation:
Given baby alligator is \(\frac{3}{4}\) long and baby gecko
is \(\frac{2}{15}\), So baby alligator is more long than the
baby gecko by \(\frac{3}{4}\) ÷ \(\frac{2}{15}\),
we write reciprocal of the fraction \(\frac{2}{15}\) as
\(\frac{15}{2}\) and multiply as \(\frac{3}{4}\) X
\(\frac{15}{2}\) = \(\frac{3 X 15}{4 X 2}\) =
\(\frac{45}{8}\),as numerator is greater than denominator
we can write in mixed fraction as (5 X 8 + 5 by 8) = 5\(\frac{5}{8}\).
Therefore 5\(\frac{5}{8}\) times longer is the baby alligator
than the baby gecko.

OPEN-ENDED
Write a real-life problem for the expression. Then solve the problem.

Question 43.
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 146
The rope of length \(\frac{5}{6}\)cm is cut into 4 equal pieces,
how long is each piece?
\(\frac{5}{24}\) cm long is each piece

Explanation:
Wrote a real -life problem, the rope of length \(\frac{5}{6}\)cm
is cut into 4 equal pieces, how long is each piece and solution is
\(\frac{5}{6}\)÷ 4,we write reciprocal of 4 as \(\frac{1}{4}\)
and multiply as \(\frac{5}{6}\) X \(\frac{1}{4}\) =
\(\frac{5 X 1}{4 X 6}\) = \(\frac{5}{24}\),
therefore each piece is of length \(\frac{5}{24}\) cm long is each piece.

Question 44.
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 147
A jug contains \(\frac{2}{5}\) of orange juice,
I drank \(\frac{3}{8}\) portion
in it, So how much portion of juice I drank?
So I drank 1\(\frac{1}{15}\) portion of orange juice.

Explanation:
Wrote a real -life problem, a jug contains \(\frac{2}{5}\) of
orange juice, I drank \(\frac{3}{8}\) portion in it,
So portion of juice I drank is \(\frac{5}{6}\) ÷ \(\frac{3}{8}\)
we write reciprocal of fraction \(\frac{3}{8}\) as
\(\frac{8}{3}\) and multiply as \(\frac{2}{5}\) X
\(\frac{8}{3}\) = \(\frac{2 X 8}{5 X 3}\) =
\(\frac{16}{15}\),as numerator is greater than denominator
we can write in mixed fraction as (1 X 15 + 1 by 15) = 1\(\frac{1}{15}\),
So I drank 1\(\frac{1}{15}\) portion of orange juice.

Question 45.
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 148

I need \(\frac{2}{3}\)ltrs of milk to feed one dog,
How many dogs I do can feed in 10 ltrs of milk.
I can feed 15 dogs.

Explanation:
Wrote a real -life problem,I need \(\frac{2}{3}\)ltrs of
milk to feed one dog, How many dogs I do can feed in 10 ltrs of milk.
So number of dogs I can feed are 10 ÷ \(\frac{2}{3}\)
we write reciprocal of fraction \(\frac{2}{3}\) as \(\frac{3}{2}\)
and multiply as 10 X \(\frac{3}{2}\) = \(\frac{30}{2}\) as
both goes in 2, 2 x 15 = 30 and 2 x 1 = 2,(15,1),So \(\frac{30}{2}\)=15,
So I can feed 15 dogs.

Question 46.
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 149
In a class there are \(\frac{2}{7}\) girls in that \(\frac{4}{9}\)
like playing chess, how many girls like playing chess ?
There are \(\frac{9}{14}\) girls like playing chess in a class

Explanation:
Wrote a real -life problem, In a class there are \(\frac{2}{7}\)
girls in that \(\frac{4}{9}\) like playing chess, how many girls like
playing chess so \(\frac{2}{7}\) ÷ \(\frac{4}{9}\),
we write reciprocal of fraction \(\frac{4}{9}\) as \(\frac{9}{4}\)
and multiply as \(\frac{2}{7}\) X \(\frac{9}{4}\) = \(\frac{2 X 9}{7 X 4}\)= \(\frac{18}{28}\) we can further simplify
as both goes 2, 2 X 9 = 18, 2 X 14 = 28,(9,14)= \(\frac{9}{14}\)
girls like playing chess in a class.

NUMBER SENSE
Copy and complete the statement.

Question 47.
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 150
The statement is \(\frac{5}{12}\) X \(\frac{12}{5}\) = 1.

Explanation:
Lets take the missing number as x, \(\frac{5}{12}\) X x = 1,
therefore when \(\frac{5}{12}\) goes other side it becomes
reciprocal, So x= 1 X \(\frac{12}{5}\), The statement is
\(\frac{5}{12}\) X \(\frac{12}{5}\) = 1.

Question 48.
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 151
The statement is 3 X \(\frac{1}{3}\) = 1.

Explanation:
Lets take the missing number as x, 3 X x = 1,therefore
when 3 goes other side it becomes reciprocal as
\(\frac{1}{3}\), So x= 1 X \(\frac{1}{3}\),
The statement is 3 X \(\frac{1}{3}\) = 1.

Question 49.
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 152
7 ÷ \(\frac{1}{8}\) = 56

Explanation:
Lets take the missing number as x, So 7 ÷ x = 56 , x = 7÷56 =
7 X \(\frac{1}{56}\)\(\frac{7}{56}\) as both goes in 7,
7 X 1 = 7, 7 X 8 = 56,(1,8) = \(\frac{1}{8}\),
x =\(\frac{1}{8}\), The statement is  7 ÷ \(\frac{1}{8}\) = 56.

REASONING
Without finding the quotient, copy and complete the statement using <, >, or =. Explain your reasoning.

Question 50.
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 153
Is <
\(\frac{7}{9}\) ÷ 5 < \(\frac{7}{9}\)

Explanation:
L.H.S is \(\frac{7}{9}\) ÷ 5  and R.H.S is \(\frac{7}{9}\)
when compared both sides we have \(\frac{7}{9}\) and
L.H.S is still 1 divided by 5 so it becomes less than R.H.S 1.
So \(\frac{7}{9}\) ÷ 5 < \(\frac{7}{9}\).

Question 51.
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 154
Is =
\(\frac{3}{7}\) ÷ 1 = \(\frac{3}{7}\)
Explanation:
L.H.S is \(\frac{3}{7}\) ÷ 1  and R.H.S is \(\frac{3}{7}\)
when compared both sides we have \(\frac{3}{7}\) and
L.H.S is still 1 divided by 1 so it becomes equal to R.H.S 1.
So \(\frac{3}{7}\) ÷ 1 < \(\frac{3}{7}\).

Question 52.
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 155
Is >
8 ÷ \(\frac{3}{4}\) > 8

Explanation:
L.H.S is 8 ÷ \(\frac{3}{4}\)  and R.H.S is 8 when compared
both sides we have 8 and L.H.S is still 3 divided by 4 so it becomes
1 ÷ \(\frac{3}{4}\) ,we write reciprocal of  \(\frac{3}{4}\) as
\(\frac{4}{3}\) ,1 X \(\frac{4}{3}\) now numerator is
greater than denominator, we get 1 whole plus number in L.H.S,
as L.H.S is greater than R.H.S 1. So 8 ÷ \(\frac{3}{4}\) > 8.

Question 53.
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 156

Is >
\(\frac{5}{6}\) ÷ \(\frac{7}{8}\) > \(\frac{5}{6}\)

Explanation:
L.H.S is \(\frac{5}{6}\) ÷ \(\frac{7}{8}\)  and
R.H.S is \(\frac{5}{6}\) when compared both sides we
have \(\frac{5}{6}\) and L.H.S is still 7 divided by 8 so it
becomes 1 ÷ \(\frac{7}{8}\)  we write reciprocal of
\(\frac{7}{8}\) as  \(\frac{8}{7}\)= 1 X \(\frac{8}{7}\)
now numerator is greater than denominator, so we get 1 whole plus
number in L.H.S as L.H.S is greater than R.H.S 1.
So \(\frac{5}{6}\) ÷ \(\frac{7}{8}\) > \(\frac{5}{6}\).

ORDER OF OPERATIONS
Evaluate the expression. Write the answer in simplest form.

Question 54.
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 157
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 157= \(\frac{1}{6}\)
Explanation:
Given expression as \(\frac{1}{6}\) ÷ 6 ÷ 6, first
we write as multiplication as \(\frac{1}{6}\) ÷ (6 X \(\frac{1}{6}\)) = \(\frac{1}{6}\) ÷ 1 = \(\frac{1}{6}\).

Question 55.
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 158
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 158= \(\frac{1}{144}\)
Explanation:
Given expression as \(\frac{7}{12}\) ÷ 14 ÷ 6,14
and 6 becomes reciprocals and multiplied as \(\frac{7}{12}\) X
\(\frac{1}{14}\) X \(\frac{1}{6}\)
= \(\frac{7 X 1 X 1 }{12 X 14 X 6 }\) =  \(\frac{7}{1008}\),
we now further simplify as both goes in 7, 7 X 1 = 7, 7 X 144 = 1008,
(1, 144) = \(\frac{1}{144}\).

Question 56.
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 159
\(\frac{3}{5}\) ÷ \(\frac{4}{7}\) ÷ \(\frac{9}{10}\) = \(\frac{7}{6}\)
Explanation:
Given expression as \(\frac{3}{5}\) ÷ \(\frac{4}{7}\) ÷
\(\frac{9}{10}\) ,\(\frac{4}{7}\),\(\frac{9}{10}\)
becomes reciprocals and been multipled as = \(\frac{3}{5}\) X
\(\frac{7}{4}\) X \(\frac{10}{9}\)=
\(\frac{3 X 7 X 10}{5 X 4 X 9}\) = \(\frac{210}{180}\)
we can further simplify as both goes in 30,
30 X 7 = 210, 30 X 6 = 180,(7,6) = \(\frac{7}{6}\),therefore
\(\frac{3}{5}\) ÷ \(\frac{4}{7}\) ÷ \(\frac{9}{10}\) =
\(\frac{7}{6}\).

Question 57.
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 160
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 160= 4
Explanation:
Given expression as 4 ÷ \(\frac{8}{9}\) – \(\frac{1}{2}\) ,
first we calculate 4 ÷ \(\frac{8}{9}\) we write \(\frac{8}{9}\)
as reciprocal \(\frac{9}{8}\) and multiply with 4 as
4 X \(\frac{9}{8}\) = \(\frac{4 x 9}{8}\) =
\(\frac{36}{8}\) we simplifies as both goes in 4 as 4 X 9 = 36,
4 x 2 = 8, (9,2) = \(\frac{9}{2}\) now we subtract \(\frac{1}{2}\)
as both denominators are 2 we subtract numerators as (9 -1)= 8,
we get \(\frac{8}{2}\) = 4.

Question 58.
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 161
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 161= 2
Explanation:
Given expression as \(\frac{3}{4}\) + \(\frac{5}{6}\) ÷
\(\frac{2}{3}\), first we calculate \(\frac{5}{6}\) ÷
\(\frac{2}{3}\) we write \(\frac{2}{3}\) reciprocal and
multiply as \(\frac{5}{6}\) X \(\frac{3}{2}\) =
\(\frac{5 X 3}{6 X 2}\) =\(\frac{15}{12}\)
we simplify further as both goes in 3, 3 X 5 = 15, 3 X 4 = 12, (5,4)=
\(\frac{5}{4}\) now we add with \(\frac{3}{4}\) ,
\(\frac{3}{4}\) + \(\frac{5}{4}\) as both have same
denominator we add numerators as 3 + 5 = 8 and write as
\(\frac{8}{4}\) as both goes in 4,we get 2.
Therefore \(\frac{3}{4}\) + \(\frac{5}{6}\) ÷
\(\frac{2}{3}\) = 2.

Question 59.
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 162

Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 162= \(\frac{5}{6}\)
Explanation:
Given expression as \(\frac{7}{8}\) – \(\frac{3}{8}\) ÷ 9 ,
first we calculate \(\frac{3}{8}\) ÷ 9,now we write 9 reciprocal and multiply as \(\frac{3}{8}\) X \(\frac{1}{9}\) = \(\frac{3 X 1}{8 X 9}\) =\(\frac{3}{72}\)  we can simplify as both goes in 3,
3 X 1 =3, 3 X 24 =72, (1,24) = \(\frac{1}{24}\)  now we subtract from
\(\frac{7}{8}\) – \(\frac{1}{24}\) as we need both
to have same denominators 24 we multiply \(\frac{7}{8}\) X
\(\frac{3}{3}\) = \(\frac{7 X 3}{8 X 3}\) =\(\frac{21}{24}\),
Now \(\frac{21}{24}\) – \(\frac{1}{24}\)
as both have same 24 denominators now we can subtract
numerators as 21 – 1 = 20 and write as \(\frac{20}{24}\)
as both goes in 4, 4 X 5 = 20, 4 X 6 = 24, (5, 6) = \(\frac{5}{6}\),
Therefore \(\frac{7}{8}\) – \(\frac{3}{8}\) ÷ 9 = \(\frac{5}{6}\).

Question 60.
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 163
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 163= 4 \(\frac{7}{8}\)
Explanation:
Given expression as \(\frac{9}{16}\) ÷ \(\frac{3}{4}\) X
\(\frac{2}{13}\), first we calculate \(\frac{3}{4}\) X
\(\frac{2}{13}\) = \(\frac{3 X 2}{4 X 13}\) = \(\frac{6}{52}\)
we can simplify as both goes in 2, 2 X 3 =6, 2 X 26 = 52, (3, 26) =
\(\frac{3}{26}\) now  \(\frac{9}{16}\) ÷ \(\frac{3}{26}\)
we write reciprocal and multiply \(\frac{9}{16}\) X
\(\frac{26}{3}\) = \(\frac{9 x 26}{16 X 3}\) =
\(\frac{234}{48}\) as both can go in 6, 6 X 39 = 234
and 6 X 8 = 48, (39,8) = \(\frac{234}{48}\) = \(\frac{39}{8}\)
as numerator is greater than denominator we can write as (4 X 8 + 7 by 8)  =
4\(\frac{7}{8}\).

Question 61.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 164

Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 164= \(\frac{1}{10}\)

Explanation:
Given expression as \(\frac{3}{14}\) X \(\frac{2}{5}\) ÷
\(\frac{6}{7}\), We write \(\frac{6}{7}\) as
reciprocal \(\frac{7}{6}\) and multiply as \(\frac{3}{14}\)
X \(\frac{2}{5}\) X \(\frac{7}{6}\)=
\(\frac{3 X 2 X 7}{14 X 5 X 6}\) = \(\frac{42}{420}\)
as both goes in 42, 42 X 1= 42 and 42 X 10 = 420, (1, 10) =
\(\frac{1}{10}\), therefore \(\frac{3}{14}\) X
\(\frac{2}{5}\) ÷ \(\frac{6}{7}\) = \(\frac{1}{10}\).

Question 62.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 165
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 165= \(\frac{2}{3}\)
Explanation:
Given expression as \(\frac{10}{27}\) X (\(\frac{3}{8}\) ÷
\(\frac{5}{24}\)), first we calculate \(\frac{3}{8}\) ÷
\(\frac{5}{24}\), we write  \(\frac{5}{24}\) reciprocal
as \(\frac{24}{5}\) and multiply as \(\frac{3}{8}\) X
\(\frac{24}{5}\) = \(\frac{3 X 24}{8 X 5}\) = \(\frac{72}{40}\),
we can simplify as both goes in 8 as 8 x 9 = 72 and
8 X 5 = 40, (9,5) = \(\frac{9}{5}\), now we multiply
\(\frac{10}{27}\) X \(\frac{9}{5}\) =
\(\frac{10 X 9}{27 X 5}\) = \(\frac{90}{135}\)
we can further simplify as both goes in 45, 45 X 2 = 90,
45 x 3 = 135, (2, 3) = \(\frac{2}{3}\).

Question 63.
NUMBER SENSE
When is the reciprocal of a fraction a whole number? Explain.

When the simplified fraction has a 1 in the numerator,
Th
e reciprocal will have a 1 in the denominator. We get a whole number

Explanation:
When the simplified fraction has a 1 in the numerator,
Th
e reciprocal will have a 1 in the denominator.
Example : If we have numerator as 1 in the simplified fraction like
\(\frac{1}{7}\) then the reciprocal becomes \(\frac{7}{1}\)
so the reciprocal will have a 1 in the denominator and becomes 7,
So we get a whole number.

Question 64.
MODELING REAL LIFE
You use \(\frac{1}{8}\) of your battery for every
\(\frac{2}{5}\) of an hour that you video chat.
You use \(\frac{3}{4}\) of your battery video chatting.
How long did you video chat?

I had video chat for 2\(\frac{2}{5}\) hours

Explanation:
If I use \(\frac{1}{8}\) of my battery for every
\(\frac{2}{5}\) of an hour of video chat, then I
use \(\frac{1}{8}\) ÷ \(\frac{2}{5}\) =
\(\frac{1}{8}\) X \(\frac{5}{2}\) =
\(\frac{1 x 5}{2 X 16}\) = \(\frac{5}{16}\) of
my battery per hour of video chat.  If I use \(\frac{3}{4}\) of
my battery for video chatting, I used for \(\frac{3}{4}\) ÷
\(\frac{5}{16}\) now \(\frac{3}{4}\) X
\(\frac{16}{5}\)=\(\frac{3 X 16}{4 x 5}\) =
\(\frac{48}{20}\) on further simplification as
both goes in 4 we get 4 X 12 =48, 4 X 5 = 20, (12,5)=
\(\frac{12}{5}\) as numerator is greater
we write as ( 2 X 5 + 2 by 5) = 2\(\frac{2}{5}\) hours.
Therefore I had video chat for 2\(\frac{2}{5}\) hours.

Question 65.
PROBLEM SOLVING
The table shows the portions of a family budget that are spent on several expenses.
a. How many times more is the expense for housing than for automobiles?
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 166
b. How many times more is the expense for food than for recreation?
c. The expense for automobile fuel is \(\frac{1}{60}\) of the total expenses.
What portion of the automobile expense is spent on fuel?

a. 6 times more is the expense for housing than for automobiles,
b. 17\(\frac{7}{9}\) times more is the expense for
food than for recreation,
c. \(\frac{1}{4}\) portion of the automobile
expense is spent on fuel.

Explanation:
a. Given we have Portions of Budget for housing is \(\frac{2}{5}\)
and for automobiles as \(\frac{1}{15}\), So the expense
for housing than for automobiles is \(\frac{2}{5}\) ÷
\(\frac{1}{15}\) we write \(\frac{1}{15}\)
as reciprocal and multiply \(\frac{2}{5}\) X \(\frac{15}{1}\) =
\(\frac{2 X 15}{5 X 1}\) = \(\frac{30}{5}\)
we can simplify as both goes in 5 , 5 X 6 = 30, 5 X 1 = 5,
(6,1),\(\frac{30}{5}\) = 6, therefore 6 times more is
the expense for housing than for automobiles.
b. Given we have Portions of Budget for food is \(\frac{4}{9}\)
and for recreation is \(\frac{1}{40}\), So the expense for
food than for recreation is \(\frac{4}{9}\) ÷ \(\frac{1}{40}\)
we write \(\frac{1}{40}\)  as reciprocal and multiply
\(\frac{4}{9}\) X \(\frac{40}{1}\) = \(\frac{4 X 40}{9 X 1}\) = \(\frac{160}{9}\) as numerator is greater than
denominator we write as (17 X 9 + 7 by 9) =
17\(\frac{7}{9}\), therefore 17\(\frac{7}{9}\)
times more is the expense for food than for recreation.
c. Given the expense for automobile fuel is \(\frac{1}{60}\)
of the total expenses.
So the portion of the automobile expense spent on fuel is
\(\frac{1}{60}\) ÷ \(\frac{1}{15}\)
we write \(\frac{1}{15}\)  as reciprocal and
multiply \(\frac{1}{60}\) X \(\frac{15}{1}\) =
\(\frac{1 X 15}{60 X 1}\) = \(\frac{15}{60}\)
we can simplify further as both goes in 15, 15 X 1 = 15, 15 X 4 = 60,
(1, 4) = \(\frac{1}{4}\), therefore \(\frac{1}{4}\)
portion of the automobile expense is spent on fuel.

Question 66.
CRITICAL THINKING
A bottle of juice is \(\frac{2}{3}\) full. The bottle contains
\(\frac{4}{5}\) of a cup of juice.
a. Write a division expression that represents the capacity of the bottle.
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 167
b. Write a related multiplication expression that represents the
capacity of the bottle.
c. Explain how you can use the diagram to verify the expression in part(b).
d. Find the capacity of the bottle.
a. \(\frac{4}{5}\) ÷ \(\frac{2}{3}\) represents the
capacity of the bottle,
b. \(\frac{4}{5}\) X \(\frac{3}{2}\) is the expression
that represents the capacity of the bottle.
c. Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals-35

d. The capacity of the bottle is \(\frac{6}{5}\) cups of juice.

Explanation:
a. Given a bottle of juice is \(\frac{2}{3}\) full and bottle
contains \(\frac{4}{5}\) of a cup of juice, So the division
expression that represents the capacity of bottle is
\(\frac{4}{5}\) ÷ \(\frac{2}{3}\) .
b. Given a bottle of juice is \(\frac{2}{3}\) full and
bottle contains \(\frac{4}{5}\) of a cup of juice,
So the multiplication expression that represents the capacity of
bottle is we write \(\frac{2}{3}\) as reciprocal \(\frac{3}{2}\)
and multiply \(\frac{4}{5}\) X \(\frac{3}{2}\).
c. We can use the diagram as \(\frac{2}{3}\) = \(\frac{4}{5}\)
and each part in blue represents half of \(\frac{4}{5}\) =
\(\frac{1}{2}\) X \(\frac{4}{5}\) =
\(\frac{1 X 4}{2 X 5}\) = \(\frac{4}{10}\) =
\(\frac{2}{5}\) each part so total parts are 3 X \(\frac{2}{5}\) =
\(\frac{6}{5}\), as shown in the figure, Now we solve with part (b)
as the multiplication expression that represents the capacity of bottle is
\(\frac{4}{5}\) X \(\frac{3}{2}\),we solve
\(\frac{4 X 3}{5 X 2}\) = \(\frac{12}{10}\)
we further simplify as both goes in 2, 2 X 6 = 12, 2 X 5 = 10,(6,5),
\(\frac{6}{5}\). So by using the diagram we got same results
as \(\frac{6}{5}\) cups of juice is the total capacity of the
bottle so our expression in part(b) is verified.
d. The capacity of the bottle is \(\frac{4}{5}\) X
\(\frac{3}{2}\),we solve \(\frac{4 X 3}{5 X 2}\) =
\(\frac{12}{10}\) we further simplify as both goes in 2,
2 X 6 = 12, 2 X 5 = 10,(6,5),\(\frac{6}{5}\) cups of juice.

Question 67.
DIG DEEPER!
You have 6 pints of glaze. It takes \(\frac{7}{8}\) of a pint to
glaze a bowl and \(\frac{9}{16}\) of a pint to glaze a plate.
a. How many bowls can you completely glaze? How many plates
can you completely glaze?
b. You want to glaze 5 bowls, and then use the rest for plates.
How many plates can you completely glaze? How much glaze will be left over?
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 168
c. How many of each object can you completely glaze so
that there is no glaze left over? Explain how you found your answer.
a. 6 bowls I can completely glaze, 10 plates I can completely glaze,
b. 2 plates , \(\frac{1}{2}\) is left over,
c. 3 bowls and 6 plates,

Explanation:
Given 6 pints of glaze, It takes \(\frac{7}{8}\) of a pint to glaze
a bowl so number of bowls I can completely glaze are
6 ÷ \(\frac{7}{8}\) , we write reciprocal of \(\frac{7}{8}\)
and multiply as 6 X \(\frac{8}{7}\) = \(\frac{6 X 8}{1 X 7}\) =
\(\frac{48}{7}\), as numerator is greater we write as
(6 X 7 + 6 by 7) So  \(\frac{48}{7}\) =
6\(\frac{6}{7}\)≈ 6, so 6 bowls I can completely glaze.
It takes \(\frac{9}{16}\) of a pint to glaze a plate so
number of bowls I can completely glaze are 6 ÷ \(\frac{9}{16}\) ,
we write reciprocal of \(\frac{9}{16}\) and multiply as
6 X \(\frac{16}{9}\) = \(\frac{6 X 16}{1 X 9}\) =
\(\frac{96}{9}\) as numerator is greater we write as
( 10 X 9 + 6 by 9) = 10\(\frac{6}{9}\) ≈ 10, so 10 plates
I can completely glaze.
b. So to glaze 5 bowls from 6 pints of glaze it will take,
as we know for 1 bowl it is \(\frac{7}{8}\) of a pint to glaze,
for 5 bowls it is 5 X \(\frac{7}{8}\) = \(\frac{5 X 7}{8}\) =
\(\frac{35}{8}\) pints to glaze so we use from 6 pints of glaze,

Question 68.
REASONING
A water tank is \(\frac{1}{8}\) full. The tank is
\(\frac{3}{4}\) full when 42 gallons of water are added to the tank.
a. How much water can the tank hold?
b. How much water was originally in the tank?
c. How much water is in the tank when it is \(\frac{1}{2}\) full?
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 169
a. The tank can hold 67.2 gallons of water,
b. 8.4 gallons of water was originally in the tank,
c. 33.6 gallons of water is in the tank when it is \(\frac{1}{2}\) full.

Explanation:
Given a water tank is \(\frac{1}{8}\) full. The tank is
\(\frac{3}{4}\) full when 42 gallons of water are
added to the tank. Since the tank started at \(\frac{1}{8}\)
full and reached \(\frac{3}{4}\) full , To find 1 tank full we
have the difference is \(\frac{3}{4}\) – \(\frac{1}{8}\) =
42 gallons, So first we make denominators common we
multiply and divide by 2 to \(\frac{3}{4}\) =
\(\frac{6}{8}\) now we subtract as
\(\frac{6}{8}\) – \(\frac{1}{8}\) = 42;
as denominators are same we minus numerators (6-1)= 5
making \(\frac{5}{8}\) = 42,therefore 1 tank full is
42 X \(\frac{8}{5}\) = \(\frac{336}{8}\) = 67.2 gallons of water.
b. Initially we had \(\frac{1}{8}\) full of water
means we had 67.2 X \(\frac{1}{8}\) = 8.4 gallons
of water was originally in the tank,
c. Water in the tank when it is \(\frac{1}{2}\) full is
67.2 X \(\frac{1}{2}\) = \(\frac{67.2}{2}\) =
33.6 gallons of water is in the tank when it is \(\frac{1}{2}\) full.

Lesson 2.3 Dividing Mixed Numbers

EXPLORATION 1
Dividing Mixed Numbers
Work with a partner. Write a real-life problem that represents each division expression described. Then solve each problem using a model. Check your answers.
a. How many three-fourths are in four and one-half?
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 170
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 170.1
b. How many three-eighths are in two and one-fourth?
c. How many one and one-halves are in six?
d. How many seven-sixths are in three and one-third?
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 171
e. How many one and one-fifths are in five?
f. How many three and one-halves are in two and one-half?
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 172
g. How many four and one-halves are in one and one-half?

a. You have four and one-half of rice, you feed three-fourth to each person,
So how many persons you can feed .
There are six, three-fourths in four and one-half.

Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals-36
b. You have two and one-forth bottles of orange juice in that you
added three -eights cups of water, So how much water is there in
two and one-fourth.
There are six, three-eighths are in two and one-fourth
Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals-37
c. You have 6 meter rope in that how many pieces of one-halves meters
length ropes you can make.
There are four, one and one-halves are in six
Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals-38

d. You have three and one-third packets of balloons in that
seven sixth are green color balloons, So how many green balloons are there.
There are \(\frac{20}{7}\) or 2\(\frac{6}{7}\),
seven-sixths are in three and one-third
Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals-39
e. I have 5 books in which I have completed reading of
one and one fifths of books, So how much portions of book
readings I have completed.
There are 4\(\frac{1}{6}\) – one and one-fifths are in five
Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals-40
f. I have two and one half of bowls of sweet, I used three and
one halves cups of milk to prepare sweet, How many cups of
three and one halves are there in two and one half of bowls of sweet,
There are \(\frac{5}{7}\)– three and one-halves
are in two and one-half
Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals-41
g. I have one and one-half bag of apples with me,
I gave four and one-halves portion to my friends,
How many four and one halves portions are there in one and half bag.
There are \(\frac{1}{3}\) – four and one-halves are in one and one-half
Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals-42

Explanation:
a. Three-fourths in four and one-half are 4\(\frac{1}{2}\) ÷
\(\frac{3}{4}\) = 4\(\frac{1}{2}\) = (4  X 2 + 1 by 2)
= \(\frac{9}{2}\) ÷ \(\frac{3}{4}\) =
now we write \(\frac{3}{4}\) as reciprocal and
multiply \(\frac{9}{2}\) X \(\frac{4}{3}\) =
\(\frac{9 X 4}{2 X 3}\) = \(\frac{36}{6}\)
as both goes in 6 we get 6 X 6 = 36 and 6 X 1= 6,(6,1),
so \(\frac{36}{6}\) = 6, there are six,
three-fourths in four and one-half.

b. Three-eighths in two and one-fourth are 2\(\frac{1}{4}\)
÷ \(\frac{3}{8}\) = 2\(\frac{1}{4}\) =
(2  X 4 + 1 by 4) = \(\frac{9}{4}\) ÷ \(\frac{3}{8}\) =
now we write \(\frac{3}{8}\) as reciprocal and multiply
\(\frac{9}{4}\) X \(\frac{8}{3}\) =
\(\frac{9 X 8}{4 X 3}\) = \(\frac{72}{12}\)
as both goes in 12 we get 12 X 6 = 72 and 12 X 1= 12,(6,1),
so \(\frac{72}{12}\) = 6,there are six,
three-eighths are in two and one-fourth.

c. One and one-halves are in six are 6 ÷ 1\(\frac{1}{2}\),
first we write \(\frac{1}{2}\) as (1 x 2 + 1 by 2) =
\(\frac{3}{2}\) now we write reciprocal as \(\frac{2}{3}\)
and multiply with 6 as 6 X \(\frac{2}{3}\) =
\(\frac{6 X 2}{1 X 3}\) =\(\frac{12}{3}\) = 4,
there are 4, one and one-halves are in six.

d. Seven-sixths are in three and one-third are 3\(\frac{1}{3}\) ÷
\(\frac{7}{6}\) , First we write mixed fraction in
fraction as (3 X 3 + 1 by 3) =  \(\frac{10}{3}\) now we
write reciprocal and multiply as \(\frac{10}{3}\) X
\(\frac{6}{7}\) =\(\frac{10 X 6}{3 X 7}\) =
\(\frac{60}{21}\)  as both goes in 3,3 X 20 = 60, 3 X 7 = 21,
(20,7) = \(\frac{20}{7}\) as numerator is greater
we can write as (2 X 7 + 6 by 7)=2 \(\frac{6}{7}\)
there are 2 \(\frac{6}{7}\)– seven-sixths are
in three and one-third .

e. One and one-fifths in five are 5 ÷1\(\frac{1}{5}\)
first we write mixed fraction as (1 X 5 +1 by 5) =
\(\frac{6}{5}\) now we write reciprocal and
multiply 5 X \(\frac{5}{6}\) =
\(\frac{5 X 5}{6}\) =\(\frac{25}{6}\)
as numerator is greater we can write as (4 X 6 + 1 by 6) =
4\(\frac{1}{6}\) there are 4\(\frac{1}{6}\) –
one and one-fifths in five.

f. Three and one-halves are in two and one-half are
2\(\frac{1}{2}\)÷ 3\(\frac{1}{2}\)
first we write 2\(\frac{1}{2}\) as (2 X 2 + 1 by 2) =
\(\frac{5}{2}\) and 3\(\frac{1}{2}\)
as (3 X 2 + 1 by 2) = \(\frac{7}{2}\).
Now \(\frac{5}{2}\) ÷\(\frac{7}{2}\),
We write reciprocal and multiply as \(\frac{5}{2}\)  X
\(\frac{2}{7}\) = \(\frac{5 X 2}{2 X 7}\) =
\(\frac{5}{7}\) there are \(\frac{5}{7}\)
– three and one-halves are in two and one-half.

g. Four and one-halves are in one and one-half are
1\(\frac{1}{2}\)÷ 4\(\frac{1}{2}\)
first we write 1\(\frac{1}{2}\) as (1 X 2 + 1 by 2) =
\(\frac{3}{2}\) and 4\(\frac{1}{2}\)
as (4 X 2 + 1 by 2) = \(\frac{9}{2}\).
Now \(\frac{3}{2}\) ÷\(\frac{9}{2}\),
We write reciprocal and multiply as \(\frac{3}{2}\)  X
\(\frac{2}{9}\) = \(\frac{3 X 2}{2 X 9}\) =
\(\frac{6}{18}\) as both goes in 6 we get 6 x 1= 6,
6 X 3= 18, (1,3) = \(\frac{1}{3}\) there are
\(\frac{1}{3}\) – four and one-halves are in one and one-half.

2.3 Lesson

Key Idea
Dividing Mixed Numbers
Write each mixed number as an improper fraction.
Then divide as you would with proper fractions.

Try It

Divide. Write the answer in simplest form.

Question 1.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 173
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 173= 11
Explanation:
Given expression as 3\(\frac{2}{3}\) ÷ \(\frac{1}{3}\),
first we write mixed fraction as (3 X 3 + 2 by 3) = \(\frac{11}{3}\) ÷
\(\frac{1}{3}\), we write  \(\frac{1}{3}\) reciprocal
as \(\frac{3}{1}\) and multiply as \(\frac{11}{3}\) X
\(\frac{3}{1}\) = \(\frac{11 X 3}{3 X 1}\) =
\(\frac{11}{1}\) = 11.

Question 2.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 174
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 174= \(\frac{15}{7}\) or 2\(\frac{1}{7}\)
Explanation:
Given expression as 1\(\frac{3}{7}\) ÷ \(\frac{2}{3}\),
first we write mixed fraction as (1 X 7 + 3 by 7) = \(\frac{10}{7}\) ÷
\(\frac{2}{3}\), we write  \(\frac{2}{3}\) reciprocal
as \(\frac{3}{2}\) and multiply as \(\frac{10}{7}\) X
\(\frac{3}{2}\) = \(\frac{10 X 3}{7 X 2}\) =
\(\frac{30}{14}\), as both goes in 2, 2 X 15 = 30, 2 X 7 = 14,
(15,7) = \(\frac{15}{7}\) as numerator is greater
we write as (2 X 7 + 1 by 7), so \(\frac{15}{7}\)= 2\(\frac{1}{7}\).

Question 3.
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 175
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 175= \(\frac{26}{9}\) or 2\(\frac{8}{9}\)
Explanation:
Given expression as 2\(\frac{1}{6}\) ÷ \(\frac{3}{4}\),
first we write mixed fraction as (2 X 6 + 1 by 6) = \(\frac{13}{6}\) ÷
\(\frac{3}{4}\), we write  \(\frac{3}{4}\) reciprocal
as \(\frac{4}{3}\) and multiply as \(\frac{13}{6}\) X
\(\frac{4}{3}\) = \(\frac{13 X 4}{6 X 3}\) =
\(\frac{52}{18}\), as both goes in 2, 2 X 26 = 52,
2 X 9 = 18, (26,9) = \(\frac{26}{9}\) as numerator is
greater we write as (2 X 9 + 8 by 9),
so \(\frac{26}{9}\)= 2\(\frac{8}{9}\).

Question 4.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 176
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 176= \(\frac{13}{4}\) or 3\(\frac{1}{4}\)
Explanation:
Given expression as 6\(\frac{1}{2}\) ÷ 2, first
we write mixed fraction as (6 X 2 + 1 by 2) = \(\frac{13}{2}\) ÷ 2,
we write 2 reciprocal as \(\frac{1}{2}\) and multiply as
\(\frac{13}{2}\) X \(\frac{1}{2}\) =
\(\frac{13 X 1}{2 X 2}\) = \(\frac{13}{4}\),
as numerator is greater we write as (3 X 4 + 1 by 4),
so \(\frac{13}{4}\)= 3\(\frac{1}{4}\).

Question 5.
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 177
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 177= 4
Explanation:
Given expressions as 10\(\frac{2}{3}\) ÷ 2\(\frac{2}{3}\),
First we write mixed fractions into fractions as 10\(\frac{2}{3}\)
= (10 X 3 + 2 by 3) = \(\frac{32}{3}\)  and 2\(\frac{2}{3}\)
= (2 X 3 + 2 by 3) = \(\frac{8}{3}\),
Now we write as we write \(\frac{32}{3}\) ÷
\(\frac{8}{3}\)now reciprocal of the fraction \(\frac{8}{3}\)
as \(\frac{3}{8}\) and multiply as \(\frac{32}{3}\) X
\(\frac{3}{8}\) = \(\frac{32 X 3}{3 X 8}\) =
\(\frac{96}{24}\), we can further simplify as both goes in 24,
24 X 4 = 96 and 24 X 1 = 24, (4,1)= 4.Therefore 10\(\frac{2}{3}\) ÷
2\(\frac{2}{3}\) = 4.

Question 6.
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 178
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 178= \(\frac{11}{2}\) or 5\(\frac{1}{2}\)
Explanation:
Given expressions as 8\(\frac{1}{4}\) ÷ 1\(\frac{1}{2}\),
First we write mixed fractions into fractions as 8\(\frac{1}{4}\) =
(8 X 4 + 1 by 4) = \(\frac{33}{4}\)  and 1\(\frac{1}{2}\) =
(1 X 2 + 1 by 2) = \(\frac{3}{2}\), Now we write
\(\frac{33}{4}\) ÷ \(\frac{3}{2}\) now reciprocal of
the fraction \(\frac{3}{2}\) as \(\frac{2}{3}\) and
multiply as \(\frac{33}{4}\) X \(\frac{2}{3}\) =
\(\frac{33 X 2}{4 X 3}\) = \(\frac{66}{12}\),
we can further simplify as both goes in 6, 6 X 11 = 66 and 6 X 2 = 12,
(11,2)=\(\frac{11}{2}\) as numerator is greater
we write as (5 X 2 + 1 by 2) = 5\(\frac{1}{2}\).
Therefore 8\(\frac{1}{4}\) ÷ 1\(\frac{1}{2}\) =
\(\frac{11}{2}\) or 5\(\frac{1}{2}\).

Question 7.
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 179
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 179= \(\frac{12}{7}\) or 1\(\frac{5}{7}\)
Explanation:
Given expressions as 3 ÷ 1\(\frac{1}{2}\),First we write
mixed fractions into fractions as 1\(\frac{3}{4}\) =
(1 x 4 + 3 by 4) = \(\frac{7}{4}\) now 3 ÷ \(\frac{7}{4}\)
now reciprocal of the fraction \(\frac{7}{4}\) as
\(\frac{4}{7}\) and multiply as 3 X \(\frac{4}{7}\) =
\(\frac{3 X 4}{7}\) = \(\frac{12}{7}\)
as numerator is greater we write as (1 X 7 + 5 by 7) = 1\(\frac{5}{7}\).
Therefore 3 ÷ 1\(\frac{3}{4}\) = \(\frac{12}{7}\) or 1\(\frac{5}{7}\).

Question 8.
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 180
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 180= \(\frac{3}{10}\)
Explanation:
Given expressions as \(\frac{3}{4}\) ÷ 2\(\frac{1}{2}\),
First we write mixed fractions into fractions as 2\(\frac{1}{2}\) =
(2 x 2 + 1 by 2) = \(\frac{5}{2}\) now \(\frac{3}{4}\) ÷
\(\frac{5}{2}\) now reciprocal of the fraction
\(\frac{5}{2}\) as \(\frac{2}{5}\) and multiply as
\(\frac{3}{4}\) X \(\frac{2}{5}\) =
\(\frac{3 X 2}{4 X 5}\) = \(\frac{6}{20}\)
as both goes in 2, 2 X 3 = 6, 2 X 10= 20, (3,10)= \(\frac{3}{10}\).
Therefore \(\frac{3}{4}\) ÷ 2\(\frac{1}{2}\) = \(\frac{3}{10}\).

Try It

Evaluate the expression. Write the answer in simplest form.

Question 9.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 181
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 181= \(\frac{65}{8}\) = 8\(\frac{1}{8}\)
Explanation:
Given expressions as (1\(\frac{1}{2}\) ÷ \(\frac{1}{6}\)) –
\(\frac{7}{8}\) ,First we write mixed fractions into fractions as
1\(\frac{1}{2}\) = (1 x 2 + 1 by 2) = \(\frac{3}{2}\)
now \(\frac{3}{2}\) ÷ \(\frac{1}{6}\) now reciprocal
of the fraction \(\frac{1}{6}\) as 6 and multiply as
\(\frac{3}{2}\) X 6 = \(\frac{3 X 6}{2 X 1}\) =
\(\frac{18}{2}\) as both goes in 2, 2 X 9 = 18, 2 X 1= 2, (9,1)=9.
Now 9 – \(\frac{7}{8}\) =(9 X 8 – 7 by 8) = \(\frac{65}{8}\)
as numerator is greater we write as (8 X 8 + 1 by 8) = \(\frac{65}{8}\) =
8\(\frac{1}{8}\). Therefore (1\(\frac{1}{2}\) ÷ \(\frac{1}{6}\)) –
\(\frac{7}{8}\) = \(\frac{65}{8}\) = 8\(\frac{1}{8}\).

Question 10.
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 182
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 182= \(\frac{44}{9}\) = 4\(\frac{8}{9}\)
Explanation:
Given expressions as (3\(\frac{1}{3}\) ÷ \(\frac{5}{6}\)) +
\(\frac{8}{9}\) ,First we write mixed fractions into fractions as
3\(\frac{1}{3}\) = (3 x 3 + 1 by 3) = \(\frac{10}{3}\)
now \(\frac{10}{3}\) ÷ \(\frac{5}{6}\) now reciprocal
of the fraction \(\frac{5}{6}\) as \(\frac{6}{5}\) and multiply as
\(\frac{10}{3}\) X \(\frac{6}{5}\) = \(\frac{10 X 6}{3 X 5}\) =
\(\frac{60}{15}\) as both goes in 15, 15 X 4 = 60, 15 X 1= 15, (4,1)=4.
Now 4 + \(\frac{8}{9}\) =(4 X 9 + 8 by 9) = \(\frac{44}{9}\)
as numerator is greater we write as (4 X 9 + 8 by 9) = \(\frac{44}{9}\) =
4\(\frac{8}{9}\). Therefore (3\(\frac{1}{3}\) ÷ \(\frac{5}{6}\)) +
\(\frac{8}{9}\) = \(\frac{44}{9}\) = 4\(\frac{8}{9}\).

Question 11.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 183
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 183= \(\frac{8}{5}\) = 1\(\frac{3}{5}\)
Explanation:
Given expressions as \(\frac{2}{5}\) + 2\(\frac{4}{5}\) ÷ 2,
First we write mixed fractions into fractions as
2\(\frac{4}{5}\) = (2 x 5 + 4 by 5) = \(\frac{14}{5}\)
now \(\frac{2}{5}\) + \(\frac{14}{5}\) as denominators
are same we add numerators as 2 + 14 and write as \(\frac{16}{5}\),
now we divide with 2, \(\frac{16}{5}\) ÷ 2, we write 2 as
reciprocal and multiply \(\frac{16}{5}\) X \(\frac{1}{2}\) =
\(\frac{16 X 1}{5 X 2}\) = \(\frac{16}{10}\) as both goes
in 2, 2 x 8 = 16, 2 X 5 = 10, (8,5) = \(\frac{8}{5}\) as numerator is
greater we write as (1 X 5 + 3 by 5) = 1\(\frac{3}{5}\) ,Therefore
\(\frac{2}{5}\) + 2\(\frac{4}{5}\) ÷ 2 =
\(\frac{8}{5}\) = 1\(\frac{3}{5}\) .

Question 12.
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 184
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 184= \(\frac{1}{3}\)
Explanation:
Given expressions as \(\frac{2}{3}\) – (1\(\frac{4}{7}\) ÷ 4\(\frac{5}{7})\),First we write mixed fractions into fractions as
1\(\frac{4}{7}\) = (1 X 7 + 4 by 7) = \(\frac{11}{7}\) and
4\(\frac{5}{7}\) = (4 X 7 + 5 by 7) = \(\frac{33}{7}\),
Now first we divide \(\frac{11}{7}\) ÷ \(\frac{33}{7}\) =
we write \(\frac{33}{7}\) as reciprocal and multiply \(\frac{7}{33}\) ,
\(\frac{11}{7}\) X \(\frac{7}{33}\) = \(\frac{11 X 7}{7 X 33}\) =
\(\frac{77}{231}\) as both goes in 77, 77 X 1= 77, 77 X 3 = 231,(1,3),
\(\frac{1}{3}\) now we subtract from \(\frac{2}{3}\) – \(\frac{1}{3}\)
as denominators are same we subtract from numerators as
(2-1) we get \(\frac{1}{3}\), Therefore \(\frac{2}{3}\) – (1\(\frac{4}{7}\) ÷ 4\(\frac{5}{7})\) = \(\frac{1}{3}\).

Self-Assessment for Concepts & Skills
Solve each exercise. Then rate your understanding of the success criteria in your journal.

EVALUATING EXPRESSIONS
Evaluate the expression. Write the answer in simplest form.

Question 13.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 185
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 185= \(\frac{224}{28}\) = 8.
Explanation:
Given expressions as 4\(\frac{4}{7}\) ÷ \(\frac{4}{7}\),
First we write mixed fractions into fractions 4\(\frac{4}{7}\) as
(4 X 7 + 4 by 7) = \(\frac{32}{7}\), Now \(\frac{4}{7}\) we
write as reciprocal and multiply \(\frac{7}{4}\) as
\(\frac{32}{7}\) X \(\frac{7}{4}\) = \(\frac{32 X 7}{7 X 4}\)=
\(\frac{224}{28}\) as both goes in 28, 28 X 8 = 224,
28 X 1 = 28, (8,1) = 8, therefore 4\(\frac{4}{7}\) ÷ \(\frac{4}{7}\)=
\(\frac{224}{28}\) = 8.

Question 14.
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 186
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 186= \(\frac{2}{21}\)
Explanation:
Given expressions as \(\frac{1}{2}\) ÷ 5\(\frac{1}{4}\),
First we write mixed fractions into fractions 5\(\frac{1}{4}\) as
(5 X 4 + 1 by 4) = \(\frac{21}{4}\) we write as reciprocal and
multiply \(\frac{4}{21}\) as \(\frac{1}{2}\) X \(\frac{4}{21}\) =
\(\frac{1 X 4}{2 X 21}\) = \(\frac{4}{42}\) as both goes in
2, 2 X 2 = 4, 2 X 21 = 42, (2, 21) = \(\frac{2}{21}\), therefore
\(\frac{1}{2}\) ÷ 5\(\frac{1}{4}\) = \(\frac{2}{21}\).

Question 15.
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 187
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 187= \(\frac{19}{4}\) = 4\(\frac{3}{4}\)
Explanation:
Given expressions as \(\frac{3}{4}\) + (6\(\frac{2}{5}\) ÷
1\(\frac{3}{5}\)), First we write mixed fractions into fractions as
6\(\frac{2}{5}\) = (6 X 5 + 2 by 5) = \(\frac{32}{5}\) and
1\(\frac{3}{5}\) = (1 X 5 + 3 by 5) = \(\frac{8}{5}\),
we write reciprocal \(\frac{5}{8}\) and multiply \(\frac{32}{5}\) X \(\frac{5}{8}\) = \(\frac{32 X 5}{5 X 8}\) = \(\frac{160}{40}\)
as both goes in 40 as 40 X 4 = 160, 40 X 1 = 40, (4,1), \(\frac{160}{40}\) =4,
Now \(\frac{3}{4}\) + 4 = ( 3 + 4 X 4 by 4 ) = \(\frac{19}{4}\),
as numerator is greater we write as ( 4 X 4 + 3 by 4) = 4\(\frac{3}{4}\),
therefore \(\frac{3}{4}\) + (6\(\frac{2}{5}\) ÷
1\(\frac{3}{5}\)) = \(\frac{19}{4}\) = 4\(\frac{3}{4}\).

Question 16.
NUMBER SENSE
Is 2\(\frac{1}{2}\) ÷ 1\(\frac{1}{4}\) the same as 1\(\frac{1}{4}\) ÷ 2\(\frac{1}{2}\)? Use models to justify your answer.
No,Is 2\(\frac{1}{2}\) ÷ 1\(\frac{1}{4}\) is not the same as
1\(\frac{1}{4}\) ÷ 2\(\frac{1}{2}\)
Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals-46

Explanation:
Given expressions as 2\(\frac{1}{2}\) ÷ 1\(\frac{1}{4}\) and
1\(\frac{1}{4}\) ÷ 2\(\frac{1}{2}\) first we write mixed fractions,
into fractions 2\(\frac{1}{2}\) = (2 X 2 + 1 by 2) = \(\frac{5}{2}\),
1\(\frac{1}{4}\) = (1 X 4 + 1 by 4) = \(\frac{5}{4}\), now
write as reciprocal as \(\frac{4}{5}\), now we multiply as
\(\frac{5}{2}\) X \(\frac{4}{5}\) = \(\frac{5 x 4}{2 X 5}\) =
\(\frac{20}{10}\) as both goes in 10, 10 X 2 = 20, 10 X 1= 10,(2,1)=2.
Now 1\(\frac{1}{4}\) ÷ 2\(\frac{1}{2}\) we write mixed fractions,
into fractions 1\(\frac{1}{4}\) = (1 X 4 + 1 by 4) = \(\frac{5}{4}\),
2\(\frac{1}{2}\) = (2 X 2 + 1 by 2) = \(\frac{5}{2}\), now
write as reciprocal as \(\frac{2}{5}\), now we multiply as
\(\frac{5}{4}\) X \(\frac{2}{5}\) = \(\frac{5 x 2}{4 X 5}\) =
\(\frac{10}{20}\) = as both goes in 10, 10 X 1 = 10, 10 X 2 = 20, (1,2) =
\(\frac{1}{2}\), therefore 2 ≠ \(\frac{1}{2}\) So no,
2\(\frac{1}{2}\) ÷ 1\(\frac{1}{4}\) is not the same as
1\(\frac{1}{4}\) ÷ 2\(\frac{1}{2}\).

Question 17.
DIFFERENT WORDS, SAME QUESTION
Which is different? Find “both” answers.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 188
d. Different one is What is \(\frac{1}{8}\) of 5 \(\frac{1}{2}\) =
5\(\frac{1}{2}\) multiplied by \(\frac{1}{8}\).
All 3 results are in whole only \(\frac{1}{8}\) of 5 \(\frac{1}{2}\)
is in fraction. So it is different from other 3 ones.

Explanation:
We calculate first 5\(\frac{1}{2}\) = (5 X 2 + 1 by 2) = \(\frac{11}{2}\)
a. As what is 5\(\frac{1}{2}\) divided by \(\frac{1}{8}\) means
5\(\frac{1}{2}\) ÷ \(\frac{1}{8}\),
\(\frac{11}{2}\) ÷  \(\frac{1}{8}\),
we write reciprocal and multiply as
\(\frac{11}{2}\) X  8 = \(\frac{88}{2}\) = 44.
b. What is quotient of 5\(\frac{1}{2}\) and \(\frac{1}{8}\) means
5\(\frac{1}{2}\) ÷ \(\frac{1}{8}\),
\(\frac{11}{2}\) ÷  \(\frac{1}{8}\),
we write reciprocal and multiply as
\(\frac{11}{2}\) X  8 = \(\frac{88}{2}\) = 44.
c. What is 5\(\frac{1}{2}\) times of 8 means
5\(\frac{1}{2}\) X 8=  \(\frac{88}{2}\) = 44.
d. What is \(\frac{1}{8}\) of 5\(\frac{1}{2}\) means
5\(\frac{1}{2}\) X \(\frac{1}{8}\),
\(\frac{11}{2}\) X \(\frac{1}{8}\) = \(\frac{11}{16}\)=0.6875,
So only different one is d. What is \(\frac{1}{8}\) of 5 \(\frac{1}{2}\)
value is different from the other 3.

Self-Assessment for Problem Solving
Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 18.
A water cooler contains 160 cups of water. During practice, each person on a team fills a water bottle with 3\(\frac{1}{3}\) cups of water from the cooler. Is there enough water for all 45 people on the team to fill their water bottles? Explain.
150 cups  of water is required for 45 people and cooler contains
160 cups of water, therefore we have sufficient enough water
for all 45 people on the team to fill their water bottles.

Explanation:
Given a water cooler contains 160 cups of water, During practice,
each person on a team fills a water bottle with 3\(\frac{1}{3}\) cups
of water from the cooler. To find is there enough water for all
45 people on the team to fill their water bottles we calculate as
45 persons X 3\(\frac{1}{3}\) =
first we write mixed fraction as fraction 3\(\frac{1}{3}\) =
(3 X 3 +1 by 3) = \(\frac{10}{3}\) now we multiply as
45  X \(\frac{10}{3}\) = \(\frac{450}{3}\) as both
goes in 3, 3 X 150 = 450,3 x 1= 3, (150,1) so \(\frac{450}{3}\)= 150 cups
of water is required for 45 people and cooler contains 160 cups of water,
therefore we have sufficient enough water for all 45 people on the team to
fill their water bottles.

Question 19.
A cyclist is 7\(\frac{3}{4}\) kilometers from the finish line of a race. The cyclist rides at a rate of 25\(\frac{5}{6}\) kilometers per hour. How many minutes will it take the cyclist to finish the race?
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 189

It will take 18 minutes for the cyclist to finish the race,

Explanation:
Given a cyclist is 7\(\frac{3}{4}\) kilometers from the finish line of a race,
The cyclist rides at a rate of 25\(\frac{5}{6}\) kilometers per hour,
means we write as 25\(\frac{5}{6}\) ÷  7\(\frac{3}{4}\)
first we write mixed fractions in fractions as 25\(\frac{5}{6}\) =
(25 X 6 + 5 by 6) = \(\frac{155}{6}\) now 7\(\frac{3}{4}\) =
(7 X 4 + 3 by 4) = \(\frac{31}{4}\) now reciprocal of \(\frac{31}{4}\) =
\(\frac{4}{31}\) and multiply as \(\frac{155}{6}\) X \(\frac{4}{31}\)=
\(\frac{155 X 4}{6 X 31}\) = \(\frac{620}{186}\) kilometers
in an hour. Now we convert into minutes as \(\frac{186 X 60}{620}\) =
\(\frac{11160}{620}\) as both goes in 620, 620 X 18 = 11160, 620 X 1 = 620,
\(\frac{11160}{620}\) = 18 minutes.Therefore it will take
18 minutes for the cyclist to finish the race.

Dividing Mixed Numbers Homework & Practice 2.3

Review & Refresh

Divide. Write the answer in simplest form.

Question 1.
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 190
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 190= \(\frac{7}{8}\)
Explanation:
Given expression as \(\frac{1}{8}\) ÷ \(\frac{1}{7}\)
we write reciprocal of the fraction \(\frac{1}{7}\) as
\(\frac{7}{1}\) and multiply as \(\frac{1}{8}\) X
\(\frac{7}{1}\) = \(\frac{1 X 7}{8 X 1}\) =
\(\frac{7}{8}\), Therefore \(\frac{1}{8}\) ÷ \(\frac{1}{7}\) = \(\frac{7}{8}\).

Question 2.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 191
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 191= \(\frac{7}{6}\) or 1\(\frac{1}{6}\)
Explanation:
Given expression as \(\frac{7}{9}\) ÷ \(\frac{2}{3}\)
we write reciprocal of the fraction \(\frac{2}{3}\) as
\(\frac{3}{2}\) and multiply as \(\frac{7}{9}\) X
\(\frac{3}{2}\) = \(\frac{7 X 3}{9 X 2}\) =
\(\frac{21}{18}\) as both goes in 3, 3 X 7 =21, 3 X 6 = 18,(7,6)=
\(\frac{7}{6}\) as numerator is greater we write as (1 X 6 + 1 by 6)=
1\(\frac{1}{6}\). Therefore \(\frac{7}{9}\) ÷ \(\frac{2}{3}\) = \(\frac{7}{6}\) or 1\(\frac{1}{6}\).

Question 3.
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 192
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 192= \(\frac{1}{12}\)
Explanation:
Given expression as \(\frac{5}{6}\) ÷ 10
we write reciprocal for 10 as \(\frac{1}{10}\)
and multiply as \(\frac{5}{6}\) X \(\frac{1}{10}\) =
\(\frac{5 X 1}{6 X 10}\) = \(\frac{5}{60}\) as both goes in 5,
5 X 1 = 5, 5 X 12 = 60 ,(1,12) = \(\frac{1}{12}\),
Therefore \(\frac{5}{6}\) ÷ 10 = \(\frac{1}{12}\).

Question 4.
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 193
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 193= 32
Explanation:
Given expression as 12 ÷ \(\frac{3}{8}\), we write reciprocal
for \(\frac{3}{8}\) as \(\frac{8}{3}\) and multiply as
12 X \(\frac{8}{3}\) = \(\frac{12 X 8}{1 X 3}\) = \(\frac{96}{3}\),
as both goes in 3, 3 X 32 = 96, 3 X 1 = 3, (32,1), So \(\frac{96}{3}\) = 32,
Therefore 12 ÷ \(\frac{3}{8}\) = 32.

Find the LCM of the numbers.

Question 5.
8, 14
The LCM of 8, 14 is 56

Explanation:
The prime factorization of 8 is  2 X 2 X 2,
The prime factorization of 14 is 2 X 7,
Eliminate the duplicate factors of the two lists,
(we have 2 in common both ),then multiply them once
with the remaining factors of the lists to get
LCM(8, 14) = 2 X 2 X 2 X 7 = 56.

Question 6.
9, 11, 12
The LCM of 9, 11, 12  is 396

Explanation:
The factors of 9 is  3 X 3,
The factors of 11 is 11,
The factors of 12 is 3 X 4
Eliminate the duplicate factors of the two lists,
(we have 3 in common),then multiply them once
with the remaining factors of the lists to get
LCM(9, 11, 12) = 3 X 3 X 11 X 4 = 396.

Question 7.
12, 27, 30
The LCM of 12, 27, 30  is 540

Explanation:
The factors of 12 is  3 X 4 =  3 X 2 X 2
The factors of 27 is 3 X 9 = 3 X 3 X 3,
The factors of 30 is 3 X 10 = 3 X 2 X 5
Eliminate the duplicate factors of the two lists,
(we have 2, 3 in common),then multiply them once
with the remaining factors of the lists to get
LCM(12, 27, 30) = 2 X 3 X 9 X 10 = 540.

Find the volume of the rectangular prism.

Question 8.
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 194
The volume of the rectangular prism is 96 mt3

Explanation:
Given length as 4 m, width as 3 m and height as 8 m,
we know the volume of the rectangular prism is
length X width X height =  4 X 3 X 8 = 96, Therefore
the volume of the rectangular prism is 96 mt3.

Question 9.
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 195
The volume of the rectangular prism is 70 in3

Explanation:
Given length as 5 in, width as 2 in and height as 7 in,
we know the volume of the rectangular prism is
length X width X height =  5 X 2 X 7 = 70, Therefore
the volume of the rectangular prism is 70 in3.

Question 10.
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 196
The volume of the rectangular prism is 240 yd3

Explanation:
Given length as 10 yd, width as 8 yd and height as 3 yd,
we know the volume of the rectangular prism is
length X width X height =  10 X 8 X 3 = 240, Therefore
the volume of the rectangular prism is 240 yd3.

Question 11.
Which number is not a prime factor of 286?
A. 2
B. 7
C. 11
D. 13

B. 7 is not a prime factor of 286,

Explanation:
286 is a composite number.
Prime factorization of 286 = 2 x 11 x 13 and
Factors of 286 are 1, 2, 11, 13, 22, 26, 143, 286.
Therefore 7 is not a prime factor of 286.

Concepts, Skills, & Problem Solving
CHOOSE TOOLS
Write a real-life problem that represents the division expression described. Then solve the problem using a model. Check your answer algebraically. (See Exploration 1, p. 61.)

Question 12.
How many two-thirds are in three and one-third?
I have three and one- third rose flowers I have, I distributed among
two-thirds children, how man children are there?
Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals-47

Five, two-thirds are in three and one-third

Explanation:
Given to find two-thirds are in three and one-third means
3\(\frac{1}{3}\) ÷ \(\frac{2}{3}\), First we write
mixed fraction as ( 3 X 3 + 1 by 3) = \(\frac{10}{3}\)÷ \(\frac{2}{3}\)
we write reciprocal of the fraction \(\frac{2}{3}\)
as \(\frac{3}{2}\) and multiply \(\frac{10}{3}\) X
\(\frac{3}{2}\) = \(\frac{10 X 3}{3 X 2}\)= \(\frac{30}{6}\)
as both goes in 6 as 6 X 5 = 30, 6 X 1= 6,(5, 1), \(\frac{30}{6}\) = 5.

Question 13.
How many one and one-sixths are in five and five-sixths?
A golden statue is five and five-sixths tall and a silver statue is
one and one-sixths tall, How many times is the golden statue
taller than silver statue.
Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals-48

Five, one and one-sixths are in five and five-sixths

Explanation:
Given to find one and one-sixths are in five and five-sixths means
5\(\frac{5}{6}\) ÷ 1\(\frac{1}{6}\), First we write
5\(\frac{5}{6}\) mixed fraction as ( 5 X 6 + 5 by 6) = \(\frac{35}{6}\)
and 1\(\frac{1}{6}\) mixed fraction as (1 X 6 + 1 by 6)= \(\frac{7}{6}\).
Now \(\frac{35}{6}\) ÷ \(\frac{7}{6}\) we write reciprocal
of the fraction \(\frac{7}{6}\) as \(\frac{6}{7}\) and multiply
\(\frac{35}{6}\) X \(\frac{6}{7}\)= \(\frac{35 X 6}{6 X 7}\)= \(\frac{210}{42}\) as both goes in 42 as 42 X 5 = 210, 42 X 1= 42,(5, 1),
\(\frac{210}{42}\) = 5.

Question 14.
How many two and one-halves are in eight and three-fourths?
I have eight and three-fourths of pens with me, My friend
has two and one-halves pens with him, How many more pens
do I have more than my friend.
Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals-49
\(\frac{7}{2}\) or 3\(\frac{1}{2}\) two and one-halves
are in eight and three-fourths
Explanation:
Given to find two and one-halves are in eight and three-fourths means
8\(\frac{3}{4}\) ÷ 2\(\frac{1}{2}\), First we write
8\(\frac{3}{4}\) mixed fraction as ( 8 X 4 + 3 by 4) = \(\frac{35}{4}\)
and 2\(\frac{1}{2}\) mixed fraction as (2 X 2 + 1 by 2)= \(\frac{5}{2}\).
Now \(\frac{35}{4}\) ÷ \(\frac{5}{2}\) we write reciprocal
of the fraction \(\frac{5}{2}\) as \(\frac{2}{5}\) and multiply
\(\frac{35}{4}\) X \(\frac{2}{5}\)= \(\frac{35 X 2}{4 X 5}\)= \(\frac{70}{20}\) as both goes in 10 as 10 X 7 = 70, 10 X 2= 20,(7, 2),
\(\frac{70}{20}\) = \(\frac{7}{2}\) as numerator is greater
we write as (3 X 2 + 1 by 2) = 3\(\frac{1}{2}\).

DIVIDING WITH MIXED NUMBERS
Divide. Write the answer in simplest form.

Question 15.
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 197
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 197= 3
Explanation:
Given expressions as 2\(\frac{1}{4}\) ÷ \(\frac{3}{4}\),
First we write mixed fraction into fraction 2\(\frac{1}{4}\) as
(2 X 4 + 1 by 4) = \(\frac{9}{4}\), Now \(\frac{3}{4}\) we
write as reciprocal and multiply \(\frac{4}{3}\) as
\(\frac{9}{4}\) X \(\frac{4}{3}\) = \(\frac{9 X 4}{4 X 3}\)=
\(\frac{36}{12}\) as both goes in 12, 12 X 3 = 36,
12 X 1 = 12, (3, 1) = 3, therefore 2\(\frac{1}{4}\) ÷ \(\frac{3}{4}\)=
\(\frac{36}{12}\) = 3.

Question 16.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 198
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 198= \(\frac{19}{2}\) = 9\(\frac{1}{2}\)
Explanation:
Given expressions as 3\(\frac{4}{5}\) ÷ \(\frac{2}{5}\),
First we write mixed fraction into fraction 3\(\frac{4}{5}\) as
(3 X 5 + 4 by 5) = \(\frac{19}{5}\), Now \(\frac{2}{5}\) we
write as reciprocal and multiply \(\frac{5}{2}\) as
\(\frac{19}{5}\) X \(\frac{5}{2}\) = \(\frac{19 X 5}{5 X 2}\)=
\(\frac{95}{10}\) as both goes in 5, 5 X 19 = 95,
5 X 2 = 10, (19, 2) = \(\frac{19}{2}\) as numerator is greater
we write as (9 X 2 + 1 by 2)= 9\(\frac{1}{2}\).
Therefore 3\(\frac{4}{5}\) ÷ \(\frac{2}{5}\) = \(\frac{19}{2}\) = 9\(\frac{1}{2}\).

Question 17.
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 199
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 199= \(\frac{39}{4}\) = 9\(\frac{3}{4}\)
Explanation:
Given expressions as 8\(\frac{1}{8}\) ÷ \(\frac{5}{6}\),
First we write mixed fraction into fraction 8\(\frac{1}{8}\) as
(8 X 8 + 1 by 8) = \(\frac{65}{8}\), Now \(\frac{5}{6}\) we
write as reciprocal and multiply \(\frac{6}{5}\) as
\(\frac{65}{8}\) X \(\frac{6}{5}\) = \(\frac{65 X 6}{8 X 5}\) =
\(\frac{390}{40}\) as both goes in 10, 10 X 39 = 390,
10 X 4 = 40, (39, 4) = \(\frac{39}{4}\) as numerator is greater
we write as (9 X 4 + 3 by 4)= 9\(\frac{3}{4}\).
Therefore 8\(\frac{1}{8}\) ÷ \(\frac{5}{6}\) = \(\frac{39}{4}\) = 9\(\frac{3}{4}\).

Question 18.
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 200
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 200= \(\frac{119}{9}\) = 13\(\frac{2}{9}\)
Explanation:
Given expressions as 7\(\frac{5}{9}\) ÷ \(\frac{4}{7}\),
First we write mixed fraction into fraction 7\(\frac{5}{9}\) as
(7 X 9 + 5 by 9) = \(\frac{68}{9}\), Now \(\frac{4}{7}\) we
write as reciprocal and multiply \(\frac{7}{4}\) as
\(\frac{68}{9}\) X \(\frac{7}{4}\) = \(\frac{68 X 7}{9 X 4}\) =
\(\frac{476}{36}\) as both goes in 4, 4 X 119 = 476,
4 X 9 = 36, (119, 9) = \(\frac{119}{9}\) as numerator is greater
we write as (13 X 9 + 2 by 9)= 13\(\frac{2}{9}\).
Therefore 7\(\frac{5}{9}\) ÷ \(\frac{4}{7}\) = \(\frac{119}{9}\) = 13\(\frac{2}{9}\).

Question 19.
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 201
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 201= \(\frac{75}{19}\) = 3\(\frac{18}{19}\)
Explanation:
Given expressions as 7\(\frac{1}{2}\) ÷ 1\(\frac{9}{10}\),
First we write mixed fractions into fractions 7\(\frac{1}{2}\) as
(7 X 2 + 1 by 2) = \(\frac{15}{2}\), 1\(\frac{9}{10}\)=
(1 x 10 + 9 by 10) = \(\frac{19}{10}\),Now \(\frac{19}{10}\) we
write as reciprocal and multiply \(\frac{10}{19}\) as
\(\frac{15}{2}\) X \(\frac{10}{19}\) = \(\frac{15 X 10}{2 X 19}\) =
\(\frac{150}{38}\) as both goes in 2, 2 X 75 = 150,
2 X 19 = 38, (75, 19) = \(\frac{75}{19}\) as numerator is greater
we write as (3 X 19 + 18 by 9)= 3\(\frac{18}{19}\).
Therefore 7\(\frac{1}{2}\) ÷ 1\(\frac{9}{10}\) = \(\frac{75}{19}\) = 3\(\frac{18}{19}\).

Question 20.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 202
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 202= \(\frac{9}{5}\) = 1\(\frac{4}{5}\)
Explanation:
Given expressions as 3\(\frac{3}{4}\) ÷ 2\(\frac{1}{12}\),
First we write mixed fractions into fractions 3\(\frac{3}{4}\) as
(3 X 4 + 3 by 4) = \(\frac{15}{4}\), 2\(\frac{1}{12}\)=
(2 x 12 + 1 by 12) = \(\frac{25}{12}\),Now \(\frac{25}{12}\) we
write as reciprocal and multiply \(\frac{12}{25}\) as
\(\frac{15}{4}\) X \(\frac{12}{25}\) = \(\frac{15 X 12}{4 X 25}\) =
\(\frac{180}{100}\) as both goes in 20, 20 X 9 = 180,
20 X 5 = 100, (9, 5) = \(\frac{9}{5}\) as numerator is greater
we write as (1 X 5 + 4 by 5)= 1\(\frac{4}{5}\).
Therefore 3\(\frac{3}{4}\) ÷ 2\(\frac{1}{12}\) = \(\frac{9}{5}\) = 1\(\frac{4}{5}\).

Question 21.
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 203
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 203= \(\frac{9}{10}\)
Explanation:
Given expressions as 7\(\frac{1}{5}\) ÷ 8,
First we write mixed fraction into fraction 7\(\frac{1}{5}\) as
(7 X 5 + 1 by 5) = \(\frac{36}{5}\),Now 8 we
write as reciprocal and multiply \(\frac{1}{8}\) as
\(\frac{36}{5}\) X \(\frac{1}{8}\) = \(\frac{36 X 1}{5 X 8}\) =
\(\frac{36}{40}\) as both goes in 4, 4 X 9 = 36,
4 X 10 = 40, (9, 10) = \(\frac{9}{10}\).
Therefore 7\(\frac{1}{5}\) ÷ 8 = \(\frac{9}{10}\).

Question 22.
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 204= \(\frac{4}{7}\)
Explanation:
Given expressions as 8\(\frac{4}{7}\) ÷ 15,
First we write mixed fraction into fraction 8\(\frac{4}{7}\) as
(8 X 7 + 4 by 7) = \(\frac{60}{7}\),Now 15 we
write as reciprocal and multiply \(\frac{1}{15}\) as
\(\frac{60}{7}\) X \(\frac{1}{15}\) = \(\frac{60 X 1}{7 X 15}\) =
\(\frac{60}{105}\) as both goes in 15, 15 X 4 = 60,
15 X 7 = 105, (4, 7) = \(\frac{4}{7}\).
Therefore 8\(\frac{4}{7}\) ÷ 15 = \(\frac{4}{7}\).

Question 23.
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 205
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 205= \(\frac{25}{2}\) = 12\(\frac{1}{2}\)
Explanation:
Given expressions as 8\(\frac{1}{3}\) ÷ \(\frac{2}{3}\),
First we write mixed fraction into fraction 8\(\frac{1}{3}\) as
(8 X 3 + 1 by 3) = \(\frac{25}{3}\),Now \(\frac{2}{3}\) we
write as reciprocal and multiply \(\frac{3}{2}\) as
\(\frac{25}{3}\) X \(\frac{3}{2}\) = \(\frac{25 X 3}{3 X 2}\) =
\(\frac{75}{6}\) as both goes in 3, 3 X 25 = 75,
3 X 2 = 6, (25, 2) = \(\frac{25}{2}\) as numerator is greater
we write as (12 X 2 + 1 by 2)= 12\(\frac{1}{2}\).
Therefore 8\(\frac{1}{3}\) ÷ \(\frac{2}{3}\) = \(\frac{25}{2}\) = 12\(\frac{1}{2}\).

Question 24.
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 206
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 206= 11
Explanation:
Given expressions as 9\(\frac{1}{6}\) ÷ \(\frac{5}{6}\),
First we write mixed fraction into fraction 9\(\frac{1}{6}\) as
(9 X 6 + 1 by 6) = \(\frac{55}{6}\),Now \(\frac{5}{6}\) we
write as reciprocal and multiply \(\frac{6}{5}\) as
\(\frac{55}{6}\) X \(\frac{6}{5}\) = \(\frac{55 X 6}{6 X 5}\) =
\(\frac{330}{30}\) as both goes in 30, 30 X 11 = 330,
3 X 1 = 30, (11, 1) = \(\frac{330}{30}\) = 11,
Therefore 9\(\frac{1}{6}\) ÷ \(\frac{5}{6}\) = 11.

Question 25.
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 207
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 207= \(\frac{6}{5}\) = 1\(\frac{1}{5}\)
Explanation:
Given expressions as 13 ÷  10\(\frac{5}{6}\)
First we write mixed fraction into fraction 10\(\frac{5}{6}\) as
(10 X 6 + 5 by 6) = \(\frac{65}{6}\),Now \(\frac{65}{6}\) we
write as reciprocal and multiply \(\frac{6}{65}\) as
13 X \(\frac{6}{65}\) = \(\frac{13 X 6}{1 X 65}\) =
\(\frac{78}{65}\) as both goes in 13, 13 X 6 = 78,
13 X 5 = 65, (6, 5) = \(\frac{6}{5}\) as numerator is greater
we write as (1 X 5 + 1 by 5)= 1\(\frac{1}{5}\).
Therefore 13 ÷  10\(\frac{5}{6}\) = \(\frac{6}{5}\) = 1\(\frac{1}{5}\).

Question 26.
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 208
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 208= \(\frac{33}{16}\) = 2\(\frac{1}{16}\)

Explanation:
Given expressions as 12 ÷ 5\(\frac{9}{11}\)
First we write mixed fraction into fraction 5\(\frac{9}{11}\) as
(5 X 11 + 9 by 11) = \(\frac{64}{11}\),Now \(\frac{65}{11}\) we
write as reciprocal and multiply \(\frac{11}{64}\) as
12 X \(\frac{11}{64}\) = \(\frac{12 X 11}{1 X 64}\) =
\(\frac{132}{64}\) as both goes in 4, 4 X 33 = 132,
4 X 16 = 64, (33, 16) = \(\frac{33}{16}\) as numerator is greater
we write as (2 X 16 + 1 by 16)= 2\(\frac{1}{16}\).
Therefore 12 ÷ 5\(\frac{9}{11}\) = \(\frac{33}{16}\) = 2\(\frac{1}{16}\).

Question 27.
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 209
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 209= \(\frac{2}{7}\)

Explanation:
Given expressions as \(\frac{7}{8}\) ÷ 3\(\frac{1}{16}\),
First we write mixed fraction into fraction 3\(\frac{1}{16}\) as
(3 X 16 + 1 by 16) = \(\frac{49}{16}\),Now \(\frac{49}{16}\) we
write as reciprocal and multiply \(\frac{16}{49}\) as
\(\frac{7}{8}\) X \(\frac{16}{49}\) = \(\frac{7 X 16}{8 X 49}\) =
\(\frac{112}{392}\) as both goes in 56, 56 X 2 = 112,
56 X 7 = 392, (2, 7) = \(\frac{2}{7}\),
Therefore \(\frac{7}{8}\) ÷ 3\(\frac{1}{16}\) = \(\frac{2}{7}\).

Question 28.
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 210
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 210= \(\frac{10}{33}\)

Explanation:
Given expressions as \(\frac{4}{9}\) ÷ 1\(\frac{7}{15}\),
First we write mixed fraction into fraction 1\(\frac{7}{15}\) as
(1 X 15 + 7 by 15) = \(\frac{22}{15}\),Now \(\frac{22}{15}\) we
write as reciprocal and multiply \(\frac{15}{22}\) as
\(\frac{4}{9}\) X \(\frac{15}{22}\) = \(\frac{4 X 15}{9 X 22}\) =
\(\frac{60}{198}\) as both goes in 6, 6 X 10 = 60,
6 X 33 = 198, (10, 33) = \(\frac{10}{33}\), therefore
\(\frac{4}{9}\) ÷ 1\(\frac{7}{15}\) = \(\frac{10}{33}\).

Question 29.
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 211
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 211= \(\frac{23}{18}\) or 1\(\frac{5}{18}\)
Explanation:
Given expressions as 4\(\frac{5}{16}\) ÷ 3\(\frac{3}{8}\),
First we write mixed fractions into fractions as 4\(\frac{5}{16}\) =
(4 X 16 + 5 by 16) = \(\frac{69}{16}\) and 3\(\frac{3}{8}\) =
(3 X 8 + 3 by 8) = \(\frac{27}{8}\), Now we write
\(\frac{69}{16}\) ÷ \(\frac{27}{8}\) now reciprocal of
the fraction \(\frac{27}{8}\) as \(\frac{8}{27}\) and
multiply as \(\frac{69}{16}\) X \(\frac{8}{27}\) =
\(\frac{69 X 8}{16 X 27}\) = \(\frac{552}{432}\),
we can further simplify as both goes in 24, 24 X 23 = 552 and 24 X 18 = 432,
(23,18)=\(\frac{23}{18}\) as numerator is greater
we write as (1 X 18 + 5 by 18) = 1\(\frac{5}{18}\).
Therefore 4\(\frac{5}{16}\) ÷ 3\(\frac{3}{8}\) = \(\frac{23}{18}\) or 1\(\frac{5}{18}\).

Question 30.
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 212
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 212= \(\frac{16}{15}\) or 1\(\frac{1}{15}\)
Explanation:
Given expressions as 6\(\frac{2}{9}\) ÷ 5\(\frac{5}{9}\),
First we write mixed fractions into fractions as 6\(\frac{2}{9}\) =
(6 X 9 + 2 by 9) = \(\frac{56}{9}\) and 5\(\frac{5}{6}\) =
(5 X 6 + 5 by 6) = \(\frac{35}{6}\), Now we write
\(\frac{56}{9}\) ÷ \(\frac{35}{6}\) now reciprocal of
the fraction \(\frac{35}{6}\) as \(\frac{6}{35}\) and
multiply as \(\frac{56}{9}\) X \(\frac{6}{35}\) =
\(\frac{56 X 6}{9 X 35}\) = \(\frac{336}{315}\),
we can further simplify as both goes in 21, 21 X 16 = 336 and 21 X 15 = 315,
(16,15)=\(\frac{16}{15}\) as numerator is greater
we write as (1 X 15 + 1 by 15) = 1\(\frac{1}{15}\), therefore
6\(\frac{2}{9}\) ÷ 5\(\frac{5}{9}\) = \(\frac{16}{15}\) or 1\(\frac{1}{15}\).

Question 31.
YOU BE THE TEACHER
Your friend finds the quotient of 3\(\frac{1}{2}\) and 1\(\frac{2}{3}\). Is your friend correct? Explain your reasoning.
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 213
No, Friend is in correct, As 3\(\frac{1}{2}\) ÷ 1\(\frac{2}{3}\) =
\(\frac{21}{10}\) or 2\(\frac{1}{10}\) ≠ 8\(\frac{3}{4}\)

Explanation:
Given expressions are 3\(\frac{1}{2}\) ÷ 1\(\frac{2}{3}\),
First we write mixed fractions into fractions as 3\(\frac{1}{2}\) =
(3 X 2 + 1 by 2) = \(\frac{7}{2}\) and 1\(\frac{2}{3}\) =
(1 X 3 + 2 by 3) = \(\frac{5}{3}\), Now we write
\(\frac{7}{2}\) ÷ \(\frac{5}{3}\) now reciprocal of
the fraction \(\frac{5}{3}\) as \(\frac{3}{5}\) and
multiply as \(\frac{7}{2}\) X \(\frac{3}{5}\) =
\(\frac{7 X 3}{2 X 5}\) = \(\frac{21}{10}\),
as numerator is greater we write as (2 X 10 + 1 by 10) = 2\(\frac{1}{10}\), therefore
3\(\frac{1}{2}\) ÷ 1\(\frac{2}{3}\) = \(\frac{21}{10}\) or
2\(\frac{1}{10}\) ≠ 8\(\frac{3}{4}\) so friend is incorrect.

Question 32.
PROBLEM SOLVING
A platinum nugget weighs 3\(\frac{1}{2}\) ounces. How many \(\frac{1}{4}\) ounce pieces can be cut from the nugget?
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 214
14 pieces of \(\frac{1}{4}\) ounce can be cut from the nugget
which weighs 3\(\frac{1}{2}\) ounces

Explanation:
Given a platinum nugget weighs 3\(\frac{1}{2}\) ounces,
number of pieces of \(\frac{1}{4}\) ounce pieces can be
cut from the nugget are 3\(\frac{1}{2}\) ÷ \(\frac{1}{4}\) =
First we write mixed fraction into fraction as 3\(\frac{1}{2}\) =
(3 X 2 + 1 by 2) = \(\frac{7}{2}\), Now
\(\frac{7}{2}\) ÷ \(\frac{1}{4}\) now reciprocal of
the fraction \(\frac{1}{4}\) as \(\frac{4}{1}\) and
multiply as \(\frac{7}{2}\) X \(\frac{4}{1}\) =
\(\frac{7 X 4}{2 X 1}\) = \(\frac{28}{2}\),we can further simplify
as both goes in 2, 2 X 14 = 28 and 2 X 1 = 2, (14,1)= 14, therefore
14 pieces of \(\frac{1}{4}\) ounce can be cut from the nugget
which weighs 3\(\frac{1}{2}\) ounces.

ORDER OF OPERATIONS
Evaluate the expression. Write the answer in simplest form.

Question 33.
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 215
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 215= 3
Explanation:
Given expression as (3 ÷ 1\(\frac{1}{5}\))+ \(\frac{1}{2}\)
First we write mixed fraction into fraction as 1\(\frac{1}{5}\) =
(1 X 5 + 1 by 5) = \(\frac{6}{5}\) Now we calculate first
3 ÷ \(\frac{6}{5}\)  now reciprocal of
the fraction \(\frac{6}{5}\) as \(\frac{5}{6}\) and
multiply as 3 X \(\frac{5}{6}\) =
\(\frac{3 X 5}{1 X 6}\) = \(\frac{15}{6}\),as both goes in 3,
3 X 5 = 15 and 3 X 2 = 6, (5,2)= \(\frac{5}{2}\), now we add with
\(\frac{1}{2}\), \(\frac{5}{2}\) + \(\frac{1}{2}\) as
both have same denominators we add numerators as (5 + 1) and write as
\(\frac{6}{2}\) now as both goes in 2, 2 X 3 = 6, 2 X 1 = 2, (3,1) = 3,
therefore (3 ÷ 1\(\frac{1}{5}\))+ \(\frac{1}{2}\) = 3.

Question 34.
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 216
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 216= \(\frac{5}{3}\) = 1\(\frac{2}{3}\)

Given expression as (4\(\frac{2}{3}\) – 1\(\frac{1}{3}\)) ÷ 2
First we write mixed fractions into fractions as 4\(\frac{2}{3}\) =
(4 X 3 + 2 by 3) = \(\frac{14}{3}\) and 1\(\frac{1}{3}\) =
(1 X 3 + 1 by 3) = \(\frac{4}{3}\), Now first we calculate
\(\frac{14}{3}\) – \(\frac{4}{3}\)  as denominators are same
numerators become (14 – 4 ) = 10, we get  \(\frac{10}{3}\)
now \(\frac{10}{3}\)  ÷ 2 we write 2
as reciprocal and multiply \(\frac{10}{3}\) X \(\frac{1}{2}\) =
\(\frac{10 X 1}{3 X 2}\) = \(\frac{10}{6}\),
we can further simplify as both goes in 2, 2 X 5 = 10 and 2 X 3 = 6,
(5,3)=\(\frac{5}{3}\) as numerator is greater
we write as (1 X 3 + 2 by 3) = 1\(\frac{2}{3}\), therefore
(4\(\frac{2}{3}\) – 1\(\frac{1}{3}\)) ÷ 2 = \(\frac{5}{3}\) or 1\(\frac{2}{3}\).

Question 35.
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 217
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 217= 3
Explanation
Given expression as \(\frac{2}{5}\) + (2\(\frac{1}{6}\) ÷ \(\frac{1}{6}\)) First we write mixed fraction into fraction as 2\(\frac{1}{6}\) =
(2 X 6 + 1 by 6) = \(\frac{13}{6}\),Now first we calculate
\(\frac{13}{6}\) ÷ \(\frac{5}{6}\)  we write \(\frac{5}{6}\)
as reciprocal and multiply \(\frac{13}{6}\) X \(\frac{6}{5}\) =
\(\frac{13 X 6}{6 X 5}\) = \(\frac{78}{30}\),
we can further simplify as both goes in 6, 6 X 13 = 78 and 6 X 5 = 30,
(13,5)=\(\frac{13}{5}\), Now we add as
\(\frac{2}{5}\) + \(\frac{13}{5}\) as both have same denominators we add numerators and write as \(\frac{15}{5}\), now as both goes in 5,
5 X 3= 15, 5 X 1 = 5, (3, 1) = 3, therefore \(\frac{2}{5}\) +
(2\(\frac{1}{6}\) ÷ \(\frac{1}{6}\)) = 3.

Question 36.
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 218
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 218= \(\frac{4}{3}\) or 1\(\frac{1}{3}\)
Explanation:
Given expression as (5\(\frac{5}{6}\) ÷ 3\(\frac{3}{4}\)) – \(\frac{2}{9}\),First we write mixed fractions into fractions as 5\(\frac{5}{6}\) =
(5 X 6 + 5 by 6) = \(\frac{35}{6}\) and 3\(\frac{3}{4}\)=
(3 X 4 + 3 by 4) = \(\frac{15}{4}\). Now first we calculate
\(\frac{35}{6}\) ÷ \(\frac{15}{4}\)  we write \(\frac{15}{4}\)
as reciprocal and multiply \(\frac{35}{6}\) X \(\frac{4}{15}\) =
\(\frac{35 X 4}{6 X 15}\) = \(\frac{140}{90}\),
we can further simplify as both goes in 10, 10 X 14 = 140 and 10 X 9 = 90,
(14,9)=\(\frac{14}{9}\), Now we subtract as \(\frac{14}{9}\) – \(\frac{2}{9}\) as both have same denominators are same we subtract numerators and write as
\(\frac{12}{9}\) now as both goes in 3,
3 X 4= 12, 3 X 3 = 9, (4, 3) = \(\frac{4}{3}\) as numerator is greater
we write as (1 X 3 + 1 by 3), Therefore (5\(\frac{5}{6}\) ÷ 3\(\frac{3}{4}\)) – \(\frac{2}{9}\) = \(\frac{4}{3}\) or 1\(\frac{1}{3}\).

Question 37.
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 219
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 219= \(\frac{165}{26}\) or 6\(\frac{9}{26}\)
Explanation:
Given expression as 6\(\frac{1}{2}\) – (\(\frac{7}{8}\) ÷ 5\(\frac{11}{16}\)), First we write mixed fraction into fraction as 5\(\frac{11}{16}\) =
(5 X 16 + 11 by 16) = \(\frac{91}{16}\), Now first we calculate
\(\frac{7}{8}\) ÷ \(\frac{91}{16}\), we write \(\frac{91}{16}\)
as reciprocal and multiply \(\frac{7}{8}\) X \(\frac{16}{91}\) =
\(\frac{7 X 16}{8 X 91}\) = \(\frac{112}{728}\),
we can further simplify as both goes in 28, 28 X 4 = 112 and 28 X 26 = 728,
(4,26)=\(\frac{4}{26}\), We write mixed fraction into fraction of 6\(\frac{1}{2}\) as (6 X 2 + 1 by 2) = \(\frac{13}{2}\) Now we subtract as \(\frac{13}{2}\) – \(\frac{4}{26}\) as both should have same denominator we multiply numerator and denominator by 13 for \(\frac{13}{2}\) X \(\frac{13}{13}\) =
\(\frac{13 X 13}{2 X 13}\) = \(\frac{169}{26}\) now we subtract as
\(\frac{169}{26}\) – \(\frac{4}{26}\)as denominators are same we
subtract numerators (169 – 4 ) and write as \(\frac{165}{26}\),
as numerator is greater we write as (6 X 26 + 9 by 26) = 6\(\frac{9}{26}\).
Therefore 6\(\frac{1}{2}\) – (\(\frac{7}{8}\) ÷ 5\(\frac{11}{16}\)) =
\(\frac{165}{26}\) or 6\(\frac{9}{26}\).

Question 38.
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 220
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 220= \(\frac{11}{10}\) or 1\(\frac{1}{10}\)
Explanation:
Given expression as 9\(\frac{1}{6}\) ÷ (5 + 3\(\frac{1}{3}\)),
First we write mixed fraction into fraction for 3\(\frac{1}{3}\) as
(3 X 3 + 1 by 3) = \(\frac{10}{3}\), Now first we calculate
5 + \(\frac{10}{3}\) as (5 X 3 +10 by 3) = \(\frac{25}{3}\),
We write mixed fraction into fraction of 9\(\frac{1}{6}\)
as (9 X 6 + 1 by 6) = \(\frac{55}{6}\) now we divide with \(\frac{25}{3}\),
now we write as reciprocal and multiply \(\frac{55}{6}\) X \(\frac{3}{25}\) =
\(\frac{55 X 3}{6 X 25}\) = \(\frac{165}{150}\),
we can further simplify as both goes in 15, 15 X 11 = 165 and 15 X 10 = 150,
(11,10)=\(\frac{11}{10}\), as numerator is greater we write as
(1 X 10 + 1 by 10) = 1\(\frac{1}{10}\). Therefore 9\(\frac{1}{6}\) ÷
(5 + 3\(\frac{1}{3}\)) = \(\frac{11}{10}\) or 1\(\frac{1}{
10}\).

Question 39.
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 221
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 221= \(\frac{66}{5}\) or 13\(\frac{1}{5}\)

Given expression as 3\(\frac{3}{5}\) + (4\(\frac{4}{15}\) ÷
\(\frac{4}{9}\)), First we write mixed fraction into fraction
for 4\(\frac{4}{15}\) as
(4 X 15 + 4 by 15) = \(\frac{64}{15}\), Now first we calculate
\(\frac{64}{15}\) ÷ \(\frac{4}{9}\)now we write as reciprocal
and multiply \(\frac{64}{15}\) X \(\frac{9}{4}\) =
\(\frac{64 X 9}{15 X 4}\) = \(\frac{576}{60}\),
we can further simplify as both goes in 12, 12 X 48 = 576 and 12 X 5 = 60,
(48, 5)=\(\frac{48}{5}\), we write mixed fraction into fraction for
3\(\frac{3}{5}\) as (3 X 5 + 3 by 5) = \(\frac{18}{5}\)
Now \(\frac{18}{5}\) + \(\frac{48}{5}\)  as both have
same denominators we add numerators and write as \(\frac{66}{5}\)
as numerator is greater we write as (13 X 5 + 1 by 5) = 13\(\frac{1}{5}\).
Therefore 3\(\frac{3}{5}\) + (4\(\frac{4}{15}\) ÷ \(\frac{4}{9}\)) =
\(\frac{66}{5}\) or 13\(\frac{1}{5}\).

Question 40.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 222
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 222= \(\frac{7}{54}\)

Given expression as (\(\frac{3}{5}\) X \(\frac{7}{12}\)) ÷
2\(\frac{7}{10}\), First we multiply \(\frac{3}{5}\) X
\(\frac{7}{12}\) = \(\frac{3 X 7}{5 X 12}\) = \(\frac{21}{60}\),
we can further simplify as both goes in 3, 3 X 7 = 21 and 3 X 20 = 60,
(7, 20)=\(\frac{7}{20}\), Now we have \(\frac{7}{20}\) ÷ 2\(\frac{7}{10}\), we write mixed fraction into fraction for 2\(\frac{7}{10}\) as
(2 X 10 + 7 by 10) = \(\frac{27}{10}\) now we write reciprocal
and multiply \(\frac{7}{20}\) X \(\frac{10}{27}\) =
\(\frac{7 X 10}{20 X 27}\) = \(\frac{70}{540}\) as both goes in 10,
10 X 7 = 70 and 10 X 54 = 540, (7,54) =  \(\frac{7}{54}\), Therefore
(\(\frac{3}{5}\) X \(\frac{7}{12}\)) ÷ 2\(\frac{7}{10}\) = \(\frac{7}{54}\).

Question 41.
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 223
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 223= \(\frac{10}{3}\) or 3\(\frac{1}{3}\)

Explanation:
Given expression as (4\(\frac{3}{8}\) ÷ \(\frac{3}{4}\)) X \(\frac{4}{7}\), First we write mixed fraction into fraction for 4\(\frac{3}{8}\) as
(4 X 8 + 3 by 8) = \(\frac{35}{8}\), now we divide with \(\frac{3}{4}\),
now we write as reciprocal and multiply \(\frac{35}{8}\) X \(\frac{4}{3}\) =
\(\frac{35 X 4}{8 X 3}\) = \(\frac{140}{24}\),
we can further simplify as both goes in 4, 4 X 35 = 140 and 4 X 6 = 24,
(35,6)=\(\frac{35}{6}\), now we multiply with \(\frac{35}{6}\) X \(\frac{4}{7}\) = \(\frac{35 X 4}{6 X 7}\) = \(\frac{140}{42}\),
as both goes in 14, 14 X 10 = 140, 14 X 3 = 42, (10, 3) = \(\frac{10}{3}\),
as numerator is greater we write (3 X 3 + 1 by 3) = 3\(\frac{1}{3}\),
therefore (4\(\frac{3}{8}\) ÷ \(\frac{3}{4}\)) X \(\frac{4}{7}\) =
\(\frac{10}{3}\) or 3\(\frac{1}{3}\).

Question 42.
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 224
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 224= \(\frac{25}{2}\) = 12\(\frac{1}{2}\)
Explanation:
Given expression as (1\(\frac{9}{11}\) X 4\(\frac{7}{12}\)) ÷ \(\frac{2}{3}\), First we write mixed fraction into fraction for 1\(\frac{9}{11}\) as
(1 X 11 + 9 by 11) = \(\frac{20}{11}\) and 4\(\frac{7}{12}\) as
(4 X 12 + 7 by 12) = \(\frac{55}{12}\) now we calculate first
\(\frac{20}{11}\) X \(\frac{55}{12}\) = \(\frac{20 X 55}{11 X 12}\) =
\(\frac{1100}{132}\) we further simply as both goes in 44,
44 X 25 = 1100, 44 X 3 = 132, (25, 3) = \(\frac{25}{3}\),
Now we calculate as \(\frac{25}{3}\) ÷ \(\frac{2}{3}\)
now we write as reciprocal and multiply \(\frac{25}{3}\) X \(\frac{3}{2}\) =
\(\frac{25 X 3}{3 X 2}\) = \(\frac{75}{6}\), as both goes in 3,
3 X 25 = 75, 3 X 2 = 6, (25, 2) = \(\frac{25}{2}\)
as numerator is greater we write (12 X 2 + 1 by 2) = 12\(\frac{1}{2}\),
therefore (1\(\frac{9}{11}\) X 4\(\frac{7}{12}\)) ÷ \(\frac{2}{3}\) =
\(\frac{25}{2}\) = 12\(\frac{1}{2}\).

Question 43.
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 225
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 225= \(\frac{7}{108}\)

Explanation:
Given expression as 3\(\frac{4}{15}\) ÷ (8 X 6\(\frac{3}{10}\)),
First we write mixed fraction into fraction for 6\(\frac{3}{10}\) as
(6 X 10 + 3 by 10) = \(\frac{63}{10}\) now we calculate first
8 X \(\frac{63}{10}\) = \(\frac{8 X 63}{10}\) =
\(\frac{504}{10}\) we further simply as both goes in 2,
2 X 252 = 504, 2 X 5 = 10, (252, 5) = \(\frac{252}{5}\),
Now 3\(\frac{4}{15}\) = ( 3 X 15 + 4 by 15) = \(\frac{49}{15}\)
Now we calculate as \(\frac{49}{15}\) ÷ \(\frac{252}{5}\)
now we write as reciprocal and multiply \(\frac{49}{15}\) X \(\frac{5}{252}\) =
\(\frac{49 X 5}{15 X 252}\) = \(\frac{245}{3780}\), as both goes in 35,
35 X 7 = 245, 35 X 108 = 3780, (7, 108) = \(\frac{7}{108}\).
Therefore 3\(\frac{4}{15}\) ÷ (8 X 6\(\frac{3}{10}\)) = \(\frac{7}{108}\).

Question 44.
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 226
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 226= \(\frac{11}{35}\)

Explanation:
Given expression as 2\(\frac{5}{14}\) ÷ (2\(\frac{5}{8}\) X
1\(\frac{3}{7}\)),First we write mixed fraction into fraction
for 2\(\frac{5}{8}\) as
(2 X 8 + 5 by 8) = \(\frac{21}{8}\) and 1\(\frac{3}{7}\) =
(1 X 7 + 3 by 7) = \(\frac{10}{7}\) now we calculate first
\(\frac{21}{8}\) X \(\frac{10}{7}\) = \(\frac{21 X 10}{8 X 7}\) =
\(\frac{210}{56}\) we further simply as both goes in 2,
2 X 105 = 210, 2 X 28 = 56, (105, 28) = \(\frac{105}{28}\),
we write mixed fraction as fraction 2\(\frac{5}{14}\) as
(2 X 14 + 5 by 14) = \(\frac{33}{14}\),Now we calculate as
\(\frac{33}{14}\) ÷ \(\frac{105}{28}\), now we write as reciprocal and multiply \(\frac{33}{14}\) X \(\frac{28}{105}\) =
\(\frac{33 X 28}{14 X 105}\) = \(\frac{462}{1470}\), as both goes in 42,
42 X 11 = 462, 42 X 35 = 1470, (11, 35) = \(\frac{11}{35}\).
Therefore 2\(\frac{5}{14}\) ÷ (2\(\frac{5}{8}\) X 1\(\frac{3}{7}\)) =
\(\frac{11}{35}\).

Question 45.
LOGIC
Your friend uses the model shown to state that Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 227. Is your friend correct? Justify your answer using the model.
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 227.1
No, friend is incorrect,

Explanation :
Given expression as 2\(\frac{1}{2}\) ÷ 1\(\frac{1}{6}\),
First we write mixed fraction into fraction for 2\(\frac{1}{2}\) as
(2 X 2 + 1 by 2) = \(\frac{5}{2}\) and 1\(\frac{1}{6}\) =
(1 X 6 + 1 by 6) = \(\frac{7}{6}\) now we wrtie reciprocal and multiply
\(\frac{5}{2}\) X \(\frac{6}{7}\) = \(\frac{5 X 6}{2 X 7}\) =
\(\frac{30}{14}\) further we  can simplify as both goes in 2, 2 X 15 = 30,
2 x 7 = 14, (15, 7), \(\frac{30}{14}\) =  \(\frac{15}{7}\) as numerator
is greator we write as (2 X 7 + 1 by 7) = 2\(\frac{1}{7}\) and
friend says it is 2\(\frac{1}{6}\) which is incorrect and model is showing
2, 1\(\frac{1}{6}\) and one piece is remaining but we
are getting 2\(\frac{1}{7}\) which is not matching, So freind is incorrect.

Question 46.
MODELING REAL LIFE
A bag contains 42 cups of dog food. Your dog eats 2\(\frac{1}{3}\) cups of dog food each day. Is there enough food to last 3 weeks? Explain.
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 228
No, there is not enough food to last for 3 weeks,

Explanation:
Given dog eats 2\(\frac{1}{3}\) cups of dog food each day,
now for 3 weeks means 3 X 7 X 2\(\frac{1}{3}\) , We write
mixed fraction 2\(\frac{1}{3}\)  as ( 2 X 3 + 1 by 3) =\(\frac{7}{3}\) =
3 X 7 X \(\frac{7}{3}\) = \(\frac{3 X 7 X 7}{3}\) = \(\frac{147}{3}\),
as both goes in 3, 3 X 49, 3 X 1 = 3,(49, 1) = \(\frac{147}{3}\) = 49.
We need 49 cups of dog food and bag contains 42 cups so shortage of 7 cups,
Therefore we do not have enough food to last for 3 weeks.

Question 47.
DIG DEEPER!
You have 12 cups of granola and 8\(\frac{1}{2}\) cups of peanuts to make trail mix. What is the greatest number of full batches of trail mix you can make? Explain how you found your answer.
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 229
4 is the greatest number of full batches of trail mix I can make,

Explanation:
Given I have 12 cups of granola and 8\(\frac{1}{2}\) cups of
peanuts to make trail mix, trail mix has 2\(\frac{3}{4}\) cups of granola
and 1\(\frac{1}{3}\) cups of peanuts,
First granola mix 12 ÷ 2\(\frac{3}{4}\), We write mixed fraction
as fraction as (2 X 4 + 3 by 4) = \(\frac{11}{4}\) now we write reciprocal
and multiply as 12 X \(\frac{4}{11}\) = \(\frac{12 X 4}{11}\) =
\(\frac{48}{11}\) as numerator is greater we write as
( 4 X 11 + 4 by 11) = 4\(\frac{4}{11}\), Now peanuts mix
8\(\frac{1}{2}\) ÷ 1\(\frac{1}{3}\),We write mixed fractions
as fractions as (8 X 2 + 1 by 2) = \(\frac{17}{2}\) and 1\(\frac{1}{3}\) =
(1 X 3 + 1 by 3) = \(\frac{4}{3}\), Now we write reciprocal and multiply
\(\frac{17}{2}\) X \(\frac{3}{4}\) = \(\frac{17 X 3}{2 X 4}\) =
\(\frac{51}{8}\), as numerator is greater we write as (6 X 8 + 3 by 8)=
6\(\frac{3}{8}\). Therefore I can make 4 full batches.

Question 48.
REASONING
At a track and field meet, the longest shot-put throw by a boy is 25 feet 8 inches. The longest shot-put throw by a girl is 19 feet 3 inches. How many times greater is the longest shot-put throw by the boy than by the girl?

1.33 times greater is the longest shot-put throw by the boy than by the girl,

Explanation:
Given at a track and field meet, the longest shot-put
throw by a boy is 25 feet 8 inches. The longest shot-put throw
by a girl is 19 feet 3 inches.
As we know 1 feet is equal to 12 inches we convert first into inches as
Boy throw = 25 X 12 + 8 = 308 inches
Girl throw = 19 X 12 +  3 = 228 + 3 = 231 inches.
Now  boy record to girl record is \(\frac{308}{231}\) = 1.3333,
Therefore 1.33 times greater is the longest shot-put throw by
the boy than by the girl.

Lesson 2.4 Adding and Subtracting Decimals

EXPLORATION 1
Using Number Lines
Work with a partner. Use each number line to find A + B and B – A. Explain how you know you are correct.
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 230
a. A + B = 1.1 and  B – A = 0.5
b. A + B = 2.24 and B – A = 0.02
c. A + B = 7.6 and B – A = 0.8
d. A + B = 5.1 and B – A = 0.8

Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals-50
a. A + B = 1.1 and  B- A = 0.5
Explanation:
a. Given A is at 0.3 and B is at 0.8,
now A + B = 0.3 + 0.8 = 1.1,
B – A = 0.8 – 0.3 = 0.5

b. Given A is at 1.11and B is at 1.13,
A + B = 1.11 + 1.13 = 2.24
B – A = 1.13 – 1.11 = 0.02

c. Given A is at 3.4 and b is at 4.2
A + B = 3.4  + 4.2 = 7.6
A – B = 4.2 – 3.4 = 0.8

d. Given A is at 2.15 and b is at 2.95
A + B = 2.15 + 2.95 = 5.1
A – B = 2.95 – 2.15 = 0.8

Explained in the picture above.

EXPLORATION 2
Work with a partner. Explain how you can use the place value chart below to add and subtract decimals beyond hundredths. Then find each sum or difference.
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 231
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 232
a. 16.05 + 2.945
b. 7.421 + 8.058
c. 38.72 – 8.618
d. 64.968 – 51.167
e. 225.1 + 85.0465
f. 1107.20592 – 102.3056
a. 16.05 + 2.945 = 18.995
b. 7.421 + 8.058 = 15.479
c. 38.72 – 8.618 = 30.102
d. 64.968 – 51.167 = 13.801
e. 225.1 + 85.0465 = 310.1465
f. 1107.20592 – 102.3056 = 1004.90032
Explanation:
a. 16.05 + 2.945 = 18.995
Here we have thousandths value as 0.000 + 0.005 =0.005
and wrote as below shown at thousandths place,
Big-Ideas-Math-Book-6th-Grade-Answer-Key-Chapter-2-Fractions-and-Decimals-51

b. 7.421 + 8.058 = 15.479
Here we have thousandths value as 0.001 + 0.0008 = 0.0009 and
wrote as below shown at thousandths place, 0.0009,
Big-Ideas-Math-Book-6th-Grade-Answer-Key-Chapter-2-Fractions-and-Decimals-52
c. 38.72 – 8.618 = 30.102
Here we have thousandths s value as 0.000 here we take 1
from tenths place and we get 10 at thousandths place and
subtract 0.008 we get 0.002 at thousandths place
and wrote as below shown at thousandths place,
Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals-53

d. 64.968 – 51.167 = 13.801
Here we have thousandths s value as 0.008 and
subtract 0.007 we get 0.001 at thousandths place
and wrote as below shown at thousandths place,
Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals-54e. 225.1 + 85.0465 = 310.1465
Here we have thousandths value as 0.000 + 0.005 =0.005
and wrote as below shown at thousandths place,
Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals-55
f. 1107.20592 – 102.3056 = 1004.90032
Here we have ten thousandths value as 0.0002 – 0.0000 =0.00002
and thousandths values as 0.0009 – 0.0006 = 0.0003,
wrote as below shown at thousandths places,
Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals-56

2.4 Lesson

Key Idea
Adding and Subtracting Decimals
To add or subtract decimals, write the numbers vertically and line up the decimal points. Then bring down the decimal point and add or subtract as you would with whole numbers.

Try It

Add.

Question 1.
4.206 + 10.85
4.206
(+) 10.850 
1
15.056     

Explanation:
4.206
(+) 10.850 
1
15.056     

here we add according to their unit place values respectively.

Question 2.
15.5 + 8.229
15.500
(+)  8.229
       1
     23.729   
Explanation:
15.500
(+)  8.229
     1        


     23.729   
here we add according to their unit place values respectively.

Question 3.
78.41 + 90.99
78.41
(+)90.99
1   11
169.40
Explanation:
78.41
(+)90.99
1   11
169.40   

here we add according to their unit place values respectively.

Subtract

Question 4.
6.34 – 5.33
6.34
(-)5.33
1.01     

Explanation:
6.34
(-)5.33
1.01    

here we subtract according to their unit place values respectively.

Question 5.
27.9 – 0.905
6,18,9,10
27.900
(-)  0.905
26.995
Explanation:
  6,18,9,10
27.900
(-)  0.905
26.995
here we subtract according to their unit place values respectively.

Question 6.
18.626 – 13.88
7,15,12
18.626
(-)13.880
4.746
Explanation:
7,15,12
18.626
(-)13.880
4.746
here we subtract according to their unit place values respectively.

Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals-57
Try It
Question 7.
WHAT IF?
The execution score is adjusted and has 1.467 in deductions. What is the gymnast’s score?

The gymnast’s score is 15.133

Explanation:
Evaluating the expression to solve the problem,
= 6.9 + (10 – 1.467) -0.3,
= 6.9 + (8.533) – 0.3
= 15.433 – 0.3 = 15.133,
therefore the gymnast’s score is 15.133.

Self-Assessment for Concepts & Skills
Solve each exercise. Then rate your understanding of the success criteria in your journal.

ADDING AND SUBTRACTING DECIMALS
Evaluate the expression.

Question 8.
23.557 – 17.601 + 5.216

23.557 – 17.601 + 5.216 = 11.172
Explanation:
1,12,15
2 3 . 5 5  7
(-) 1 7.  6 0  1
5.   9  5  6
Now add 5.956 with 5.216
5.956
(+)  5.216 
1    1
11.172
Therefore we get 23.557 – 17.601 + 5.216 = 11.172.

Question 9.
16.5263 + 12.404 – 11.73

16.5263 + 12.404 – 11.73 = 17.2003
Explanation:
16.5263
(+)12.4040
           1      
     28.9303
Now we subtract 11.73 from 28.9303
28.9303
(-)11.7300 
    17.2003
Therefore we get 16.5263 + 12.404 – 11.73 = 17.2003.

Question 10.
CHOOSE TOOLS
Why is it helpful to estimate the answer before adding or subtracting decimals?

Estimation is useful when we don’t need an exact answer.
It also lets you check to be sure an exact answer is close
to being correct. Estimating with decimals works
just the same as estimating with whole numbers.
When rounding the values to be added, subtracted,
multiplied, or divided, it helps to round to numbers
that are easy to work with.

Question 11.
WRITING
When adding or subtracting decimals, how can you be sure to add or subtract only digits that have the same place value?
When adding and subtracting decimals we use few steps as below:
1) We line up the place values of the numbers by lining up the decimals.
2) We Add in filler zeros if needed.
3) Adding or Subtracting the numbers in the same place value positions.
Examples
Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals-58

Question 12.
OPEN-ENDED
Describe two real-life examples of when you would need to add and subtract decimals.

We deal with decimal addition and subtraction in everyday
life while dealing with:
Money
Measurements (Length, Mass, Capacity)
Temperature

Decimals can be added or subtracted in the same way as we add or subtract whole numbers.

Example:
a. Kate had $ 368.29. Her mother gave her $ 253.46 and her
sister gave her $ 57.39. How much money does she has now?
Answer:
Money Kate had                                        =             $ 368.29
Money gave her mother                           =              $ 253.46
Money gave her sister                              =        +   $   57.39
Total money she has now                         =             $ 679.14

b. Kylie had 25 m of ribbon. She uses 8 m and 13 cm to
decorate a skirt. How much ribbon is remaining with Kylie?
Answer:
Length of ribbon Kylie had = 25 m = 25.00
Length of ribbon used by Kylie = 8 m 13 cm = 8.13
The remaining length of ribbon = 25.00 – 8.13 = 16.87

Big-Ideas-Math-Book-6th-Grade-Answer-Key-Chapter-2-Fractions-and-Decimals-60

Question 13.
STRUCTURE
You add 3.841 + 29.999 as shown. Describe a method for adding the numbers using mental math. Which method do you prefer? Explain.
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 232.1
I prefer the way as we add or subtract whole numbers as shown above
than the mental math method.

Explanation:
Rounding is a mental math strategy for adding and subtracting numbers.
When you round, you will likely need to adjust your answer to get
the exact answer.
For Example:
23 + 58 can be rounded to 20 + 60 = 80.
23 is 3 more than 20 and 58 is 2 less than 60.(3 -2)
So adjust answer by adding 1. (80 + 1), Answer is 81.
Example:
76 – 40 can be rounded to 80 – 40 = 40.
76 is 4 less than 80. So adjust answer by subtracting 4.
(40- 4) , Answer is 36.

Question 14.
OPEN-ENDED
Write three decimals that have a sum of 27.905.
The three decimals are 15.812, 4.670, 7.423 to have a sum of 27.905

Explanation:
We take any decimals randomly as 15.812, 4.670, 7.423,
we have sum of 27.905 as
15.812
4.670
(+)7.423
  11  1
27.905
Therefore, the three decimals are 15.812, 4.670, 7.423
to have a sum of 27.905.

Self-Assessment for Problem Solving
Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 15.
A field hockey field is rectangular. Its width is 54.88 meters, and its perimeter is 289.76 meters. Find the length of the field.
The length of hockey field is 90 meters

Explanation:
Given a field hockey field is rectangular. Its width is 54.88 meters
and its perimeter is 289.76 meters.
We know perimeter of rectangle is 2( length + width),
So, 289.76 = 2(length + 54.88)
length + 54.88 = \(\frac{289.76}{2}\),
length + 54.88 = 144.88,
length = 144.88 – 54.88 = 90 meters,
Therefore the length of hockey field is 90 meters.

Question 16.
DIG DEEPER!
You mix 23.385 grams of sugar and 12.873 grams of baking soda
in a glass container for an experiment. You place the container on a
scale to find that the total mass is 104.2 grams.
What is the mass of the container?

The mass of the container is 67.942 grams

Explanation:
Given I mix 23.385 grams of sugar and 12.873 grams of baking soda
in a glass container for an experiment. You place the container on a
scale to find that the total mass is 104.2 grams. So the mass of the container
is 104.2 – (23.385 + 12.873) =104.2 – 36.258 = 67.942 grams.
Therefore the mass of the container is 67.942 grams.

Question 17.
One molecule of water is made of two hydrogen atoms and one oxygen atom. The masses (in atomic mass units) for one atom of hydrogen and oxygen are shown. What is the mass (in atomic mass units) of one molecule of water?
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 232.2
The mass of one molecule of water is 18.0148 amu
(amu = atomic mass units)

Explanation:
Given One molecule of water is made of two hydrogen
atomsand one oxygen atom. The mass of one atom of
hydrogen is 1.0079 amu and oxygen is 15.999 amu.
So the mass of one molecule of water is (2 X 1.0079) + 15.999,
= 2.0158 + 15.999 = 18.0148 amu, therefore the mass of
one molecule of water is 18.0148 amu.

Adding and Subtracting Decimals Homework & Practice 2.4

Review & Refresh

Divide. Write the answer in simplest form.

Question 1.
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 233
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 233= \(\frac{13}{3}\) = 4\(\frac{1}{3}\)
Explanation:
Given expressions as 3\(\frac{1}{4}\) ÷ \(\frac{3}{4}\),
First we write mixed fraction into fraction 3\(\frac{1}{4}\) as
(3 X 4 + 1 by 4) = \(\frac{13}{4}\),Now \(\frac{3}{4}\) we
write as reciprocal and multiply \(\frac{4}{3}\) as
\(\frac{13}{4}\) X \(\frac{4}{3}\) = \(\frac{13 X 4}{4 X 3}\) =
\(\frac{52}{12}\) as both goes in 4, 4 X 13 = 52,
4 X 3 = 12, (13, 3) = \(\frac{13}{3}\) as numerator is greater
we write as (4 X 3 + 1 by 3)= 4\(\frac{1}{3}\).
Therefore 3\(\frac{1}{4}\) ÷ \(\frac{3}{4}\) = \(\frac{13}{3}\) = 4\(\frac{1}{3}\).

Question 2.
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 234
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 234= \(\frac{5}{6}\)
Explanation:
Given expressions as 4\(\frac{1}{6}\) ÷ 5
First we write mixed fraction into fraction 4\(\frac{1}{6}\) as
(4 X 6 + 1 by 6) = \(\frac{25}{6}\), Now 5 we
write as reciprocal and multiply \(\frac{1}{5}\) as
\(\frac{25}{6}\) X \(\frac{1}{5}\) = \(\frac{25 X 1}{6 X 5}\) =
\(\frac{25}{30}\) as both goes in 5, 5 X 5 = 25,
5 X 6 = 30, (5, 6) = \(\frac{5}{6}\),
Therefore 4\(\frac{1}{6}\) ÷ 5 = \(\frac{5}{6}\).

Question 3.
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 235
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 235= \(\frac{25}{12}\) = 2\(\frac{1}{12}\)
Explanation:
Given expressions as 6\(\frac{2}{3}\) ÷ 3\(\frac{1}{5}\),
First we write mixed fractions into fractions 6\(\frac{2}{3}\) as
(6 X 3 + 2 by 3) = \(\frac{20}{3}\) and 3\(\frac{1}{5}\) as
(3 X 5 + 1 by 5) = \(\frac{16}{5}\),Now \(\frac{16}{5}\) we
write as reciprocal and multiply \(\frac{5}{16}\) as
\(\frac{20}{3}\) X \(\frac{5}{16}\) = \(\frac{20 X 5}{3 X 16}\) =
\(\frac{100}{48}\) as both goes in 4, 4 X 25 = 100,
4 X 12 = 48, (25, 12) = \(\frac{25}{12}\) as numerator is greater
we write as (2 X 12 + 1 by 12)= 2\(\frac{1}{12}\).
Therefore 6\(\frac{2}{3}\) ÷ 3\(\frac{1}{5}\) = \(\frac{25}{12}\) = 2\(\frac{1}{12}\).

Find the GCF of the numbers.

Question 4.
16, 28, 40

The GCF of 16, 28, 40 is 4

Explanation:
Factors of 16 are 1, 2, 4, 8, 16
Factors of 28 are 1, 2, 4, 7, 14, 28
Factors of 40 are 1, 2, 4, 5, 8, 10, 20, 40
as 4 is the greatest common number in all the three,
therefore GCF (16, 28, 40) is 4.

Question 5.
39, 54, 63

The GCF of 39, 54, 63 is 3

Explanation:
Factors of 39 are 1, 3, 13 and 39
Factors of 54 are 1, 2, 3, 6, 9, 18, 27 and 54
Factors of 63 are 1, 3, 7, 9, 21 and 63
as 3 is the greatest common number in the three,
therefore GCF (39, 54, 63) is 3.

Question 6.
24, 72, 132

The GCF of 24, 72, 132 is

Explanation:
Factors of 24 are 1, 2, 3, 4, 6, 8, 12 and 24
Factors of 72 are 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36 and 72
Factors of 132 are 1, 2, 3, 4, 6, 11, 12, 22, 33, 44, 66 and 132
as 12 is the greatest common number in all the three,
therefore GCF (24, 72, 132) is 12.

Find the value of the power.

Question 7.
112

112 = 1

Explanation:
1 to the power of 12 means we multiply 1 by
12 times as 1 X 1 X 1 X 1 X 1 X 1 X 1 X 1 X 1 X 1 X 1 X 1 = 1.

Question 8.
24

24 = 16

Explanation:
2 to the power of 4 means we multiply 2 by
4 times as 2 X 2 X 2 X 2 = 16.

Question 9.
36

36 = 729

Explanation:
3 to the power of 6 means we multiply 3 by
6 times as 3 X 3 X 3 X 3 X 3 X 3 = 729.

Question 10.
54
54 = 625

Explanation:
5 to the power of 4 means we multiply 5 by
4 times as 5 X 5 X 5 X 5 = 625.

Classify the quadrilateral.

Question 11.
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 236
Square

Explanation:
Quadrilaterals can be classified into Parallelograms,
Squares, Rectangles , Trapezoids and Rhombuses.
As given in the picture above it is a square because,
All sides are equal,
Each angle is a right angle,
Opposite sides are equal.
So, it is a square.

Question 12.
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 237
Trapezoid

Explanation:
Quadrilaterals can be classified into Parallelograms,
Squares, Rectangles , Trapezoids and Rhombuses.
As given in the picture above it is a trapezoid because,
Opposite sides are parallel.
Adjacent angles add up to 180°.
So, it is a Trapezoid.

Question 13.
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 238
Parallelogram

Explanation:
Quadrilaterals can be classified into Parallelograms,
Squares, Rectangles , Trapezoids and Rhombuses.
As given in the picture above it is a parrelogram because,
Opposite sides are parallel.
Opposites sides are equal.
Opposite angles are equal.
All sides are equal,
So it is a parallelogram.

Concepts, Skills, & Problem Solving
USING TOOLS
Use a place value chart to find the sum or difference. (See Exploration 2, p. 67.)

Question 14.
4.63 + 8.547
4.63 + 8.547 = 13.177

Explanation:
Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals-61

Question 15.
3.6257 – 2.98 = 0.6457

Explanation:
Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals-62

Question 16.
14.065 + 13.8542 = 27.9192

Explanation:
Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals-63

ADDING DECIMALS
Add.

Question 17.
7.82 + 3.209
7.82 + 3.209 = 11.029

Explanation:
1,1
7.820
(+)3.209
11.029
We added digits according to their place values.

Question 18.
3.7 + 2.77
3.7 + 2.77 = 6.47

Explanation
1,
3.70
(+)2.77
6.47
We added digits according to their place values.

Question 19.
12.829 + 10.07
12.829 + 10.07 = 22.899

Explanation:
12.829
(+)10.070
22.899
We added digits according to their place values.

Question 20.
20.35 + 13.748
20.35 + 13.748 = 34.098

Explanation:
1
20.350
(+)13.748
34.098
We added digits according to their place unit values.

Question 21.
11.212 + 7.36
11.212 + 7.36 = 18.572

Explanation:
11.212
(+)07.360
18
.572
We added digits according to their place unit values.

Question 22.
14.91 + 2.095
14.91 + 2.095 = 17.005

Explanation:
      1,1
14.910
(+)02.095
17.005
We added digits according to their place unit values. 

Question 23.
31.994 + 8.006
31.994 + 8.006 = 40

Explanation:
1,1,1,1
31.994
(+)08.006
40.000
We added digits according to their place unit values.

Question 24.
3.946 + 6.052
3.946 + 6.052 = 9.998

Explanation:
3.946
(+)6.052
9.998
We added digits according to their place unit values.

Question 25.
41.226 + 102.774
41.226 + 102.774 = 144

Explanation:
1,1,1,1
41.226
(+)102.774
144.000
We added digits according to their place unit values.

Question 26.
122.781 + 19.228
122.781 + 19.228 = 142.009

Explanation:
   1,1,1,
122.781
(+)   19.228
142.009
We added digits according to their place unit values.

Question 27.
17.440 + 12.497
17.440 + 12.497 = 29.937

Explanation:
1,
17.440
(+)12.497
29.937
We added digits according to their place unit values.

Question 28.
15.255 + 19.058
15.255 + 19.058 = 34.313

Explanation:
1,  1,1
15.255
(+)19.058
34.313
We added digits according to their place unit values.

SUBTRACTING DECIMALS
Subtract.

Question 29.
4.58 – 3.12
4.58 – 3.12 = 1.46

Explanation:
4.58
(-)3.12
1.46
We subtract each digit according to their place unit values.

Question 30.
8.629 – 5.309
8.629 – 5.309 = 3.320

Explanation:
8.629
(-)5.309
3.320
We subtract each digit according to their place unit values.

Question 31.
6.98 – 2.614
6.98 – 2.614 = 4.366

Explanation:
  7,10    
6.980
(-) 2.614
4.366
We subtract each digit according to their place unit values.

Question 32.
15.131 – 11.57
15.131 – 11.57 = 3.561

Explanation:
       10,13  
15.1   3   1
(-)11.5   7   0
3.5   6   1
We subtract each digit according to their place unit values.

Question 33.
13.5 – 10.856
13.5 – 10.856 = 2.644

Explanation:
     15,9,10
13.5 0 0
(-)10.8 5 6
2.6  4 4
We subtract each digit according to their place unit values.

Question 34.
25.82 – 22.936 = 2.884

Explanation:
    4,17,11,10
25.820
(-)22.936
2.884
We subtract each digit according to their place unit values.

Question 35.
17.651 – 12.04
17.651 – 12.04 = 5.611

Explanation:
17.651
(-)12.040
05.611
We subtract each digit according to their place unit values.

Question 36.
19.255 – 6.194
19.255 – 6.194 = 13.061

Explanation:
         15
19.255
(-)06.194
13.061
We subtract each digit according to their place unit values.

Question 37.
56.217 – 35.8
56.217 – 35.8 = 20.417

Explanation:
        12
56.217
(-)35.800
20.417
We subtract each digit according to their place unit values.

Question 38.
62.486 – 18.549
62.486 – 18.549 =  43.937

Explanation:
   11, 14, 16  
62.486
(-)18.549
43.937
We subtract each digit according to their place unit values.

Question 39.
152.883 – 35.6247
152.883 – 35.6247 = 117.2583

Explanation:
      4,12,7,12,10  
152.8830
(-) 035.6247
117.2583
We subtract each digit according to their place unit values.

Question 40.
129.343 – 125.0372
129.343 – 125.0372 = 4.3058

Explanation:
             12,10  
129.3430
(-)125.0372
4.3058
We subtract each digit according to their place unit values.

YOU BE THE TEACHER
Your friend finds the sum or difference. Is your friend correct? Explain your reasoning.

Question 41.
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 239
Yes, Friend is correct as the sum is 10.008

Explanatation:
    1,1    
6.058
+3.950
10.008
Friend aslo got the same value so friend is
correct as 6.058 + 3.950 = 10.008.

Question 42.
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 240
No, Friend is incorrect as the difference is 2.32 ≠ 2.48

Explanation:
     4,10    
9.50
-7.18
2.32
Friend got different value so incorrect as
9.50 – 7.18 = 2.32 ≠ 2.48.

Question 43.
PROBLEM SOLVING
Vehicles must weigh no more than 10.75 tons to cross a bridge.
A truck weighs 11.638 tons. By how many tons does the truck
exceed the weight limit?
0.888 ton weigh is the truck exceed the weight limit

Explanation:
Given vehicles must weigh no more than 10.75 tons to cross a bridge.
A truck weighs 11.638 tons. So truck weighs more,
we calculate the difference as
   15,13  
11.638
-10.750
0.888 

Therefore by 0.888 ton weighs is the truck exceed the weight limit.

ADDING AND SUBTRACTING DECIMALS
Evaluate the expression.

Question 44.
6.105 + 10.4 + 3.075
6.105 + 10.4 + 3.075 = 19.58

Explanation:
       1  
6.105
10.400
+3.075
19.580
We add digit according to the place values.

Question 45.
22.6 – 12.286 – 3.542
22.6 – 12.286 – 3.542 = 6.772

Explanation:
First we subtract 12.286 from 22.6 as
     5,9,10 
22.600
-12.286
10.314
Now we subtract 3.542 from 10.314 as
9,12,11
10.314
-3.542
6.772
Therefore 22.6 – 12.286 – 3.542 = 6.772.

Question 46.
15.35 + 7.604 – 12.954
15.35 + 7.604 – 12.954 = 10

Explanation:
First we add 15.350 + 7.604 as

 1          
15.350
+7.604
22.954
Now we subtract 12.954 from 22.954 as
22.954
-12.954
10.000
Therefore 15.35 + 7.604 – 12.954 = 10.

Question 47.
16.5 – 13.45 + 7.293
16.5 – 13.45 + 7.293 = 10.343

Explanation:
First we subtract 13.45 from 16.5 as
      4,10 
16.500
-13.450
03.050
Now we add 3.05 with 7.293 as
1, 1     
3.050
+7.293
10.343
Therefore 16.5 – 13.45 + 7.293 = 10.343.

question 48.
25.92 – 18.478 + 8.164
25.92 – 18.478 + 8.164 = 15.606

Explanation:
First we subtract 18.478 from 25.920 as
  15, 11,10 
25.920
-18.478
7.442
Now we add 7.442 with 8.164 as
1,   1     
7.442
+8.164
15.606
Therefore 25.92 – 18.478 + 8.164 = 15.606.

Question 49.
23.45 + 17.75 – 19.618
23.45 + 17.75 – 19.618 = 21.582

Explanation:
First we add 23.45 and 17.75 as
  1,1,1  
23.45
+17.75
41.20
Now we subtract 19.618 from 41.200 as
3,10,11,9,10
41.200
-19.618
21.582
Therefore 23.45 + 17.75 – 19.618 = 21.582.

Question 50.
14.549 – (8.131 + 3.7024)
14.549 – (8.131 + 3.7024) = 2.7156

Explanation:
First we add (8.131 + 3.7024 ) as
1               
8.1310
+3.7024
11.8334
Now we subtract 11.8334 from 14.549 as
  3,15, 8,10
14.5490
-11.8334
2.7156
Therefore 14.549 – (8.131 + 3.7024) = 2.7156.

Question 51.
41.563 – (18.65 + 15.9214) + 9.6
41.563 – (18.65 + 15.9214) + 9.16 =16.5916

Explanation:
First we add (8.131 + 3.7024 ) as
   1,1         
18.6500
+15.9214
34.5714
Now we subtract 34.5714 from 41.563 as
 10,14,16,10
41.5630
-34.5714
6.9916
Now we add 6.9916 and 9.6 as
1,1
  6.9916
+9.6000
16.5916
Therefore 41.563 – (18.65 + 15.9214) + 9.16 =16.5916.

Question 52.
MODELING REAL LIFE
A day-care center is building a new outdoor play area.
The diagram shows the dimensions in meters. How much
fencing is needed to enclose the play area?
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 241
34.995 meters fencing is needed to enclose the play area.

Explanation:
Given sides of the outdoor play area, Fencing needed is
sum of the all three sides, So 10.6 + 11.845 + 12.55,
10.600
11.845
+12.550
    1
34. 995
therefore 34.995 meters fencing is needed to enclose the play area.

Question 53.
PROBLEM SOLVING
On a fantasy football team, a tight end scores 11.15 points and a
running back scores 11.75 points. A wide receiver scores 1.05 points
less than the running back. How many total points do the three players score?

33.6 points all the three players score,

Explanation:
Given a tight end scores 11.15 points and a
running back scores 11.75 points. A wide receiver scores 1.05 points
less than the running back. So a wide receiver scores 11.75 – 1.05 point,
11.75
-1.05
10.70
So a wide receiver scores 10.70 points,
Now total points do the three players score are
11.15
11.75
+10.70
    1  1
33.60
Therefore 33.6 points all the three players score.

MODELING REAL LIFE
An astronomical unit (AU) is the average distance between Earth and the Sun. In Exercises 54–57, use the table that shows the average distance of each planet in our solar system from the Sun.

Question 54.
How much farther is Jupiter from the Sun than Mercury?

Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 242
4.816 AU farther is Jupiter from the Sun than Mercury,

Explanation:
Given 5.203 AU is Jupiter average distance from the sun and
mercury is 0.387 AU, Now farther distance Jupiter from
the sun than mercury is
 4,11,9,13   
5.203
-0.387
4.816
Therefore 4.816 AU farther is Jupiter from the Sun than Mercury.

Question 55.
How much farther is Neptune from the Sun than Mars?
28.546 AU farther is Neptune from the Sun than Mars,

Explanation:
Given 30.07 AU is Neptune average distance from the sun and
mars is 1.524 AU, Now farther distance Neptune from
the sun than mars is
2, 9,10,6,10  
30.070
-1.524
28.546
Therefore 28.546 AU farther is Neptune from the Sun than Mars.

Question 56.
Estimate the greatest distance between Earth and Uranus.
20.189 AU is the greatest distance between Earth and Uranus,

Explanation:
Given 1.000 AU is earth average distance from the sun and
Uranus is 19.189 AU, Now the greatest distance between earth
and Uranus, is
  1          
1.000
+19.189
20.189
Therefore 20.189 AU is the greatest distance between Earth and Uranus.

Question 57.
Estimate the greatest distance between Venus and Saturn.
10.26 AU is the greatest distance between Venus and Saturn,

Explanation:
Given 0.723 AU is the Venus average distance from the sun and
Saturn is 9.537 AU, Now the greatest distance between Venus
and Saturn, is
   1,  1   
0.723
+9.537
10.260
Therefore 10.26 AU is the greatest distance between Venus and Saturn.

Question 58.
STRUCTURE
When is the sum of two decimals equal to a whole number?
When is the difference of two decimals equal to a whole number? Explain.

The sum of two decimal numbers is a whole number if the
sum of two fractions can be simplified to a whole number.
or this will happen if the fractional parts of the numbers sum to 1.

The difference of two decimal numbers (real numbers
written in decimal digit expansion form) will be an whole number
if and only if their fractional parts are equal and
any whole numbers a & b, a should be always greater or
equal to b then only a-b will be whole number.

Explanation:
When we do adding both numbers the fractional parts of the
numbers so become whole example 1.5 and 3.5 when we add we
get 5 which is a whole number.

In difference example we take 1.17 subtract from 4.17,
here fractional parts are equal 0.17 respectively, we get
the  difference as 4.17 -1.17 = 3 which is whole number.

Lesson 2.5 Multiplying Decimals

EXPLORATION 1
Multiplying Decimals
Work with a partner.
a. Write the multiplication expression represented by each area model. Then find the product. Explain how you found your answer.
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 243
Big Ideas Math Answers 6th Grade Chapter 2 Fractions and Decimals 244
b. How can you find the products in part(a) without using a model? How do you know where to place the decimal points in the answers?
c. Find the product of 0.55 and 0.45. Explain how you found your answer.

a. i. The multiplication expression is \(\frac{5}{10}\) X \(\frac{8}{10}\),
The product is \(\frac{40}{100}\),
ii. The multiplication expression is \(\frac{9}{10}\) X \(\frac{4}{10}\),
The product is \(\frac{36}{100}\),
iii. The multiplication expression of whole part is
\(\frac{10}{10}\) X \(\frac{5}{10}\) and
decimal part is \(\frac{5}{10}\) X \(\frac{5}{10}\)
then the product results is \(\frac{50}{100}\) + \(\frac{25}{100}\),
iv. The multiplication expression of whole part is
\(\frac{10}{10}\) X \(\frac{7}{10}\) and
decimal part is \(\frac{7}{10}\) X \(\frac{7}{10}\)
then the product results is \(\frac{70}{100}\) + \(\frac{49}{100}\),
c. The product of 0.55 and 0.45 is 0.2475,

Explanation:
a. i. By counting the blocks in the area model found
the multiplication expression as \(\frac{5}{10}\) X \(\frac{8}{10}\)
and Step I: We multiply the numerators as 5 X 8 = 40
Step II: We multiply the denominators as 10 X 10 =100
Step III: We write the fraction in the simplest form as
\(\frac{40}{100}\), So \(\frac{5 X 8}{10 X 10}\) = \(\frac{40}{100}\).
If we see the area model the purple color blocks show
40 out of 100 blocks.
ii. By counting the blocks in the area model found
the multiplication expression as \(\frac{9}{10}\) X \(\frac{4}{10}\)
and Step I: We multiply the numerators as 9 X 4 = 36
Step II: We multiply the denominators as 10 X 10 =100
Step III: We write the fraction in the simplest form as
\(\frac{36}{100}\), So \(\frac{10 X 10}{10 X 10}\) = \(\frac{36}{100}\).
If we see the area model the purple color blocks show
36 out of 100 blocks.
iii. By counting the blocks in the area model found
the multiplication expression as \(\frac{10}{10}\) X \(\frac{5}{10}\)
and Step I: We multiply the numerators as 10 X 5 = 50
Step II: We multiply the denominators as 10 X 10 =100
Step III: We write the fraction in the simplest form as
\(\frac{50}{100}\), So \(\frac{10 X 5}{10 X 10}\) = \(\frac{50}{100}\).
Now in decimal part we have \(\frac{5}{10}\) X \(\frac{5}{10}\),
similar to whole part we do multiplication
Step I: We multiply the numerators as 5 X 5 = 25
Step II: We multiply the denominators as 10 X 10 =100
Step III: We write the fraction in the simplest form as
\(\frac{25}{100}\), therefore then the product results is
\(\frac{50}{100}\) + \(\frac{25}{100}\),
iv. By counting the blocks in the area model found
the multiplication expression as \(\frac{10}{10}\) X \(\frac{7}{10}\)
and Step I: We multiply the numerators as 10 X 7 = 70
Step II: We multiply the denominators as 10 X 10 =100
Step III: We write the fraction in the simplest form as
\(\frac{70}{100}\), So \(\frac{10 X 7}{10 X 10}\) = \(\frac{70}{100}\).
Now in decimal part we have \(\frac{7}{10}\) X \(\frac{7}{10}\),
similar to whole part we do multiplication
Step I: We multiply the numerators as 7 X 7 = 49
Step II: We multiply the denominators as 10 X 10 =100
Step III: We write the fraction in the simplest form as
\(\frac{49}{100}\), therefore then the product results is
\(\frac{70}{100}\) + \(\frac{49}{100}\).

b. We use normal multiplication of numbers without any model,
Decimals are a shorthand way to write fractions and
mixed numbers with denominators that are powers of 10, like 10,100,1000,10000,

10,100,1000,10000,

etc.
If a number has a decimal point, then the first digit to the
right of the decimal point indicates the number of tenths.

For example, the decimal 0.3 is the same as the fraction \(\frac{3}{10}\)
The second digit to the right of the decimal point indicates
the number of hundredths.

For example, the decimal 3.26 is the same as the mixed number
3 \(\frac{26}{100}\) . (Note that the first digit to the left of
the decimal point is the ones digit.)

We can write decimals with many places to the right of the decimal point.
As shown in below example
Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals-64
c. The product of 0.55 and 0.45 is
   2   
   2    
0.55————– 2 decimal places
X0.45 ———— 2 decimal places
0275         (first 0.05 X 0.55)
2200         (0.4 X 0.55)
0.2475 ———- 4 decimal places

Therefore the product of 0.55 and 0.45 is 0.2475.

2.5 Lesson

Key Idea
Multiplying Decimals by Whole Numbers
Words
Multiply as you would with whole numbers. Then count the number of decimal places in the decimal factor. The product has the same number of decimal places.
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 245

Try It
Multiply. Use estimation to check your answer.

Question 1.
12.3 × 8
12.3 X 8 = 98.4
Is not reasonable 96 ≠ ≈ 98,

Explanation:
  2    
12.3          1 decimal place
X  8
98.4  

Estimate : 12 X 8 = 96,
Reasonable 98.4 ≈ 98,
Is not reasonable 96 ≠ ≈ 98.

Question 2.
5 × 14.51
5 × 14.51 = 72.55
Is not reasonable 75 ≠ ≈ 73.

Explanation:
2,2    
14.51          2 decimal places
X   5
72.55     

Estimate : 5 X 15 = 75,
Reasonable 72.55 ≈ 73,
Is not reasonable 75 ≠ ≈ 73.

Question 3.
2.3275 X 90
2.3275 X 90 = 209.475
Is not reasonable 180 ≠ ≈ 210.

Explanation:
2,4    
2.3275      4 decimal places
X    90
0000000
209475
209.4750

Estimate : 2 X 90 = 180,
Reasonable 209.4750 ≈ 210,
Is not reasonable 180 ≠ ≈ 210.

The rule for multiplying two decimals is similar to the rule for multiplying a decimal by a whole number.

Key Idea
Multiplying Decimals by Decimals
Words
Multiply as you would with whole numbers. Then add the number of decimal places in the factors. The sum is the number of decimal places in the product.
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 246

Try It
Multiply. Use estimation to check your answer.

Question 5.
8.1 × 5.6
8.1 × 5.6 = 45.36
Is not reasonable 45 ≠ ≈ 48,

Explanation:
8.1 ——-1 decimal place
X 5.6 —–1 decimal place
   486  
4050
45.36   —– 2 decimal places
Estimate : 8 X 6 = 48,
Reasonable 45.36 ≈ 45,
Is not reasonable 45 ≠ ≈ 48.

Question 6.
2.7 × 9.04
2.7 × 9.04 = 24.408
Is not reasonable 24 ≠ ≈ 27,

Explanation:
    2    
9.04 ——-2 decimal places
X 2.7 ——1 decimal places
   6328
18080
24.408   —– 3 decimal places
Estimate : 3 X 9 = 27,
Reasonable 24.408 ≈ 24,
Is not reasonable 24 ≠ ≈ 27.

Question 7.
6.32 × 0.09
6.32 × 0.09 = 0.5688
Is reasonable 0.6 = 0.6

Explanation:
  2,1    
6.32 ——-2 decimal places
X 0.09 ——2 decimal places
0.5688 —– 4 decimal places
Estimate : 6 X 0.1 = 0.6,
Reasonable 0.5688≈ 0.6,
Is reasonable 0.6= 0.6.

Question 8.
1.785 × 0.2
1.785 × 0.2 = 0.357
Is reasonable 0.4= 0.4,

Explanation:
1,1,1    
1.785 ——-3 decimal places
X 0.2 ——  1 decimal places
0.3570—– 4 decimal places
Estimate : 2 X 0.2 = 0.4,
Reasonable 0.3570≈ 0.4,
Is reasonable 0.4= 0.4.

Self-Assessment for Concepts & Skills
Solve each exercise. Then rate your understanding of the success criteria in your journal.

EVALUATING AN EXPRESSION
Evaluate the expression.

Question 11.
8 × 11.215
8 × 11.215 = 89.72

Explanation:
  1,   4   
11.215 —– 3 decimal places
X 8
89.720—— 3 decimal places
Therefore, 8 × 11.215 = 89.72

Question 12.
9.42 . 6.83
9.42 X 6.83 = 64.3386

Explanation:
5,1
3,1
  1,     
9.42 —– 2 decimal places
X6.83 ——2 decimal places
002826
075360
565200
64.3386—— 4 decimal places
Therefore, 9.42 X 6.83 = 64.3386

Question 13.
0.15(4.3 – 2.417)
0.15 X (4.3 – 2.417) = 0.28245

Explanation:
First we calculate subtracting 2.417 from 4.3 as
  12,9,10
4.300
-2.417
1.883

now we multiply 0.15 and 1.883 as
4,4,5
1.883—- 3 decimal places
X 0.15—-2 decimal places
  9415
18830
0.28245— 5 decimal places
Therefore, 0.15 X (4.3 – 2.417) = 0.28245.

Question 14.
NUMBER SENSE
If you know 12 × 24 = 288, how can you find 0.12 × 0.24?
We have 4 deciml places so 0.12 X 0.24 = 0.0288

Explanation:
Given I know 12 X 24 = 288,
we find 0.12 X 0.24 as we have 2 + 2 = 4 decimals
we write 244 with four decimals as 0.0288,
therefore 0.12 X 0.24 = 0.0288.

Question 15.
NUMBER SENSE
Is the product 1.23 × 8 greater than or less than 8? Explain.
The product of 1.23 X 8 is greater than 8,

Explanation:
1.23 X 8 = 9.84,

Now comparing 9.84 with 8, 9.84 is greater than 8,
or we have the whole part 1 plus 0.23 when multiplied by 8 it’s
value becomes more only.
therefore 1.23 X 8 >8.

Question 16.
REASONING
Copy the problem and place the decimal point in the product.
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 247
8.722 we have 3 decimal places so
1.78 X 4.9 = 8.722
Explanation:
1.78 —-2 decimal places
X 4.9—1 decimal place
8.722—3 decimal places
Thefore, 8.722 has decimal point before 3 digits.

Self-Assessment for Problem Solving

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 17.
You earn $9.15 per hour painting a fence. It takes 6.75 hours to paint the fence.
Did you earn enough money to buy the jersey shown? If so, how much money
do you have left? If not, how much money do you need to earn?
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 248

Yes, I earned enough money $61.7625 to buy the jersey shown,
Money left is $1.7725.

Explanation:
Given I earn $9.15 per hour painting a fence.
It takes 6.75 hours to paint the fence. So money earned is
$9.15 X 6.75 = $61.7625
 3
   3

9.15 ——2 decimal values
X6.75—– 2 decimal values
 004575
064050
+549000
61.7625—-4 decimal places
Therefore, yes, I earned enough money $61.7625 to buy the
jersey shown, because given the cost of jersey’s is $59.99 each
which is less compared to what i have earned.
Now money left is $61.7625 – $59.99 = $1.7725,
 10,16,16
61.7625
-59.9900
   1.7725
So, money left is $1.7725.

Question 18.
A sand volleyball court is a rectangle that has a length of 52.5 feet and
a width that is half of the length. In case of rain, the court is covered
with a tarp. How many square feet of tarp are needed to cover the court?

1378.125 square feet of trap is needed to cover the court,

Explanation:
Given a sand volleyball court is a rectangle that has a length
of 52.5 feet and a width that is half of the length means width is
52.5 ÷ 2 = 26.25 feet, Now to cover the court with a trap we need
52.5 X 26.25 = 1378.125 square feet,
52.5————2 decimal places
X26.25——– 2 decimal places
2625 —-(52.5 X 0.05)
1050 —(52.5 X0.2)
31500—(52.5 X 6)
+105000—(52.5 X20)
1378.0125——4 decimal places
Therefore,1378.125 square feet of trap is needed to cover the court.

Question 19.
DIG DEEPER!
You buy 4 cases of bottled water and 5 bottles of fruit punch for a
birthday party. Each case of bottled water costs $2.75, and each bottle
of fruit punch costs $1.35. You hand the cashier a $20 bill.
How much change will you receive?

I will receive $2.25 as a change,

Explanation:
Given I buy 4 cases of bottled water and 5 bottles of fruit punch for a
birthday party. Each case of bottled water costs $2.75, and each bottle
of fruit punch costs $1.35, Now 4 cases of bottled costs is
4 X $2.75=$11,
 30,2 
2.75—— 2 decimal places
x 4
11.00——2 decimal places
and 5 bottles of fruit punch costs is 5 X $1.35 = $6.75
1,2   
1.35—- 2 decimal places
X 5
6.75—-2 decimal places
Now total bill amount is $11 + $ 6.75= $17.75,
11.00
+6.75
17.75
I hand the cashier $20 , The change I will receive is
  9,9,10
20.00
-17.75
  2.25
Therefore, I will receive $2.25 as a change.

Multiplying Decimals Homework & Practice 2.5

Review & Refresh

Add or subtract.

Question 1.
12.29 – 6.15
12.29 – 6.15 = 6.14

Explanation:
 ,12
12.29
-6.15
06.14
Therefore, 12.29 – 6.15 = 6.14.

Question 2.
4.6 + 11.81
4.6 + 11.81 = 16.41

Explanation:
 ,    1    
04.60
+11.81
16.41
Therefore, 4.6 + 11.81 = 16.41.

Question 3.
9.34 + 17.009
9.34 + 17.009 = 26.349
Explanation:
 , 1    
09.340
+17.009
26.349
Therefore, 9.34 + 17.009 = 26.349.

Question 4.
18.247 – 16.262
18.247 – 16.262 = 1.985

Explanation:
   11,14
18.247
-16.262
1.985
Therefore, 18.247 – 16.262 = 1.985.

Divide.

Question 5.
78 ÷ 3
78 ÷ 3 = 26

Explanation:
Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals-65
here the first number 78 is called the dividend
and t
he second number 3 is called the divisor.
We will get 0 as remainder and top 26 is the answer,
as shown in the picture,
Therefore 78 ÷ 3 = 26.

Question 6.
65 ÷ 13
65 ÷ 13 = 5

Explanation:
Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals-66
here the first number 65 is called the dividend
and t
he second number 13 is called the divisor.
We will get 0 as remainder and top 5 is the answer
as shown in the picture,
Therefore 65 ÷ 13 = 5.

Question 7.
57 ÷ 19
57 ÷ 19 = 3

Explanation:
Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals-67 19 X 3 = 57
here the first number 57 is called the dividend
and t
he second number 19 is called the divisor.
We will get 0 as remainder top 3 is the answer
as shown in the picture,
Therefore 57 ÷ 19 = 3.

Question 8.
84 ÷ 12
84 ÷ 12 = 7

Explanation:
Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals-68
here the first number 84 is called the dividend
and t
he second number 12 is called the divisor.
We will get 0 as remainder top 7 is the answer
as shown in the picture,
Therefore 84 ÷ 12 = 7.

Question 9.
What is Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 249
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 249 = \(\frac{52}{15}\) = 3\(\frac{7}{15}\)

Explanation:
Given expression as 4\(\frac{1}{3}\) X \(\frac{4}{5}\) ,
First we write mixed fraction as (4 X 3 + 1 by 3) =\(\frac{13}{3}\), ,
\(\frac{13}{3}\) X \(\frac{4}{5}\) =\(\frac{13 X 4}{3 X 5}\) =
\(\frac{52}{3}\) as numerator is greater we write in
mixed fraction as (3 X 15 + 7 by 15) = 3\(\frac{7}{15}\),
Therefore, 4\(\frac{1}{3}\) X \(\frac{4}{5}\) =
\(\frac{52}{15}\) = 3\(\frac{7}{15}\).

Evaluate the expression.

Question 10.
4 + 62 ÷ 2
4 + 62 ÷ 2 = 2

Explanation:
Given expression as 4 + 62 ÷ 2,
first we evaluate 62 = 6 X 6 = 36, then
4 + 36 = 40, Now 40 ÷ 2 = 20 (2 X 20 = 40),
therefore, 4 + 62 ÷ 2 = 2.

Question 11.
(35 + 9) ÷ 4 – 32
(35 + 9) ÷ 4 – 32 = 2

Explanation:
Given expression as (35 + 9) ÷ 4 – 32,
first we evaluate 35 + 9 = 44,
then 44 ÷ 4  = 11 (4 X 11 = 44),
and 11 – 32 = 11 – (3 X 3) = 11 – 9 = 2,
therefore, (35 + 9) ÷ 4 – 32 = 2.

Question 12.
82 ÷ [(14 – 12) × 23]
82 ÷ [(14 – 12) × 23] = 4

Explanation:
Given expression as 82 ÷ [(14 – 12) × 23]
First we calculate [(14 – 12) × 23] = (14 -12) X (2 X 2 X 2),
2 X 8 = 16, Now 82 ÷ 16 = (8X 8) ÷ 16 = 64 ÷ 16 = 4 (16 x 4 =64),
therefore,82 ÷ [(14 – 12) × 23] = 4.

Concepts, Skills, & Problem Solving

USING TOOLS
Use an area model to find the product. (See Exploration 1, p. 73.)

Question 13.
2.1 × 1.5
2.1 × 1.5 = 3.15
Explanation:
Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals-69
2.1 × 1.5 = 3.15,
We took 2 full and 0.1 then multiplied by
1 full and 0.5 we got 3 full and 0.15 as shown in the picture.

Question 14.
0.6 × 0.4
0.6 X 0.4 = 0.24

Explanation:
Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals-70
The purple color shows the result 0.24.

Question 15.
0.7 × 0.3
0.7 X 0.3 = 0.21

Explanation:
Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals-71
The purple color shows the result 0.21.

Question 16.
2.7 × 2.3
2.7 X 2.3 = 6.21

Explanation:
Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals-72
The purple color shows the result 6.21.

MULTIPLYING DECIMALS AND WHOLE NUMBERS
Multiply. Use estimation to check your answer.

Question 17.
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 249.1
33.6

Explanation:
 5   
4.8 ——1 decimal place
X 7
33.6—– 1 decimal place
Therefore, 4.8 X 7 = 33.6.

Question 18.
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 250
  31.5

Explanation:
1
6.3 ——1 decimal place
X 5
31.5—– 1 decimal place
Therefore, 6.3 X 5 = 31.5.

Question 19.
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 251
115.04

Explanation:
1, 5   
7.19 ——2 decimal places
X 16
4314
7190
115.04—– 2 decimal places
Therefore, 7.19 X 16 = 115.04.

Question 20.
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 252
  18.27

Explanation:
1, 1
0.87 ——2 decimal places
X 21
    87
1740
18.27—– 2 decimal places
Therefore, 8.87 X 21 = 18.27.

Question 21.
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 253
 21.45

Explanation:
1.95 ——2 decimal places
X 11
  195
1950
21.45—– 2 decimal places
Therefore, 1.95 X 11 = 21.45.

Question 22.
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 254
  29.45

Explanation:
4,4
5.89 ——2 decimal places
X 5
29.45—– 2 decimal places
Therefore, 5.89 X 5 = 29.45.

Question 23.
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 255
  13.888

Explanation:
1,2
3.472 ——3 decimal places
X   4
13.888—– 3 decimal places
Therefore, 3.472 X 4 = 13.888.

Question 24.
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 256
  98.256

Explanation:
1,1
8.188 ——3 decimal places
X  12
16376
81880
98.256—– 3 decimal places
Therefore, 8.188 X 12 = 98.256.

Question 25.
100 × 0.024
100 × 0.024 = 2.4

Explanation:
0.024 — 3 decimal places
X100
0000
00000
002400
002.400—–3 decimal places
Therefore, 0.02 X 100 = 2.4.

Question 26.
19 × 0.004
19 × 0.004 = 0.076

Explanation:
0.004 — 3 decimal places
X19   
0036
00040
0.076—–3 decimal places
Therefore, 0.004 X 19 = 0.076.

Question 27.
3.27 × 14
3.27 × 14 = 45.78

Explanation:
8
3.27—– 2 decimal places
X 14
1308
3270
45.78—- 2 decimal places
Therefore, 3.27 X 14 = 45.78.

Question 28.
46 . 5.448
46 X 5.448 = 250.608

Explanation:
 1,1,3
2, 2,4
5.448 —– 3 decimal places
X 46
  32688
217920
250.608 —-3 decimal places
Therefore, 46 X 5.448 = 250.608.

Question 29.
50 × 12.21
50 X 12.21 = 610.5

Explanation:
1,1
12.21—-2 decimal places
X 50
 0000
6105
610.50—-2 decimal places
Therefore 50 X 12.21 = 610.5.

Question 30.
104 . 4.786
104 X 4.786 =497.744

Explanation:
3,3,2
4.786—-3 decimal places
X104
019144
00000
478600
497.744—-3 decimal places
Therefore 104 X 4.786 = 497.744.

Question 31.
0.0038 × 9
0.0038 X 9 = 0.0342

Explanation:
3
0.0038—-4 decimal places
X       9
0.0342—-4 decimal places
Therefore, 0.0038 X 9 = 0.0342

Question 32.
10 × 0.0093
10 × 0.0093 = 0.093

Explanation:
3
0.0093—-4 decimal places
X     10
  00000
000930
0.0930 —-4 decimal places
Therefore, 10 × 0.0093 = 0.093.

YOU BE THE TEACHER
Your friend finds the product. Is your friend correct? Explain your reasoning.

Question 33.
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 257
Yes, Friend is correct as 0.0045 X 9 = 0.0405,

Explanation:
      4,   
0.0045—– 4 decimal places
X       9
0.0405—–4 decimal places
Therefore friend is correct as 0.0405 is equal to friends value 0.0405.
Question 34.
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 258
No, Friend is incorrect as 0.32 X 5 = 1.6 ≠ 0.160

Explanation:
  1   
0.32—–2 decimal places
X   5
1.60  —-2 decimal places
Therefore, friend is incorrect as 0.32 X 5 = 1.6 ≠ 0.160.

Question 35.
MODELING REAL LIFE
The weight of an object on the Moon is about 0.167 of its weight on Earth.
How much does a 180-pound astronaut weigh on the Moon?
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 259
I80-pound astronaut weigh on the Moon is 30.06 pounds

Explanation:
Given the weight of an object on the Moon is about 0.167
of its weight on Earth. So, 180-pound astronaut weigh on the Moon is
0.167 X 180 = 30.06 pounds
   5,5   
0.167 —– 3 decimal places
X 180  
0000
13360
016700
30.060—- 3 decimal places
Therefore, I80-pound astronaut weigh on the Moon is 30.06 pounds.

MULTIPLYING DECIMALS
Multiply.

Question 36.
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 260
  0.14

Explanation:
1
0.7 ——1 decimal place
X 0.2——1 decimal place
   14
000
0.14—– 2 decimal places
Therefore, 0.7 X 0.2 = 0.14.

Question 37.
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 261
 0.024

Explanation:
  2
0.08 ——–2 decimal places
X 0.3——–1 decimal place
0024
0000
0.024——–3 decimal places
Therefore, 0.08 X 0.3 = 0.024.

Question 38.
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 262
0.00021

Explanation:
     2
0.007 ——–3 decimal places
X 0.03——–2 decimal places
  0021
00000
000000
0.00021——-5 decimal places
Therefore, 0.007 X 0.03 = 0.00021.

Question 39.
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 263
0.000072

Explanation:
     7
0.0008 ——-4 decimal places
X 0.09——–2 decimal places
  00072
000000
0000000
0.000072—–6 decimal places
Therefore, 0.0008 X 0.09 = 0.000072.

Question 40.
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 264
0.0036

Explanation:
    3   
0.004 ——-3 decimal places
X  0.9——-1 decimal place
  0036
00000
0.0036——-4 decimal places
Therefore, 0.004 X 0.9 = 0.0036.

Question 41.
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 265
   0.03

Explanation:
   3
0.06—–2 decimal places
X 0.5—–1 decimal place
0030
0000
0.030—-3 decimal places
Therefore, 0.06 X 0.5 = 0.03.

Question 42.
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 266
0.0000032

Explanation:
          3
0.0008——-4 decimal places
X 0.004——3 decimal places
   00032
000000
0000000
0.0000032—-7 decimal places
Therefore, 0.0008 X 0.004 = 0.0000032.

Question 43.
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 267
0.000012

Explanation:
1
0.0002——-4 decimal places
X 0.06——  2 decimal places
00012
000000
0000000
0.000012—-6 decimal places
Therefore, 0.0002 X 0.06 = 0.000012.

Question 44.
12.4 × 0.2
12.4 X 0.2 = 2.48

Explanation:
12.4——–1 decimal place
X 0.2——-1 decimal place
248
0000
2.48——-2 decimal places
Therefore, 12.4  X 0.2 = 2.48.

Question 45.
18.6 . 5.9
18.6 X 5.9 = 109.74

Explanation:
 4,3
7,5
18.6—-1 decimal place
X5.9— 1 decimal place
1674
9300
109.74—-2 decimal places
Therefore, 18.6 X 5.9 = 109.74.

Question 46.
7.91 × 0.72
7.91 X 0.72 = 5.6952

Explanation:
 6
 1
7.91—–2 decimal places
X0.72—-2 decimal places
01582
55370
00000
5.6952—-4 decimal places
Therefore, 7.91 X 0.72 = 5.6952.

Question 47.
1.16 × 3.35
1.16 × 3.35 = 3.886

Explanation:
     1  
   1  
    3
1.16——-2 decimal places
X 3.35—–2 decimal places
00580
03480
34800
3.8860—–4 decimal places
Therefore, 1.16 X 3.35 = 3.8860.

Question 48.
6.478 × 18.21
6.478 × 18.21 = 117.96438

Explanation:
  3,5,6 
     1,1
6.478——3 decimal places
X 18.21—-2 decimal places
0006478
0129560
5182400
6478000
117.96438—5 decimal places
Therefore, 6.478 × 18.21 = 117.96438.

Question 49.
1.9 × 7.216
1.9 X 7.216 = 13.7104

Explanation:
   1,1,5
7.216——3 decimal places
X 1.9——1 decimal place
 64944
 72160
13.7104——4 decimal places
Therefore, 7.216 X 1.9 = 13.7104.

Question 50.
0.0021 × 18.2
0.0021 × 18.2 = 0.03822

Explanation:
      1  
0.0021——3 decimal places
X 18.2——-1 decimal place
0000042
0001680
0002100
0.03822——4 decimal places
Therefore, 0.0021 X 18.2 = 0.03822.

Question 51.
6.109 . 8.4
6.109 X 8.4 = 51.3156

Explanation:
     7   
     3   
6.109——3 decimal places
X 8.4——1 decimal place
024436
488720
51.3156——4 decimal places
Therefore, 6.109 X 8.4 = 51.3156.

Question 52.
YOU BE THE TEACHER
Your friend finds the product of 4.9 and 3.8. Is your friend correct? Explain your reasoning.
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 268
No, Friend is incorrect as 4.9 X 3.8 = 18.62 ≠ 186.2

Explanation:
   2
  7
4.9 ——1 decimal place
X3.8——1 decimal place
0392
1470
18.62——2 decimal places
Therefore, Friend is incorrect as 4.9 X 3.8 = 18.62 ≠ 186.2.

Questi1on 53.
PROBLEM SOLVING
A Chinese restaurant offers buffet takeout for $4.99 per pound.
How much does your takeout meal cost?
The takeout meal cost $3.5429

Explanation:
Given a Chinese restaurant offers buffet takeout for $4.99 per pound
and 1 dollar in pound = 0.71 pound, So the takeout meal costs
$4.99 X 0.71 = $3.5429
4.99——2 decimal places
X0.71——2 decimal places
00499
34930
00000
3.5429——4 decimal places
Therefore, the takeout meal cost $3.5429.

Question 54.
PROBLEM SOLVING
On a tour of an old gold mine, you find a nugget containing
0.82 ounce of gold. Gold is worth $1323.80 per ounce.
How much is your nugget worth?
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 269
The nugget worth is $1085.516,

Explanation:
Given I found a nugget containing 0.82 ounce of gold.
and gold is worth $1323.80 per ounce. So my nugget worth is
$1323.80 X 0.82
2,1,2,6
       1 
1323.80—–2 decimal places
X 0.82——-2 decimal places
00264760
10590400
00000000
1085.5160—-4 decimal places
Therefore, the nugget worth is $1085.516.

Question 55.
PRECISION
One meter is approximately 3.28 feet. Find the height of each building in feet.
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 270
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 271
Carlton Centre is 731.44 feet,
Burj Khalifa is 2715.84 feet,
Q1 is 1059.44 feet,
Federation Tower is 1226.72 feet,
One World Trade Centre is  1774.48 feet,
Gran Torre Santiago is 984 feet,

Explanation:
Given one meter is approximately 3.28 feet and
heightsof each building in meters in the table above,
now we will find height in feet as
a. Carlton Centre is 223 meters, So in feet it  is
223 X 3.28 = 731.44
 1,2
223
X3.28———–2 decimal places
01784
04460
66900
731.44——-2 decimal places
So, Carlton Centre is 731.44 feet,
b. Burj Khalifa is 828 meters , In feet is
828 X 3.28 = 2715.84
2        
1,        
1,6   
828
X3.28 ——– 2 decimal places
006624
016560
248400
2715.84——-2 decimal places
So, Burj Khalifa is 2715.84 feet,
c. Q1 is 323 meters, In feet is
323 X 3.28 = 1059.44
1,2
323
X3.28——-2 decimal places
02584
06460
96900
1059.44——-2 decimal places
So, Q1 is 1059.44 feet,
d. Federation Tower is 374 meters, In feet is
374 X 3.28 = 1226.72
 5,3
374
X 3.28——-2 decimal places
002992
007480
112200
1226.72——-2 decimal places
So, Federation Tower is 1226.72 feet,
e. One World Trade Centre is 541 meters, In feet is
541 X 3.28 = 1774.48
1
 3
541
X3.28——-2 decimal places
004328
010820
162300
1774.48——-2 decimal places
So, One World Trade Centre is  1774.48 feet,
f. Gran Torre Santiago is 300 meters, Now in feet is
300 X 3.28 = 984
300
X 3.28——-2 decimal places
02400
06000
90000
984.00——-2 decimal places
So, Gran Torre Santiago is 984 feet.

Question 56.
REASONING
Show how to evaluate (7.12 × 8.22) × 100 without multiplying two decimals.
(7.12 X 8.22) X 100 = 5852.64

Explanation:
First multiply as whole numbers only we get 712 X 822 = 585264,
We have 4 decimal places but it is multiplied by 100,
We get 2 decimal places so we put as 5852.64,
Therefore (7.12 × 8.22) × 100 = 5852.64.

EVALUATING AN EXPRESSION

Evaluate the expression.

Question 57.
2.4 × 16 + 7
2.4 × 16 + 7 = 45.4

Explanation:
Given expression as 2.4 × 16 + 7, So first we calculate
2.4 X 16 as
 2
2.4——1 decimal place
x16
144
240
38.4—- 1 decimal place
Now 38.4 + 7 = 45.4
Therefore, 2.4 × 16 + 7 = 45.4

Question 58.
6.85 × 2 × 10
6.85 × 2 × 10 = 137

Explanation:
Given expression as 6.85 × 2 × 10,
First we calculate 6.85 X 2 as
1,1
6.85—–2 decimal places
X 2
13.70—–2 decimal places
Now, 13.70 X 10 we move 1 decimal place as 137.0,
Therefore 6.85 × 2 × 10 = 137.

Question 59.
1.047 × 5 – 0.88
1.047 × 5 – 0.88 = 4.355

Explanation:
Given expression as 1.047 × 5 – 0.88, First we calculate
1.047 X 5 as
  2,3
1.047— 3 decimal places
X   5
5.235— 3 decimal places

Now 5.235 – 0.88
5.235
0.880
4.355
Therefore, 1.047 × 5 – 0.88 = 4.355.

Question 60.
4.32(3.7 + 1.65)
4.32 X (3.7 + 1.65) = 23.112

Explanation:
Given expression as 4.32(3.7 + 1.65), first we calculate
(3.7 + 1.65) = 5.35
3.70
+1.65
5.35
Now, we multiply as 4.32 X 5.35 as
  1,1
   1,1
4.32 ——2 decimal places
X5.35—–2 decimal places
02160
12960
21600
23.1120—–4 decimal places
Therefore, 4.32 X (3.7 + 1.65) = 23.112

Question 61.
23.98 – 1.72 . 7.6
23.98 – 1.72 . 7.6 = 2.016

Explanation:
Given expression as 23.98 – 1.72 . 7.6, first we calculate
(1.72 X 7.6) = (1.7 X 1.7 X 7.6) again here first we
multiply 1.7 X 1.7 as
4    
1.7—— 1 decimal place
X1.7—– 1 decimal place
119
170
2.89—– 2 decimal places
Now we multiply with 2.89 with 7.6 as
6,6
  5,5
2.89—— 2 decimal places
X7.6—— 1 decimal place
01734
20230
21.964—-3 decimal places
Now further we find 23.98 minus 21.964 as
23.980
-21.964
2.016

Therefore 23.98 – 1.72 . 7.6 = 2.016.

Question 62.
12 . 5.16 + 10.064
12 . 5.16 + 10.064 = 182.688

Explanation:
Given expression as 12 X (5.16 + 10.064), first we calculate
5.16 + 10.064 as

05.160
+10.064
15.224
Now we calculate 12 X 15.224 as
1
15.224———3 decimal places
X    12
030448
152240
182.688———3 decimal places
Therefore 12 . 5.16 + 10.064 = 182.688.

Question 63.
0.9(8.2 . 20.35)
0.9(8.2 . 20.35) = 150.183

Explanation:
Given expression as 0.9(8.2 . 20.35), First we calculate
8.2 X 20.35 as
  2,4  
     1  
20.35—– 2 decimal places
X8.2 ——1 decimal place
004070
162800
166.870—–3 decimal places
Now we multiply 0.9 with 166.870 as
   6, 6,7,6
166.870—–3 decimal places
X      0.9—–1 decimal place
150.183—–3 decimal places
Therefore 0.9(8.2 . 20.35) = 150.183.

Question 64.
7.52(6.084 – 5.44)
7.52(6.084 – 5.44) = 36.225

Explanation:
Given expression as 7.52(6.084 – 5.44), First we will find
6.084 – 5.44 as
6.084
-5.440
0.644
Now 7.52 X 0.644 as (7.5 X 7.5 X 0.644) as
3
  2
7.5—–1 decimal place
X7.5—-1 decimal place
0375
5250
56.25—2 decimal places
Now 56.25 X 0.644 as
56.25—–2 decimal places
X0.644—–3 decimal places
00022500
00225000
03375000
00000000
36.22500—-5 decimal places
Therefore 7.52(6.084 – 5.44) = 36.225.

Question 65.
0.629[81 ÷ (10 × 2.7)]
0.629[81 ÷ (10 × 2.7)] = 1.887

Explanation:
Given expression as 0.629[81 ÷ (10 × 2.7)], First we will find
10 X 2.7 = 27.0 1 decimal place = 27,
Now 81 ÷  27 = 3(as 27 X 3 = 81), Now
0.629 X 3 is
     2    
0.629—- 3 decimal places
X    3
1.887—- 3 decimal places
So, 0.629[81 ÷ (10 × 2.7)] = 1.887

Question 66.
REASONING
Without multiplying, how many decimal places does 3.42 have? 3.433? 3.44? Explain your reasoning.
3.42 will have 2 decimal places
3.433 will have 6 decimal places
3.44 will have 4 decimal places

Explanation:
We have 3.42 ,3.433,3.44 without multiplying we write
decimal places as 3.42 will have 2 decimal places
3.433 will have 6 decimal places
3.44 will have 4 decimal places,
We take as 3.42 , 1 +1 decimal places = 2 decimal places,
3.433 , 2 decimals X 3 = 6 decimal places and
3.44 , 1 decimal X 4 = 4 decimal places and so on like
number of decimal places multiplied by powers.

Question 67.
MODELING REAL LIFE
You buy 2.6 pounds of apples and 1.475 pounds of peaches. You hand the cashier a $20 bill. How much change will you receive?
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 272
I will receive a change of $14.03,

Explanation:
Given 2.6 pounds of apples and 1.475 pounds of peaches I buyed,
Apples are $1.23 per pound means the cost of apples
bought are 1.23 X 2.6 =3.198
  1,1
1.23—–2 decimal places
X2.6—–1 decimal place
0738
2460
3.198—–3 decimal places
Now as cost of peaches are $1.88 per pound
total cost of peaches are 1.475 X 1.88 = 2.77300
3,6,4
3,6,4
1.475—–3 decimal places
X1.88—–2 decimal places
011800
118000
147500
2.77300—–5 decimal places
So, Total cost for apples and peaches are 3.198 plus 2.77300
3.198
+2.773
5.971
We got total cost for fruits as $5.971, I hand
the cashier $20, I will receive a change of 20-5.971 = 14.029
20.000
-05.971
14.029
Therefore I will receive a change of $14.029 ≈$14.03 in return.

PATTERNS
Describe the pattern. Find the next three numbers.

Question 68.
1, 0.6, 0.36, 0.216, . . .
1, 0.6, 0.36, 0.216, 0.1296, 0.07776,0.046656

Explanation:
Given series as 1, 0.6, 0.36, 0.216, . . .
each number is multiplied by 0.6
because 0.6 X 0.6 = 0.36,
0.36 X 0.6 = 0.216,
Next number is 0.216 X 0.6 = 0.1296,
the next number is 0.1296 X 0.6 = 0.07776 and
the next number is 0.0776 X 0.6 = 0.046656 and so on,
Therefore 1, 0.6, 0.36, 0.216, . . . the next three numbers are
1, 0.6, 0.36, 0.216, 0.1296, 0.07776, 0.046656 .

Question 69.
15, 1.5, 0.15, 0.015, . . .
15, 1.5, 0.15, 0.015, 0.0015, 0.00015, 0.000015

Explanation:
Given series as 15, 1.5, 0.15, 0.015, . . .
each number is multiplied by 0.1
because 15 X 0.1 = 1.5,
1.5 X 0.1 = 0.15,
0.15 X 0.1 = 0.015,
the next number is 0.015 X 0.1 = 0.0015 and
the next number is 0.0015 X 0.1 = 0.00015 and
the next number is 0.00015 X 0.1 = 0.000015 so on,
Therefore 15, 1.5, 0.15, 0.015, . . . the next three numbers are
15, 1.5, 0.15, 0.015, 0.0015, 0.00015, 0.000015.

Question 70.
0.04, 0.02, 0.01, 0.005, . . .
0.04, 0.02, 0.01, 0.005, 0.0025, 0.00125, 0.000625

Explanation:
Given series as 0.04, 0.02, 0.01, 0.005 each number is
divided by 2 as 0.04 ÷ 2 = 0.02,
0.02 ÷ 2 = 0.01, 0.01 ÷ 2 = 0.005,
the next number is 0.005 ÷ 2 = 0.0025,
the next number is 0.0025 ÷ 2 = 0.00125,
and the next number is 0.00125 ÷ 2 = 0.000625
and so on, therefore 0.04, 0.02, 0.01, 0.005, . . .
the next three numbers are
0.04, 0.02, 0.01, 0.005, 0.0025, 0.00125, 0.000625 respectively.

Question 71.
5, 7.5, 11.25, 16.875, . . .
5, 7.5, 11.25, 16.875, 25.3125, 37.96875, 56.953125

Explanation:
Given series are 5, 7.5, 11.25, 16.875, each number
is multiplied by 1.5 of the previous number as shown, So
5 X 1.5 = 7.5, 7.5 X 1.5 = 11.25, 11.25 X 1.5 = 16.875,
the next number is 16.875 X 1.5 = 25.3125,
the next number after 25.3125 X 1.5 = 37.96875 and
the next number is 37.96875 X 1.5 = 56.953125.
therefore 5, 7.5, 11.25, 16.875, the next three numbers are
5, 7.5, 11.25, 16.875, 25.3125, 37.96875, 56.953125.

Question 72.
DIG DEEPER!
You are preparing for a trip to Canada. At the time of your trip,
each U.S dollar is worth 1.293 Canadian dollars and each
Canadian dollar is worth 0.773 U.S dollar.
a. You exchange 150 U.S dollars for Canadian dollars. How many
Canadian dollars do you receive?
b. You spend 120 Canadian dollars on the trip. Then you exchange
the remaining Canadian dollars for U.S dollars.
How many U.S dollars do you receive?
a. I receive 193.95 Canadian dollars,
b. I receive 57.16335 U.S dollars,

Explanation:
Given I am preparing for a trip to Canada. At the time of my trip,
each U.S dollar is worth 1.293 Canadian dollars and each
Canadian dollar is worth 0.773 U.S dollar.
a. I exchange 150 U.S dollars for Canadian dollars ,
So i receive Canadian dollars as 1.293 X 150 =
 1,    4,1
1.293—– 3 decimal places
x150
000000
064650
129300
193.950—– 3 decimal places
Therefore, I receive 193.95 Canadian dollars.
b. I spend 120 Canadian dollars on the trip. Then I exchange
the remaining Canadian dollars for U.S dollars.
So I have Canadian dollars left after my trip are
193.950 – 120 = 73.950, Now i will convert
Canadian dollars into U.S dollars as 73.950 X 0.773 =
2,6,3   
    2,6,3  
     2,1
73.950—- 3 decimal places
X0.773—- 3 decimal places
00221850
05176500
51765000
00000000
57.163350—-6 decimal places
Therefore, I receive 57.16335 U.S dollars.

Question 73.
OPEN-ENDED
You and four friends have dinner at a restaurant.
a. Draw a restaurant menu that has main items, desserts, and beverages, with their prices.
b. Write a guest check that shows what each of you ate. Find the subtotal.
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 273
c. Multiply by 0.07 to find the tax. Then find the total.
d. Round the total to the nearest whole number.
Multiply by 0.20 to estimate a tip. Including the tip,
how much did the dinner cost?
a.
Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals-73

b. Guest Check I ate Passion Mousse, Iced Coffee and
Chicken Breasts Quarters only,
My four friends ate Pecan Cheese Cake, Expresso and
Meatloaf with Gravy,
Subtotal is $75.95,
c. Including Tax amount the Total cost is $81.2665,
d. Including the tip, the dinner cost is $96,

Explanation:
a. Shown restaurant menu that has main items, desserts and
beverages, with their prices.
b. I ate Passion Mousse, Iced Coffee and Chicken Breasts Quarters only,
my friend ate Pecan Cheese Cake, Expresso and Meatloaf with Gravy,
Now Passion Mousse, Iced Coffee and Chicken Breasts Quarters only
costs $5.90 + $5.90 + $3.75 =
5.90
5.90
+3.75
15.55
My cost is $15.55
and friend costs are Pecan Cheese Cake, Expresso and
Meatloaf with Gravy  $4.95 + $5.90  + $4.25 =
4.95
5.90
+4.25

15.10
and friend costs are  $15.10, and for 4 friends it is
4 X 15.10 =
2
15.10—-2 decimal values
X 4
60.40 —-2 decimal values
For four friends it is $60.40,
Now Subtotal of mine and my four freind’s are
$15.55 + $60.40 =
15.55
+60.40
75.95
Subtotal costs to $75.95
c. Given tax is 0.07, So tax on amount is
$75.95 X 0.07 =
4,6,3
75.95—-2 decimal places
X 0.07—2 decimal places
5.3165–4 decimal places
So, The total amount after paying tax is $75.95 + $5.3165,
75.9500
+5.3165
81.2665
Therefore, including tax amount the Total cost is $81.2665.
d. Rounding  the total to the nearest whole number as
$81.2665 ≈ $80, now multiplying  by 0.20 we get
$80 X 0.20 = $16 is the tip.
Including the tip, the dinner cost is $80 + $16 = $96.

Question 74.
GEOMETRY
A rectangular painting has an area of 9.52 square feet.
a. Draw three different ways in which this can happen.
b. The cost of a frame depends on the perimeter of the painting.
Which of your drawings from part(a) is the least expensive to frame?
Explain your reasoning.
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 274
c. The thin, black framing costs $1 per foot. The fancy framing costs $5 per foot. Will the fancy framing cost five times as much as the black framing? Explain why or why not.
d. Suppose the cost of a frame depends on the outside perimeter of the frame. Does this change your answer to part(c)? Explain why or why not.
a.
Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals-74
b. Photoframe i (3.2, 2.975)  is the least expensive to frame.
c. Yes, the fancy framing cost five times as much as the black framing,
d. Yes, It will change if it depends on the outside
perimeter of the frame, not on the material used.

Explanation:
a. Given a rectangular painting has an area of 9.52 square feet,
We know area of rectangle is length X width and it is 9.52 square feet,
Drawn three different ways in which this can happen in the above picture,
as i. 3.2 X 2.975 = 9.52 square feet, ii. 3.4  X 2.8 = 9.52 square feet and
3.8 X 2.505 = 9.52 square feet .
b. The perimeter of paintings are
i. 2(3.2 + 2.975) = 2 X 6.175 = 12.35 feet
ii. 2(3.4 + 2.8) = 2 X 6.2 = 12.4 feet
iii. 2( 3.8 + 2.505) = 2 X 6.305 = 12.61 feet
among the three the least perimeter is 12.35 feet,
So, photoframe i(3.2, 2.975)  is the least expensive to frame.
c. The thin, black framing costs $1 per foot.
The fancy framing costs $5 per foot.
So for black framing it is 9.52 X 1 =9.52 dollars,
and for fancy framing it costs 9.52 X 5 = 47.6 dollars,
which is 5 times more than black framing, therefore
fancy framing costs five times as much as the black framing.
d. If the cost of a frame depends on the outside perimeter of the frame
not the material used then black framing and fancy frame will have
same cost, which will differ from part c results.

Lesson 2.6 Dividing Whole Numbers

EXPLORATION 1
Using a Double Bar Graph
Work with a partner. The double bar graph shows the history of a citywide cleanup day.
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 275
a. Make five conclusions from the graph.
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 276
b. Compare the results of the city cleanup day in 2016 to the results in 2014.
c. What is the average combined amount of trash and recyclables collected each year over the four-year period?
d. Make a prediction about the amount of trash collected in a future year.

a. Five conclusions from the graph:
1. Every year the amount of trash and recyclables
quantity is increasing.
2. Every year the trash quantity is more
than the quantity of recyclables.
3. In the Year 2017 the amount of trash
and recyclables collected in pounds is more
than 7000 pounds.
4. In the year 2014 he amount of trash
and recyclables collected in pounds is less
than 3000 pounds.
5. We have collected information of data
for 4 consecutive years.

b.  In 2016 amount collected is 4970 + 732 = 5702 pounds and
in 2014 it is 2310 + 183 = 2493 pounds, Therefore in 2016 the
amount collected  is more compared to 2014.

c. The average combined amount of trash and recyclables
collected each year over the four-year period is 5052.5 pounds.

d. We can predict about the amount of trash collected
in a future year 2018 will be approximately 7500 pounds.

Explanation:
a. Five conclusions from the graph are written as below
1. Every year the amount of trash and recyclables
quantity is increasing in 2014 it is 2310, 183,
in 2015 it is 3975,555, in 2016 it is 4970, 732 and
in year 2017 it is 6390, 1095 pounds.
2. Every year the trash quantity is more
than the quantity of recyclables.
In 2014 – 2310 > 183,
In 2015 – 3975 > 555,
In 2016 – 4970 > 732,
In 2017 – 6390 > 1095.
3. In the Year 2017 the amount of trash
and recyclables collected in pounds is more
than 7000 pounds because in 2017 it is 6390 + 1095 =
7485 pounds which is more than 7000 pounds.
4. In the year 2014 he amount of trash
and recyclables collected in pounds is less
than 3000 pounds because in 2014 it is 2310 + 183
= 2493 pounds which is less than 3000 pounds.
5. We have collected information of data
for 4 consecutive years as 2014 , 2015 , 2016 and 2017.

b. In 2016 it is 4970 pounds of trash  and 732 pounds of recyclables,
In 2014 it is 2310 pounds of trash and 183 pounds of recyclables,
Now in n 2016 amount collected is 4970 + 732 = 5702 pounds and
in 2014 it is 2310 + 183 = 2493 pounds, Therefore in 2016 the
amount collected  is more compared to 2014.

c. The average combined amount of trash and recyclables
collected each year is
In 2014 = 2310 + 183 = 2493 pounds
In 2015 = 3975 + 555 = 4530 pounds
In 2016 = 4970 + 732 = 5702 pounds
In 2017 = 6390 + 1095 = 7485 pounds
Total amount of collection in 4 years is
2493 + 4530 + 5702 + 7485 = 20210 pounds
and average for 4 years is 20210 ÷ 4 = 5052.5 pounds,
Therefore the average combined amount of trash and recyclables
collected each year over the four-year period is 5052.5 pounds.

d. Every year we see how much it increased
2015 & 2014- trash is 3975 – 2310 = 1665 pounds,
recyclables is 555 – 183 = 372 pounds
2016 & 2015 – trash 4970 – 3975 = 995 pounds
recyclables is 732 – 555 = 177 pounds
2017 & 2016 – trash 6390 – 4970 = 1420
recyclables is 1095 – 732 = 363 pounds,
as we see every year trash has increased
So we can predict about the amount of trash collected
in a future year 2018 will be approximately 7500 pounds.

Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 277
2.6 Lesson

You have used long division to divide whole numbers. When the divisor divides evenly into the dividend, the quotient is a whole number.
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 278
When the divisor does not divide evenly into the dividend, you obtain a remainder. When this occurs, you can write the quotient as a mixed number.

Try It
Divide. Use estimation to check your answer.

Question 1.
234 ÷ 9
234 ÷ 9 = 26
Estimation is reasonable,

Explanation:
Given 234 ÷ 9 =
Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals-75
234 ÷ 9 = 26,
Estimation is 230 ÷ 9 = 25.555 ≈ 26,
So estimation is reasonable.

Question 2.
\(\frac{6096}{30}\)
\(\frac{6096}{30}\) = 203 and 6 remainder,
Estimation is reasonable,

Explanation:
Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals-76
\(\frac{6096}{30}\) = 203 and 6 remainder,
Estimation is \(\frac{6100}{30}\) = 203.33 ≈ 203,
So estimation is reasonable,

Question 3.
45,691 ÷ 28
45,691 ÷ 28 = 1631 and remainder is 23,
Estimation is reasonable,

Explanation:
Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals-77
45,691 ÷ 28 = 1631 and remainder is 23,
Estimation is 45700 ÷ 28 = 1632 ≈ 1631,
So estimation is reasonable.

Question 4.
Find the quotient of 9920 and 320.
The quotient of 9920 and 320 is 31,
Estimation is reasonable,

Explanation:
Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals-78
The quotient of 9920 and 320 is 31,
Estimation is 9900 ÷ 320 = 30.93 ≈ 31,
So estimation is reasonable.

Try It

Question 5.
WHAT IF?
You make 30 equal payments for the go-kart.
Total is $1380. How much is each payment?
Each payment is $46.

Explanation:
Given I make 30 equal payments for the go-kart.
total is $1380.
Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals-79
So each payment is $46.

Self-Assessment for Concepts & Skills

Solve each exercise. Then rate your understanding of the success criteria in your journal.

DIVIDING WHOLE NUMBERS
Divide. Use estimation to check your answer.

Question 6.
876 ÷ 12
876 ÷ 12 = 73,
Estimation is reasonable.

Explanation:
Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals-80
876 ÷ 12 = 73,
Estimation is 900 ÷ 12 = 75 ≈ 73,
So estimation is reasonable.

Question 7.
3024 ÷ 7
3024 ÷ 7 = 432
Estimation is not reasonable,

Explanation:
Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals-81
3024 ÷ 7 = 432,
Estimation is 3000 ÷ 7 = 428.5 ≈ 429 ≠ 432
So, estimation is not reasonable.

Question 8.
1043 ÷ 22
1043 ÷ 22 = 47 and remainder is 9,
Estimation is not reasonable.

Explanation:
Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals-82
1043 ÷ 22 = 47 and remainder is 9,
Estimation is 1000 ÷ 22 = 45 and remainder is 10 ≠
47 and remainder is 9, So, estimation is not reasonable.

Question 9.
VOCABULARY
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 279
Use the division problem shown to tell whether the number is the
divisor, dividend, or quotient.
a. 884 — dividend
b. 26 —quotient
c. 34 — divisor

Explanation:
Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals-83

In division, we divide a number by any other number to
get another number as a result. So, the number which is
getting divided here is called the dividend.
The number which divides a given number is the divisor.
And the number which we get as a result is known as the quotient.
Divisor Formula: The operation of division in the form of:
Dividend ÷ Divisor = Quotient,
The above expression can also be written as:
Divisor = Dividend ÷ Quotient
Here, ‘÷’ is the symbol of division. But sometimes,
it is also represented by the ‘/’ symbol, such as
Dividend / Divisor = Quotient.

Question 10.
NUMBER SENSE
Without calculating, decide which is greater:
3999 ÷ 129 or 3834 ÷ 142. Explain.
3999 ÷ 129 is greater,

Explanation:
As given to find  which is greater among
3999 ÷ 129 or 3834 ÷ 142 if we compare
numerators 3999 > 3834 and denominators
129 < 142 , So obviously 3999 ÷ 129 is greater.

Question 11.
REASONING
In a division problem, can the remainder be greater than the divisor? Explain.

No, the remainder cannot be greater than the divisor,

Explanation:
Remainder means something which is ‘left over’ or ‘remaining’.
When one number divides another number completely,
the remainder is 0.
The remainder is always less than the divisor.
If the remainder is greater than the divisor,
it means that the division is incomplete,
therefore the remainder cannot be greater than the divisor,

Self-Assessment for Problem Solving

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 12.
In a movie’s opening weekend, 879,575 tickets are sold in 755 theaters.
The average cost of a ticket is $9.50. What is the average amount of
money earned by each theater?
The average amount of money earned by each theater is
$11067.5,

Explanation:
Given in a movie’s opening weekend, 879,575 tickets are sold
in 755 theaters. So each theaters it is 879,575 ÷ 755 = 1165
Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals-85
Now the average cost of a ticket is $9.50 and we have
each theaters tickets sold are 1165, So average amount earned
is 1165 X $9.50 =
1165
X9.50—- 2 decimal places
0000000
0058250
1048500
11067.50— 2 decimal places
Therefore, the average amount of money earned by each theater is
$11067.5.

Question 13.
A boat can carry 582 passengers to the base of a waterfall.
A total of 13,105 people ride the boat today.
All the rides are full except for the first ride. How many rides are given?
How many people are on the first ride?
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 280
Rides given are 22 and passengers on first ride are 301,

Explanation:
Given a boat can carry 582 passengers to the base of a waterfall.
A total of 13,105 people ride the boat today.
So rides given are 13,105 ÷ 582 =
Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals-86
Given all the rides are full except for the first ride,
means total 22 rides and for first ride there are
301 passengers.

Question 14.
DIG DEEPER!
A new year begins at 12:00 A.M. on January 1.
What is the date and time 12,345 minutes after
the start of a new year?

The date is January 9th , time is 23:45 pm,

Explanation:
Given new year begins at 12:00 A.M. on January 1.
The date and time 12,345 minutes after the start of a new year
will be, first we convert into hours as 1 hour means
60 minutes, 12,345 ÷ 60 =
Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals-87
205 hours and 45 minutes, Now each day has 24 hours,
So 205 ÷ 24 =
Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals-88
So therefore it will be 8 complete days with
23 hours and 45 minutes,
the date is January 9th , time is 23:45 pm.

Dividing Whole Numbers Homework & Practice 2.6

Review & Refresh

Multiply.

Question 1.
8 × 3.79
8 X 3.79 = 30.32

Explanation:
Given expression as 8 × 3.79 =
 6,7
3.79—— 2 decimal places
X  8
30.32—— 2 decimal places

Therefore, 8 X 3.79 = 30.32.

Question 2.
12.1 × 2.42
12.1 × 2.42 = 29.282

Explanation:
Given expression as 12.1 × 2.42 =
12.1—— 1 decimal place
X 2.42—–2 decimal places
00242
04840
24200
29.282—–3 decimal places

Therefore, 12.1 × 2.42 = 29.282.

Question 3.
6.43 × 0.28
6.43 × 0.28 = 1.8004

Explanation:
Given expression as 6.43 × 0.28 =
6.43—— 2 decimal places
X 0.28—–2 decimal places
005144
012860
000000
1.8004—–4 decimal places

Question 4.
9.526 . 6.61
9.526 X  6.61 = 62.96686

Explanation:
Given expression as 9.526 × 6.61 =
9.526—– 3 decimal places
X 6.61—–2 decimal places
0009526
0571560
5715600
62.96686—–5 decimal places

List the factor pairs of the number.

Question 5.
26.
Factors pairs of 26 = (1,26), (2, 13),

Explanation:
Factors of 26 : 1, 2, 13, 26,
So factor pairs of 26 are 1 x 26 or 2 x 13,
(1,26), (2, 13).

Question 6.
72
Factors pairs of 72 are (1, 72) or (2, 36) or (3, 24) or
(4, 18) or (6, 12) or (8, 9).

Explanation:
Factors of 72 : 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36 and 72,
So factors pairs of 72 are (1, 72) or (2, 36) or (3, 24) or
(4, 18) or (6, 12) or (8, 9).

Question 7.
50
Factors pairs of 50 are (1, 50), (2, 25) ,(5,10),

Explanation:
Factors of 50 : 1, 2, 5,10, 25, and 50,
Factor pairs of 50 are (1, 50), (2, 25) ,(5,10).

Question 8.
98
Factors pairs of 98 are (1, 98), (2, 49), (7, 14),

Explanation:
Factors of 98 : 1, 2, 7, 14, 49, and 98,
Factors pairs of 98 are (1, 98), (2, 49), (7, 14).

Match the expression with its value.

Question 9.
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 281
\(\frac{6}{7}\)  ÷ \(\frac{3}{5}\) = 1\(\frac{3}{7}\),
matches with B,

Explanation:
Given expressions as \(\frac{6}{7}\)  ÷ \(\frac{3}{5}\),
we write reciprocal of the fraction \(\frac{3}{5}\) as
\(\frac{5}{3}\) and multiply as \(\frac{6}{7}\) X \(\frac{4}{7}\) =
\(\frac{6 X 5}{7 X 3}\) = \(\frac{30}{21}\),
as both goes in 3, 3 X 10 = 30, 3 X 7 = 21, \(\frac{30}{21}\) =
\(\frac{10}{7}\) as numerator is greater we write as
(1 X 7 + 3 by 7) = 1\(\frac{3}{7}\) matches with B.

Question 10.
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 282
\(\frac{3}{7}\)  ÷ \(\frac{6}{5}\) = \(\frac{5}{14}\),
matches with C,

Explanation:
Given expressions as \(\frac{3}{7}\)  ÷ \(\frac{6}{5}\),
we write reciprocal of the fraction \(\frac{6}{5}\) as
\(\frac{5}{6}\) and multiply as \(\frac{3}{7}\) X \(\frac{5}{6}\) =
\(\frac{3 X 5}{7 X 6}\) = \(\frac{15}{42}\),
as both goes in 3, 3 X 5 = 15, 3 X 14 = 42, \(\frac{15}{42}\) =
\(\frac{5}{14}\) matches with C.

Question 11.
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 283
\(\frac{6}{5}\)  ÷ \(\frac{3}{7}\) = 2\(\frac{4}{5}\),
matches with D,

Explanation:
Given expressions as \(\frac{6}{5}\)  ÷ \(\frac{3}{7}\),
we write reciprocal of the fraction \(\frac{3}{7}\) as
\(\frac{7}{3}\) and multiply as \(\frac{6}{5}\) X \(\frac{7}{3}\) =
\(\frac{6 X 7}{5 X 3}\) = \(\frac{42}{15}\),
as both goes in 3, 3 X 14 = 72, 3 X 5 = 15, \(\frac{42}{15}\) =
\(\frac{14}{5}\) as numerator is greater we write as
(2 X 5 + 4 by 5) = 2\(\frac{4}{5}\) matches with D.

Question 12.
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 284
\(\frac{3}{5}\)  ÷ \(\frac{6}{7}\) = \(\frac{7}{10}\),,
matches with A,

Explanation:
Given expressions as \(\frac{3}{5}\)  ÷ \(\frac{6}{7}\),
we write reciprocal of the fraction \(\frac{6}{7}\) as
\(\frac{7}{6}\) and multiply as \(\frac{3}{5}\) X \(\frac{7}{6}\) =
\(\frac{3 X 7}{5 X 6}\) = \(\frac{21}{30}\),
as both goes in 3, 3 X 7 = 21, 3 X 10 = 30, \(\frac{21}{30}\) =
\(\frac{7}{10}\) matches with A.

Concepts, Skills, & Problem Solving
OPERATIONS WITH WHOLE NUMBERS
The bar graph shows the attendance at a food festival. Use the graph to answer the question. (See Exploration 1, p. 81.)

Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 285

Question 13.
What is the total attendance at the food festival from 2014 to 2017?

The total attendance at the food festival from
2014 to 2017 is 16,648,

Explanation:
As shown in bar graph we add attendance from
year 2014 to 2017 as 2118 + 3391 + 4785 + 6354 =
2118
3391
4785
+6354
16,648
Therefore ,the total attendance at the food festival from
2014 to 2017 is 16,648.

Question 14.
How many times more people attended the food festival in 2017 than in 2014?

4,236 more people attended the food festival in 2017,

Explanation:
In Year 2017 people attended the food festival are 6354 and
in year 2014 it is 2118, So more people attended the food festival
in 2017 are 6354 – 2118 =
6354
-2118
4236
So, 4,236 more people attended the food festival in 2017.

Question 15.
What is the average attendance at the festival each year over the four-year period?

The average attendance at the festival each year
over the four-year period is 4162,

Explanation:
We have total attendance at the food festival from
2014 to 2017 is 16,648, Now the average attendance is
16648 ÷ 4 = 4162
Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals-89
Therefore, The average attendance at the festival each year
over the four-year period is 4162.

Question 16.
The festival projects that the attendance for 2018 will be
twice the attendance in 2016. What is the
projected attendance for 2018?

The projected attendance for 2018 is 9570,

Explanation:
Given the attendance in 2016 is 4785,
the projected attendance for 2018 is
twice the attendance in 2016 is 2 X 4785 = 9570.
Therefore, the projected attendance for 2018 is 9570.

DIVIDING WHOLE NUMBERS
Divide. Use estimation to check your answer.

Question 17.
837 ÷ 27
837 ÷ 27 = 31,

Explanation:
Given expression as 837 ÷ 27 =
Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals-90
So, 837 ÷ 27 = 31.

Question 18.
1088 ÷ 34
1088 ÷ 34 = 32,

Explanation:
Given expression as 1088 ÷ 34 =
Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals-91
Therefore, 1088 ÷ 34 = 32.

Question 19.
903 ÷ 72
903 ÷ 72 = 12\(\frac{13}{24}\),

Explanation:
Given \(\frac{903}{72}\) as both goes in 3,
3 X 301, 3 X 24 = 72, (301, 24) = \(\frac{301}{24}\) as
numerator is greater we write as (12 X 24 + 13 by 24) =
12\(\frac{13}{24}\).

Question 20.
6409 ÷ 61
6409 ÷ 61 = 105\(\frac{4}{61}\),

Explanation:
Given \(\frac{6409}{61}\) as
numerator is greater we write as (105 X 61 + 4 by 61) =
105\(\frac{4}{61}\).

Question 21.
\(\frac{5986}{82}\)
\(\frac{5986}{82}\) = 73,

Explanation:
Given \(\frac{5986}{82}\) as both goes in 82,
82 X 73 = 5986 we get \(\frac{5986}{82}\) = 73.

Question 22.
6200 ÷ 163
6200 ÷ 163 = 38\(\frac{6}{163}\)

Explanation:
Given expression as \(\frac{6200}{163}\) here
numerator is greater we write as ( 38 X 163  + 6 by 163) =
38\(\frac{6}{163}\).

Question 23.
6255 ÷ 118
6255 ÷ 118 = 53\(\frac{1}{118}\)

Explanation:
Given expression as \(\frac{6255}{118}\), here
numerator is greater we write as (53 X 118 + 1 by 118) =
53\(\frac{1}{118}\).

Question 24.
\(\frac{588}{84}\)
\(\frac{588}{84}\) = 7

Explanation:
Given expression as \(\frac{588}{84}\) as
both goes in 84, 84 X 7 = 588, 84 X 1 = 84,
So \(\frac{588}{84}\) = 7.

Question 25.
7440 ÷ 124
7440 ÷ 124 = 60

Explanation:
Given expression as \(\frac{7440{124}\), here
both goes in 124 as 124 X 60 = 7440, 124 X 1 = 124, (60, 1),
therefore \(\frac{7440{124}\) = 60.

Question 26.
26,862 ÷ 407
26,862 ÷ 407 = 66

Explanation:
Given expression as \(\frac{26862}{407}\), here
both goes in 407 as 407 X 66 = 26862, 407 X 1 = 407, (66, 1),
therefore \(\frac{26862}{407}\) = 66.

Question 27.
8241 ÷ 173
8241 ÷ 173 = 47\(\frac{110}{173}\),

Explanation:
Given expression as \(\frac{8241}{173}\),
numerator is greater so we write as (47 X 173 + 110 by 173),
\(\frac{8241}{173}\) = 47\(\frac{110}{173}\).

Question 28.
\(\frac{33,505}{16}\)
\(\frac{33,505}{16}\) = 2094\(\frac{1}{16}\),

Explanation:
Given expression as \(\frac{33505}{16}\),
numerator is greater so we write as (2094 X 16 + 1 by 16),
therefore \(\frac{33505}{16}\) = 2094\(\frac{1}{16}\).

Question 29.
MODELING REAL LIFE
A pharmacist divides 364 pills into prescription bottles.
Each bottle contains 28 pills. How many bottles does
the pharmacist fill?

The pharmacist fills 13 bottles,

Explanation:
Given a pharmacist divides 364 pills into prescription bottles.
Each bottle contains 28 pills, number of bottles the pharmacist fill
are 364 ÷ 28 = 13,
Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals-92
therefore, the pharmacist fills 13 bottles.

YOU BE THE TEACHER
Your friend finds the quotient. Is your friend correct? Explain your reasoning.

Question 30.
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 286
No, friend is incorrect as 963 ÷ 8 = 120 and remainder is 3
≠ 12 remainder 3,

Explanation:
Given expression as \(\frac{963}{8}\),
numerator is greater so we write as ( 120 X 8 + 3 by 8),
therefore \(\frac{963}{16}\) = 120\(\frac{3}{8}\) ≠
12 remainder 3, So friend is incorrect.

Question 31.
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 287
No, friend is incorrect as 1308 ÷ 12 = 109 ≠ 19,

Explanation:
Given expression as \(\frac{1308}{12}\),
numerator is greater and 12 X 109 = 1308
therefore \(\frac{1308}{12}\) = 109 ≠ 19,
So friend is incorrect.

GEOMETRY
Find the perimeter of the rectangle.

Question 32.
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 288
The perimeter of the rectangle is 24 inches,

Explanation:
Given area of rectangle as 35 in2 and width as 7 in,
we know area of rectangle is width X length,
35 = 7 X length, So length = 35 ÷ 7 = 5 ( 7 X 5 = 35),
So length of rectangle is 5 in, Now perimeter of
the rectangle is 2( length + width) = 2, (5 + 7) =
2 X 12 = 24 inches, therefore the perimeter of the
rectangle is 24 inches.

Question 33.
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 289
The perimeter of the rectangle is 42 ft,

Explanation:
Given area of rectangle as 108 ft2 and width as 12 ft,
we know area of rectangle is width X length,
108 = 12 X length, So length = 108 ÷ 12 = 9, ( 9 X 12 = 108),
So length of rectangle is 9 ft, Now perimeter of
the rectangle is 2( length + width) = 2 (9 + 12) = 2 X 21 = 42 ft,
therefore the perimeter of the rectangle is 42 ft.

Question 34.
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 290

The perimeter of the rectangle is 36 m,

Explanation:
Given area of rectangle as 80 m2 and width as 10 m,
we know area of rectangle is width X length,
80 = 10 X length, So length = 80 ÷ 10 = 8, (8 X 10 = 80),
So length of rectangle is 8 m, Now perimeter of
the rectangle is 2( length + width) = 2 (8 + 10) = 2 X 18 = 36 m,
therefore the perimeter of the rectangle is 36 m.

Question 35.
REASONING
You borrow bookcases like the one shown to display 943 books
at a book sale. You plan to put 22 books on each shelf.
No books will be on top of the bookcases.
a. How many bookcases must you borrow to display all the books?
b. You fill the shelves of each bookcase in order, starting
with the top shelf. How many books are on each shelf of the
last bookcase?
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 291
a. 9 bookcases I must borrow to display all the books,
b. Books in  each shelf of the last bookcase are
shelf 1-22, shelf  2- 22, shelf 3-19, shelf 4-0, shelf 5 -0 books,

Explanation:
Given I borrow bookcases like the one shown to display
943 books at a book sale.I plan to put 22 books on each
shelf and no books will be on top of the bookcases.
a. Number of  bookcases I must borrow to display
all the books are each bookcases have 5 shelves and in each
shelf I can put 22 books so in each bookcase I can put
5 X 22 = 110 books, Now 943 ÷ 110 =
\(\frac{943}{110}\) as numerator is greater we
write as ( 8 X 110 + 63 by 110) = 8\(\frac{63}{110}\) ,
means 8 bookcases and 63 books are still there,
so I require 9 bookcases to completely keep the books
in the shelves.
b. Now I fill the shelves of each bookcase in order, starting
with the top shelf. I have 63 books to arrange in 9 bookcase,
So books are on each shelf of the last bookcase are
shelf 1 – 22 books, now we are left with  63 – 22 = 41 books
shelf 2 = 22 books, now we are left with 41 – 22 = 19 books
shelf 3 = 19 books,
shelf 4 = 0 books,
shelf 5 = 0 books respectively.

Question 36.
DIG DEEPER!
The siding of a house is 2250 square feet. The siding needs two coats of paint.
a. What is the minimum cost of the paint needed to complete the job?
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 292
b. How much paint is left over when you spend the minimum amount?

a. The minimum cost of the paint needed to complete the job is $435,
b. \(\frac{30}{32}\) gallon paint is left over
when I spend the minimum amount,

Explanation:
Given the siding of a house is 2250 square feet.
The siding needs two coats of paint. So the siding
becomes 2 X 2250 = 4500 square feet,
Now we check with the paints needed If we take
1 quart it will cover 80 square feet and 1 quart cost is $ 18 means
we require 4500 ÷ 80 = \(\frac{4500}{80}\) as both goes
in 10, 10 X 450 = 4500, 10 X 8 = 80, (450 , 8) = \(\frac{450}{8}\)
as numerator is greater we write as ( 56 X 8 + 2 by 8) = 56\(\frac{2}{8}\) ,
we need approximately 57 quart, the cost will be
57 X $18 = $1026 to complete the job,
If we take 1 gallon it will cover 320 square feet and
1 gallon costs is $29 means we require 4500 ÷ 320 =
\(\frac{4500}{320}\) as both goes in 10, 10 X 450 = 4500,
10 X 32 = 320, (450 , 32) = \(\frac{450}{32}\)
as numerator is greater we write as ( 14 X 32 + 2 by 32) = 14\(\frac{2}{32}\) ,
we need approximately 15 gallons, the cost will be
15 X $29 = $435 to complete the job,
So the minimum cost of the paint needed to complete the job is $435.
b. The paint left over when I spend the minimum amount is
15 –  \(\frac{450}{32}\) = (15 X 32 – 450 by 32) =
(480-450 by 32) = \(\frac{30}{32}\) gallon paint is left.

Question 37.
CRITICAL THINKING
Use the digits 3, 4, 6, and 9 to complete the division problem. Use each digit once.
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 293
36,000 ÷ 900 = 40 or 36,000 ÷ 400 = 90,

Explanation:
To complete the division problem we use digits
3, 4, 6, 9 each digit once as if we take first 3,4,
we are left with digits 6, 9,
34,000 ÷ 900 ≈ 38 not matches to complete the division problem,
34,000 ÷ 600 ≈ 57 not matches to complete the division problem,
next 36,000 ÷ 900 = 40 matches to complete the division problem,
36,000 ÷ 400 = 90 matches to complete the division problem,
next we take 4, 3 we are left with digits 6 ,9,
43,000 ÷ 900 ≈ 47 not matches to complete the division problem,
43,000 ÷ 600 ≈ 71 not matches to complete the division problem,
next we take 6, 4  we are left with digits 3 ,9,
64,000 ÷ 300 ≈ 213 not matches to complete the division problem,
64,000 ÷ 900 ≈ 71 not matches to complete the division problem,
next we take 9, 6 we are left with digits 3,4,
96,000 ÷ 300 ≈ 320 not matches to complete the division problem,
96,000 ÷ 400 ≈ 240 not matches to complete the division problem,
So to complete the division problem we write as
36,000 ÷ 900 = 40 or 36,000 ÷ 400 = 90.

Lesson 2.7 Dividing Decimals

EXPLORATION 1
Dividing Decimals
Work with a partner.
a. Write two division expressions represented by each area model. Then find the quotients. Explain how you found your answer.
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 294
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 295
b. Use a calculator to find 119 ÷ 17, 11.9 ÷ 1.7, 1.19 ÷ 0.17
and 0.119 ÷ 0.017. What do you notice? Explain how you can
use long division to divide any pair of multi-digit decimals.

a. i. Two division expressions are \(\frac{40}{10}\) ÷ \(\frac{8}{10}\) =
\(\frac{5}{10}\) or \(\frac{40}{100}\)  ÷ \(\frac{5}{10}\) =
\(\frac{8}{10}\),

ii. The division expression of whole part is
\(\frac{50}{100}\) ÷ \(\frac{5}{10}\) =  \(\frac{10}{10}\) or
and \(\frac{50}{100}\) ÷ \(\frac{10}{10}\) = \(\frac{5}{10}\)
decimal part is \(\frac{25}{10}\) ÷ \(\frac{5}{10}\) = \(\frac{5}{10}\)
or \(\frac{25}{10}\) ÷ \(\frac{5}{10}\) = \(\frac{5}{10}\),

iii. The division expression of whole part is
\(\frac{70}{100}\) ÷  \(\frac{7}{10}\) = \(\frac{10}{10}\) or
\(\frac{70}{100}\) ÷  \(\frac{10}{10}\) = \(\frac{7}{10}\) and
decimal part is \(\frac{49}{10}\) ÷ \(\frac{7}{10}\) = \(\frac{7}{10}\),

b. 19 ÷ 17 = 7, 11.9 ÷ 1.7 = 7, 1.19 ÷ 0.17 = 7
and 0.119 ÷ 0.017 = 7,
To multiply decimals, first multiply as
if there is no decimal. Next, count the number of digits after
the decimal in each factor. Finally, put the same number of digits
behind the decimal in the product.

Explanation:
a. Two division expressions are \(\frac{40}{10}\) ÷ \(\frac{8}{10}\) =
\(\frac{5}{10}\) or \(\frac{40}{100}\)  ÷ \(\frac{5}{10}\) =
\(\frac{8}{10}\), or
By counting the blocks in the area model found
the two division expressions are \(\frac{5}{10}\) ÷ \(\frac{10}{8}\)
first we write the reciprocal \(\frac{10}{8}\) and multiply as
\(\frac{5}{10}\) X \(\frac{8}{10}\)
and Step I: We multiply the numerators as 5 X 8 = 40
Step II: We multiply the denominators as 10 X 10 =100
Step III: We write the fraction in the simplest form as
\(\frac{40}{100}\), So \(\frac{5 X 8}{10 X 10}\) = \(\frac{40}{100}\).
If we see the area model the purple color blocks show
40 out of 100 blocks.
ii. The division expression of whole part is
\(\frac{50}{100}\) ÷ \(\frac{5}{10}\) =  \(\frac{10}{10}\) or
and \(\frac{50}{100}\) ÷ \(\frac{10}{10}\) = \(\frac{5}{10}\)
decimal part is \(\frac{25}{10}\) ÷ \(\frac{5}{10}\) = \(\frac{5}{10}\)
or \(\frac{25}{10}\) ÷ \(\frac{5}{10}\) = \(\frac{5}{10}\), or
By counting the blocks in the area model found
the multiplication expression as \(\frac{10}{10}\) ÷ \(\frac{5}{10}\)
first we write the reciprocal \(\frac{10}{5}\) and multiply as
\(\frac{10}{10}\) X \(\frac{5}{10}\)
and Step I: We multiply the numerators as 10 X 5 = 50
Step II: We multiply the denominators as 10 X 10 =100
Step III: We write the fraction in the simplest form as
\(\frac{50}{100}\), So \(\frac{10 X 5}{10 X 10}\) = \(\frac{50}{100}\).
Now in decimal part we have \(\frac{5}{10}\) ÷ \(\frac{10}{5}\),
first we write the reciprocal \(\frac{10}{5}\) and multiply as
\(\frac{5}{10}\) X \(\frac{5}{10}\)
similar to whole part we do multiplication
Step I: We multiply the numerators as 5 X 5 = 25
Step II: We multiply the denominators as 10 X 10 =100
Step III: We write the fraction in the simplest form as
\(\frac{25}{100}\), therefore then the product results is
\(\frac{50}{100}\) + \(\frac{25}{100}\),

iii. The division expression of whole part is
\(\frac{70}{100}\) ÷  \(\frac{7}{10}\) = \(\frac{10}{10}\) or
\(\frac{70}{100}\) ÷  \(\frac{10}{10}\) = \(\frac{7}{10}\) and
decimal part is \(\frac{49}{10}\) ÷ \(\frac{7}{10}\) = \(\frac{7}{10}\), or By counting the blocks in the area model found
the division expression as \(\frac{10}{10}\) ÷ \(\frac{10}{7}\)
first we write the reciprocal \(\frac{10}{7}\) and multiply as
\(\frac{10}{10}\) X \(\frac{7}{10}\)
and Step I: We multiply the numerators as 10 X 7 = 70
Step II: We multiply the denominators as 10 X 10 =100
Step III: We write the fraction in the simplest form as
\(\frac{70}{100}\), So \(\frac{10 X 7}{10 X 10}\) = \(\frac{70}{100}\).
Now in decimal part we have \(\frac{7}{10}\) ÷ \(\frac{10}{7}\),
first we write the reciprocal \(\frac{10}{7}\) and multiply as
\(\frac{7}{10}\) X \(\frac{7}{10}\),
similar to whole part we do multiplication
Step I: We multiply the numerators as 7 X 7 = 49
Step II: We multiply the denominators as 10 X 10 =100
Step III: We write the fraction in the simplest form as
\(\frac{49}{100}\), therefore then the product results is
\(\frac{70}{100}\) + \(\frac{49}{100}\).

b. 19 ÷ 17 = 7, 11.9 ÷ 1.7 = 7, 1.19 ÷ 0.17 = 7
and 0.119 ÷ 0.017 = 7,
To multiply decimals, first multiply as
if there is no decimal. Next, count the number of digits after
the decimal in each factor. Finally, put the same number of digits
behind the decimal in the product.
Example: 7.6)19.76(
The first thing that we want to do when dividing decimals is to turn
the divisor into a whole number. We do this by moving the decimal
place to the right:
7.6 —–> 76

If we move the decimal over one place in the divisor,
we must also move the decimal over one place in the dividend:
19.76 —–>197.6

The new division problem should look as follows: 
76 )197.6(
We’ve already placed the decimal in our answer.
When we divide decimals, we place the decimal directly above
the decimal in the dividend, but only after we’ve completed the
first two steps of moving the decimal point in the divisor and dividend.

Now we can divide like normal: 76)197.6(
Think: how many times can 76 go into 197
76 can go into 197 two times so we write a 2 over the 7
in the dividend:
76) 197.6(2.
Next, we multiply 2 and 76 and write that product underneath
the 197 and subtract:
76)197.6(2.
  -152   
45

Now we bring down the 6 from the dividend to make the 45 into a 456.

Think: how many times can 76 go into 456?

76 can go into 465 six times so we write a 6 above the 6 in the dividend:
76)197.6(2.6
  -152   
456

Next, we multiply 6 and 76 and write that product underneath
the 456 and subtract:
Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals-93
We are left with no remainder and a final quotient of 2.6.

2.7 Lesson

Key Idea
Dividing Decimals by Whole Numbers
Words
Place the decimal point in the quotient above the decimal point in the dividend.
Then divide as you would with whole numbers. Continue until there is no remainder.
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 296

Try It

Divide. Use estimation to check your answer.

Question 1.
36.4 ÷ 2
36.4 ÷ 2 = 18.2
Estimation is reasonable,

Explanation:
Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals-94
Therefore, 36.4 ÷ 2 = 18.2,
Estimation is 36 ÷ 2 = 18 is reasonable.

Question 2.
22.2 ÷ 6
22.2 ÷ 6 = 3.7,
Estimation is reasonable,

Explanation:
    3.7 
6)22.2               6 X 3 = 18
   18    
4.2               6 X 0.7 = 4.2
     4.2
0
Therefore,  22.2 ÷ 6 = 3.7.
Estimation is 22 ÷ 6 = 3.66 is reasonale.

Question 3.
59.64 ÷ 7
59.64 ÷ 7 = 8.52,
Estimation is reasonable,

Explanation:
    8.52  
7)59.64          7 X 8 = 56
   56    
3.6             7 X 0.5 = 3.5
     3.5   
0.14          7 X 0.02 = 0.14
       0.14
           0
Therefore,  59.64 ÷ 7 = 8.52,
Estimation is 60 ÷ 7 = 8.57 is reasonale.

Question 4.
3.12 ÷ 16
3.12 ÷ 16 = 0.195,
Estimation is reasonable,

Explanation:
    0.195
16)3.12         16 X 0.1 = 1.6
     1.6
      1.52        16 X 0.09 = 1.44
      1.44
      0.08        16 X 0.005 = 0.08
       0.08
           0
Therefore, 3.12 ÷ 16 = 0.195,
Estimation is 3 ÷ 16 = 0.187 is reasonale.

Question 5.
6.224 ÷ 4
6.224 ÷ 4 = 1.556,
Estimation is reasonable,

Explanation:
    1.556
4)6.224         4 X 1 = 4
   4
    2               4 X 0.5 =2
    2   
0.2           4 X 0.05 = 0.2      
0.2
    0.024       4 X 0.006 = 0.024
    0.024
         0
Therefore, 6.224 ÷ 4 = 1.556,
Estimation is 6 ÷ 4 = 1.5 is reasonale.

Question 6.
43.407 ÷ 14
43.407 ÷ 14 = 3.1005,
Estimation is reasonable,

Explanation:
    3.1005 
14)43.407        14 X 3 = 42
     42
     1.4             14 X 0.1 = 1.4
     1.4   
0.007            14 X 0.0005 = 0.007      
0.007
          0
Therefore, 43.407 ÷ 14 = 3.1005,
Estimation is 43 ÷ 14 = 3.0714 is reasonale.

Key Idea

Dividing Decimals by Decimals
Words
Multiply the divisor and the dividend by a power of 10 to make
the divisor a whole number. Then place the decimal point in the
quotient above the decimal point in the dividend and divide as
you would with whole numbers. Continue until there is no remainder.
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 297

Try It

Divide. Check your answer.

Question 7.
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 298
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 298 = 8

Explanation:
Given expression is 9.6 ÷ 1.2 =
    8     
1.2) 9.6      1.2 X 8 = 9.6
       9.6
        0
Therefore, 9.6 ÷ 1.2 = 9.6.

Question 8.
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 299
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 299= 17
Explanation:
      17        
3.4)57.8        3.4 X 17 = 57.8
      57.8
        0
Therefore, 57.8 ÷ 3.4 = 17.

Question 9.
21.643 ÷ 2.3
21.643 ÷ 2.3 = 9.41

Explanation:
      9.41    
2.3)21.643      2.3 X 9 = 20.7
      20.7
        0.94          2.3 X 0.4 = 0.92
        0.92     
0.023         2.3 X 0.01 = 0.023      
0.023
          0
Therefore, 21.643 ÷ 2.3 = 9.41.

Question 10.
0.459 ÷ 0.51
0.459 ÷ 0.51 =0.9

Explanation:
      0.9    
0.51)0.459      0.51 X 0.9 = 0.459
        0.459
          0

Therefore, 0.51 ÷ 0.9 = 0.459

Try It

Divide. Check your answer.

Question 11.
3.8 ÷ 0.16
3.8 ÷ 0.16 = 23.75

Explanation:
      23.75    
0.16)3.8          0.16 X 23 = 3.68
        3.68
        0.120      0.16 X 0.7 = 0.112
        0.112     
0.008        0.16 X 0.05 = 0.008      
0.008
          0
Therefore, 3.8 ÷ 0.16 = 23.75.

Question 12.
15.6 ÷ 0.78
15.6 ÷ 0.78 = 20

Explanation:
      20    
0.78)15.6          0.78 X 20 = 15.6
        15.6
          0
Therefore, 15.6 ÷ 0.78 = 20.

Question 13.
7.2 ÷ 0.048
7.2 ÷ 0.048 = 150

Explanation:
         150    
0.048)7.2          0.048 X 150 = 7.2
         7.2
          0
Therefore, 7.2 ÷ 0.048 = 7.2.

Question 14.
42 ÷ 3.75
42 ÷ 3.75 = 11.2

Explanation:
        11.2    
3.75) 42          3.75 X 11= 41.25
         41.25
        0.75        3.75 X 0.2 = 0.75
        0.75     
0
Therefore, 42 ÷ 3.75 = 11.2.

Self-Assessment for Concepts & Skills
Solve each exercise. Then rate your understanding of the success criteria in your journal.

DIVIDING DECIMALS
Divide. Check your answer.

Question 15.
37.7 ÷ 13
37.7 ÷ 13 = 2.9

Explanation:
       2.9    
13)37.7          13 X 2 = 26
     26
       11.7        13 X 0.9 = 11.7
       11.7    
0
Therefore, 37.7 ÷ 13 = 2.9.

Question 16.
33 ÷ 4.4
33 ÷ 4.4 = 7.5

Explanation:
     7.5    
4.4)33           4.4 X 7 = 30.8
      30.8
        2.2        4.4 X 0.5 = 2.2
        2.2    
0
Therefore, 33 ÷ 4.4 = 7.5.

Question 17.
2.16 ÷ 0.009
2.16 ÷ 0.009 = 240

Explanation:
         240       
0.009)2.16           0.009 X 240 = 2.16
          2.16
             0
Therefore, 33 ÷ 4.4 = 7.5.

Question 18.
NUMBER SENSE
Fix the one that is not correct.
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 300
The one is not correct,
    6.1   
4)2.44

Explanation:
    0.61  
4)2.44            4 X 0.61 = 2.44
   2.44
      0
2.44 ÷4 = 0.61 ≠ 6.1
    6.1   
4)2.44    is the one which is incorrect.

Question 19.
NUMBER SENSE
Rewrite Big Ideas Math Solutions Grade 6 Chapter 2 Fractions and Decimals 301 so that the divisor is a whole number.

18.5 ÷ 2, Here the divisor is a whole number

Explanation:
Here 2.16 is the divisor,
We rewrite divisor as a whole number as 2.16 ≈ 2,
18.5 ÷ 2.

Question 20.
STRUCTURE
Write 1.8 ÷ 6 as a multiplication problem with a missing factor.
Explain your reasoning.

1.8 ÷ 6 as a multiplication problem with a missing factor is
6 X _____ = 1.8

Explanation:
Given expression as 1.8 ÷ 6 now we write as
a multiplication problem as
we know 1.8 ÷ 6 = 0.3 means , So 1.8 = 6 X 0.3,
Now we write 1.8 = 6 X 0.3, with as missing factor as
6 X _____ = 1.8.
Therefore, 1.8 ÷ 6 as a multiplication problem with a
missing factor is 6 X _____ = 1.8.

Self-Assessment for Problem Solving

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 21.
A magazine subscription costs $29.88 for 12 issues or $15.24 for 6 issues.
Which subscription costs more per issue? How much more?

$15.24 for 6 issues subscription costs more per issue and
more it costs is $0.05,

Explanation:
Given a magazine subscription costs $29.88 for 12 issues,
means 1 issue it is $29.88 ÷ 12 =
     2.49   
12)29.88      12 X 2.49 = 29.33
     29.88
0
So each issue it is $2.49,
Now we have $15.24 for 6 issues means for 1 issue it is
$15.24 ÷ 6 =
    2.54   
6)15.24      6 X 2.54 = 15.24
    15.24 
0
So each issue it is $2.54.
Now how much costs more is $2.54 minus $ 2.49 =
2.54
-2.49
0.05
Therefore, $15.24 for 6 issues subscription costs more per issue and
more it costs is $0.05.

Question 22.
The track of a roller coaster is 1.265 miles long. The ride lasts for 2.3 minutes.
What is the average speed of the roller coaster in miles per hour?

33 miles per hour is the average speed of the roller coaster,

Explanation:
Given the track of a roller coaster is 1.265 miles long.
The ride lasts for 2.3 minutes.
Now the average speed of the roller coaster in miles per hour is
we know speed = distance ÷ time = 1.265 ÷ 2.3 =
       0.55    
2.3)1.265     2.3 X 0.55 = 1.265
      1.265
          0
As speed = 0.55 miles per minute we convert it into hours as
0.55 X 60 = 33 miles per hour.
therefore, 33 miles per hour is the average speed of the roller coaster.

Question 23.
DIG DEEPER!
The table shows the number of visitors to a website each year for 4 years.
Does the number of visitors increase more from Year 1 to Year 2 or
from Year 3 to Year 4? How many times greater is the increase?
Big Ideas Math Solutions Grade 6 Chapter 2 Fractions and Decimals 302
Yes, the number of visitors increased more from Year 1 to Year 2
compared to Year 3 to Year 4,
2.1 times greater is the increase.

Explanation:
Given the table shows the number of visitors to a website
each year for 4 years,
the number of visitors increase more from Year 1 to Year 2 is
Year 2 – 32.22 millions and Year 1- 2.4 millions,
increased in more are 32.22 -2.4 =
32.22
-2.4
29.82 millions
Now the number of visitors increase more from Year 3 to Year 4 is
Year 4 – 102.6 millions and Year 3 – 88.4 millions
increased in more are 102.6 – 88.4 =
102.6
-88.4
14.2 millions
The number of visitors increased more from Year 1 to Year 2 is when
compared to Year 3 to Year 4 , by more times is 29.82 ÷ 14.2 =
         2.1     
14.2)29.82     14.2 X 2 =
        28.40
          1.42     14.2 X 0.1 = 1.42
          1.42
0
So it is 2.1 times greater in the increase from Year 1 to Year 2.

Dividing Decimals Homework & Practice 2.7

Review & Refresh

Divide.

Question 1.
84 ÷ 14
84 ÷ 14 = 6

Explanation:
Given expression as 84 ÷ 14 =
     6    
14)84       14 X 6 = 84
     84
0
Therefore, 84 ÷ 14 = 6.

Question 2.
391 ÷ 23
391 ÷ 23 = 17

Explanation:
Given expression as 391 ÷ 23 =
     17    
23)391       23 X 1 = 23
     23
     161       23 X 7 = 161
     161
0
Therefore, 391 ÷ 23 = 17.

Question 3.
1458 ÷ 54
1458 ÷ 54 = 27

Explanation:
Given expression as 1458 ÷ 54 =
      27    
54)1458       54 X 2 = 108
     108
       378       54 X 7 = 378
       378
0
Therefore, 1458 ÷ 54 = 27.

Question 4.
\(\frac{68,134}{163}\)
\(\frac{68,134}{163}\) = 418,

Explanation:
Given expression as \(\frac{68,134}{163}\) =
        418    
163)68134        163 X 4 = 652
       652
         293         163 X 1 = 163
         163
         1304       163 X 8 = 1304
          1304
0
Therefore, \(\frac{68,134}{163}\) = 418.

Question 5.
What is the value of 18 + 32 ÷ [3 × (8 − 5)]?
A. 3
B. 19
C. 27
D. 49

18 + 32 ÷ [3 × (8 − 5)] = 19, B,

Explanation:
Given 18 + 32 ÷ [3 × (8 − 5)] =
[3 X (8-5)] = 3 X 3 = 9
32 = 3 X 3 = 9,
Now 32 ÷ [3 × (8 − 5)] = 9 ÷ 9 = 1
so, 18 + 1 = 19,
therefore, 18 + 32 ÷ [3 × (8 − 5)] = 19, matches with B.

Add or subtract.

Question 6.
7.635 – 5.046
7.635 – 5.046 = 2.589

Explanation:
Given expression as 7.635 – 5.046 =
  15,12,15
7.635
-5.046
2.589
therefore 7.635 – 5.046 = 2.589.

Question 7.
12.177 + 3.09
12.177 + 3.09 = 15.267,

Explanation:
       1  
12.177
+ 3.090
15.267
therefore 12.177 + 3.09 = 15.267.

Question 8.
14.008 – 9.433
14.008 – 9.433 = 4.575,

Explanation:
 14,9,10 
14.008
– 9.433
4.575
therefore 14.008 – 9.433 = 4.575.

Concepts, Skills, & Problem Solving

DIVIDING DECIMALS
Write two division expressions represented by the area model. Then find the quotients. Explain how you found your answer. (See Exploration 1, p. 87.)

Question 9.
Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals 302.1

Two division expressions are \(\frac{54}{100}\) ÷ \(\frac{6}{10}\) =
\(\frac{9}{10}\) or \(\frac{54}{100}\)  ÷ \(\frac{9}{10}\) =
\(\frac{6}{10}\), we rewrite  the expression as
\(\frac{54}{10}\) = \(\frac{6}{10}\) X \(\frac{9}{10}\),

Explanation:
By counting the blocks in the area model found
the two division expressions are \(\frac{54}{100}\) ÷ \(\frac{6}{10}\)
= \(\frac{9}{10}\)
If we see the area model the purple color blocks show
54 out of 100 blocks.
now we write as \(\frac{54}{100}\) = \(\frac{6}{10}\) X
\(\frac{9}{10}\), So,\(\frac{54}{100}\) = \(\frac{6 X 9}{10 X 10}\).

Question 10.
Big Ideas Math Solutions Grade 6 Chapter 2 Fractions and Decimals 303
Two division expressions are \(\frac{16}{100}\) ÷ \(\frac{2}{10}\) =
\(\frac{8}{10}\) or \(\frac{16}{100}\)  ÷ \(\frac{8}{10}\) =
\(\frac{2}{10}\), we rewrite  the expression as
\(\frac{16}{100}\) = \(\frac{8}{10}\) X \(\frac{2}{10}\),
So \(\frac{16}{100}\) = \(\frac{8 X 2}{10 X 10}\).

Explanation:
By counting the blocks in the area model found
the two division expressions are \(\frac{16}{100}\) ÷ \(\frac{6}{10}\)
= \(\frac{2}{10}\) or \(\frac{16}{100}\)  ÷ \(\frac{8}{10}\) =
\(\frac{2}{10}\)
If we see the area model the purple color blocks show
16 out of 100 blocks.
now we write as \(\frac{16}{100}\) = \(\frac{6}{10}\) X
\(\frac{2}{10}\), So,\(\frac{16}{100}\) = \(\frac{6 X 2}{10 X 10}\).

DIVIDING DECIMALS BY WHOLE NUMBERS
Divide. Use estimation to check your answer.

Question 11.
Big Ideas Math Solutions Grade 6 Chapter 2 Fractions and Decimals 304
Big Ideas Math Solutions Grade 6 Chapter 2 Fractions and Decimals 304= 4.2

Explanation:
Given expression as 25.2 ÷ 6
    4.2            
6)25.2          6 X 4 = 24
   24
     1.2         6 X 0.2 = 1.2
     1.2
        0
Therefore 25.2 ÷ 6 = 4.2.

Question 12.
Big Ideas Math Solutions Grade 6 Chapter 2 Fractions and Decimals 305
Big Ideas Math Solutions Grade 6 Chapter 2 Fractions and Decimals 305= 6.7

Explanation:
Given expression as 33.5 ÷ 5
    6.7            
5)33.5          5 X 6 = 30
   30
     3.5         5 X 0.7 = 3.5
     3.5
        0
Therefore 33.5 ÷ 5 = 6.7.

Question 13.
Big Ideas Math Solutions Grade 6 Chapter 2 Fractions and Decimals 306
Big Ideas Math Solutions Grade 6 Chapter 2 Fractions and Decimals 306 = 0.5

Explanation:
Given expression as 3.5 ÷ 7
    0.5            
7)3.5              7 X 0.5 = 3.5
   3.5
    0
Therefore 3.5 ÷ 7 = 0.5.

Question 14.
Big Ideas Math Solutions Grade 6 Chapter 2 Fractions and Decimals 307
Big Ideas Math Solutions Grade 6 Chapter 2 Fractions and Decimals 307= 1.3
Explanation:
Given expression as 3.5 ÷ 7
    0.5            
7)3.5              7 X 0.5 = 3.5
   3.5
    0
Therefore 3.5 ÷ 7 = 0.5.

Question 15.
38.79 ÷ 9
38.79 ÷ 9 = 4.31

Explanation:
Given expression as  38.79 ÷ 9 =
      4.31    
9)38.79        9 X 4 = 36
   36
       2.7        9 X 0.3 = 2.7
       2.7
        0.09      9 X 0.01 = 0.09
        0.09
0
Therefore, 38.79 ÷ 9 = 4.31.

Question 16.
37.72 ÷ 4
37.72 ÷ 4 = 9.43

Explanation:
Given expression as  37.72 ÷ 4 =
     9.43    
4)37.72        4 X 9 = 36
   36
      1.7         4 X 0.4 = 1.6
      1.6
        0.12     4 X 0.03 = 0.12
        0.12
0
Therefore, 37.72 ÷ 4 = 9.43.

Question 17.
43.4 ÷ 7
43.4 ÷ 7 = 6.2

Explanation:
Given expression as  43.4 ÷ 7 =
    6.2    
7)43.4        7 X 6 = 42
   42
      1.4        7 X 0.2 = 1.4
      1.4
        0
Therefore, 43.4 ÷ 7 = 6.2.

Question 18.
22.505 ÷ 7
22.505 ÷ 7 = 3.215

Explanation:
Given expression as  22.505 ÷ 7 =
   3.215    
7)22.505        7 X 3 = 21
   21
     1.5            7 X 0.2 = 1.4
     1.4
     0.10          7 X 0.01 = 0.07
     0.07
     0.035        7 X 0.005 = 0.035
     0.035
        0 
Therefore, 22.505 ÷ 7 = 3.215.

Question 19.
44.64 ÷ 8
44.64 ÷ 8 = 5.58

Explanation:
Given expression as  44.64 ÷ 8 =
   5.58    
8)44.64        8 X 5 = 40
   40
      4.6        8 X 0.5 = 4.0
      4.0
      0.64      8 X  0.08 = 0.64
      0.64
        0
Therefore, 44.64 ÷ 8 = 5.58.

Question 20.
0.294 ÷ 3
0.294 ÷ 3 = 0.098

Explanation:
Given expression as 0.294 ÷ 3 =
   0.098    
3)0.294        3 X 0.09 = 0.27
   0.27
   0.024        3 X 0.008 = 0.024
   0.024
       0
Therefore, 0.294 ÷ 3 = 0.098.

Question 21.
3.6 ÷ 24
3.6 ÷ 24 = 0.15

Explanation:
Given expression as 3.6÷ 24 =
   0.15    
24)3.6        24 X 0.1 = 2.16
     2.4
     1.2       24 X 0.05 = 1.2
     1.2
       0
Therefore, 3.6 ÷ 24 = 0.15

Question 22.
52.014 ÷ 20
52.014 ÷ 20 = 2.6007

Explanation:
Given expression as  52.014 ÷ 20 =
    2.6007    
20)52.014        20 X 2 = 40
     40
      12              20X 0.6 = 12
      12
      0.014         20 X  0.0007 = 0.014
      0.014
        0
Therefore, 520.14 ÷ 20 = 2.6007.

YOU BE THE TEACHER
Your friend finds the quotient. Is your friend correct? Explain your reasoning.

Question 23.
Big Ideas Math Solutions Grade 6 Chapter 2 Fractions and Decimals 308

Yes, Friend is correct,

Explanation:
Given expression as  28.08 ÷ 9 =
    3.12  
9)28.08      9 X 3 = 27
   27
     1.0        9 X 0.1 = 0.9
     0.9
      0.18       9 X  0.02 = 0.18
      0.18
        0
Therefore, 28.08 ÷ 9 = 3.12,
which is same as friends findings,  So friend is correct.

Question 24.
Big Ideas Math Solutions Grade 6 Chapter 2 Fractions and Decimals 309
No, Friend is incorrect,

Explanation:
Given expression as  28.08 ÷ 9 =
    0.086  
6)0.516      6 X 0.08 = 0.48
   0.48
   0.036       6 X 0.006 = 0.036
   0.036
        0
Therefore, 0.516 ÷ 6 = 0.086,
which is not same as friends findings of 0.86,
So friend is incorrect.

Question 25.
PROBLEM SOLVING
You buy the same pair of pants in 3 different colors for $89.85.
How much does each pair of pants cost?

Each pair of pants cost $29.95,

Explanation:
Given I buy the same pair of pants in 3 different colors for $89.85.
means cost is same, now each pair of pants cost $89.85 ÷ 3 =
29.95
3)89.85    3 X 2 = 6
   6
   29        3 X 9 = 27
   27
     2.8     3 X 0.9 = 2.7
     2.7
0.15   3 X 0.05 = 0.15
     0.15
0
therefore, each pair of pants cost $29.95.

Question 26.
REASONING
Which pack of fruit punch is the best buy? Explain.
Big Ideas Math Solutions Grade 6 Chapter 2 Fractions and Decimals 310

12 pack fruit punch is the best buy,

Explanation:
Given 4-pack is $2.95, 12 pack is $8.65 and 24 pack is $17.50,
now we will see how much each pack will cost seperatley as
i. 4 pack – $2.95 means 2.95 ÷ 4 =
    0.7375
4)2.95      4 X 0.7 = 2.8
   2.8
0.15    4 X 0.03 = 0.12
    0.12
0.030   4 X 0.007 = 0.028
    0.028
0.002   4 X 0.0005 = 0.002
    0.002
        0
So if we take 4 pack it will cost for 1 pack as 0.7375,
ii. 12 pack – $8.65 means 8.65 ÷ 12 =
    0.72082
12)8.65      12 X 0.7 = 8.4
     8.4
0.25       12 X 0.02 = 0.24
    0.24 
0.01000   12 X 0.0008 = 0.0096
    0.00968
0.00032   12 X 0.00002 = 0.00024
    0.00024
    0.00008 remainder
So if we take 12 pack it will cost for 1 pack as 0.72082,
iii. 24 pack – $17.50 means 17.50 ÷ 24 =
    0.72916
24)17.50      24 X 0.7 = 16.8
     16.80
0.70      24 X 0.02 = 0.48
      0.48 
0.220       24 X 0.009 = 0.216
      0.216
0.0040   24 X 0.0001 = 0.0024
      0.0024
    0.00160    24 X 0.00006 = 0.001444
    0.00144
    0.00016 remainder
So if we take 24 pack it will cost for 1 pack as 0.72916,
Now we compare we get 0.72082< 0.72916 < 0.7375,
therefore 12 pack fruit punch is the best buy.

DIVIDING DECIMALS
Divide. Check your answer.

Question 27.
Big Ideas Math Solutions Grade 6 Chapter 2 Fractions and Decimals 311
Big Ideas Math Solutions Grade 6 Chapter 2 Fractions and Decimals 311 = 12,

Explanation:
Given expression as  25.2 ÷ 2.1 =
    12  
2.1)25.2      2.1 X 12 = 25.2
     25.2
       0
Therefore, 25.2 ÷ 2.1 = 12.

Question 28.
Big Ideas Math Solutions Grade 6 Chapter 2 Fractions and Decimals 312
Big Ideas Math Solutions Grade 6 Chapter 2 Fractions and Decimals 312= 9,

Explanation:
Given expression as  34.2 ÷ 3.8 =
       9     
3.8)34.2      3.8 X 9 = 34.2
      34.2
       0
Therefore, 34.2 ÷ 3.8 = 9.

Question 29.
36.47 ÷ 0.7
36.47 ÷ 0.7 = 52.1,

Explanation:
Given expression as  36.47 ÷ 0.7 =
      52.1     
0.7)36.47      0.7 X 52 = 36.4
      36.4
       0.07      0.7 X 0.1 = 0.07
       0.07
        0
Therefore, 36.47 ÷ 0.7 = 52.1.

Question 30.
0.984 ÷ 12.3
0.984 ÷ 12.3 = 0.08,

Explanation:
Given expression as  0.984 ÷ 12.3 =
        0.08     
12.3)0.984      12.3 X 0.08  = 0.984
       0.984
           0
Therefore, 0.984 ÷ 12.3 = 0.08.

Question 31.
6.64 ÷ 8.3
6.64 ÷ 8.3 = 0.8,

Explanation:
Given expression as  6.64 ÷ 8.3 =
      0.8     
8.3)6.64      8.3 X  0.8 = 6.64
     6.64
        0
Therefore, 6.64 ÷ 8.3 = 0.8.

Question 32.
83.266 ÷ 13.43
83.266 ÷ 13.43 = 6.2,

Explanation:
Given expression as 83.266 ÷ 13.43 =
          6.2        
13.43)83.266      13.43 X 6 = 80.58
          80.58
           2.686       13.43 X 0.2 = 2.686
           2.686
            0
Therefore, 83.266 ÷ 13.43 = 6.2.

Question 33.
Big Ideas Math Solutions Grade 6 Chapter 2 Fractions and Decimals 313
Big Ideas Math Solutions Grade 6 Chapter 2 Fractions and Decimals 313= 11.7,

Explanation:
Given expression as 1.053 ÷ 0.09 =
         11.7        
0.09)1.053       0.09 X 11 = 0.99
        0.99
        0.063       0.09 X 0.7 = 0.063
        0.063
         0
Therefore, 1.053 ÷ 0.09 = 11.7.

Question 34.
35.903 ÷ 16.1
35.903 ÷ 16.1 = 2.23,

Explanation:
Given expression as 35.903 ÷ 16.1 =
        2.23        
16.1)35.903       16.1 X 2 = 32.2
        32.2
          3.70         16.1 X 0.2 = 3.22
          3.22
         0.483        16.1 X 0.03 = 0.483
         0.483
           0
Therefore, 35.903 ÷ 16.1 = 2.23.

Question 35.
0.996 ÷ 0.12
0.996 ÷ 0.12 = 8.3,

Explanation:
Given expression as 0.996 ÷ 0.12 =
        8.3        
0.12)0.996          0.12 X 8 = 0.96
        0.96
        0.036            0.12 X 0.3 = 0.036
        0.036
          0
Therefore, 0.996 ÷ 0.12 = 8.3.

Question 36.
Big Ideas Math Solutions Grade 6 Chapter 2 Fractions and Decimals 314
Big Ideas Math Solutions Grade 6 Chapter 2 Fractions and Decimals 314 = 2.7,

Explanation:
Given expression as 12.501 ÷ 4.63 =
        2.7        
4.63)12.501          4.63 X 2 = 9.26
          9.26
          3.241            4.63 X 0.7 = 3.241
          3.241
             0
Therefore, 12.501 ÷ 4.63 = 2.7.

Question 37.
Big Ideas Math Solutions Grade 6 Chapter 2 Fractions and Decimals 315
Big Ideas Math Solutions Grade 6 Chapter 2 Fractions and Decimals 315= 0.23,

Explanation:
Given expression as 0.00115 ÷ 0.005 =
          0.23        
0.005)0.00115          0.005 X 0.2 = 0.001
          0.001
          0.00015           0.005 X 0.03 = 0.00015
          0.00015
             0
Therefore, 0.00115 ÷ 0.005 = 0.23.

Question 38.
56.7175 ÷ 4.63
56.7175 ÷ 4.63 = 12.25,

Explanation:
Given expression as 56.7175 ÷ 4.63 =
        12.25        
4.63)56.7175          4.63 X 12 = 55.56
        55.56
          1.1575          4.63 X 0.25 = 1.1575
          1.1575
             0
Therefore, 56.7175 ÷ 4.63 = 12.25.

Question 39.
4.23 ÷ 0.012
4.23 ÷ 0.012 = 352.5,

Explanation:
Given expression as 4.23 ÷ 0.012 =
        352.5    
0.012)4.23        0.012 X 352 = 4.224
          4.224
          0.006      0.012 X 0.5 = 0.006
          0.03
             0
Therefore, 4.23 ÷ 0.012 = 0.006.

Question 40.
0.52 ÷ 0.0013
0.52 ÷ 0.0013 = 400,

Explanation:
Given expression as 0.52 ÷ 0.0013 =
             400    
0.0013)0.52        0.0013 X 400 = 0.52
           0.52
             0
Therefore, 0.52 ÷ 0.0013 = 400.

Question 41.
95.04 ÷ 0.0132
95.04 ÷ 0.0132 = 7,200,

Explanation:
Given expression as 95.04 ÷ 0.0132 =
           7200    
0.0132)95.04        0.0132 X 700 = 95.04
            95.04
              0
Therefore, 95.04 ÷ 0.0132 = 7,200.

Question 42.
32.2 ÷ 0.07
32.2 ÷ 0.07 = 460,

Explanation:
Given expression as 32.2 ÷ 0.07 =
         460      
0.07)32.2        0.07 X 460 = 32.2
        32.2
            0
Therefore, 32.2 ÷ 0.07 = 460.

Question 43.
Big Ideas Math Solutions Grade 6 Chapter 2 Fractions and Decimals 316
Big Ideas Math Solutions Grade 6 Chapter 2 Fractions and Decimals 316= 40,

Explanation:
Given expression as 54.8 ÷ 1.37 =
         40      
1.37)54.8        1.37 X 40 = 54.8
        54.8
            0
Therefore, 54.8 ÷ 1.37 = 40.

Question 44.
44.2 ÷ 3.25
44.2 ÷ 3.25 = 13.6,

Explanation:
Given expression as 44.2 ÷ 3.25 =
        13.6      
3.25)44.2        3.25 X 13 = 42.25
        42.25
           1.95     3.25 X 0.6 = 1.95
Therefore, 44.2 ÷ 3.25 = 13.6.

Question 45.
Big Ideas Math Solutions Grade 6 Chapter 2 Fractions and Decimals 317
Big Ideas Math Solutions Grade 6 Chapter 2 Fractions and Decimals 317= 12.5,

Explanation:
Given expression as 50.5 ÷ 4.04 =
        12.5      
4.04)50.5        4.04 X 12 = 48.48
        48.48
          2.02     4.04 X 0.5 = 2.02
          2.02
            0
Therefore, 50.5 ÷ 4.04 = 12.5.

Question 46.
250 ÷ 0.008
250 ÷ 0.008 = 31250,

Explanation:
Given expression as 250 ÷ 0.008 =
         31250      
0.008)250        0.008 X 31250 = 250
         250
            0
Therefore, 250 ÷ 0.008 = 31250.

Question 47.
11.16 ÷ 0.062
11.16 ÷ 0.062 = 180,

Explanation:
Given expression as 11.16 ÷ 0.062 =
         180     
0.062)11.16        0.062 X 180 = 11.16
          11.16
            0
Therefore, 11.16 ÷ 0.062 = 180.

Question 48.
Big Ideas Math Solutions Grade 6 Chapter 2 Fractions and Decimals 318
Big Ideas Math Solutions Grade 6 Chapter 2 Fractions and Decimals 318= 66.8,

Explanation:
Given expression as 835 ÷ 12.5 =
         66.8      
12.5)835        12.5 X 66 = 825
        825
          10       12.5 X 0.8 = 10
           10
            0
Therefore, 835 ÷ 12.5 = 66.8.

Question 49.
597.6 ÷ 12.45
597.6 ÷ 12.45 = 48,

Explanation:
Given expression as 597.6 ÷ 12.45 =
          48      
12.45)597.6        12.45 X 48 = 597.6
          597.6
            0
Therefore, 597.6 ÷ 12.45 = 48.

Question 50.
Big Ideas Math Solutions Grade 6 Chapter 2 Fractions and Decimals 319
Big Ideas Math Solutions Grade 6 Chapter 2 Fractions and Decimals 319 = 272,

Explanation:
Given expression as 118.32 ÷ 0.435 =
          272      
0.435)118.32        0.435 X 272 = 118.32
          118.32
            0
Therefore, 118.32 ÷ 0.435 = 272.

Question 51.
80.89 ÷ 8.425
80.89 ÷ 8.425 ≈ 9.60,

Explanation:

Explanation:
Given expression as 80.89 ÷ 8.425 =
          9.60      
8.425)80.89        8.425 X 9 = 75.825
          75.825
            5.065     8.425 X 0.6 = 5.055
            5.055
             0.010  remainder
Therefore, 80.89 ÷ 8.425 ≈ 9.60.

Question 52.
0.8 ÷ 0.6
0.8 ÷ 0.6 ≈ 1.33,

Explanation:
Given expression as 0.8 ÷ 0.6 =
      1.33   
0.6)0.8        0.6 X 1 = 75.825
      0.6
      0.20      0.6 X 0.3 = 0.18
      0.18
      0.020    0.6 X 0.03 = 0.018
      0.018
      0.002  remainder
Therefore, 0.8 ÷ 0.6 ≈ 1.33.

Question 53.
38.9 ÷ 6.44
38.9 ÷ 6.44 = 6.04,

Explanation:
Given expression as 38.9 ÷ 6.44 =
        6.04   
6.44)38.9        6.44 X 6 = 38.64
        38.64
          0.26      6.44 X 0.04 = 0.2576
          0.2576
          0.0024  remainder
Therefore, 38.9 ÷ 6.44 = 6.04.

Question 54.
11.6 ÷ 0.95
11.6 ÷ 0.95 = 12.2,

Explanation:
Given expression as 11.6 ÷ 0.95 =
        12.2   
0.95)11.6       0.95 X 12 = 11.4
        11.4
          0.2      0.95 X 0.2 = 0.19
          0.19
          0.01  remainder
Therefore, 11.6 ÷ 0.95 = 12.2.

Question 55.
YOU BE THE TEACHER
Your friend rewrites the problem. Is your friend correct? Explain your reasoning.
Big Ideas Math Solutions Grade 6 Chapter 2 Fractions and Decimals 320
No, Friend is incorrect, 146.4 ÷ 0.32 ≠ 1.464 ÷ 32,
We rewrite as 1.464 ÷ 32 as 14640 ÷ 32,

Explanation:
Given 146.4 ÷ 0.32 —> 1.464 ÷ 32,
friend is incorrect as 146.4 ÷ 0.32 = 146.4 X 100 by 32  =
14640 by 32 ≠ 1.464 by 32, So friend is incorrect,
we rewrite 1.464 ÷ 32 as 14640 ÷ 32 which is correct to
146.4 ÷ 0.32 = 14640 ÷ 32.

ORDER OF OPERATIONS
Evaluate the expression.

Question 56.
7.68 + 3.18 ÷ 12
7.68 + 3.18 ÷ 12 = 7.945,

Explanation:
Given expression is 7.68 + 3.18 ÷ 12,
according to order of operations we take division first then
addition as 7.68 +(3.18 ÷ 12) =
First we calculate 3.18 ÷ 12 =
     0.265
12)3.18    12 X 0.2 = 2.4
     2.4
     0.78      12 X 0.06 = 0.72
     0.72
     0.06    12 X 0.005 = 0.06
     0.06
         0
We got 3.18 ÷ 12 = 0.265, Now we add 7.68 as
7.680
+0.265
7.945
therefore, 7.68 + 3.18 ÷ 12 = 7.945.

Question 57.
10.56 ÷ 3 – 1.9
10.56 ÷ 3 – 1.9 = 1.62,

Explanation:
Given expression is 10.56 ÷ 3 – 1.9,
according to order of operations we take division first then
subtraction as  (10.56 ÷ 3) – 1.9 =
First we calculate 10.56 ÷ 3 =
    3.52   
3)10.56        3 X 3 = 9
     9.00
     1.5          3 X 0.5 = 1.5
     1.5
     0.06        3 X 0.02 = 0.06
     0.06
       0
We got 10.56 ÷ 3 = 3.52, Now we subtract 1.9 from 3.52 as
3.52
-1.90
1.62
therefore 10.56 ÷ 3 – 1.9 = 1.62.

Question 58.
19.6 ÷ 7 × 9
19.6 ÷ 7 X 9 = 25.2,

Explanation:
Given expression is 19.6 ÷ 7 X 9,
according to order of operations we take division first then
multiplication as  (19.6 ÷ 7) X 9 =
First we calculate 19.6 ÷ 7 =
     2.8   
7)19.6       7 X 2 = 14
   14
      5.6      7 X 0.8 = 5.6
      5.6
        0
We got 19.6 ÷ 7 = 2.8, Now we multiply 2.8 with 9 as
  7  
2.8  —- 1 decimal place
X 9
25.2 —- 1 decimal place
therefore 19.6 ÷ 7 X 9 = 25.2.

Question 59.
5.5 × 16.56 ÷ 9
5.5 × 16.56 ÷ 9 = 10.12,

Explanation:
Given expression is 5.5 × 16.56 ÷ 9
according to order of operations we take multiplication first then
division as  (5.5 X 16.56) ÷ 9 =
First we calculate 5.5 X 16.56 =
  2,2,3
  2,2,3
16.56 —- 2 decimal places
X 5.5
0828
8280
91.08—- 2 decimal places

We got 5.5 X 16.56 = 91.08, Now we divide 91.08 with 9 as
   10.12
9)91.08     9 X 10 = 90
   90
      1.0     9 X 0.1 = 0.9
      0.9
      0.18    9 X 0.02 = 0.18
      0.18
         0
therefore 5.5 × 16.56 ÷ 9 = 10.12.

Question 60.
35.25 ÷ 5 ÷ 3
35.25 ÷ 5 ÷ 3 = 2.35,

Explanation:
Given expression is 35.25 ÷ 5 ÷ 3,
according to order of operations we take division first then
division as  (35.25 ÷ 5) ÷ 3 =
First we calculate 35.25 ÷ 5 =
    7.05
5) 35.25      5 X 7 = 35
    35
       0.25     5 X 0.05 = 0.25
       0.25
0
We got 35.25 ÷ 5 = 7.05, Now we divide 7.05 with 3 as
   2.35
3) 7.05      3 X 2 = 6
    6
    1.0        3 X 0.3 = 0.9
    0.9
    0.15     3 X 0.05 = 0.15
    0.15
      0
therefore,  35.25 ÷ 5 ÷ 3 = 2.35.

Question 61.
13.41 × (5.4 ÷ 9)
13.41 × (5.4 ÷ 9) = 8.046,

Explanation:
Given expression is 13.41 X (5.4 ÷ 9),
according to order of operations we take division first then
multiplication so first we calculate 5.4 ÷ 9 =
     0.6   
9)5.4        9 X 0.6 = 5.4
   5.4   
0
We got 5.4 ÷ 9 = 0.6, Now we multiply 13.41 with 0.6 as
  1, 2    
13.41——2 decimal places
X 0.6 —– 1 decimal place
8.046——3 decimal places
therefore, 13.41 × (5.4 ÷ 9) = 8.046.

Question 62.
6.2 . (5.16 ÷ 6.45)
6.2 X (5.16 ÷ 6.45) = 4.96,

Explanation:
Given expression is 6.2 X (5.16 ÷ 6.45),
according to order of operations we take division first then
multiplication so first we calculate 5.16 ÷ 6.45 =
         0.8   
6.45)5.16        6.45 X 0.8 = 5.16
        5.16   
0
We got 5.16 ÷ 6.45 = 0.8, Now we multiply 6.2 with 0.8 as
1
6.2 —- 1 decimal place
X 0.8 —- 1 decimal place
496
000
4.96 —- 2 decimal places
therefore, 6.2 X (5.16 ÷ 6.45) = 4.96.

Question 63.
132.06 ÷ (42 + 2.6)
132.06 ÷ (42 + 2.6) = 7.1,

Explanation:
Given expression as 132.06 ÷ (42 + 2.6) , first we calculate
(42 + 2.6) = 4 X 4 + 2.6 = 16 + 2.6 =
16
+2.6
18.6
Now, we divide 132.06 with 18.6 as
        7.1        
18.6)132.06   18.6 X 7 = 130.2
        130.2
            1.86   18.6 X 0.1 = 1.86
            1.86
               0
therefore, 132.06 ÷ (42 + 2.6) = 7.1.

Question 64.
4.8[23.9841 ÷ (1.16 + 1.27)]
4.8[23.9841 ÷ (1.16 + 1.27)] = 47.376,

Explanation:
Given expression as 4.8[23.9841 ÷ (1.16 + 1.27)],
first we calculate (1.16 + 1.27) =
      1    
1.16
+1.27
   2.43
next we calculate 23.9841 ÷ 2.43 as
        9.87      
2.43)23.9841     2.43 X 9 = 21.87
        21.87
          2.114        2.43 X 0.8 = 1.944
          1.944
          0.1701        2.43 X 0.07 = 0.1701
          0.1701
             0
Now we multiply 4.8 with 9.87 as
9.87 —– 2 decimal places
X4.8 —– 1 decimal place
 7.896
39480
47.376—-3 decimal places
therefore, 4.8[23.9841 ÷ (1.16 + 1.27)] = 47.376.

Question 65.
MODELING REAL LIFE
A person’s running stride is about 1.14 times the person’s height.
Your friend’s stride is 5.472 feet. How tall is your friend?

My friend is 4.8 feet tall,

Explanation:
Given a person’s running stride is about 1.14 times the
person’s height. My friend’s stride is 5.472 feet so my friend
height is 5.472 ÷ 1.14 =
        4.8     
1.14)5.472     1.14 X 4 = 4.56
        4.56
        0.912     1.14 X 0.8 = 0.912
        0.912
          0
therefore, my friend is 4.8 feet tall.

Question 66.
PROBLEM SOLVING
You have 3.4 gigabytes available on your tablet.
A song is about 0.004 giga byte. How many songs can
you download onto your tablet?
Big Ideas Math Solutions Grade 6 Chapter 2 Fractions and Decimals 321

I can download 850 songs onto my tablet,

Explanation:
Given I have 3.4 gigabytes available on my tablet.
A song is about 0.004 giga byte. I can download
3.4 ÷ 0.004 songs as
        850  
0.004)3.4    0.004 X 850 = 3.4
          3.4
            0
Therefore, I can download 850 songs onto my tablet.

REASONING
Without finding the quotient, copy and complete the
statement using <, >, or =. Explain your reasoning.

Question 67.
Big Ideas Math Solutions Grade 6 Chapter 2 Fractions and Decimals 322
6.66 ÷ 0.74 = 66.6 ÷ 7.4,

Explanation:
Given expressions as 6.66 ÷ 0.74 , 66.6 ÷ 7.4, now
we write 66.6 ÷ 7.4 as 6.66 X 10 ÷ 7.4 = 6.66 ÷ 0.74
now both sides are equal, therefore
6.66 ÷ 0.74 = 66.6 ÷ 7.4.

Question 68.
Big Ideas Math Solutions Grade 6 Chapter 2 Fractions and Decimals 323
32.2 ÷ 0.7 > 3.22 ÷ 7,

Explanation:
Given expressions as 32.2 ÷ 0.7,  3.22 ÷ 7,
now 32.2 ÷ 0.7 we write as 32.2 X 10 ÷ 7 = 32.2 ÷ 7,
now we compare 32.2 ÷ 7 and 3.22 ÷ 7 if we see 32.2
is greater than 3.22 and both are divided by 7 only,
the quotient of 32.2 will be greater than that of 3.22,
So 32.2 ÷ 0.7 > 3.22 ÷ 7.

Question 69.
Big Ideas Math Solutions Grade 6 Chapter 2 Fractions and Decimals 324
160.72 ÷ 16.4 < 160.72 ÷ 1.64 ,

Explanation:
Given expressions as 160.72 ÷ 16.4 ,160.72 ÷ 1.64 ,
if we compare we have 160.72 ÷ 1.64 we can write as
160.72 X 10 ÷ 1.64 X 10 = 1607.2 ÷ 16.4 now if we compare
1607.2 is more than 160.72 and both are divided by 16.4,
So 160.72 ÷ 16.4 < 160.72 ÷ 1.64

Question 70.
Big Ideas Math Solutions Grade 6 Chapter 2 Fractions and Decimals 325
75.6 ÷ 63 < 7.56 ÷ 0.63,

Explanation:
Given expressions as 75.6 ÷ 63 , 7.56 ÷ 0.63 ,
if we compare we have 75.6 ÷ 63 we can write as
0.756 X 100 ÷ 0.63 X 100 = 0.756 ÷ 0.63 now if we compare
7.56 is more than 0.756 and both are divided by 0.63,
So 75.6 ÷ 63 < 7.56 ÷ 0.63.

Question 71.
DIG DEEPER!
The table shows the top three times in a swimming event at the Summer Olympics.
The event consists of a team of four women swimming 100 meters each.
Big Ideas Math Solutions Grade 6 Chapter 2 Fractions and Decimals 326
a. Suppose the times of all four swimmers on each team were the same.
For each team, how much time does it take a swimmer to swim 100 meters?
b. Suppose each U.S. swimmer completed 100 meters a quarter second faster.
Would the U.S. team have won the gold medal? Explain your reasoning.

a. For each team the time it will take a swimmer to
swim 100 meters is, for Australia it will be 52.6625 seconds,
for United states it will be 52.9725 seconds and
for Canada it will be 53.2225 seconds.
b. No, because even if U.S swimmer completes a quarter second faster
it becomes 210.89 which is more time than Australia team, So U.S team
would not have won the gold medal.

Explanation:
Given table shows the top three times in a swimming event
at the Summer Olympics.
a. For each team the time it will take a swimmer to
swim 100 meters is for Australia it will be 210.65 ÷ 4 =
   52.6625 
4)210.65      4 X 5 = 20
   20
    10           4 X 2 = 8
8
       2.6       4 X 0.6 = 2.4
2.4
       0.25     4 X 0.06 = 0.24
0.24
        0.010    4 X 0.002 = 0.008
0.008
        0.002    4 X 0.0005 = 0.002 
0.002
           0
for Australia it will be 52.6625 seconds,
Now for United States 211.89 ÷ 4 =
   52.9725 
4)211.89      4 X 5 = 20
   20
    11           4 X 2 = 8
08
       3.8       4 X 0.9 = 3.6
3.6
       0.29     4 X 0.07 = 0.28
0.28
       0.010    4 X 0.002 = 0.008
0.008
        0.002    4 X 0.0005 = 0.002 
0.002
           0
for United States it will be 52.9725 seconds,
Now for Canada it will be 212.89 ÷ 4 =
   53.2225  
4)212.89      4 X 5 = 20
   20
     12           4 X 3 = 12
12
       0.8       4 X 0.2 = 0.8
0.8
       0.09     4 X 0.02 = 0.08
0.08
        0.010    4 X 0.002 = 0.008
0.008
        0.002    4 X 0.0005 = 0.002 
0.002
           0
for Canada it will be 53.2225 seconds,

b. If each U.S. swimmer completed 100 meters a quarter second faster,
we will see the U.S. team have won the gold medal or not as
quarter second faster 4 women in each team means 1 second,
so it would have finished in 211.89 – 1 =
211.89
–   1.00
210.89
Now comparing with Australia it will be 210.65 even then it is
210.89
-210.65
0.24
So U.S team would have taken 0.24 seconds more,
therefore U.S team would not have won the gold medal.

Question 72.
PROBLEM SOLVING
To approximate the number of bees in a hive, multiply the number
of bees that leave the hive in one minute by 3 and divide by 0.014.
You count 25 bees leaving a hive in one minute. How many bees are in the hive?
Big Ideas Math Solutions Grade 6 Chapter 2 Fractions and Decimals 327There are 5357 bees are in the hive,

Explanation:
Given to approximate the number of bees in a hive,
multiply the number of bees that leave the hive in one
minute by 3 and divide by 0.014. I count 25 bees leaving
a hive in one minutes so number of bees in the hive are
(25 X 3) ÷ 0.014 = 75 ÷ 0.014 =
          5357
0.014)75.000     0.014 X 5357 = 74.998
          74.998
0.002  remainder
therefore there are 5357 bees are in the hive.

Question 73.
PROBLEM SOLVING
You are saving money to buy a new bicycle that costs $155.75.
You have $30 and plan to save $5 each week. Your aunt decides
to give you an additional $10 each week.
a. How many weeks will you have to save until you
have enough money to buy the bicycle?
b. How many more weeks would you have to save to
buy a new bicycle that costs $203.89? Explain how you found your answer.

a. 9 weeks I have to save until I have enough money to buy
the bicycle.
b. 3 weeks more I have to save to buy a new bicycle that
costs $203.89.

Explanation:
a. Given  I am  saving money to buy a new bicycle that costs $155.75.
I have $30 and plan to save $5 each week. my aunt decides
to give me an additional $10 each week. So number of weeks
will I have to save until I have enough money to buy the bicycle is,
I alreday have $30 means I need to save is  $155.75 – $30 = $125.75,
every week I can save $5 + $10 = $15,Now weeks it will take is
$125.75 ÷ 15 =
     8.383
15)125.75     15 X 8 = 120
     120
          5.75    15 X 0.3 = 4.5
4.50
           1.25   15 X 0.08 = 1.20
1.20
           0.050   15 X 0.003 = 0.045
           0.045
0.0050   remainder
therefore approximately I need to wait 9 weeks,

b. Number of more weeks would you have to save to
buy a new bicycle that costs $203.89 , $ 203.89 – $30 =
$173.89 I need to save, So number of weeks now is
$173.89 ÷ 15 =
     11.59   
15)173.89       15 X 11 = 165
     165
         8.89       15 X 0.5 = 7.5
         7.50
1.39       15 X 0.09 =1.35
         1.35
         0.04 remainder
Therefore approximately I need to wait 12 weeks,
So more number of  weeks would I have to save to
buy a new bicycle that costs $203.89 is 3 weeks.

Question 74.
PRECISION
A store sells applesauce in two sizes.
a. How many bowls of applesauce fit in a jar?
Round your answer to the nearest hundredth.
b. Explain two ways to find the better buy.
c. Which is the better buy?
Big Ideas Math Solutions Grade 6 Chapter 2 Fractions and Decimals 327.1

a. Number of bowls of applesauce fit in a jar,
rounding answer to the nearest hundredth is 6.15.
b. i. First way is Dividing to find 1 ounce of applesauce
seperately,
ii. Now we have unitary method,
c. The better buy is 24 ounce jar,

Explanation:
Given 3.9 ounce is bowl, So number of bowls of
applesauce fit in a jar is 24 ÷ 3.9 =
      6.153
3.9)24          3.9 X 6 = 23.4
      23.4
        0.6      3.9 X 0.1 = 0.39
        0.39
        0.21    3.9 X 0.05 = 0.195
        0.195
0.015  3.9 X 0.003 = 0.0117
        0.011
0.004  remainder
therefore numberof bowls of applesauce fit in a jar,
rounding answer to the nearest hundredth is 6.15.
b. First way :
1 ounce bowl is $0.52 ÷ 3.9
0.133    
3.9)0.52    3.9 X 0.1 = 0.39
      0.39
      0.130   3.9 X 0.03 = 0.117
0.117
      0.0130   3.9 X 0.003 = 0.0117
0.0117
0.0013  remainder
So 1 ounce bowl is $0.133
Now 1 ounce jar is $2.63 ÷ 24
0.109
24)2.63      24 X 0.1 = 2.4
     2.4
     0.230      24 X 0.009 = 0.216
     0.216
0.014   remainder

So 1 ounce jar is $0.109
Second way :
We know unitary methods we have 3.9 ounce bowl is $0.52,
24 once jar is $2.63 as both goes in 3 (3.9,24) we have
1 ounce bowl will be 3.9 ÷ 3 = 0.52 ÷ 3; 1.3 = 0.173,
Now 0.173 ÷ 1.3 =
0.133
1.3)0.173   1.3 X 0.133 = 0.1729
      0.1729
0.0001 remainder
So it is $0.1333 for 1 ounce bowl
Now 1 ounce jar is 24 ÷ 3 = 2.63 ÷ 3;  8 = 0.876
Now 0.876 ÷ 8 =
0.1095
8)0.876  8 X 0.1095 = 0.876
   0.876
    0
So now 1 ounce jar is $0.1095,
C. We have 1 ounce bowl is $0.133 and 1 ounce jar is
$0.109, so on comparing $0.109< $0.133  jar is the better buy.

Question 75.
GEOMETRY
The large rectangle’s dimensions are three times the dimensions
of the small rectangle.
a. How many times greater is the perimeter of the large rectangle than the
perimeter of the small rectangle?

b. How many times greater is the area of the large rectangle than the
area of the small rectangle?
c. Are the answers to parts (a) and (b) the same? Explain why or why not.
d. What happens in parts (a) and (b) if the dimensions of the large rectangle
are two times the dimensions of the small rectangle?
a. It will be 3 times greater is the perimeter of the large rectangle than the
perimeter of the small rectangle,

b. It will be 9 times greater is the area of the large rectangle than the
area of the small rectangle.

c. No, because perimeter will be measured in feet and
area is measured in  square feet.

d. Perimeter will be 2 times more in large rectangle and
4 times in area in large rectangle than the small rectangle.

Explanation:
Given the large rectangle’s dimensions are three times the
dimensions of the small rectangle means lets take the
smaller rectangle length as x and width y , Then the larger ones
will be length 3x and width will be 3y,
a. Now perimeter of both we know perimeter is
length + breadth for smaller it will be (x+y) and for
larger it will be 3x + 3y = 3(x + y), So It will be 3 times
greater is the perimeter of the large rectangle than the
perimeter of the small rectangle.
b. Now we know area of rectangle is length X breadth,
we have for small rectangle as x X y = xy and for large it
is 3x X 3y  = 9xy which is 9 times greater is the area of the large
rectangle than the area of the small rectangle.
c. The answers are not the same in part a and part b,
because one will be measured in feet and other is square feet,
( one is adding and other is multiplying) both differs.
d. Now if the dimensions of the large rectangle
are two times the dimensions of the small rectangle
then perimeter becomes as for small it is x + y and for
large it is 2x + 2y = 2(x + y) and area for small is xy and large
2x X 2y = 4xy, means perimeter will be 2 times more in large rectangle
and  4 times in area in large rectangle than the small rectangle.

Fractions and Decimals Connecting Concepts

2 Connecting Concepts

Using the Problem-Solving Plan

Question 1.
You change the water jug on the water cooler.
How many glasses can be completely filled before you need to
change the water jug again?
Understand the problem
You know the capacities of the water jug and the glass.
You are asked to determine how many glasses the water jug can fill.
Make a plan.2
First, use what you know about converting measures to
find the number of fluid ounces in 5 gallons. Then divide this amount
by the capacity of the glass to find the number of glasses that can be filled.
Solve and check.
Use the plan to solve the problem. Then check your solution.
Big Ideas Math Solutions Grade 6 Chapter 2 Fractions and Decimals 329
64 glasses can be completely filled before we need to
change the water jug again,

Explanation:
We know 1 gallons is equal to 128 fluid ounce,
So in 5 gallons we have 5 X 128 = 640 fluid ounces water
is there in 5 gallons water jug.
Now we have 1 glass has 10 fluid ounces of water then in
640 it will be 640 ÷ 10 = 64,
therefore 64 glasses can be completely filled before we need to
change the water jug again.

Question 2.
Two ferries just departed from their docks at the same time.
Ferry A departs from its dock every 1.2 hours. Ferry B departs
from its dock every 1.8 hours. How long will it be until both ferries
depart from their docks at the same time again?
Big Ideas Math Solutions Grade 6 Chapter 2 Fractions and Decimals 330

It will be at 3.6 hours until both ferries depart from their
docks at the same time again.

Explanation:
Given two ferries just departed from their docks at the same time.
Ferry A departs from its dock every 1.2 hours. Ferry B departs
from its dock every 1.8 hours. So the ratio is 1.2 : 1.8 = 2 : 3,
we use cross product for equal hours, ratios as
both meet at 1.2 X 3 = 3.6 hours or 1.8 X 2 = 3.6 hours,
So,It will be at 3.6 hours until both ferries depart from their
docks at the same time again.

Question 3.
You want to paint the ceiling of your bedroom. The ceiling has
two square skylights as shown. Each skylight has a side length of
1\(\frac{7}{8}\) feet. How many square feet will you paint?

Justify your answer.
Big Ideas Math Solutions Grade 6 Chapter 2 Fractions and Decimals 331
I will paint 132\(\frac{62}{64}\) square feet,

Explanation:
Given  the length and width of rectangle as
10 ft and 14 ft, Area of rectangle is length X breadth=
10 X 14 = 140 square feet,
Now we have 2 square sky light with 1\(\frac{7}{8}\) feet,
so area of skylight is 1\(\frac{7}{8}\)  X 1\(\frac{7}{8}\) =
frist we write mixed fraction into fraction as (1 X 8 + 7 by 8) =
\(\frac{15}{8}\) X \(\frac{15}{8}\) =\(\frac{15 X 15}{8 X 8}\) =
\(\frac{225}{64}\) square feet, So 2 sky lighta are there so
\(\frac{225}{64}\) + \(\frac{225}{64}\) as
denominators are same we add numerators as
(225 +225 = 450) we get \(\frac{450}{64}\) square feet,
We have area of rectangle as 140 square feet and 2 skylights as
\(\frac{450}{64}\) square feet, So we will paint
140 – \(\frac{450}{64}\) = \(\frac{8960-450}{64}\) =
\(\frac{8510}{64}\) as numerator is greater we write as
(132 X 64 + 62 by 64) = 132\(\frac{62}{64}\) square feet
i will paint.

Performance Task
Space Explorers
At the beginning of this chapter, you watched a STEAM video called “Space is Big.”
You are now ready to complete the performance task for this video, available at BigIdeasMath.com. Be sure to use the problem-solving complan as you work through the performance task.
Big Ideas Math Solutions Grade 6 Chapter 2 Fractions and Decimals 332

Fractions and Decimals Chapter Review

2 Chapter Review

Review Vocabulary

Write the definition and give an example of each vocabulary term.
Big Ideas Math Solutions Grade 6 Chapter 2 Fractions and Decimals 333
Reciprocals or Multiplicative Invesrses :
A reciprocal, or multiplicative inverse, is simply one of a pair of
numbers that, when multiplied together, equal 1.
If you can reduce the number to a fraction, finding the
reciprocal is simply a matter of transposing the numerator and
the denominator.
Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals-98

Graphic Organizers

You can use a Summary Triangle to explain a concept. Here is an example of a Summary Triangle for dividing fractions.
Big Ideas Math Solutions Grade 6 Chapter 2 Fractions and Decimals 334

Choose and complete a graphic organizer to help you study the concept.

  1. multiplying fractions
  2. multiplying mixed numbers
  3. reciprocals
  4. dividing mixed numbers
  5. adding and subtracting decimals
  6. multiplying decimals by decimals
  7. dividing whole numbers
  8. dividing decimals by decimals

Big Ideas Math Solutions Grade 6 Chapter 2 Fractions and Decimals 335

1. multiplying fractions:

To multiply fractions :
a. Simplify the resulting fraction if possible.
b. Simplify the fractions if not in lowest terms.
c. Multiply the numerators of the fractions to get the new numerator,
Multiply the denominators of the fractions to get the new denominator.

Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals-97

2. multiplying mixed numbers :
A mixed number is a number that contains a whole number and a fraction,
for instance 2\(\frac{1}{2}\) is a mixed number.
Mixed numbers can be multiplied by first converting them
to improper fractions. Below are the general rules for
multiplying mixed numbers:
* Convert the mixed numbers to improper fractions first.
* Multiply the numerators from each fraction to each other
and place the product at the top.
* Multiply the denominators of each fraction by each other
(the numbers on the bottom). The product is the denominator
of the new fraction.
* Simplify or reduce the final answer to the lowest terms possible.
Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals-99

3. reciprocals :

The reciprocal or multiplicative inverse of a number

x

is the number which, when multiplied by x , gives 1,
So, the product of a number and its reciprocal is

1

.
(This is sometimes called the property of reciprocals),
Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals-100

4. dividing mixed numbers :
Steps for dividing mixed numbers.
a. Change each mixed number to an improper fraction.
b. Multiply by the reciprocal of the divisor, simplifying if possible.
c. Put answer in lowest terms.
Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals-101

5. adding and subtracting decimals :
Step 1: Line up the numbers vertically so that the decimal
points all lie on a vertical line.
Step 2: Add extra zeros to the right of the number so that
each number has the same number of digits to the right
of the decimal place.
Step 3: Subtract the numbers as you would whole numbers.
Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals-102

 

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 Solutions Grade 6 Chapter 2 Fractions and Decimals 336

2.1 Multiplying Fractions (pp. 45-52)

Multiply. Write the answer in simplest form.

Question 1.
Big Ideas Math Solutions Grade 6 Chapter 2 Fractions and Decimals 337
Big Ideas Math Solutions Grade 6 Chapter 2 Fractions and Decimals 337= \(\frac{16}{99}\),
Explanation:
Given expression as \(\frac{2}{9}\) X\(\frac{8}{11}\) =
Step I: We multiply the numerators as 2 X 8 = 16
Step II: We multiply the denominators as 9 X 11 =99
Step III: We write the fraction in the simplest form as
\(\frac{16}{99}\),
So \(\frac{2}{9}\) X\(\frac{8}{11}\) = \(\frac{16}{99}\).

Question 2.
Big Ideas Math Solutions Grade 6 Chapter 2 Fractions and Decimals 338
Big Ideas Math Solutions Grade 6 Chapter 2 Fractions and Decimals 338= \(\frac{6}{25}\),
Explanation:
Given expression as \(\frac{3}{10}\) X \(\frac{4}{5}\) =
Step I: We multiply the numerators as 3 X 4 = 12
Step II: We multiply the denominators as 10 X 5 =50
Step III: We write the fraction in the simplest form as
\(\frac{12}{50}\), as both goes in 2 we
can further simplify as 2 X 6 = 12, 2 X 25 = 50, (6,25) = \(\frac{6}{25}\),
So \(\frac{3}{10}\) X \(\frac{4}{5}\) = \(\frac{6}{25}\).

Question 3.
Big Ideas Math Solutions Grade 6 Chapter 2 Fractions and Decimals 339
Big Ideas Math Solutions Grade 6 Chapter 2 Fractions and Decimals 339= \(\frac{184}{15}\) or 12\(\frac{4}{15}\),
Explanation:
Given expression as 2\(\frac{3}{10}\) X 5\(\frac{1}{3}\),
we write mixed fraction as fraction 2\(\frac{3}{10}\) =
(2 x 10 + 3 by 10)  = \(\frac{23}{10}\) and 5\(\frac{1}{3}\) as
( 5 X 3 + 1 by 3) = \(\frac{16}{3}\), Now we multiply as
\(\frac{23}{10}\) X \(\frac{16}{3}\)
Step I: We multiply the numerators as 23 X 16 = 368
Step II: We multiply the denominators as 10 X 3 =30
Step III: We write the fraction in the simplest form as
\(\frac{368}{30}\), as both goes in 2,
we  can further simplify as 2 X 184 = 368, 2 X 15 = 30, (184,15) = \(\frac{184}{15}\),
as numerator is greater we write as (12 X 15 + 4 by 15) = 12\(\frac{4}{15}\),
So 2\(\frac{3}{10}\) X 5\(\frac{1}{3}\) = \(\frac{184}{15}\) or 12\(\frac{4}{15}\).

Question 4.
Big Ideas Math Solutions Grade 6 Chapter 2 Fractions and Decimals 340
Big Ideas Math Solutions Grade 6 Chapter 2 Fractions and Decimals 340= \(\frac{80}{63}\) or 1\(\frac{17}{63}\),
Explanation:
Given expression as \(\frac{2}{7}\) X 4\(\frac{4}{9}\),
we write mixed fraction as fraction 4\(\frac{4}{9}\) =
(4 x 9 + 4 by 9)  = \(\frac{40}{9}\), Now we multiply as
\(\frac{2}{7}\) X \(\frac{40}{9}\)
Step I: We multiply the numerators as 2 X 40 = 80
Step II: We multiply the denominators as 7 X 9 =63
Step III: We write the fraction in the simplest form as
\(\frac{80}{63}\), as numerator is greater we write as
(1 X 63 + 17 by 63) = 1\(\frac{17}{63}\),
So, \(\frac{2}{7}\) X 4\(\frac{4}{9}\) = \(\frac{80}{63}\) or 1\(\frac{17}{63}\).

Question 5.
Write two fractions whose product is \(\frac{21}{32}\).

The two fractions can be \(\frac{7}{8}\) or \(\frac{3}{4}\),
whose product is \(\frac{21}{32}\),

Explanation:
Given to find two fractions whose product is \(\frac{21}{32}\),
We can take factors of 21 are 1, 3, 7, 21, So numerators can be
(1,21) or (3, 7), now factors of 32 are 1, 2, 4, 8, 16, and 32,
denominators can be (1, 32) or (2, 16) or (4, 8), So the two fractions
can be \(\frac{7}{8}\) or \(\frac{3}{4}\) whose
product is  \(\frac{7 X 3 }{8 X 4}\) or \(\frac{21}{32}\).

Question 6.
A costume designer needs to make 12 costumes for the school play.
Each costume requires 2\(\frac{2}{3}\) yards of fabric.
How many yards of fabric does the costume designer need to
make all the costumes?
Big Ideas Math Solutions Grade 6 Chapter 2 Fractions and Decimals 341
The costume designer need 32 yards of fabric to
make all the costumes,

Explanation:
Given a costume designer needs to make 12 costumes
for the school play. Each costume requires 2\(\frac{2}{3}\) yards of fabric.
So for 12 costumes we require 12 X 2\(\frac{2}{3}\) yards of fabric=
first we write mixed fraction into fraction as 2\(\frac{2}{3}\) =
(2 X 3 + 2 by 3) = \(\frac{8}{3}\) So 12 X \(\frac{8}{3}\) =
\(\frac{12 X 8}{3}\) = \(\frac{96}{3}\)  as it goes in 3,
3 X 32 = 96 we get \(\frac{96}{3}\) = 32.
therefore the costume designer need 32 yards of fabric to
make all the costumes.

Question 7.
You spend \(\frac{4}{5}\) of an hour on your homework.
You spend \(\frac{1}{2}\) of that time working on your
science homework. How many minutes do you spend working on
science homework?

I have spent 24 minutes working on science homework,

Explanation:
Given I spend \(\frac{4}{5}\) of an hour on my homework.
I have  spend \(\frac{1}{2}\) of that time working on my
science homework. So time spent for working on science homework is
\(\frac{4}{5}\) X \(\frac{1}{2}\) = \(\frac{4 X 1}{5 X 2}\) =
\(\frac{4}{10}\) as further can be simplified as both
goes in 2, 2 X 2 = 4 ,  2 X 5 = 10, (2, 5) = \(\frac{2}{5}\) hours,
now we convert into minutes as 1 hour is equal to
60 minutes, So \(\frac{2}{5}\) X 60 = \(\frac{2 X 60}{5}\)  =
\(\frac{120}{5}\)  as 5 X 24 =120 we get \(\frac{120}{5}\) = 24,
therefore I have spent 24 minutes working on science homework.

2.2 Dividing Fractions (pp. 53-60)

Divide. Write the answer in simplest form.

Question 8.
Big Ideas Math Solutions Grade 6 Chapter 2 Fractions and Decimals 342
Big Ideas Math Solutions Grade 6 Chapter 2 Fractions and Decimals 342 = \(\frac{9}{10}\),
Explanation:
Given expressions as \(\frac{3}{4}\) ÷ \(\frac{5}{6}\),
we write reciprocal of the fraction \(\frac{5}{6}\) as \(\frac{6}{5}\)
and multiply as \(\frac{3}{4}\) X \(\frac{6}{5}\) =
\(\frac{3 X 6}{4 X 5}\) = \(\frac{18}{20}\),
we can further simplify as both goes in 2, 2 X 9 = 18 and 2 X 10 = 20,
(9,10)=\(\frac{9}{10}\). Therefore \(\frac{3}{4}\) ÷
\(\frac{5}{6}\) = \(\frac{9}{10}\).

Question 9.
Big Ideas Math Solutions Grade 6 Chapter 2 Fractions and Decimals 343
Big Ideas Math Solutions Grade 6 Chapter 2 Fractions and Decimals 343= \(\frac{1}{20}\),
Explanation:
Given expressions as \(\frac{2}{5}\) ÷ 8
we write reciprocal of the fraction 8 as \(\frac{1}{8}\)
and multiply as \(\frac{2}{5}\) X \(\frac{1}{8}\) =
\(\frac{2 X 1}{5 X 8}\) = \(\frac{2}{40}\),
we can further simplify as both goes in 2, 2 X 1 = 2 and 2 X 20 = 40,
(1,20)=\(\frac{1}{20}\). Therefore \(\frac{2}{5}\) ÷ 8
= \(\frac{1}{20}\).

Question 10.
Big Ideas Math Solutions Grade 6 Chapter 2 Fractions and Decimals 344
Big Ideas Math Solutions Grade 6 Chapter 2 Fractions and Decimals 344= 15
Explanation:
Given expressions as 5 ÷ \(\frac{1}{3}\),
we write reciprocal of the fraction \(\frac{1}{3}\) as 3
and multiply as 5 X 3 =15, Therefore 5 ÷ \(\frac{1}{3}\) = 15.

Question 11.
Big Ideas Math Solutions Grade 6 Chapter 2 Fractions and Decimals 345
Big Ideas Math Solutions Grade 6 Chapter 2 Fractions and Decimals 345= \(\frac{80}{27}\) = 2\(\frac{26}{27}\),
Explanation:
Given expressions as \(\frac{8}{9}\) ÷ \(\frac{3}{10}\),
we write reciprocal of the fraction \(\frac{8}{9}\) as \(\frac{10}{3}\)
and multiply as \(\frac{8}{9}\) X \(\frac{10}{3}\) =
\(\frac{8 X 10}{9 X 3}\) = \(\frac{80}{27}\),
as numerator is greater we write as ( 2 X 27 + 26 by 27)= 2\(\frac{26}{27}\).
Therefore \(\frac{8}{9}\) ÷ \(\frac{3}{10}\) = \(\frac{80}{27}\) = 2\(\frac{26}{27}\).

Question 12.
A box contains 10 cups of pancake mix. You use \(\frac{2}{3}\) cup
each time you make pancakes. How many times can you make pancakes?
Big Ideas Math Solutions Grade 6 Chapter 2 Fractions and Decimals 346
We can make 15 times pancakes,

Explanation:
Given a box contains 10 cups of pancake mix. I use \(\frac{2}{3}\) cup
each time you make pancakes, So number of times I
can make pancakes are 10 ÷ \(\frac{2}{3}\) =
we write reciprocal of the fraction \(\frac{2}{3}\) as \(\frac{3}{2}\)
and multiply as 10 X \(\frac{3}{2}\) = \(\frac{10 X 3}{2}\) =
\(\frac{30}{2}\) as 30 goes in 2 we get \(\frac{30}{2}\) = 15,
therefore we can make 15 times pancakes.

Question 13.
Write two fractions whose quotient is \(\frac{28}{45}\).

Explanation:

The two fractions can be \(\frac{4}{9}\) ÷  \(\frac{5}{7}\) whose
quotient is \(\frac{28}{45}\).

Explanation:
Given to find two fractions whose quotient is \(\frac{28}{45}\),
We can take factors of 28 are  1, 2, 4, 7, 14 and 28, So numerators can be
(1,28) or (2, 14) or (4, 7), now factors of 45 are 1, 3, 5, 9, 15, and 45,
denominators can be (1, 45) or (3, 15) or (5, 9) and 45
factors will be reciprocal, So the two fractions can be
\(\frac{4}{9}\) ÷ \(\frac{5}{7}\) now we calculate
as \(\frac{4}{9}\) X \(\frac{7}{5}\)
quotient is \(\frac{4 X 7}{9 X 5}\) = \(\frac{28}{45}\).

2.3 Dividing Mixed Numbers (pp. 61–66)

Divide. Write the answer in simplest form.

Question 14.
Big Ideas Math Solutions Grade 6 Chapter 2 Fractions and Decimals 347
Big Ideas Math Solutions Grade 6 Chapter 2 Fractions and Decimals 347= 2\(\frac{9}{20}\),
Explanation:
Given expressions as 1\(\frac{2}{5}\) ÷ \(\frac{4}{7}\),
we write mixed fraction as ( 1 X 5 + 2 by 5) = \(\frac{7}{5}\),
we write reciprocal of the fraction \(\frac{4}{7}\) as \(\frac{7}{4}\)
and multiply as \(\frac{7}{5}\) X \(\frac{7}{4}\) =
\(\frac{7 X 7}{5 X 4}\) = \(\frac{49}{20}\),
as numerator is greater we write as ( 2 X 20 + 9 by 20)= 2\(\frac{9}{20}\).
Therefore 1\(\frac{2}{5}\) ÷ \(\frac{4}{7}\) = \(\frac{49}{20}\) = 2\(\frac{9}{20}\).

Question 15.
Big Ideas Math Solutions Grade 6 Chapter 2 Fractions and Decimals 348
Big Ideas Math Solutions Grade 6 Chapter 2 Fractions and Decimals 348= 1\(\frac{7}{8}\),
Explanation:
Given expressions as 5\(\frac{5}{8}\) ÷ 3,
we write mixed fraction as ( 5 X 8 + 5 by 8) = \(\frac{45}{8}\),
we write reciprocal of 3 as \(\frac{1}{3}\)
and multiply as \(\frac{45}{8}\) X \(\frac{1}{3}\) =
\(\frac{45 X 1}{8 X 3}\) = \(\frac{45}{24}\), as both
goes in 3, we get 3 X 15 = 45, 3 X 8 = 24,(15,8) = \(\frac{15}{8}\)
as numerator is greater we write as ( 1 X 8 + 7 by 8)= 1\(\frac{7}{8}\),
Therefore 5\(\frac{5}{8}\) ÷ 3 = 1\(\frac{7}{8}\).

Question 16.
Big Ideas Math Solutions Grade 6 Chapter 2 Fractions and Decimals 349
Big Ideas Math Solutions Grade 6 Chapter 2 Fractions and Decimals 349 = 1\(\frac{3}{4}\),
Explanation:
Given expressions as 5 ÷ 2\(\frac{6}{7}\) ,
we write mixed fraction as ( 2 X 7 + 6 by 7) = \(\frac{20}{7}\),
we write reciprocal of \(\frac{20}{7}\) as \(\frac{7}{20}\)
and multiply as 5 X \(\frac{7}{20}\) =
\(\frac{5 X 7}{1 X 20}\) = \(\frac{35}{20}\), as both
goes in 5, we get 5 X 7 = 35, 5 X 4 = 20,(7,4) = \(\frac{7}{4}\)
as numerator is greater we write as ( 1 X 4 + 3 by 4)= 1\(\frac{3}{4}\),
Therefore 5 ÷ 2\(\frac{6}{7}\) = 1\(\frac{3}{4}\).

Question 17.
Big Ideas Math Solutions Grade 6 Chapter 2 Fractions and Decimals 350
Big Ideas Math Solutions Grade 6 Chapter 2 Fractions and Decimals 350= 1\(\frac{5}{6}\),
Explanation:
Given expressions as 4\(\frac{1}{8}\) ÷ 2\(\frac{1}{4}\),
we write mixed fraction 4\(\frac{1}{8}\) as (4 X 8 + 1 by 8) =
\(\frac{33}{8}\) and 2\(\frac{1}{4}\) as (2 X 4 + 1 by 4) =
\(\frac{9}{4}\) now we write reciprocal of the fraction
\(\frac{9}{4}\) as \(\frac{4}{9}\)
and multiply as \(\frac{33}{8}\) X \(\frac{4}{9}\) =
\(\frac{33 X 4}{8 X 9}\) = \(\frac{132}{72}\),
as both goes in 12, 12 X 11 = 132, 12 X 6 = 72, (11,6) = \(\frac{11}{6}\)
as numerator is greater we write as ( 1 X 6 + 5 by 6)= 1\(\frac{5}{6}\).
Therefore 4\(\frac{1}{8}\) ÷ 2\(\frac{1}{4}\) = 1\(\frac{5}{6}\).

Question 18.
Evaluate Big Ideas Math Solutions Grade 6 Chapter 2 Fractions and Decimals 351. Write the answer in simplest form.
5\(\frac{5}{7}\) ÷ 1\(\frac{3}{5}\)X 4\(\frac{2}{3}\) =
16\(\frac{2}{3}\),

Explanation:
Given expressions as 5\(\frac{5}{7}\) ÷ 1\(\frac{3}{5}\) X
4\(\frac{2}{3}\), first we write mixed fraction 5\(\frac{5}{7}\) as
(5 X 7 + 5 by 7) = \(\frac{40}{7}\) and 1\(\frac{3}{5}\)
as (1 X 5 + 3 by 5) = \(\frac{8}{5}\) now we write reciprocal of the fraction
\(\frac{8}{5}\) as \(\frac{5}{8}\)
and multiply as \(\frac{40}{7}\) X \(\frac{5}{8}\) =
\(\frac{40 X 5}{7 X 8}\) = \(\frac{200}{56}\),
as both goes in 8, 8 X 25 = 200, 8 X 7 = 56, (25,7) = \(\frac{25}{7}\),
Now we write 4\(\frac{2}{3}\)  as ( 4 X 3 + 2 by 3) = \(\frac{14}{3}\),
we multiply as \(\frac{25}{7}\) X \(\frac{14}{3}\) =
\(\frac{25 X 14}{7 X 3}\) = \(\frac{350}{21}\), as both goes
in 7 as 7 X 50 = 350, 7 X 3 = 21, (50, 3) = \(\frac{50}{3}\),
as numerator is greater we write as ( 16 X 3 + 2 by 3)= 16\(\frac{2}{3}\),
Therefore 5\(\frac{5}{7}\) ÷ 1\(\frac{3}{5}\) X
4\(\frac{2}{3}\) = 16\(\frac{2}{3}\).

Question 19.
You have 23\(\frac{1}{2}\) pounds of blueberries to store in freezer bags.
Each bag holds 3\(\frac{3}{4}\) pounds of blueberries.
What is the minimum number of freezer bags needed to store all the blueberries?

7 freezer bags minimum are needed to store all the blueberries.

Explanation:
Given I have 23\(\frac{1}{2}\) pounds of blueberries to
store in freezer bags. Each bag holds 3\(\frac{3}{4}\) pounds
of blueberries. Minimum number of freezer bags needed to
store all the blueberries 23\(\frac{1}{2}\) ÷ 3\(\frac{3}{4}\) =
We write 23\(\frac{1}{2}\) as  (23 X 2 + 1 by 2) = \(\frac{47}{2}\),
and 3\(\frac{3}{4}\) as ( 3 X 4 + 3 by 4) = \(\frac{15}{4}\),
now we write \(\frac{15}{4}\) as reciprocal \(\frac{4}{15}\)
and multiply as \(\frac{47}{2}\) X \(\frac{4}{15}\)  =
\(\frac{47 X 4}{2 X 15}\)  = \(\frac{188}{30}\) as both
goes in 2, 2 X 94 = 188, 2 X 15 = 30, (94, 15) = \(\frac{94}{15}\) ,
as numerator is greater we write as (6 X 15 + 4 by 15) = 6\(\frac{4}{15}\),
So approximately we require 7 freezer bags minimum
needed to store all the blueberries.

Question 20.
A squirrel feeder holds 4\(\frac{1}{2}\) cups of seeds.
Another squirrel feeder holds 6\(\frac{7}{8}\) cups of seeds.
One scoop of seeds is 1\(\frac{5}{8}\) cups.
How many scoops of seeds do you need to fill both squirrel feeders?
Big Ideas Math Solutions Grade 6 Chapter 2 Fractions and Decimals 351.1

7 scoops of seeds we need to fill both squirrel feeders,

Explanation:
Given a squirrel feeder holds 4\(\frac{1}{2}\) cups of seeds.
Another squirrel feeder holds 6\(\frac{7}{8}\) cups of seeds.
One scoop of seeds is 1\(\frac{5}{8}\) cups.
So first squirrel feeder needs 4\(\frac{1}{2}\) ÷ 1\(\frac{5}{8}\),
We write 4\(\frac{1}{2}\) as (4 X 2 + 1 by 2) = \(\frac{9}{2}\)
and 1\(\frac{5}{8}\) = (1X 8 + 5 by 8) = \(\frac{13}{8}\),
now we write \(\frac{13}{8}\) reciprocal and multiply as
\(\frac{9}{2}\) X \(\frac{8}{13}\) = \(\frac{9 x 8}{2 X 13}\) =
\(\frac{72}{26}\) as both goes in 2, 2 X 36 = 72, 2 X 13 = 26, (36,13) =
\(\frac{36}{13}\),
Now second squirrel feeder needs 6\(\frac{7}{8}\) ÷ 1\(\frac{5}{8}\),
We write 6\(\frac{7}{8}\) as (6 X 8 + 7 by 8) = \(\frac{55}{8}\),
now we have 1\(\frac{5}{8}\) = (1X 8 + 5 by 8) = \(\frac{13}{8}\),
now we write \(\frac{13}{8}\) reciprocal and multiply as
\(\frac{55}{8}\) X \(\frac{8}{13}\) = \(\frac{55 X 8}{8 X 13}\) =
\(\frac{440}{104}\) as both goes in 8, 8 X 55 = 440, 8 X 13 = 104,
(55, 13) = \(\frac{55}{13}\), Now we have feeder 1 and feeder 2,
so total is \(\frac{36}{13}\) + \(\frac{55}{13}\) as
both have same denomiators we add numerators and write as
\(\frac{36 + 55}{13}\) = \(\frac{91}{13}\) as 91 goes
in 13( 13 X 7 = 91) we get \(\frac{91}{13}\) = 7,therefore
7 scoops of seeds we need to fill both squirrel feeders.

2.4 Adding and Subtracting Decimals (pp. 67–72)

Add.

Question 21.
3.78 + 8.94
3.78 + 8.94 = 12.72,

Explanation:
Given expression 3.78 + 8.94 we add as
   1,1  
3.78
+8.94
12.72
Therefore, 3.78 + 8.94 = 12.72.

Question 22.
19.89 + 4.372
19.89 + 4.372 = 24.262,

Explanation:
Given expression 19.89 + 4.372 we add as
  1,1,1  
19.890
+ 4.372
24.262
Therefore, 19.89 + 4.372 = 24.262.

Question 23.
24.916 + 17.385
24.916 + 17.385 = 42.301,

Explanation:
Given expression 24.916 + 17.385  we add as
    1,1,1,1  
24.916
+ 17.385
42.301
Therefore, 24.916 + 17.385 = 42.301.

Subtract.

Question 24.
7.638 – 2.365
7.638 – 2.365 = 5.273,

Explanation:
Given expression 7.638 – 2.365  we subtract as
      5,13  
7.638
– 2.365
5.273
Therefore, 7.638 – 2.365 = 5.273.

Question 25.
14.21 – 4.103
14.21 – 4.103 = 10.107,

Explanation:
Given expression 14.21 – 4.103 we subtract as
   14,13,10  
14.210
–  4.103
10.107
Therefore, 14.21 – 4.103 = 10.107.

Question 26.
5.467 – 2.736
5.467 – 2.736 = 2.731,

Explanation:
Given expression 5.467 – 2.736 we subtract as
  4,14  
5.467
– 2.736
2.731
Therefore, 5.467 – 2.736 = 2.731.

Question 27.
Write three decimals that have a sum of 10.806.

The three decimals are 4.203, 3.769, 2.834  which
will have a sum of 10.806,

Explanation:
To make sum of 10.806 we can take decimals as
4.203, 3.769, 2.834  and check as
  1,1,1  
4.203
3.769
+2.834
10.806
therefore the three decimals are 4.203, 3.769, 2.834
which will have a sum of 10.806.

Question 28.
To make fuel for the main engines of a space shuttle,
102,619.377 kilograms of liquid hydrogen and 616,496.4409 kilograms
of liquid oxygen are mixed together in the external tank.
How much fuel is stored in the external tank?
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 352

Fuel stored in the external tank is 719,115.786 kilograms,

Explanation:
Given to make fuel for the main engines of a space shuttle,
102,619.377 kilograms of liquid hydrogen and 616,496.4409 kilograms
of liquid oxygen are mixed together in the external tank.
So fuel is stored in the external tank is
102,619.377 + 616,496.4409 =
      1,1,1,  1      
102,619.3770
+616,496.4409
719115.8179
Therefore, fuel stored in the external tank is 719,115.786 kilograms.

2.5 Multiplying Decimals (pp. 73–80)

Multiply. Use estimation to check your answer.

Question 29.
26.174 × 79
26.174 X 79 = 2067.746,

Explanation:
Given expression 26.174 X 79 we add as
4,1,5,2
5,1,6,3  
26.174 —— 3 decimal places
X   79
  235566
1832180
2067.746—— 3 decimal places
Therefore, 26.174 X 79 = 2067.746.

Question 30.
9.475 × 8.03
9.475 × 8.03 = 76.08425,

Explanation:
Given expression 9.475 X 8.03 we add as
3,6,4
1,2,1  
9.475 —— 3 decimal places
X 8.03 —– 2 decimal places
0028425
0000000
7580000
76.08425——5 decimal places
Therefore, 9.475 × 8.03 = 76.08425.

Question 31.
0.051 × 0.244
0.051 × 0.244 = 0.012444,

Explanation:
Given expression 9.475 X 8.03 we add as
 1        
   2     
     2     
0.051 —— 3 decimal places
X 0.244 —– 3 decimal places
0000204
0002040
0110200
0000000
0.012444——6 decimal places
Therefore, 0.051 × 0.244 = 0.012444.

Question 32.
Evaluate 3.76(2.43 + 9.8).
3.76(2.43 + 9.8) = 45.9848,

Explanation:
Given expression as 3.76(2.43 + 9.8),
First we calculate (2.43 + 9.8) =
2.43
+9.8
12.23
Now 3.76 X 12.23 =
 1,1,2  
  1,1,1
12.23 —— 2 decimal places
X3.76 —— 2 decimal places
0073.38
085610
36.6900
45.9848 —-4 decimal places
therfeore, 3.76(2.43 + 9.8) = 45.9848.

Question 33.
Hair grows about 1.27 centimeters each month.
How much does hair grow in 4 months?

5.08 centimeters hair will grow in 4 months,

Explanation:
Given hair grows about 1.27 centimeters each month.
so hair growth in 4 months is 4 X 1.27 =
1,2  
1.27
X 4
5.08
Therefore, 5.08 centimeters hair will grow in 4 months.

2.6 Dividing Whole Numbers (pp. 81–86)

Divide. Use estimation to check your answer.

Question 34.
7296 ÷ 38
7296 ÷ 38 = 192,
Estimation is reasonable,

Explanation:
Given 7296 ÷ 38 =
      192
38)7296    38 X 1 = 38
   38
     349     38 X 9 = 342
     342
         76   38 X 2 = 76
         76
          0
Therefore 7296 ÷ 38 = 192,
Estimation is 7300 ÷ 38 = 192.1,
So estimation is reasonable.

Question 35.
5081 ÷ 203
5081 ÷ 203 = 25.02,
Estimation is reasonable,

Explanation:
Given 5081 ÷ 203 =
      25.02 
203)5081     203 X 2 = 406
     406
       1021    203 X 5 = 1015
       1015
            6     203 X 0.02 = 4.06
            4.06
1.94  reaminder
Therefore 5081 ÷ 203 = 25.02,
Estimation is 5100 ÷ 203 = 25.123
So estimation is reasonable.

Question 36.
\(\frac{17,264}{128}\)
\(\frac{17,264}{128}\) = 134.875,
Estimation is approximately reasonable,

Explanation:
Given 17264 ÷ 128 =
      134.875 
128)17264    128 X 1 = 128
     128
         446     128 X 3 = 384
         384
            624  128 X 4 = 512
            512
   112   128 X 0.8 =
   102.4
9.6             128 X 0.07 = 8.96
8.96
0.64  128 X 0.005 = 0.64
0.64
 0
Therefore \(\frac{17,264}{128}\) = 134.875
Estimation is 17,000 ÷ 128 = 132.8125
So estimation is approximately reasonable.

Question 37.
Your local varsity basketball team offers bus transportation for a
playoff game. Each bus holds 56 people. A total of 328 people sign up.
All buses are full except for the last bus. How many buses are used?
How many people are in the last bus?

6 buses are used and in the last bus
there are 48 people.

Explanation:
Given my local varsity basketball team offers bus transportation for a
playoff game. Each bus holds 56 people. A total of 328 people sign up.
So, buses used are 328 ÷ 56 =
     5
56)328    56 X 5 = 280
     280
48 remainder
we got 5 buses + 48 people,
Therefore we need total 6 buses and in the last bus
there are 48 people.

Question 38.
You have 600 elastic bands to make railroad bracelets.
How many complete bracelets can you make?
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 353

15 bracelets I can make,

Explanation:
Given 600 elastic bands to make railroad bracelets
and we need for one bracelet 28 elastic bands for
outer rails and 10 elastic bands for inner track,
means 28 + 10 = 38 elastic bands for one bracelet,
From 600 we can make 600 ÷ 38 =
     15.78
38)600     38 X 1 = 38
     38
     220     38 X 5 = 190
     190
30     38 X 0.7 = 26.6
26.6
3.4   38 X 0.08 = 3.04
 3.04
         0.36 remainder
therefore, it means I can make 15 railroad bracelets
from 600 elastic bands.

2.7 Dividing Decimals (pp. 87–94)

Divide. Check your answer.

Question 39.
0.498 ÷ 6
0.498 ÷ 6 = 0.083,

Explanation:
Given 0.498÷ 6 =
  0.083
6)0.498      6 X 0.08 = 0.48
   0.48
    0.018     6 X 0.003 = 0.018
    0.018
       0
Therefore 0.498 ÷ 6 = 0.083.

Question 40.
8.9 ÷ 0.356
8.9 ÷ 0.356 = 25

Explanation:
Given 8.9÷ 0.356 =
       25     
0.356)8.9      0.356 X  25 = 8.9
          8.9
          0
Therefore 8.9 ÷ 0.356 = 8.9.

Question 41.
21.85 ÷ 3.8
21.85 ÷ 3.8 = 5.75

Explanation:
Given 21.85 ÷ 3.8 =
       5.75     
3.8)21.85      3.8 X 5 = 19
      19
        2.85       3.8 X 0.7 = 2.66
        2.66
       0.19        3.8 X 0.05 = 0.19
0.19
Therefore 21.85 ÷ 3.8 = 5.75.

Question 42.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 254
(14.075 + 24.67) ÷  3.15 = 12.3,

Explanation:
Given expression as (14.075 + 24.67) ÷  3.15,
first we calculate (14.075 + 24.67) = 38.745 now
38.745 ÷  3.15 =
        12.3 
3.15)38.745    3.15 X 12 = 37.8
        37.8
          0.945    3.15 X 0.3 = 0.945
          0.945
           0
Therefore, (14.075 + 24.67) ÷  3.15 = 12.3.

Question 43.
Your beginning balance on your lunch account is $42.
You buy lunch for $1.80 every day and sometimes buy a snack for $0.85.
After 20 days, you have a balance of $0.05. How many snacks did you buy?

I buyed 7 snacks in 20 days,

Explanation:
Given my beginning balance on lunch account is $42.
I buy lunch for $1.80 every day and sometimes buy a snack
for $0.85. After 20 days, I have a balance of $0.05.
So for 20 days I buyed for lunch is 20 X $1.80 =
$36 and have balance left of $0.05 means 36 + 0.05 = $36.05,
Now I spent on snacks is $42 – $36.05 = $5.95 and snack
costs $0.85, So number of snacks I did buyed is
$5.95 ÷  $0.85 =
         7   
0.85)5.95   0.85 X 7 = 5.95
        5.95
          0
therefore, I buyed 7 snacks in 20 days.

Fractions and Decimals Practice Test

2 Practice Test

Evaluate the expression. Write the answer in simplest form.

Question 1.
5.138 + 2.624
5.138 + 2.624 = 7.762

Explanation:
Given expression 24.916 + 17.385  we add as
        1   
5.138
+2.624
7.762
Therefore, 5.138 + 2.624 = 7.762.

Question 2.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 354
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 354= \(\frac{7}{4}\) or 1\(\frac{3}{4}\),

Explanation:
Given expressions as \(\frac{5}{6}\) ÷ \(\frac{10}{21}\),
we write reciprocal of the fraction \(\frac{10}{21}\) as \(\frac{21}{10}\)
and multiply as \(\frac{5}{6}\) X \(\frac{21}{10}\) =
\(\frac{5 X 21}{6 X 10}\) = \(\frac{105}{60}\),
we can further simplify as both goes in 15, 15 X 7 = 105 and 15 X 4 = 60,
(7, 4)= \(\frac{7}{4}\) as numerator is greater we write as
(1 X 4 + 3 by 4) = 1\(\frac{3}{4}\), Therefore \(\frac{5}{6}\) ÷
\(\frac{10}{21}\) = \(\frac{7}{4}\) = 1\(\frac{3}{4}\).

Question 3.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 355
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 355 = 21.84,
Explanation:
Given 5.46 ÷ 0.25 =
      21.84     
0.25)5.46      0.25 X 21 = 5.25
        5.25
        0.21      0.25 X 0.8 = 0.2
        0.20
       0.01       0.25 X 0.04 = 0.01
       0.01
         0
Therefore 5.46 ÷ 0.25 = 21.84.

Question 4.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 356
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 356= \(\frac{3}{8}\),
Explanation:
Given expression as \(\frac{9}{16}\) X \(\frac{2}{3}\) =
Step I: We multiply the numerators as 9 X 2 = 18
Step II: We multiply the denominators as 16 X 3 =48
Step III: We write the fraction in the simplest form as
\(\frac{18}{48}\), as both goes in 6 we
can further simplify as 6 X 3 = 18, 6 X 8 = 48, (3, 8) = \(\frac{3}{8}\),
So \(\frac{9}{16}\) X \(\frac{2}{3}\) = \(\frac{3}{8}\).

Question 5.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 357
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 357= \(\frac{70}{23}\) = 3\(\frac{1}{23}\),
Explanation:
Given expressions as 8\(\frac{3}{4}\) ÷ 2\(\frac{7}{8}\),
we write mixed fraction 8\(\frac{3}{4}\) as (8 X 4 + 3 by 4) =
\(\frac{35}{4}\) and 2\(\frac{7}{8}\) as (2 X 8 + 7 by 8) =
\(\frac{23}{8}\) now we write reciprocal of the fraction
\(\frac{23}{8}\) as \(\frac{8}{23}\)
and multiply as \(\frac{35}{4}\) X \(\frac{8}{23}\) =
\(\frac{35 X 8}{4 X 23}\) = \(\frac{280}{92}\),
as both goes in 4, 4 X 70 = 280, 4 X 23 = 92, (70, 23) = \(\frac{70}{23}\)
as numerator is greater we write as (3 X 23 + 1 by 23)= 3\(\frac{1}{23}\).
Therefore 8\(\frac{3}{4}\) ÷ 2\(\frac{7}{8}\) = \(\frac{70}{23}\) = 3\(\frac{1}{23}\),

Question 6.
4.87 × 7.23
4.87 × 7.23 = 35.2101,

Explanation:
Given expression 4.87 X 7.23 we add as
6,4
1,1
2,2,2  
4.87 —— 2 decimal places
X7.23—-  2 decimal places
001461
009740
340900
35.2101——4 decimal places
Therefore, 4.87 × 7.23 = 35.2101,

Question 7.
1875 ÷ 125
1875 ÷ 125 = 15,

Explanation:
        15   
125)1875    125 X 1 = 125
       125
         625    125 X 5 = 625
         625
          0
Therefore, 1875 ÷ 125 = 15.

Question 8.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 358
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 358= 25,
Explanation:
Given expressions as 10 ÷ \(\frac{2}{5}\),
now we write reciprocal of the fraction
\(\frac{2}{5}\) as \(\frac{5}{2}\)
and multiply as 10 X \(\frac{5}{2}\) =
\(\frac{10 X 5}{1 X 2}\) = \(\frac{50}{2}\),
as both goes in 2, 2 X 25 = 50, 2 X 1 = 2, (25, 1) = \(\frac{50}{2}\) = 25,
Therefore 10 ÷ \(\frac{2}{5}\) = 25.

Question 9.
57.82 ÷ 0.784
57.82 ÷ 0.784 = 73.75,

Explanation:
Given expression as 57.82 ÷ 0.784 =
73.75
0.784)57.82   0.784 X 7 = 57.232
         57.232
          0.588   0.784 X 0.7 = 0.5488
0.5488
          0.0392  0.784 X 0.05 = 0.0392
          0.0392
            0
Therefore, 57.82 ÷ 0.784 = 73.75.

Question 10.
5.316 ÷ 1.942
5.316 ÷ 1.942 = 2.737,

Explanation:
Given expression as 5.316 ÷ 1.942 =
   2.737
1.942)5.316    1.942 X 2 = 3.884
          3.884
          1.432    1.942 X 0.7 = 1.3594
1.3594
          0.0726   1.942 X 0.03 = 0.05826
0.05826
         0.01434  1.942 X 0.007 = 0.013594
0.013594
0.000746 remainder
therefore 5.316 ÷ 1.942 = 2.737.

Question 11.
6.729 × 8.3
6.729 × 8.3 = 55.8507,

Explanation:
Given expression as 6.729 X 8.3 =
5,2,7
  2,2
6.729—— 3 decimal places
X 8.3 —— 1 decimal place
020187
538320
55.8507—— 4 decimal places

Question 12.
\(\frac{13,376}{248}\)
\(\frac{13,376}{248}\) = 53.935,

Explanation:
Given expression as \(\frac{13,376}{248}\),
       53.935
248)13376    248 X 5 = 1240
       1240 
976    248 X 3 = 744
           744
232    248 X 0.9 = 223.2
           223.2
             8.8   248 X 0.03 = 7.44
             7.44
1.36  248 X 0.005 = 1.24
             1.24
0.12 remainder
therefore \(\frac{13,376}{248}\) = 53.935.

Question 13.
On a road trip, you notice that the gas tank is \(\frac{1}{4}\) full.
The gas tank can hold 18 gallons, and the vehicle averages 22 miles per gallon.
Will you make it to your destination 110 miles away before you run out of gas?
Explain.

No, I cannot make it to destination, I will be running out of gas,

Explanation:
Given on a road trip, you notice that the gas tank is \(\frac{1}{4}\) full.
The gas tank can hold 18 gallons, and the vehicle averages
22 miles per gallon. Will I make it to my destination 110 miles
away before I run out of gas, First we calculate tank has
18 X \(\frac{1}{4}\) = \(\frac{18}{4}\) , as both
goes in 2 we get 2 X 9 = 18, 2 X 2 = 4, (9, 2) = \(\frac{9}{2}\) gallon,
now we have the vehicle averages 22 miles per gallon means
22 X \(\frac{9}{2}\) = \(\frac{198}{2}\) as
it goes in 2 we get 2 X 99 = 198, 2 X 1 = 2, \(\frac{198}{2}\) =99,
as 99 is less than 110 miles no, I cannot make it to destination,
I will be running out of gas.

Question 14.
For a diving event, the highest and the lowest of seven scores
are discarded. Next, the total of the remaining scores is multiplied
by the degree of difficulty of the dive. That value is then multiplied
by 0.6 to determine the final score. Find the final score for the dive.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 359

The final score for the dive is 73.47,

Explanation:
Given for a diving event, the highest and the lowest of seven scores
are discarded. Next, the total of the remaining scores is multiplied
by the degree of difficulty of the dive. That value is then multiplied
by 0.6 to determine the final score.
After discarding the highest and the lowest of seven scores-
9.0, 7.0 we have remaining scores as 8, 7.5, 8, 8.5, 7.5,
now we add as 8 +7.5 + 8 +8.5 + 7.5 = 39.5, So the total of the
remaining scores is multiplied by the degree of difficulty
of the dive, given the difficulty of the dive as 3.1,
So 39.5 X 3.1 =
2,1
39.5 —— 1 decimal place
X3.1 —— 1 decimal place
00395
11850
122.45—— 2 decimal places
Now 122.45 is multiplied by 0.6 to determine the final score as
122.45 X 0.6 =
 1,1,2,3
122.45—— 2 decimal places
X 0.6  —— 1 decimal place
073470
000000
73.470—— 3 decimal places,
So, The final score for the dive is 73.47.

Question 15.
You are cutting as many 20\(\frac{1}{2}\)-inch pieces
from the board to make ladder steps for a tree fort.
How many steps can you make? How much wood is left over?
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 360
I can make 5 steps and \(\frac{35}{41}\)-inch piece
wood will be left over.

Explanation:
Given I am cutting as many 20\(\frac{1}{2}\)-inch pieces
from the board of 120 in to make ladder steps for a tree fort.
Number of steps I can make are 120 ÷ 20\(\frac{1}{2}\),
first we write mixed fraction into fraction and write
reciprocal then multiply as 20\(\frac{1}{2}\) = ( 20 X 2 + 1 by 2) =
\(\frac{41}{2}\), now reciprocal as \(\frac{2}{41}\)
and multiply as 120 X \(\frac{2}{41}\) = \(\frac{240}{41}\),
as numerator is greater we write as (5 X 41 + 35 by 41) = 5\(\frac{35}{41}\),
So, I can make 5 steps and \(\frac{35}{41}\)-inch piece
wood will be left over.

Question 16.
You spend 2\(\frac{1}{2}\)-hours online.
You spend \(\frac{1}{5}\) of that time writing a blog.
How long do you spend writing your blog?

I have spent 12\(\frac{1}{2}\) hours for writing blog,

Explanation:
Given I spend 2\(\frac{1}{2}\)-hours online and
spent \(\frac{1}{5}\) of that time writing a blog.
So I spend writing my blog for 2\(\frac{1}{2}\) ÷ \(\frac{1}{5}\) =
first we write mixed fraction as fraction 2\(\frac{1}{2}\) =
(2 X 2 + 1 by 2) = \(\frac{5}{2}\) ÷ \(\frac{1}{5}\) now
we write  \(\frac{1}{5}\) as reciprocal we get 5 and multiply,
\(\frac{5}{2}\) X 5 = \(\frac{5 X 5 }{2}\) = \(\frac{25}{2}\),
as numerator is greater we write as ( 2 X 12 + 1 by 2) = 12\(\frac{1}{2}\),
Therefore I have spent 12\(\frac{1}{2}\) hours for writng blog.

Question 17.
You and a friend take pictures at a motocross event.
Your camera can take 24 pictures in 3.75 seconds.
Your friend’s camera can take 36 pictures in 4.5 seconds.
Evaluate the expression (36 ÷ 4.5) (24 ÷ 3.75) to find how many
times faster your friend’s camera is than your camera.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 361

51.2 seconds times faster my friend’s camera is
than my camera,

Explanation:
Given expression as (36 ÷ 4.5) (24 ÷ 3.75) =
First we will find (36 ÷ 4.5) =
      8 
4.5)36    4.5 X 8 = 36
      36
        0
So, (36 ÷ 4.5) = 8,
Now we will find (24 ÷ 3.75)=
       6.4
3.75)24         3.75 X 6 = 22.5
       22.5
1.5        3.75 X 0.4 = 1.5
       1.5
        0
So, (24 ÷ 3.75) = 6.4,
Now 8 X 6.4 =
3,
6.4 —– 1 decimal place
X 8
51.2 —– 1 decimal place
Therefore, 51.2 seconds times faster my friend’s camera is
than my camera.

Fractions and Decimals Cumulative Practice

2 Cumulative Practice

Question 1.
Which number is equivalent to the expression below?
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 362

Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 363
6 X 8 – 2 X 32 = 30, matches with B,

Explanation:
Given expression as (6 X 8 – 2 X 32 ) =
First we will find (2 X 32 ) = 2 X 3 X 3 = 18,
Now 6 X 8 = 48, So 48 -18 = 30,
Therefore (6 X 8 – 2 X 32 ) = 30 which matches with B.

Question 2.
What is the greatest common factor of 48 and 120?
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 364
Greatest common factor of 48 and 120 is 24,

Explanation:
Given to find greatest common factor of 48 and 120,
We have factors of 48 are 1, 2, 3, 4, 6, 8, 12, 16, 24 and
factors of 120 are 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60,
Here we have the biggest common factor number is 120,
therefore greatest common factor of 48 and 120 is 24.

Question 3.
Which number is equivalent to 5.139 – 2.64?
F. 2.499
G. 2.599
H. 3.519
I. 3.599
5.139 – 2.64 = 2.499 equivalent to F,

Explanation:
Given to evaluate 5.139 – 2.64 =
5.139
-2.640
2.499
Therefore, 5.139 – 2.64 = 2.499 which equivalent to F.

Question 4.
Which number is equivalent to Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 365
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 366
\(\frac{4}{9}\) ÷ \(\frac{5}{7}\) = \(\frac{28}{45}\),
matches with B.
Explanation:
Given expressions as \(\frac{4}{9}\) ÷ \(\frac{5}{7}\),
we write reciprocal of the fraction \(\frac{5}{7}\) as \(\frac{7}{5}\)
and multiply as \(\frac{4}{9}\) X \(\frac{7}{5}\) =
\(\frac{4 X 7}{9 X 5}\) = \(\frac{28}{45}\),
Therefore \(\frac{4}{9}\) ÷ \(\frac{5}{7}\) = \(\frac{28}{45}\),
which matches with B.

Question 5.
You buy orange and black streamers for a party. The orange
streamers are 9 feet long, and the black streamers are 12 feet long.
What are the least numbers of streamers you should buy in order for
the total length of the orange streamers to be the same as the total
length of the black streamers?
F. 36 orange streamers and 36 black streamers
G. 12 orange streamers and 9 black streamers
H. 3 orange streamers and 4 black streamers
I. 4 orange streamers and 3 black streamers

The total length of the orange streamers to be the same
as the total length of the black streamers is
4 orange streamers and 3 black streamers matches with I.

Explanation:
Given I buy orange and black streamers for a party. The orange
streamers are 9 feet long, and the black streamers are 12 feet long.
What are the least numbers of streamers I should buy in order for
the total length of the orange streamers to be the same as the total
length of the black streamers are  we first list the multiples of
9 and 12 and then we find the smallest multiple they have in common.
Multiples of 9: 9, 18, 27, 36, 45, 54, etc.
Multiples of 12: 12, 24, 36, 48, 60, 72, etc.
The least multiple on the two lists that they have in common
is the LCM of 9 and 12. Therefore, the LCM of 9 and 12 is 36.
So orange streamers are 36 ÷ 9 =
   4
9)36     9 X 4 =36
   36
    0
36 ÷ 9 = 4,
and black streamers are 36 ÷ 12 =
   3
12)36    12 X 3 = 36
     36
0
36 ÷ 12 = 3,
Therefore, The total length (36) of the orange streamers
to be the same as the total length of the black streamers is
4 orange streamers and 3 black streamers matches with I.

Question 6.
Which number is a prime factor of 572?
A. 4
B. 7
C. 13
D. 22
13 number is prime factor of 572 ,matches with C,

Explanation:
Given to find the number which is prime factor of 572,
We have prime factors os 572 as 2, 2, 11, 13,
Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals-95
So from given numbers C.13 – matches.

Question 7.
Which number is equivalent to 7059 ÷ 301?
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 367

\(\frac{7059}{301}\) = 23\(\frac{136}{301}\) which matches with H.

Explanation:
Given expressions as \(\frac{7059}{301}\) =
        23  
301)7059    301 X 2 = 602
       602
1039   301 X 3 = 903
         903
136 remainder
So, \(\frac{7059}{301}\) = 23\(\frac{136}{301}\) which matches with H.

Question 8.
A square wall tile has side lengths of 4 inches. You use 360 of the tiles.
What is the area of the wall covered by the tiles?
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 368
The area of the wall covered by the tiles is 3760 in.2
matches with D.

Explanation:
Given a square wall tile has side lengths of 4 inches, we
use 360 of the tiles, so the area of the wall covered by the tiles is
(we have area of sqaure is (Side X Side)) = 360 X 4 X 4 = 5760,
Therefore the area of the wall covered by the tiles is 3760 in.2
which matches with D.

Question 9.
Which expression is equivalent to a perfect square?
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 369

Expression I. 32 + 6 X 5 ÷ 3 is equivalent to a perfect square,

Explanation:
Given to find an expressions equivalent to a perfect square is
the square of a number is that number times itself.
The perfect squares are the squares of the whole numbers:
1, 4, 9, 16, 25, 36, 49, 64, 81, 100,
Now square roots are
Big Ideas Math Book 6th Grade Answer Key Chapter 2 Fractions and Decimals-96
First we the expressions as 3 + 22 X 7 = 3 + ((2 X 2) X 7) =
3 + ( 4 X 7) = 3 + 28 = 31, as 31 lies between 25 and 36, 52 and 62
not a perfect square,
Next expression is 34 + (18 ÷ 22 ) =
34 + (18 ÷ (2 X 2)) = 34 + 18 ÷ 4 = 34 + 4.5 =39.5,
It lies between 36 and 49, 62 and 72
not a perfect square,
Next expression is (80 + 4) ÷ 4 = 84 ÷ 4 = 21,
It lies between 16 and 25, 42 and 52
not a perfect square,
Next expression is 32 + 6 X 5 ÷ 3 = (((3 X 3) + 6) X 5) ÷ 3 =
((9 + 6) X 5 ÷ 3) = (15 X 5) ÷ 3 = 75 ÷ 3 =
   25
3)75    3 X 2 = 6
   6  
15   3 X 5 = 15
    15
0
So, 75 ÷ 3 = 25, 25 is perfect square 52 ,
therefore expression I. 32 + 6 X 5 ÷ 3 is equivalent to a perfect square.

Question 10.
What is the missing denominator in the expression below?
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 370
A. 1
B. 2
C. 3
D. 8
3 is the missing denominator in the expression,
matches with C,

Explanation :
let us take the missing denominator as x,
given expression as \(\frac{4}{8}\) ÷ \(\frac{2}{x}\) = \(\frac{3}{4}\) ,
Now we have \(\frac{2}{x}\) we write reciprocal \(\frac{x}{2}\)
and multiply as \(\frac{4}{8}\) X \(\frac{x}{2}\) =
\(\frac{4 X x}{8 X 2}\) = \(\frac{4x}{16}\) both goes in 4,
4 X x = 4x, 4 X 4 = 16, (x, 4) = \(\frac{x}{4}\) now we equate with
\(\frac{3}{4}\) we get x = 3, Therefore 3 is the missing denominator in
the expression which matches with C.

Question 11.
What is 4.56 × 0.7?
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 371

Explanation:
Given expression as 4.56 X 0.7 =
 3,4
4.56  —- 2 decimal places
X 0.7—- -1  decimal place
3.192 — 3 decimal places

Therfore, 4.56 X 0.7 = 3.192.

Question 12.
The area of the large rectangle is how many times the area of the small rectangle?
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 372
F. 4.4515
G. 5.915
H. 17.2575
I. 111.2875
The area of the large rectangle is 17.2575 cm2
times the area of the small rectangle matches with H.

Explanation:
Given large rectangle width 7.25 cm and length as 3.07 cm,
So area of large rectangle is 7.25 X 3.07 =
 1
1,3
7.25 —- 2 decimal places
X3.07—- 2 decimal places
005075
000000
217500
22.2575 —- 4 decimal places

therefore area of lare rectangle is 22.2575 cm2,
Given small rectangle width 4 cm and length as 1.25 cm,
So area of small rectangle is 4 X 1.25 =
   2
1.25
X 4
5
therefore area of small rectangle is 5 cm2,
now we will find the area of the large rectangle is how many
times the area of the small rectangle as 22.2575 – 5 = 17.2575 cm2,
So, matches with H.

Question 13.
Which expression is equivalent to 5 × 5 × 5 × 5?
A. 5 × 4
B. 45
C. 54
D. 55
The expression which is equivalent to 5 X 5 X 5 X 5 = 54,
matches with C,

Explanation:
Given expression as 5 X 5 X 5 X 5 means 5 is multiplied by 4 times,
So 5 X 5 X 5 X 5 = 54 which matches with C.

Question 14.
A walkway is built using identical concrete blocks.
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 372.1
Part A
How much longer, in inches, is the length of the walkway than the
width of the walkway? Show your work and explain your reasoning.
Part B
How many times longer is the length of the walkway than the
width of the walkway? Show your work and explain your reasoning.
Part A :
\(\frac{11}{4}\) inches in the length of the
walkway more than the width of the walkway.
Part B :
2 times longer is the length of the walkway than the
width of the walkway,

Explanation:
Part A :
Given length as 5\(\frac{1}{2}\) inches and width as
2\(\frac{3}{4}\) of walkway. Now longer, in inches,
is the length of the walkway than the width of the walkway is
5\(\frac{1}{2}\) – 2\(\frac{3}{4}\) first we write
mixed fractions into fractions before subtraction as
5\(\frac{1}{2}\) = (5 x 2 + 1 by 2) = \(\frac{11}{2}\) and
2\(\frac{3}{4}\) = (2 X 4 + 3 by 4) = \(\frac{11}{4}\),
So \(\frac{11}{2}\) – \(\frac{11}{4}\) ,now we make
both denominators same as 4 so  we multiply numerator and
denominator by 2 for \(\frac{11}{2}\)  = \(\frac{11 x 2}{2 X 2}\) =
\(\frac{22}{4}\) ,Now \(\frac{22}{4}\) – \(\frac{11}{4}\),
as denominators are same we subtract numerators and write as
\(\frac{22- 11}{4}\) = \(\frac{11}{4}\) inches.
Therefore, \(\frac{11}{4}\) inches in the length of the
walkway more than the width of the walkway.
Part B :
Now we will find many times longer in the length of the walkway than the
width of the walkway as \(\frac{11}{2}\) ÷ \(\frac{11}{4}\),
now we write reciprocal and multiply as \(\frac{11}{2}\) X \(\frac{4}{11}\) =
\(\frac{11 X 4}{2 X 11}\) as both have 11 they get cancel,
we will get \(\frac{4}{2}\), as both goes in 2 we get 2 X 2 = 4,
2 X 1 = 2, (2, 1), \(\frac{4}{2}\) = 2,
therefore 2 times longer is the length of the walkway than the
width of the walkway.

Question 15.
A meteoroid moving at a constant speed travels 6\(\frac{7}{8}\) miles
in 30 seconds. How far does the meteoroid travel in 1 second?
Big Ideas Math Answer Key Grade 6 Chapter 2 Fractions and Decimals 373
The meteoroid travels in 1 second is \(\frac{11}{48}\) mile,
matches with G.

Explanation:
Given a meteoroid moving at a constant speed travels 6\(\frac{7}{8}\) miles
in 30 seconds, Now in 1 second the meteoroid travel is
6\(\frac{7}{8}\) ÷ 30 =
First we write mixed fraction as 6\(\frac{7}{8}\)  = (6 X 8 + 7 by 8)
\(\frac{55}{8}\) now we divide with 30 as \(\frac{55}{8}\) ÷ 30 =
Now we write 30 as reciprocal as \(\frac{1}{30}\) and multiply as
\(\frac{55}{8}\) X \(\frac{1}{30}\) = \(\frac{55 X 1}{8 X 30}\)  =
\(\frac{55}{240}\), now both goes in 5 as 5 X 11 = 55
and 5 X 48 = 240, (11, 48) = \(\frac{11}{48}\) miles,
therefore, the meteoroid travels in 1 second is \(\frac{11}{48}\) mile,
which matches with G.

Final Words:

We hope that the details of Big Ideas Math Answers Grade 6 Chapter 2 Fractions and Decimals pdf are really helpful for all the candidates. Check all the concepts of decimals and fractions in detail from here. While practicing, if you have any doubts regarding any of the concepts, feel free to ask them in the comments section. Follow or keep in touch with us to get all the updates regarding any of the topics. BIM Grade 6 Answer Key will clear all your doubts regarding the exam.

Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20

Big Ideas Math Answers Grade 1 Chapter 4

In Order to attempt the 1st Grade Exam, you need to have strong fundamentals. Download Big Ideas Math Book 1st Grade Answer Key Chapter 4 Add Numbers within 20 free of cost and get the homework help you need. Attempt the practice test to know your preparation level within the Chapter. Practice as much as you can using the Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 and clear the exam with better scores.

Big Ideas Math Book 1st Grade Answer Key Chapter 4 Add Numbers within 20

You can make the most out of the Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 through the quick links available. Simply tap on the topic you want and solve various questions involved in it easily. There are various models in addition and subtraction. We have listed all of them by considering enough examples and explained every problem step by step.

Vocabulary

Lesson: 1 Add Doubles from 6 to 10

Lesson: 2 Use Doubles within 20

Lesson: 3 Count on to Add within 20

Lesson 4: Add Three Numbers

Lesson 5 Add Three Numbers by Making a 10

Lesson: 6 Add 9

Lesson 7 Make a 10 to Add

Lesson: 8 Problem Solving: Addition within 20

Performance Task

Add Numbers within 20 Vocabulary

Organize It

Review Words:
addend
doubles
doubles minus 1
doubles plus 1
sum

Use the review words to complete the graphic organizer.

Big Ideas Math Answer Key Grade 1 Chapter 4 Add Numbers within 20 1

Answer:

Explanation:
By Looking at graphic organizer
1.  Here we have 2 blocks and in second block we have one more added,
so we write the review words as doubles plus 1.
2. Here we have only two block so we write the review words as doubles.
3. Here also we have 2 blocks, but in second block on one dot it is crossed,
so we write the review words as doubles minus 1.

Define It

Match the review word to its definition.

Big Ideas Math Answer Key Grade 1 Chapter 4 Add Numbers within 20 2
Answer:

Explanation:
Addend means any of the numbers that are added together.
Sum can be defined as the result or answer we get on adding two or more numbers or terms.
Here 1. 5 + 3 = 8  numbers 5, 3 are addends and 8 is sum
2. 4 + 3 = 7 number 4,3 are addends and 7 is sum.

Lesson 4.1 Add Doubles from 6 to 10

Explore and Grow

Use counters to model the story.

You have 7 marbles. Your friend has 7 marbles. How many marbles are there in all?

Big Ideas Math Answer Key Grade 1 Chapter 4 Add Numbers within 20 3
Answer:
In all, there are total of 14 marbles.

Explanation:
Given I have 7 marbles and my friend has 7 marbles,
The total number of marbles available are ( my marbles and friend’s marble),
7 + 7 = 14 marbles. So 14 marbles are there in all.

Show and Grow

Question 1.
Big Ideas Math Answer Key Grade 1 Chapter 4 Add Numbers within 20 4
___8____ + ___8___ = __16_____
Answer:
8 + 8 = 16 dots

Explanation:
Given in block 1 we have 8 dots and in block 2 we have 8 dots,
So in total we have 8 + 8 = 16 dots.

Question 2.
Big Ideas Math Answer Key Grade 1 Chapter 4 Add Numbers within 20 5
____7___ + __7____ = ___14____
Answer:
7 + 7 = 14 dots

Explanation:
Given in block 1 we have 7 dots and in block 2 we have 7 dots,
So in total we have 7 + 7 = 14 dots.

Question 3.
Big Ideas Math Answer Key Grade 1 Chapter 4 Add Numbers within 20 6 =
Answer:
9 + 9 = 18 dots

Explanation:
Given in block 1 we have 9 dots and in block 2 we have 9 dots,
So in total we have 9 + 9 = 18 dots.

Question 4.
Big Ideas Math Answer Key Grade 1 Chapter 4 Add Numbers within 20 7

Answer:
10 + 10 = 20 dots

Explanation:
Given in block 1 we have 10 dots and in block 2 we have 10 dots,
So in total we have 10 + 10 = 20 dots.

Question 5.
Big Ideas Math Answer Key Grade 1 Chapter 4 Add Numbers within 20 8
____8___ + _8_____ = _______
Answer:
8 + 8 = 16 dots

Explanation:
Given in block 1 we have 8 dots and in block 2 we have 8 dots,
So in total we have 8 + 8 = 16 dots.

Question 6.
Big Ideas Math Answer Key Grade 1 Chapter 4 Add Numbers within 20 9
___6____ + __6___ = ___12____
Answer:
6 + 6 = 12 dots

Explanation:
Given in block 1 we have 6 dots and in block 2 we have 6 dots,
So in total we have 6 + 6 = 12 dots.

Question 7.
7 + 7 = ___14____
Answer:
7 + 7 = 14 dots

Explanation:
Given first number is 7 and second number is 7,
So in total we have 7 + 7 = 14 dots.

Question 8.
10 + 10 = ___20_____
Answer:
10 + 10 = 20

Explanation:
Given first number is 10 and second number is 10,
So in total we have 10 + 10 = 20.

Question 9.
Big Ideas Math Answer Key Grade 1 Chapter 4 Add Numbers within 20 10
Answer:

Explanation:
Given first number as 9 and second number as 9,
So in total we have 9 + 9 = 18.

Question 10.
Big Ideas Math Answer Key Grade 1 Chapter 4 Add Numbers within 20 11
Answer:

Explanation:
Given first number as 8 and second number as 8,
So in total we have 8 + 8 = 16.

Question 11.
MP Reasoning
You and your friend each read 6 books. How many books do you and your friend read in all?
____12______ books
Answer:
Total number of books I and my friend read are 12.

Explanation:
Given I and my friend each read 6 books, that means I read 6 books,
my friend read 6 books, So in total we read 6 + 6 = 12.

Think and Grow: Modeling Real Life

You and your friend have the same number of video games. There are 16 in all. How many video games do you have?
Big Ideas Math Answer Key Grade 1 Chapter 4 Add Numbers within 20 12
Draw a Picture:

Addition equation: X + X = 16,

___8_______ video games
Answer:
I have 8 number of video games.

Explanation:
Given I and my friend have same number of video games,
So in all both have 16 video games , Let us take I have X number of video games,
means friend will also have X number of video games,
Now addition equation is X + X = 16, 2 X = 16 , So X = 16 ÷ 2 = 8,
So I have 8 video games.

Show and Grow

Question 12.
2 friends give you the same number of pictures. You have 12 in all. How many pictures does each friend give you?
Big Ideas Math Answer Key Grade 1 Chapter 4 Add Numbers within 20 13
Draw a Picture:

Addition equation:
X + X = 12,

____6______ pictures
Answer:
Each friend gives 6 pictures to me.

Explanation:
Given 2 friends gave me same number of pictures,
So in all both gave 12 pictures , Let us take each gave X number of pictures,
So we write addition equation as X + X = 12, 2 X = 12 , So X = 12 ÷ 2 = 6,
So each friend gave me 6 number of pictures.

Add Doubles from 6 to 10 Practice 4.1

Question 1.
Big Ideas Math Answer Key Grade 1 Chapter 4 Add Numbers within 20 14
____10______ + ____10______ = ___20______
Answer:
10 + 10 = 20 dots

Explanation:
Given in block 1 we have 10 dots and in block 2 we have 10 dots,
So in total we have 10 + 10 = 20 dots.

Question 2.
Big Ideas Math Answer Key Grade 1 Chapter 4 Add Numbers within 20 15
_____6_____ + ____6______ = ____12_____
Answer:
6 + 6 = 12 dots

Explanation:
Given in block 1 we have 6 dots and in block 2 we have 6 dots,
So in total we have 6 + 6 = 12 dots.

Question 3.
Big Ideas Math Answer Key Grade 1 Chapter 4 Add Numbers within 20 16
Answer:


Explanation:
Given first number as 9 and second number as 9,
So in total we have 9 + 9 = 18.

Question 4.
Big Ideas Math Answer Key Grade 1 Chapter 4 Add Numbers within 20 17
Answer:

Explanation:
Given first number as 10 and second number as 10,
So in total we have 10 + 10 = 20.

Question 5.
MP Reasoning
You and your friend each do 7 jumping jacks. How many jumping jacks do you and your friend do in all?

____14______ jumping jacks
Answer:
14 jumping jacks I and my friend do in all .

Explanation:
Given I and my friend each do 7 jumping jacks,
means I did 7 jumping jacks, My friend did 7 jumping jacks,
So both overall did 7 + 7 = 14 jumping jacks.

Question 6.
Modeling Real Life
Newton and Descartes each have the same number of treats. They have 18 treats in all.
How many treats does Newton have?
Big Ideas Math Answer Key Grade 1 Chapter 4 Add Numbers within 20 18

____9______ treats
Answer:
Newton have 9 treats.

Explanation:
Given Newton and Descartes each have the same number of treats,
Suppose if  Newton have  X treats means Descartes have same X treats
and in all both have 18 treats, So X + X = 18 , 2X = 18 , X = 18 ÷ 2 = 9,
So Newton has 9 treats.

Review & Refresh

Use the picture to write an equation.

Question 7.
Big Ideas Math Answer Key Grade 1 Chapter 4 Add Numbers within 20 19
____3______ + ____2_____ = ____5_____
Answer:
3 + 2 = 5

Explanation:
By seeing the picture there are 3 children sitting and reading books on the chair and
2 children are having bags and going to school, So we write the equation as
3 + 2 = 5.

Question 8.
Big Ideas Math Answer Key Grade 1 Chapter 4 Add Numbers within 20 20
___5_______ + ____1_____ = ____6_____
Answer:
5 + 1 = 6

Explanation:
By seeing the picture there are 5 children sitting on the ground and reading books,
1 child having book is waving hand, So we write the equation as
5 + 1 = 6.

Lesson 4.2 Use Doubles within 20

Explore and Grow

Use counters to model the story.

You collect 8 leaves. Your friend collects 8 leaves. How many leaves are there in all?

____16_____ leaves
Explanation:
Both I and friend collect same leaves ,
8 + 8 = 16 leaves we call the sum as doubles as both addends are 8.

You collect 1 more leaf. How many leaves are there now?

_____17_____ leaves
Explanation:
We got doubles as 16, Now the required sum is one more than doubles,
That is 17,Therefore 8+8+1= 17 so we call as double plus 1.

Big Ideas Math Answers 1st Grade 1 Chapter 4 Add Numbers within 20 21

Show and Grow

Question 1.
Use the double 6 + 6 to find each sum.
6 + 7 = ___13____
Big Ideas Math Answers 1st Grade 1 Chapter 4 Add Numbers within 20 22
Answer:
6 + 7= 13

Explanation:
We Know the double  6 + 6 = 12 , The number 7 is one more than 6,
So we can write 7 as 6 + 1, The required sum is one more than double,
12 + 1 =13, Sum is 13, therefore 6 + 7 = 13, Double plus 1.

6 + 5 = ___11____
Big Ideas Math Answers 1st Grade 1 Chapter 4 Add Numbers within 20 23
Answer:
6+5 = 11

Explanation:
We Know the double of 6 + 6 = 12 , The number 5 is one less than 6,
So we can write 5 as 6 – 1, The required sum is one less than double,
12 – 1 = 11, Sum is 11, therefore 6 + 5 = 11, Double minus 1.

Apply and Grow: Practice

Use the double 7 + 7 to find each sum.

Question 2.
7 + 8 = ___15_____
Big Ideas Math Answers 1st Grade 1 Chapter 4 Add Numbers within 20 24
Answer:
7 + 8 = 15

Explanation:
We know the double of 7 + 7 = 14 , The number 8 is one more than 7,
So we can write 8 as 7 + 1, The required sum is one more than double,
14 + 1 =15, Sum is 15, therefore 7 + 8 = 15, Double plus 1.

7 + 6 = ____13_____
Big Ideas Math Answers 1st Grade 1 Chapter 4 Add Numbers within 20 25
Answer:
7 + 6 = 13

Explanation:
We know the double of 7+ 7 = 14 , The number 6 is one less than 7,
So we can write 6 as 7 – 1, The required sum is one less than double,
14 – 1 = 13, Sum is 13, therefore 7 + 6 = 13, Double minus 1.

Find the sum. Write the double you used.

Question 3.
5 + 6 = ___11______

___5_+5__+_ ○ ___1___ = ___11____
Answer:
5 + 6 = 11, we use double 5 + 5,

Explanation:
As the first addend is 5 and we know the double is 5 + 5 ,
So 5 + 5 = 10 , The number 6 is one more than 5,
So we can write 6 as 5 + 1, The required sum is one more than double,
10 + 1 =11, Sum is 11, therefore 5 + 6 = 11, Double plus 1.
We use double 5 + 5.

Question 4.
9 + 8 = ___17______

___9+9 -____ ○ ___1____ = __17_____
Answer:
9 + 8 = 17, we use double 9 + 9 ,

Explanation:
As the first addend is 9 and we know the double is 9 + 9,
9 + 9 = 18 , The number 8 is one less than 9,
So we can write 8 as 9 – 1, The required sum is one less than double,
18 – 1 = 17, Sum is 17, therefore 9 + 8 = 17, Double minus 1.
We use double 9 + 9

Question 5.
DIG DEEPER!
Circle two ways you can solve 9 + 10. Show how you know.
9 + 9 and 1 more
9 + 9 and 1 less
10 – 10 and 1 more
10 + 10 and 1 less
Answer:
To solve 9 + 10 the two ways  which we can solve are
9 + 9 and 1 more and other is 10 + 10 and 1 less.

Explanation:
We know 9 + 10 = 19 now we will see which way we can solve,
1. 9 + 9 and 1 more means , it is 18 + 1 which is equal to 19,
Similar to 9 + 10 = 19 , therefore we can solve with 9 + 9 and 1 more .
2. 9 + 9 and 1 less means it is 18 – 1 which is 17,
As 17 is not equal to 9 + 10 = 19, So 9 + 9 and 1 less way we will not solve 9 + 10 .
3. 10 – 10 and 1 more means 0 + 1 = 1 , here this is not equal to 9 + 10,
So by 10 – 10 and 1 more way we will not solve for 9 + 10.
4. 10 + 10 and 1 less means 10 + 9 = 19 which is equal to 9 + 10 ,
So 10 + 10 and 1 less way we will solve 9 + 10.
Therefore to solve 9 + 10 the two ways which we can solve are
9 + 9 and 1 more and other is 10 + 10 and 1 less.

Think and Grow: Modeling Real Life

A music room has 7 keyboards. There is 1 more recorder than keyboards. How many instruments are there?
Big Ideas Math Answers 1st Grade 1 Chapter 4 Add Numbers within 20 26
Which doubles can you use to find the sum?
7 + 7          8 + 8            6 + 6
Equation:

__15_____ instruments
Answer:
There are 15 instruments in the music room.
We use doubles 7 + 7  to find the sum 7 + 8 = 15.

Explanation:
Given a music room has 7 keyboards and there are  1 more recorder than keyboards,
means number of recorders are 7 + 1 = 8, so total number of instruments in room are
7+ 8 = 7 + 7 + 1 = 15, Here we have used doubles 7 + 7.
Therefore  to find the sum we use doubles 7 + 7 and the sum is 7 + 8 = 15.

Show and Grow

Question 6.
A museum has lo paintings. There is 1 fewer sculpture than paintings. How many art pieces are there?
Big Ideas Math Answers 1st Grade 1 Chapter 4 Add Numbers within 20 27
Which doubles can you use to find the sum?
8 + 8        9 + 9      10 + 10
Equation:

____19______ art pieces
Answer:
There are 19 art pieces, We use doubles 10 + 10  to find the sum 10 + 9 = 19.

Explanation:
Given a museum has 10 paintings and there is 1 fewer sculpture than paintings,
means number of sculptures are one less than paintings 10 – 1 = 9,
So total number of art pieces are 10 + 9 = 10 + 10 – 1 = 19,
Here we have used doubles 10 + 10.
Therefore  to find the sum we use doubles 10 + 10 and the sum is 10 + 9 = 19.

Use Doubles within 20 Practice 4.2

Question 1.
9 + 10 = ____19____
Big Ideas Math Answers 1st Grade 1 Chapter 4 Add Numbers within 20 28
Answer:
9 + 10 = 19, We use double 9 + 9

Explanation:
As the first addend is 9 and we know the double is 9 + 9 ,
So 9 + 9 = 18 , The number 10 is one more than 9,
So we can write 10 as 9 + 1, The required sum is one more than double,
18 + 1 =19, Sum is 19, therefore 9 + 10 = 19, Double plus 1.
Here we have use double 9 + 9

9 + 8 = ____17____
Big Ideas Math Answers 1st Grade 1 Chapter 4 Add Numbers within 20 29
Answer:
9 + 8 = 17, We use double 9 + 9

Explanation:
As the first addend is 9 and we know the double is 9 + 9,
9 + 9 = 18 , The number 8 is one less than 9,
So we can write 8 as 9 – 1, The required sum is one less than double,
18 – 1 = 17, Sum is 17, therefore 9 + 8 = 17, Double minus 1.
We have use double 9 + 9

Find the sum. Write the double you used.

Question 2.
10 + 9 = ____19_____

____10+10___ -○ ___1____ = __19_____
Answer:
10 + 9 = 19, we use double 10 + 10,

Explanation:
As the first addend is 10 and we know the double is 10 + 10,
10 + 10 = 20 , The number 9 is one less than 10,
So we can write 9 as 10 – 1, The required sum is one less than double,
20 – 1 = 19, Sum is 19, therefore 10 + 9 = 17, Double minus 1.
We have use double 10 + 10 .

Question 3.
7 + 8 = ____15_____

_7_+ 7_+____ ○ __1_____ = ___15___
Answer:
7 + 8 = 15, We use double 7 + 7
Explanation:
As the first addend is 7 and we know the double is 7 +7  ,
So 7 + 7 = 14 , The number 8 is one more than 7,
So we can write 8 as 7 + 1, The required sum is one more than double,
14 + 1 =15, Sum is 15, therefore 7 + 8 = 15, Double plus 1.
We use double 7 + 7.

Question 4.
DIG DEEPER!
Circle two ways you can solve 8 + 9. Show how you know.
8 + 8 and 1 more
8 + 8 and 1 less
9 + 9 and 1 more
9 + 9 and 1 less
Answer:
To solve 8 + 9 the two ways  which we can solve are
8 + 8 and 1 more and other is 9 + 9 and 1 less.

Explanation:
We know 8 + 9 = 17 now we will see which way we can solve,
1. 8 + 8 and 1 more means , it is 16 + 1 which is equal to 17,
Similar to 8 + 9 = 17 , therefore we can solve with 8 + 8 and 1 more .
2. 8 + 8 and 1 less means it is 16 – 1 which is 15,
As 15 is not equal to 8 + 9 = 17, So 8 + 8 and 1 less way we will not solve 8 + 9 .
3. 9 + 9 and 1 more means 18 + 1 = 19 , here this is not equal to 17,
So by 9 + 9 and 1 more way we will not solve for 8 + 9.
4. 9 + 9 and 1 less means 18 -1 = 17 which is equal to sum of 8 + 9  ,
So 9 + 9 and 1 less way we will solve 8 + 9.
Therefore to solve 8 + 9 the two ways which we can solve are
8 + 8 and 1 more and other is 9 + 9 and 1 less.

Question 5.
Modeling Real Life
There are 6 balls. There is 1 more toy hoop than balls. How many toys are there?
Which doubles can you use to find the sum?
Big Ideas Math Answers 1st Grade 1 Chapter 4 Add Numbers within 20 30
7 + 7           8 + 8          6 + 6

___13_____ toys
Answer:
There are 13 toys in total,
We use doubles 6 + 6 to find the sum 6 + 7 = 13.

Explanation:
Given there are 6 balls and there is  1 more toy hoop than balls means
number of toy hoops =  6 + 1 = 7, so total number of toys are
6+ 7 = 6+ 6 + 1 = 13, Here we have used doubles 6 + 6.
Therefore to find the sum we use doubles 6 + 6 and the sum is 6 + 7 = 13.

Review & Refresh

Use the picture to write an equation.

Question 6.
Big Ideas Math Answers 1st Grade 1 Chapter 4 Add Numbers within 20 31
____7____ – ___1____ = ____6____
Answer:
Equation is 7 – 1 = 6

Explanation:
By seeing the picture there are total 7 children in the picture,
6 children are sitting and reading books and one boy with bag on shoulder’s is
walking away from other children, So we write the equation as 7 -1 = 6.

Question 7.
Big Ideas Math Answers 1st Grade 1 Chapter 4 Add Numbers within 20 32
____5____ – ___1____ = ___4_____
Answer:
Equation is 5 – 1 = 4

Explanation:
By seeing the picture there are total 5 children ,
in them four are sitting and playing with toy,
one boy is walking away from others,
So we write the equation as 5 – 1 =4.

Lesson 4.3 Count on to Add within 20

Explore and Grow

Model the story.

There are 8 coins in a piggy bank. You put in 5 more. How many coins are in the bank now?
Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 33
Answer:
There are 13 coins in the piggy bank.

Explanation:
Given there are 8 coins in a piggy bank and I have put 5 more means,
we start from 8 and move 5 units from 8 on number line we get 13, 8 + 5 = 13,
There are total 13 coins in the piggy bank.

Show and Grow

Question 1.
8 + 4 = _____12_____
Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 34
Answer:
8 + 4 = 12

Explanation:
First we start from 8 and move 4 units we get 12 on the number line,
So sum is 8 + 4 = 12 .

Question 2.
5 + 10 = ___15______
Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 35
Answer:
5 + 10 = 15

Explanation:
First we start from 5 and move 10 units we get 15 on the number line,
So sum is 5 + 10 = 15 .

Apply and Grow: Practice

Question 3.
Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 36
6 + 8 = _____14_____
Answer:
6 + 8 = 14

Explanation:
First we start from 6 and move 8 units we get 14 on the number line,
So sum is 6 + 8 = 14 .

Question 4.
Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 36
12 + 5 = __17______
Answer:
12 + 5 = 17

Explanation:
First we start from 12 and move 5 units we get 17 on the number line,
So sum is 12 + 5 = 17 .

Question 5.
Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 36
7 + 8 = ___15______
Answer:
7 + 8 = 15

Explanation:
First we start from 7 and move 8 units we get 15 on the number line,
So sum is 7 + 8 = 15 .

Question 6.
Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 36
10 + 9 = ___19_____
Answer:
10 + 9 = 19

Explanation:
First we start from 10 and move 9 units we get 19 on the number line,
So sum is  10 + 9 = 19 .

Question 7.
Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 36
____11___ = 0 + 11
Answer:
11 = 0 + 11 or 0 + 11 = 11

Explanation:
First we start from 0 and move 11 units we get 11 on the number line,
So sum is 11= 0 + 11 or 0 + 11 = 11.

Question 8.
Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 36
___13_____ = 4 + 9
Answer:
13 = 4 + 9

Explanation:
First we start from 4 and move 9 units we get 13 on the number line,
So sum is 13= 4 + 9  or 4 + 9 = 13.

Question 9.
MP Structure
Write the equation shown by the number line. Then write the equation another way.
Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 37
Answer:
7 + 4 = 11     or  4 + 7 = 11 or 11 = 7 + 4

Explanation:
As shown in the number line we are starting from 7
and moving 4 units we get 11,
So the equation is 7 + 4 = 11, we can also write  the equation as 4 + 7 =11,
there is no change in the value of sum.
We get sum 11 if we add 7 and 4.

Think and Grow: Modeling Real Life

You have 7 train cars. Newton has 5 more than you. Descartes has 4 more than you. Who has more train cars, Newton or Descartes?
Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 38
Model:
Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 39

Equations:
Newton               Descartes
7 + 5 = 12          7 + 4 = 11

Who has more?         Newton             Descartes
Newton has more trains.

Explanation:
Given I have 7 trains, Newton has 5 more than you,
So the equation is 7 + 5 = 12, Newton has 12 trains,
and Descartes has 4 more than me , the equation is
7 + 4 = 11, Descartes has 11 trains. Now we compare with
Newton and Descartes so in between 12 and 11 we compare,
as 12 is more or greater than 11, means Newton has more trains = 12.

Show and Grow

Question 10.
You have 6 comic books. Newton has 5 more than you. Descartes has 6 more than you. Who has more comic books, Newton or Descartes?
Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 40
Model:
Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 39

Equations:
Newton             Descartes
6 + 5 = 11          6 + 6 = 12

Who has more?         Newton             Descartes
Descartes has more comic books.

Explanation:
Given I have 6 comic books , Newton has 5 more than you,
So the equation is 6 + 5 = 11, Newton has 11 comic books,
and Descartes has 6 more than me , the equation is
6 + 6 = 12, Descartes has 12 comic books. Now we compare with
Newton and Descartes so in between 11 and 12 we compare,
as 12 is more or greater than 11, means Descartes has more comic books = 12.

Count on to Add within 20 Practice 4.3

Question 1.
Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 41
10 + 7 = ___17_______
Answer:
10 + 7 = 17

Explanation:
First we start from 10 and move 7 units we get 17 on the number line,
So sum is 10 + 7 = 17.

Question 2.
Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 41
8 + 9 = ____17_____
Answer:
8 + 9 = 17

Explanation:
First we start from 8 and move 9 units we get 17 on the number line,
So sum is 10 + 7 = 17.

Question 3.
Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 41
_____15_____ = 6 + 9
Answer:
6 + 9 = 15 or 15 = 6 + 9

Explanation:
First we start from 6 and move 9 units we get 15 on the number line,
So sum is 6 + 9 = 15 or 15 = 6 + 9.

Question 4.
Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 41
______16_____ = 12 + 4
Answer:
16= 12 + 4 or 12 + 4 = 16

Explanation:
First we start from 12 and move 4 units we get 16 on the number line,
So sum is  12 + 4 = 16 or 16 = 12 + 4.

Question 5.
MP Structure
Write the equation shown by the number line. Then write the equation another way.
Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 42
Answer:
5 + 8 = 13  or  8 + 5 = 13 or 13 = 5 + 8

Explanation:
As shown in the number line we are starting from 5
and moving 8 units we get 13,
So the equation is 5 + 8 = 13, we can also write  the equation as 8 + 5 =13,
there is no change in the value of sum.
We get sum 13 if we add 5 and 8.

Question 6.
Modeling Real Life
You have 11 toys. Newton has 3 more than you. Descartes has 6 more than you. Who has more toys, Newton or Descartes?
Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 43
Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 44
Answer:
Equations:     Newton             Descartes
11 + 3 = 14          11 + 6 = 17
Descartes has more toys than Newton

Explanation:
Given I have 11 toys , Newton has 3 more than you,
So the equation is 11 + 3 = 14, Newton has 14 toys,
and Descartes has 6 more than me , the equation is
11 + 6 = 17, Descartes has 17 toys. Now we compare with
Newton and Descartes so in between 14 and 17 toys if  we compare
we get 17 is more or greater than 14, means Descartes has more toys = 17.

Review & Refresh

Question 7.
Circle the triangle. Draw a rectangle around the hexagon.
Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 45
Answer:

Explanation:
We know triangle means a closed, two-dimensional shape with three straight sides.
In shapes given in figure we have triangle in green so we will circle it,
and we know hexagon can be defined as a polygon with six sides.
In shapes given in the figure we have hexagon in yellow,
So we draw a rectangle around the hexagon.

Question 8.
Circle the cube. Draw a rectangle around the sphere.
Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 46
Answer:

Explanation:
Cube means a symmetrical three-dimensional shape,
either solid or hollow, contained by six equal squares,
In figure we have cube in purple so we circle the cube.
Sphere means a round solid figure, or its surface,
with every point on its surface equidistant from its center.
In the shapes given in figure we have sphere in yellow,
So we draw a rectangle around hexagon.

Lesson 4.4 Add Three Numbers

Explore and Grow

Use linking cubes to model the story.

You have 5 red pencils, 4 yellow pencils, and 3 blue pencils. How many pencils do you have in all?
Big Ideas Math Solutions Grade 1 Chapter 4 Add Numbers within 20 47
Answer:

Explanation:
Given 5 red pencils , 4 yellow pencils  and 3 blue pencils,
a. Now by using linking the cubes first we link with 5 red and 4 yellow pencils
we get 5 + 4 = 9 pencils now we link with 3 blue pencils 9 + 3 = 12 pencils.
b. Now by using linking the cubes first we link with 4 yellow pencils and 3 blue pencils
we get 4 + 3 = 7 pencils now we link with 5 red pencils 7 + 5 = 12 pencils.
c. Now by using linking the cubes first we link with 5 red and 3 blue pencils
we get 5 + 3 = 8 pencils now we link with 4 yellow pencils 8 + 4 = 12 pencils.
In all we have 12 pencils.

Show and Grow

Question 1.
Big Ideas Math Solutions Grade 1 Chapter 4 Add Numbers within 20 48
Answer:

Explanation:
By using linking the cubes
a. First we add 2 + 5 = 7 and now we add 7+ 6 we get 13.
b. Here we add 5 + 6 =11, now we add 11 + 2 =13
c. Now here we add first 2 with 6 we get 8 to 8 we add 5
we will get 8 + 5 = 13, So we get sum as 13.

Question 2.
Big Ideas Math Solutions Grade 1 Chapter 4 Add Numbers within 20 49
Answer:

Explanation:
By using linking the cubes
a. First we add 4 + 4 = 8 and now we add 8+ 3 we get 11.
b. Here we add 3 + 4 =7, now we add 7 + 4 =11
c. Now here we add first 4 with 3 we get 7 to 7 we add 4
we will get 7 + 4 = 11, So we get sum as 11.

Apply and Grow: Practice

Question 3.
Big Ideas Math Solutions Grade 1 Chapter 4 Add Numbers within 20 50
Answer:

Explanation:
By using linking the cubes
a. First we add 3 + 3 = 6 and now we add 6+ 5 we get 11.
b. Here we add 3 + 5 =8, now we add 8 + 3 =11
c. Now here we add first 5 with 3 we get 8 to 8 we add 3
we will get 8 + 3 = 11, So we get sum as 11.

Question 4.
Big Ideas Math Solutions Grade 1 Chapter 4 Add Numbers within 20 51
Answer:

Explanation:
By using linking the cubes
a. First we add 3 + 1 = 4 and now we add 4+ 2 we get 6.
b. Here we add 1 + 2 =3, now we add 3 + 3 =6,
c. Now here we add first 3 with 2 we get 5 to 5 we add 1
we will get 5 + 1 = 6, So we get sum as 6.

Question 5.
Big Ideas Math Solutions Grade 1 Chapter 4 Add Numbers within 20 52
Answer:

Explanation:
By using linking the cubes
a. First we add 7 + 5 = 12 and now we add 12+ 5 we get 17.
b. Here we add 5 + 5 =10, now we add 10 + 7 = 17,
c. Now here we add first 7 with 5 we get 12 to 12 we add 5
we will get 12 + 5 = 17, So we get sum as 17.

Question 6.
Big Ideas Math Solutions Grade 1 Chapter 4 Add Numbers within 20 53
Answer:

Explanation:
By using linking the cubes
a. First we add 4 + 5 = 9 and now we add 9 + 2 we get 11.
b. Here we add 5 + 2 = 7, now we add 7 + 4 = 11,
c. Now here we add first 4 with 2 we get 6 to 6 we add 5
we will get 6 + 5 = 11, So we get sum as 11.

Question 7.
7 + 8 + 1 = _________
Answer:
7 + 8 +1 = 16

Explanation:
By using linking the cubes
a. First we add 7 + 1 = 8 and now we add 8 + 8 we get 16.
b. Here we add 7 + 8 = 15, now we add 15 + 1 = 16,
c. Now here we add first 8 with 1 we get 9 to 9 we add 7
we will get 9 + 7 = 16, So we get sum as 16.

Question 8.
8 + 6 + 2 = _________
Answer:
8 + 6 + 2 = 16

Explanation:
By using linking the cubes
a. First we add 8 + 2 = 10 and now we add 10 + 6 we get 16.
b. Here we add 8 + 6 = 14, now we add 14 + 2 = 16,
c. Now here we add first 6 with 2 we get 8 to 8 we add
8 we will get 8 + 8 = 16, So we get sum as 16.

Question 9.
4 + 9 + 4 = __________
Answer:
4 + 9 + 4 = 17

Explanation:
By using linking the cubes
a. First we add 4 + 4 = 8 and now we add 8 + 9 we get 17.
b. Here we add 4 + 9 = 13, now we add 13 + 4 = 17,
c. Now here we add first 9 with 4 we get 13 to 13 we add 4
we will get 13 + 4 = 17, So we get sum as 17.

Question 10.
DIG DEEPER
Complete the number puzzle so that each branch has a sum of 14.
Big Ideas Math Solutions Grade 1 Chapter 4 Add Numbers within 20 54
Answer:

Explanation:
Let us take the missing number as X, Y Z, P
We take 6 + 1 + ___X___ = 14 , means X = 7 + X = 14,
therefore X = 14 – 7 = 7, So 6 + 1 + 7 = 14,
Next  Y, 5 + Y + 6 = 14 , 11 + Y = 14 , Y = 14 -11,
We get Y = 3,Making 5 + 3 + 6 = 14
Now Z, 6 + 7 + Z = 14, 13 + Z = 14, Z = 14 – 13,
Z = 1 Making 6 + 7 + 1 = 14
Now P, 4 + P + 3 = 14, 7 + P = 14 ,P = 14 – 7,
We get P = 7 making  4 + 7 + 3 = 14.

Think and Grow: Modeling Real Life

You have 6 goldfish, 7 minnows, and 6 guppies. Will the tank hold all of your fish?
Big Ideas Math Solutions Grade 1 Chapter 4 Add Numbers within 20 55
Equation:
Big Ideas Math Solutions Grade 1 Chapter 4 Add Numbers within 20 56

Show how you know:
Answer:

Explanation:
Given I have 6 goldfish, 7 minnows and 6 guppies, Total Number of
Fishes with me are  6 + 7 + 6 = 19 , No the tank can not hold
all of fishes as given in the figure a tank can hold 15 Fish and
I have 19 which are more than 15 ,
No, it is not possible for the tank to hold 19 Fish.

Show and Grow

Question 11.
You have 2 lovebirds, 4 canaries, and 3 finches. Will the cage hold all of your birds?
Big Ideas Math Solutions Grade 1 Chapter 4 Add Numbers within 20 57
Equation:
Big Ideas Math Solutions Grade 1 Chapter 4 Add Numbers within 20 58

Show how you know:
Answer:

Explanation:
Given I have 2 lovebirds, 4 canaries, and 3 finches ,
Total Number of birds I have are 2 + 4 + 3 = 9,
Now as shown in the picture a cage can holds 10 birds,
We have total 9 birds which is less than 10 means,
Yes, the cage can hold all of my birds .

Add Three Numbers Practice 4.4

Question 1.
Big Ideas Math Solutions Grade 1 Chapter 4 Add Numbers within 20 59
Answer:

Explanation:
By using linking the cubes
a. First we add 2 + 2 = 4 and now we add 6 + 4 we will get 10.
b. Here we add first 6 + 2 = 8, now we add 8 + 2 = 10,
c. Now here we add 6 with 2 we get 8 to 8 we add 2,
we will get  8 + 2 = 10, So we get sum as 10.

Question 2.
Big Ideas Math Solutions Grade 1 Chapter 4 Add Numbers within 20 60
Answer:

Question 3.
Big Ideas Math Solutions Grade 1 Chapter 4 Add Numbers within 20 61
Answer:

Explanation:
By using linking the cubes
a. First we add 8 + 7 = 15 and now we add 15 + 4 we will get 19.
b. Here we add first 7 + 4 = 11, now we add 8 + 11 = 19,
c. Now here we add 8 with 4 we get 12 to 12 we add 7,
we will get  12 + 7 = 19, So we get sum as 19.

Question 4.
Big Ideas Math Solutions Grade 1 Chapter 4 Add Numbers within 20 62
Answer:

Explanation:
By using linking the cubes
a. First we add 9 + 1 = 10 and now we add 10 + 10 we will get 20.
b. Here we add first 9 + 10 = 19, now we add 19 + 1 = 20,
c. Now here we add 10 with 1 we get 11 to 11 we add 9,
we will get  11 + 9 = 20, So we get sum as 20.

Question 5.
3 + 4 + 5 = _________
Answer:

Explanation:
By using linking the cubes
a. First we add 3 + 5 = 8 and now we add 8 + 4 we will get 12.
b. Here we add first 3 + 4 = 7, now we add  7 and 5 as 7 + 5 = 12,
c. Now here we add 4 with 5 we get 9 to 9 we add 3,
we will get  3 + 9 = 12, So we get sum as 12.

Question 6.
7 + 5 + 5 = ________
Answer:

Explanation:
By using linking the cubes
a. First we add 7 + 5 = 12 and now we add 12 + 5 we will get 17.
b. Here we add first 7 + 5 = 12, now we add  12 and 5 as 12 + 5 = 17,
c. Now here we add 5 and 5 we get 10, we add 7 to 10,
we will get  7 + 10 = 17, So we get sum as 17.

Question 7.
8 + 4 + 8 = ________
Answer:

Explanation:
By using linking the cubes
a. First we add 8 + 8 = 16 and now we add 16 + 4 we will get 20.
b. Here we add first 8 + 4 = 12, now we add 12 and 8 as 12 + 8 = 20,
c. Now here we add 4 with 8 we get 12 ,we add 8 to 12,
we will get  8 + 12 = 20, So we get sum as 20.

Question 8.
DIG DEEPER!
Complete the number puzzle so that each branch has the sum of 13.
Big Ideas Math Solutions Grade 1 Chapter 4 Add Numbers within 20 63
Answer:

Explanation:
Let us take the missing number as X, Y Z, P
We take 5 + 1 + ___X___ = 13 , means X = 6 + X = 13,
therefore X = 13 – 6 = 7, So 5 + 1 + 7 = 13,
Next  Y, 1 + Y + 6 = 13 , 7 + Y = 13, Y = 13 – 7,
We get Y = 6,Making 1 + 6 + 6 = 13
Now Z, 4 + 5 + Z = 13, 9 + Z = 13, Z = 13 – 9,
Z = 4 Making 4 + 5 + 4 = 13
Now P, 7 + P + 2 = 13, 9 + P = 13 ,P = 13 – 9,
We get P = 4 making  7 + 4 + 2 = 13.

Question 9.
Modeling Real Life
You have 7 white chickens, I black chicken, and 8 brown chickens. Will the chicken coop hold all of your chickens?
Big Ideas Math Solutions Grade 1 Chapter 4 Add Numbers within 20 64
Big Ideas Math Solutions Grade 1 Chapter 4 Add Numbers within 20 65

Show how you know:
Answer:

Explanation:
Given I have 7 white chickens, 1 black chicken and 8 brown chickens,
Total Number of  chickens with me are  7 + 1 + 8 = 16 ,
Given in the figure a chicken coop can hold 15 chickens and
I have 16 which is more than 15.
No, it is not possible for the chicken coop to hold all 16 chickens.

Review & Refresh

Question 10.
5 + __5__ = 10
Answer:
5 + 5 = 10

Explanation:
Let us take the missing number as X,
It is given in the equation that 5 + X = 10,
So X = 10 – 5 = 5, means 5 + 5 = 10.

Question 11.
10 = 6 + ___4___
Answer:
10 = 6 + 4

Explanation:
Let us take the missing number as X,
It is given in the equation that 10 = 6 + X,
So X = 10 – 6 = 4, means 10 = 6 + 4 .

Question 12.
8 + __2___ = 10
Answer:
8 + 2 = 10

Explanation:
Let us take the missing number as X,
It is given in the equation that 8 + X = 10,
So X = 10 – 8 = 2, means 8 + 2 = 10.

Question 13.
10 = 3 + __7___
Answer:
10 = 3 + 7

Explanation:
Let us take the missing number as X,
It is given in the equation that 10 = 3 + 7,
So X = 10 – 3 = 7, means 10 = 3 + 7.

Lesson 4.5 Add Three Numbers by Making a 10

Explore and Grow

Show three ways to find a sum.
4 + 3 + 7 = ________

4 + 3 + 7 = ________

4 + 3 + 7 = ________
Answer:
4 + 3 + 7 = 14

Explanation:
By using the three ways we find sum,
Way 1. First we add 4 + 7 = 11 and now we add 11 + 3 we will get 14.
Way 2. Here we add first 4 + 3 = 7, now we add 7 and 7 as 7 + 7 = 14,
Way 3. Now here we add 3 with 7 we get 10 to 10 we add 4,
we will get  4 + 10 = 14, So we get sum as 14.

Show and Grow

Make a 10 to add.

Question 1.

Big Ideas Math Answer Key Grade 1 Chapter 4 Add Numbers within 20 66

Answer:

Add 9,1 first to make 10 then add 3, sum = 9 + 1 + 3 = 13

Explanation:
To make 10 first we see adding 9 + 1 = 10 is equal to 10,
next we add 1 +3 =4 not equal to 10, 9 + 3 = 12 which
is not equal to 10 ,So to make 10 we add  9,1 first, then add 3,
So the sum is 9 + 1 + 3 = 13.

Question 2.
Big Ideas Math Answer Key Grade 1 Chapter 4 Add Numbers within 20 67
Answer:

Add 2,8 first to make 10 then add 4, sum = 4 + 2 + 8 = 14

Explanation:
To make 10 first we see adding 4 + 2 = 6,
6 is not equal to 10, 2 + 8 = 10 which is equal to 10 ,
and now we add 4 + 8 = 12 which is not equal to 10,
So to make 10 we add 2,8 first then add 4
So the sum is 4 + 2 + 8 = 14.

Question 3.
Big Ideas Math Answer Key Grade 1 Chapter 4 Add Numbers within 20 68
Answer:

Add 3,7 first to make 10 then add 2, sum = 2 + 3 + 7 = 12

Explanation:
To make 10 first we see adding 2 + 3 = 5,
5 is not equal to 10, 3 + 7 = 10 which is equal to 10 ,
and now we add 2 + 7 = 9 which is not equal to 10,
So to make 10 we add 3,7 first and then add 2
So the sum is 2 + 3 + 7 = 12.

Question 4.
Big Ideas Math Answer Key Grade 1 Chapter 4 Add Numbers within 20 69
Answer:

Add 5,5 first to make 10 then add 9, sum = 5 + 5 + 9 = 19

Explanation:
To make 10 first we see adding 5 + 5 = 10,
so is equal to 10, now we add 5 + 9 = 14  which is not equal to 10 ,
and now we add other 5 + 9 = 14 which is also not equal to 10,
So to make 10 we add 5,5 first and then add 9,
So the sum is 5 + 5 + 9 = 19.

Apply and Grow: Practice

Make a 10 add.

Question 5.
Big Ideas Math Answer Key Grade 1 Chapter 4 Add Numbers within 20 70
Answer:

Add 3,7 or 7,3 first to make 10 then add 3, sum = 3 + 7 + 3 = 13

Explanation:
To make 10 first we see adding 3 + 7 = 10,
so is equal to 10, now we add 7 + 3 = 10  which is also equal to 10 ,
and now we add other 3 + 3 = 6 which is not equal to 10,
So to make 10 we add 3,7 or 7,3 first and then add 3
So the sum is 3 + 7 + 3 = 13.

Question 6.
Big Ideas Math Answer Key Grade 1 Chapter 4 Add Numbers within 20 71
Answer:

Add 8,2 first to make 10 then add 6, sum = 8 + 6 + 2 = 16

Explanation:
To make 10 first we see adding 8 + 6 = 14,
14 is not equal to 10, now we add 6 + 2 = 8  which is not equal to 10 ,
and now we add 8 + 2 = 10 which is equal to 10,
So to make 10 we add 8,2 first and then add 6
So the sum is 8 + 6 + 2 = 16.

Question 7.
Big Ideas Math Answer Key Grade 1 Chapter 4 Add Numbers within 20 72
Answer:

Add 6,4 first to make 10 then add 7, sum = 6 + 7 + 4 = 17

Explanation:
To make 10 first we see adding 6 + 7 = 13,
It is not equal to 10, now we add 7 + 4 = 11  which is not equal to 10 ,
and now we add 6 + 4 = 10 which is equal to 10,
So to make 10 we add 6,4 first then add 7
So the sum is 6 + 7 + 4 = 17.

Question 8.
Big Ideas Math Answer Key Grade 1 Chapter 4 Add Numbers within 20 73
Answer:

Add 5,5 first to make 10 then add 8, sum = 8 + 5 + 5 = 18

Explanation:
To make 10 first we see adding 8 + 5 = 13,
It is not equal to 10, now we add 5 + 5 = 10  which equal to 10 ,
and now we add 8 + 5 = 13 which is not equal to 10,
So to make 10 we add 5,5 first then add 8
So the sum is 8 + 5 + 5 = 18.

Question 9.
4 + 6 + 2 = ________
Answer:

Add 4,6 first to make 10 then add 2, sum = 4 + 6 + 2 = 12

Explanation:
To make 10 first we see adding 4 + 6 = 10,
which is equal to 10, now we add 6 + 2 = 8  which is not equal to 10 ,
and now we add 4 + 2 = 6 which is not equal to 10,
So to make 10 we add 4,6 first then add 2
So the sum is 4 + 6 + 2 = 12.

Question 10.
5 + 9 + 1 = ________
Answer:

Add 9,1 first to make 10 then add 5, sum = 5 + 9 + 1 = 15

Explanation:
To make 10 first we see adding 5 + 9 = 14,
which is not equal to 10, now we add 9 + 1 = 10  which is equal to 10 ,
and now we add 5 + 1 = 6 which is not equal to 10,
So to make 10 we add 9,1 first then add 5,
So the sum is 5 + 9 + 1 = 15.

Question 11.
DIG DEEPER!
What do you know about the missing addends and the sum?
Big Ideas Math Answer Key Grade 1 Chapter 4 Add Numbers within 20 74
Answer:
We know the sum of the missing addends is 10 and the total sum is 10 + 3 = 13.

Explanation:
From the picture it is showing that if we add the missing
addends we get sum as 10 and the total sum is 10 + 3 = 13.
So whenever we are adding two missing addends we should always check,
their sum should be always 10, Example if one number is 1
other numbers will be 10-1 = 9. So the missing numbers are (1,9).
It can be even (2,8), (3,7),(4,6),(5,5),(6,4),(7,3),(8,2) or (9,1).

Think and Grow: Modeling Real Life

You need to knock down 20 pins to win. Do you win?
Big Ideas Math Answer Key Grade 1 Chapter 4 Add Numbers within 20 75
Equation:
Big Ideas Math Answer Key Grade 1 Chapter 4 Add Numbers within 20 76
Show how you know:
Answer:

6 + 3 + 7 = 16, No we will not win.

Explanation:
Given pins as 6,3,7 if we add all we get 6 + 3 + 7 = 16,
we need 20 pins to knock down to win and we got 16 which are less than 20,
So no we will not win.

Show and Grow

Question 12.
Your hockey team needs 12 goals to break a record. Does your team break the record?
Big Ideas Math Answer Key Grade 1 Chapter 4 Add Numbers within 20 77
Big Ideas Math Answer Key Grade 1 Chapter 4 Add Numbers within 20 78
Equation:
Big Ideas Math Answer Key Grade 1 Chapter 4 Add Numbers within 20 79
Show how you know:
Answer:

5 + 6 + 4 = 15, Yes my hockey team will break the record.

Explanation:
Given the goals made are 5,6,4 so the total number of goals
made are 5 + 6 + 4 = 15, to break hockey record
we need 12 goals as my team made 15 goals which is more than 12,
Yes my hockey team will break the record.

Add Three Numbers by Making a 10 Practice 4.5

Make a 10 to add.

Question 1.
Big Ideas Math Answer Key Grade 1 Chapter 4 Add Numbers within 20 80
Answer:

Add 6,4 first to make 10 then add 8, sum = 6 + 8 + 4 = 18

Explanation:
To make 10 first we see adding 6 + 8 = 14,
which is not equal to 10, now we add 8 + 4 = 12
which is not equal to 10 ,
and now we add 6 + 4 = 10 which is equal to 10,
So to make 10 we add 6,4 first then add 8,
So the sum is 6 + 8 + 4 = 18.

Question 2.
Big Ideas Math Answer Key Grade 1 Chapter 4 Add Numbers within 20 81
Answer:

Add 3,7 first to make 10 then add 4, sum = 3 + 7 + 4 = 14

Explanation:
To make 10 first we see adding 3 + 7 = 10,
which is equal to 10, now we add 7 + 4 = 11
which is not equal to 10 ,
and now we add 3 + 4 = 7 which is not equal to 10,
So to make 10 we add 3,7 first then add 4,
So the sum is 3 + 7 + 4 = 14.

Question 3.
5 + 8 + 2 = _______
Answer:

Add 8,2 first to make 10 then add 5, sum = 5 + 8 + 2 = 15

Explanation:
To make 10 first we see adding 5 + 8 = 13,
which is not equal to 10, now we add 8 + 2 = 10
which is equal to 10 ,
and now we add 5 + 2 = 7 which is not equal to 10,
So to make 10 we add 8,2 first then add 5,
So the sum is 5 + 8 + 2 = 15.

Question 4.
1 + 9 + 7 = _________
Answer:

Add 1,9 first to make 10 then add 7, sum = 1 + 9 + 7 = 17

Explanation:
To make 10 first we see adding 1 + 9 = 10,
which is equal to 10, now we add 9 + 7 = 16
which is not equal to 10 ,
and now we add 1 + 7 = 8 which is not equal to 10,
So to make 10 we add 1,9 first then add 7,
So the sum is 1 + 9 + 7 = 17.

Question 5.
DIG DEEPER!
What do you know about the missing addends and the sum?
Big Ideas Math Answer Key Grade 1 Chapter 4 Add Numbers within 20 82
Answer:
we know the sum of the missing addends is 10 and the total sum is 10 + 2 = 12.

Explanation:
From the picture it is showing that if we add the missing
addends we get sum as 10 and the total sum is 10 + 2 = 12.
So whenever we are adding two missing addends we should always check,
their sum should be always 10, Example if one number is 2
other number will be 10-2 = 8. So the missing addends here are (2,8).
It can be (1,9),(3,7),(4,6),(5,5),(6,4),(7,3),(8,2) or (9,1).

Question 6.
Modeling Real Life
Your baseball team needs 15 runs to break a record. Does your team break the record?
Big Ideas Math Answer Key Grade 1 Chapter 4 Add Numbers within 20 83
Big Ideas Math Answer Key Grade 1 Chapter 4 Add Numbers within 20 84
Equation:
Big Ideas Math Answer Key Grade 1 Chapter 4 Add Numbers within 20 85
Show how you know:
Answer:

6 + 4 + 6 = 16, Yes my team will break the record.

Explanation:
Given my team makes 6,4,6 runs so total runs made are
6 + 4 + 6 = 16, to break the record we require 15 runs
but the team made 16 as 16 is more than 15,
Yes, my team will break the record.

Review & Refresh

Circle the heavier object.

Question 7.
Big Ideas Math Answer Key Grade 1 Chapter 4 Add Numbers within 20 86
Answer:

Explanation:
By looking at the pictures book is heavier than a pencil.
So the book is circled.

Question 8.
Big Ideas Math Answer Key Grade 1 Chapter 4 Add Numbers within 20 87
Answer:

Explanation:
By looking at the pictures apple is heavier than the feather,
So apple is circled.

Lesson 4.6 Add 9

Explore and Grow

Use counters and the ten frames to find the sum. Show how you can make a lo to solve.

Big Ideas Math Answers 1st Grade 1 Chapter 4 Add Numbers within 20 88
Answer:
9 + 1 + 4 = 14

Explanation:
First we make sum 10 , For 9 to make 10 we need 10-9=1,
as we have add 1 to 9 now we will minus 1 from 5 we get 5 – 1 = 4,
Now we write the total sum as 9 + 1 + 4 = 14 .

Show and Grow

Make a 10 to add.

Question 1.
Big Ideas Math Answers 1st Grade 1 Chapter 4 Add Numbers within 20 89
Answer:

Sum is 9 + 1 + 2 = 12  or  9 + 3 = 12.

Explanation:
First we make sum 10 , For 9 to make 10 we need 10-9=1,
as we have to  add 1 to 9 now  we will minus 1 from 3 we get 3 – 1 = 2,
Now we write the total sum as 9 + 1 + 2 = 12  or 9 + 3 = 12.

Question 2.
Big Ideas Math Answers 1st Grade 1 Chapter 4 Add Numbers within 20 90
Answer:

Sum is 9 + 1 + 5 = 15 or 9 + 6 = 15.
Explanation:
First we make sum 10 , For 9 to make 10 we need 10-9=1,
as we have to  add 1 to 9 now  we will minus 1 from 6 we get 6 – 1 = 5,
Now we write the total sum as 9 + 1 + 5 = 15 or 9 + 6 = 15.

Apply and Grow: Practice

Make a 10 to add.

Question 3.
Big Ideas Math Answers 1st Grade 1 Chapter 4 Add Numbers within 20 91
Answer:

Sum is 9 + 1 + 1 = 11  or 9 + 2 = 11.

Explanation:
First we make sum 10 , For 9 to make 10 we need 10-9=1,
as we have to  add 1 to 9 now  we will minus 1 from 2 we get 2 – 1 = 1,
Now we write the total sum as 9 + 1 + 1 = 11  or 9 + 2 = 11.

Question 4.
Big Ideas Math Answers 1st Grade 1 Chapter 4 Add Numbers within 20 92
Answer:

Sum is 9 + 1 + 8 = 18  or 9 + 9 = 18.

Explanation:
First we make sum 10 , For 9 to make 10 we need 10-9=1,
as we have to add 1 to 9 now  we will minus 1 from 9 we get 9 – 1 = 8,
Now we write the total sum as 9 + 1 + 8 = 18  or 9 + 9 = 18.

Question 5.
Big Ideas Math Answers 1st Grade 1 Chapter 4 Add Numbers within 20 93
Answer:

Sum is 9 + 1 + 4 = 14 or 9 + 5 = 14.

Explanation:
First we make sum 10 , For 9 to make 10 we need 10-9=1,
as we have to add 1 to 9 now  we will minus 1 from 5 we get 5 – 1 = 4,
Now we write the total sum as 9 + 1 + 4 = 14 or 9 + 5 = 14.

Question 6.
DIG DEEPER!
Use the ten frame to complete the equations.
9 + ? = 17
Big Ideas Math Answers 1st Grade 1 Chapter 4 Add Numbers within 20 94
10 + _______ = 17
So, 9 + ________ = 17
Answer:
10 + 7 = 17,
So  9 + 1 + 7 = 17

Explanation:
Given 10 + ____ = 17, to find the missing number we take it as X,
10 + X = 17 , X = 17 – 10 =7. So the equation is 10 + 7 = 17.
Now we write 10 as 9 + 1 by using ten frame and
add 7 so the total sum is 9 + 1 + 7 = 17 or 10 + 7 = 17.

Think and Grow: Modeling Real Life

You have 9 stickers and earn 5 more. Your friend has 6 stickers and earns 10 more. Do you and your friend have the same number of stickers?
Big Ideas Math Answers 1st Grade 1 Chapter 4 Add Numbers within 20 95
Addition equations:
you                Friend
9 + 5 = 14          6 + 10 = 16

Yes       No

Show how you know:
Answer:
My equation is 9 + 5 = 14 and friend has 6 + 10 =16
as 14 is not equal to 16, so I and my friend don’t
have the same number of stickers.

Explanation:
Given I have 9 stickers and earn 5 more means in total I have 9 + 5 = 14 stickers and friend has 6 stickers and earns 10 more means in total friend have 6 + 10 = 16 stickers,
Now comparing between me and my friend as 14 and 16 are not same, So No, I and my friend do not have the same number of stickers.

Show and Grow

Question 7.
Your friend has 9 magnets and finds 6 more. You have 5 magnets and find 10 more. Do you and your friend have the same number of magnets?
Big Ideas Math Answers 1st Grade 1 Chapter 4 Add Numbers within 20 96
Addition equations:
 you             Friend
5 + 10 =15      9 + 6 = 15

Yes       No

Show how you know:

Answer:
My equation is 5 + 10 = 15 and friend equation is 9 + 6 = 15,
as both are equal, So Yes, I and my friend have the same number of magnets.

Explanation:
Given I have 5 magnets and find 10 more means in total, I have 5 + 10 = 15 magnets and friend has 9 magnets and finds 6 more means in total friend have 9 + 6 = 15 magnets, Now comparing between me and my friend as 15 and 15 are same, So Yes, I and my friend have the same number of magnets.

Add 9 Practice 4.6

Make a 10 to add

Question 1.
Big Ideas Math Answers 1st Grade 1 Chapter 4 Add Numbers within 20 97
Answer:

Sum is 9 + 1 + 2 = 12 or 9 + 3 = 12.

Explanation:
First we make sum 10, For 9 to make 10 we need 10-9=1,
as we have to add 1 to 9 now we will minus 1 from 3 we get 3 – 1 = 2,
Now we write the total sum as 9 + 1 + 2 = 12 or 9 + 3 = 12.

Question 2.
DIG DEEPER!
Use the ten frame to complete the equations.
16 = ? + ?
Big Ideas Math Answers 1st Grade 1 Chapter 4 Add Numbers within 20 98
10 + ___6_____ = ___16____
So, 16 = __9 +__1____ + __6___.
Answer:
We write the equation as 16 = 10 + 6
So 16 =  9 + 1 + 6 or 16 = 10 + 6

Explanation:
Given 10 + ____ = 16, to find the missing number we take it as X,
10 + X = 16 , X = 16 – 10 =6. So the equation is 10 + 6 = 16.
Now we write 10 as 9 + 1 by using the ten frame
and add 6 so the total sum is 9 + 1 + 6 = 16 or 10 + 6 = 16.

Question 3.
Modeling Real Life
You have 9 rocks and collect 7 more. Your friend has 8 rocks and collects 8 more. Do you and your friend have the same number of rocks?
Big Ideas Math Answers 1st Grade 1 Chapter 4 Add Numbers within 20 99
Yes             No
Show how you know:
Answer:
I have 9 + 7 = 16 rocks,
Friend have 8 + 8 = 16 rocks,
Yes, I and my friend have the same number of rocks.

Explanation:
Given I have 9 rocks and collected 7 more,
So I have total 9 + 7 = 16 rocks and my friend have 8 rocks and collects 8 rocks total he have 16 rocks,
Now comparing both me and my friend both have the equal number of rocks, Yes, I and my friend have the same number of rocks.

Review & Refresh

Circle the longer object.

Question 4.
Big Ideas Math Answers 1st Grade 1 Chapter 4 Add Numbers within 20 100
Answer:

Explanation:
By seeing the picture we can identify yellow spectacles are longer than the red spectacles, So circled the yellow spectacles.

Question 5.
Big Ideas Math Answers 1st Grade 1 Chapter 4 Add Numbers within 20 101
Answer:

Explanation:
By seeing the picture we can identify tooth paste is longer than the toothbrush, So circled the toothpaste.

Lesson 4.7 Make a 10 to Add

Explore and Grow

Use counters and the ten frame to find the sum.
Show how you can make a lo to solve.

Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 102
Answer:

Explanation:
First we make sum 10, For 8 to make 10 we need 10-8=2,
as we have to add 2 to 8 now we will minus 2 from 6 we get 6 – 2 = 4,
Now we write the total sum as 8 + 2 + 4 = 14.

Show and Grow

Make a 10 to add.

Question 1.
Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 103
Answer:

Sum as 8 + 2 + 3 = 13 or 8 + 5 = 13.

Explanation:
First we make sum 10, For 8 to make 10 we need 10-8=2,
as we have to add 2 to 8 now we will minus 2 from 5 we get 5 – 2 = 3,
Now we write the total sum as 8 + 2 + 3 = 13 or 8 + 5 = 13.

Question 2.
Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 104
Answer:

Sum as 7 + 3 + 4  = 14 or 7 + 7 = 14.

Explanation:
First, we make sum 10, For 7 to make 10 we need 10-7=3,
as we have to add 3 to 7 now we will minus 3 from 7 we get 7 – 3 = 4,
Now we write the total sum as 7 + 3 + 4  = 14 or 7 + 7 = 14.

Apply and Grow: Practice

Make a 10 to add.

Question 3.
Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 105
Answer:

Sum as 7 + 3 + 3  = 13 or 7 + 6 = 13.

Explanation:
First we make sum 10, For 7 to make 10 we need 10-7=3,
as we have to add 3 to 7 now we will minus 3 from 6 we get 6 – 3 = 3,
Now we write the total sum as 7 + 3 + 3  = 13 or 7 + 6 = 13.

Question 4.
Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 106
Answer:

Sum as 7 + 3 + 1  = 11 or 7 + 4 = 11.

Explanation:
First we make sum 10, For 7 to make 10 we need 10-7=3,
as we have to add 3 to 7 now we will minus 3 from 4 we get 4 – 3 = 1,
Now we write the total sum as 7 + 3 + 1  = 11 or 7 + 4 = 11.

Question 5.
Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 107
Answer:

Sum as 8 + 2 + 5 = 15 or 8 + 7 = 15.

Explanation:
First we make sum 10, For 8 to make 10 we need 10-8=2,
as we have to add 2 to 8 now we will minus 2 from 7 we get 7 – 2 = 5,
Now we write the total sum as 8 + 2 + 5 = 15 or 8 + 7 = 15.

Question 6.
Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 108
Answer:

Sum as 6 + 4 + 2 = 12 or 6 + 6 = 12.
Explanation:
First we make sum 10, For 8 to make 10 we need 10-6=4,
as we have to add 4 to 6 now we will minus 4 from 6 we get 6 – 4 = 2,
Now we write the total sum as 6 + 4 + 2 = 12 or 6 + 6 = 12.

Question 7.
MP Number Sense
Use 4, 7, and 10 to complete the sentence.
7 + ___7_____ has the same sum as ___4___ + __10_____.
Answer:
7 + 7  has the same sum as  4 + 10

Explanation:
Given to use numbers 4,7,10 to make the sentence
7 +  _____ = ________ + _______ both sides sum is same
As left side 7 is given we substitute and check first we use 4, then 7 and 10
7 +  4 = 11 now we have numbers 7 and 10, sum is 7+10 = 17
So (7 + 4)= 11  ≠ (7 + 10) =17, next we take 7 and check 7 + 7 = 14
now we have numbers 4 , 10 and sum is 4 + 10 = 14
So (7 + 7) = 14 = (4 + 10) = 14 here if we take for 7 left side and 4,10 right
we get same sum both side, now we  even check taking 10 first,
7 + 10 = 17 in left side and we have numbers 7 and 4 for right as 7 + 4 =11,
So (7 + 10)= 17 ≠ (7 + 4) =11, So to make both the same sum
we take left side number as 7 and right side numbers 4, 10
making the sentence as (7 + 7) = (4 + 10) = 14.

Think and Grow: Modeling Real Life

There are 8 crabs. 7 more join them. There are 10 turtles. 5 more join them.
Is the number of crabs the same as the number of turtles?
Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 109
Addition equations:
Yes              No
Show how you know:
Answer:
Addition equations:
Crabs                 Turtles
8 + 7 = 15          10 + 5 =15
Yes, the number of crabs are the same as the number of turtles.

Explanation:
Given 8 crabs and 7 more joined them so the number of crabs are 8 + 7 = 15 crabs and there are 10 turtles and 5 more joined them
So the total number of turtles are 10 + 5 = 15 turtles, as both are 15,
Yes, the number of crabs are the same as the number of turtles.

Show and Grow

Question 8.
There are 6 squirrels. 6 more join them. There are 10 birds. 2 more join them. Is the number of squirrels the same as the number of birds?
Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 110
Addition equations:
Yes               No
Show how you know:
Answer:
Squirrels          Birds
6 + 6 = 12    10 + 2 = 12
Yes, the number of squirrels are the same as the number of birds.

Explanation:
Given 6 squirrels and 6 more joined them so total squirrels are 6 + 6 = 12 and 10 birds, 2 more birds joined making total as 10 + 2 = 12 birds as both are 12,
Yes, the number of squirrels are the same as the number of birds.

Make a 10 to Add Practice 4.7

Make a 10 to add.

Question 1.
Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 111
Answer:

Sum as 6 + 4 + 1  = 11 or 6 + 5 = 11.

Explanation:
First we make sum 10, For 6 to make 10 we need 10-6=4,
as we have to add 4 to 6 now we will minus 4 from 5 we get 5 – 4 = 1,
Now we write the total sum as 6 + 4 + 1  = 11 or 6 + 5 = 11.

Question 2.
MP Number Sense
Use 2, 6, and 10 to complete the sentence.
________ + ________ has the same sum as ______ + 6.
Answer:

2 + 10   has the same sum as  6 + 6

Explanation:
Given to use numbers 2,6,10 to make the sentence
_______ +  _____ = ________ +  6 both sides sum is same
As right side 6 is given we substitute and check first we use 2, then 6 and 10,
2 +  6 = 8 now we have numbers 6 and 10, sum is 6 + 10 = 16
So (6 + 10) = 16  ≠ (2 + 6) = 8, next we take 6 and check 6 + 6 = 12
now we have numbers 2 , 10 and sum is 2 + 10 = 12
So (2 + 10) = 12 = ( 6 + 6) = 12 here if we take for 6 right side and 2,10 left
we get same sum both side, now we  even check taking 10 first,
10 + 6 = 16 in right side and we have numbers 2 and 6 for left as 2 + 6 =8,
So (2 + 6)= 8 ≠ (10 + 6) =16, So to make both the same sum
we take left side numbers as 2,10 and right side number 6
making the sentence as (2 + 10) = (6 + 6) = 12.

Question 3.
Modeling Real Life
There are 7 monkeys. 4 more join them. There are 6 birds. 5 more join them. Is the number of monkeys the same as the number of birds?
Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 112
Yes            No
Show how you know:
Answer:
Equations are:   Monkeys       Birds
7 + 4 = 11     6 + 5 = 11
Yes, the number of monkeys are the same as the number of birds.

Explanation:
Given 7 monkeys and 4 more joined them so total
monkeys are 7 + 4 = 11 and 6 birds, 5 more birds joined
making total as 6 + 5 = 11 birds as both are 11,
Yes, the number of monkeys are the same as the number of birds.

Review & Refresh

Question 4.
There are 3 Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 113
2 more Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 113 join them.
How many Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 113 are there now?
____3____ + ___2____ = ____5____ Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 113
Answer:
3 + 2 = 5, there are 5 aero planes now.

Explanation:
Given there are 3 aero planes and 2 more join them making total
as 3 + 2 = 5 , therefore there are 5 aero planes now.

Question 5.
There are 4 Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 114
6 more Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 114 join them.
How many Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 114 are there now?
___4_____ + ___6_____ = ___10_____ Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 114
Answer:
4 + 6 = 10, there are 10 airplanes now.

Explanation:
Given there are 4 aero planes and 6 more join them making total as  4 + 6 = 10, therefore there are 10 aero planes now.

Lesson 4.8 Problem Solving: Addition within 20

Explore and Grow

Newton has 9 red crayons, 3 blue crayons, and 1 yellow crayon. How many crayons does he have in all?

________ crayons
Answer:
Newton has 9 + 3 + 1 = 13 crayons in all.

Explanation:
Given Newton has 9 red crayons, 3 blue crayons and 1 yellow crayon,
So the total number of crayons Newton has are 9 + 3 + 1 = 13 crayons,
therefore Newton has 13 crayons in all.

Show and Grow

Question 1.
You have 6 notebooks. You buy 5 more. How many notebooks do you have now?
Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 115
Circle what you know.
Underline what you need to find.
Solve:

_______ ○ _______ = _______                           _______ notebooks
Answer:
notebooks
I have 11 notebooks in all.

Explanation:
Given I have 6 note books and buy 5 more , So I know
I have 6 & 5 books, So I circled what I know, I need to find
how many total notebooks I do have now, so I underlined it,
making 6 + 5 = 11 notebooks. So I have  total 11 notebooks now.

Apply and Grow: Practice

Question 2.
There are 6 soccer balls, 10 basketballs, and 4 volleyballs. How many balls are there in all?
Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 116
Circle what you know.
Underline what you need to find.
Solve:
Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 117
Answer:
balls
I have 20 balls in all.

Explanation:
Given there are  6 soccer balls, 10 basketballs, and 4 volleyballs, So I know 6 , 10 & 4 balls, So I circled what I know, I need to find
how many total balls there in all so I underlined it, making 6 + 10 + 4 = 20 balls, therefore there are total 20 balls in all.

Question 3.
You do 8 push-ups. Your friend does I fewer than you. How many push-ups do you and your friend do in all?

____15_____ push-ups
Answer:
In total, I and my friend does 15 push-ups in all.

Explanation:
Given I do 8 push-ups and friend does 1 fewer than me
means 8-1 = 7 push-ups now total number of push-ups are
8 + 7 = 15, So in total I and my friend does 15 push-ups in all.

Question 4.
YOU BE THE TEACHER
Newton has 9 magnets. Descartes has 8 more than Newton. Your friend uses a bar model to show how many magnets Descartes has. Is your friend correct? Show how you know.
Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 118Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 118
Answer:
No, your friend is not correct as in bar model it is showing Descartes has 8 magnets.

Descartes has 9 + 8 =  17 magnets.
Explanation :
Given Newton has 9 magnets and Descartes has 8 more than Newton means Descartes has 9 + 8 = 17 magnets. But the bar model is showing Descartes has only 8 magnets, I have shown in the figure Descartes have 9 + 8 = 17 Magnets.

Think and Grow: Modeling Real Life

You have 5 bracelets. You have 7 fewer than your friend. How many bracelets does your friend have?
Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 119
Circle what you know.
Underline what you need to find.
Solve:
Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 120

________ + ________ = _________                                       ____12____ bracelets
Answer:

Friend have 12 bracelets.

Explanation:
Given I have 5 bracelets, I have 7 fewer than my friend means
friend have in total as 5 + 7 = 12 bracelets, So circled what I know
5, 7 and underlined the number how many bracelets does friend have,
So friend have 12 bracelets.

Show and Grow

Question 5.
Your friend finds 9 seashells. You find 6 more than your friend. How many seashells do you find?
Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 121
Circle what you know.
Underline what you need to find.
Solve:

____6_____ + ___9____ = __15_____                                          ___15_____ seashells
Answer:

I found 15 seashells.

Explanation:
Given friend finds 9 seashells and I found 6 more than my friend,
So the total number of shells I found are 9 + 6 = 15 seashells.
I Know friend has 9 and I have more 6 seashells,
So I circled it and underlined to find how many seashells I found.

Problem Solving: Addition within 20 Practice 4.8

Question 1.
You have 5 robots. Your friend gives you more. Now you have 14. How many robots did your friend give you?
Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 122
Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 123
Answer:

Friend has 9 robots.

Explanation :
Given I have 5 robots and now I have 14 robots so I circled it,
I need to find how many friend gives I have underlined it.
Let us take friend gives X robots and we write equation as
5 + X = 14, therefore X = 14 – 5 = 9 means friend gave 9 robots.

Question 2.
There are some plates on a table. You add 8 more. Now there are 12. How many plates were on the table to start?

______4____ plates
Answer:
There were 4 plates on the table to start.

Explanation:
Let us take X as some plates on the table, I added 8 more and
there are 12 now means X + 8 = 12, Therefore X = 12 – 8 = 4,
So 4 plates are there on the table to start.

Question 3.
YOU BE THE TEACHER
Newton has 8 tickets. Descartes has 5 more than Newton. Your friend uses a number line to show how many tickets Descartes has. Is your friend correct? Show how you know.
Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 124
Answer:
Descartes has 13 Tickets,
Yes, Friend is correct.

Explanation:
Given Newton has 8 tickets and Descartes has 5 more than Newton means 8 + 5 = 13 tickets, Descartes has 13  tickets, Now my friend is showing this on number line he is starting from 8 because Newton has 8 and moving 5 units to right as Descartes has 5 more tickets till 13
which is the same as we did sum 8 +5 = 13, So yes, friend is correct.

Question 4.
Modeling Real Life You have 9 medals. You have 9 fewer than your friend. How many medals does your friend have?
Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 125
Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 126

____18_______ medals
Answer:

Friend have 18 medals.

Explanation:
Given I have 9 medals and 9 fewer than friend, So friend have
9 + 9 what I have, So friend have 18 medals.

Review & Refresh

Question 5.
4 – 3 = ________
Answer:
4- 3 = 1

Explanation:
Given the equation as 4 – 3 = _____ means subtract 3 from 4 we get 1
making equation as 4- 3 = 1.

Question 6.
5 – 2 = ________
Answer:
5 – 2 = 3

Explanation:
Given the equation as 5 – 2 = _____ means subtract 2 from 5 we get 3
making equation as 5 – 2 = 3.

Question 7.
9 – 9 = _________
Answer:
9 – 9 = 0

Explanation:
Given the equation as 9 – 9 = _____ means subtract 9 from 9 we get 0
making equation as 9 – 9 = 0.

Question 8.
7 – 0 = _________
Answer:
7 – 0 = 7

Explanation:
Given the equation as 7 – 0 = _____ means subtract 0 from 7 we get 7
making equation as 7- 0 = 7.

Add Numbers within 20 Performance Task

Question 1.
You track the weather for a few weeks. Each week has 7 days.
Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 127

a. You track the weather every day for the first week. But you miss 1 day in the second week. How many days do you track the weather?

___13_______ days
Answer:
I have track of 13 days weather.

Explanation:
Given each week has 7 days, I have the track for every day for the first week,
Means the first week = 7 days, Now in the second week I missed 1 day means
7 – 1 = 6 days track in the second week, So the total number of days are
first week track + second week track = 7 + 6 = 13 , total I have track of 13 days weather.

b. You track the weather for 1 more week. How many days in all do you track the weather?

____20______ days
Answer:
I have track of 20 days weather in all.

Explanation :
Given that I have tracked for 1 more week, We got 13 days before
and now we have 1 more week means 7 days to 13 making total as
13 +  7 = 20 days, So now I have track of 20 days weather in all.

Question 2.
Your friend also tracks the weather. She records 9 sunny days and 5 cloudy days.
How many days does your friend track the weather?

____14______ days
Answer:
My friend has 14 days track record of the weather.

Explanation:
Given my friend records 9 sunny days and 5 cloudy days
track of the weather so in all my friend has track of
9 + 5 = 14 days. Therefore my friend has 14 days track record of the weather.

Question 3.
You record 10 rainy days in the first three weeks. Is the number of rainy days the same as the number of sunny days?
Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20 128
Yes                                No
Show how you know:
Answer:
No, Number of rainy days are not the same as the number of sunny days.

Explanation:
Given I record 10 rainy days in first three weeks and number of
sunny days for three weeks are 4 + 3 + 4 = 11 days , now comparing
with rainy days as rainy days are less to number of sunny days we say
No, number of rainy days are not the same as the number of sunny days.

Add Numbers within 20 Chapter Practice

Add Doubles from 6 to 10 Homework & Practice 4.1

Question 1.
Big Ideas Math Solutions Grade 1 Chapter 4 Add Numbers within 20 129
____10______ + ____10____ = ___20______
Answer:
10 + 10 = 20 dots

Explanation:
Given in block 1 we have 10 dots and in block 2 we have 10 dots,
So in total we have 10 + 10 = 20 dots.

Question 2.
Big Ideas Math Solutions Grade 1 Chapter 4 Add Numbers within 20 130
Answer:

9 + 9 = 18 dots

Explanation:
Given in block 1 we have 9 dots and in block 2 we have 9 dots,
So in total we have 9 + 9 = 18 dots.

Use Doubles within 20 Homework & Practice 4.2

Use the doubles 6 + 6 to find each sum.

Question 3.
6 + 7 = ___13_____
Big Ideas Math Solutions Grade 1 Chapter 4 Add Numbers within 20 131
Answer:
6 + 7  = 13

Explanation:
We Know the double  6 + 6 = 12 , The number 7 is one more than 6,
So we can write 7 as 6 + 1, The required sum is one more than double,
12 + 1 =13, Sum is 13, therefore 6 + 7 = 13, Double plus 1.

6 + 5 = _______
Big Ideas Math Solutions Grade 1 Chapter 4 Add Numbers within 20 132
Answer:
6 + 5 = 11

Explanation:
We Know the double of 6 + 6 = 12 , The number 5 is one less than 6,
So we can write 5 as 6 – 1, The required sum is one less than double,
12 – 1 = 11, Sum is 11, therefore 6 + 5 = 11, Double minus 1.

Find the sum. Write the double you used.

Question 4.
7 + 8 = _______

____7_____ ○ ____8____ = ___15____
Answer:

7 + 8 = 15, 7 + 7 double is used

Explanation:
As the first addend is 7 and we know the double of 7 + 7 = 14 ,
The number 8 is one more than 7, So we can write 8 as 7 + 1,
The required sum is one more than double,
14 + 1 =15, Sum is 15, therefore 7 + 8 = 15, we used the double 7 +7.

Question 5.
10 + 9 = ___19______

_________ ○ ________ = _______
Answer:
10 + 9 = 19, we use double 10 + 10,

Explanation:
As the first addend is 10 and we know the double is 10 + 10,
10 + 10 = 20 , The number 9 is one less than 10,
So we can write 9 as 10 – 1, The required sum is one less than double,
20 – 1 = 19, Sum is 19, therefore 10 + 9 = 17, Double minus 1.
We have use double 10 + 10 .

Count on to Add within 20 Homework & Practice 4.3

Question 6.
11 + 6 = ___17____
Big Ideas Math Solutions Grade 1 Chapter 4 Add Numbers within 20 133
Answer:

11 + 6 = 17

Explanation:
First we start from 11 and move 6 units we get 17 on the number line,
So sum is 11 +6 = 17 .

Add Three Numbers Homework & Practice 4.4

Question 7.
Big Ideas Math Solutions Grade 1 Chapter 4 Add Numbers within 20 134
Answer:

6 + 4 + 4 = 14

Explanation:
By using linking the cubes
a. First we add 4 + 4 = 8 and now we add 6 + 8 we get 14.
b. Here we add 6 + 4 = 10, now we add 10 + 4 =14,
c. Now here we add first 6 with 4 we get 10 to 10 we add 4
we will get 10 + 4 = 14, So we get sum as 14.

Question 8.
Big Ideas Math Solutions Grade 1 Chapter 4 Add Numbers within 20 135
Answer:

7 + 8 + 5 = 20

Explanation:
By using linking the cubes
a. First we add 7 + 8 = 15 and now we add 15+ 5 we get 20.
b. Here we add 8 + 5 = 13, now we add 13 + 7 =20,
c. Now here we add first 7 with 5 we get 12 to 12 we add 8
we will get 12 + 8 = 20, So we get sum as 20.

Question 9.
Big Ideas Math Solutions Grade 1 Chapter 4 Add Numbers within 20 136
Answer:

9 + 8 + 1 = 18

Explanation:
By using linking the cubes
a. First we add 9 + 8 = 17 and now we add 17 + 1 we get 18.
b. Here we add 8 + 1 =9, now we add 9 + 9 =18,
c. Now here we add first 9 with 1 we get 10 to 10 we add 8
we will get 10 + 8 = 18, So we get sum as 18.

Question 10.
Big Ideas Math Solutions Grade 1 Chapter 4 Add Numbers within 20 137
Answer:

2 + 3 + 5 = 10

Explanation:
By using linking the cubes
a. First we add 2 + 3 = 5 and now we add 5 + 5 we get 10.
b. Here we add 3 + 5 =8, now we add 8 + 2 =10,
c. Now here we add first 2 with 5 we get 7 to 7 we add 3
we will get 7 + 3 = 10, So we get sum as 10.

Add Three Numbers by Making a 10 Homework & Practice 4.5

Make a 10 to add.

Question 11.
Big Ideas Math Solutions Grade 1 Chapter 4 Add Numbers within 20 138
Answer:

Add 7,3 first to make 10 then add 9, sum = 7 + 9 + 3 = 19

Explanation:
To make 10 first we see adding 7 + 9 = 16 is not equal to 10,
next we add 9 + 3 = 12 not equal to 10, 7 + 3 = 10 which
is equal to 10 ,So to make 10 we add  7,3 first, then add 9,
So the sum is 7 + 9 + 3 = 19.

Question 12.
Big Ideas Math Solutions Grade 1 Chapter 4 Add Numbers within 20 139
Answer:

Add 9,1 first to make 10 then add 4, sum = 4 + 9 + 1 = 14

Explanation:
To make 10 first we see adding 4 + 9 = 13 is not equal to 10,
next we add 9 + 1 = 10 is equal to 10, 4 + 1 = 5 which
is not equal to 10 ,So to make 10 we add  9,1 first, then add 4,
So the sum is 4 + 9 + 1 = 14.

Question 13.
MP Number Sense
What do you know about the missing addends and the sum?
Big Ideas Math Solutions Grade 1 Chapter 4 Add Numbers within 20 140
Answer:
We know the sum of the missing addends is 10 and the total sum is 10 + 5 = 15.

Explanation:
From the picture it is showing that if we add the missing
addends we get sum as 10 and the total sum is 10 + 5 = 15.
So whenever we are adding two missing addends we should always check,
their sum should be always 10, Example if one number is 1
other number will be 10-1 = 9. So the missing numbers are (1,9).
It can be even (2,8), (3,7),(4,6),(5,5),(6,4),(7,3),(8,2) or (9,1).

Add 9 Homework & Practice 4.6

Question 14.
Make a 10 to add 9 + 4.
Big Ideas Math Solutions Grade 1 Chapter 4 Add Numbers within 20 141
Big Ideas Math Solutions Grade 1 Chapter 4 Add Numbers within 20 142
Answer:

Sum as 9 + 1 + 3 = 13  or 9 + 4 = 13.

Explanation:
First we make sum 10 , For 9 to make 10 we need 10-9=1,
as we have to  add 1 to 9 now  we will minus 1 from 4 we get 4 – 1 = 3,
Now we write the total sum as 9 + 1 + 3 = 13  or 9 + 4 = 13.

Make a 10 to Add Homework & Practice 4.7

Question 15.
Make a 10 to add 8 + 5.
Big Ideas Math Solutions Grade 1 Chapter 4 Add Numbers within 20 143
Answer:

Sum as 8 + 2 + 3 = 13  or 8 + 5 = 13.

Explanation:
First we make sum 10 , For 8 to make 10 we need 10-8=2,
as we have to  add 2 to 8 now  we will minus 2 from 5 we get 5 – 2 = 3,
Now we write the total sum as 8 + 2 + 3 = 13  or 8 + 5 = 13.

Question 16.
7 + 7 = ?
Big Ideas Math Solutions Grade 1 Chapter 4 Add Numbers within 20 144
Answer:

Sum as 7 + 3 + 4  = 14 or 7 + 7 = 14.

Explanation:
First we make sum 10, For 7 to make 10 we need 10-7=3,
as we have to add 3 to 7 now we will minus 3 from 7 we get 7 – 3 = 4,
Now we write the total sum as 7 + 3 + 4  = 14 or 7 + 7 = 14.

Question 17.
9 + 8 = ?
Big Ideas Math Solutions Grade 1 Chapter 4 Add Numbers within 20 145
Answer:

Sum as 9 + 1 + 7  = 17 or 9 + 8 = 17.

Explanation:
First we make sum 10, For 9 to make 10 we need 10-9=1,
as we have to add 1 to 9 now we will minus 1 from 8 we get 8 – 1 = 7,
Now we write the total sum as 9 + 1 + 7  = 17 or 9 + 8 = 17.

Problem Solving: Addition within 20 Homework & Practice 4.8

Question 18.
Modeling Real Life
Your friend finds 7 insects. You find 9 more than your friend. How many insects do you find?
Big Ideas Math Solutions Grade 1 Chapter 4 Add Numbers within 20 146

_____16____ insects
Answer:
I found 16 insects.

Explanation:
Given a friend finds 7 insects, I find 9 more than my friend means
I have in total as 7 + 9 = 16 insects,
So I found 16 insects.

Conclusion:

Learn the Big Ideas Math Book 1st Grade Answer Key Chapter 4 Add Numbers within 20 provided and improve your math as well as problem solving skills. You can achieve greater heights and fall in love with Math with our Big Ideas Math Answers Grade 1 Chapter 4 Add Numbers within 20. Stay with us to get the solutions of all Big Ideas Math Grade 1 Chapters from 1 to 14.

Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20

Big Ideas Math Answers Grade 1 Chapter 5

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Big Ideas Math Book 1st Grade Answer Key Chapter 5 Subtract Numbers within 20

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Vocabulary

Lesson: 1 Count Back to Subtract within 20

Lesson: 2 Use Addition to Subtract within 20

Lesson: 3 Subtract 9

Lesson: 4 Get to 10 to Subtract

Lesson: 5 More True or False Equations

Lesson: 6 Make True Equations

Lesson: 7 Problem Solving: Subtraction within 20

Chapter 5 Subtract Numbers within 20

Chapter 5 Subtract Numbers within 20 Vocabulary

Organize It

Review Words:
bar model
difference
minus
part-part-whole model
subtraction equation

Use the review words to complete the graphic organizer.
Big Ideas Math Answer Key Grade 1 Chapter 5 Subtract Numbers within 20 1
Answer:

Explanation:
Given 8 – 3 = 5, Here – represents minus sign minus
represents the arithmetic operation of subtraction between
two numbers, 5 represents difference is the result of subtracting
one number from another and 8 – 3 = 5 is called subtraction equation
 is a statement that shows a subtraction operation using the minuend,
subtrahend, difference.

Define It

Match the review word to its definition.

Big Ideas Math Answer Key Grade 1 Chapter 5 Subtract Numbers within 20 2
Answer:

Explanation:
In the above picture, the first one is a bar model  a pictorial representation
of a problem or concept where bars or boxes are used to represent the
known and unknown quantities, the second picture is the part-part-whole model
is a simple pictorial representation of a problem that helps us
see the relationships between numbers.

Lesson 5.1 Count Back to Subtract within 20

Explore and Grow

Model the story.

You are on floor number 12. You go down 4 floors. What floor are you on now?
Big Ideas Math Answer Key Grade 1 Chapter 5 Subtract Numbers within 20 3

Big Ideas Math Answer Key Grade 1 Chapter 5 Subtract Numbers within 20 4

floor number ____8______
Answer:

I am in 8 floors now.

Explanation:
Given I am on floor number 12 and go down 4 floors means
now I am on floor 12 – 4 = 8 . on the number line we start from 12
since subtracting makes numbers smaller, we move left to subtract.
So from 12 we move 4 parts back or left to 8. Therefore now I am at
8 floor now.

Show and Grow

Question 1.
15 – 8 = ___7____
Big Ideas Math Answer Key Grade 1 Chapter 5 Subtract Numbers within 20 5
Answer:

15 – 8 = 7

Explanation:
We subtract 15 – 8 = 7, on number line we start at 15 and
move 8 steps back or left we get 7.

Question 2.
12 – 3 = ___9______
Big Ideas Math Answer Key Grade 1 Chapter 5 Subtract Numbers within 20 6
Answer:

12- 3 = 9

Explanation:
We subtract 12 – 3 = 9, on number line we start at 12 and
move 3 steps back or left we get 9.

Question 3.
17 – 9 = ____8_____
Big Ideas Math Answer Key Grade 1 Chapter 5 Subtract Numbers within 20 6
Answer:

17 – 9 = 8

Explanation:
We subtract 17 – 9 = 8, on the number line we start at 17 and
move 9 steps back or left we get 8.

Apply and Grow: Practice

Question 4.
11 – 6 = ____5____
Big Ideas Math Answer Key Grade 1 Chapter 5 Subtract Numbers within 20 6
Answer:

11 – 6 = 5

Explanation:
We subtract 11 – 6 = 5, on the number line we start at 11 and
move 6 steps back or left we get 5.

Question 5.
13 – 7 = ___6_____
Big Ideas Math Answer Key Grade 1 Chapter 5 Subtract Numbers within 20 6
Answer:

13 – 7 = 6

Explanation:
We subtract 13 – 7 = 6, on the number line we start at 13 and
move 7 steps back or left we get 6.

Question 6.
18 – 9 = ___9_____
Big Ideas Math Answer Key Grade 1 Chapter 5 Subtract Numbers within 20 6
Answer:

18- 9 = 9

Explanation:
We subtract 18 – 9 = 9, on number line we start at 18 and
move 9 steps back or left we get 9.

Question 7.
20 – 10 = ____10_____
Big Ideas Math Answer Key Grade 1 Chapter 5 Subtract Numbers within 20 6
Answer:

20 – 10 = 10

Explanation:
We subtract 20 – 10 = 10, on the number line we start at 20 and
move 10 steps back or left we get 10.

Question 8.
17 – 8 = ____9_____
Big Ideas Math Answer Key Grade 1 Chapter 5 Subtract Numbers within 20 6
Answer:

17- 8 = 9

Explanation:
We subtract 17 – 8 = 9, on the number line we start at 17 and
move 8 steps back or left we get 9.

Question 9.
18 – 3 = ____15______
Big Ideas Math Answer Key Grade 1 Chapter 5 Subtract Numbers within 20 6
Answer:

18 – 3 = 15

Explanation:
We subtract 18 – 3 = 15, on the number line we start at 18 and
move 3 steps back or left we get 15.

Question 10.
___12____ = 16 – 4
Big Ideas Math Answer Key Grade 1 Chapter 5 Subtract Numbers within 20 6
Answer:

16- 4 = 12

Explanation:
We subtract 16 – 4 = 12, on the number line we start at 16 and
move 4 steps back or left we get 12.

Question 11.
___6_____ = 14 – 8
Big Ideas Math Answer Key Grade 1 Chapter 5 Subtract Numbers within 20 6
Answer:

14- 8 = 6

Explanation:
We subtract 14 – 8 = 6, on the number line we start at 14 and
move 8 steps back or left we get 6.

Question 12.
DIG DEEPER!
Write the equation shown by the number line.
Big Ideas Math Answer Key Grade 1 Chapter 5 Subtract Numbers within 20 7
___15_____ – __9______ = ___6_____
Answer:
The equation on the number line is 15 – 9 = 6.

Explanation:
As shown on the number line we start at 15 and move
9 steps or 9 units back or left we reached at 6 so the equation shown by the number line is 15 – 9 = 6.

Think and Grow: Modeling Real Life

You collect 6 gems. Your friend collects 14. How many fewer gems do you collect?
Big Ideas Math Answer Key Grade 1 Chapter 5 Subtract Numbers within 20 8
Model:

Big Ideas Math Answer Key Grade 1 Chapter 5 Subtract Numbers within 20 9
Subtraction equation:

____8____ fewer gems

Subtraction equation is 14 – 6 = 8,
I have 8 fewer gems.

Explanation:
Given I have collected 6 gems and friend collects 14 gems
now how many less or fewer gems I have than my friend are
14 – 6 = 8 gems, On number line we start at 14 and
move 6 steps or 6 units back or left we reach at 8.
So the Subtraction equation is 14 – 6 = 8,
I have 8 fewer gems.

Show and Grow

Question 13.
Your friend finds 16 gold bars. You find 9. How many fewer gold bars do you find?
Big Ideas Math Answer Key Grade 1 Chapter 5 Subtract Numbers within 20 10
Model:

Big Ideas Math Answer Key Grade 1 Chapter 5 Subtract Numbers within 20 11
Subtraction equation:

_____7_____ gold bars
Answer:

Subtraction equation is 16 – 9 = 7,
I have 7 fewer gold bars.

Explanation:
Given friend finds 16 gold bars and I found 9,
now how many less or fewer gold bars I have than my friend are
16 – 9 = 7 gold bars, On number line we start at 16 and
move 9 steps or 9 units back or left we reach at 7.
So the Subtraction equation is 16 – 9 = 7,
I have 7 fewer gold bars.

Count Back to Subtract within 20 Practice 5.1

Question 1.
12 – 5 = __7____
Big Ideas Math Answer Key Grade 1 Chapter 5 Subtract Numbers within 20 12
Answer:

12 – 5 = 7

Explanation:
We subtract 12 – 5 = 7, on number line we start at 12 and
move 5 steps back or left we get 7.

Question 2.
15 – 7 = ____8____
Big Ideas Math Answer Key Grade 1 Chapter 5 Subtract Numbers within 20 12
Answer:

15 – 7 = 8

Explanation:
We subtract 15 – 7 = 8, on number line we start at 15 and
move 7 steps back or left we get 8.

Question 3.
13 – 6 = ___7_____
Big Ideas Math Answer Key Grade 1 Chapter 5 Subtract Numbers within 20 12
Answer:

13 – 6 = 7

Explanation:
We subtract 13 – 6 = 7, on number line we start at 13 and
move 6 steps back or left we get 7.

Question 4.
14 – 5 = ___9_______
Big Ideas Math Answer Key Grade 1 Chapter 5 Subtract Numbers within 20 12
Answer:

14 – 5 = 9

Explanation:
We subtract 14 – 5 = 9, on number line we start at 14 and
move 5 steps back or left we get 9.

Question 5.
11 – 5 = ____6______
Big Ideas Math Answer Key Grade 1 Chapter 5 Subtract Numbers within 20 12
Answer:

11- 5 = 6

Explanation:
We subtract 11 – 5 = 6, on number line we start at 11 and
move 5 steps back or left we get 6.

Question 6.
15 – 4 = ___11_____
Big Ideas Math Answer Key Grade 1 Chapter 5 Subtract Numbers within 20 12
Answer:

15- 4 = 11

Explanation:
We subtract 15 – 4 = 11, on number line we start at 15 and
move 4 steps back or left we get 11.

Question 7.
__10____ = 18 – 8
Big Ideas Math Answer Key Grade 1 Chapter 5 Subtract Numbers within 20 12
Answer:

18 – 8 = 10

Explanation:
We subtract 18 – 8 = 10, on number line we start at 18 and
move 8 steps back or left we get 10.

Question 8.
____4_____ = 12 – 8
Big Ideas Math Answer Key Grade 1 Chapter 5 Subtract Numbers within 20 12
Answer:

12 – 8 = 4

Explanation:
We subtract 12 – 8 = 4, on number line we start at 12 and
move 8 steps back or left we get 4.

Question 9.
DIG DEEPER
Write the equation shown by the number line.
Big Ideas Math Answer Key Grade 1 Chapter 5 Subtract Numbers within 20 13
___14_____ – ___9_____ = ___5_____
Answer:

The equation on the number line is 14 – 9 = 5.

Explanation:
As shown on the number line we start at 14 and move
9 steps or 9 units back or left we reach at 5, so the
equation shown by the number line is 14 – 9 = 5.

Question 10.
Modeling Real Life
You play soccer. The visiting team scores 12 goals. Your team scores 4 fewer. How many goals does your team score?
Big Ideas Math Answer Key Grade 1 Chapter 5 Subtract Numbers within 20 14

Big Ideas Math Answer Key Grade 1 Chapter 5 Subtract Numbers within 20 15

_____8_____ goals
Answer:

Subtraction equation is 12 – 4 = 8,
My team scored 8 goals

Explanation:
Given visiting team scores 12 goals, My team scores 4 fewer goals
My team have scored 12 – 4 = 8 goals, On number line we start at
12 and move 4 steps or 4 units back or left we reach at 8.
So the Subtraction equation is 12 – 4 = 8,
therefore my team score 8 goals.

Review & Refresh

Question 11.
6 + 2 = ___8_____
Answer:
6 + 2 = 8

Explanation :
The addition equation is 6 + 2 = 8,
We add 2 to 6 we get 8.

Question 12.
4 + 3 = ___7_____
Answer:
4 + 3 = 7

Explanation :
The addition equation is 4 + 3 = 7,
We add 3 to 4 we get 7.

Question 13.
8 + 2 = ___10____
Answer:
8 + 2 = 10

Explanation :
The addition equation is 8 + 2 = 10,
We add 2 to 8 we get 10.

Question 14.
5 + 4 = ____9___
Answer:
5 + 4 = 9

Explanation :
The addition equation is 5 + 4 = 9,
We add 4 to 5 we get 9.

Lesson 5.2 Use Addition to Subtract within 20

Explore and Grow

Model the story.
Your class needs to make 15 scrapbook pages. 8 are already made. How many more pages does your class need to make?
Big Ideas Math Answers 1st Grade 1 Chapter 5 Subtract Numbers within 20 16

Big Ideas Math Answers 1st Grade 1 Chapter 5 Subtract Numbers within 20 17

____7_____ more pages

15 – 8 = 7,
We need 7 more pages to make.

Explanation:
Given my class needs to make 15 scrapbook pages,
8 are already made ,We need 15 – 8 = 7 more pages
are needed  to make scrapbook.

Show and Grow

Question 1.
11 – 7 = ?
Big Ideas Math Answers 1st Grade 1 Chapter 5 Subtract Numbers within 20 18
Think 7 + ___4____ = 11.
So, 11 – 7 = ___4____ .
Answer:

7 + 4 = 11
11 – 7 = 4

Explanation:
Explanation:
First we move from 7 on number line and start moving till
11 we count how many units right we have moved we get
4 units so 7 + 4 = 11 or we start at 11 on number line and
move 7 steps back or left we get 4, 11 – 7 = 4.

Question 2.
16 – 8 = ?
Big Ideas Math Answers 1st Grade 1 Chapter 5 Subtract Numbers within 20 19
Think 8 + ___8____ = 16.
So, 16 – 8 = ___8____ .
Answer:

8 + 8 = 16
16 – 8 = 8

Explanation:
First we move from 8 on number line and start moving till
16 we count how many units right we have moved we get
8 units so 8 + 8 = 16 or we start at 16 on number line and
move 8 steps back or left we get 8, 16 – 8 = 8.

Question 3.
13 – 10 = ?
Big Ideas Math Answers 1st Grade 1 Chapter 5 Subtract Numbers within 20 19
Think 10 + __3_____ = 13.
So, 13 – 10 = ___3____ .
Answer:

10 + 3 = 13
13 – 10 = 3

Explanation:
First we move from 10 on number line and start moving till
13 we count how many units right we have moved we get
3 units so 10 + 3 = 13 or we start at 13 on number line and
move 10 steps back or left we get 3, 13 – 10 = 3.

Apply and Grow: Practice

Question 4.
Big Ideas Math Answers 1st Grade 1 Chapter 5 Subtract Numbers within 20 19
12 – 7 = ?
Think 7 + ___5____ = 12.
So, 12 – 7 = ___5____ .
Answer:

7 + 5 = 12
12 – 7 = 5

Explanation:
First we move from 7 on number line and start moving till
12 we count how many units right we have moved we get
5 units so 7 + 5 = 12 or we start at 12 on number line and
move 7 steps back or left we get 5, 12 – 7 = 5.

Question 5.
Big Ideas Math Answers 1st Grade 1 Chapter 5 Subtract Numbers within 20 19
17 – 8 = ?
Think 8 + ___9____ = 17.
So, 17 – 8 = __9____ .
Answer:

8 + 9 = 17
17 – 8 = 9

Explanation:
First we move from 8 on number line and start moving till
17 we count how many units right we have moved we get
9 units so 8 + 9 = 17 or we start at 17 on number line and
move 8 steps back or left we get 9, 17 – 8 = 9.

Question 6.
Big Ideas Math Answers 1st Grade 1 Chapter 5 Subtract Numbers within 20 19
11 – 2 = ___9_____
Answer:

11 – 2 = 9

Explanation:
We subtract 11 – 2 = 9, on number line we start at 11 and
move 2 steps back or left we get 9.

Question 7.
Big Ideas Math Answers 1st Grade 1 Chapter 5 Subtract Numbers within 20 19
_____14____ = 19 – 5
Answer:

19 – 5 = 14 or 14 = 19 – 5

Explanation:
We subtract 19 – 5 = 14, on number line we start at 19 and
move 5 steps back or left we get 14.

Question 8.
DIG DEEPER!
Tell what subtraction problem Newton and Descartes solved. Think: What strategies did they use?
Big Ideas Math Answers 1st Grade 1 Chapter 5 Subtract Numbers within 20 20
Answer:
Newton and Descartes solved the subtraction equation as
First  by 5 + = 13
Second by 13 – 5 =_

Explanation :
Newton and Descartes solved the subtraction equation on the number line
first they solved equation 5 + _______ = 13 as started from 5 and moved
8 units right reached till 13 so the missing number is 8, 5 + 8 = 13.
In second method they solved by starting from 13 and moved 5 units
left till they reached 8 on the number line so they solved
the subtraction equation as 13 – 5 = 8.

Think and Grow: Modeling Real Life

There are 15 people in an elevator. Some of them exit. There are 7 left. How many people exit?
Big Ideas Math Answers 1st Grade 1 Chapter 5 Subtract Numbers within 20 21
Model:

Big Ideas Math Answers 1st Grade 1 Chapter 5 Subtract Numbers within 20 22
Subtraction equation:
15 – 7 = 8

____8_____ people
Answer:

15 – 7 = 8,
8 people exit.

Explanation:
Given 15 people are there in elevator, some of them exit,
There are 7 left, So people exit are 15 – 7 = 8,
So 8 people exit from elevator.

Show and Grow

Question 9.
There are 18 people in a subway car. Some of them exit. There are 9 left. How many people exit?
Big Ideas Math Answers 1st Grade 1 Chapter 5 Subtract Numbers within 20 23
Model:

Big Ideas Math Answers 1st Grade 1 Chapter 5 Subtract Numbers within 20 22
Subtraction equation:
18 -9 = 9

_____9_____ people
Answer:

18 – 9 = 9,
9 people exit.

Explanation:
Given there are 18 people in a subway car , some of them exit,
There are 9 left, So people exit are 18 – 9 = 9,
So 9 people exit from subway car.

Use Addition to Subtract within 20 Practice 5.2

Question 1.
Big Ideas Math Answers 1st Grade 1 Chapter 5 Subtract Numbers within 20 24
13 – 9 = ?
Think 9 + __4____ = 13
So, 13 – 9 = ___4____.
Answer:

9 + 4 = 13
13 – 9 = 4

Explanation:
First we move from 9 on number line and start moving till
13 we count how many units right we have moved we get
4 units so 9 + 4 = 13 or we start at 13 on number line and
move 9 steps back or left we get 4, 13 – 9 = 4.

Question 2.
Big Ideas Math Answers 1st Grade 1 Chapter 5 Subtract Numbers within 20 24
17 – 8 = ?
Think 8 + __9____ = 17
So, 17 – 8 = ___9____.
Answer:

8 + 9 = 17
17 – 8 = 9

Explanation:
First we move from 8 on number line and start moving till
17 we count how many units right we have moved we get
9 units so 8 + 9 = 17 or we start at 17 on number line and
move 8 steps back or left we get 9, 17 – 8 = 9.

Question 3.
Big Ideas Math Answers 1st Grade 1 Chapter 5 Subtract Numbers within 20 24
14 – 7 = ____7_____
Answer:

14 – 7 = 7

Explanation:
We subtract 14 – 7 = 7, on number line we start at 14 and
move 7 steps back or left we get 7.

Question 4.
Big Ideas Math Answers 1st Grade 1 Chapter 5 Subtract Numbers within 20 24
____13_____ = 17 – 4
Answer:

17 – 4 = 13

Explanation:
We subtract 17 – 4 = 13, on number line we start at 17 and
move 4 steps back or left we get 13.

Question 5.
DIG DEEPER!
Tell what subtraction problem Newton and Descartes solved. Think: What strategies did they use?
Big Ideas Math Answers 1st Grade 1 Chapter 5 Subtract Numbers within 20 25
Answer:
Newton and Descartes solved the subtraction equation as
First by 11 – 7 =_4 
Second  by 7 +  4  =  11

Explanation :
Newton and Descartes solved the subtraction equation on the number line
first method they solved by starting from 11 and moved 7 units
left till they reached 4 on the number line so they solved
the subtraction equation as 11 – 7 = 4.
Second they solved equation 7 + ______ = 11 as started from 7 and moved
4 units right reached till 11 so the missing number is 4, 7 + 4 = 11.

Question 6.
Modeling Real Life
There are 13 people on a train. Some of them exit. There are 6 left. How many people exit?
Big Ideas Math Answers 1st Grade 1 Chapter 5 Subtract Numbers within 20 26

Big Ideas Math Answers 1st Grade 1 Chapter 5 Subtract Numbers within 20 27

_____7_______ people
Answer:
Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20-1

13 – 6 = 7,
So, 7 people exit.

Explanation:
Given there are 13 people on a train , some of them exit,
There are 7 left, So people exit are 13 – 6 = 7,
So 7 people exit from the train.

Review & Refresh

Question 7.
10 – 9 = ____1______
Answer:
10 – 9 = 1

Explanation :
The subtraction equation is 10 – 9 = 1,
We subtract 9 from 10 we get 1.

Question 8.
9 – 9 = ____0______
Answer:
9 – 9 = 0

Explanation :
The subtraction equation is 9 – 9 = 0,
We subtract  9 from 9 we get 0.

Lesson 5.3 Subtract 9

Explore and Grow

Use counters to find each difference.

Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20 28

Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20-2
Explanation:
To find 15 -10 =  here we use two counters  first
we take total 15 circles in that we strike 10 numbers
the left number of circles are 5, So 15 – 10 = 5.
To find 15 – 9 = here we use two counters and we take
15 circles in that we will strike out 9 numbers the left are
6, So 15 – 9 = 6.

Show and Grow

Get to 10 to subtract.

Question 1.
17 – 9 = ?
Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20 29
17 – __7____ = 10
10 – ___2____ = ___8___
So, 17 – 9 = ___8_____
Answer:
17 – 7 = 10 we get 10
10 – 2 = 8,
So 17 – 9 = 8

Explanation:
We have 2 counters in counter one we have 10 circles
and in next counter we have 7 circles so in total we have 17,
First we see out of 17 if we consider counter-2 we have 7 strike outs
means 17 – 7 = 10 we got 10 in counter-1 and if we see first counter
we have 2 strike outs 10 – 2 = 8, So total number of circles strike out
from 2-counters are 7 + 2 = 9, So making equation as 17 – 9 = 8.

Question 2.
14- 9 = ?
Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20 30
14 – __4____ = 10
10 – ___5____ = ___5___
So, 14 – 9 = _____5___
Answer:
14 – 4  = 10 we get 10
10 – 5 = 5
So 14 – 9 = 5

Explanation:
We have  2 counters in counter one we have 10 circles
and in next counter we have 4 circles so in total we have 14,
First we see out of 14 if we consider counter-2 we have 4 strike outs
means 14 – 4 = 10 we got 10 in counter-1 and if we see first counter
we have 5 strike outs 10 – 5 = 5,So total number of circles strike out
from 2-counters are 4 + 5 = 9, So making equation as 14 – 9 = 5.

Apply and Grow: Practice

Question 3.
16 – 9 = ?
Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20 31
16 – ___6___ = 10
10 – ____3___ = __7____
So, 16 – 9 = ___7_____
Answer:
16 – 6 = 10 we get 10
10 – 3 = 7
So 16 – 9 = 7

Explanation:
We have  2 counters in counter one we have 10 circles
and in next counter we have 6 circles so in total we have 16,
First we see out of 16 if we consider counter-2 we have 6 strike outs
means 16 – 6 = 10 we got 10 in counter-1 and if we see first counter
we have 3 strike outs 10 – 3 = 7, So total number of circles strike out
from 2-counters are 6 + 3 = 9, So making equation as 16 – 9 = 7.

Question 4.
11 – 9 = ?
11 – ___1____ = 10
10 – ___8____ = ___2____
S0, 11 – 9 = ____2____ .
Answer:
Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20-3
11 – 1 = 10 we get 10
10 – 8 = 2
So 11 – 9 = 2

Explanation:
To solve 11 – 9  and to get 10,
We solve by using 2 counters in counter one we have 10 circles
and in next counter we have 1 circle so in total we have 11,
Now in all we will strike 9 circles, 1 circle from counter-2 making equation
as 11 – 1 = 10 we got 10 in counter-1 and we strike out  8 circles from counter-1
we are left with 2 making it as 10 – 8 = 2, So in total we have taken 8 + 1 = 9
from 11 in 2-counters, making equation as 11 -9 = 2 as shown in the picture.

Question 5.
15 – 9 = ?
15 – __5_____ = 10
10 – __4______ = __6______
So, 15 – 9 = ___6_____
Answer:
Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20-4
15- 5 = 10 we get 10
10 – 4 = 6
So 15 – 9 = 6

Explanation:
To solve 15 – 9  and to get 10,
We solve by using 2 counters in counter one we have 10 circles
and in next counter we have 5 circles so in total we have 15,
Now in all we will strike 9 circles, 5 circles from counter-2 making equation
as 15 – 5 = 10 we got 10 in counter-1 and we strike 4 circles from counter-1
we are left with 6 making it as 10 – 4 = 6, So in total we have taken 5 + 4 = 9
from 15 in 2-counters, making equation as 15 -9 = 6 as shown in the picture.

Question 6.
MP Structure
Which models show 12 – 9?
Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20 32
Answer:
Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20-5 Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20-6

First model and third model shows 12 – 9.

Explanation:
If we see all models are having 12 now we will see which model is subtracting 9,
For this we consider both counters and count total strike outs
Model 1 have 9 strike outs , Model 2 have 10 strike outs,
Model 3 have 9 strike outs, So first model and third model shows 12 – 9.

Think and Grow: Modeling Real Life

You have 12 eggs. You use 9 of them. How many eggs are left?
Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20 33
Model:
Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20 34
Subtraction equation:

___3_______ eggs
Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20-7
12 – 9 = 3,
So 3 eggs are left.

Explanation:
Given I have 12 eggs and use 9 eggs,
By using counters first I take total 12 circles then I strike out 9,
So left are 3 so the subtraction equation is 12 – 9 = 3,
So 3 eggs are left.

Show and Grow

Question 7.
An egg carton has 18 eggs. You crack 9 of them. How many eggs are not cracked?
Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20 35
Model:
Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20 34
Subtraction equation:

_____9_______ eggs
Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20-8
18 – 9 = 9,
So 9 eggs are not cracked.

Explanation:
Given egg carton has 18 eggs and crack 9 of them,
By using counters first I take total 18 circles then I strike out 9,
So left are 9 so the subtraction equation is 18 – 9 = 9,
So 9 eggs are not cracked.

Subtract 9 Practice 5.3

Get to 10 to subtract.

Question 1.
18 – 9 = ?
Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20 36
18 – ___8_____ = 10
10 – __1_____ = __9_____
So, 18 – 9 = ____9_____ .
Answer:

18- 8 = 10 we get 10
10 – 1 = 9
So 18 – 9 = 9

Explanation:
To solve 18 – 9  and to get 10,
We solve by using 2 counters in counter one we have 10 circles
and in next counter we have 8 circles so in total we have 18,
Now in all we will strike 9 circles, 8 circles from counter-2 making equation
as 18 – 8 = 10 we got 10 in counter-1 and we strike 1 circle from counter-1
we are left with 9 making it as 10 – 1 = 9, So in total we have taken 8 + 1 = 9
from 18 in 2-counters, making equation as 18 – 9 = 9 as shown in the picture.

Question 2.
12 – 9 = ?
Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20 37
12 – ___2_____ = 10
10 – ___7____ = ___3____
So, 12 – 9 = _____3____ .
Answer:
12 – 2 = 10 we get 10
10 – 7 = 3
12 – 9 = 3

Explanation:
To solve 12 – 9  and to get 10,
We solve by using 2 counters in counter one we have 10 circles
and in next counter we have 2 circles so in total we have 12,
Now in all we will strike 9 circles, 2 circles from counter-2 making equation
as 12 – 2 = 10 we got 10 in counter-1 and we strike 7 circles from counter-1
we are left with 3 making it as 10 – 7 = 3, So in total we have taken 2 + 7 = 9
from 12 in 2-counters, making equation as 12 – 9 = 3 as shown in the picture.

Question 3.
19 – 9 = ?
19 – ____9____ = 10
10 – ___0____ = ___10____
So, 19 – 9 = ____10_____ .
Answer:
Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20-9
19 – 9 = 10,
10 – 0 = 10
So, 19 -9 = 10

Explanation:
To solve 19 – 9  and to get 10,
We solve by using 2 counters in counter one we have 10 circles
and in next counter we have 9 circles so in total we have 19,
Now in all we will strike 9 circles, 9 circles from counter-2 making equation
as 19 – 9 = 10 we got 10 in counter-1, we will not strike in counter-1 as we have
already taken 9 ,So 10 – 0 = 10, We are left with 10 and the equation is 19 – 9 = 10.

Question 4.
MP Structure
Which models show 16 – 9?
Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20 38
Answer:

Second model and third model shows 16 – 9.

Explanation:
If we see all models are having 16 now we will see which model is subtracting 9,
For this we consider both counters and count total strike outs
Model 1 have 10 strike outs , Model 2 have 9 strike outs,
Model 3 have 9 strike outs, So Second model and third model shows 16 – 9.

Question 5.
Modeling Real Life
You have 14 water balloons. You break 9 of them. How many water balloons are left?
Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20 39
Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20 40

____5______ Water balloons
Answer:
Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20-10
14 – 9 = 5
5 water balloons are left.

Explanation:
Given I have 14 water balloons. I break 9 of them,
So I will take 2 counters counter-1 has 10 circles and counter-2 has
4 circles in all I have 14 number of circles, now over all I will
strike out 9 from the 2-counters. As counter-2 has only 4 I will take 5 from
counter-1 so that in total 4 + 5 = 9 circles are strike out so we are left with 5,
So the subtraction equation is 14  – 9 = 5, So 5 water balloons are left.

Review & Refresh

Question 6.
Use the picture to complete the number bond.
Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20 41
Answer:
Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20-11

Explanation:
Number bonds are pictures that show the bond between parts and whole.
So here we know that 10 is the whole number
and 8 is one of the numbers from the pair of number bonds.
The other number that combines with 8 to give 10 is 2.

Lesson 5.4 Get to 10 to Subtract

Explore and Grow

Use counters to find the difference. Show how you can make a 10 to solve.

Big Ideas Math Solutions Grade 1 Chapter 5 Subtract Numbers within 20 42
Answer:

Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20-12

14 – 4 = 10
10 – 2 = 8
14 – 6 = 8

Explanation:
To solve 14 – 6  and to get 10,
We solve by using 2 counters in counter one we have 10 circles
and in next counter we have 4 circles so in total we have 14,
Now in all we will strike 6 circles, 4 circles from counter-2 making equation
as 14 – 4 = 10 we got 10 in counter-1 and we strike 2 circles from counter-1
we are left with 8 making it as 10 – 2 = 8, So in total we have taken 4 + 2 = 6
from 14 in 2-counters, making equation as 14 – 6 = 8 as shown in the picture.

Show and Grow

Get to 10 to subtract.

Question 1.
12 – 5 = ?
Big Ideas Math Solutions Grade 1 Chapter 5 Subtract Numbers within 20 43
12 – __2____ = 10
10 – __3___ = ___7_____
So, 12 – 5 = ____7______
Answer:

12 – 2 = 10 we get 10
10 – 3 = 7
12 – 5 = 7

Explanation:
To solve 12 – 2  and to get 10,
We solve by using 2 counters in counter one we have 10 circles
and in next counter two we have 2 circles so in total we have 12,
Now in all we will strike 5 circles, 2 circles from counter-2 making equation
as 12 – 2 = 10 we got 10 in counter-1 and we strike 3 circles from counter-1
we are left with 7 making it as 10 – 3 = 7, So in total we have taken 2 + 3 = 5
from 12 in 2-counters, making equation as 12 – 5 = 7 as shown in the picture.

Question 2.
17 – 8 = ?
Big Ideas Math Solutions Grade 1 Chapter 5 Subtract Numbers within 20 44
17 – ___7___ = 10
10 – __1___ = ___9_____
So, 17 – 8 = ___9_______
Answer:
17 – 7 = 10
10 – 1 = 8
So 17 – 8 = 9

Explanation:
To solve 17 -8  and to get 10,
We solve by using 2 counters in counter one we have 10 circles
and in next counter two we have 7 circles so in total we have 17,
Now in all we will strike 8 circles, 7 circles from counter-2 making equation
as 17 – 7 = 10 we got 10 in counter-1 and we strike 1 circle from counter-1
we are left with 9 making it as 10 – 1 = 9, So in total we have taken 7 + 1 = 8
from 17 in 2-counters, making equation as 17 – 8 = 9 as shown in the picture.

Apply and Grow: Practice

Get to 10 to subtract

Question 3.
16 – 7 = ?
Big Ideas Math Solutions Grade 1 Chapter 5 Subtract Numbers within 20 45
I6 – __6______ = 10
10 – ___1_____ = ____9____
So, 16 – 7 = _____9_____ .
Answer:
16 – 6 = 10
10 – 1 = 9
So 16 – 7 = 9

Explanation:
To solve 16 -7  and to get 10,
We solve by using 2 counters in counter one we have 10 circles
and in next counter two we have 6 circles so in total we have 16,
Now in all we will strike 7 circles, 6 circles from counter-2 making equation
as 16 – 6 = 10 we got 10 in counter-1 and we strike 1 circle from counter-1
we are left with 9 making it as 10 – 1 = 9, So in total we have taken 6 + 1 = 7
from 16 in 2-counters, making equation as 16 – 7 = 9 as shown in the picture.

Question 4.
11 – 4 = ?
11 – ____1___ = 10
10 – ___3____ = ___7_____
So, 11 – 4 = ___7_____ .
Answer:
Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20-13

Answer:
11 – 1 = 10
10 – 3 = 7
So 11 – 4 = 7

Explanation:
To solve 11 -4  and to get 10,
We solve by using 2 counters in counter one we have 10 circles
and in next counter two we have 1 circles so in total we have 11,
Now in all we will strike 4 circles, 1 circles from counter-2 making equation
as 11 – 1 = 10 we got 10 in counter-1 and we strike 3 circles from counter-1
we are left with 9 making it as 10 – 3 = 7, So in total we have taken 1 + 3 = 4
from 11 in 2-counters, making equation as 11 – 4 = 7 as shown in the picture.

Question 5.
13 – 5 = ?
13 – ____3_____ = 10
10 – ____2_____ = ____8______
So, 13 – 5 = ____8____ .
Answer:
Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20-14
13 – 3 = 10
10 – 2 = 8
So 13 – 5 = 8

Explanation:
To solve 13 -5  and to get 10,
We solve by using 2 counters in counter one we have 10 circles
and in next counter two we have 3 circles so in total we have 13,
Now in all we will strike 5 circles, 3 circles from counter-2 making equation
as 13 – 3 = 10 we got 10 in counter-1 and we strike 3 circles from counter-1
we are left with 8 making it as 10 – 2 = 8, So in total we have taken 3 + 2 = 5
from 13 in 2-counters, making equation as 13 – 5 = 8 as shown in the picture.

Question 6.
MP Number Sense
Which equations did Newton use to solve the problem?
Big Ideas Math Solutions Grade 1 Chapter 5 Subtract Numbers within 20 46
Answer:
Newton uses Equation 3 : 12 – 2 =10 , 10 – 4 = 6 so we get 12 – 6 = 6

Explanation:
If we see picture we have two counters in total we have total 12 circles,
In counter-2 we have 2 strike out means 12 – 2 = 10 and from counter-1
we have 4 strike out so 10 – 4 = 6 so in total out of 12 we strike 6 making
the equation as 12 – 6 = 6.
Out of three equations Newton uses equation 12 – 2 = 10 , 10 – 4 = 6,
other equations have 17 count in total and only 3rd equation is having
12 count as per the given picture, So newton use Equation 3.

Think and Grow: Modeling Real Life

Your friend checks out 13 books. You check out 4 fewer. How many books do you check out?
Big Ideas Math Solutions Grade 1 Chapter 5 Subtract Numbers within 20 47
Model:
Big Ideas Math Solutions Grade 1 Chapter 5 Subtract Numbers within 20 48
Subtraction equation:

____9___ books
Answer:
Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20-15
I have checked 9 books.

Explanation:
Given my friend checks out 13 books and I check out 4 fewer,
Means I will take 2 counters in counter-1 I will have 10 books and counter-2
3 books in total 13 books, now I have 4 fewer means I will strike out 4,
3 from counter-2 and 1 from counter-1 in total 4 from both the counters,
So 13 – 4 = 9, I checked 9 books.

Show and Grow

Question 7.
Your friend skips 16 stones. You skip 8 fewer. How many stones do you skip?
Big Ideas Math Solutions Grade 1 Chapter 5 Subtract Numbers within 20 49
Model:
Big Ideas Math Solutions Grade 1 Chapter 5 Subtract Numbers within 20 50
Subtraction equation:

_____8_______ stones
Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20-16
16 – 8 = 8
So I skip 8 stones.

Explanation:
Given my friend skips 16 stones and I skip 8 fewer,
Means I will take 2 counters in counter-1 I will have 10 stones and counter-2
6 stones in total 16 stones, now I have 8 fewer means I will strike out 8 ,
6 from counter-2 and 2 from counter-1 in total 8 from both the counters,
So 16 – 8 = 8, I skip 8 stones.

Get to 10 to Subtract Practice 5.4

Get to 10 to Subtract

Question 1.
15 – 6 = ?
Big Ideas Math Solutions Grade 1 Chapter 5 Subtract Numbers within 20 51
15 – __5_____ = 10
10 – ___1____ = __9______
So, 15 – 6 = ____9____ .
Answer:
15 – 5 = 10
10 -1 = 9
So, 15 – 6 = 9.

Explanation:
To solve 15 -6  and to get 10,
We solve by using 2 counters in counter one we have 10 circles
and in next counter two we have 5 circles so in total we have 15,
Now in all we will strike 6 circles, 5 circles from counter-2 making equation
as 15 – 5 = 10 we got 10 in counter-1 and we strike 1 circle from counter-1
we are left with 9 making it as 10 – 1 = 9, So in total we have taken 5 + 1 = 6
from 15 in 2-counters, making equation as 15 – 6 = 9 as shown in the picture.

Question 2.
12 – 4 = ?
Big Ideas Math Solutions Grade 1 Chapter 5 Subtract Numbers within 20 52
12 – ___2____ = 10
10 – ___2____ = ___8_____
So, 12 – 4 = ___8_____ .
Answer:
12 – 2 = 10
10 -2 = 8
So, 12 – 4 = 8.

Explanation:
To solve 12 -4  and to get 10,
We solve by using 2 counters in counter one we have 10 circles
and in next counter two we have 2 circles so in total we have 12,
Now in all we will strike 4 circles, 2 circles from counter-2 making equation
as 12 – 2 = 10 we got 10 in counter-1 and we strike 2 circles from counter-1
we are left with 8 making it as 10 – 2 = 8, So in total we have taken 2 + 2 = 4
from 12 in 2-counters, making equation as 12 – 4 = 8 as shown in the picture.

Question 3.
13 – 7 = ?
13 – ___3____ = 10
10 – ___4____ = ___6_____
So, 13 – 7 = ____6____ .
Answer:
Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20-17
13 – 3 = 10
10 – 4 = 6
13 – 7 = 6

Explanation:
To solve 13 -7  and to get 10,
We solve by using 2 counters in counter one we have 10 circles
and in next counter two we have 3 circles so in total we have 13,
Now in all we will strike 7 circles, 3 circles from counter-2 making equation
as 13 – 3 = 10 we got 10 in counter-1 and we strike 4 circles from counter-1
we are left with 6 making it as 10 – 4 = 6, So in total we have taken 3 + 4 = 7
from 13 in 2-counters, making equation as 13 – 7 = 6 as shown in the picture.

Question 4.
14 – 8 = ?
14 – ___4____ = 10
10 – ___4____ = ___6_____
So, 14 – 8 = ____6____ .
Answer:
Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20-18
14 – 4 = 10
10 – 4 = 6
14 – 8 = 6

Explanation:
To solve 14 -8  and to get 10,
We solve by using 2 counters in counter one we have 10 circles
and in next counter two we have 4 circles so in total we have 14,
Now in all we will strike 8 circles, 4 circles from counter-2 making equation
as 14 – 4 = 10 we got 10 in counter-1 and we strike 4 circles from counter-1
we are left with 6 making it as 10 – 4 = 6, So in total we have taken 4 + 4 = 8
from 14 in 2-counters, making equation as 14 – 8 = 6 as shown in the picture.

Question 5.
MP Number Sense
Which equations did Descartes use to solve the problem?
Big Ideas Math Solutions Grade 1 Chapter 5 Subtract Numbers within 20 53
Answer:

Descartes uses Equation 1 : 15 – 5 =10 , 10 – 4 = 6 so we get 15 – 9 = 6

Explanation:
If we see picture we have two counters in total we have total 15 circles,
In counter-2 we have 5 strike out means 15 – 5 = 10 and from counter-1
we have 4 strike out so 10 – 4 = 6 so in total out of 15 we strike 9 making
the equation as 15 – 9 = 6.
Out of three equations Newton uses equation 15 – 5 = 10 , 10 – 4 = 6,
other equations have 20 count in total and only 1st equation is having
15 count as per the given picture, So newton use 1 Equation.

Question 6.
Modeling Real Life
Your friend recycles 14 cans. You recycle 7 fewer. How many cans do you recycle?
Big Ideas Math Solutions Grade 1 Chapter 5 Subtract Numbers within 20 54
Big Ideas Math Solutions Grade 1 Chapter 5 Subtract Numbers within 20 55

____7_____ cans
Answer:
Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20-19

I recycle 7 cans.

Explanation:
Given my friend recycle 14 cans and I recycle 7 fewer,
Means I will take 2 counters in counter-1 I will have 10 cans and counter-2
4 cans in total 14 cans, now I have 7 fewer means I will strike out 7,
4 from the counter-2 and 3 from counter-1 in total 7 from both the counters,
So 14 – 7 = 7, I recycle 7 cans.

Review & Refresh

Is the equation true or false?

Question 7.
Big Ideas Math Solutions Grade 1 Chapter 5 Subtract Numbers within 20 56
Answer:
Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20-20
3 + 2 = 5 + 0
5 = 5 ,
So True.

Explanation:
We take 3 + 2 means adding 3 and 2 we get 3 + 2 = 5
and 5 + 0 is 5 so both sides are equal therefore
3 + 2 = 5 + 0 is True.

Question 8.
Big Ideas Math Solutions Grade 1 Chapter 5 Subtract Numbers within 20 57
Answer:
Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20-21
8 – 2 = 6 ≠ 5 + 5 = 10
6 ≠ 10
So False

Explanation:
We take 8 – 2 means subtracting 2 from 8 we get 8 – 2 = 6
and 5 + 5 means adding 5 and 5 we get 5 + 5  =10,
If we see both sides they are not equal as 6 ≠ 10,
So 8 – 2 = 5 + 5 is False.

Lesson 5.5 More True or False Equations

Explore and Grow

Color the stars that have a sum or difference equal to 19 – 5.

Big Ideas Math Answer Key Grade 1 Chapter 5 Subtract Numbers within 20 58
Answer:
https://ccssmathanswers.com/wp-content/uploads/2021/01/Big-Ideas-Math-Answers-Grade-1-Chapter-5-Subtract-Numbers-within-20-22.bmp
Explanation:
We color the stars that have a sum or difference equal to 19 – 5,
means  19 – 5 = 14, We check each star if equal to 14 we will color.
a. The sum is 9 + 5 = 14, so color it with green,
b. The difference is 20 – 6 = 14, we color it with blue
c. The sum is 12 + 1 + 1 = 14, we color it with green
d. The difference is 16 – 2 = 14, we color it with blue
e. The difference is 13 – 1 = 12 , we will not color it
as this is not equal to 14.
f. The sum is 14 + 0 = 14 , we will color it with green

Show and Grow

Is the equation true or false ?

Question 1.
Big Ideas Math Answer Key Grade 1 Chapter 5 Subtract Numbers within 20 59
Answer:
Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20-23
As 17 – 9 = 8  ≠ 14 -5 = 9
So False

Explanation:
We take 17 – 9 means subtracting 9 from 17 we get 17 – 9 = 8
and 14 – 5 means subtracting 5 from 14 we get 14 – 5 = 9,
If we see both sides they are not equal as 8 ≠ 9,
So 17 – 9 = 14 – 5 is False.

Question 2.
Big Ideas Math Answer Key Grade 1 Chapter 5 Subtract Numbers within 20 60
Answer:
Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20-24
As 6 + 5 = 11 = 18 – 7 = 11
So True.

Explanation:
We take 6 + 5 means adding 6 and 5 we get 6 + 5 = 11,
and 18 – 7 means subtracting 7 from 18 we get 18 – 7 = 11,
As both sides are equal therefore
6 + 5 = 11 = 18 – 7 = 11 is True

Apply and Grow: Practice

Is the equation true or false?

Question 3.
Big Ideas Math Answer Key Grade 1 Chapter 5 Subtract Numbers within 20 61
Answer:
Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20-25

As 5 + 7 = 12  ≠  3 + 8 = 11
So False

Explanation:
We take 5 + 7 means adding 5 and 7 we get 5 + 7 = 12,
and 3 + 8 means adding 3 and 8 we get 3 + 8 = 11,
As both sides are not equal therefore
5 + 7 = 12 ≠ 3 + 8 = 11 is False.

Question 4.
Big Ideas Math Answer Key Grade 1 Chapter 5 Subtract Numbers within 20 62
Answer:
Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20-26
As 4 + 9 = 13 = 5 + 3 + 5 = 13
Is True.

Explanation:
We take 4 + 9 means adding 4 and 9 we get 4 + 9 = 13,
and 5 + 3 + 5  means adding 5 , 3 and 5 we get 5 + 3 + 5 = 13,
As both sides are equal therefore
4 + 9 = 13 = 5 + 3 + 5 = 13 is True.

Question 5.
Big Ideas Math Answer Key Grade 1 Chapter 5 Subtract Numbers within 20 63
Answer:
Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20-27

As 12 – 7 = 5 ≠ 13 – 5 = 8
So False.

Explanation:
We take 12 – 7 means subtracting 7 from 12 we get 12 – 7 = 5,
and 13 – 5 means subtracting 5 from 13 we get 13 – 5 = 8,
As both sides are not equal therefore
12 – 7  = 5 ≠ 13 – 5 = 8 is False.

Question 6.
Big Ideas Math Answer Key Grade 1 Chapter 5 Subtract Numbers within 20 66
Answer:
Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20-28

As 14 – 8 = 6 = 12 – 6 = 6
So True.

Explanation:
We take 14 – 8 means subtracting 8 from 14 we get 14 – 8 = 6,
and 12 – 6 means subtracting 6 from 12 we get 12 – 6 = 6,
As both sides are equal therefore
14 – 8  = 6 = 12 – 6 = 6 is True.

Question 7.
Big Ideas Math Answer Key Grade 1 Chapter 5 Subtract Numbers within 20 64
Answer:
Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20-29
1 + 8 = 9 = 16 -7 = 9
So True.

Explanation:
We take 1 + 8  means adding 1 and 8 we get 1 + 8  = 9,
and 16 – 7 means subtracting 7 from 16 we get 16 – 7 = 9,
As both sides are equal therefore
1 + 8  = 9 = 16 – 7 = 9 is True.

Question 8.
Big Ideas Math Answer Key Grade 1 Chapter 5 Subtract Numbers within 20 65
Answer:
Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20-30
18 – 9 = 9 ≠  5 + 1 + 4 = 10
So False.

Explanation:
We take 18 – 9  means subtracting 9 from 18 we get 18 – 9  = 9,
and 5 + 1 + 4 means adding 5,1 and 4 we get 5 + 1 + 4 = 10,
As both sides are not equal therefore
18 – 9  = 9 ≠ 5 + 1 + 4 = 10 is False.

Question 9.
MP Number Sense
Circle all of the equations that are true.
Big Ideas Math Answer Key Grade 1 Chapter 5 Subtract Numbers within 20 67
Answer:
Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20-31

Explanation:
Given equations are a. 20 = 2 as 20 is not equal to 2,false so we will not circle,
b. 19 – 7 = 12 as 19 – 7 is 12 this is true we will circle it.
c. 6 + 6 + 3 = 7 + 8 as 6 + 6 + 3 = 15 and 7 + 8 = 15, Both are same
and true we will circle the equation.

Think and Grow: Modeling Real Life

You hove 12 lemons. You use 4 of them. Your friend has 3 lemons and buys 4 more.
Do you each have the same number of lemons?
Big Ideas Math Answer Key Grade 1 Chapter 5 Subtract Numbers within 20 68
Equation:
Big Ideas Math Answer Key Grade 1 Chapter 5 Subtract Numbers within 20 69
Answer:
Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20-32
No each do not have the same number of lemons as 8 ≠ 7 .

Explanation:
Given I have 12 lemons and use 4 of them means 12 – 4
My friend have 3 lemons and buys 4 more means 3 + 4
now checking both of them 12 – 4 = 8 and 3 + 4 = 7 as 8 ≠ 7
both are not equal each do not have the same number of lemons.

Show and Grow

Question 10.
You have 14 grapes. You eat 7 of them. Your friend has 10 grapes and eats 3 of them.
Do you each have the same number of grapes?
Equation:
Big Ideas Math Answer Key Grade 1 Chapter 5 Subtract Numbers within 20 70
Answer:
Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20-34
Yes each have the same number of grapes as 7 = 7.

Explanation:
Given I have 14 grapes and I eat 7 of them means 14 -7
My friend has 10 grapes and eats 3 of them means 10 – 3,
Now checking both we have 14 – 7 = 7 and 10 – 3 = 7,
Both are equal, So, Yes each of them have the same number of grapes 7= 7.

More True or False Equations Practice 5.5

Is the equation true or false?

Question 1.
Big Ideas Math Answer Key Grade 1 Chapter 5 Subtract Numbers within 20 71
Answer:
Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20-35
13 – 9 = 4 = 11 – 7 = 4
So True.

Explanation:
Checking 13 – 9 and 11 – 7 , subtracting 9 from 13 we get 4
and subtracting 7 from 11 we get 4
as both equations values are equal, true.

Question 2.
Big Ideas Math Answer Key Grade 1 Chapter 5 Subtract Numbers within 20 72
Answer:
Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20-36
9 + 8 = 17 = 17 + 0 = 17
So True.

Explanation:
Checking 9 + 8 and 17 + 0, as 9 + 8 = 17 and
17 + 0 = 17 as both equations values are equal to 17 so true.

Question 3.
Big Ideas Math Answer Key Grade 1 Chapter 5 Subtract Numbers within 20 73
Answer:
Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20-37
12 – 4 = 8 ≠ 15 – 6 = 9,
So False.

Explanation:
Checking 12 – 4 and 15 – 6, We get 12 – 4 = 8 and
15 – 6 = 9 as both equations values are not equal 8 ≠ 9,
So false.

Question 4.
Big Ideas Math Answer Key Grade 1 Chapter 5 Subtract Numbers within 20 74
Answer:
Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20-38
4 + 5 = 9 ≠ 11 – 3 = 8
Is False

Explanation:
Checking 4 + 5 and 11 – 3, We get 4 + 5 = 9 and
11 – 3 = 8 as both equations values are not equal 9 ≠ 8,
So false.

Question 5.
Big Ideas Math Answer Key Grade 1 Chapter 5 Subtract Numbers within 20 75
Answer:
Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20-39
15 – 7 = 8 = 4 + 4 = 8
So True.

Explanation:
Checking 15 – 7 and  4 + 4  we get 15 – 7 = 8
and  4 + 4 = 8 as both equations values are same 8 = 8
so true.

Question 6.
Big Ideas Math Answer Key Grade 1 Chapter 5 Subtract Numbers within 20 76
Answer:
Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20-40
0 + 5 + 2 = 7 = 12 – 5 =  7
So True.

Explanation:
Checking 0 + 5 + 2 = 12 – 5, We get 0 + 5 + 2 = 7
and 12 – 5 = 7 as both equations values are same 7 = 7
so true.

Question 7.
Big Ideas Math Answer Key Grade 1 Chapter 5 Subtract Numbers within 20 77
Answer:
Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20-41
16 – 8 = 8  ≠  2 + 3 + 2 = 7
So False

Explanation:
Checking 16 – 8 and 2 + 3 + 2 we see 16 – 8 = 8
and  2 + 3 + 2 = 7 as both equations are not equal
8 ≠ 7 so false.

Question 8.
MP Number Sense
Circle all of the equations that are false.
Big Ideas Math Answer Key Grade 1 Chapter 5 Subtract Numbers within 20 78
Answer:
Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20-42
Explanation:
Checking equations
a. 7 + 2 = 9 and 11 -2 = 9 both are equal so true we will not circle.
b. 3 , 12 – 8 = 4 as 3 ≠  4, So false we will circle it.
c. 4 + 1 + 4 = 9 and 14 -6 = 12 as 9 ≠ 12 , So false we will circle it.

Question 9.
Modeling Real Life
You have 9 badges. You earn 3 more. Your friend has 5 badges and earns 7 more. Do you each have the same number of badges?
Big Ideas Math Answer Key Grade 1 Chapter 5 Subtract Numbers within 20 79
Big Ideas Math Answer Key Grade 1 Chapter 5 Subtract Numbers within 20 80
Answer:
Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20-43
Yes , each have the same number of badges.

Explanation:
Given I have 9 badges and  earn 3 more means 9 + 3 = 12,
My friend has 5 badges and earns 7 more means 5 + 7 = 12,
as both are having 12 badges Yes, I and my friend each
have same number of badges .

Review & Refresh

Question 10.
_____4____ + 2 = 6
Answer:
4 + 2 = 6

Explanation:
Given _______ + 2 = 6, Let us take missing number as X,
X + 2 = 6 means X = 6 – 2 = 4, X = 4 So missing number is 4,
the equation is 4 + 2 = 6.

Question 11.
____6____ + 3 = 9
Answer:
6 + 3 = 9

Explanation:
Given _______ + 3 = 9, Let us take missing number as X,
X + 3 = 9 means X = 9 – 3 = 6, X = 6 So missing number is 6,
the equation is 6 + 3 = 9.

Lesson 5.6 Make True Equations

Explore and Grow

Complete the equation.

14 – 5 = 3 + __________

Big Ideas Math Answers 1st Grade 1 Chapter 5 Subtract Numbers within 20 81
Answer:
Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20-44
14 – 5 = 3 + 6

Explanation:
Given 14 – 5 = 3 + ________ Let us take missing number as X,
First we see 14 – 5 on number line we move
5 steps left from 14 we get 9, means 3 + X = 9
so to get X we move on number line from 3 till 9
we got 6 units means X = 6,
therefore the true equation is 14 – 5 = 3 + 6.

Show and Grow

Question 1.
? + 6 = 10 – 3
? + 6 = __7______
___1_____ + 6 = ___7____

So, ____1____ + 6 = 10 – 3.
Answer:
Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20-45
1 + 6 = 10 – 3

Explanation:
Given _____ + 6 = 10 – 3
Now first we start from 10 and move 3 units back on the
number line we get 7, 10 – 3 = 7 now from 6 we will move
right till we get 7 we moved 1 unit means 6 + 1  = 7,
So the true equation is 1 + 6 = 10 – 3.

Question 2.
12 – 9 = ? – 5
____3____ = ? – 5
____3____ = __8_____ – 5

So, 12 – 9 = ____8____ – 5.
Answer:
Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20-46
12 – 9 = 8 – 5

Explanation:
First we find 12 – 9 we start from 12 on number line
and move 9 units back we get 3 means 12 – 9 = 3 and ____ – 5 = 3,
Let us take missing number as X and X – 5 = 3 , therefore X = 3 + 5
on number line we move 3 units right from 5 we get 8,
making the true equation as 12 – 9 = 8 – 5.

Apply and Grow: Practice

Question 3.
? + 8 = 9 + 6
? + 8 = __15______
____7____ + 8 = _15___

So, ___7_____ + 8 = 9 + 6.
Answer:
_ + 8 = 9 + 6
_ + 8 = 15
7 + 8 = 15
So 7 + 8 = 9 + 6

Explanation :

Given ___ +  8 = 9 + 6, First we see 9 + 6 = 15 means
____ + 8 = 15 So let us take missing number as X ,
X + 8 = 15 therefore X = 15 – 8 = 7 making true equation as
7 + 8 = 9 + 6.

Question 4.
13 – 8 = ? – 6
___5_____ = ? – 6
____5____ = __11_ – 6

So, 13 – 8 = __11_ – 6.
Answer:
13 – 8 = _______ – 6
5 = ______ – 6
5 = 11 – 6
So 13 – 8 = 11 – 6

Explanation:
Given 13 – 8 = ____ – 6 means first we see 13 – 8 we get 5
means _____ – 6 = 5 let us take missing number as X
So X – 6 = 5, X = 5 + 6 = 11 making the true equation as
13 – 8  = 11 – 6.

Question 5.
14 – 6 = ? + 2
___8_____ = ? + 2
____8____ = ___6____ + 2

So, 14 – 6 = ____6____ + 2.
Answer:
14 – 6 = _____ + 2
8 = ________ + 2
8 = 6 + 2
So 14 – 6 = 6 + 2

Explanation:
Given 14 – 6 is equal to ________ + 2 , first we see 14 – 6 we get
8 means _______ + 2 = 8 , let us take missing number as X
so 8 = X + 2 means X = 8 – 2 = 6 making the true equation as
14 – 6 = 6 + 2.

Question 6.
15 – ? = 3 + 3
15 – ? = __6______
15 – ___9_____ = ___6____

So, 15 – ____9____ = 3 + 3.
Answer:
15 – ___ = 3 + 3
15 – ___ =  6
15 – 9 = 6
15 – 9 = 3 + 3

Explanation:
Given 15 – ___ = 3 + 3 means first we see 3 + 3 = 6,
means 15 – __ = 6 , Let us take X as missing number means
15 – X = 6 we get X = 15 – 6 = 9  therefore 15- 9 = 6,
making true equation as 15 – 9 = 3 + 3.

Question 7.
_____12_____ – 6 = 3 + 2 + 1
Answer:
12 – 6 = 3 + 2 + 1

Explanation:
Given __________ – 6 = 3 + 2 + 1 we see first
3 + 2 + 1 = 6 means ________ – 6 = 6,
let us take missing number as X , X – 6 = 6,
X = 6 + 6 = 12 making the true equation as 12 – 6 = 3 + 2 + 1.

Question 8.
6 + 4 + 4 = 6 + _______
Answer:
6 + 4 + 4 = 6 + 8

Explanation:
Given 6 + 4 + 4 = 6 + _______, First we add 6 + 4 + 4 we get 14
means 14 = 6 + ____, we will take missing number as X So that
6 + X = 14 , X = 14 – 6 = 8 making the true equation as 6 + 4 + 4 = 6 + 8.

Question 9.
YOU BE THE TEACHER
Newton says 2 makes the equation true. Is Newton correct? Show how you know.
Big Ideas Math Answers 1st Grade 1 Chapter 5 Subtract Numbers within 20 82
Answer:
Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20-47
Yes, Newton is correct, as 7 + 2 = 9 = 11 – 2.

Explanation:
Given the equation as 7 + 2 = 11 -____ and newton says 2
makes the equation true, First we see 7 + 2 = 9
on number line we move 2 steps from 7 we get 9 so
now on number line we need to reach 9  from 11
so we will move 2 steps back or left we reach 9 so 11 – 2 = 9,
Both equations are equal so Newton is true.

Think and Grow: Modeling Real Life

You catch 11 butterflies. 4 fly away. Your friend catches 3 butterflies.
How many more butterflies must your friend catch to have the same number as you?
Big Ideas Math Answers 1st Grade 1 Chapter 5 Subtract Numbers within 20 83
Equation:
___11____ – ___4_____ = ____3___ + ?

____4______ butterflies

11 – 4  = 3 + 4
My friend should catch 4 butterflies to have same number as me.

Explanation :
Given I catch 11 butterflies. 4 fly away means I have
11 – 4 = 7 butterflies now and my friend catches 3 butterflies
to have same number of butterflies let my friend should catch
X butterflies as 3 + X = 7 so X = 7 – 3 = 4 making equation as
11 – 4 = 3 + 4. Therefore my friend should catch 4 butterflies
to have same number as me.

Show and Grow

Question 10.
You catch 15 leaves. 6 of them blow away. Your friend catches 12 leaves and some of them blow away. Now you each have the same number of leaves. How many of your friend’s leaves blow away?
Big Ideas Math Answers 1st Grade 1 Chapter 5 Subtract Numbers within 20 84
Equation:
___15____ – ___6_____ = ___12____ – ?

______3____ leaves
15 – 6 = 12 – 3,
3 leaves of my friend’s blow away.

Explanation:
Given I catch 15 leaves and 6 of them blow away means I have
15 – 6 = 9, Your friend catches 12 leaves and some of them blow away
lets take it as X number and given I & my friend have same number of leaves now
12 – X = 9 , X = 12 – 9 = 3. So 3 leaves of my friend’s blow away.

Make True Equations Practice 5.6

Question 1.
12 – ? = 8 – 4
12 – ? = __4______
12 – ___8_____ = ___4____

So, 12 – ___4_____ = 8 – 4.
Answer:
12 – ____ = 8 – 4
12 – ___ = 4
12 – 8 = 4
So, 12 – 8 =  8 – 4

Explanation:
Given equation as 12 – ___ = 8 – 4, first we see 8 – 4
we get as 4, 8-4 = 4 So 12 – ___ = 4,
Let us take missing number as X so 12 – X = 4,
X = 12 – 4= 8, Now the true equation is 12 – 8 = 8 – 4.

Question 2.
7 + 9 = 8 + ?
____16____ = 8 + ?
_____16___ = 8 + ___8____

So, 7 + 9 = 8 + ___8_____ .
7 + 9 = 8 + ___
16 = 8 + ___
16 = 8 + 8
So, 7 + 9 = 8 + 8

Explanation:
Given equation as 7 + 9 = 8 + __,  So first we see
7 + 9 = 16 means 8 + ___ = 16 ,Let us take X as missing number
8 + X = 16, X = 16 – 8  = 8, Now the true equation is 7 + 9 = 8 + 8

Question 3.
4 + 3 = __________ – 7
Answer:
4 + 3 = 14 – 7

Explanation:
Given 4 + 3 = ___ – 7, first we see 4 + 3 we get 7,
means 7  = ____ – 7, Let us take X as missing number
so 7 = X – 7, X = 7 + 7 = 14 making true equation as
4 + 3 = 14 – 7.

Question 4.
7 + _________ = 17 – 0
Answer:
7 + 10 = 17 – 0

Explanation:
Given as 7 + ___ = 17 – 0 first we see 17 – 0 we get 17
means 7 + ___ = 17, Now we take missing number as X,
7 + X = 17, X = 17 – 7 = 10 therefore the true equation is
7 + 10 = 17 – 0.

Question 5.
__________ + 20 = 8 + 10 + 2
Answer:
0 + 20 = 8 + 10 + 2

Explanation:
Given ____ + 20 = 8 + 10 + 2, We solve first 8 + 10 + 2
we get 20 so  ____ + 20 = 20, Let us take missing number as X,
X + 20 = 20 therefore X = 20 -20 = 0 , Making true equation as
0 + 20 = 8 + 10 + 2.

Question 6.
3 + 1 + 1 = 14 – __________
Answer:
3 + 1 + 1 = 14 – 9

Explanation:

Given 3 + 1 + 1 = 14 – ____, first we see 3 + 1 + 1 we get
5 and 14 – ___ = 5 , Let us take missing number as X, 14 – X = 5,
X = 14 – 5 = 9, making true equation as 3 + 1 + 1 = 14 – 9.

Question 7.
YOU BE THE TEACHER
Descartes says 5 makes the equation true. Is Descartes correct? Show how you know.
Big Ideas Math Answers 1st Grade 1 Chapter 5 Subtract Numbers within 20 85
Answer:
Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20-48
17 – 9 ≠ 5 – 2 ,So Descartes is in correct.

Explanation:
Given equation is 17 – 9 = ___ – 2 and 5 makes the equation true ,
we see first 17 – 9 on number line we start from 17 and move 9 units back
we get 8 means 17 – 9 = 8, Now we take as per Descartes 5 – 2 on number line
we start from 5 and move 2 units back we get 3 which is not equal to 8,
So 17 – 9 ≠ 5 – 2 ,So Descartes is in correct.

Question 8.
Modeling Real Life
You catch 14 fireflies. You lose 8 of them. Your friend catches 11 fireflies and loses some of them.
Now you each have the same number of fireflies. How many fireflies does your friend lose?
Big Ideas Math Answers 1st Grade 1 Chapter 5 Subtract Numbers within 20 86
___14_____ – ___8____ = ___11____ – ?

__5_____ fireflies
14 – 8 = 11 – 5, My friend loses 5 fireflies.

Explanation:
Given I catch 14 fireflies and lose 8 of them, means I have
14 – 8 = 6 with me, My friend catches 11 fireflies and
loses some of them let us take my friend loses X fireflies and
each have the same number of fireflies means 11 – X = 6, So
X = 11 – 6 = 5, making the equation as 14 – 8 = 11 – 5,
therefore my friend loses 5 fireflies.

Review & Refresh

Question 9.
color the rectangles.
Big Ideas Math Answers 1st Grade 1 Chapter 5 Subtract Numbers within 20 87
Answer:
Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20-49

Explanation:
Given in the picture we have four shapes and we color the third shape
which is rectangle means a quadrilateral with four right angles, or a parallelogram
containing a right angle, so colored it with purple color.

Question 10.
color the squares.
Big Ideas Math Answers 1st Grade 1 Chapter 5 Subtract Numbers within 20 88
Answer:
Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20-50
Explanation:
Given in the picture we have four shapes and we color the first shape
which is rectangle means a quadrilateral with four right angles, or a parallelogram
containing a right angle, so colored it with orange color.

Lesson 5.7 Problem Solving: Subtraction within 20

Explore and Grow

Model the story.

There are 18 seagulls. Some of them fly away. There are 9 left. How many seagulls flew away?

_____9___ seagulls
Answer:
Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20-51
18 – 9 = 9
9 seagulls flew away.

Explanation:
Given there are 18 seagulls. Some of them fly away.
There are 9 left means out of 18 we are left with 9,
Let us take X number of them fly away means
18 – X  = 9, X = 18 – 9 = 9, We show in bar model as
total we have 18, 9 left and 9 seagulls flew away.

show and Grow

Question 1.
You have some stuffed animals. You give 3 away. You have 8 left. How many stuffed animals did you have to start?
Circle what you know.
Underline what you need to find.
Solve:
____11____ ○ ___3______ = ____8______

_____11_____ stuffed animals
Answer:
Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20-52
11 – 3 = 8
I have to start with 11 stuffed animals.

Explanation:
Given I have some stuffed animals and I give 3 away so have 8 left,
let I have X number of stuffed animals so From X  gave 3 so
X – 3 = 8 , X = 8 + 3 = 11 I know I gave 3 and left with 8 so I circled them,
I need to know how many stuffed animals I had so underlined it
as shown above in the picture. making equation as 11 – 3 = 8,
therefore I have to start with 11 stuffed animals.

Apply and Grow: Practice

Question 2.
A group of students are at an arcade. 8 of them leave. There are 3 left. How many students were at the arcade to start?
Circle what you know.
Underline what you need to find.
Solve:
Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20 89
Answer:
Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20-53
11 – 8 = 3,
11 students were at the arcade to start.

Explanation:
Given a group of students are at an arcade let us take as X,
8 of them leave and there are 3 left means X – 8 = 3,
X = 3 + 8 = 11 as shown in bar model  8 leave and 3 left total 11.
the equation is 11 – 8 = 3, I circled 8 and 3 I know, underlined 11
to find. So there are 11 students at the arcade to start.

Question 3.
You have 15 trucks. Your friend has 7. How many more trucks do you have?

_____8____ more trucks
Answer:
Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20-54
15 – 7 = 8
I have 8 more trucks.

Explanation:
Given I have 15 trucks and friend have 7, So more trucks I have than my
friend are 15 – 7 = 8, as shown in bar model I have 15 and friend have
7 and 8 more trucks I have than my friend.

Question 4.
DIG DEEPER
You have 16 stickers. Your friend has 7 fewer than you. Which bar model shows how many stickers your friend has?
Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20 90
Answer:
Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20-55
16 – 7 = 9
circled the equation shown in bar model which
shows how many stickers my friend has.

Explanation:
Given I have 16 stickers and my friend has 7 fewer than me
means my friend have 16 – 7 = 9 stickers ,So I have circled the
equation shown in bar model which shows how many stickers my friend has.

Think and Grow: Modeling Real Life

Your friends mask has 13 feathers. Your mask has 7 feathers. How many fewer feathers does your mask have?
Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20 91
Circle what you know.
Underline what you need to find.
Solve:
Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20 92
_________ – ________ = _________ fewer feathers
Answer:
Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20-56
Explanation:
Given friends mask has 13 feathers and my mask has 7 feathers,
So I have 13 – 7 = 6 feathers fewer in my mask as shown in the bar model
I have circled 13, 7 in the equation as I know how many my friend
and I have and underlined 6 so that how many fewer feathers my mask have.

Show and Grow

Question 5.
There are 12 party hats. There are 5 fewer noisemakers than party hats. How many noisemakers are there?
Circle what you know.
Underline what you need to find.
Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20 93
Solve:
____12_____ – ____5____ = ____7_____ noisemakers
Answer:
Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20-57
12 – 5 = 7
So there are 7 noisemakers.

Explanation:
Given there are 12 party hats and there are 5 fewer
noisemakers than party hats means 12 – 5 = 7 noisemakers
are available, as I know how many party hats and how many fewer
noisemakers are there so I circled them 12, 5 and I need to find how
many noisemakers are there so I underlined 7 in the equation.

Problem Solving: Subtraction within 20 Practice 5.7

Question 1.
You have 12 erasers. Your friend takes some of them. You have 5 left. How many erasers does your friend take?
Circle what you know.
Underline what you need to find.
Solve:
Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20 94
Answer:
Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20-58
12 – 7 = 5
My friend have taken 7 erasers.

Explanation:
Given I have 12 erasers and my friend takes some of them
and I have 5 left. Let my friend takes X number so
12 – X = 5, X = 12 – 5 = 7 , So 12 – 7 = 5 as shown in bar model,
As I know 12 and 5 I have circled it and I need to find
how many my friend have taken so I have underlined 7 in the
equation.

Question 2.
You have 17 glitter pens. Your friend has 9 fewer than you. How many does your friend have?

_____8______ glitter pens
Answer:
Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20-59
17 – 9 = 8,
My friend have 8 glitter pens.

Explanation:
Given I have 17 glitter pens and my friend has 9 fewer than me,
Now let us take my friend have X number of glitter pens so
X = 17 – 9, X = 8 therefore my friend have 8 glitter pens.

Question 3.
DIG DEEPER!
A group of students are at a library. 6 of them leave. There are 7 left. Which part-part whole model shows how many students were at the library to start?
Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20 95
Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20 96
Answer:
Second One Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20-60shows
13 number students were there at the library to start.

Explanation:
Given a group of students are at a library 6 of them leave and
there are 7 left. So a number of students were there at the library
to start are 6 + 7 = 13, so the second one part-part whole model shows
how many students were there at the library to start.

Question 4.
Modeling Real Life
Your friend’s shirt has 14 buttons. Your shirt has 7 fewer. How many buttons does your shirt have?
Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20 97
Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20 98

___________ buttons
Answer:
Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20-61
14 – 7 = 7
My shirt have 7 buttons.

Explanation:
Given my friend’s shirt has 14 buttons and my shirt has 7 fewer,
So my shirt have 14 – 7 = 7 buttons as shown in the bar model
therefore my shirt has 7 buttons.

Review & Refresh

Question 5.
There are 8 Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20 99.
3 Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20 99 swim away.
How many Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20 99 are left?
____8____ – ___3____ = ___5_____ Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20 99
Answer:
8 – 3 = 5,
There are 5 seals left.

Explanation:
Given there are 8 seals and 3 of them swim away means
from 8 we will take 3 making equation as 8 – 3 = 5,
So there are 5 seal left.

Chapter 5 Subtract Numbers within 20 Performance Task

Question 1.
You keep track of the number of honeybees and bumblebees you see.
Big Ideas Math Solutions Grade 1 Chapter 5 Subtract Numbers within 20 100

a. How many more honeybees did you see on Monday than on Tuesday?
Big Ideas Math Solutions Grade 1 Chapter 5 Subtract Numbers within 20 101

_____6_____ more honeybees
Answer:

12 – 6 = 6,
I have 6 more honeybees seen on Monday than on Tuesday.

Explanation:
Given track of the number of honeybees seen
in that on Monday it is 12 and on Tuesday it is 6,
more number of honeybees seen are 12 – 6 = 6 on Monday than on Tuesday.

b. How many fewer bees did you see on Monday than on Tuesday?
Big Ideas Math Solutions Grade 1 Chapter 5 Subtract Numbers within 20 101

_____9_____ fewer bees
Answer:
14 – 5 = 9
9 fewer bees I have seen on Monday than on Tuesday.

Explanation:
Given track of the number of bees  seen
in that on Monday it is 5 and on Tuesday it is 14, now on Monday
it is less than Tuesday  so out of 14 we have seen 5 means
we have 14 – 5  = 9  fewer bees seen on Monday than on Tuesday.

c. How many bumblebees must you see on Wednesday so that the numbers of bees you see on
Tuesday and Wednesday are the same?
Big Ideas Math Solutions Grade 1 Chapter 5 Subtract Numbers within 20 101

____14______ bumblebees
Answer:
14 number of bumblebees I must see on Wednesday so that the numbers
of bees I see on Tuesday and Wednesday are the same.

Explanation:
Given track of the number of bumblebees seen on Tuesday are 14, So
bumblebees I must  see on Wednesday so that the numbers of bees I see on
Tuesday and Wednesday are the same are 14 only as both days is should be the same.

Chapter 5 Subtract Numbers within 20 Chapter Practice

Count Back to Subtract within 20 Homework & Practice 5.1

Question 1.
Big Ideas Math Solutions Grade 1 Chapter 5 Subtract Numbers within 20 102
11 – 3 = ____8______
Answer:
Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20-62

11 – 3 = 8

Explanation:
We subtract 11 – 3 = 9, on number line we start at 11 and
move 3 steps back or left we get 8.

Question 2.
Big Ideas Math Solutions Grade 1 Chapter 5 Subtract Numbers within 20 102
13 – 4 = _____9_____
Answer:
Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20-63
13 – 4 = 9

Explanation:
We subtract 13 – 4 = 9, on number line we start at 13 and
move 4 steps back or left we get 9.

Question 3.
Big Ideas Math Solutions Grade 1 Chapter 5 Subtract Numbers within 20 102
____12______ = 15 – 3
Answer:
Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20-64
12 = 15 – 3

Explanation:
We subtract 15 – 3 = 12, on number line we start at 15 and
move 3 steps back or left we get 12.

Question 4.
Big Ideas Math Solutions Grade 1 Chapter 5 Subtract Numbers within 20 102
____8______ = 16 – 8
Answer:
Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20-65
8 = 16 – 8

Explanation:
We subtract 16 – 8 = 8, on number line we start at 16 and
move 8 steps back or left we get 8.

Use Addition to Subtract within 20 Homework & Practice 5.2

Question 5.
Big Ideas Math Solutions Grade 1 Chapter 5 Subtract Numbers within 20 102
11 – 9 = ?
Think 9 + ___2____ = 11
So, 11 – 9 = ____2_____ .
Answer:
Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20-66

9 + 2 = 11
11 – 9 = 2

Explanation:
First we move from 9 on number line and start moving till
11 we count how many units right we have moved we get
2 units, so 9 + 2 = 11 or we start at 11 on number line and
move 9 steps back or left we get 2, 11 – 9 = 2.

Question 6.
Big Ideas Math Solutions Grade 1 Chapter 5 Subtract Numbers within 20 102
13 – 8 = ?
Think 8 + __5_____ = 13
So, 13 – 8 = ____5_____ .
Answer:
Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20-67
8 + 5 = 13
13 – 8 = 5

Explanation:
First we move from 8 on number line and start moving till
13 we count how many units right we have moved we get
5 units, so 8 + 5 = 13 or we start at 13 on number line and
move 8 steps back or left we get 5 so 13 – 8 = 5.

Question 7.
Big Ideas Math Solutions Grade 1 Chapter 5 Subtract Numbers within 20 102
____13_____ = 20 – 7
Answer:
Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20-68
13 + 7 = 20
20 – 7 = 13

Explanation:
First we move from 13 on number line and start moving till
20 we count how many units right we have moved we get
7 units, so 13 + 7 = 20 or we start at 20 on number line and
move 7 steps back or left we get 13 so 20 – 7 = 13.

Question 8.
Big Ideas Math Solutions Grade 1 Chapter 5 Subtract Numbers within 20 102
____9____ = 12 – 3
Answer:
Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20-69
9 + 3 = 12
9 = 12 – 3

Explanation:
First we move from 9 on number line and start moving till
12 we count how many units right we have moved we get
3 units so 9 + 3 = 12 or we start at 12 on number line and
move 3 steps back or left we get 9 so 12 – 3 = 9.

Subtract 9 Homework & Practice 5.3

Get to 10 to subtract

Question 9.
15 – 9 = ?
Big Ideas Math Solutions Grade 1 Chapter 5 Subtract Numbers within 20 103
15 – ____5____ = 10
10 – _____4_____ = ____6_____
So, 15 – 9 = ___6_____
Answer:
15 – 5 = 10
10 – 4 = 6
So, 15 – 9 = 6

Explanation:
We have 2 counters in counter one we have 10 circles
and in the next counter we have 5 circles so in total we have 15,
First we see out of 15 if we consider counter-2 we have 5 strike outs
means 15 – 5 = 10 we got 10 in counter-1 and if we see first counter
we have 4 strike outs 10 – 4 = 6, So total number of circles strike out
from 2-counters are 5 + 4 = 9, So making equation as 15 – 9 = 6.

Question 10.
17 – 9 =?
Big Ideas Math Solutions Grade 1 Chapter 5 Subtract Numbers within 20 104
17 – ___7_____ = 10
10 – _____2_____ = ____8_____
So, 17 – 9 = ____8____
Answer:
17 – 7 = 10
10 – 2 = 8
So, 17 – 9 = 8

Explanation:
We have 2 counters in counter one we have 10 circles
and in next counter we have 7 circles so in total we have 17,
First we see out of 17 if we consider counter-2 we have 7 strike outs
means 17 – 7 = 10 we got 10 in counter-1 and if we see first counter
we have 2 strike outs 10 – 2 = 8, So total number of circles strike out
from 2-counters are 7 + 2 = 9, So making equation as 17 – 9 = 8.

Get to 10 to subtract Homework & Practice 5.4

Get to 10 to subtract

Question 11.
12 – 7 = ?
Big Ideas Math Solutions Grade 1 Chapter 5 Subtract Numbers within 20 105
12 – ____2_____ = 10
10 – ____5______ = ____5_____
So, 12 – 7 = _____5_____ .
Answer:
12 – 2 = 10
10 – 5 = 5
So, 12 – 7 = 5.

Explanation:
We have 2 counters in counter one we have 10 circles
and in the next counter, we have 2 circles so in total we have 12,
First, we see out of 12 if we consider counter-2 we have 2 strike outs
means 12 – 2 = 10 we got 10 in counter-1 and if we see first counter
we have 5 strike outs 10 – 5 = 5, So the total number of circles strike out
from 2-counters are 5 + 2 = 7, So making equation as 12 – 7 = 5.

Question 12.
17 – 8 = ?
17 – ____7____ = 10
10 – ____1_____ = ___9_____
So, 17 – 8 = ___9____ .
Answer:
Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20-70
17 – 7 = 10
10 – 1 = 9
So, 17 – 8 = 9

Explanation:
We have 2 counters in counter one we have 10 circles
and in the next counter, we have 7 circles so in total we have 17,
First, we see out of 17 if we consider counter-2 we have 7 strike outs
means 17 – 7 = 10 we got 10 in counter-1 and if we see the first counter
we have 1 strike outs 10 – 1 = 9, So the total number of circles strike out
from 2-counters are 7+ 1 = 8, So making equation as 17 – 8 = 9.

More True or False Equations Homework & Practice 5.5

Is the equation true or false?

Question 13.
Big Ideas Math Solutions Grade 1 Chapter 5 Subtract Numbers within 20 106
Answer:
Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20-71

8 + 3 + 8  = 13 + 6
19 = 19 ,
So True.

Explanation:
We take 8 + 3 + 8 means adding 8 to 3 and
again adding 8 we get 8 + 3 + 8 = 19
Now 13 + 6 is 19 so both sides are equal therefore
8 + 3 +8 = 13 + 6 is True.

Question 14.
Big Ideas Math Solutions Grade 1 Chapter 5 Subtract Numbers within 20 107
Answer:
Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20-72
16 – 8 = 6 + 2
8 = 8
So True.

Explanation:
We take 16 – 8 means subtracting 8 from 16
we get 16 – 8 = 8,
Now 6 + 2 is 8 so both sides are equal therefore
16 – 6  = 6 + 2  is True.

Question 15.
Big Ideas Math Solutions Grade 1 Chapter 5 Subtract Numbers within 20 108
Answer:
Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20-73

14 – 7 = 7 ≠ 12 – 9 = 3
7 ≠ 3
So False

Explanation:
We take 14 – 7 means subtracting 7 from 14 we get 14 – 7 = 7
and 12 -9 means subtracting  9 from 12 we get 12 – 9 = 3,
If we see both sides they are not equal as 7 ≠ 3,
So 14 – 7 = 12 -9 is False.

Make True Equations Homework & Practice 5.6

Question 16.
11 – 7 = 10 – ?
___4____ = 10 – ?
___4____ = 10 – ___6_____
So, 11 – 7 = 10 – ___6_____ .
Answer:

11 – 7 = 10 – ____
___4____ = 10 – ___
___4____ = 10 – ___6_____
So, 11 – 7 = 10 – ___6_____ .

Explanation :
Given 11 – 7 = 10 –  ___, First we see 11 – 7 = 4 means
10 – ___ = 4 So let us take missing number as X ,
10 – X = 4 therefore X = 10 – 4 = 6 making true equation as
11 – 7 = 10 – 6.

Question 17.
2 + 0 + 6 = ? – 5
____8____ = ? – 5
____8____ = ___13___ – 5
So, 2 + 0 + 6 = ____13____ – 5
Answer:
2 + 0 + 6 = ____ – 5
____8____ = ___ – 5
____8____ = ___13___ – 5
So, 2 + 0 + 6 = ____13____ – 5

Explanation :
Given 2 + 0 + 6 = ___- 5 , First we see 2 + 0 + 6 = 8 means
___ – 5 = 8 So let us take missing number as X ,
X – 5 = 8 therefore X = 8 + 5 = 13 making true equation as
2 + 0 + 6 = 13 – 5.

Problem Solving: Subtraction within 20 Homework & Practice 5.7

Question 18.
There are 13 people on a train. Some of them exit. There are 5 left. How many people exit the train?
Big Ideas Math Solutions Grade 1 Chapter 5 Subtract Numbers within 20 109
Circle what you know.
Underline what you need to find.
Solve:
Big Ideas Math Solutions Grade 1 Chapter 5 Subtract Numbers within 20 110

_____13_____ ○- ____8______ = _____5______, ____8_____ people
Answer:
Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20-74
13 – 8 = 5
8 people exit the train.

Explanation:
Given there are 13 people on a train, some of them exit and
there are 5 left. On number line we start at 13 and move 5 units
left or back so we reach at 8 means 8 people exit the train,
As we know 13 people are on the train and 5 left we circle
in the equation and we need to know how many exit the train,
so we underline 8 in the equation  13 – 8 =5 above.

Question 19.
A group of students are at a museum. 8 of them leave. There are 7 left. How many students were there to start ?
Big Ideas Math Solutions Grade 1 Chapter 5 Subtract Numbers within 20 111

_____15_______ Students
Answer:
Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20-75
8 + 7 = 15 or 15 – 8 = 7
There were 15 students at the start.

Explanation:
Given a group of students are at a museum, 8 of them leave,
There are 7 left. We start from 8 and move 7 units rights we get 15,
8 + 7 = 15, So there are 15 students at the start.

Question 20.
You and a friend play basketball. Your friend scores 17 points. You score 8 fewer. How many points do you score?
Big Ideas Math Solutions Grade 1 Chapter 5 Subtract Numbers within 20 112

_____9_______ points
Answer:
Big Ideas Math Answers Grade 1 Chapter 5 Subtract Numbers within 20-76
17 – 8 = 9
I scored 9 points.

Explanation:
Given I and a friend play basketball my friend scores 17 points
and I score 8 fewer. On the number line, I start from 17 and move
8 steps left I reach at 9 means 17 – 8 = 9, so I scored 9 points.

Conclusion:

Students can find Big Ideas Math Solution Keys for all the chapters at one place. All the Concepts in the Big Ideas Math Answer are prepared by the subject experts. Resolve all your queries and assess your preparation standard using the Common Core Big Ideas Math Grade 1 Answer Key.

Big Ideas Math Answers Grade 1 Chapter 13 Two – and Three – Dimensional Shapes

Big Ideas Math Answers Grade 1 Chapter 13

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Big Ideas Math Book 1st Grade Answer Key Chapter 13 Two – and Three – Dimensional Shapes

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Vocabulary

Lesson: 1 Sort Two-Dimensional Shapes

Lesson: 2 Describe Two-Dimensional Shapes

Lesson: 3 Combine Two-Dimensional Shapes

Lesson: 4 Create More Shapes

Lesson: 5 Take Apart Two-Dimensional Shapes

Lesson: 6 Sort Three-Dimensional Shapes

Lesson: 7 Describe Three-Dimensional Shapes

Lesson: 8 Combine Three-Dimensional Shapes

Lesson: 9 Take Apart Three-Dimensional Shapes

Chapter – 13: Two – and Three – Dimensional Shapes

Two – and Three – Dimensional Shapes Vocabulary

Organize It

Review Words:
hexagon
square

Use the review words to complete the graphic organizer.
Big Ideas Math Answer Key Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 1
Answer:
Big Ideas Math Book 1st Grade Answer Key Chapter 13 Two - and Three - Dimensional Shapes-1
Explanation:
Above given figures are
Square : A Square is a flat shape with 4 equal sides and every angle is a right angle (90°).
Hexagon : A regular hexagon is a closed shape polygon which has six equal sides and six equal angles. In case of any regular polygon, all its sides and angles are equal.
Define It

Use your vocabulary cards to identify the words. Find each word in the word search.
Big Ideas Math Answer Key Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 2
Answer:
Big Ideas Math Book 1st Grade Answer Key Chapter 13 Two - and Three - Dimensional Shapes-2

Explanation:
Cone : A cone is a three- dimensional geometric shape that tapers smoothly from a flat base (frequently, though not necessarily, circular) to a point called the apex or vertex. In the above figure, the arrow is pointing towards the apex or vertex.
Cube: A line segment formed where two edges meet. A cube has 12 edges. Because all faces are squares and congruent to each other, all 12 edges are the same length. In the above figure, the arrow is pointing towards edge.
Triangle: In geometry, a triangle is a closed two- dimensional shape with three straight sides. A triangle is also a polygon. In the above figure the arrow is pointing towards side.

Lesson 13.1 Sort Two-Dimensional Shapes

Explore and Grow

Sort the Shape Sort Cards. Explain how you sorted.
Answer:
Big Ideas Math Book 1st Grade Answer Key Chapter 13 Two - and Three - Dimensional Shapes-3
Explanation:
Sort the Shape Sort Cards means we need some shapes, we take print out shapes(in color) and then cut them out, or print shapes(black and white) on different colors of paper or card, cut them out, then mix them up. Or we might like to color them by our self.  Now we have our shapes, let’s sort them by color. In the above figure we are given with shapes and shapes in color now we sort shapes according to colors as shown above.

Show and Grow

Question 1.
Circle the closed shapes with 4 vertices.
Big Ideas Math Answer Key Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 3
Answer:
Big Ideas Math Book 1st Grade Answer Key Chapter 13 Two - and Three - Dimensional Shapes-4
Explanation:
In the above given figures we have closed shapes with 4 vertices, so I have circled them. 2 are rectangles, one without color, one with blue color, one is square with no color, One is rhombus with blue in color these are closed shapes with 4 vertices.

Question 2.
Circle the closed shapes with no straight sides.
Big Ideas Math Answer Key Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 4
Answer:
Big Ideas Math Book 1st Grade Answer Key Chapter 13 Two - and Three - Dimensional Shapes-5
Explanation:
In the above given figures we have closed shapes with no straight sides so I have circled them.
3 are circles, one without color, one with green color small in shape, One more circle in green color with little big in shape, these are closed shapes with no straight sides.

Apply and Grow: Practice

Question 3.
Circle the closed shapes with only 3 vertices.
Big Ideas Math Answer Key Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 5
Answer:
Big Ideas Math Book 1st Grade Answer Key Chapter 13 Two - and Three - Dimensional Shapes-6
Explanation:
In the above given figures we have closed shapes with 3 vertices, so I have circled them. 3 are triangles, one without color, one with green color and one shape is right angled triangle these are closed shapes with 3 vertices.

Question 4.
Circle the closed shapes with only L-shaped vertices.
Big Ideas Math Answer Key Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 6
Answer:
Big Ideas Math Book 1st Grade Answer Key Chapter 13 Two - and Three - Dimensional Shapes-7
Explanation:
In the above given figures we have closed shapes with only L-shaped vertices, so I have circled them. one rectangle without color, one square with purple color and one shape is right angled triangle with purple color these are closed shapes with only L-shaped vertices.

Question 5.
Circle the shapes with more than 4 straight sides.
Big Ideas Math Answer Key Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 7
Answer:
Big Ideas Math Book 1st Grade Answer Key Chapter 13 Two - and Three - Dimensional Shapes-8
Explanation:
In the above given figures we have shapes with more than 4 straight sides, so I have circled them. one rectangle without color, one trapezium with no color
and one square with orange color are shapes with more than 4 straight sides.

Question 6.
Circle the shapes with 6 straight sides.
Big Ideas Math Answer Key Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 8
Answer:
Big Ideas Math Book 1st Grade Answer Key Chapter 13 Two - and Three - Dimensional Shapes-9
Explanation:
In the above given figures we have shapes with 6 straight sides, so I have circled them. one little big hexagon without color  and one small hexagon with blue color are shapes with 6 straight sides.

Question 7.
DIG DEEPER!
Draw 2 different two-dimensional shapes that have only 4 straight sides.
Answer:
Big Ideas Math Book 1st Grade Answer Key Chapter 13 Two - and Three - Dimensional Shapes-10

Explanation:
I have drawn 2 different two-dimensional shapes that have only 4 straight sides are square shape with no color and rectangle shape in green color both are two-dimensional shapes with only 4 straight sides.

Think and Grow: Modeling Real Life

Use the clues to color the picture.
Only 3 straight sides: blue
Only 4 straight sides: green
No straight sides: yellow
More than 4 vertices: red
Big Ideas Math Answer Key Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 9
Answer:
Big Ideas Math Book 1st Grade Answer Key Chapter 13 Two - and Three - Dimensional Shapes-11

Explanation:
I have used the given clues to color the picture.
Only 3 straight sides triangles with blue,
Only 4 straight sides square and rhombus with green,
No straight sides circles with yellow,
More than 4 vertices hexagon with red.

Show and Grow

Question 8.
Use the clues to color the picture.
Only 3 vertices: green
All L-shaped vertices: orange
Only 4 straight sides and
no L-shaped vertices: blue
6 straight sides: yellow
Big Ideas Math Answer Key Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 10
Answer:
Big Ideas Math Book 1st Grade Answer Key Chapter 13 Two - and Three - Dimensional Shapes-12

Explanation:
I have used the given clues to color the picture.
Only 3 vertices triangles with green,
All L-shaped vertices rectangle with orange,
Only 4 straight sides rhombus and
no L-shaped vertices with blue,
6 straight sides hexagon with yellow.

Sort Two-Dimensional Shapes Practice 13.1

Question 1.
Circle the closed shapes with no straight sides.
Big Ideas Math Answer Key Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 11
Answer:
Big Ideas Math Book 1st Grade Answer Key Chapter 13 Two - and Three - Dimensional Shapes-13
Explanation:
In the above given figures we have shapes with no straight sides, so I have circled them. one small circle with blue color, one little big circle with no color and one curve are shapes with no straight sides.

Question 2.
Circle the closed shapes with 4 sides of the same length.
Big Ideas Math Answer Key Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 12
Answer:
Big Ideas Math Book 1st Grade Answer Key Chapter 13 Two - and Three - Dimensional Shapes-14Big Ideas Math Book 1st Grade Answer Key Chapter 13 Two - and Three - Dimensional Shapes-14
Explanation:
In the above given figures we have shapes with 4 sides of the same length, so I have circled them. one small rhombus with no color, one little big rhombus with green color and one square with no color are shapes with 4 sides of the same length.

Question 3.
Circle the shapes with no vertices.
Big Ideas Math Answer Key Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 13
Answer:
Big Ideas Math Book 1st Grade Answer Key Chapter 13 Two - and Three - Dimensional Shapes-15
Explanation:
In the above given figures we have shapes with no vertices, so I have circled them. one small circle with no color, one little big circle with red color and one big circle with red color are shapes with vertices.

Question 4.
Circle the shapes with more than 4 vertices.
Big Ideas Math Answer Key Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 14
Answer:
Big Ideas Math Book 1st Grade Answer Key Chapter 13 Two - and Three - Dimensional Shapes-16
Explanation:
In the above given figures we have shapes with more than 4 vertices, so I have circled them.
One is Pentagon with no color, one hexagon with orange color are shapes with more than 4 vertices.

Question 5.
DIG DEEPER!
Draw 2 different two-dimensional shapes with
2 long straight sides and 2 short straight sides.
Answer:
Big Ideas Math Book 1st Grade Answer Key Chapter 13 Two - and Three - Dimensional Shapes-17
Explanation:
In the above figures we have drawn 2 different two-dimensional shapes one triangle  shape with 2 long straight sides and one rectangle shape  with 2 short straight sides.

Question 6.
Modeling Real Life
Use the clues to color the picture.
Only 3 vertices: yellow
Only 4 sides: black
Only 1 straight side: orange
No straight sides: blue
Big Ideas Math Answer Key Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 15
Answer:
Big Ideas Math Book 1st Grade Answer Key Chapter 13 Two - and Three - Dimensional Shapes-18
Explanation:
Using the clues to colored the picture,
Only 3 vertices triangle with yellow,
Only 4 sides square with black,
Only 1 straight side semi circles with orange,
and no straight sides circles with blue.

Review & Refresh

Question 7.
Circle the longer object.
Big Ideas Math Answer Key Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 16
Answer:
Big Ideas Math Book 1st Grade Answer Key Chapter 13 Two - and Three - Dimensional Shapes-19
Explanation:
In the given images tennis bat is longer than the color brush so circled it with circle.

Lesson 13.2 Describe Two-Dimensional Shapes

Explore and Grow

Which shape has three sides?
Big Ideas Math Answers 1st Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 17

Which shapes have 4 sides and 4 L-shaped vertices?
Big Ideas Math Answers 1st Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 18

Which shapes have 4 sides and no L-shaped vertices?
Big Ideas Math Answers 1st Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 19
Use your materials to build each shape you circled.
Answer:
Big Ideas Math Book 1st Grade Answer Key Chapter 13 Two - and Three - Dimensional Shapes-20
Big Ideas Math Book 1st Grade Answer Key Chapter 13 Two - and Three - Dimensional Shapes-21
Big Ideas Math Book 1st Grade Answer Key Chapter 13 Two - and Three - Dimensional Shapes-22

Explanation:
Triangle shape has three sides,
Rectangle and Square shapes have 4 sides as 1,2,3,4 and
4 L-shaped vertices as a, b, c, d.
1 Trapezium, 2 rhombus are shapes have 4 sides as 1,2,3,4 and
no L-shaped vertices pointed them.

Show and Grow

Question 1.
Big Ideas Math Answers 1st Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 20
Answer:
Big Ideas Math Book 1st Grade Answer Key Chapter 13 Two - and Three - Dimensional Shapes-23
Explanation:
Given Hexagon shape which has 6 straight sides as 1,2,3,4,5,6 and 6 vertices as a, b, c, d ,e, f.

Question 2.
Big Ideas Math Answers 1st Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 20
Answer:
Big Ideas Math Book 1st Grade Answer Key Chapter 13 Two - and Three - Dimensional Shapes-24
Explanation:
Given Hexagon shape which has 6 straight sides as 1,2,3,4,5,6 and 6 vertices as a, b, c, d ,e, f.

Question 3.
Big Ideas Math Answers 1st Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 22
Answer:
Big Ideas Math Book 1st Grade Answer Key Chapter 13 Two - and Three - Dimensional Shapes-25
Explanation:
Given right angled triangle has 3 straight sides as
1,2,3 and 3 vertices as a, b, c.

Question 4.
Big Ideas Math Answers 1st Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 23
Answer:
Big Ideas Math Book 1st Grade Answer Key Chapter 13 Two - and Three - Dimensional Shapes-26
Explanation:
Given rhombus has 4 straight sides as 1,2,3,4 and
4 vertices as a, b, c, d.

Apply and Grow: Practice

Question 5.
Big Ideas Math Answers 1st Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 24
Answer:
Big Ideas Math Book 1st Grade Answer Key Chapter 13 Two - and Three - Dimensional Shapes-27
Explanation:
Given rhombus has 4 straight sides as 1,2,3,4 and
4 vertices as a, b, c, d.

Question 6.
Big Ideas Math Answers 1st Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 25
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-26
Explanation:
Given right angled triangle has 3 straight sides 1,2,3 and 3 vertices as a, b, c.

Question 7.
Big Ideas Math Answers 1st Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 26
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-27

Question 8.
Big Ideas Math Answers 1st Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 27
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-28
Explanation:
Given Hexagon shape which has 6 straight sides as 1,2,3,4,5,6 and 6 vertices as a, b, c, d ,e, f.

Circle the attributes of the shape.

Question 9.
Big Ideas Math Answers 1st Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 28
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-29

Explanation:
Triangle has 3 straight sides and 3 vertices so circled the attributes of the shape.

Question 10.
Big Ideas Math Answers 1st Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 29
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-30
Explanation:
Square has 4 straight sides of the same length , 4 vertices and its closed shape so circled the attributes of the shape.

Question 11.
MP Precision
Match each shape with an attribute that describes it.
Circle                           Rectangle                   Hexagon
6 straight sides           0 vertices                    only 4 vertices
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-31
Explanation:
We know Circle has 0 vertices,
Rectangle has only 4 vertices and
Hexagon has 6 straight sides so matched
according to their attributes that describe shape about them.

Think and Grow: Modeling Real Life

A board game has ‘4 sides and 4 L-shaped vertices.
Name and draw two shapes for the board game.
Big Ideas Math Answers 1st Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 30
Circle: Square        Hexagon        Trapezoid      Rectangle
Draw shapes:
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-32
Explanation :
Circled Square and Rectangle.
Given a board game has ‘4 sides and 4 L-shaped vertices.
we can draw two shapes for the board game as
Square and Rectangle
with Square shape of 4 sides as 1,2,3,4 and
4 L-shaped vertices as a, b, c, d
and Rectangle shape of 4 sides as 1,2,3,4 and
4 L-shaped vertices as a, b, c, d.

Show and Grow

Question 12.
A board game has 4 sides and no L-shaped vertices.
Name and draw two shapes for the board game.
Circle: Triangle      Trapezoid       Rhombus      Square
Draw shapes:
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-33
Explanation:
Given a board game has 4 sides and no L-shaped vertices.
we can draw two shapes for the board game as
Trapezoid and Rhombus with as Trapezoid shape has 4 sides
as 1,2,3,4 and no L-shaped vertices
and Rhombus shape of 4 sides as 1,2,3,4 and
no L-shaped vertices.

Describe Two-Dimensional Shapes Practice 13.2

Question 1.
Big Ideas Math Answers 1st Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 31
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-34
Explanation:
Given Hexagon shape which has 6 straight sides as 1,2,3,4,5,6 and 6 vertices as a, b, c, d ,e, f.

Question 2.
Big Ideas Math Answers 1st Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 32
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-36
Explanation:
Given rhombus has 4 straight sides as 1,2,3,4 and 4 vertices as a, b, c, d.

Question 3.
Big Ideas Math Answers 1st Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 33
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-37
Explanation:
Given circle has no straight sides and no vertices.

Question 4.
Big Ideas Math Answers 1st Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 34
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-38
Explanation:
Given triangle has 3 straight sides and 3 vertices.

Circle the attributes of the shape.

Question 5.
Big Ideas Math Answers 1st Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 35
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-39
Explanation:
Trapezoid has 4 straight sides and its closed shape so circled the attributes of the Trapezoid shape.

Question 6.
Big Ideas Math Answers 1st Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 36
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-40
Explanation:
Rectangle has 4 straight sides
and 4 vertices circled the attributes of the  Rectangle shape.

Question 7.
MP Precision
Match each shape with an attribute that describes it.
Triangle                             Trapezoid               Circle
only 3 straight sides          0 straight sides       4 vertices
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-41
Explanation:
We know Triangle has only 3 straight sides,
Trapezoid has 4 vertices and
Circle has 0 straight sides, so matched
according to their attributes that describe shape about them.

Question 8.
Modeling Real Life
A photograph has 4 straight sides of the same length and 4 vertices. Draw and name two possible shapes for the photograph.
Big Ideas Math Answers 1st Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 37

_Square___ , __Rhombus__
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-42
Explanation:
Given a photograph has 4 straight sides of the same length and 4 vertices.
So two possible shapes for the photograph are one is Square and one
is Rhombus both has 4 straight sides of the same length and 4 vertices.
So drawn as shown in the figure above.

Review & Refresh

Question 9.
MP Reasoning
Which sentences are correct?
Big Ideas Math Answers 1st Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 38
There are 5 ponies.
There are 7 deer.
There are more goats than ponies.
The numbers of deer and goats are the same.
Answer:
Sentences  2. There are 7 deer and
3. There are more goats than ponies are correct.

Explanation:
Given 4 sentences as
There are 5 ponies.
There are 7 deer.
There are more goats than ponies.
The numbers of deer and goats are the same.
1.There are 5 ponies. which is incorrect as there are only 4 ponies,
2. There are 7 deer yes correct as there are 7 deer,
3. There are more goats than ponies correct because 8 goats are there
and 4 ponies so goats are more,
4. The numbers of deer and goats are the same is incorrect
as number of deer are 7 and goats are 8 both are not the same,
Therefore Sentences  2. There are 7 deer and
3. There are more goats than ponies are correct.

Lesson 13.3 Combine Two-Dimensional Shapes

Explore and Grow

Use 2 triangles to make a new two-dimensional shape. Draw your shape.
Big Ideas Math Answers Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 39
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-43
Explanation:
Using 2 triangles to make a new two-dimensional shape
We got Rhombus shape.

Use 3 triangles to make a new two-dimensional shape. Draw your shape.
Big Ideas Math Answers Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 40
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-44
Explanation:
Using 3 triangles to make a new two-dimensional shape
We got Trapezoid shape.

Show and Grow

Question 1.
Big Ideas Math Answers Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 41
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-45
Explanation:
2 Trapezoids make a Hexagon.

Question 2.
Big Ideas Math Answers Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 42
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-46
Explanation:
3 Rhombus make a Hexagon.

Question 3.
Big Ideas Math Answers Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 43
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-47
Explanation:
2 Semi Circles make a one Circle.

Question 4.
Big Ideas Math Answers Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 44
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-48
Explanation:
2 Right Angled Triangles make a Rectangle.

Apply and Grow: Practice

Question 5.
Big Ideas Math Answers Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 45
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-49

Explanation:
3 Triangles make a Trapezoid.

Question 6.
Big Ideas Math Answers Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 46
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-50

Explanation:
2 Squares make a Rectangle.

Question 7.
Draw the shape you can use 2 times to make a Big Ideas Math Answers Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 47.
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-51
Explanation:
We take 2 times trapezoid to make a hexagon as
shown in the figure above.

Question 8.
Draw the shape you can use 3 times to make a Big Ideas Math Answers Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 47.
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-52

Explanation:
We take 3 times rhombus to make a hexagon as
shown in the figure above.

Question 9.
MP Choose Tools
Which shape can you use 2 times to make a Big Ideas Math Answers Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 48 ?
Big Ideas Math Answers Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 49
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-53
Explanation:
Triangle shape can be used 2 times to make a Rhombus
as shown in the figure.

Question 10.
DIG DEEPER!
Draw to show 2 different ways you can use pattern blocks to make the shape.
Big Ideas Math Answers Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 50
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-54

Explanation:
We can use 2 different ways pattern blocks to make the shape
one is of triangle and another is square.

Think and Grow: Modeling Real Life

Use the number of pattern blocks to fill the shape on the sign. How many of each block do you use? Draw to show your work.
Big Ideas Math Answers Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 51
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-55

Explanation:
First time we use 3  blocks of trapezoid pattern blocks to fill the shape on the sign, second time we use 4 blocks in that 3 blocks of triangles and 1 block of hexagon to fill the shape on the sign.

Show and Grow

Question 11.
Use 3 pattern blocks to fill the shape on the sign. How many of each block do you use? Draw to show your work.
Big Ideas Math Answers Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 52
Big Ideas Math Answers Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 53
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-56
Explanation:
We use 3 pattern blocks to fill the shape on the sign.
Each block of triangle, one trapezoid and one rhombus to fill the shape on the sign.

Combine Two-Dimensional Shapes Practice 13.3

Question 1.
Big Ideas Math Answers Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 54
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-57
Explanation:
2 triangles make a rhombus as shown in the figure.

Question 2.
Big Ideas Math Answers Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 55
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-58
Explanation:
As shown above in the figure 2 rhombus fill the shape of
the given sign.

Question 3.
Big Ideas Math Answers Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 56
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-59
Explanation:
6 triangles make a hexagon as shown in the figure.

Question 4.
Big Ideas Math Answers Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 57
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-60
Explanation:
2 quadrant makes a half semi circle as shown
in the figure above.

MP Choose Tools
Which 2 pattern blocks can you use to make the shape?
Question 5.
Big Ideas Math Answers Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 58
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-61
Explanation:
We use one rhombus and one triangle blocks to make
the trapezoid shape as shown in the figure.

Question 6.
Big Ideas Math Answers Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 59
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-62
Explanation:
We use one rhombus and one triangle blocks to make
the triangle shape as shown in the figure.

Question 7.
Modeling Real Life
Use 5 pattern blocks to fill the shape on the sign. How many of each block do you use? Draw to show your work.
Big Ideas Math Answers Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 60
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-63
Explanation:
We have used 5 pattern blocks to fill the shape on the sign and 4 blocks of triangles,1 block of square, 2 trapezoids, 1 rhombus and 1 hexagon to complete the Newton’s house.

Review & Refresh

Write the name

Question 8.
Big Ideas Math Answers Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 61
______6____: ____00______ :__00________
Answer:
6:00:00

Explanation:
The clock is showing 6 Hours, Zero Minutes and Zero Seconds.

Question 9.
Big Ideas Math Answers Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 62
___8_______: ___30_______: ___00_______
Answer:
8:30:00

Explanation:
The clock is showing 8 Hours, Thirty Minutes and Zero Seconds.

Lesson 13.4 Create More Shapes

Explore and Grow

Use two or more shapes to make the center of the flower. Use more shapes to fill in the rest of the flower.
Big Ideas Math Solutions Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 63
Big Ideas Math Solutions Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 64
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-64
Explanation:
Used trapezoid, rhombus and triangle shapes to make the center of the flower
and used more shapes to fill in the rest of the flower.

Show and Grow

Question 1.
Use Big Ideas Math Solutions Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 65 to make a Big Ideas Math Solutions Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 66. Draw to show your work.
Big Ideas Math Solutions Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 67
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-65
Explanation:
4 quadrants makes a circle as shown in the figure,
Step 1: Take one quadrant place it,
Step 2 : Take one more quadrant and place it makes semi circle,
Step 3 : Place one more quadrant as shown,
Step 4 : Now placing one more quadrant
and joining all together makes a full circle as shown.

Apply and Grow: Practice

Question 2.
Use Big Ideas Math Solutions Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 68 to make a Big Ideas Math Solutions Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 69. Draw to show your work.
Big Ideas Math Solutions Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 70
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-66
Explanation:
4 right angled triangles makes rectangle as shown in the figure,
Step 1: Take one right angled triangle place it,
Step 2 : Take one more right angled triangle and place it
as shown it becomes square
Step 3 : Place one right angled triangle as shown,
Step 4 : Now place one more right angled triangle as shown
now joining all together makes a rectangle full as shown.

Question 3.
Use Big Ideas Math Solutions Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 71 and Big Ideas Math Solutions Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 72 to make a Big Ideas Math Solutions Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 73. Draw to show your work.
Big Ideas Math Solutions Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 74
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-67
Explanation:
4 triangles makes rhombus as shown in the figure,
Step 1: Take one triangle place it,
Step 2 : Take one more triangle and place it
as shown it becomes rhombus
Step 3 : Place one triangle as shown, it becomes trapezoid
Step 4 : Now place one more triangle as shown
now joining all together makes a rhombus full as shown.

Question 4.
Draw the shape you can use 3 times to make a Big Ideas Math Solutions Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 75.
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-68
Explanation:
We take 3 quadrants to make a shape as shown in the figure,
Step 1: Take one quadrant place it,
Step 2 : Take one more quadrant and place it makes semi circle,
Step 3 : Place one more quadrant as shown,
and joining all together makes a shape as required.

Question 5.
Draw the shape you can use 4 times to make a Big Ideas Math Solutions Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 76.
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-68
Explanation:
We can use triangle 4 times to make a Triangle as
shown in four steps above.
Step 1: Take one triangle place it,
Step 2 : Take one more triangle and place it beside
the first one we get rhombus,
Step 3 : Place one more triangle beside rhombus as shown,
we get trapezoid,
Step 4: Now we place one more triangle on top of
trapezoid after joining all together makes a triangle shape.

Question 6.
DIG DEEPER!
Draw to show two different ways you con join the shapes on the left to make the larger shape.
Big Ideas Math Solutions Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 77
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-69
Explanation:
First Way
Step 1: We take rectangle as shown
Step 2: We take one right angled triangle and join
Step 3: We take one more right angled triangle and place
with another right angled triangle as shown above we get the larger shape.
Second Way:
Step 1: Here we first place right angled triangle,
Step 2 : Take another right angled triangle and place as shown
beside the previous right angled triangle,
Step 3 : Now we join with a rectangle as shown ,
we get the larger shape.

Think and Grow: Modeling Real Life

Use pattern blocks to complete the puzzle. How many of each block do you use? Draw to show your work.
Big Ideas Math Solutions Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 78
Big Ideas Math Solutions Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 79
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-70
Explanation:
As shown above used 2 triangles, 1 square,
1 trapezoid, 3 rhombus and 1 more rhombus pattern
to complete the puzzle.

Show and Grow

Question 7.
Use pattern blocks to complete the puzzle. How many of each block do you use? Draw to show your work.
Big Ideas Math Solutions Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 80
Big Ideas Math Solutions Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 81
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-71

Explanation:
As shown above used 5 triangles, 1 square,
2 rhombus, 2 trapezoid and 1 more rhombus pattern
blocks to complete the puzzle.

Create More Shapes Practice 13.4

Question 1.
Use Big Ideas Math Solutions Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 82 to make a larger Big Ideas Math Solutions Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 82. Draw to show your work.
Big Ideas Math Solutions Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 83
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-72

Explanation:
We can use square 4 times to make a Square as
shown in four steps above.
Step 1: Take one  square place it,
Step 2 : Take one more square and place it below
the first one we get rectangle,
Step 3 : Place one more square beside as shown,
Step 4: Now we place one more square as shown
after joining all together makes a square shape.

Question 2.
Use Big Ideas Math Solutions Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 84 to make a Big Ideas Math Solutions Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 85. Draw to show your work.
Big Ideas Math Solutions Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 86
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-73
Explanation:
We can use right angled triangles 4 times to make a Rhombus as
shown in four steps above.
Step 1: Take one right angled triangle place it,
Step 2 : Take one more right angled  triangle and place it beside
the first one as shown
Step 3 : Place one more right angled triangle as shown,
Step 4: Now we place one more right angled triangle
after joining all together makes a rhombus shape.

Question 3.
DIG DEEPER!
Draw to show two ways you can combine the 3 shapes on the left to make the larger shape.
Big Ideas Math Solutions Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 87
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-74
Explanation:
First Way
Step 1: We take square as shown,
Step 2: We take one rectangle and join with square,
Step 3: We take one more square and place
as shown above we get the larger shape.
Second Way:
Step 1: Here we first place rectangle,
Step 2 : Take square and place as shown,
Step 3 : Now we take one more square and join
as shown in figure, we get the larger shape.

Question 4.
Modeling Real Life
Use pattern blocks to complete the puzzle. How many of each block do you use? Draw to show your work.
Big Ideas Math Solutions Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 88
Big Ideas Math Solutions Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 89
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-75
Explanation:
As shown above used 1 triangle, 2 squares,
3 rhombus, 2 trapezoid and 1 more rhombus pattern
blocks to complete the puzzle.

Review & Refresh

Question 5.
Big Ideas Math Solutions Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 90
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-76
17 is less than 19.

Explanation:
As 17 is less than < 19,  we select less than.

Question 6.
Big Ideas Math Solutions Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 91
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-77
12 is greater than 11.

Explanation:
As 12 is greater than > 11, we select greater than

Lesson 13.5 Take Apart Two-Dimensional Shapes

Explore and Grow

Draw lines to take apart each figure.
Big Ideas Math Answer Key Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 92
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-78
Explanation:
1. We draw lines to show two rectangles as
shown in the above figure,
2.  We draw lines to show four squares as
shown in the above figure,
3. We draw lines to show two triangles and one
rectangle as shown in the above figure,
4. We draw lines to show two triangles and two
square as shown in the above figure.

Show and Grow

Draw one line to show the parts.

Question 1.
2 triangles
Big Ideas Math Answer Key Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 93
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-79
Explanation:
Drawn one line in between rectangle,
as shown we get 2 triangles.

Question 2.
2 triangles
Big Ideas Math Answer Key Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 94
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-80
Explanation:
Drawn one line in between rectangle,
as shown we get 2 triangles.

Question 3.
2 trapezoids
Big Ideas Math Answer Key Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 95
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-81
Explanation:
Drawn one line in between trapezoid,
as shown we get 2 trapezoids.

Question 4.
1 triangle and 1 trapezoid
Big Ideas Math Answer Key Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 96
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-82
Explanation:
Drawn one line in between triangle
as shown we get 1 triangle and  1 trapezoid.

Apply and Grow: Practice

Draw one line to show the parts.

Question 5.
2 triangles
Big Ideas Math Answer Key Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 97
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-83

Explanation:
Drawn one line in between triangle
as shown we get 2 triangles.

Question 6.
2 squares
Big Ideas Math Answer Key Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 98
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-84
Explanation:
Drawn one line in between rectangle
as shown we get 2 squares.

Draw two lines to show the parts.

Question 7.
2 triangles and 1 trapezoid
Big Ideas Math Answer Key Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 99
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-86
Explanation:
Drawn two lines in between rectangle
as shown we get 2 triangles and 1 trapezoid.

Question 8.
2 triangles and 1 rectangle
Big Ideas Math Answer Key Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 100
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-87

Explanation:
Drawn two lines in between hexagon
as shown we get 2 triangles and 1 rectangle.

Question 9.
MP Reasoning
Show how to use the shapes to make the hexagon.
Big Ideas Math Answer Key Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 101
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-88
Explanation:
In first hexagon shape we made 1 trapezoid and
3 triangles are made as shown in the figure.
In second hexagon shape we made 2 rhombus and
2 triangles as shown in the figure.

Question 10.
MP Reasoning
Show how to use the shapes to make a circle. How many of each shape do you use?
Big Ideas Math Answer Key Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 102
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-89
Explanation:
In the first circle we need 4 quadrants to make a circle,
In second circle we need 2 semi circles to make a circle.

Think and Grow: Modeling Real Life

How many squares can you find on the Four Square court?
Big Ideas Math Answer Key Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 103
Big Ideas Math Answer Key Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 104
_____5_______ squares
Answer:
We have 5 squares in a Four Square Court.

Explanation:
We are given with a Four Square Court,
in that we have 4 squares and the outer boundary has
one square making in total 5 squares..

Show and Grow

Question 11.
How many squares can you find on the magic square?
Big Ideas Math Answer Key Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 105
_____14_______ squares
Answer:
We have 14 squares on the magic square

Explanation:
We are given the magic square on that we have first 9 squares plus
one in the outer boundary, 4 squares of 4 blocks from the 9 squares.

Take Apart Two-Dimensional Shapes Practice 13.5

Draw one line to show the parts.

Question 1.
2 trapezoids
Big Ideas Math Answer Key Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 106
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-90

Explanation:
Drawn one line in between Hexagon,
as shown we get 2 trapezoids.

Question 2.
1 rectangle and 1 square
Big Ideas Math Answer Key Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 107
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-91

Explanation:
Drawn one line in between Rectangle,
as shown we get 1 rectangle and 1 square .

Draw two lines to show the parts.

Question 3.
3 triangles
Big Ideas Math Answer Key Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 108
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-92

Explanation:
Drawn two lines in between Trapezoid
as shown we get 3 triangles.

Question 4.
1 rectangle and 2 triangles
Big Ideas Math Answer Key Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 109
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-93

Explanation:
Drawn two lines in between Square
as shown we get 1 rectangle and 2 triangles.

Question 5.
MP Reasoning
Show how to use the shapes to make the Big Ideas Math Answer Key Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 110.
Big Ideas Math Answer Key Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 111
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-94

Explanation:
In first rhombus shape we made 1 rhombus and
2 triangles are made as shown in the figure.
In second rhombus shape we made 1 trapezoid and
1 triangle as shown in the figure.

Question 6.
MP Reasoning
Show how to use the shapes to make the hexagon. How many of each shape do you use?
Big Ideas Math Answer Key Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 112
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-95

Explanation:
In first we use 3 rhombus to make a hexagon
as shown in the figure.
In second we use 6 triangles to make a hexagon
as shown in the figure.

Question 7.
Modeling Real Life
How many triangles are in Descartes’s design?

Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-96
Explanation:
In Descartes’s design we have total 5 triangles,
4 inside the triangle and one outer boundary,
Making a total of 5 triangles in Descartes’s design.

Review & Refresh

Question 8.
Circle the three-dimensional shapes. Draw rectangles around the two-dimensional shapes.
Big Ideas Math Answer Key Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 114
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-97
Explanation:
Circled the three -dimensional shapes cube, books and sphere. Drawn rectangle around the dollar because it is a two-dimensional shape.

Lesson 13.6 Sort Three-Dimensional Shapes

Explore and Grow

Sort the Three-Dimensional Shape Cards. Explain how you sorted.
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-98

Explanation:
Sort the Shape Cards means we need some shapes,
we take print out shapes(in color) and then cut them out,
or print shapes(black and white) on different colors of paper or card,
cut them out, then mix them up. Or we might like to color them
by our self.  Now we have our three dimensional shapes, let’s sort them by
shape and name. In the above figure we are given with shape name,
now we sort shapes according to their names as shown above.

Show and Grow

Question 1.
Circle the shapes with flat surfaces that are circles.
Big Ideas Math Answer Key Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 115
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-99
Explanation:

Circled the shapes 2 cones, 3 cylinders as they have
flat surfaces that are circles.

Question 2.
Circle the shapes with both flat and curved surfaces.
Big Ideas Math Answer Key Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 116
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-100
Explanation:
Circled the shapes with both flat and curved surfaces
as 2 cones, 2 cubes , 1 cuboid and 2 cylinders.

Apply and Grow: Practice

Question 3.
Circle the shapes with 1 or more flat surfaces.
Big Ideas Math Answer Key Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 117
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-101
Explanation:
Circled the shapes with 1 or more flat surfaces,
as 2 cubes, 1 cone, 1 cuboid and 2 cylinders.

Question 4.
Circle the shapes with a curved surface.
Big Ideas Math Answer Key Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 118
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-102

Explanation:Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-103
We know Cylinder, Cone, Sphere have curved
face so circled 2 cones, 2 spheres and one cylinder
shapes as they have curved surface.

Question 5.
Circle the shapes with only 2 flat surfaces.
Big Ideas Math Answer Key Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 119
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-104

Explanation:
Circled the shapes with only 2 flat surfaces are
2 cylinders as shown in picture above.

Question 6.
MP Structure
Match each shape to its group.
Big Ideas Math Answer Key Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 120
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-105

Explanation:
Matched each shape according to its group as
Cube and Cuboid has only flat surfaces, Cone &
Cylinder have flat and curved surfaces and Sphere
have only a curved surface.

Think and Grow: Modeling Real Life

You need to find an object that has no flat surfaces for a scavenger hunt. Circle the objects you can use.
Big Ideas Math Answer Key Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 121
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-106
Explanation:
Objects that has no flat surfaces for a scavenger hunt are
Big Ball, Cool Drink Tin and Cricket Ball circled them as
shown in the figure above.

Show and Grow

Question 7.
You need to find an object that has only two flat surfaces for a scavenger hunt. Circle the objects you can use.
Big Ideas Math Answer Key Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 122
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-107
Explanation:
Object that has only two flat surfaces for a scavenger hunt are
1 Musical Drum, 1 Log and Colored Cylinder so circled them as
shown in the figure above.

Sort Three-Dimensional Shapes Practice 13.6

Question 1.
Circle the shapes with no flat surface.
Big Ideas Math Answer Key Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 123
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-108
Explanation:
Circled the shapes with no flat surface
as 3 spheres as shown in the figure.

Question 2.
Circle the shapes with flat surfaces that are rectangles.
Big Ideas Math Answer Key Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 124
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-109
Explanation:
Circled the shapes with flat surfaces that are rectangles
as 3 cuboids as shown in the figure above.

Question 3.
Circle the shapes with more than 2 flat surfaces
Big Ideas Math Answer Key Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 125
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-110
Explanation:
Circled the shapes with more than 2 flat surfaces as
2 cuboids, 1 cube and one cylinder as shown in the figure above.

Question 4.
MP Structure
Match each shape to its group.
Big Ideas Math Answer Key Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 126
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-111

Explanation:
Matched each shape according to its group as
Sphere have no flat surface, Cuboid, Cylinder and
Cube have more than 1 flat surface, Cone have only 1
flat surface.

Question 5.
Modeling Real Life
You need to find an object that has both flat and curved surfaces for a scavenger hunt. Circle the objects you can use.
Big Ideas Math Answer Key Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 127
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-112
Explanation:
Objects that has both flat and curved surfaces for a scavenger hunt
are one cuboid shaped, one cone shaped, one clay, one glass jar,
one glue stick, one tin so circled them as shown in the figure above.

Review & Refresh

Question 6.
30 + 30 = ___60__
Answer:
30 + 30 = 60

Explanation:
Given expression as 30 + 30 we add we get
result as 60.

Question 7.
60 + 20 = ____80__
Answer:
60 + 20 = 80

Explanation:
Given expression as 60 + 20 we add we get
result as 80.

Question 8.
50 + 10 = ___60____
Answer:
50 + 10 = 60

Explanation:
Given expression as 50 + 10 we add we get
result as 60.

Question 9.
30 + 40 = ____70___
Answer:
30 + 40 = 70

Explanation:
Given expression as 30 + 40 we add we get
result as 70.

Lesson 13.7 Describe Three-Dimensional Shapes

Explore and Grow

Use your materials to build one of the three dimensional shapes shown. Circle the shape you make. How many flat surfaces does your shape have? How many vertices does your shape have?

____6_____ flat surfaces

_____8_____ vertices
Big Ideas Math Answers 1st Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 128
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-113
Explanation:
We have build three dimensional shape of a Cube as shown above,
A cube has 6 flat surfaces as (HEAD, GFBC, ABFE, DCGH, HEFG, ABCD),
A cube has 8 vertices as 8 (A, B, C, D, E, F, G, H).

Show and Grow

Question 1.
Big Ideas Math Answers 1st Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 129
__6_______ flat surfaces ____8_____ vertices ____12_____ edges
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-114
Explanation:
A cuboid is a three dimensional shape has 6 flat surfaces,
8 vertices and 12 edges as shown above.

Apply and Grow: Practice

Question 2.
Big Ideas Math Answers 1st Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 130
__2______ flat surfaces _____0____ vertices ____0_____ edges
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-115
Explanation:
A cylinder is a three dimensional shape has 2 flat surfaces,
0 vertices and 0 edges as shown above.

Question 3.
Big Ideas Math Answers 1st Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 131
___6______ flat surfaces ____8_____ vertices ______12___ edges
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-116
Explanation:
A cuboid three dimensional shape  has 6 flat surfaces,
8 vertices and 12 edges as shown above.

Circle the attributes of the shape.

Question 4.
Big Ideas Math Answers 1st Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 132
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-117
Explanation:
Cone is a three dimensional shape,
A cone has 1 flat surfaces, 1 vertices so circled attributes as
1 flat surface.

Question 5.
Big Ideas Math Answers 1st Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 133
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-118
Explanation:
Cube is a three dimensional shape,
A cube has 6 flat surfaces , 8 vertices and 12 edges so
circled the attributes as 6 flat surfaces and 12 edges.

Question 6.
I am a three-dimensional shape that has no flat surfaces, no vertices, and no edges. What am I?
_Sphere.___
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-119
Explanation:
A three-dimensional shape that has no flat surfaces,
no vertices, and no edges is a Sphere.

Question 7.
I am a three-dimensional shape that has I flat surface, I vertex, and no edges. What am I?
______Cone__.
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-120
Explanation:
A three-dimensional shape that has 1 flat surface,
1 vertex and no edges is Cone.

Question 8.
DIG DEEPER!
Newton buys an item that has 2 more flat surfaces than edges. Which item does he buy?
Big Ideas Math Answers 1st Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 134
Answer:
Newton buys Dog Food which has 2 more
flat surfaces than edges.

Explanation:
As in given images we have Dog Food which is in cylinder
shape and has 2 more flat surfaces than edges So Newton
buys Dog Food.

Think and Grow: Modeling Real Life

Circle the object below the table that has 0 flat surfaces. Draw a line through the object above the basketball that has 12 edges.
Big Ideas Math Answers 1st Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 135
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-121
Explanation:
Circled the object basket ball below the table that has 0 flat surfaces
and I have drawn a line through the tissue paper box
above the basketball that has 12 edges which is in cuboid shape.

Show and Grow

Question 9.
Circle the object in front of the campers that has more than 2 flat surfaces. Draw a line through the object behind the logs that has I vertex and I flat surface.
Big Ideas Math Answers 1st Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 136
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-122

Explanation:
Circled the objects in front of the campers that has more than 2 flat surfaces are First Aid Box, Dust Bin Box and Logs , drawn a line through the object
Foot Ball behind the logs that has I vertex and I flat surface are tent, 1 light display in cone shape as shown in the image above.

Describe Three-Dimensional Shapes Practice 13.7

Question 1.
Big Ideas Math Answers 1st Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 137
____0_____ flat surfaces ____0_____ vertices ____0___ edges
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-123
0 flat surfaces, 0 vertices and 0 edges.

Explanation:
A sphere has 0 flat surfaces, 0 vertices and 0 edges.

Question 2.
Big Ideas Math Answers 1st Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 138
__6___ flat surfaces ____8___ vertices ___12__ edges
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-124
Explanation:
Cube is a three dimensional shape,
A cube has 6 flat surfaces , 8 vertices and 12 edges.

Question 3.
Circle the shape that has the same number of vertices as edges.
Big Ideas Math Answers 1st Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 139
Answer:
Cylinder
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-125
Explanation:
Cylinder has the same number of vertices as edges that
is zero. So we circled the shape cylinder.

Question 4.
Circle the shape that has the same number of faces as vertices.
Big Ideas Math Answers 1st Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 140
Answer:
Cone
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-126
Explanation:
Cone has the same number of faces as vertices as it has
1 flat surface and 1 vertices so we circled the shape cone.

Circle the attributes of the shape.

Question 5.
Big Ideas Math Answers 1st Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 141
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-127
Explanation:
Cylinder has 2 flat surfaces so circled the attributes of it.

Question 6.
Big Ideas Math Answers 1st Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 142
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-128
Explanation:
Given Rectangular Prism it is a three – dimensional shape,
with 12 edges so circled the attributes of them as shown above.

Question 7.
DIG DEEPER!
Descartes buys an item that has 2 fewer flat surfaces than vertices.
Which item does he buy?
Big Ideas Math Answers 1st Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 143
Answer:
Descartes buy CAT TREATS that has 2 fewer flat surfaces than vertices.

Explanation:
As CAT TREATS is in shape of cuboid shape,
as we know a cuboid has 6 flat surfaces and 8 vertices,
So Descartes buy CAT TREATS which has 2 fewer
flat surfaces than vertices.

Question 8.
Modeling Real Life
Circle the object next to the hat that has 6 square flat surfaces. Draw a line through the object in front of the hat that has 0 edges and 1 vertex.
Big Ideas Math Answers 1st Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 144
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-129
Explanation:
Circled the objects next to the hat that has 6 square flat surfaces
are cube object and cuboid object which have 6 flat surfaces,
The objects in front of the hat is paint bucket so drawn a line to
objects that has 0 edges and 1 vertex are traffic cones, one small cone
as shown in the picture above.

Review & Refresh

Question 9.
20 + 18 = ______38_____
Answer:
20 + 18 = 38

Explanation:
Given expression as 20 + 18 we add we get
result as 38.

Question 10.
40 + 25 = ____65______
Answer:
40 + 25 = 65

Explanation:
Given expression as 40 + 25 we add we get
result as 65.

Question 11.
36 + 69 = ___105________
Answer:
36 + 69 = 105

Explanation:
Given expression as 36 + 69 we add we get
result as 105.

Question 12.
9 + 90 = _____99______
Answer:
9 + 90 = 99

Explanation:
Given expression as 9 + 90 we add we get
result as 99.

Question 13.
18 + 70 = _____88______
Answer:
18 + 70 = 88

Explanation:
Given expression as 18 + 70  we add we get
result as 88.

Question 14.
27 + 50 = ____77_______
Answer:
27 + 50 = 77

Explanation:
Given expression as 27 + 50 we add we get
result as 77.

Lesson 13.8 Combine Three-Dimensional Shapes

Explore and Grow

Which three-dimensional shapes can you make using cubes? Build one of the shapes.
Big Ideas Math Answers Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 145
Answer:
Cuboid
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-130
Explanation:
Using cubes we came make three-dimensional shape as cuboid
as shown above.

Show and Grow

Circle the new shape that you can make.

Question 1.
Big Ideas Math Answers Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 146
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-131
Explanation:
Given shapes are cube and cone using them we get new shape
cube with cone circled as shown in the picture above.

Question 2.
Big Ideas Math Answers Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 147
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-132
Explanation:
Given shapes are cylinder and cuboid using them we get new shape
cylinder with cuboid circled as shown in the picture above.

Apply and Grow: Practice

Circle the new shape that you can make.

Question 3.
Big Ideas Math Answers Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 148
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-133
Explanation:
Given shapes are cubes using them we get new shape
as cube on cube circled as shown in the picture above.

Question 4.
Big Ideas Math Answers Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 149
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-134
Explanation:
Given shapes are cone and cuboid using them we get new shape
as cone with cuboid circled as shown in the picture above.

Question 5.
Big Ideas Math Answers Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 150
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-135
Explanation:
Given shapes are cylinder and cube using them we get new shape
as cylinder with cube circled as shown in the picture above.

Question 6.
DIG DEEPER!
How many cubes do you need in all to make the next shape?
Big Ideas Math Answers Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 151
_____9_____ cubes
Answer:
9 cubes are needed to make the next shape

Explanation:
Given in first shape it is 3 cubes, next shape it is 5 cubes,
in 3rd shape it has 7 cubes as it is increasing by two
or  plus two so the next shape will have 7 + 2 = 9 cubes.
Therefore we need 9 cubes to make the next shape.

Think and Grow: Modeling Real Life

You build a wall. It is 5 cubes long and 2 cubes tall.
Your friend builds a wall. It is 4 cubes long and 2 cubes tall.
How many more cubes do you use than your friend?
Draw Pictures:     You                     Friend

Equation:

_____1_____ more cubes
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-136
I used 1 more cube than my friend.

Explanation:
Given I build a wall with 5 cubes long and 2 cubes tall
makes in total 7 blocks and friend builds a wall which has
4 cubes long and 2 cubes tall makes in total of 6 blocks,
So I used one more cube than my friend.

Show and Grow

Question 7.
You build a wall. It is 3 cubes long and 3 cubes tall. Your friend builds a wall.
It is 5 cubes long and 3 cubes tall. How many more cubes does your friend use than you?
Draw Pictures:      You                     Friend

Equation:

____2______ more cubes
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-137
Friend used 2 more cubes than mine.

Explanation:
Given I build a wall with 3 cubes long and 3 cubes tall makes in total 6 blocks and friend builds a wall which has 5 cubes long and 3 cubes tall makes in total of 8 blocks, So my friend used two more cubes than me.

Combine Three-Dimensional Shapes Practice 13.8

Circle the new shape that you can make.

Question 1.
Big Ideas Math Answers Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 152
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-138
Explanation:
Given shapes are cone and cylinder using them we get new shape
as cone with cylinder circled as shown in the picture above.

Question 2.
Big Ideas Math Answers Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 153
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-139
Explanation:
Given shapes are 2 cylinders using them we get new shape
as cylinder with cylinder circled as shown in the picture above.

Question 3.
DIG DEEPER!
How many cubes do you need in all to make the next shape?
Big Ideas Math Answers Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 154
______13_______ cubes
Answer:
13 cubes are needed to make the next shape

Explanation:
Given in first shape it is 1 cube, next shape it is 5 cubes,
in 3rd shape it has 9 cubes as it is increasing by four
or  plus four so the next shape will have 9 + 4 = 13 cubes.
Therefore we need 13 cubes to make the next shape.

Question 4.
Modeling Real Life
You build a wall that is 2 cubes long and ‘4 cubes tall. Your friend builds a wall that is 4 cubes long and 3 cubes tall. How many more cubes does your friend use than you?

_____one________ more cubes
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-140
Friend used 1 more cube than mine.

Explanation:
Given I build a wall with 2 cubes long and 4 cubes tall makes in total 6 blocks and friend builds a wall which has 4 cubes long and 3 cubes tall makes in total of 7 blocks, So my friend used one more cube than me.

Review & Refresh

Question 5.
Order from shortest to longest.
Big Ideas Math Answers Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 155
___black______, __green_______, __yellow_______
Answer:
Order from shortest to longest is black, green, yellow

Explanation:
Given color brushes we sort according to the order as among three colors green is least small next small is green brush and the most longest is yellow so order from shortest to longest is black, green, yellow.

Lesson 13.9 Take Apart Three-Dimensional Shapes

Explore and Grow

Circle the three-dimensional shapes used to build the castle.
Big Ideas Math Solutions Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 156
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-141
Cone, Cylinder and Cuboid.

Explanation:
We circled the three-dimensional shapes cone, cylinder and
cuboid which are used to build the castle.

Show and Grow

Circle the shapes that make up the structure.

Question 1.
Big Ideas Math Solutions Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 157
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-142
One Cylinder and One Cone

Explanation:
As shown we use 1 cylinder and 1 cone three-dimensional
shapes to make up the given structure.

Question 2.
Big Ideas Math Solutions Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 158
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-143
Two Cylinders

Explanation:
As shown we use 2 cylinders three-dimensional
shape to make up the given structure.

Question 3.
Big Ideas Math Solutions Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 159
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-144
Six Cubes and One Cylinder

Explanation:
As shown we use 6 cubes and 1 cylinder three-dimensional
shapes to make up the given structure.

Question 4.
Big Ideas Math Solutions Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 160
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-145
Three Cones, One Cuboid and Two Cylinders

Explanation:
As shown we use 3 cones, 1 cuboid and 2 cylinders
three-dimensional shapes to make up the given structure.

Apply and Grow: Practice

Circle the shapes that make up the structure.

Question 5.
Big Ideas Math Solutions Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 161
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-146
Four Cubes, One Cuboid and One Cone

Explanation:
As shown we use 4 cubes, 1 cuboid and 1 cone
three-dimensional shapes to make up the given structure.

Question 6.
Big Ideas Math Solutions Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 162
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-147
Four Cubes, Two Cuboids and One Cylinder

Explanation:
As shown we use 4 cubes, 2 cuboid and 1 cylinder
three-dimensional shapes to make up the given structure.

Question 7.
Big Ideas Math Solutions Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 163
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-148
Three Cubes, Three Cylinders and One Cone

Explanation:
As shown we use 3 cubes, 3 cylinders and 1 cone
three-dimensional shapes to make up the given structure.

Question 8.
Big Ideas Math Solutions Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 164
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-149
Three Cubes, Three Cylinders, Three Cones and One Cuboid

Explanation:
As shown we use 3 cubes, 3 cylinders, 3 cones and 1 cuboid
three-dimensional shapes to make up the given structure.

Question 9.
MP Reasoning
Which two structures are the same?
Big Ideas Math Solutions Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 165
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-150
Structure 1 and structure 3 are the same

Explanation:
Structure 1 has six cubes and structure 3 has three cubes with 1 cuboid
both make the same structure, So Structure 1 and  Structure 3 are the same.

Think and Grow: Modeling Real Life

How many of each shape make up the gate?
Big Ideas Math Solutions Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 166
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-151
6 cubes, 1 rectangular prism, 2 cylinders and 2 cones.

Explanation:
The gate is made up of 6 cubes, 1 rectangular prism,
2 cylinders and 2 cones.

Show and Grow

Question 10.
How many of each shape make up the bridge?
Big Ideas Math Solutions Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 167
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-152
4 cubes and  7 rectangular prisms

Explanation:
The bridge is made up of 4 cubes and  7 rectangular prisms.

Take Apart Three-Dimensional Shapes Practice 13.9

Circle the shapes that make up the structure.

Question 1.
Big Ideas Math Solutions Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 168
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-153
Three Cubes

Explanation:
As shown we use 3 cubes three-dimensional shape
to make up the given structure.

Question 2.
Big Ideas Math Solutions Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 169
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-154
Two Cylinders and Two Cuboids

Explanation:
As shown we use 2 cylinders and 2 cuboids
three-dimensional shapes to make up the given structure.

Question 3.
Big Ideas Math Solutions Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 170
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-155
Three Cubes, Two Cylinders and Three Cones

Explanation:
As shown we use 3 cubes , 2 cylinders and 3 cones
three-dimensional shapes to make up the given structure.

Question 4.
Big Ideas Math Solutions Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 171
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-156
Two Cubes, One Cylinder, One Cuboid and One Cone

Explanation:
As shown we use 2 cubes , 1 cylinder, 1 Cuboid and 1 cone
three-dimensional shapes to make up the given structure.

Question 5.
MP Reasoning
Which two structures are the same?
Big Ideas Math Solutions Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 173
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-157
Structure 2 and structure 3 are the same

Explanation:
Structure 2 has six cubes and 1 cuboid and
structure 3 has three cuboids both make the same structure,
So Structure 2 and  Structure 3 are the same.

Question 6.
Modeling Real Life
How many of each shape make up the castle?
Big Ideas Math Solutions Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 173
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-158
6 cubes, 5 rectangular prisms, 8 cylinders and 4 cones.

Explanation:
The gate is made up of 6 cubes, 5 rectangular prisms,
8 cylinders and 4 cones.

Review & Refresh

Question 7.
12 + 7 = _____19______
Answer:
12 + 7 = 19

Explanation:
Given expression as 12 + 7 we add we get
result as 19.

Question 8.
42 + 14 = ___56________
Answer:
42 + 14 = 56

Explanation:
Given expression as 42 + 14 we add we get
result as 56.

Question 9.
25 + 32 = _____57______
Answer:
25 + 32 = 57

Explanation:
Given expression as 25 + 32 we add we get
result as 57.

Question 10.
68 + 11 = _____79______
Answer:
68 + 11 = 79

Explanation:
Given expression as 68 + 11 we add we get result as 79.

Two – and Three – Dimensional Shapes Performance Task

Question 1.
Use the clues to finish the two-dimensional sand castle drawing.
Big Ideas Math Solutions Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 174

  • The flag on the castle is a closed shape with only 3 straight sides.
  • The handle of the shovel is a closed shape with L-shaped vertices and 4 sides of the same length.
  • The window on the castle is a closed shape with only 6 straight sides.
  • The door on the castle is a closed shape with 4 sides that you can use 2 times to make a square.

Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-159
Explanation:
Used the given clues to finish the two-dimensional sand castle drawing.
* The flag on the castle is a closed shape with only 3 straight sides is
triangle,
* The handle of the shovel is a closed shape with L-shaped vertices and
4 sides of the same length is a square,
* The window on the castle is a closed shape with only 6 straight sides
is a hexagon,
* The door on the castle is a closed shape with 4 sides so used
2 times rectangle to make a square.

Question 2.
You are building a sand castle using these three-dimensional shapes.

  • 4 shapes that have square flat surfaces
  • 5 shapes that have 2 flat surfaces and no vertices
  • 3 shapes that have the same number of flat surfaces as vertices

a. Which shape is missing from the sand castle?
Big Ideas Math Solutions Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 175
Answer:
Hexagon three-dimensional shape is missing from the sand castle.

Explanation:
We can use Hexagon three-dimensional shape as missing from the sand castle. We can use for the windows on the castle as a closed shape
with only 6 straight sides.
So hexagon three-dimensional shape is missing from the sand castle.

b. Color a flat surface to show where you would stack the missing shape to
complete the sand castle.
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-160
Colored hexagon shape.

Explanation:
Colored a flat surface windows to show where I would stack the missing shape hexagon to complete the sand castle.

Two – and Three – Dimensional Shapes Chapter Practice

Sort Two-Dimensional Shapes Homework & Practice 13.1

Question 1.
Circle the closed shapes with only 3 straight sides.
Big Ideas Math Answer Key Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 176
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-161
Explanation:
In the above given figures we have closed shapes
with only 3 straight sides. so I have circled them.
2 are triangles, one triangle with orange color, one right angle triangle
with blue color these are triangles with closed shapes with only 3 straight sides.

Question 2.
MP Structure
Draw 2 different two-dimensional shapes that have 1 or more L-shaped vertices.
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-162
Square and Rectangle

Explanation:
Drawn 2 different two-dimensional shapes one is Square and
other is Rectangle that have 1 or more L-shaped vertices as shown in
the picture above.

Describe Two-Dimensional Shapes Homework & Practice 13.2

Circle the attributes of the shape.

Question 3.
Big Ideas Math Answer Key Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 177
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-163

Explanation:
Hexagon has 6 straight sides of the
same length, its closed shape so
circled the attributes of the shape.

Question 4.
Big Ideas Math Answer Key Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 178
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-164
Explanation:
Rhombus shape has 4 straight sides of the
same length and 4 vertices, so
circled the attributes of the shape.

Combine Two-Dimensional Shapes Homework & Practice 13.3

Question 5.
Big Ideas Math Answer Key Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 179
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-165

Explanation:
6 triangles make a hexagon as shown in the figure.

Question 6.
Big Ideas Math Answer Key Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 180
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-166
Explanation:
3 Triangles make a Trapezoid.

Create More Shapes Homework & Practice 13.4

Question 7.
Big Ideas Math Answer Key Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 181
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-167
Explanation:
We can use right angled triangles 4 times to make a Rectangle as
shown in four steps above.
Step 1: Take one right angled triangle place it,
Step 2 : Take one more right angled  triangle and place it beside
the first one as shown,
Step 3 : Place one more right angled triangle as shown,
Step 4: Now we place one more right angled triangle
after joining all together makes a Rectangle shape.

Question 8.
Big Ideas Math Answer Key Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 182
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-168
Explanation:
We can use 2 times of quadrants and 2 squares to make the given
shape as shown in four steps above.
Step 1: Take one quadrants place it,
Step 2 : Take one more quadrants and place it beside
the first one as shown,
Step 3 : Place one square as shown,
Step 4: Now we place one more square beside the previous one
after joining all together makes a required shape as shown.

Take Apart Two-Dimensional Shapes Homework & Practice 13.5

Draw two lines to show the parts.

Question 9.
1 square and 2 triangles
Big Ideas Math Answer Key Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 183
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-169
1 square and 2 triangles

Explanation:
Drawn one line in between rectangle
as shown we get 2 squares in one square
we draw one more line as shown we get 2 triangles
we get now 1 square and 2 triangles.

Question 10.
2 triangles and 1 square
Big Ideas Math Answer Key Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 184
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-170
2 triangles and 1 square

Explanation:
Drawn one line in between trapezoid
as shown we get 1 right angled triangle and
we draw one more line as shown we get 1 more
right angled triangle and center we get a square
we get now 2 right angled triangles and 1 square.

Sort Three-Dimensional Shapes Homework & Practice 13.6

Question 11.
Circle the shapes with flat surfaces that are all squares.
Big Ideas Math Answer Key Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 185
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-171
Explanation:
Circled the shapes with flat surfaces that are all squares are
2 cubes and 2 cuboids as shown in the figure above.

Question 12.
Modeling Real Life
You need to find an object that has only flat surfaces for a scavenger hunt. Circle the objects you can use.
Big Ideas Math Answer Key Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 186
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-172
Explanation:
Objects that has only flat surfaces for a scavenger hunt are
Box and  Match Box so circled them as shown in the figure above.

Describe Three-Dimensional Shapes Homework & Practice 13.7

Circle the attributes of the shape.

Question 13.
Big Ideas Math Answer Key Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 187
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-173
Explanation:
Given Rectangular Prism shape it has 6 flat surfaces,
with 12 edges so circled the attributes of them as shown above.

Question 14.
Big Ideas Math Answer Key Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 188
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-174
Explanation:
Given Sphere shape it has 0 flat surfaces,
with 0 edges so circled the attributes of them as shown above.

Combine Three-Dimensional Shapes Homework &
Practice 13.8

Question 15.
Circle the new shape that you can make.
Big Ideas Math Answer Key Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 189
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-175
Explanation:
Given shapes are 1 cylinder and 1 cube using them we get new shape
as cylinder with cube so circled it as shown in the picture above.

Take Apart Three-Dimensional Shapes Homework & Practice 13.9

Question 16.
Circle the shapes that make up the structure.
Big Ideas Math Answer Key Grade 1 Chapter 13 Two-and Three-Dimensional Shapes 190
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-13-Two-and-Three-Dimensional-Shapes-176

Explanation:
As shown we use 6 cubes, 1 cylinder and 2 cuboids three-dimensional shapes to make up the given structure.

Conclusion:

Hoping that Big Ideas Math Answers Grade 1 Chapter 13 Two – and Three-Dimensional Shapes has helped you clear your queries. Access the BIM Grade 3 Answer Key Ch 13 for free of cost prepared by math experts. Stay glued to get the Solutions for all Big Ideas Math Grade 1 Chapters.

Big Ideas Math Answers Grade 4 Chapter 9 Multiply Whole Numbers and Fractions

Big Ideas Math Answers Grade 4 Chapter 9

Big Ideas Math Answers Grade 4 Chapter 9 Multiply Whole Numbers and Fractions is an essential study material for students. Basic concepts of Multiplication of Whole Numbers and Fractions are clearly explained to score good marks for students. Also, in-depth knowledge of Chapter 9 Multiplies Whole Numbers and Fractions Big Ideas Grade 4 Math Answer Key mentioned with a clear explanation. Furthermore, BIM Grade 4 Multiply Whole Numbers and Fractions Answer Key is very convenient for students who are struggling to learn that concept. We have also given extra practice in small groups for the best practice of the students.

Big Ideas 4th Grade Chapter 9 Multiply Whole Numbers and Fractions Math Book Answer Key

Every problem is solved and given with a detailed explanation to help the students while their preparation. We have included the BIM Grade 4 Chapter 9 Solution Key in all the ways such as quiz, formative assessment, homework, or extra practice, etc. Practice in all the ways to get a complete grip on the concept. Provide a bright future for students, by giving the Chapter 9 Multiplies Whole Numbers and Fractions Big Ideas Grade 4 Math Answers as a reference.

Lesson: 1 Understand Multiples of Unit Fractions

Lesson: 2 Understand Multiples of Fractions

Lesson: 3 Multiply Whole Numbers and Fractions

Lesson: 4 Multiply Whole Numbers and Mixed Numbers

Lesson: 5 Problem Solving: Fraction Operations

Performance Task

Lesson 9.1 Understand Multiples of Unit Fractions

Explore and Grow

Draw a model of any fraction using unit fractions. Then write an addition equation to represent your model.
Reasoning
How can you rewrite the equation using multiplication? Explain.
Answer:
Let us take the fraction as \(\frac{3}{5}\) we draw a model using fractions as below

Addition equation to represent \(\frac{3}{5}\)  model is \(\frac{3}{5}\) = \(\frac{1}{5}\) +
\(\frac{1}{5}\) +\(\frac{1}{5}\) to rewrite the equation using multiplication,
We write the equation as \(\frac{3}{5}\) = \(\frac{1}{5}\) +\(\frac{1}{5}\) +
\(\frac{1}{5}\) = 3 X \(\frac{1}{5}\) as the fraction \(\frac{3}{5}\)
represents 3 parts that are each \(\frac{1}{5}\) of the whole.

Think and Grow: Multiples of Unit Fractions

Any fraction can be written as a multiple of a unit fraction with a like denominator.
Example
Write \(\frac{5}{8}\) as a multiple of a unit fraction.
The fraction \(\frac{5}{8}\) represents 5 parts that are each \(\frac{1}{8}\) of the whole.
Big Ideas Math Answers Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 500
Answer:

Show and Grow

Write the fraction as a multiple of a unit fraction.

Question 1.
Big Ideas Math Answers Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 1
Answer:
\(\frac{2}{3}\) = \(\frac{1}{3}\) +\(\frac{1}{3}\)
=  2 X \(\frac{1}{3}\).

Explanation:
We write \(\frac{2}{3}\) in addition equation as \(\frac{2}{3}\) =\(\frac{1}{3}\) +
\(\frac{1}{3}\) and multiplication as fraction \(\frac{2}{3}\) represents 2 parts that are each
\(\frac{1}{3}\) of the whole, \(\frac{2}{3}\) =2 X \(\frac{1}{3}\).

Question 2.
Big Ideas Math Solutions Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 2
Answer:
\(\frac{4}{8}\) = \(\frac{1}{8}\) +\(\frac{1}{8}\) + \(\frac{1}{8}\) +
\(\frac{1}{8}\) = 4 X \(\frac{1}{8}\).
Explanation:
We write \(\frac{4}{8}\) in addition equation as \(\frac{4}{8}\) = \(\frac{1}{8}\) +
\(\frac{1}{8}\)  + \(\frac{1}{8}\) +\(\frac{1}{8}\) and multiplication as
fraction \(\frac{4}{8}\) represents 4 parts that are each \(\frac{1}{8}\) of the whole,
\(\frac{4}{8}\) =4 X \(\frac{1}{8}\).

Question 3.
\(\frac{6}{5}\)
Answer:
\(\frac{6}{5}\) = \(\frac{1}{5}\)+\(\frac{1}{5}\) +\(\frac{1}{5}\)+
\(\frac{1}{5}\)+\(\frac{1}{5}\)+\(\frac{1}{5}\) = 6 X \(\frac{1}{5}\).

Explanation:
We write \(\frac{6}{5}\) in addition equation as \(\frac{6}{5}\)= \(\frac{1}{5}\)+
\(\frac{1}{5}\) +\(\frac{1}{5}\)+\(\frac{1}{5}\)+\(\frac{1}{5}\)+
\(\frac{1}{5}\) and multiplication as fraction \(\frac{6}{5}\) represents 6 parts
that are \(\frac{1}{5}\) of the whole \(\frac{6}{5}\) = 6 X \(\frac{1}{5}\).

Question 4.
\(\frac{7}{100}\)
Answer:
\(\frac{7}{100}\) = \(\frac{1}{100}\) + \(\frac{1}{100}\) + \(\frac{1}{100}\) +
\(\frac{1}{100}\)+ \(\frac{1}{100}\)+ \(\frac{1}{100}\)+ \(\frac{1}{100}\)
= 7 X \(\frac{1}{100}\).

Explanation:
We write \(\frac{7}{100}\) in addition equation as \(\frac{7}{100}\)= \(\frac{1}{100}\) +
\(\frac{1}{100}\) + \(\frac{1}{100}\) + \(\frac{1}{100}\)+ \(\frac{1}{100}\)+
\(\frac{1}{100}\)+ \(\frac{1}{100}\) and multiplication as fraction \(\frac{7}{100}\) represents
7 parts that are \(\frac{1}{100}\) of the whole, \(\frac{7}{100}\)= 7 X \(\frac{1}{100}\).

Apply and Grow: Practice

Write the fraction as a multiple of a unit fraction.

Question 5.
\(\frac{2}{6}\)
Answer:
\(\frac{2}{6}\) = \(\frac{1}{6}\) + \(\frac{1}{6}\)
= 2 X \(\frac{1}{6}\).

Explanation:
We write \(\frac{2}{6}\) in addition equation as \(\frac{2}{6}\)= \(\frac{1}{6}\) +
\(\frac{1}{6}\) and multiplication as fraction \(\frac{2}{6}\) represents 2 parts that are
\(\frac{1}{6}\) of the whole,\(\frac{2}{6}\) = 2 X \(\frac{1}{6}\).

Question 6.
\(\frac{6}{8}\)
Answer:
\(\frac{6}{8}\) = \(\frac{1}{8}\) + \(\frac{1}{8}\) + \(\frac{1}{8}\)+
\(\frac{1}{8}\)+ \(\frac{1}{8}\)+ \(\frac{1}{8}\) = 6 X \(\frac{1}{8}\).

Explanation:
We write \(\frac{6}{8}\) in addition equation as \(\frac{6}{8}\) = \(\frac{1}{8}\) +
\(\frac{1}{8}\) + \(\frac{1}{8}\)+ \(\frac{1}{8}\)+ \(\frac{1}{8}\)+
\(\frac{1}{8}\) and multiplication as fraction \(\frac{6}{8}\) represents 6 parts that are
\(\frac{1}{8}\) of the whole, \(\frac{6}{8}\) =6 X \(\frac{1}{8}\).

Question 7.
\(\frac{5}{4}\)
Answer:
\(\frac{5}{4}\) = \(\frac{1}{4}\) + \(\frac{1}{4}\) +\(\frac{1}{4}\) +
\(\frac{1}{4}\) +\(\frac{1}{4}\) = 5 X \(\frac{1}{4}\).

Explanation:
We write \(\frac{5}{4}\) in addition equation as \(\frac{5}{4}\)=\(\frac{1}{4}\) +
\(\frac{1}{4}\) +\(\frac{1}{4}\) +\(\frac{1}{4}\) + \(\frac{1}{4}\)
and multiplication as fraction \(\frac{5}{4}\) represents 5 parts that are \(\frac{1}{4}\)
of the whole, \(\frac{5}{4}\) = 5 X \(\frac{1}{4}\).

Question 8.
\(\frac{3}{12}\)
Answer:
\(\frac{3}{12}\)= \(\frac{1}{12}\) + \(\frac{1}{12}\) +\(\frac{1}{12}\) =
3 X \(\frac{1}{12}\)

Explanation :
We write \(\frac{3}{12}\) in addition equation as \(\frac{3}{12}\)= \(\frac{1}{12}\) +
\(\frac{1}{12}\) + \(\frac{1}{12}\) and multiplication as fraction \(\frac{3}{12}\)
represents 3 parts that are \(\frac{1}{12}\) of the whole, \(\frac{3}{12}\) =
3 X \(\frac{1}{12}\).

Question 9.
\(\frac{8}{100}\)
Answer:
\(\frac{8}{100}\) = \(\frac{1}{100}\) + \(\frac{1}{100}\)+\(\frac{1}{100}\)+
\(\frac{1}{100}\)+\(\frac{1}{100}\)+\(\frac{1}{100}\)+\(\frac{1}{100}\)+
\(\frac{1}{100}\) = 8 X \(\frac{1}{100}\).

Explanation :
We write \(\frac{8}{100}\) in addition equation as \(\frac{8}{100}\) =
\(\frac{1}{100}\) + \(\frac{1}{100}\)+\(\frac{1}{100}\)+
\(\frac{1}{100}\)+\(\frac{1}{100}\)+ \(\frac{1}{100}\)+
\(\frac{1}{100}\)+\(\frac{1}{100}\) and multiplication as fraction
\(\frac{8}{100}\) represents 8 parts that are \(\frac{1}{100}\) of the whole,
\(\frac{8}{100}\)= 8 X \(\frac{1}{100}\).

Question 10.
\(\frac{7}{10}\)
Answer:
\(\frac{7}{10}\) = \(\frac{1}{10}\) +\(\frac{1}{10}\) +\(\frac{1}{10}\) +
\(\frac{1}{10}\) +\(\frac{1}{10}\) +\(\frac{1}{10}\) +\(\frac{1}{10}\)
= 7 X \(\frac{1}{10}\).

Explanation:
We write \(\frac{7}{10}\) in addition equation as \(\frac{7}{10}\) = \(\frac{1}{10}\) +
\(\frac{1}{10}\) +\(\frac{1}{10}\) +\(\frac{1}{10}\) +\(\frac{1}{10}\) +
\(\frac{1}{10}\) +\(\frac{1}{10}\) and multiplication as fraction \(\frac{7}{10}\)
represents 7 parts that are \(\frac{1}{10}\) of the whole,\(\frac{7}{10}\) = 7 X \(\frac{1}{10}\).

Question 11.
Structure
When a fraction is written as a multiple of a unit fraction, what is the relationship between the numerator of the
fraction and the number that is multiplied by the unit fraction?
Answer:
When a fraction is written as a multiple of a unit fraction, the relationship between the numerator of the fraction
and the number that is multiplied by the unit fraction, is the whole number times the unit fraction.

Explanation : We know that the numerator of a unit fraction is always one. So When a fraction is
written as a multiple of a unit fraction the number will be always the whole number (whole number of times the unit fraction)
and is multiplied by unit fraction. For example \(\frac{3}{8}\) we write it as 3 X \(\frac{1}{8}\)
here it is 3 and this 3 is multiplied by unit fraction.

DIG DEEPER!
Write the mixed number as a multiple of a unit fraction.

Question 12.
1\(\frac{1}{8}\)
Answer:
1\(\frac{1}{8}\) = \(\frac{9}{8}\) = \(\frac{1}{8}\)+\(\frac{1}{8}\)+
\(\frac{1}{8}\)+\(\frac{1}{8}\)+\(\frac{1}{8}\)+\(\frac{1}{8}\)+
\(\frac{1}{8}\)+\(\frac{1}{8}\)+\(\frac{1}{8}\)= 9 X \(\frac{1}{8}\) .

Explanation:
First we convert a mixed number to an improper fraction, It is to multiply the bottom number 8 by the
“regular” number 1, add in the top number 1, and then put this on top of the original bottom number as 9 by 8.
1\(\frac{1}{8}\) = 1+\(\frac{1}{8}\) = \(\frac{9}{8}\)
We write \(\frac{9}{8}\) in addition equation as \(\frac{9}{8}\)= \(\frac{1}{8}\)+
\(\frac{1}{8}\)+\(\frac{1}{8}\)+\(\frac{1}{8}\)+\(\frac{1}{8}\)+
\(\frac{1}{8}\)+\(\frac{1}{8}\)+\(\frac{1}{8}\) +\(\frac{1}{8}\)
and multiplication as fraction \(\frac{9}{8}\) represents 9 parts that are \(\frac{1}{10}\)
of the whole, \(\frac{9}{8}\) = 9 X \(\frac{1}{8}\).

Question 13.
3\(\frac{4}{5}\)
Answer:
3\(\frac{4}{5}\) = \(\frac{19}{5}\)=\(\frac{1}{5}\)+\(\frac{1}{5}\)+
\(\frac{1}{5}\) +\(\frac{1}{5}\)+\(\frac{1}{5}\)+\(\frac{1}{5}\)+
\(\frac{1}{5}\)+\(\frac{1}{5}\)+\(\frac{1}{5}\)+\(\frac{1}{5}\)+
\(\frac{1}{5}\)+\(\frac{1}{5}\)+\(\frac{1}{5}\)+\(\frac{1}{5}\)+
\(\frac{1}{5}\)+\(\frac{1}{5}\)+\(\frac{1}{5}\)+\(\frac{1}{5}\)+
\(\frac{1}{5}\)= 19 X \(\frac{1}{5}\).

Explanation:
First we convert a mixed number to an improper fraction, It is to multiply the bottom number 5 by the “regular”
number 3, add in the top number 4, and then put this on top of the original bottom number as 19 by 5.
3\(\frac{4}{5}\)=3+\(\frac{4}{5}\) = \(\frac{19}{5}\)
We write \(\frac{19}{5}\) in addition equation as \(\frac{19}{5}\) = \(\frac{1}{5}\)+
\(\frac{1}{5}\)+\(\frac{1}{5}\) +\(\frac{1}{5}\)+\(\frac{1}{5}\)+
\(\frac{1}{5}\)+\(\frac{1}{5}\)+\(\frac{1}{5}\)+\(\frac{1}{5}\)+
\(\frac{1}{5}\)+\(\frac{1}{5}\)+\(\frac{1}{5}\)+\(\frac{1}{5}\)+
\(\frac{1}{5}\)+\(\frac{1}{5}\)+\(\frac{1}{5}\)+\(\frac{1}{5}\)+
\(\frac{1}{5}\)+\(\frac{1}{5}\) and multiplication as fraction \(\frac{19}{5}\)
represents 19 parts that are \(\frac{1}{5}\) of the whole,\(\frac{19}{5}\)= 19 X \(\frac{1}{5}\).

Question 13.

Think and Grow: Modeling Real Life

Example
A juice stand worker uses \(\frac{1}{2}\) of an orange to garnish each drink.
The worker has \(\frac{7}{12}\) of an orange left. How many more drinks can the worker garnish?
Write the fraction of orange left as a multiple of \(\frac{1}{12}\).
Big Ideas Math Solutions Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 3
Big Ideas Math Answers Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 501
Answer:
The worker can garnish \(\frac{1}{12}\) more drinks,
\(\frac{7}{12}\) minus \(\frac{1}{2}\) means \(\frac{7}{12}\) – \(\frac{1}{2}\)
= \(\frac{1}{12}\), Orange left is 7 X \(\frac{1}{12}\).

Explanation:
Given a juice stand worker uses \(\frac{1}{2}\) of an orange to garnish each drink.
The worker has \(\frac{7}{12}\) of an orange left. So the worker can garnish more \(\frac{7}{12}\)
minus \(\frac{1}{2}\) means \(\frac{7}{12}\) – \(\frac{1}{2}\) = \(\frac{1}{12}\),
Given \(\frac{7}{12}\) of an orange is left , to write orange left as a multiple of \(\frac{1}{12}\)
we have orange left as \(\frac{7}{12}\) so write it as 7 parts that are \(\frac{1}{12}\) of the whole,
\(\frac{7}{12}\)= 7 X \(\frac{1}{12}\).

Show and Grow

Question 14.
A piece of rope is \(\frac{8}{5}\) meters long. You cut the rope into \(\frac{1}{5}\) meter long pieces.
How many pieces do you cut?
Answer:
8 pieces I have cut.

Explanation:
Given a piece of rope is \(\frac{8}{5}\) meters long and it has been cut into \(\frac{1}{5}\)
meter long pieces, therefore pieces of rope cut is \(\frac{8}{5}\) divided by \(\frac{1}{5}\)
= \(\frac{8}{5}\) X latex]\frac{5}{1}[/latex] = 8.

Question 15.
A restaurant serves \(\frac{4}{10}\) of a meatloaf to 4 customers. Each customer receives
the same amount of meatloaf. What fraction of the meatloaf does each customer receive?
Answer:
The fraction of the meatloaf each customer receives is \(\frac{1}{10}\)

Explanation:
Given a restaurant serves \(\frac{4}{10}\) of a meatloaf to 4 customers and each customer receives the same
amount of meatloaf so fraction of the meatloaf each customer receives is  \(\frac{4}{10}\)  divided by 4 =
\(\frac{4}{10}\) x \(\frac{1}{4}\) = \(\frac{1}{10}\).

Question 16.
You use 3\(\frac{3}{4}\) pounds of trail mix to make treat bags. You put \(\frac{1}{4}\)
pound of trail mix into each bag. How many treat bags do you make?
Big Ideas Math Solutions Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 4
Answer:
Total number of treat bags made are 15

Explanation:
I use 3\(\frac{3}{4}\) pounds of trail mix to make treat bags, We convert a mixed number to an
improper fraction,It is to multiply the bottom number 4 by the “regular” number 3, add in the top number 3,
and then put this on top of the original bottom number as 15 by 4, it becomes \(\frac{15}{4}\),Each bag i kept
\(\frac{1}{4}\) pound of trail mix so total number of treat bags required to make are \(\frac{15}{4}\)
divided by \(\frac{1}{4}\)=\(\frac{15}{4}\) X \(\frac{4}{1}\)=15 treat bags.

Question 17.
DIG DEEPER!
You walk from home to school and then back home again each day for 5 days. Altogether, you walk \(\frac{10}{8}\) miles. What is the distance from your home to school? Explain.
Answer:
The distance from home to school is \(\frac{1}{8}\) miles

Explanation:
Total number of miles walked is \(\frac{10}{8}\) in 5 days that to from home to school and
then back home again means it is twice a day , Let us take X as the distance from home to school is,
Given 2X x 5=\(\frac{10}{8}\) miles, X = Then each day number of miles walked is  \(\frac{10}{8}\)
divided by 10 which is  equal to  \(\frac{10}{8}\) x \(\frac{1}{10}\) =\(\frac{1}{8}\),
Therefore X = \(\frac{1}{8}\) miles.

Understand Multiples of Unit Fractions Homework & practice 9.1

Write the fraction as a multiple of a unit fraction.

Question 1.
Big Ideas Math Solutions Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 5
Answer:
\(\frac{2}{12}\)= \(\frac{1}{12}\)+\(\frac{1}{12}\) = 2 X \(\frac{1}{12}\)

Explanation:
We write \(\frac{2}{12}\) in addition equation as \(\frac{2}{12}\)= \(\frac{1}{12}\) +
\(\frac{1}{12}\) and multiplication as fraction \(\frac{2}{12}\) represents 2 parts that are
\(\frac{1}{12}\) of the whole, \(\frac{2}{12}\)= 2 X \(\frac{1}{12}\).

Question 2.
Big Ideas Math Solutions Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 6
Answer:
\(\frac{5}{100}\)= \(\frac{1}{100}\)+\(\frac{1}{100}\)+\(\frac{1}{100}\)+
\(\frac{1}{100}\)+\(\frac{1}{100}\) = 5 X \(\frac{1}{100}\).

Explanation:
We write \(\frac{5}{100}\) in addition equation as \(\frac{5}{100}\)= \(\frac{1}{100}\)+
\(\frac{1}{100}\)+\(\frac{1}{100}\)+\(\frac{1}{100}\)+\(\frac{1}{100}\)
and multiplication as fraction \(\frac{5}{100}\) represents 5 parts that are \(\frac{1}{100}\) of the whole, \(\frac{5}{100}\) = 5 X \(\frac{1}{100}\).

Question 3.
\(\frac{3}{6}\)
Answer:
\(\frac{3}{6}\)=\(\frac{1}{6}\)+\(\frac{1}{6}\)+\(\frac{1}{6}\)
= 3 X \(\frac{1}{6}\).

Explanation:
We write \(\frac{3}{6}\) in addition equation as \(\frac{3}{6}\)=\(\frac{1}{6}\)+
\(\frac{1}{6}\)+\(\frac{1}{6}\) and multiplication as fraction \(\frac{3}{6}\)
represents 3 parts that are \(\frac{1}{6}\) of the whole,\(\frac{3}{6}\) = 3 X \(\frac{1}{6}\).

Question 4.
\(\frac{7}{10}\)
Answer:
\(\frac{7}{10}\) =\(\frac{1}{10}\)+\(\frac{1}{10}\)+\(\frac{1}{10}\)+
\(\frac{1}{10}\)+\(\frac{1}{10}\)+\(\frac{1}{10}\)+\(\frac{1}{10}\)
= 7 X \(\frac{1}{10}\)

Explanation:
We write \(\frac{7}{10}\) in addition equation as \(\frac{7}{10}\)= \(\frac{1}{10}\)+
\(\frac{1}{10}\)+\(\frac{1}{10}\)+\(\frac{1}{10}\)+\(\frac{1}{10}\)+
\(\frac{1}{10}\)+\(\frac{1}{10}\) and multiplication as \(\frac{7}{10}\) represents
7 parts that are \(\frac{1}{10}\) of the whole,\(\frac{7}{10}\) = 7 X \(\frac{1}{10}\).

Question 5.
\(\frac{4}{5}\)
Answer:
\(\frac{4}{5}\) = \(\frac{1}{5}\)+\(\frac{1}{5}\)+
\(\frac{1}{5}\)+\(\frac{1}{5}\) = 4 X \(\frac{1}{5}\).

Explanation:
We write \(\frac{4}{5}\) in addition equation as \(\frac{4}{5}\)=\(\frac{1}{5}\)+
\(\frac{1}{5}\)+\(\frac{1}{5}\)+\(\frac{1}{5}\) and multiplication as \(\frac{4}{5}\)
represents 4 parts that are \(\frac{1}{5}\) of the whole,\(\frac{4}{5}\)= 4 X \(\frac{1}{5}\).

Question 6.
\(\frac{9}{2}\)
Answer:
\(\frac{9}{2}\) = \(\frac{1}{2}\)+\(\frac{1}{2}\)+\(\frac{1}{2}\)+
\(\frac{1}{2}\)+\(\frac{1}{2}\)+\(\frac{1}{2}\)+\(\frac{1}{2}\)+
\(\frac{1}{2}\)+\(\frac{1}{2}\) = 9 X \(\frac{1}{2}\).

Explanation:
We write \(\frac{9}{2}\) in addition equation as \(\frac{9}{2}\) = \(\frac{1}{2}\)+
\(\frac{1}{2}\)+\(\frac{1}{2}\)+\(\frac{1}{2}\)+\(\frac{1}{2}\)+
\(\frac{1}{2}\)+\(\frac{1}{2}\)+\(\frac{1}{2}\)+\(\frac{1}{2}\)
and multiplication as \(\frac{9}{2}\) represents 9 parts that are \(\frac{1}{2}\) of the whole,
\(\frac{9}{2}\)= 9 X \(\frac{1}{2}\).

Question 7.
\(\frac{8}{8}\)
Answer:
\(\frac{8}{8}\)=1=\(\frac{1}{8}\)+\(\frac{1}{8}\)+\(\frac{1}{8}\)+
\(\frac{1}{8}\)+\(\frac{1}{8}\)+\(\frac{1}{8}\)+\(\frac{1}{8}\)+
\(\frac{1}{8}\) = \(\frac{8}{8}\) = 8 X \(\frac{1}{8}\).

Explanation:
\(\frac{8}{8}\)= 1 ,but we write the fraction as a addition of a unit fraction and multiple of a unit fraction,
We write \(\frac{8}{8}\) in addition equation as \(\frac{8}{8}\) = \(\frac{1}{8}\)+
\(\frac{1}{8}\)+\(\frac{1}{8}\)+\(\frac{1}{8}\)+\(\frac{1}{8}\)+
\(\frac{1}{8}\)+\(\frac{1}{8}\)+\(\frac{1}{8}\) and multiplication as \(\frac{8}{8}\)
represents 8 parts that are \(\frac{1}{8}\) of the whole,\(\frac{8}{8}\)= 8 X \(\frac{1}{8}\).

Question 8.
\(\frac{6}{3}\)
Answer:
\(\frac{6}{3}\)=2,\(\frac{6}{3}\)= \(\frac{1}{3}\)+\(\frac{1}{3}\)+
\(\frac{1}{3}\)+\(\frac{1}{3}\)+\(\frac{1}{3}\)+\(\frac{1}{3}\)
= 6 X \(\frac{1}{3}\).

Explanantion:
\(\frac{6}{3}\)= 2 ,but we write the fraction as a addition of a unit fraction and multiple of a unit fraction,
We write \(\frac{6}{3}\) in addition equation as \(\frac{6}{3}\)= \(\frac{1}{3}\)+
\(\frac{1}{3}\)+\(\frac{1}{3}\)+\(\frac{1}{3}\)+\(\frac{1}{3}\)+
\(\frac{1}{3}\) and multiplication as \(\frac{6}{3}\) represents 6 parts that are
\(\frac{1}{3}\) of the whole,\(\frac{6}{3}\)= 6 X \(\frac{1}{3}\).

Question 9.
Logic
What is Newton’s fraction? Write the fraction as a multiple of a unit fraction.
Big Ideas Math Solutions Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 7
Answer:
Newton’s fraction = \(\frac{9}{12}\)=\(\frac{1}{12}\) +\(\frac{1}{12}\) +
\(\frac{1}{12}\) +\(\frac{1}{12}\) +\(\frac{1}{12}\) +\(\frac{1}{12}\) +
\(\frac{1}{12}\) +\(\frac{1}{12}\) +\(\frac{1}{12}\) = 9 X \(\frac{1}{12}\).

Explanation:
Given that Newton’s fraction as sum of 9 unit fractions and each unit fraction is one twelfth of a whole.
So Newton’s fraction is \(\frac{9}{12}\) we write \(\frac{9}{12}\) in addition equation as
\(\frac{9}{12}\)=  \(\frac{1}{12}\) +\(\frac{1}{12}\) +\(\frac{1}{12}\) +
\(\frac{1}{12}\) +\(\frac{1}{12}\) +\(\frac{1}{12}\) +\(\frac{1}{12}\) +
\(\frac{1}{12}\) +\(\frac{1}{12}\) and multiplication as \(\frac{9}{12}\) represents 9 parts
that are \(\frac{1}{12}\) of the whole,\(\frac{9}{12}\)= 9 X \(\frac{1}{12}\).

Question 10
Structure
Write a multiplication equation that represents the grapefruit halves.
Big Ideas Math Solutions Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 8
Answer:
\(\frac{6}{2}\)=\(\frac{1}{2}\)+\(\frac{1}{2}\)+\(\frac{1}{2}\)+
\(\frac{1}{2}\)+\(\frac{1}{2}\)+\(\frac{1}{2}\)
= 6 X \(\frac{1}{2}\).

Explanation:
Given 6 number of grapefruit halves so there are total grapefruit as \(\frac{6}{2}\) we write
\(\frac{6}{2}\) in addition equation as \(\frac{6}{2}\)=\(\frac{1}{2}\)+
\(\frac{1}{2}\)+\(\frac{1}{2}\)+\(\frac{1}{2}\)+\(\frac{1}{2}\)+
\(\frac{1}{2}\),so the multiplication equation is \(\frac{6}{2}\) represents 6 parts that are
\(\frac{1}{2}\) of the whole,\(\frac{6}{2}\)= 6 X \(\frac{1}{2}\).

Question 11.
Which One Doesn’t Belong?
Which expression does not belong with the other three?
Big Ideas Math Solutions Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 9
Answer:
4 X \(\frac{1}{5}\) expression does not belong with the other three, as all the other expressions values results are same =\(\frac{5}{4}\) only 4 X \(\frac{1}{5}\) expression is not.

Explanation:
1. 5 X \(\frac{1}{4}\) = \(\frac{5}{4}\)
2. \(\frac{1}{4}\)+\(\frac{1}{4}\)+\(\frac{1}{4}\)+\(\frac{1}{4}\)=
5 X \(\frac{1}{4}\)= \(\frac{5}{4}\)
3. \(\frac{5}{4}\) =\(\frac{5}{4}\)
4. 4 X \(\frac{1}{5}\)= \(\frac{4}{5}\)≠  \(\frac{5}{4}\)
So in the above expressions only 4 X \(\frac{1}{5}\)  does not belong with the other three expressions.

Question 12.
Modeling Real Life
You are making blueberry pancakes. You have \(\frac{6}{8}\) cup of blueberries. You put \(\frac{1}{8}\) cup of blueberries in each pancake. How many pancakes do you make?
Big Ideas Math Solutions Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 10
Answer:
6 pancakes i can make.

Explanation:
I have \(\frac{6}{8}\) cup of blueberries and i can put \(\frac{1}{8}\) cup of blueberries
in each pan cake so total number of pancakes made are \(\frac{6}{8}\) divided by \(\frac{1}{8}\) =
\(\frac{6}{8}\) X \(\frac{8}{1}\)= 6 pancakes.

Question 13.
DIG DEEPER!
You cut a loaf of zucchini bread into 20 equal slices. You and your friends eat \(\frac{3}{10}\) of the slices.
You want to put each leftover slice into its own bag. How many bags do you need?
Answer:
14 bags are needed.

Explanation :
Given a loaf of zucchini bread is cut  into 20 equal slices and i ,friend eat \(\frac{3}{10}\) of the slices,
Total number of slices of zucchini bread =20, Number of slices ate = \(\frac{3}{10}\) means it says
out of 10 slices 3 were ate, so out of 20 slices  2 x 3= 6 slices were eaten. Number of slices left = 20 – 6 =14 slices
and one bag contains only one slice. Therefore number of bags needed are 14.

Question 14.
Is 46 a multiple of 45.
Answer:
No 46 is not a multiple of 45.

Explanation:
A multiple is a number that can be divided by another number a certain number of times without a remainder.
45 when divided by 46 a certain times will leave a remainder, So 46 is not a multiple of 45
or a factor is one of two or more numbers that divides a given number without a remainder.
As factors of 45 are 1,3,5,9,15 and factors of 46 are 1,2,23, both do not have common factors so 46 is not a multiple of 45.

Question 15.
Is 3 a factor of 75?
Answer:
Yes 3 is a factor of 75
Explanation:
We know that a factor is one that divides a given number without a remainder and 75 when divided by 3
leaves no remainder, So 3 is a factor of 75.

Lesson 9.2 Understand Multiples of Fractions

Draw a model for each expression. Then write a multiplication expression to represent each model.
Big Ideas Math Solutions Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 11
Big Ideas Math Solutions Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 12
Answer:

The multiplication expression is \(\frac{2}{6}\) +\(\frac{2}{6}\) +\(\frac{2}{6}\) +
\(\frac{2}{6}\) = 4 X \(\frac{2}{6}\) =
\(\frac{1}{6}\) +\(\frac{1}{6}\)+ \(\frac{1}{6}\) +\(\frac{1}{6}\)+
\(\frac{1}{6}\) +\(\frac{1}{6}\)+ \(\frac{1}{6}\) +\(\frac{1}{6}\) =
8 X \(\frac{1}{6}\).

Explanation:
\(\frac{2}{6}\) +\(\frac{2}{6}\) +\(\frac{2}{6}\) +\(\frac{2}{6}\)
the multiplication equation is  4 parts that are \(\frac{2}{6}\) of the whole as  4 X \(\frac{2}{6}\)
\(\frac{1}{6}\) +\(\frac{1}{6}\)+ \(\frac{1}{6}\) +\(\frac{1}{6}\)+
\(\frac{1}{6}\) +\(\frac{1}{6}\)+ \(\frac{1}{6}\) +\(\frac{1}{6}\) =
8 X \(\frac{1}{6}\) the multiplication equation is 8 parts that are \(\frac{1}{6}\) of the whole as
8 X \(\frac{1}{6}\).

Structure
Compare your expressions. How are they the same? How are they different?
Same means the result of both expressions is same = \(\frac{8}{6}\)
Different means in one expression we take value of \(\frac{2}{6}\) 4 times and in another
expression we take \(\frac{1}{6}\) 8 times.

Explanation:
Same means \(\frac{2}{6}\) +\(\frac{2}{6}\) +\(\frac{2}{6}\) +\(\frac{2}{6}\) =
4 X \(\frac{2}{6}\) = \(\frac{8}{6}\) and \(\frac{1}{6}\) +\(\frac{1}{6}\)+
\(\frac{1}{6}\) +\(\frac{1}{6}\)+\(\frac{1}{6}\) +\(\frac{1}{6}\)+
\(\frac{1}{6}\) +\(\frac{1}{6}\) = 8 X \(\frac{1}{6}\) =\(\frac{8}{6}\) ,
both have results similar as \(\frac{8}{6}\), Different means here we are taking expressions as addition values of  \(\frac{2}{6}\) by 4 times and in other expression we are taking addition values of \(\frac{1}{6}\) by 8 times.

Think and Grow: Multiples of Fractions

A multiple of any fraction can be written as a multiple of a unit fraction with a like denominator.
Big Ideas Math Solutions Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 13
Answer:


Explanation:
To write 2 X \(\frac{3}{5}\) multiple of unit fraction first we write \(\frac{3}{5}\) as multiple of
3 X \(\frac{1}{5}\) then we multiply it by 2 as 2 X 3 X \(\frac{1}{5}\)=6 X \(\frac{1}{5}\).

Show and Grow

Write the product as a multiple of a unit fraction. Then find the product.

Question 1.
Big Ideas Math Solutions Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 14
Answer:
2 X \(\frac{4}{5}\) = 2 X 4 X \(\frac{1}{5}\) = 8 X \(\frac{1}{5}\) = \(\frac{8}{5}\).

Explanation:
First we write \(\frac{4}{5}\)  as  \(\frac{1}{5}\) + \(\frac{1}{5}\) +
\(\frac{1}{5}\) + \(\frac{1}{5}\) = 4 X \(\frac{1}{5}\) then multiply it by 2
which is  2 X 4 X \(\frac{1}{5}\) = 8 X \(\frac{1}{5}\) = \(\frac{8}{5}\).

Question 2.
Big Ideas Math Solutions Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 15
Answer:
3 X \(\frac{2}{10}\) = 3 X 2 X  \(\frac{1}{10}\) = 6 X \(\frac{1}{10}\) =\(\frac{6}{10}\).

Explanation:
First we write \(\frac{2}{10}\) as \(\frac{1}{10}\) +\(\frac{1}{10}\) = 2 X \(\frac{1}{10}\)
then multiply it by 3 which is 3 X 2 X \(\frac{1}{10}\) = 6 X \(\frac{1}{10}\) =\(\frac{6}{10}\).

Question 3.
Big Ideas Math Solutions Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 16
Answer:
4 X \(\frac{3}{2}\) = 4 X 3 X \(\frac{1}{2}\) = 12 X \(\frac{1}{2}\) = \(\frac{12}{2}\) = 6.

Explanation:
First we write \(\frac{3}{2}\) as \(\frac{1}{2}\) +\(\frac{1}{2}\) + \(\frac{1}{2}\) = 3 X \(\frac{1}{2}\) then we multiply it by 4 which is 4 X 3 X \(\frac{1}{2}\)= 12 X \(\frac{1}{2}\)= \(\frac{12}{2}\)=6.

Apply and Grow: Practice

Write the product as a multiple of a unit fraction. Then find the product.

Question 4.
Big Ideas Math Solutions Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 17
Answer:
5 X \(\frac{2}{3}\) = 5 X 2 X \(\frac{1}{3}\) = 10 X \(\frac{1}{3}\)=
\(\frac{10}{3}\).

Explanation:
First we write \(\frac{2}{3}\) as \(\frac{1}{3}\)+\(\frac{1}{3}\)=
2 X \(\frac{1}{3}\) then multiply it by 5 which is 5 X 2 X \(\frac{1}{3}\)=
10 X \(\frac{1}{3}\)= \(\frac{10}{3}\).

Question 5.
Big Ideas Math Solutions Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 18
Answer:
6 X \(\frac{5}{8}\)= 6 X 5 X \(\frac{1}{8}\)= 30 X \(\frac{1}{8}\) =
\(\frac{30}{8}\).

Explanation:
First we write \(\frac{5}{8}\) as \(\frac{1}{8}\)+ \(\frac{1}{8}\)+
\(\frac{1}{8}\) + \(\frac{1}{8}\)+ \(\frac{1}{8}\) = 5 X \(\frac{1}{8}\)
then multiply it by 6 which is 6 X 5 X \(\frac{1}{8}\)=30 X \(\frac{1}{8}\)= \(\frac{30}{8}\).

Question 6.
Big Ideas Math Solutions Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 19
Answer:
9 X \(\frac{7}{4}\) = 9 X 7 X\(\frac{1}{4}\)= 63 X \(\frac{1}{4}\) =
\(\frac{63}{4}\).

Explanation:
First we write \(\frac{7}{4}\) as \(\frac{1}{4}\)+\(\frac{1}{4}\)+
\(\frac{1}{4}\)+ \(\frac{1}{4}\)+\(\frac{1}{4}\)+\(\frac{1}{4}\)+
\(\frac{1}{4}\) = 7 x \(\frac{1}{4}\)
then we multiply it by 9 which is 9 X 7 X \(\frac{1}{4}\)= 63 X \(\frac{1}{4}\) =
\(\frac{63}{4}\).

Question 7.
Big Ideas Math Solutions Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 20
Answer:
7 X \(\frac{4}{12}\)=7 X 4 X \(\frac{1}{12}\)= 28 X \(\frac{1}{12}\)=
\(\frac{28}{12}\) also 7 X \(\frac{4}{12}\)= 7 X  \(\frac{1}{3}\)= \(\frac{7}{3}\).

Explanation:
First we write \(\frac{4}{12}\) as \(\frac{1}{12}\)+ \(\frac{1}{12}\)+
\(\frac{1}{12}\)+\(\frac{1}{12}\)= 4 X \(\frac{1}{12}\) then we multiply it by 7
which is 7 X 4 X \(\frac{1}{12}\)= 28 X \(\frac{1}{12}\)=\(\frac{28}{12}\)
we can also simplify 7 X \(\frac{4}{12}\) as 4,12 can be divided by 4 we can also write
7 X \(\frac{4}{12}\) as 7 X  \(\frac{1}{3}\)= \(\frac{7}{3}\).

Question 8.
Big Ideas Math Solutions Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 21
Answer:
\(\frac{9}{6}\) X 8 = 9 X \(\frac{1}{6}\) X 8 = 72 X \(\frac{1}{6}\) =
\(\frac{72}{6}\) =12.

Explanation:
First we write \(\frac{9}{6}\) as \(\frac{1}{6}\) +\(\frac{1}{6}\) +
\(\frac{1}{6}\) + \(\frac{1}{6}\) +\(\frac{1}{6}\) +\(\frac{1}{6}\) +
\(\frac{1}{6}\) + \(\frac{1}{6}\) +\(\frac{1}{6}\) =9 X \(\frac{1}{6}\)
then we multiply it by 8 which is 9 X 8 X \(\frac{1}{6}\) = 72 X \(\frac{1}{6}\) =
\(\frac{72}{6}\)  further as 6 can divide 72 and leaving no remainder we write \(\frac{72}{6}\) =12.

Question 9.
Big Ideas Math Solutions Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 22
Answer:
10 X \(\frac{20}{100}\) = 10 X \(\frac{2}{10}\)=2 X 10 X \(\frac{1}{10}\)=
20 \(\frac{1}{10}\)=\(\frac{20}{10}\)=2.

Explanation:
First we write \(\frac{20}{100}\) = \(\frac{2}{10}\) as \(\frac{1}{10}\) +
\(\frac{1}{10}\) = 2 X \(\frac{1}{10}\) then multiply it by 10 which is
2 X 10 X \(\frac{1}{10}\)= 20 X \(\frac{1}{10}\)=\(\frac{20}{10}\)
and 10 can divide 20 leaving no remainder further we can write \(\frac{20}{10}\) as 2.

Number Sense
Find the unknown number.

Question 10.
Big Ideas Math Solutions Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 23
Answer:
The unknown number is 2.

Explanation:
___ X \(\frac{8}{10}\) = \(\frac{16}{10}\)= to get the unknown number we take it as P,
P X \(\frac{8}{10}\) = \(\frac{16}{10}\) means P = \(\frac{16}{10}\) X
\(\frac{10}{8}\)  we get 2 , so P=2.

Question 11.
Big Ideas Math Solutions Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 24
Answer:
The unknown number is 5.

Explanation:
Let us take unknown letter as P so 4 X \(\frac{P}{2}\) = \(\frac{20}{2}\) ,
4 X P= 2 X \(\frac{20}{2}\) to get value of P we take 4 to other side ,
Therefore P=20 X \(\frac{1}{4}\) = 5, so the unknown number is 5.

Question 12.
Big Ideas Math Solutions Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 25
Answer:
The unknown number is 100.

Explanation:
Let us take unknown number as P given 3 X \(\frac{9}{P}\) = \(\frac{27}{100}\) ,
27 X[ latex]\frac{1}{P}[/latex] = \(\frac{27}{100}\), Therefore P= 27 X \(\frac{100}{27}\)=100.

Question 13.
Reasoning
Without calculating, would you plot the product of 5 and \(\frac{3}{6}\) to the left or to the right of 5 on a number line? Explain.
Big Ideas Math Solutions Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 26
Answer:
Yes , without calculating we can plot 5 and \(\frac{3}{6}\) to the left of 5

Explanation:
5 X \(\frac{3}{6}\)= 5 X  \(\frac{1}{2}\) since the denominator of each given fraction is 2
divide the space between every pairs of two consecutive integers (on the number line) in 2 equal parts each part
so obtained will represent the fraction \(\frac{1}{2}\) , so on the number line we move 5 parts on the
right-side of zero as shown in the figure.

Question 14.
Patterns
Describe and complete the pattern.
Big Ideas Math Solutions Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 27
Answer:

Explanation:
We first write the expressions as multiple of unit fraction and find the product so
1. 3 X \(\frac{1}{5}\) can be written as  product of 3 X 1 X  \(\frac{1}{5}\)=
3 X \(\frac{1}{5}\) = \(\frac{3}{5}\)
2. 3 X \(\frac{2}{5}\)  can be written as product of  3 X 2 X \(\frac{1}{5}\) =
6 X \(\frac{1}{5}\) = \(\frac{6}{5}\)
3. 3 X \(\frac{3}{5}\) can be written as product of 3 X 3 X \(\frac{1}{5}\) =
9 X \(\frac{1}{5}\) = \(\frac{9}{5}\)
4. 3 X \(\frac{4}{5}\) can be written as product of  3 X 4 X \(\frac{1}{5}\) =
12 X \(\frac{1}{5}\) = \(\frac{12}{5}\)
5. 3 X \(\frac{5}{5}\) can be written as product of 3 X 5 X \(\frac{1}{5}\) =
15 X \(\frac{1}{5}\) = \(\frac{15}{5}\) further can be simplified as 15,5 goes in 5,
so dividing by 5 we get \(\frac{15}{5}\) =  3.

Think and Grow: Modeling Real Life

A bird keeper uses a \(\frac{1}{3}\) cup scoop to feed 3 birds. He feeds each bird \(\frac{2}{3}\) cup of birdseed. How many times does he fill the scoop?
Three birds each need \(\frac{2}{3}\) cup of birdseed, so find 3 × \(\frac{2}{3}\).
Big Ideas Math Solutions Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 28
Big Ideas Math Solutions Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 29
Answer:

Explanation :
Given A bird keeper uses a \(\frac{1}{3}\) cup scoop to feed 3 birds each bird he feeds
\(\frac{2}{3}\) cup of birdseed, Three birds each need \(\frac{2}{3}\) cup of birdseed
so the bird keeper fills the scoop as 3 X \(\frac{2}{3}\) = 3 X _____P______ X \(\frac{1}{3}\) here
P is the  number of times the bird keeper fills the cup scoop.
3 X 2 X \(\frac{1}{3}\) =  3 X P X \(\frac{1}{3}\) , when we equate both sides,
we get 3 X 2 = 3 X P , therefore P = 2.
Show and Grow

Question 15.
A chef makes 4 servings of honey1oatmeal. She uses a \(\frac{1}{2}\) tablespoon measuring spoon to measure \(\frac{3}{2}\) tablespoons of honey for each serving. How many times does she fill the measuring spoon?
Answer:
3 times the chef fill the measuring spoon.

Explanation:
Chef uses a \(\frac{1}{2}\) table spoon as measuring spoon, To measure honey for each serving
she needed for each serving is \(\frac{3}{2}\) , so 4 X \(\frac{3}{2}\) ,
Let us take P as number of times of measuring spoon , Given 4 X \(\frac{3}{2}\) =
4 X _P_ X \(\frac{1}{2}\), 4 X ___3___ X \(\frac{1}{2}\)= 4 X _P_ X \(\frac{1}{2}\) ,
we equate both sides  we get value of P as 3 , therefore 3 times the chef fill the measuring spoon.

Question 16.
You have 7 magnetic blocks that are each \(\frac{9}{100}\) meter long. You connect the ends of the blocks to make a snake. What fraction of a meter is the block snake?
Answer:
\(\frac{63}{100}\) meter is the block snake.

Explanation:
Given 7 magnetic blocks and each are \(\frac{9}{100}\) meter long. After connecting the ends
of the blocks to make a snake is 7 X \(\frac{9}{100}\) = 7 X 9 X \(\frac{1}{100}\) =
63 X \(\frac{1}{100}\) = \(\frac{63}{100}\) meter.

Question 17.
Your friend roller-skates \(\frac{15}{10}\) miles each day for 5 days. How many miles does your friend roller-skate in all?
Big Ideas Math Solutions Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 30
Answer:
My friend roller skate \(\frac{15}{2}\) miles in all.

Explanation:
Given friend roller-skates \(\frac{15}{10}\) miles each day means \(\frac{3}{2}\) miles
each day so for 5 days it is 5 X \(\frac{3}{2}\) = 5 X 3 X \(\frac{1}{2}\) =
15 X \(\frac{1}{2}\) = \(\frac{15}{2}\) miles.

Understand Multiples of Fractions Homework & Practice 9.2

Write the product as a multiple of a unit fraction. Then find the product.

Question 1.
Big Ideas Math Solutions Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 31
Answer:
2 X \(\frac{2}{3}\) = 2 X 2 X \(\frac{1}{3}\) = 4 X (\(\frac{1}{3}\)) =
\(\frac{4}{3}\)

Explanation:
To write the product as a multiple of a unit fraction we write it \(\frac{2}{3}\)
as 2 X \(\frac{1}{3}\) and multiply by 2 = 2 X 2 X \(\frac{1}{3}\) =
4 X \(\frac{1}{3}\) = \(\frac{4}{3}\)

Question 2.
Big Ideas Math Solutions Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 32
Answer:
3 X \(\frac{5}{8}\) = 3 X 5 X \(\frac{1}{8}\) = 15 X (\(\frac{1}{8}\)) =
\(\frac{15}{8}\).

Explanation:
To write the product as a multiple of a unit fraction we write it \(\frac{5}{8}\) as
5 X \(\frac{1}{8}\) and multiply by 3 = 3 X 5 X \(\frac{1}{8}\) = 15 X \(\frac{1}{8}\)
= \(\frac{15}{8}\)

Question 3.
Big Ideas Math Solutions Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 33
Answer:
4 X \(\frac{8}{2}\) = 4 X 8 X \(\frac{1}{2}\) = 32 X (\(\frac{1}{2}\)) =
\(\frac{32}{2}\) =16.

Explanation:
To write the product as a multiple of a unit fraction we write it \(\frac{8}{2}\) as
8 X \(\frac{1}{2}\) and multiply by 4 = 4 X 8 X \(\frac{1}{2}\) =
32 X \(\frac{1}{2}\) = \(\frac{32}{2}\) = 16.

Question 4.
Big Ideas Math Solutions Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 34
Answer:
5 X \(\frac{9}{10}\) = 5 X 9 X \(\frac{1}{10}\) = 45 X (\(\frac{1}{10}\)) =
\(\frac{45}{10}\) = \(\frac{9}{2}\)

Explanation:
To write the product as a multiple of a unit fraction we write it \(\frac{9}{10}\) as
9 X \(\frac{1}{10}\) and multiply by 5 = 5 X 9 X \(\frac{1}{10}\) =
45 X \(\frac{1}{10}\) = \(\frac{45}{10}\) = \(\frac{9}{2}\).

Question 5.
Big Ideas Math Solutions Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 35
Answer:
8 X \(\frac{6}{5}\) = 8 X 6 X \(\frac{1}{5}\) = 48X (\(\frac{1}{5}\)) =
\(\frac{48}{5}\)

Explanation:
To write the product as a multiple of a unit fraction we write it \(\frac{6}{5}\) as
6 X \(\frac{1}{5}\) and multiply by 8 = 8 X 6 X \(\frac{1}{5}\) =
48 X \(\frac{1}{5}\) = \(\frac{48}{5}\).

Question 6.
Big Ideas Math Solutions Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 36
Answer:
10 X \(\frac{2}{4}\) = 10 X 2 X \(\frac{1}{4}\) = 20 X (\(\frac{1}{4}\)) =
\(\frac{20}{4}\) = \(\frac{10}{2}\).

Explanation:
To write the product as a multiple of a unit fraction we write it \(\frac{2}{4}\) as 2 X \(\frac{1}{4}\)
and multiply by 10 = 2 X 10 X \(\frac{1}{4}\) = 20 X \(\frac{1}{4}\) = \(\frac{20}{4}\) =
\(\frac{10}{2}\).

Question 7.
Number Sense
Which expression are equivalent to Big Ideas Math Solutions Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 37 ?
Big Ideas Math Solutions Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 38
Answer:
The expressions equivalent to 4 X \(\frac{7}{8}\) are  (4 X 7) x \(\frac{1}{8}\) ,
\(\frac{28}{8}\),\(\frac{7}{8}\)+\(\frac{7}{8}\)+\(\frac{7}{8}\) +
\(\frac{7}{8}\).

Explanation:
We have the value of expression of 4 X \(\frac{7}{8}\) as \(\frac{28}{8}\) So
1.(4 X 7) x \(\frac{1}{8}\)= \(\frac{28}{8}\)
2. \(\frac{28}{8}\)=\(\frac{28}{8}\)
3. 4 X 7 ≠ \(\frac{28}{8}\)
4. \(\frac{32}{7}\) ≠ \(\frac{28}{8}\)
5. \(\frac{7}{8}\)+\(\frac{7}{8}\)+\(\frac{7}{8}\)+\(\frac{7}{8}\) =
\(\frac{28}{8}\), Therefore expressions 1,2,5 are equivalent to \(\frac{28}{8}\).
The expressions equivalent to 4 X \(\frac{7}{8}\) are  (4 X 7) x \(\frac{1}{8}\) , \(\frac{28}{8}\),\(\frac{7}{8}\)+\(\frac{7}{8}\)+\(\frac{7}{8}\)+\(\frac{7}{8}\)

Question 8.
Which is greater, Big Ideas Math Solutions Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 39 ? Explain.
Answer:
6 X \(\frac{6}{2}\) is greater

Explanation:
6 X \(\frac{6}{2}\)= 6 X 6 X \(\frac{1}{2}\)=36 X \(\frac{1}{2}\)=
\(\frac{36}{2}\) and 5 X \(\frac{7}{2}\) =5 X 7 X \(\frac{1}{2}\)=
35 X \(\frac{1}{2}\), Now comparing both sides 36 X \(\frac{1}{2}\) and
35 X \(\frac{1}{2}\) we see 36 X \(\frac{1}{2}\) > 35 X \(\frac{1}{2}\)
so 6 X \(\frac{6}{2}\) is greater.

Question 9.
Structure
Your friend fills a \(\frac{3}{4}\) – cup measuring cup with rice 2 times.
Write an equation to show how much rice she uses.

Answer:
Friend uses \(\frac{3}{2}\) cups.

Explanation:
Friend fills a \(\frac{3}{4}\) – cup with rice 2 times, so she uses 2 X \(\frac{3}{4}\) =
2 X 3 x \(\frac{1}{4}\) = 6 X \(\frac{1}{4}\)= \(\frac{6}{4}\)=
\(\frac{3}{2}\).

Question 10.
Modeling Real Life
You are making a tornado in each of 2 bottles. Each bottle needs to contain \(\frac{20}{4}\) cups of water.
You only have a \(\frac{1}{4}\) – cup measuring cup. How many times do you need to fill the measuring cup?
Big Ideas Math Solutions Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 39.1
Answer:
To make a tornado we require 40 cups of water.

Explanation:
Each bottle needs \(\frac{20}{4}\) cups of water and have a \(\frac{1}{4}\) – cup
measuring cup, Now we take P as number of times for filling so \(\frac{20}{4}\) =
P X \(\frac{1}{4}\), 20 X \(\frac{1}{4}\)=P X \(\frac{1}{4}\),
Equating both sides we get  P = 20,  so it is 20 times to fill the measuring cups of water
for one bottle, Given that to make a tornado we require 2 bottles, 2 X 20 cups of water = 40 cups of water.

Question 11.
DIG DEEPER!
You and your friend are each selling 12 coupon books. So far, you have sold \(\frac{2}{6}\)
of your books. Your friend has sold 3 times as many as you. What fraction of your friend’s coupon books has she sold?

Answer:
We will get the fraction as whole 1 so friend has sold all the 12 coupon books.

Explanation:
So far i sold \(\frac{2}{6}\) books out of 12 coupon books. Lets us take P number of books,
i have sold P out of 12 coupon books, P= 12 X \(\frac{2}{6}\) =12 X 2 X \(\frac{1}{6}\) =
24 X \(\frac{1}{6}\) = \(\frac{24}{6}\) =4 , So i have sold 4 coupon books,
Given friend has sold 3 times as many as I, Means 3 X 4 =12 means friend has sold all the coupon books

Review & Refresh

Find the product.

Question 12.
12 × 47 = ___
Answer:
12 X 47 = 564

Explanation:
When 12 is multiplied by 47 we get the product as 564.

Question 13.
35 × 31 = ___
Answer:
35 X 31 =1,085

Explanation:
When 35 is multiplied by 31 we get the product as 1,085.

Question 14.
58 × 49 = ___
Answer:
58 X 49 = 2,842

Explanation:
When 58 is multiplied by 49 we get the product as 2,842.

Lesson 9.3 Multiply Whole Numbers and 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 Solutions Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 40
Construct Arguments
Explain how to find the product of a whole number and a fraction without using models.
Answer:

Explanation:

We notice that each expression can be written as the product as a multiple of a unit fraction,
The product of a number and a counting number is a multiple of the number. We can find multiples
of unit fractions. To find the product of a whole number and a fraction without using models,
First we multiply the numerator by the whole number, then rewrite the expression as a multiple
of a unit fraction and denominator stays the same we get the product results.

Think and Grow: Multiply Whole Numbers and Fractions

You can find the product of a whole number and a fraction by multiplying the numerator by the whole number.
The denominator stays the same.
Big Ideas Math Solutions Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 41
Answer:

Explanation:
To find 4 X \(\frac{2}{3}\)
Method 1:  Multiply the numerator by the whole number as 4 X 2 and divide by 3= \(\frac{8}{3}\)
Method 2 : Then writing the expression as a multiple of unit fraction \(\frac{2}{3}\)=
4 X ( 2 X \(\frac{1}{3}\) ) applying associative property of multiplication
(4 X 2) X \(\frac{1}{3}\)= 8 X \(\frac{1}{3}\)=\(\frac{8}{3}\)

Show and Grow

Multiply.

Question 1.
Big Ideas Math Solutions Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 42
Answer:
4 X \(\frac{1}{6}\) = \(\frac{4}{6}\) also equals to \(\frac{2}{3}\)
Explanation :
Already the expression is in multiple of unit fraction so 4 X \(\frac{1}{6}\) = \(\frac{4}{6}\) and further simplification it is equal to \(\frac{2}{3}\)

Question 2.
Big Ideas Math Solutions Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 43
Answer:
3 X \(\frac{2}{4}\) = 3 X \(\frac{1}{2}\) = \(\frac{3}{2}\).

Explanation:
3 X \(\frac{2}{4}\) can be simplified as 3 X \(\frac{1}{2}\) now the expression is in
multiple of unit fraction so  3 X \(\frac{1}{2}\) =\(\frac{3}{2}\).

Question 3.
Big Ideas Math Solutions Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 44
Answer:
7 X  \(\frac{3}{2}\) = 7 x 3 X  \(\frac{1}{2}\) = 21 X \(\frac{1}{2}\) =
\(\frac{21}{2}\).

Explanation:
7 X \(\frac{3}{2}\)  first we write the expression as multiple of unit fraction so
7 X ( 3 X \(\frac{1}{2}\) ) applying associative property of multiplication
( 7 X  3) X \(\frac{1}{2}\) ) = 21 X \(\frac{1}{2}\) = \(\frac{21}{2}\)

Apply and Grow: Practice

Multiply.

Question 4.
Big Ideas Math Solutions Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 45
Answer:
2 X \(\frac{1}{5}\) = \(\frac{2}{5}\).

Explanation :
Already the expression 2 X \(\frac{1}{5}\) is a multiple of unit fraction so we multiply
the whole with the numerator and numerator is being same, 2 X \(\frac{1}{5}\) =
\(\frac{2}{5}\).

Question 5.
Big Ideas Math Solutions Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 46
Answer:
5 X \(\frac{3}{10}\) = \(\frac{15}{10}\)=\(\frac{3}{2}\).

Explanation:
5 X \(\frac{3}{10}\), we write the expression as multiple of unit fraction so
5 X 3 X \(\frac{1}{10}\)= 15 X \(\frac{1}{10}\)= \(\frac{15}{10}\)
further it can be simplified, as both numerator and denominator can be divided by 5 so
\(\frac{15}{10}\)=\(\frac{3}{2}\).

Question 6.
Big Ideas Math Solutions Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 47
Answer:
6 X \(\frac{7}{8}\)= \(\frac{42}{8}\)= \(\frac{21}{4}\).

Explanation:
6 X  \(\frac{7}{8}\), we write the expression as multiple of unit fraction so
6 X 7 X \(\frac{1}{8}\) applying associative property of multiplication
6 X ( 7 X \(\frac{1}{8}\) )= ( 6 X 7 ) X \(\frac{1}{8}\)= 42 X \(\frac{1}{8}\)=
\(\frac{42}{8}\) and further also the fraction can be simplified, as both numerator and denominator
can be divided by 2, making \(\frac{42}{8}\) as \(\frac{21}{4}\).

Question 7.
Big Ideas Math Solutions Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 48
Answer:
8 X \(\frac{9}{10}\)= \(\frac{72}{10}\)=\(\frac{36}{5}\).

Explanation:
8 X  \(\frac{9}{10}\), we write the expression as multiple of unit fraction
so 8 X 9 X \(\frac{1}{10}\) applying associative property of multiplication
8 X ( 9 X \(\frac{1}{10}\) )= (8 X 9 ) X \(\frac{1}{10}\)=
72 X \(\frac{1}{10}\)= \(\frac{72}{10}\) and further also the fraction can be simplified,
as both numerator and denominator can be divided by 2, making \(\frac{72}{10}\) as \(\frac{36}{5}\).

Question 8.
Big Ideas Math Solutions Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 49
Answer:
3 X \(\frac{60}{100}\) =\(\frac{180}{100}\) =\(\frac{18}{10}\).

Explanation:
3 X  \(\frac{60}{100}\),  we write the expression as multiple of unit fraction,
So 3 X 60 X \(\frac{1}{100}\) applying associative property of multiplication
3 X (60 X \(\frac{1}{100}\) )= (3 X 60 ) X \(\frac{1}{100}\)= 180 X \(\frac{1}{100}\)=
\(\frac{180}{100}\) and further also the fraction can be simplified,
as both numerator and denominator can be divided by 10, making \(\frac{180}{100}\)
as \(\frac{18}{10}\).

Question 9.
Big Ideas Math Solutions Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 50
Answer:
\(\frac{4}{2}\) X 10 = 20.

Explanation :
\(\frac{4}{2}\) X 10 , we write the expression as multiple of unit fraction,
So 4 X \(\frac{1}{2}\) X 10= 40 X \(\frac{1}{2}\)=\(\frac{40}{2}\)
as the fraction can be further simplified as both numerator and denominator can be divided by 2, making it as 20.

Compare.

Question 10.
Big Ideas Math Solutions Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 51
Answer:
7 X \(\frac{2}{4}\) =\(\frac{14}{4}\) is smaller <  5 X \(\frac{3}{4}\)=
\(\frac{15}{4}\).

Explanation:
7 X \(\frac{2}{4}\), First we write expression as a multiple of unit fraction
7 X 2 X \(\frac{1}{4}\) =14 X \(\frac{1}{4}\) = \(\frac{14}{4}\)
Now 5 X \(\frac{3}{4}\) is also written as a multiple of unit fraction 5 X 3 X \(\frac{1}{4}\) =
15 X \(\frac{1}{4}\) =\(\frac{15}{4}\) comparing \(\frac{14}{4}\) and
\(\frac{15}{4}\) as the denominators are same same 4 and in numerators 15 is great than 14,
So  \(\frac{14}{4}\)  is less than or smaller or < \(\frac{15}{4}\) ,
7 X \(\frac{2}{4}\) is smaller < 5 X \(\frac{3}{4}\)

Question 11.
Big Ideas Math Solutions Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 52

Answer:
4 X \(\frac{1}{8}\) is greater than > 4 X \(\frac{1}{10}\).

Explanation:
4 X \(\frac{1}{8}\) already the expression is in multiple of unit fraction,
we write as \(\frac{4}{8}\) and even the expression 4 X \(\frac{1}{10}\)
is in multiple of unit fraction we write as \(\frac{4}{10}\) because in order to compare
the fractions we must first turn their different denominators into the same denominators to make
denominators same we multiply 4 X \(\frac{1}{8}\) by 10 /10 the value will not change =
4 X 10 X \(\frac{1}{80}\)=40 X \(\frac{1}{80}\) =  \(\frac{40}{80}\)
now other expression 4 X \(\frac{1}{10}\) by 8/8 so that the value = 4 X 8 X \(\frac{1}{80}\) =
32 X \(\frac{1}{80}\) now as denominator are same we can compare only numerators
\(\frac{40}{80}\) with \(\frac{32}{80}\) as 40 is greater than 32,
So  \(\frac{4}{8}\) is greater than > \(\frac{4}{10}\) ,
4 X \(\frac{1}{8}\) is greater than > 4 X \(\frac{1}{10}\).

Question 12.
Big Ideas Math Solutions Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 53
Answer:
\(\frac{4}{3}\) X 9 = 6 X \(\frac{6}{3}\) both are equal.

Explanation:
First we write expressions as a multiple of unit fraction \(\frac{4}{3}\) X 9 =
9 X 4 X \(\frac{1}{3}\) =36 X \(\frac{1}{3}\) =\(\frac{36}{3}\)
now other expression  6 X \(\frac{6}{3}\) as = 6 X 6 X \(\frac{1}{3}\) =
36 X \(\frac{1}{3}\) = \(\frac{36}{3}\) ,
As denominator is same in the both expression we compare numerators and even
the numerators are same in both the expression so given both products are equal .
Therefore \(\frac{4}{3}\) X 9 =, is equal to 6 X \(\frac{6}{3}\).

Question 13.
YOU BE THE TEACHER
Your friend says the product of 6 and \(\frac{5}{8}\) is \(\frac{5}{48}\). Is your friend correct? Explain.
Answer:
No she is not correct as the product of  6 X \(\frac{5}{8}\) ≠ \(\frac{5}{48}\).

Explanation:
We write the expression as a product  6 and \(\frac{5}{8}\) = 6 X \(\frac{5}{8}\)
to find the product of a whole number and a fraction we multiply the numerator by the whole number.
The denominator stays the same. 6 X 5 X  \(\frac{1}{8}\) =30 X \(\frac{1}{8}\) =
\(\frac{30}{8}\) now comparing with \(\frac{5}{48}\) as denominators are different in order
to compare the fractions we must first turn their different denominators into the same denominators
to make denominators same we multiply \(\frac{30}{8}\) with numerator and denominator
as 6= \(\frac{30}{8}\) X \(\frac{6}{6}\) we get \(\frac{180}{48}\),
now comparing with \(\frac{5}{48}\) being denominators same, now numerators 180,5 both are
not same so friend is not correct as the product of  6 X \(\frac{5}{8}\) ≠ \(\frac{5}{48}\).

Question 14.
Open-Ended
The product of a whole number and a fraction is \(\frac{24}{10}\). What could the two factors be?
Answer:
The product is given as \(\frac{24}{10}\) and the two factors can be (1,24) or (2,12) or (3,8) or (4,6)
with multiple of \(\frac{1}{10}\).

Explanation:
Given the product of a whole number and a fraction is \(\frac{24}{10}\) to find the two factors,
we first write the product as whole with multiple of unit fraction,\(\frac{24}{10}\),
we write as 24 X \(\frac{1}{10}\), now the whole part is 24 we find the factors for 24,
two more numbers that multiply to make 24 are 1 x 24 = 2 x 12 = 3 x 8 = 4 x 6 = 24.
This means the factors of 24 are 1, 2, 3, 4, 6, 8, 12 and 24 so the two factors can be
(1,24) X \(\frac{1}{10}\) or  (2,12) X  \(\frac{1}{10}\) or (3,8) X \(\frac{1}{10}\)
or (4,6) X \(\frac{1}{10}\) = 24 X \(\frac{1}{10}\) = \(\frac{24}{10}\).

Think and Grow: Modeling Real Life

Example
A short roller-coaster track is \(\frac{3}{10}\) mile long. A longer roller-coaster track is about 4 times as long as the short track. About how long is the longer roller-coaster track?
Multiply the shorter roller-coaster track length by 4.
Big Ideas Math Solutions Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 54
Big Ideas Math Solutions Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 54.1
So, the longer roller-coaster track is about _\(\frac{12}{10}\) _ miles long or \(\frac{6}{5}\) miles long
Answer:
The longer roller-coaster track is about \(\frac{12}{10}\) miles long or \(\frac{6}{5}\).

Explanation:
Given a short roller-coaster track is \(\frac{3}{10}\) mile long and longer roller-coaster
track is about 4 times as long as the short track to find how long is the longer roller-coaster track
we multiply short roller-coaster track  \(\frac{3}{10}\) mile by 4, 4 X \(\frac{3}{10}\)
now we write expression as a multiple of unit fraction so 4 X 3 X \(\frac{1}{10}\) =
12 X \(\frac{1}{10}\) =\(\frac{12}{10}\)  as numerator and denominator
can be further simplified as both can be divided by 2 making it as \(\frac{6}{5}\).

Show and Grow

Question 15.
The Renaissance Tower in Dallas, Texas is \(\frac{27}{100}\) kilometer tall. The Burj Khalifa is about 3 times as tall as the Renaissance Tower. About how tall is the Burj Khalifa?
Big Ideas Math Solutions Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 55
Answer:
The Burj Khalifa is \(\frac{81}{100}\) kilometers tall.

Explanation:
Given the Renaissance Tower in Dallas, Texas is \(\frac{27}{100}\) kilometer tall and
The Burj Khalifa is about 3 times as tall as the Renaissance Tower, To find how tall is the Burj Khalifa
we multiply the Renaissance Tower in Dallas, Texas length by 3,
3 X \(\frac{27}{100}\) we write expression as a multiple of unit fraction
3 X 27 X \(\frac{1}{100}\) = 81 X \(\frac{1}{100}\) =\(\frac{81}{100}\),
So the Burj Khalifa is \(\frac{81}{100}\) kilometers tall.

Question 16.
You water 6 plants using \(\frac{3}{5}\) liter of water for each. How many liters of water do you use?
Between which two whole numbers does your answer lie?
Answer:
\(\frac{18}{5}\) liters of water is used, The answer will lie in between 3 and 4.

Explanation:
Given \(\frac{3}{5}\) liter of water for each is watered, for 6 plants total number of liters of water
used is 6 X \(\frac{3}{5},now we write as multiple of unit fraction as 6 X 3 X [latex]\frac{1}{5}\) =
18 X \(\frac{1}{5}\)= \(\frac{18}{5}\) liters of water is used.

Now we divide \(\frac{18}{5}\) we get the value as 3.6,means between 3 and 4 whole numbers our answer will lie.

Question 17.
DIG DEEPER!
You have 6 cups of strawberries. You want to make 4 strawberry-banana smoothies and 4 strawberry-kiwi smoothies. Each smoothie needs \(\frac{2}{3}\) cup of strawberries. Do you have enough? If not, how many more cups of strawberries do you need?
Answer:
Yes we have enough cups of strawberries, we dont require more cups of strawberries.

Explanation:
Given to make each smoothie we need \(\frac{2}{3}\) cup of strawberries
first to make 4 strawberry-banana smoothies we need 4 X \(\frac{2}{3}\) cup of strawberries
= 4 X 2 X \(\frac{1}{3}\)= 8 X \(\frac{1}{3}\) =\(\frac{8}{3}\),
Now similarly for making 4 strawberry-kiwi smoothies we need 4 X \(\frac{2}{3}\)
cup of strawberries = 4 X 2 X \(\frac{1}{3}\)= 8 X \(\frac{1}{3}\) =
\(\frac{8}{3}\),Now total cups of strawberries requried to make smoothie is
\(\frac{8}{3}\) + \(\frac{8}{3}\) = 2 X \(\frac{8}{3}\)=
2 X 8 X \(\frac{1}{3}\)=16 X \(\frac{1}{3}\)= \(\frac{16}{3}\) totals cups
of strawberries required to make smoothies. Now we compare \(\frac{16}{3}\) with 6,
To compare first we make both denominators same so we multiply 6 with \(\frac{3}{3}\)=
6 X \(\frac{3}{3}\)= 6 X 3 X \(\frac{1}{3}\)= 18 X \(\frac{1}{3}\)=
\(\frac{18}{3}\) as both the denominators are same we compare
\(\frac{16}{3}\) and \(\frac{18}{3}\) by seeing numerator
16 is less than 18 means we have enough cups of strawberries.

Multiply Whole Numbers and Fractions Homework & practice 9.3

Multiply

Question 1.
Big Ideas Math Solutions Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 56
Answer:
2 X \(\frac{1}{4}\) = \(\frac{2}{4}\)=\(\frac{1}{2}\)
Explanation:
As the given expression is already a unit fraction we multiply it as 2 X \(\frac{1}{4} =
\)\(\frac{2}{4}\) and further both numerator and denominator can be divided by 2
making it as \(\frac{1}{2}\)

Question 2.
Big Ideas Math Solutions Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 57
Answer:
3 X \(\frac{3}{5}\)= \(\frac{9}{5}\).

Explanation:
3 X \(\frac{3}{5}\) can be written as multiple of unit fraction as
3 X (3 X \(\frac{1}{5}\)) applying associative property of multiplication
( 3 X 3 ) X \(\frac{1}{5}\)= 9 X \(\frac{1}{5}\) = \(\frac{9}{5}\).

Question 3.
Big Ideas Math Solutions Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 58
Answer:
1 X \(\frac{6}{8}\)= \(\frac{6}{8}\)= \(\frac{3}{4}\).

Explanation:
1 X \(\frac{6}{8}\) can be written as multiple of unit fraction as 1 X ( 6 X  \(\frac{1}{8}\))
applying associative property of multiplication ( 1 X 6 ) X \(\frac{1}{8}\) = 6 X \(\frac{1}{8}\)=
\(\frac{6}{8}\) as numerator and denominator can be further simplified as both can be divided by 2
making as \(\frac{6}{8}\)= \(\frac{3}{4}\)X \(\frac{2}{2}\)=\(\frac{3}{4}\).

Question 4.
Big Ideas Math Solutions Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 59
Answer:
4 X \(\frac{10}{12}\)= \(\frac{40}{12}\)= \(\frac{10}{3}\)

Explanation:
4 X \(\frac{10}{12}\) can be written as multiple of unit fraction as 4 X ( 10 X  \(\frac{1}{12}\))
applying associative property of multiplication ( 4 X 10 ) X \(\frac{1}{12}\) = 40 X \(\frac{1}{12}\)=
\(\frac{40}{12}\) as numerator and denominator can be further simplified as both can be divided by 4
as \(\frac{40}{12}\)= \(\frac{10}{3}\) X \(\frac{4}{4}\)=\(\frac{10}{3}\) X 1 =
\(\frac{10}{3}\).

Question 5.
Big Ideas Math Solutions Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 60
Answer:
7 X \(\frac{6}{10}\) = \(\frac{42}{10}\)= \(\frac{21}{5}\)

Explanation:
7 X \(\frac{6}{10}\) can be written as multiple of unit fraction as 7 X ( 6 X  \(\frac{1}{10}\))
applying associative property of multiplication ( 7 X 6 ) X \(\frac{1}{10}\) =
42 X \(\frac{1}{10}\)= \(\frac{42}{10}\) as numerator and denominator
can be further simplified as both can be divided by 2 as \(\frac{42}{10}\)=
\(\frac{21}{5}\) X \(\frac{2}{2}\)=\(\frac{21}{5}\) X 1 = \(\frac{21}{5}\).

Question 6.
Big Ideas Math Solutions Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 61
Answer:
\(\frac{4}{6}\) X 5 =\(\frac{20}{6}\) =\(\frac{10}{3}\).

Explanation:
\(\frac{4}{6}\) X 5  can be written as multiple of unit fraction as 5 X ( 4 X  \(\frac{1}{6}\))
applying associative property of multiplication ( 5 X 4 ) X \(\frac{1}{6}\) =
20 X \(\frac{1}{6}\)= \(\frac{20}{6}\) as numerator and denominator
can be further simplified as both can be divided by 2 as \(\frac{20}{6}\) =
\(\frac{10}{3}\) X \(\frac{2}{2}\)=\(\frac{10}{3}\) X 1 =\(\frac{10}{3}\).

Question 7.
Big Ideas Math Solutions Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 62
Answer:
8 X \(\frac{5}{2}\) = \(\frac{40}{2}\) =20.

Explanation:
8 X \(\frac{5}{2}\) can be written as multiple of unit fraction as
8 X ( 5 X  \(\frac{1}{2}\)) applying associative property of multiplication
( 8 X 5 ) X \(\frac{1}{2}\) = 40 X \(\frac{1}{2}\) = \(\frac{40}{2}\)
further it can be simplified as \(\frac{40}{2}\)=20

Question 8.
Big Ideas Math Solutions Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 63
Answer:
\(\frac{70}{100}\) X 6 = \(\frac{420}{100}\) = \(\frac{42}{10}\).

Explanation:
\(\frac{70}{100}\) X 6  can be written as multiple of unit fraction as
6 X ( 70 X  \(\frac{1}{100}\)) applying associative property of multiplication
( 6 X 70 ) X \(\frac{1}{100}\) = 420 X \(\frac{1}{100}\)= \(\frac{420}{100}\)
as numerator and denominator can be further simplified as both can be divided by 10 as \(\frac{420}{100}\)= \(\frac{42}{10}\) X \(\frac{10}{10}\)=\(\frac{42}{10}\) X 1 =\(\frac{42}{10}\).

Question 9.
Big Ideas Math Solutions Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 64
Answer:
10 X \(\frac{9}{3}\)=\(\frac{90}{3}\)=30.

Explanation:
10 X \(\frac{9}{3}\) can be written as multiple of unit fraction as
10 X ( 9 X  \(\frac{1}{3}\)) applying associative property of multiplication
( 10 X 9 ) X \(\frac{1}{3}\) = 90 X \(\frac{1}{3}\)= \(\frac{90}{3}\)
further it can be simplified  because 90 can be divided by 3 without leaving remainder making
\(\frac{90}{3}\) as 30.

Compare.

Question 10.
Big Ideas Math Solutions Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 65
Answer:
2 X \(\frac{4}{2}\) = 8 X \(\frac{1}{2}\).

Explanation:
2 X \(\frac{4}{2}\) we write as multiple of unit fraction 2 X 4 X \(\frac{1}{2}\) =
8 X \(\frac{1}{2}\) =  \(\frac{8}{2}\) and
8 X \(\frac{1}{2}\) is already multiple of unit fraction = latex]\frac{8}{2}[/latex]
now we can compare both sides  as the denominators are same and numerators are same
both the expressions are equal 2 X \(\frac{4}{2}\) = 8 X \(\frac{1}{2}\).

Question 11.
Big Ideas Math Solutions Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 66
Answer:
9 X \(\frac{1}{5}\) > \(\frac{9}{12}\)

Explanation:
9 X \(\frac{1}{5}\) already it is multiple of unit fraction we write it as \(\frac{9}{5}\)
and 9 X \(\frac{1}{12}\) is also a multiple of unit fraction we write it as \(\frac{9}{12}\)
now comparing first we make both denominators same so the multiples of 5 are 5,10,15,20,25,30,35,40,45,50,55,60,65
and multiples of 12 are 12,24,36,48,60,72 so the least common multiple of both 5 and 12 is 60 ,
So to get denominator 60 we multiply top and bottom by 12 for \(\frac{9}{5}\) =
\(\frac{9}{5}\) X \(\frac{12}{12}\)= \(\frac{108}{60}\) ,
So to get denominator 60 we multiply top and bottom by 5 for \(\frac{9}{12}\)=
\(\frac{9}{12}\) X \(\frac{5}{5}\)= \(\frac{45}{60}\)
Now comparing \(\frac{108}{60}\) with \(\frac{45}{60}\)  as both denominator are same,
As we see numerator it is clear that 108 is more than 45 so \(\frac{108}{60}\) > \(\frac{45}{60}\) ,
9 X \(\frac{1}{5}\) > \(\frac{9}{12}\).

Question 12.
Big Ideas Math Solutions Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 67

Answer:
\(\frac{6}{10}\)X 4 < 9 X \(\frac{3}{10}\).

Explanation:
\(\frac{6}{10}\)X 4 we write as multiple of unit fraction 4 X 6 X \(\frac{1}{10}\)=
24 X \(\frac{1}{10}\) =\(\frac{24}{10}\) and 9 X \(\frac{3}{10}\) as
9 X 3 X \(\frac{1}{10}\)= 27 X \(\frac{1}{10}\)=\(\frac{27}{10}\),
We can compare both as denominators of both are 10 we compare numerators as 24 with 27
as 27 is great,\(\frac{24}{10}\) < \(\frac{27}{10}\),
\(\frac{6}{10}\) X 4 < 9 X \(\frac{3}{10}\).

Question 13.
Writing
Explain one way to multiply a whole number by a fraction.
Answer:
First we write fraction as multiple of unit fraction then applying associative property of multiplication
we multiply the whole numbers first then with the unit fraction,later writing product.

Explanation:
Let us take the whole number as 2 and fraction as \(\frac{3}{5}\),Here first we write fraction as
multiple of unit fraction as 2 X 3 X \(\frac{1}{5}\) now applying associative property of multiplication,
we multiply the whole numbers first then with the unit fraction,
6 X \(\frac{1}{5}\) as \(\frac{6}{5}\).

Question 14.
Number Sense
Between which two whole numbers does the product of 10 and \(\frac{5}{6}\) lie ?
Answer:
In between 8 and 9 whole numbers the product of 10 and \(\frac{5}{6}\) lies.

Explanation:
10 X \(\frac{5}{6}\) we first write it as multiple of unit fraction as
10 X ( 5 X \(\frac{1}{6}\)) applying associative property of multiplication,
We write (10 X 5) X \(\frac{1}{6}\)= 50 X \(\frac{1}{6}\) = \(\frac{50}{6}\)
if we divide \(\frac{50}{6}\))  we get 8.333 which is in between 8 and 9 whole numbers.

Question 15.
Modeling Real Life
A seamstress makes fabric dog collars. An extra small dog collar is \(\frac{1}{3}\) yard long.
A large dog collar is 2 times as long as the extra small dog collar. How long is the large dog collar?
Answer:
The large dog collar is \(\frac{2}{3}\) yards long.

Explanation:
Given the extra small dog collar is \(\frac{1}{3}\) yard long and a large dog collar is
2 times as long as the extra small dog collar, so the large dog collar = 2 X \(\frac{1}{3}\)
as the fraction is multiple of unit fraction we write as \(\frac{2}{3}\) ,
therefore the large dog collar is \(\frac{2}{3}\) yards long.

Question 16.
DIG DEEPER!
Newton has 16 baseball cards. Descartes has \(\frac{1}{4}\) as many cards as Newton, and your friend has \(\frac{1}{2}\) as many cards as Descartes. How many baseball cards do Newton, Descartes, and your friend have in all?
Big Ideas Math Solutions Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 68

Answer:
Newton has 16 baseball cards, Descartes has 4 baseball cards and my friend has 2 baseball cards,
over all there are total 22 baseball cards.

Explanation:
Given Newton has 16 baseball cards and Descartes has \(\frac{1}{4}\) as many cards as
Newton so\(\frac{1}{4}\) of 16 = 16 X \(\frac{1}{4}\) cards  it is already
as multiple of unit fraction we can write as \(\frac{16}{4}\) further can be simplified
because 16 can be divide by 4 so we get as 4 means Descartes has 4 cards.Now my friend has
\(\frac{1}{2}\) as many cards as Descartes So Descartes has 4 cards means my friend has
\(\frac{1}{2}\) out of 4= 4 X \(\frac{1}{2}\)  as it is already a multiple of unit fracrtion,
we take it as \(\frac{4}{2}\) further we can simplify as 4 is divisiblle by 2 we get 2 cards.
Newton has 16 baseball cards,Descartes has 4 baseball cards and my friend has 2 baseball cards,
over all there are total 22 baseball cards.

Review & Refresh

Divide

Question 17.
Big Ideas Math Solutions Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 69
Answer:
Big Ideas Math Solutions Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 69= 19.

Explanation:

Question 18.
Big Ideas Math Solutions Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 70
Answer:
Big Ideas Math Solutions Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 70=81 with 4 as remainder.

Explanation:
Step 1:Start by setting it up with the divisor 7 on the left side and the dividend 571 on the right side,
Big Ideas Math Solutions Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 70
Step 2: The divisor (7) does not  goes into the first digit of the dividend (5), So we take second digit also , The divisor (7) goes into the two digits of the dividend (57) by 8 time(s),there fore put 8 on the top
8
Big Ideas Math Solutions Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 70
Step 3: Multiply the divisor by the result in the previous step. Therefore, put 8 on top (7 x 9 = 56) and write that answer below the dividend.
8
Big Ideas Math Solutions Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 70
56
Step 4: Subtract the result in the previous step from the two digits of the dividend (57 -56 = 1) and write the answer below.
8
Big Ideas Math Solutions Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 70
 56
1
Step 5: Move down the 3rd digit of the dividend (1)
8
Big Ideas Math Solutions Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 70
 56
11
Step 6:The divisor (7) goes into the bottom number (11),  time(s). Therefore, put 1 on top:
81
Big Ideas Math Solutions Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 70
     56
11
Step 7 : Multiply the divisor by the result in the previous step (7 x 1 = 7) and write that answer at the bottom:
81
Big Ideas Math Solutions Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 70
     56
11
07
Step 8 : Subtract the result in the previous step from the number written above it. (11 – 7 = 04) and write the answer at the bottom
81
Big Ideas Math Solutions Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 70
 56
11
    07
4
We stop here as 4 is less than 7 and further we get in decimals , So Big Ideas Math Solutions Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 70=81 with 4 as remainder
Question 19.
Big Ideas Math Solutions Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 71
Answer:
Big Ideas Math Solutions Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 71= 184 with 3 as remainder.

Explanation:
Step 1:Start by setting it up with the divisor 5 on the left side and the dividend 923 on the right side,
Big Ideas Math Solutions Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 71
Step 2: The divisor (5)   goes into the first digit of the dividend (9)  by 8 time there fore put 1 on the top
1
Big Ideas Math Solutions Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 71
Step 3: Multiply the divisor by the result in the previous step  (5 x 1 = 5) and write that answer below the dividend.
1
Big Ideas Math Solutions Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 71
5
Step 4: Subtract the result in the previous step from the first digit of the dividend (9 -5 = 4) and write the answer below.
1
Big Ideas Math Solutions Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 71
   5
    4
Step 5: Move down the 2nd digit of the dividend (2),like below
1
Big Ideas Math Solutions Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 71
   5
     42
Step 6:The divisor (5) goes into the bottom number (40),  8 time(s). Therefore, put 8 on top:
18
Big Ideas Math Solutions Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 71
   5
     42
Step 7 : Multiply the divisor by the result in the previous step (5 x 8 = 40) and write that answer at the bottom:
18
Big Ideas Math Solutions Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 71
   5
     42
40
Step 8: Subtract the result in the previous step from the number written above it. (42 – 40 = 04) and write the answer at the bottom
18
Big Ideas Math Solutions Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 71
   5
     42
     40   
2
Step 9 :Move down the 3nd digit of the dividend (3),like below
18
Big Ideas Math Solutions Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 71
   5
     42
     40   
23
Step 10 : The divisor (5) goes into the bottom number (20),  4 time(s). Therefore, put 4 on top:
184
Big Ideas Math Solutions Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 71
     5
     42
     40   
23
20
Step 11 : Subtract the result in the previous step from the number written above it. (23 – 20 = 03) and write the answer at the bottom
184
Big Ideas Math Solutions Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 71
     5
     42
     40   
23
     20
       03
We stop here as 3 is less than 5 and further we get in decimals , So Big Ideas Math Solutions Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 71= 184 with 3 as remainder.

Lesson 9.4 Multiply Whole Numbers and Mixed Numbers

Explore and Grow

Use models to help you complete the table.
Big Ideas Math Answers Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 72

Structure
How does the Distributive Property relate to your models? Explain.

Answer:

Explanation:
According to the distributive property, multiplying the sum of two or more addends by a number
will give the same result as multiplying each addend individually by the number and then adding the products together.
We use distribute property while relating the models first we draw each multiplying addend individually and
then add the products together to give the result.
Example : 2 X 1 \(\frac{3}{8}\) using the distributive property we write
2 X 1 \(\frac{3}{8}\) as 2 X ( 1 + \(\frac{3}{8}\) )here first we draw model of
2 X 1 then we draw model of 2 X \(\frac{3}{8}\) then we add the products together as
2 + \(\frac{6}{8}\)= 2 + \(\frac{3}{4}\)=\(\frac{11}{4}\).
Think and Grow: Multiply Whole Numbers and Mixed Numbers

You can find the product of a whole number and a mixed number by writing the mixed number as a fraction or by using the Distributive Property
Big Ideas Math Answers Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 73
One Way: Write the mixed number as a fraction, then multiply.
Big Ideas Math Answers Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 74
Another Way: Use the Distributive Property.
Big Ideas Math Answers Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 75
Big Ideas Math Answers Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 76
Answer:
3 X 1 \(\frac{1}{2}\) = \(\frac{9}{2}\)
Explanation:
First we write the mixed number as a fraction then multiply, so 1 \(\frac{1}{2}\) =
1+ \(\frac{1}{2}\) = 2 x 1+1 by 2=  \(\frac{3}{2}\) now we multiply it with the whole ,
3 X \(\frac{3}{2}\) now we write in unit fraction and multiply  3 X 3 X \(\frac{1}{2}\)=
9 X  \(\frac{1}{2}\)= \(\frac{9}{2}\).

Show and Grow

Multiply

Question 1.
Big Ideas Math Answers Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 77
Answer:
Big Ideas Math Answers Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 77 = \(\frac{27}{4}\)=6 \(\frac{3}{4}\)
Explanation:
First we write the mixed number as a fraction then multiply, so 2 \(\frac{1}{4}\) =
2 + \(\frac{1}{4}\)=(2 X 4 + 1) by 4 = \(\frac{9}{4}\) now we multiply it with the whole,
3 X \(\frac{9}{4}\) now we write in unit fraction and multiply  3 X 9 X \(\frac{1}{4}\) =
27 X \(\frac{1}{4}\) = \(\frac{27}{4}\), we can write the product as mixed fraction as
4 goes in 6 times. 6 will be our whole number 4 X 6 is 24 and we have 3 left over (2724),
3 will be our numerator and our denominator will stay the same 4= \(\frac{27}{4}\)=
6 \(\frac{3}{4}\).

Question 2.
Big Ideas Math Answers Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 78
Answer:
Big Ideas Math Answers Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 78 \(\frac{38}{12}\)= 3 \(\frac{2}{12}\).

Explanation:
First we write the mixed number as a fraction then multiply, so 1 \(\frac{7}{12}\)=
1 + \(\frac{7}{12}\)= (1 X 12 + 7 ) by 12 = \(\frac{19}{12}\)
now we multiply it with the whole, 2 X \(\frac{19}{12}\) now we write in unit fraction and
multiply  2 X 19 X \(\frac{1}{12}\)= 38 X \(\frac{1}{12}\) = \(\frac{38}{12}\),
we can write the product as mixed fraction as 12 goes in 3 times. 3 will be our whole number 12 X 3 is 36 and
we have 2 left over (38−36). 2 will be our numerator and our denominator will be the same 12
= \(\frac{38}{12}\)=  3 \(\frac{2}{12}\).

Question 3.
Big Ideas Math Answers Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 79
Answer:
Big Ideas Math Answers Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 79= \(\frac{116}{8}\)= 14 \(\frac{4}{8}\)
Explanation:
First we write the mixed number as a fraction then multiply, so 3 \(\frac{5}{8}\)=
3 + \(\frac{5}{8}\)= (3 X 8 + 5) by 8 = \(\frac{29}{8}\) now we multiply it with the whole,
4 X \(\frac{29}{8}\) now we write in unit fraction and multiply  4 X 29 X \(\frac{1}{8}\)=
116 X \(\frac{1}{8}\) = \(\frac{116}{8}\), we can write the product as mixed fraction as
8 goes in 14 times. 14 will be our whole number 8 X 14 is 112 and we have 4 left over (116−112).
4 will be our numerator and our denominator will be the same 8= \(\frac{116}{8}\)=
14 \(\frac{4}{8}\).

Apply and Grow: Practice

Multiply.

Question 4.
Big Ideas Math Answers Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 80
Answer:
Big Ideas Math Answers Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 80= \(\frac{50}{3}\)=16 \(\frac{2}{3}\)
Explanation:
First we write the mixed number as a fraction then multiply, so 1 \(\frac{2}{3}\)=
1 + \(\frac{2}{3}\)= (1 X 3 + 2) by 3 = \(\frac{5}{3}\) now we multiply it with the whole,
10 X \(\frac{5}{3}\) now we write in unit fraction and multiply  10 X 5 X \(\frac{1}{3}\)=
50 X \(\frac{1}{3}\) = \(\frac{50}{3}\), we can write the product as mixed fraction as
3 goes in 16 times. 16 will be our whole number 3 X 16 is 48 and we have 2 left over (50−48).
2 will be our numerator and our denominator will be the same 3= \(\frac{50}{3}\)= 16 \(\frac{2}{3}\).

Question 5.
Big Ideas Math Answers Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 81
Answer:
Big Ideas Math Answers Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 81= \(\frac{25}{2}\)=12 \(\frac{1}{2}\)

Explanation:
First we write the mixed number as a fraction then multiply, so 2 \(\frac{1}{2}\)=
2 + \(\frac{1}{2}\)= (2 X 2 + 1) by 2 = \(\frac{5}{2}\)
now we multiply it with the whole, 5 X \(\frac{5}{2}\) .
We write in unit fraction and multiply  5 X 5 X \(\frac{1}{2}\)=
25 X \(\frac{1}{2}\) = \(\frac{25}{2}\), we can write the product as mixed fraction
as 2 goes in 12 times. 12 will be our whole number 2 X 12 is 24 and we have 1 left over (25−24).
1 will be our numerator and our denominator will be the same 2= \(\frac{25}{2}\)=
12 \(\frac{1}{2}\)

Question 6.
Big Ideas Math Answers Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 82
Answer:
Big Ideas Math Answers Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 82= \(\frac{184}{6}\)=\(\frac{92}{3}\)=30 \(\frac{4}{6}\).

Explanation:
First we write the mixed number as a fraction then multiply, so 7 \(\frac{4}{6}\)=
7 + \(\frac{4}{6}\)=(7 X 6 + 4) by 6 = \(\frac{46}{6}\) now we multiply it with the whole,
4 X \(\frac{46}{6}\) now we write in unit fraction and multiply  4 X 46 X \(\frac{1}{6}\)=
184 X \(\frac{1}{6}\) = \(\frac{184}{6}\), we can write the product as mixed fraction as
6 goes in 30 times. 30 will be our whole number 6 X 30 is 180 and we have 4 left over (184−180).
4 will be our numerator and our denominator will be the same 6= \(\frac{184}{6}\)=
30 \(\frac{4}{6}\) and \(\frac{184}{6}\) can also written as \(\frac{92}{3}\)
because both 184 and 6 can also be further divided by 2.

Question 7.
Big Ideas Math Answers Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 83
Answer:
Big Ideas Math Answers Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 83= \(\frac{243}{5}\)= 48 \(\frac{3}{5}\)
Explanation:
First we write the mixed number as a fraction then multiply, so 5 \(\frac{2}{5}\)=
5 + \(\frac{2}{5}\)= (5 X 5 + 2) by 5 = \(\frac{27}{5}\) now we multiply it with the whole,
9 X \(\frac{27}{5}\) now we write in unit fraction and multiply  9 X 27 X \(\frac{1}{5}\)=
243 X \(\frac{1}{5}\) = \(\frac{243}{5}\), we can write the product as mixed fraction as
5 goes in 48 times. 48 will be our whole number 5 X 48 is 240 and we have 3 left over (243−240).
3 will be our numerator and our denominator will be the same 5= \(\frac{243}{5}\)=
48 \(\frac{3}{5}\).

Question 8.
Big Ideas Math Answers Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 84
Answer:
Big Ideas Math Answers Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 84= \(\frac{312}{10}\)= 31 \(\frac{2}{10}\)

Explanation:
First we write the mixed number as a fraction then multiply, so 3 \(\frac{9}{10}\)=
3 + \(\frac{9}{10}\)= (3 X 10 + 9) by 10 = \(\frac{39}{10}\)
now we multiply it with the whole, 8 X \(\frac{39}{10}\) now we write in unit fraction
and multiply  8 X 39 X \(\frac{1}{10}\)= 312 X \(\frac{1}{10}\) =
\(\frac{312}{10}\), we can write the product as mixed fraction as 10 goes in 31 times.
31 will be our whole number 10 X 31 is 310 and we have 2 left over (312−310).
2 will be our numerator and our denominator will be the same 10= \(\frac{312}{10}\)=
31 \(\frac{2}{10}\).

Question 9.
Big Ideas Math Answers Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 85
Answer:
Big Ideas Math Answers Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 85= \(\frac{5466}{100}\)=54 \(\frac{66}{100}\).
Explanation:
First we write the mixed number as a fraction then multiply, so 9 \(\frac{11}{100}\)=
9 + \(\frac{11}{100}\)= (9 X 100 + 11) by 100 = \(\frac{911}{100}\)
now we multiply it with the whole, 6 X \(\frac{911}{100}\) now we write in unit fraction
and multiply  6 X 911 X \(\frac{1}{100}\)= 5466 X \(\frac{1}{100}\) =
\(\frac{5466}{100}\), we can write the product as mixed fraction as 100 goes in 54 times.
54 will be our whole number 100 X 54 is 5400 and we have 66 left over (5466−5400).
66 will be our numerator and our denominator will be the same 100= \(\frac{5466}{100}\)=
54 \(\frac{66}{100}\).

Question 10.
Reasoning
How can you check whether your answer in Exercise 9 is reasonable?
Answer:
Exercise 9 is reasonable.

Explanation:
In the above problem 6 X 9 \(\frac{11}{100}\) we got product as
\(\frac{5466}{100}\) =54 \(\frac{66}{100}\)  we can write the product as mixed fraction
as 100 goes in 54 times. 54 will be our whole number 100 X 54 is 5400 and 5466-5400,
leaving 66 as remainder so we consider the whole part 54,As 54 is reasonable close to 100.

Question 11.
YOU BE THE TEACHER
Your friend finds the product of 9 and 6\(\frac{1}{3}\). Is your friend correct? Explain.
Big Ideas Math Answers Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 86
Answer:
Yes friend is correct.

Explanation:
First we write the mixed number as a fraction then multiply, so 6 \(\frac{1}{3}\)= 6 + \(\frac{1}{3}\)= (6 X 3 + 1) by 3 = \(\frac{19}{3}\) now we multiply it with the whole, 9 X \(\frac{19}{3}\) now we write in unit fraction and multiply  9 X 19 X \(\frac{1}{3}\)= 171 X \(\frac{1}{3}\) = \(\frac{171}{3}\), As 3 goes 57 times leaving no remainder so we get 9 X 6 \(\frac{1}{3}\)=57 which matches with the friend’s product so friend is correct.

Question 12.
DIG DEEPER!
Without multiplying, can you tell which expression is greater, Big Ideas Math Answers Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 87 ? Explain.
Answer:
5 X 6 \(\frac{1}{4}\) < 6 X 5 \(\frac{1}{4}\)

Explanation:

By drawing models also we can say

If the denominators are the same, then the fraction with the greater numerator is the greater fraction.
The fraction with the lesser numerator is the lesser fraction, But first we write the mixed number as a fraction
so in 5 X 6 \(\frac{1}{4}\), 6 \(\frac{1}{4}\)= 6+ \(\frac{1}{4}\)=
(6 X 4 +1) by 4 =  \(\frac{25}{4}\) now we multiply it with the whole as
5 X \(\frac{25}{4}\) now we write the mixed fraction 5 \(\frac{1}{4}\)
as 5+ \(\frac{1}{4}\) = 5 x 4 +1 by 4=5 \(\frac{21}{4}\)
now we multiply it with  the whole as 6 X \(\frac{21}{4}\) ,
Now comparing 5 X  \(\frac{25}{4}\)  and  6 X \(\frac{21}{4}\)
in both denominators are same and numerators 5 X \(\frac{25}{4}\) is less than
6 X \(\frac{21}{4}\) so 5 X 6 \(\frac{1}{4}\) < 6 X 5 \(\frac{1}{4}\)

Think and Grow: Modeling Real Life

Example
An elephant sleeps 2\(\frac{1}{2}\) hours in 1 day. A koala sleeps 6 times as long as the elephant in the same day. How many hours does the koala sleep that day?
Big Ideas Math Answers Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 87.1
Multiply the number of hours the elephant sleeps by 6.
Big Ideas Math Answers Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 88
Answer:
The koala sleeps 5 hours that day.

Explanation:
First we write mixed fraction as fraction so 2 \(\frac{1}{2}\) =( 2 X 2 + 1 ) by 2
= \(\frac{5}{2}\) , now we multiply with whole number 2 = 2 X \(\frac{5}{2}\) =
2 X 5 X \(\frac{1}{2}\) = 10 X \(\frac{1}{2}\) = \(\frac{10}{2}\)
further it can be simplified as both 10,2 can be divided by 2 so \(\frac{10}{2}\) = 5 hours that day.

Show and Grow

Question 13.
A small bag of popcorn has 1\(\frac{7}{8}\) cups of popcorn. A large bag has 7 times as many cups of popcorn as the small bag. How many cups of popcorn are in the large bag?
Answer:
The large bag has \(\frac{105}{8}\) cups of popcorn or 13 \(\frac{1}{8}\) cups of popcorn.

Explanation:
Given A small bag of popcorn has 1\(\frac{7}{8}\) cups of popcorn so first we write mixed fraction
as fraction as 1 \(\frac{7}{8}\)= ( 1 X 8 + 7 ) by 8 =\(\frac{15}{8}\) ,as a large bag has
7 times as many cups of popcorn as the small bag so 7 times \(\frac{15}{8}\)=
7 X 15 X \(\frac{1}{8}\)=105 X \(\frac{1}{8}\)=\(\frac{105}{8}\) cups of popcorn.
We can write the product as mixed fraction as 8 goes in 13 times.
13 will be our whole number 8 X 13 is 104 and we have 1 left over (105−104).
1 will be our numerator and our denominator will be the same 8= \(\frac{105}{8}\)=
13 \(\frac{1}{8}\) cups of popcorn.

Question 14.
DIG DEEPER!
A deli worker prepares 4 packages of meat. Each package contains 3\(\frac{1}{3}\) pounds of meat.
Your friend says the deli worker uses between 12 and 13 pounds of meat. Is your friend correct? Explain.
Answer:
No, friend is not correct as deli worker uses between 13 and 14 pounds of meat,
not between 12 and 13 pounds of meat.

Explanation:
Given each package contains 3\(\frac{1}{3}\) pounds of meat.
First we write mixed fraction as fraction as ( 3 X 3 +1 ) by 3= \(\frac{10}{3}\)
now deli worker prepares 4 packages of meat so 4 times  \(\frac{10}{3}\)=
4 X 10 X \(\frac{1}{3}\)= 40 X \(\frac{1}{3}\)=\(\frac{40}{3}\) ,
we will write the product as mixed fraction as 3 goes in 13 times. 13 will be our whole number
3 X 13 is 39 and we have 1 left over (40−39). 1 will be our numerator and our denominator
will be the same 3= \(\frac{105}{8}\)= 13 \(\frac{1}{3}\),
so the whole number is greater than 13 means deli worker uses between 13 and 14 pounds of meat
not between 12 and 13 pounds of meat.

Question 15.
DIG DEEPER!
An athlete’s goal is to run at least 80 miles in 1 week. He runs 5\(\frac{3}{4}\) miles 2 times each day for 1 week. Does the athlete meet his goal? Explain.
Big Ideas Math Answers Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 89
Answer:
Athlete runs more than 80 miles in 1 week so he meets his goal.

Explanation:
Given an athlete runs 5\(\frac{3}{4}\) miles 2 times each day, First we write mixed fraction
as fraction as ( 5 X 4 +3) by 4 =\(\frac{23}{4}\) now we multiply by 2 ,
2 X \(\frac{23}{4}\)= 2 X 23 X \(\frac{1}{4}\)=46 x \(\frac{1}{4}\)=
\(\frac{46}{4}\), Now for 1 week it is 7 X \(\frac{46}{4}\)= 7 X 46 X \(\frac{1}{4}\)=
322 X \(\frac{1}{4}\)= \(\frac{322}{4}\) now we will write the product as mixed fraction
as 4 goes in 80 times. 80 will be our whole number 4 X 80 is 320 and we have 2 left over (322−320).
2 will be our numerator and our denominator will be the same 4= \(\frac{322}{4}\)= 80 \(\frac{2}{4}\) ,
As 80 \(\frac{2}{4}\) is greater than 80,  athlete runs more than 80 miles in 1 week he meets his goal.

Multiply Whole Numbers and Mixed Numbers Homework & Practice 9.4

Multiply

Question 1.
Big Ideas Math Answers Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 90
Answer:
2 X 1 \(\frac{1}{12}\) = \(\frac{26}{12}\) = \(\frac{13}{6}\)=
2 \(\frac{1}{6}\).

Explanation:
First we write the mixed number as a fraction then multiply, so 1 \(\frac{1}{12}\) =
1 + \(\frac{1}{12}\) = ( 1 X 12 + 1 )by 12 = \(\frac{13}{12}\) now we multiply it with the whole,
2 X \(\frac{13}{12}\) now we write in unit fraction and multiply  2 X 13 X \(\frac{1}{12}\)=
26 X \(\frac{1}{12}\) = \(\frac{26}{6}\), Now the fraction can be further divided by 2
as 26,6 both goes in 2 after dividing it becomes as \(\frac{13}{6}\) now we can write the product
as mixed fraction as 6 goes in 2 times. 2 will be our whole number 6 X 2 is 12 and we have 1 left over (13−12).
1 will be our numerator and our denominator will be the same 6= \(\frac{13}{6}\)=
2 \(\frac{1}{6}\).

Question 2.
Big Ideas Math Answers 4th Grade Chapter 9 Multiply Whole Numbers and Fractions 91
Answer:
2 X 3\(\frac{5}{6}\)=\(\frac{46}{6}\) = \(\frac{23}{3}\)= 7 \(\frac{2}{3}\)

Explanation:
First we write the mixed number as a fraction then multiply, so 3 \(\frac{5}{6}\) =
3 + \(\frac{5}{6}\) =(3 X 6 + 5) by 6 = \(\frac{23}{6}\) now we multiply it with the whole,
2 X \(\frac{23}{6}\) now we write in unit fraction and multiply  2 X 23 X \(\frac{1}{6}\)=
46 X \(\frac{1}{6}\) = \(\frac{46}{6}\), Now the fraction can be further divided by 2
as 46,6 both goes in 2 after dividing it becomes as 23,3 = \(\frac{23}{3}\)
now we can write the product as mixed fraction as 3 goes in 7 times. 3 X 7  is 21 ,
7 will be our whole number and  (23−21) we have 2 left over. 2 will be our numerator and
our denominator will be the same 3= \(\frac{23}{3}\)= 7 \(\frac{2}{3}\).

Question 3.
Big Ideas Math Answers 4th Grade Chapter 9 Multiply Whole Numbers and Fractions 92
Answer:
4 X 3 \(\frac{6}{10}\) = \(\frac{144}{10}\) = \(\frac{72}{5}\) = 14 \(\frac{2}{5}\).

Explanation:
First we write the mixed number as a fraction then multiply, so 3 \(\frac{6}{10}\) =
3 + \(\frac{6}{10}\) =( 3 X 10 + 6 ) by 10 = \(\frac{36}{10}\) now we multiply it with the whole,
4 X \(\frac{36}{10}\) now we write in unit fraction and multiply  4 X 36 X \(\frac{1}{10}\)=
144 X \(\frac{1}{10}\) = \(\frac{144}{10}\), Now the fraction can be further divided by 2
as 144,10 both goes in 2 after dividing it becomes as 72,5 = \(\frac{72}{5}\)
now we can write the product as mixed fraction as 5 goes in 14 times. 5 X 14  is 70 ,14 will be our whole number
and  (72−70) we have 2 left over. 2 will be our numerator and our denominator will be the same 5
= \(\frac{72}{5}\)= 14 \(\frac{2}{5}\).

Question 4.
Big Ideas Math Answers 4th Grade Chapter 9 Multiply Whole Numbers and Fractions 93
Answer:
2 \(\frac{3}{8}\) X 5 = \(\frac{95}{8}\) =11 \(\frac{7}{8}\).

Explanation:
First we write the mixed number as a fraction then multiply, so 2 \(\frac{3}{8}\) =
2 + \(\frac{3}{8}\) =(2 X 8 + 3) by 8 = \(\frac{19}{8}\) now we multiply it with the whole,
5 X \(\frac{19}{8}\) now we write in unit fraction and multiply  5 X 19 X \(\frac{1}{8}\)=
95 X \(\frac{1}{8}\) = \(\frac{95}{8}\), Now we can write the product as mixed fraction as
8 goes in 11 times. 8 X 11  is 88 ,11 will be our whole number and  (95−88) we have 7 left over.
7 will be our numerator and our denominator will be the same 8= \(\frac{95}{8}\)=
11 \(\frac{7}{8}\).

Question 5.
Big Ideas Math Answers 4th Grade Chapter 9 Multiply Whole Numbers and Fractions 94
Answer:
4 X 6 \(\frac{4}{5}\) = \(\frac{136}{5}\) = 27 \(\frac{1}{5}\)
Explanation:
First we write the mixed number as a fraction then multiply, so 6 \(\frac{4}{5}\) =
6 + \(\frac{4}{5}\) =(6 X 5 + 4) by 5 = \(\frac{34}{5}\) now we multiply it with the whole,
4 X \(\frac{34}{5}\) now we write in unit fraction and multiply  4 X 34 X \(\frac{1}{5}\) =
136 X \(\frac{1}{5}\) = \(\frac{136}{5}\), now we can write the product as mixed fraction as
5 goes in 27 times. 5 X 27  is 135 , 27 will be our whole number and  (136−135) we have 1 left over.
1 will be our numerator and our denominator will be the same 5 = \(\frac{136}{5}\)=
27 \(\frac{1}{5}\).

Question 6.
Big Ideas Math Answers 4th Grade Chapter 9 Multiply Whole Numbers and Fractions 95
Answer:
8 \(\frac{20}{100}\) X 10 = \(\frac{8200}{100}\) = 82.

Explanation:
First we write the mixed number as a fraction then multiply, so 8 \(\frac{20}{100}\) =
8 + \(\frac{20}{100}\) = ( 8 X 100 + 20 ) by 100 = \(\frac{820}{100}\)
now we multiply it with the whole, 100 X \(\frac{820}{100}\) now we write in unit fraction
and multiply  10 X 820 X \(\frac{1}{100}\) = 8200 X \(\frac{1}{100}\) =
\(\frac{8200}{100}\), Now the fraction can be further divided by 100 as 8200,100
both goes in 100 after dividing by 100  it becomes as 82.

Question 7.
Reasoning
Without calculating, is the product of 7 and 5\(\frac{3}{4}\) greater than or less than 35? Explain.
Answer:
7 X 5 \(\frac{3}{4}\) is greater than ( > ) 35.

Explanation:
Given to compare between 7 X 5 \(\frac{3}{4}\) and 35, first we take 7 X 5 \(\frac{3}{4}\)
to this expression if we apply distributive property means if multiplying the sum of two or more addends
by a number will give the same result as multiplying each addend individually by the number and
then adding the products together. example a X b \(\frac{c}{d}\) = a X b + a X \(\frac{c}{d}\)
So 7 X 5 \(\frac{3}{4}\) = 7 X 5 + 7 X \(\frac{3}{4}\) means 35 + 7 X  \(\frac{3}{4}\)
So by comparing left side it is addition of 35 plus some value ,it will be more than 35 only and right hand side
it it 35 so obiviously 7 X 5 \(\frac{3}{4}\) is greater than ( > ) 35.

Question 8.
YOU BE THE TEACHER
Your friend finds the product of 4 and 2\(\frac{8}{10}\). Is your friend correct? Explain.
Big Ideas Math Answers 4th Grade Chapter 9 Multiply Whole Numbers and Fractions 96
Answer:
Yes friend is correct, because the value is same as 11 \(\frac{2}{10}\) .

Explanation:
To caluculate the value of 4 X 2 \(\frac{8}{10}\) , First we write the mixed number as a
fraction then multiply, so 2 \(\frac{8}{10}\) = 2 + \(\frac{8}{10}\) =
( 2 X 10 + 8 ) by 10 = \(\frac{28}{10}\) now we multiply it with the whole,
4 X \(\frac{28}{10}\) now we write in unit fraction and multiply  4 X 28 X \(\frac{1}{10}\)=
112 X \(\frac{1}{10}\) = \(\frac{112}{10}\), now we can write the product as mixed fraction
as 10 goes in 11 times. 10 X 11  is 110 , 11 will be our whole number and  (112−110) we have 2 left over.
2 will be our numerator and our denominator will be the same 10= \(\frac{112}{10}\)=
11 \(\frac{2}{10}\) is same as friend, So friend is correct.

Question 9.
Number Sense
Between which two whole numbers does the product of 9 and 7 \(\frac{1}{8}\) lie?
Answer:
The product of 9 and 7 \(\frac{1}{8}\) lies in between 64 and 65.

Explanation:
First we write the mixed number 7 \(\frac{1}{8}\) as a fraction then multiply, so 7 \(\frac{1}{8}\) =
7 + \(\frac{1}{8}\) = 7 X 8 + 1 by 8 = \(\frac{57}{8}\) now we multiply it with the whole,
9 X \(\frac{57}{8}\) now we write in unit fraction and multiply  9 X 57 X \(\frac{1}{8}\) =
513 X \(\frac{1}{8}\) = \(\frac{513}{8}\), now we can write the product as mixed fraction as
8 goes in 64 times. 8 X 64  is 512 , 64 will be our whole number and  (513−512) we have 1 left over.
1 will be our numerator and our denominator will be the same 8 = \(\frac{513}{8}\)= 64
\(\frac{1}{8}\). Any how we got 64 whole number and next whole number is 65 means
the product of 9 and 7 \(\frac{1}{8}\) lies in between 64 and 65.

Question 10.
Modeling Real Life
Athlete A holds a 2\(\frac{1}{2}\) – kilogram plate while doing squats. Athlete B holds a plate that is 4 times heavier than Athlete A’s. How many kilograms is the plate held by Athlete B?
Big Ideas Math Answers 4th Grade Chapter 9 Multiply Whole Numbers and Fractions 97
Answer:
Athlete B holds 10 kilograms of plate while doing squats.

Explanation:
Given Athlete A hold a 2 \(\frac{1}{2}\) kilogram plate while doing squats,
Athlete B holds a plate that is 4 times heavier than Athlete A’s, So Athlete B holds
4 X 2 \(\frac{1}{2}\), First we write the mixed number as a fraction then multiply,
so 2 \(\frac{1}{2}\) = 2 + \(\frac{1}{2}\) = ( 2 X 2 + 1 ) by 2 = \(\frac{5}{2}\)
now we multiply it with the whole, 4 X \(\frac{5}{2}\) now we write in unit fraction and
multiply  4 X 5 X \(\frac{1}{2}\) = 20 X \(\frac{1}{2}\) = \(\frac{20}{2}\)
as 20,2 goes by 2 we further divide both by 2 we get 10,1 = \(\frac{20}{2}\) = 10,
therefore Athlete B holds 10 kilograms of plate while doing squats.

Question 11.
DIG DEEPER!
A zoo nutritionist orders 5 \(\frac{1}{4}\) tons of apples and 7\(\frac{2}{4}\)tons of bananas each year to feed the animals. She orders 6 times as many tons of herbivore pellets than tons of fruit. How many tons of herbivore pellets does the nutritionist order?
Answer:
Explanation:

Review & Refresh

Subtract.

Question 12.
Big Ideas Math Answers 4th Grade Chapter 9 Multiply Whole Numbers and Fractions 98
Answer:
Big Ideas Math Answers 4th Grade Chapter 9 Multiply Whole Numbers and Fractions 98= \(\frac{12}{4}\) = 3
Explanation:

First we write the mixed numbers into fractions and subtract, So 9 \(\frac{1}{4}\) =
9 +\(\frac{1}{4}\) = (9 x 4 + 1) by 4 = \(\frac{37}{4}\) and  6 \(\frac{1}{4}\) =
6 + \(\frac{1}{4}\) = 6 X 4 + 1 by 4 = \(\frac{25}{4}\) now subtracting
\(\frac{37}{4}\) – \(\frac{25}{4}\) to subtract a fraction from another
we need to first make sure both fractions have the same denominator as both denominators
have same value 4 now we can subtract one numerator from the other to make one fraction as 37 – 25 by 4 = \(\frac{12}{4}\) as 12, 4 goes by 4 we further can divide both by 4 we get 12,1 = \(\frac{12}{4}\) = 3 .

Question 13.
Big Ideas Math Answers 4th Grade Chapter 9 Multiply Whole Numbers and Fractions 99
Answer:
Big Ideas Math Answers 4th Grade Chapter 9 Multiply Whole Numbers and Fractions 99 = \(\frac{11}{3}\) = 3 \(\frac{2}{3}\)
Explanation:
First we write the mixed numbers into fractions and subtract, So 6 \(\frac{1}{3}\) =
6 +\(\frac{1}{3}\) =(6 x 3 + 1) by 3 = \(\frac{19}{3}\) and  2 \(\frac{2}{3}\) =
2 + \(\frac{2}{3}\) = 2 X 3 + 2 by 3 = \(\frac{8}{3}\) now subtracting
\(\frac{19}{3}\) – \(\frac{8}{3}\) to subtract a fraction from another
we need to first make sure both fractions have the same denominator as both denominators
have same value 3 now we can subtract one numerator from the other to make one fraction as 19 – 8 by 3
= \(\frac{11}{3}\) since the numerator is greater than the denominator,
we can further simplify it into a mixed fraction as 3 goes in 3 times. 3 X 3  is 9 , 3 will be our
whole number and  (11−9) we have 2 left over. 2 will be our numerator and our denominator will be the same 3
= \(\frac{11}{3}\) = 3 \(\frac{2}{3}\) .

Question 14.
Big Ideas Math Answers 4th Grade Chapter 9 Multiply Whole Numbers and Fractions 100
Answer:

Big Ideas Math Answers 4th Grade Chapter 9 Multiply Whole Numbers and Fractions 100= \(\frac{78}{12}\) = 6 \(\frac{6}{12}\) or
\(\frac{78}{12}\) = \(\frac{39}{6}\) = \(\frac{13}{2}\) =  6 \(\frac{1}{2}\).

Explanation:
First we write the mixed numbers into fractions and subtract, So 8 \(\frac{4}{12}\) =
8 +\(\frac{4}{12}\) = ( 8 x 12 + 4 ) by 12 = \(\frac{100}{12}\) and  1 \(\frac{10}{12}\) = 1 + \(\frac{10}{12}\) = 1 X 12 + 10 by 12 = \(\frac{22}{12}\) now subtracting \(\frac{100}{12}\) – \(\frac{22}{12}\) to subtract a fraction from another we need to first make sure both fractions
have the same denominator as both denominators have same value 3 now we can subtract one numerator
from the other to make one fraction as 100 – 22 by 12 = \(\frac{78}{12}\) since the numerator
is greater than the denominator , we can further simplify it into a mixed fraction as 12 goes in 6 times.
12 X 6  is 72 , 6 will be our whole number and  (78−12) we have 6 left over. 6 will be our numerator and
our denominator will be the same 12, 6 \(\frac{6}{12}\) , Further \(\frac{78}{12}\)
can be simplified as 78,12 both goes by  2 making them as 39, 6=\(\frac{39}{6}\),
even further \(\frac{39}{6}\) can be simplified as 39,6 both goes by 3 making them as 13,2
=\(\frac{13}{2}\), Since the numerator is greater than the denominator , we can further simplify
it into a mixed fraction as 2 goes in 6 times. 6 X 2  is 12 , 6 will be our whole number and  (13−12) we have 1 left over.
1 will be our numerator and our denominator will be the same 3 = \(\frac{78}{12}\) =
6 \(\frac{6}{12}\) or \(\frac{78}{12}\) = \(\frac{39}{6}\) = \(\frac{13}{2}\) =
6 \(\frac{1}{2}\).

Lesson 9.5 Problem Solving: Fraction Operations

Explore and Grow

You want to make 3 batches of the recipe. Explain how to find how much of each ingredient you need.
Big Ideas Math Answers 4th Grade Chapter 9 Multiply Whole Numbers and Fractions 101

Reasoning
Explain how you can tell whether you need more than or less than 6 cups of blueberries without calculating
Answer:
To make 3 batches of recipe we require \(\frac{21}{4}\) pounds of strawberries or
5 \(\frac{1}{4}\) pounds of strawberries required,
\(\frac{3}{2}\) cups of orange juice or 1 \(\frac{1}{2}\) cups of orange juice,
12 tablespoons of lemon juice,12 tablespoons of honey and
\(\frac{27}{4}\) cups of blue berries are required or 6 \(\frac{3}{4}\) cups of blueberries are required.
We require more than 6 cups of blueberries why because if we see product required is
3 X 2 \(\frac{1}{4}\) cups of blue berries without calculating , Applying distributive property
we get (3 X 2 ) + ( 3 X \(\frac{1}{4}\) ) = 6 + ( 3 X \(\frac{1}{4}\) ) it clearly shows that
we require more than 6 cups of blueberries.

Explanation:
Given  to make 1 batch of Recipe Ladybug Ice pops we require  1 \(\frac{3}{4}\) pounds of strawberries ,
\(\frac{1}{2}\) cup of orange juice , 4 tablespoons of lemon juice, 4 tablespoons of honey,
2 \(\frac{1}{4}\) cups of blue berries.
To find out how much of each ingredient needed to make 3 batches of recipe we multiply each ingredient with 3.
Strawberries needed are 3 X 1 \(\frac{3}{4}\), First we write the mixed number as a
fraction then multiply so 1 \(\frac{3}{4}\) = ( 1 X 4 + 3 ) by 4 = \(\frac{7}{4}\) ,
so 3  X  \(\frac{7}{4}\)  now we write in unit fraction and multiply  3 X 7 X \(\frac{1}{4}\) =
21 X \(\frac{1}{4}\) = \(\frac{21}{4}\) pounds of strawberries required,
Since the numerator is greater than the denominator , we can further simplify it into a mixed fraction as
4 goes in 5 times. 5 X 4  is 20 , 5 will be our whole number and  (21−20) we have 1 left over.
1 will be our numerator and our denominator will be the same 4 =5 \(\frac{1}{4}\)
pounds of strawberries required,
Orange juice required 3 X \(\frac{1}{2}\) already it is in unit fraction
so we require \(\frac{3}{2}\) cups of orange juice Since the numerator is greater than the denominator ,
we can further simplify it into a mixed fraction as 2 goes in 1 times. 1 X 2 is 2, 1 will be our whole number and (3-2),
we have 1 left over. 1 will be our numerator and our denominator will be the same
2 = 1 \(\frac{1}{2}\) cups of orange juice.
Lemon juice required is 3 X 4 = 12 tablespoons of lemon juice, 3 X 4 = 12 tablespoons of honey,
now blue berries required are 3 X 2 \(\frac{1}{4}\) cups, First we write the mixed number as a
fraction then multiply so 2 \(\frac{1}{4}\) = ( 2 X 4 + 1 ) by 4 = \(\frac{9}{4}\) ,
now  3  X  \(\frac{9}{4}\)  now we write in unit fraction and multiply  3 X 9 X \(\frac{1}{4}\) =
27 X \(\frac{1}{4}\) = \(\frac{27}{4}\) cups of blue berries are required,
Since the numerator is greater than the denominator , we can further simplify it into a mixed fraction as
4 goes in 6 times. 6 X 4 is 24, 6 will be our whole number and (27-24) we have 3 left over.
3 will be our numerator and our denominator will be the same 4 = 6 \(\frac{3}{4}\) cups of blueberries are required.
Now comparing blueberries we require more than 6 cups of blueberries why because we use is
3 X 2 \(\frac{1}{4}\) cups of blue berries are required without calculating,
Applying distributive property  we get (3 X 2 ) +( 3 X \(\frac{1}{4}\) )=
6 +( 3 X \(\frac{1}{4}\)), it clearly shows that we require more than 6 cups of blueberries.

Think and Grow: Problem Solving: Fraction Operations

Example
To convert a temperature from degrees Celsius to degrees Fahrenheit, multiply the Celsius temperature by \(\frac{9}{5}\), then add 32. What is the temperature shown by the thermometer in degrees Fahrenheit?
Big Ideas Math Answers 4th Grade Chapter 9 Multiply Whole Numbers and Fractions 102
Answer:
The temperature shown by the thermometer is 500 Fahrenheit.

Explanation:
We know to convert a temperature from degrees Celsius to degrees Fahrenheit is Fahrenheit =
Celsius temperature X \(\frac{9}{5}\) + 32 .
So multiply the Celsius temperature by \(\frac{9}{5}\), then add 32,
Given the Celsius temperature as 100 in the figure above, First we write \(\frac{9}{5}\) in unit fraction and
multiply by Celsius temperature so \(\frac{9}{5}\)=9 X \(\frac{1}{5}\)
now we multiply by 10 as 10 X 9 X \(\frac{1}{5}\) = 90 X \(\frac{1}{5}\) =
\(\frac{90}{5}\) this value can be further simplified as 90,5 both goes by  5 making them as 18,1,
Now we got result as 18 , We add 18 + 32  we will get 500 Fahrenheit.

Understand the Problem

What do you know?

  • To convert a temperature from degrees Celsius to degrees Fahrenheit, multiply the Celsius temperature by \(\frac{9}{5}\), then add 32.
  • The thermometer shows 10 degrees Celsius.
    What do you need to find?
  • You need to find the temperature shown by the thermometer in degrees Fahrenheit.

Make a Plan

How will you solve?

  • First, multiply the Celsius temperature, 10 degrees, by \(\frac{9}{5}\).
  • Then add 32 to the product.

Solve
Big Ideas Math Answers 4th Grade Chapter 9 Multiply Whole Numbers and Fractions 103
Big Ideas Math Answers 4th Grade Chapter 9 Multiply Whole Numbers and Fractions 104
So, the temperature shown by the thermometer is _500_ degrees Fahrenheit.
Answer:
The temperature shown by the thermometer is 500 Fahrenheit.

Explanation :
To convert a temperature from degrees Celsius to degrees Fahrenheit, Let us take  p as multiply of
Celsius temperature by \(\frac{9}{5}\), Now we take f as p+32,
First we write \(\frac{9}{5}\) in unit fraction and multiply by Celsius temperature,
so \(\frac{9}{5}\)=9 X \(\frac{1}{5}\) now multiply by 10 = 10 X 9 X \(\frac{1}{5}\) =
90 X \(\frac{1}{5}\) = \(\frac{90}{5}\) this value we take as p this can be further simplified
as 90,5 both goes by  5 making them as 18, Now we got p as 18 we add 32 to p ,
we get f, So f = 18 + 32 = 500
So the temperature shown by the thermometer is 500 Fahrenheit.

Show and Grow

Question 1.
Show how to solve the example above using one equation.
Answer:
F= C X \(\frac{9}{5}\) + 32 degrees.

Explanation:
To solve the example above using one equation, Let us take Fahrenheit as F ,
Celsius as C so to convert a temperature from degrees Celsius to degrees Fahrenheit we multiply
C with \(\frac{9}{5}\) and add 32.

Apply and Grow: Practice

Understand the problem. What do you know? What do you need to find? Explain.

Answer:
We know Celsius temperature, We need to find the temperature shown by the thermometer in degrees Fahrenheit.

Explanation:
Given the temperature in Celsius and we know the formula to convert from Celsius temperature to
Fahrenheit as F= C X \(\frac{9}{5}\) + 32 degrees so we substitute the values and find the temperature.

Question 2.
You make a friendship bracelet with 3 pink strings and 2 blue strings. Each string is 3\(\frac{3}{4}\) feet long.
How many feet of string do you use?

Answer:
\(\frac{75}{4}\) feet long strings is used or 18 \(\frac{3}{4}\) feet long strings is used.

Explanation:
Given to make a friendship bracelet we use 3 pink strings , 2 blue strings means total 5 strings are needed
and each string is 3\(\frac{3}{4}\) feet long, First we convert mixed fraction into fraction as
( 3 X 4 + 3 ) by 4 = \(\frac{15}{4}\) ,Now we will convert into unit fraction as 15 X \(\frac{1}{4}\)
now we multiply with 5 , 5 X 15 X \(\frac{1}{4}\) =\(\frac{75}{4}\),
since the numerator is greater than the denominator , we can further simplify it into a mixed fraction as
4 goes in 18 times. 18 X 4 is 72, 18 will be our whole number and (75-72) we have 3 left over.
3 will be our numerator and our denominator will be the same 4 = 18 \(\frac{3}{4}\),
So \(\frac{75}{4}\) feet long strings is used or 18 \(\frac{3}{4}\) feet long strings is used.

Question 3.
A smoothie store worker makes 4 peanut butter banana smoothies and 2 fruit smoothies. The worker uses \(\frac{2}{3}\) cup of bananas in each smoothie. How many cups of bananas does the worker need?

Answer:
The worker needed  \(\frac{12}{3}\) cups of bananas or 4 cups of bananas are needed to make smoothie.

Explanation:
Given a smoothie store worker makes 4 peanut butter banana smoothies and 2 fruit smoothies.
The worker uses \(\frac{2}{3}\) cup of bananas in each smoothie so the number of cups of bananas
the worker needed are 4 X latex]\frac{2}{3}[/latex] plus 2 X \(\frac{2}{3}\) first we convert
both into unit fractions and multiply 4 X 2 X \(\frac{1}{3}\) = 8 X \(\frac{1}{3}\) =
\(\frac{8}{3}\)  of peanut butter banana smoothies and 2 X 2 X \(\frac{1}{3}\) =
4 X \(\frac{1}{3}\) = \(\frac{4}{3}\) fruit smoothies, now we add both \(\frac{8}{3}\) +
\(\frac{4}{3}\) as both denominators are same we add numerators as 8 + 4 =12 as numerator and
denominator as 3 = \(\frac{12}{3}\) as numerator is greater than denominator further can be simplified and
12 divides by 3 gives 4 with 0 remainder so the worker needed  \(\frac{12}{3}\) cups of bananas or
4 cups of bananas are needed to make smoothie.

Understand the problem. Then make a plan. How will you solve? Explain.

Question 4.
Your friend walks her dog for \(\frac{1}{4}\) mile each day. She then runs 2\(\frac{3}{4}\) miles each day.
How many total miles does she walk her dog and run in 1 week?

Answer:
Total 21 miles my friend walks her dog and run in 1 week.

Explanation:
Given my friend walks her dog for \(\frac{1}{4}\) mile each day , She then runs
2\(\frac{3}{4}\) miles each day. So in each day she walk her dog and run  is
\(\frac{1}{4}\) mile + 2\(\frac{3}{4}\) miles, Then we multiply with
7 because we have to calculate for 1 week. First we convert mixed fraction into fraction
2 \(\frac{3}{4}\) , ( 2 X 4 + 3 ) by 4 = \(\frac{11}{4}\), Now we add with
\(\frac{1}{4}\) + \(\frac{11}{4}\) = \(\frac{12}{4}\)
as numerator is greater than denominator further can be simplified and 12 divides by 4
gives 3 with 0 remainder, so 3 miles for each day, Now for 1 week it is 7 X 3  = 21 miles.
My friend walks 21 miles her dog and run in 1 week.

Question 5.
Hair donations must be 12 inches longer. Your friend’s hair is 7 inches long. Her hair grows
about \(\frac{1}{2}\) inch each month. Can she donate her hair in 8 months?

Answer:

No friend cannot donate her hair in 8 months as in 8 months it grows to 11 inches long and for
hair donations it must be 12 inches longer.

Explanation:

Given friend’s hair is 7 inches long. Her hair grows about \(\frac{1}{2}\) inch each month.
In 8 months it grows 8  X \(\frac{1}{2}\) = \(\frac{8}{2}\) as numerator is
greater than denominator further can be simplified and 8 divides by 2 gives 4 with 0 remainder,
in 8 months it will grow 4 inches , So after 8 months it will be 7 + 4 =11 inches long as 11 is less than 12
and hair donations must be 12 inches longer so friend cannot donate her hair in 8 months.

Question 6.
Today you walk \(\frac{6}{10}\) mile from the Martin Luther King Jr. Memorial to the Washington Monument. Tomorrow you will walk about 4 times as far from the Washington Monument to the White House. About how much farther will you walk tomorrow?

Answer:

Tomorrow i will walk father \(\frac{18}{10}\) miles  or \(\frac{9}{5}\) miles or 1 \(\frac{4}{5}\) .

Explanation:

Given today i walked \(\frac{6}{10}\) mile from the Martin Luther King Jr. Memorial
to the Washington Monument, Tomorrow i will walk about 4 times as far from the Washington Monument
to the White House , So number of miles walked tomorrow will be 4 X \(\frac{6}{10}\)
first we write in unit fraction 4 X 6 X \(\frac{1}{10}\) = 24 X \(\frac{1}{10}\)  =
\(\frac{24}{10}\) as numerator is greater than denominator further can be simplified as
24,10 both can be divided by 2 gives 12,5 so \(\frac{24}{10}\) = \(\frac{12}{5}\)
now farther i will walk tomorrow will be \(\frac{24}{10}\) – \(\frac{6}{10}\) as denominator
of both are same we substract numerators , 24 – 6 =18 and denominator will remain same 10 =
\(\frac{18}{10}\) miles or further can be simplified and 18,10 both can be divided by by 2
gives 9, 5 = \(\frac{18}{10}\) miles = \(\frac{9}{5}\) miles as numerator is greater than
denominator we can write it in mixed fraction as 5 goes in 1 times. 5 X 1 is 5 , 1 will be our whole number and
(9−5) we have 4 left over. 4 will be our numerator and our denominator will be the same 5 = 1 \(\frac{4}{5}\) .

Think and Grow: Modeling Real Life

Example
A 20-pound dog and a 60-pound dog eat the recommended amounts of dog food each day. How much more food does the 60-pound dog eat in 1 week than the 20-pound dog?
Think: What do you know? What do you need to find? How will you solve?
Big Ideas Math Answers 4th Grade Chapter 9 Multiply Whole Numbers and Fractions 105
Big Ideas Math Answers 4th Grade Chapter 9 Multiply Whole Numbers and Fractions 106
Big Ideas Math Answers 4th Grade Chapter 9 Multiply Whole Numbers and Fractions 107
So, the 60-pound dog eats 10 \(\frac{1}{2}\)  more cups of food than the 20-pound dog in 1 week.

Answer:

The 60-pound dog eats 10 \(\frac{1}{2}\)  more cups of food than the 20-pound dog in 1 week.

Explanation:

Given A 20-pound dog eats 1 \(\frac{1}{2}\) cups  and a 60-pound dog eats 3 cups,
the recommended amount of food each day. Now we need to calculate how much more food does
the 60-pound dog eat in 1 week than the 20-pound dog,
So Step 1 : First we find how much food the 20-pound dog eats in 1 week,
let us take it as b = 7 X 1 \(\frac{1}{2}\) we write mixed fraction as fraction
1 \(\frac{1}{2}\) = ( 1 X 2 + 1 )by 2 = \(\frac{3}{2}\) now multiply by 7 =
7 X \(\frac{3}{2}\) we write into unit fraction 7 X 3 X \(\frac{1}{2}\) =
21 X \(\frac{1}{2}\) = \(\frac{21}{2}\) cups,
Step 2: Now we find how much food the 60-pound dog eats in 1 week let us take it as c = 7 X 3 = 21 cups,
Step 3: Subtract b from c to find how much more food the 60-pound dog eats in 1 week.
c- b = 21 – \(\frac{21}{2}\) to subtract first we make both denominators same,
we multiply and divide 21 by 2 making as (21 x 2) by 2 = \(\frac{42}{2}\) –\(\frac{21}{2}\)
as denominators are same we subtarct numerators 42-21=21 will be numerator and denominator is same 2
=\(\frac{21}{2}\) as numerator is greater than denominator and 2 goes 10 times, 2 X 10 is 20 ,
10 will be our whole number and  (21−20) we have 1 left over. 1 will be our numerator and our denominator
will be the same 2 = 10 \(\frac{1}{2}\) . Therefore the 60-pound dog eats 10 \(\frac{1}{2}\)
more cups of food than the 20-pound dog in 1 week.

Show and Grow

Question 7.
Use the table above. A 40-pound dog and a 100-pound dog eat the recommended amounts of dog food each day. How much food is needed to feed both dogs for 1 week?
Big Ideas Math Answers 4th Grade Chapter 9 Multiply Whole Numbers and Fractions 108

Answer:
To feed the food for both dogs for 1 week we require \(\frac{182}{4}\) cups or \(\frac{91}{2}\) or 45 \(\frac{1}{2}\) cups.

Explanation:
From the table 40-pound dog eats 2 \(\frac{1}{4}\) and 100-pound dog eats
4 \(\frac{1}{4}\) the recommended amounts of dog food each day,
to find how much food is needed to feed both dogs for 1 week first we calculate for
40-pound dog= 7 X 2 \(\frac{1}{4}\) we write the mixed fraction as a fraction
2 x 4 + 1 by 4 = \(\frac{9}{4}\) and multiply by 7 , 7 X \(\frac{9}{4}\)
now we write in unit fraction 7 X 9 X \(\frac{1}{4}\) = 63 X \(\frac{1}{4}\) =
\(\frac{63}{4}\) now we calculate for 100-pound dog eats 4 \(\frac{1}{4}\) in 1 week is
7 X 4 \(\frac{1}{4}\) we write the mixed fraction as a fraction ( 4 X 4 + 1 )by 4 =
\(\frac{17}{4}\) and multiply by 7 , 7 X \(\frac{17}{4}\) now we write in unit fraction
7 X 17 X \(\frac{1}{4}\) = 119 X \(\frac{1}{4}\) = \(\frac{119}{4}\).
Now to feed the food for both dogs for 1 week is \(\frac{63}{4}\) + \(\frac{119}{4}\)
as denominators are same we add numerators we get 63 + 119 = 182 as numerator, \(\frac{182}{4}\)
further can be simplifed as 182 and 4 goes by 2 we write as \(\frac{91}{2}\) numerator is greater
we can write in mixed fraction as 2 goes 45 times, 2 X 45 is 90 , 45 will be our whole number and
(91−90) we have 1 left over. 1 will be our numerator and our denominator will be the same 2,
making as 45 \(\frac{1}{2}\) cups.

Problem Solving: Fraction Operations Homework & Practice 9.5

Understand the problem. Then make a plan. How will you solve? Explain.

Question 1.
Your friend makes strawberry jam and raspberry jam. He makes enough strawberry jam to fill \(\frac{1}{2}\) of a jar. He makes 5 times as much raspberry jam as strawberry jam. How many full jars of raspberry jam does he make?
Big Ideas Math Answers 4th Grade Chapter 9 Multiply Whole Numbers and Fractions 109
Answer:
Friend makes \(\frac{5}{2}\) jars= 2 \(\frac{1}{2}\) jars in that full jars are 2,
So 2 jars of raspberry jam he makes.

Explanation:
Given friend makes strawberry jam to fill \(\frac{1}{2}\) of a jar and makes 5 times as
much raspberry jam as strawberry jam means he makes 5 X \(\frac{1}{2}\) of jars of
raspberry jam now we write as fraction \(\frac{5}{2}\) as numerator is greater,
we can write in mixed fraction as 2 goes 2 times, 2 X 2 is 4 , 2 will be our whole number and
(5−4) we have 1 left over. 1 will be our numerator and our denominator will be the same 2,
making as 2 \(\frac{1}{2}\) ,here 2 is whole there fore full jars of raspberry jam is 2.
So friend makes \(\frac{5}{2}\) jars= 2 \(\frac{1}{2}\) jars in that full jars are 2,
So 2 jars of raspberry jam he makes.

Question 2.
You buy a pair of jeans that originally cost $22. The pair of jeans is \(\frac{1}{2}\) off the original price.
You pay with a $20 bill. How much change do you receive?
Answer:
I receive $9 as change.
Explanation:
The pair of jeans is \(\frac{1}{2}\) off the original price and original cost is $ 22,
So the cost to be paid is $22 X \(\frac{1}{2}\) = \(\frac{22}{2}\)
further this can be simplified as both can be divided by 2 numerator becomes 11 and denominator as 1 = 11,
therefore the original cost for the pair of jeans is $ 11, Now i have paid $ 20bill,
i will receive a change that is $20 – $11= $9, i receive $9 as a change.

Question 3.
The observation deck of the Space Needle in Seattle is 520 feet above ground.
The tip of the Space Needle is 85 feet above the observation deck.
An artist makes a replica that is \(\frac{1}{100}\) the height of the entire Space Needle. How tall is the replica?
Big Ideas Math Answers 4th Grade Chapter 9 Multiply Whole Numbers and Fractions 110
Answer:
The height of replica is \(\frac{605}{100}\) =\(\frac{121}{20}\) or
6 \(\frac{1}{20}\) feet.

Explanation:
Given the observation deck of the Space Needle in Seattle is 520 feet above ground and
the tip of the Space Needle is 85 feet above the observation deck means the height of
the entire Space Needle is 520 + 85 = 605 feet above the ground to tip,
now an artist makes a replica that is \(\frac{1}{100}\)
the height of the entire Space Needle. So the height of the replica is
605 X \(\frac{1}{100}\) = \(\frac{605}{100}\)
further can be simplifed as 605 and 100 goes by 5 we get numerator as 121 and denominator as 20 =
\(\frac{121}{20}\) as numerator is greater we can write in mixed fraction as 20 goes
6 times, 20 X 6 is 120 , 6 will be our whole number and  (121−120) we have 1 left over.
1 will be our numerator and our denominator will be the same 20, making as 6 \(\frac{1}{20}\) feet tall.

Question 4.
Modeling Real Life
How many sheets of orange and black paper does a teacher need so that 30 students can each complete the art project?
Big Ideas Math Answers 4th Grade Chapter 9 Multiply Whole Numbers and Fractions 111
Answer:
For 30 students to complete the Art Project orange sheets required are 11 \(\frac{2}{8}\) and
18 \(\frac{6}{8}\)  black sheets of paper is required.

Explanation:
Given paper needed for Art Project orange is \(\frac{3}{8}\) and for 30 students it is
30 X \(\frac{3}{8}\) we write in unit fraction as 30 X 3 X \(\frac{1}{8}\) =
90 X \(\frac{1}{8}\) = \(\frac{90}{8}\) as numerator is greater we can write in mixed fraction
as 8 goes 11 times, 8 X 11 is 88, 11 will be our whole number and  (90−88) we have 2 left over.
2 will be our numerator and our denominator will be the same 8, making as
11 \(\frac{2}{8}\) sheets of orange is required, given black paper needed for
Art Project is \(\frac{5}{8}\) and for 30 students it is 30 X \(\frac{5}{8}\)
we write in unit fraction as 30 X 5 X \(\frac{1}{8}\) = 150 X \(\frac{1}{8}\) =
\(\frac{150}{8}\) as numerator is greater we can write in mixed fraction as 8 goes 18 times,
8 X 18 is 144, 18 will be our whole number and  (150−144) we have 6 left over.
6 will be our numerator and our denominator will be the same 8, making as
18 \(\frac{6}{8}\) black sheets are required.
Therefore for 30 students to complete the Art Project orange sheets required are
11 \(\frac{2}{8}\) and 18 \(\frac{6}{8}\) black sheets of paper is required.

Question 5.
Writing
Write and solve a two-step word problem with mixed numbers that can be solved using multiplication.

Answer:
A two-step problem is a word problem that requires two operations to solve it.
We take two mixed numbers and slove using multiplication. In step one
we use multiplication and add and in another step we use addition and again use multiplication
as explained below with an example,

Explanation:
For example: Percy walks 2 \(\frac{5}{4}\) kilometers for 2 days and
1 \(\frac{1}{4}\) kilometers for 5 days , how many kilometers does Percy walks in 1 week?
Given in first 2 days Percy walks 2 \(\frac{5}{4}\) kilometers so In first step
we change mixed fraction to fraction by using multiplication and then add 2 X 4 + 5 by 4 =
\(\frac{13}{4}\) kilometers for 1 day for 2 days it is 2 X \(\frac{13}{4}\) =
2 X 13 X \(\frac{1}{4}\) = 26 X \(\frac{1}{4}\) = \(\frac{26}{4}\) kilometers for 2 days ,
Now for rest of 5 days we calculate. Here 1 \(\frac{1}{4}\) we change the mixed fraction to fraction as
( 1 X 4 + 1 ) by 4 = \(\frac{5}{4}\) for 1 day, For 5 days it will be 5 X \(\frac{5}{4}\),
5 X 5 X \(\frac{1}{4}\) = 25 X \(\frac{1}{4}\) = \(\frac{25}{4}\) kilometers for rest 5 days ,
Now total number of kilometers Percy walks is  \(\frac{26}{4}\) + \(\frac{25}{4}\) as denominators
are same we add numerators we get 26 + 25 as 51 and denominator being same as 4 = \(\frac{51}{4}\) kilometers
Percy walks in 1 week and further can be simplified as numerator is greater we can write in mixed fraction as
4 goes 12 times, 4 X 12 is 48, 12 will be our whole number and  (51−48) we have 3 left over.
3 will be our numerator and our denominator will be the same 4, making as
12 \(\frac{3}{4}\) kilometers Percy walks in 1 week.

Question 6.
Modeling Real Life
In science class, you test how many grams different-sized bundles of spaghetti can hold. You predict that a bundle of 5 spaghetti noodles can hold 10 quarters before breaking. You predict that a bundle of 10 spaghetti noodles can hold 20 quarters before breaking. Each quarter 5\(\frac{67}{100}\) grams. According to your predictions, how many more grams can the bundle of 10 noodles hold than the bundle of 5 noodles?
Big Ideas Math Answer Key Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 112
Answer:
\(\frac{970}{100}\) grams more the bundle of 10 noodles hold than the bundle of 5 noodles
or \(\frac{97}{10}\) grams more the bundle of 10 noodles hold than the bundle of 5 noodles
or 9 \(\frac{7}{10}\) grams more the bundle of 10 noodles hold than the bundle of 5 noodles.
Explanation:
Given that bundle of 5 noodles can hold 10 quarters means 5 noodles can hold
10 X 5 \(\frac{67}{100}\) grams first we change mixed fraction to fraction by
using multiplication and then add as 5 \(\frac{67}{100}\) = 5 X 6 + 67 by 100 =
\(\frac{97}{100}\) now we multiply by 10, 10 X \(\frac{97}{100}\) =
10 X 97 X \(\frac{1}{100}\) = 970 X \(\frac{1}{100}\) = \(\frac{970}{100}\) grams ,
so 5 spaghetti noodles can hold \(\frac{970}{100}\) grams before breaking,
Now given 10 spaghetti noodles can hold 20 quarters means 10 spaghetti noodles can hold
20 X 5 \(\frac{67}{100}\) grams again first we change mixed fraction to fraction
by using multiplication and then add as 5 \(\frac{67}{100}\) = ( 5 X 6 + 67 ) by 100 =
\(\frac{97}{100}\) now we multiply by 20, 20 X \(\frac{97}{100}\) =
20 X 97 X \(\frac{1}{100}\) = 1940 X \(\frac{1}{100}\) = \(\frac{1940}{100}\) grams ,
so 10 spaghetti noodles can hold \(\frac{1940}{100}\) grams before breaking now to calculate
how many more grams can the bundle of 10 noodles hold than the bundle of 5 noodles we subtract
\(\frac{1940}{100}\) grams – \(\frac{970}{100}\) grams as denominators are same first
we substract numerators as 1940 – 970 = 970 as numerator and denominator being same 100
we get \(\frac{970}{100}\) grams more the bundle of 10 noodles hold than the bundle of 5 noodles,
further can be simplified as 970 and 100 both can be divided by 10 making it as \(\frac{97}{10}\) grams
more the bundle of 10 noodles hold than the bundle of 5 noodles and as numerator is greater ,
we can write in mixed fraction as 10 goes 9 times, 10 X 9 is 90, 9 will be our whole number and  (97−90)
we have 7 left over. 7 will be our numerator and our denominator will be the same 10,
making as 9 \(\frac{7}{10}\) grams more the bundle of 10 noodles hold than the bundle of 5 noodles.

Review & Refresh

Compare

Question 7.
Big Ideas Math Answer Key Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 113
Answer:
\(\frac{9}{10}\) is greater than > \(\frac{4}{5}\) or
\(\frac{9}{10}\) is greater than > \(\frac{8}{10}\)

Explanation:
In order to compare the fractions we must first turn their different denominators into
the same denominators to make denominators same we multiply \(\frac{4}{5}\)
with both numerator and denominator by 2 the value will not change =
\(\frac{4}{5}\)  X \(\frac{2}{2}\)= \(\frac{8}{10}\)
now as denominators are same 10 we can compare only numerators
\(\frac{9}{10}\) with \(\frac{8}{10}\) as 9 is greater than 8,
So \(\frac{9}{10}\) is greater than > \(\frac{8}{10}\) or
\(\frac{9}{10}\) is greater than > \(\frac{4}{5}\)

Question 8.
Big Ideas Math Answer Key Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 114
Answer:
\(\frac{3}{8}\) < is less than \(\frac{5}{6}\) or
\(\frac{9}{24}\) < is less than \(\frac{20}{24}\).

Explanation:
In order to compare the fractions we must first turn their different denominators into
the same denominators to make denominators same ,
we find the least common multiplier of 8 ,6 as ( 2 X 2 X 2 X 3 ) = 24,
solution

Now convert each one of \(\frac{3}{8}\) and \(\frac{5}{6}\)
into an equivalent fraction having  as denominator,
\(\frac{3}{8}\) = \(\frac{3}{8}\) X \(\frac{3}{3}\) =
\(\frac{9}{24}\) and \(\frac{5}{6}\) = \(\frac{5}{6}\) X \(\frac{4}{4}\) =
\(\frac{20}{24}\) now as denominators are same 24 we can compare only numerators
\(\frac{9}{24}\) with \(\frac{20}{24}\) as 9 is less than 20,
so \(\frac{9}{24}\) < is less than \(\frac{20}{24}\) or
\(\frac{3}{8}\) < is less than \(\frac{5}{6}\)

Question 9.
Big Ideas Math Answer Key Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 115
Answer:
\(\frac{1}{3}\) = is equal to  \(\frac{4}{12}\) or
\(\frac{1}{3}\) = is equal to  \(\frac{1}{3}\)
Explanation:
Here first we simplify \(\frac{4}{12}\) as both can be divided by 4 if we dividey
both numerator and denominator by 4 we get numerator as 1 and denominator as 4 =
\(\frac{1}{3}\) now comparing both sides, as values are same
so \(\frac{1}{3}\) = is equal to  \(\frac{4}{12}\) or
\(\frac{1}{3}\) = is equal to  \(\frac{1}{3}\).

Multiply Whole Numbers and Fractions Performance Task

Question 1.
You use identical glass jars, colored water, and a spoon to learn about sounds. When you tap a spoon on a jar, the vibrations make a sound. The jar that has the least amount of water makes the lowest sound.
Big Ideas Math Answer Key Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 116
a. You fill each glass using a \(\frac{3}{4}\)-cup measuring cup. Complete the table to find the total amount of water in each jar.
b. How much more water is in the purple jar than the green jar?
c. How many cups of water are used in all?
d. Each jar can hold 4 cups of water. Is it possible to add another \(\frac{3}{4}\) cup of water to the purple jar? Explain.
e. Which jars are more than half full?
f. You add another \(\frac{3}{4}\) cup of water to the green jar. How does that affect the sound?
Answer:
a.
b. Purple jar is 3 cups more than the Green jar or Purple jar is \(\frac{12}{4}\) cups more than the Green jar.
c. Total number of cups of water used in all is \(\frac{45}{4}\) cups or 11 \(\frac{1}{4}\) cups.
d. No, it is not possible to add another \(\frac{3}{4}\) cup of water to the purple jar as if we add it will become \(\frac{18}{4}\) cups = 4 \(\frac{2}{4}\) = 4 + \(\frac{2}{4}\) cups of water
to the purple jar cups which is more than 4 cups it will overflow so it is not possible to add another
\(\frac{3}{4}\) cup of water to the purple jar.
e. Jars which are more than half full are Orange, Red and Purple.
f. After adding add another \(\frac{3}{4}\) cup of water to the green jar it becomes
\(\frac{6}{4}\) which is equal to blue color jar, we know the jar that has the least amount of water
makes the lowest sound before it was the green jar which had least amount of water so it had lowest sound,
so now after adding \(\frac{3}{4}\) cup of water to green jar it has become equal to blue jar
so now both green and blue jar has least amount of water so they both make the lowest sound.

Explanantion:
a. Given we fill each glass \(\frac{3}{4}\)-cup measuring cup so total water cups used for each color are
Green 1 means 1 X \(\frac{3}{4}\) cup = \(\frac{3}{4}\) cup
Blue 2 of \(\frac{3}{4}\) = 2 X \(\frac{3}{4}\) = 2 X 3 X \(\frac{1}{4}\) =
6 X \(\frac{1}{4}\)= \(\frac{6}{4}\) cups , as both numerator and denominator
can be divided by 2 we get \(\frac{3}{2}\) cups
Orange 3 of \(\frac{3}{4}\) = 3 X \(\frac{3}{4}\) = 3 X 3 X \(\frac{1}{4}\) =
9 X \(\frac{1}{4}\) = \(\frac{9}{4}\) cups
Red 4 of \(\frac{3}{4}\) = 4 X \(\frac{3}{4}\) = 4 X 3 X \(\frac{1}{4}\) =
12 X \(\frac{1}{4}\) = \(\frac{12}{4}\) as both numerator and denominator can be divided by 4
we get numerator as 3 and denominator 1 making Red 4 of \(\frac{3}{4}\) = 3 cups.
Purple 5 of \(\frac{3}{4}\) = 5 X \(\frac{3}{4}\) = 5 X 3 X \(\frac{1}{4}\) =
15 X \(\frac{1}{4}\) =
\(\frac{15}{4}\) cups.

b. To calculate how much more water is in the Purple jar than the Green jar we subtract
the quantity of purple jar minus quantity in green jar as \(\frac{15}{4}\) cups – \(\frac{3}{4}\) cup,
as denominators are same for the both we will sustract numerators 15-3 = 12 and
denominator remains the same as 4, \(\frac{12}{4}\) as both numerator and
denominator can be divided by 4 we get numerator as 3 and denominator 1 so Purple Jar is 3 cups
more than the Green jar.

c. Total number of cups of water are used in all is adding Green, Blue, Orange, Red,
Purple – \(\frac{3}{4}\) + \(\frac{6}{4}\) + \(\frac{9}{4}\)  + \(\frac{12}{4}\) +
\(\frac{15}{4}\) as denominators are same we add numerators as 3 + 6 + 9 + 12 + 15 = 45 and
denominator being same as 4 = \(\frac{45}{4}\) as numerator is greater we can also write in mixed fraction
as 4 goes 11 ,X 11 is 44, 11 will be our whole number and  (45−44) we have 1 left over.
1 will be our numerator and our denominator will be the same 4, making as 11 \(\frac{1}{4}\),
therfore total number of cups of water used in all is \(\frac{45}{4}\) cups or 11 \(\frac{1}{4}\) cups.

d. We got Purple jar contains \(\frac{15}{4}\) cups to this we will  add another
\(\frac{3}{4}\) cup of water = \(\frac{15}{4}\) + \(\frac{3}{4}\) as denominator is
same we add numerators as 15 + 3 = 18 and denominator is same 4 so now the Purple jar contains
\(\frac{18}{4}\) cups , to find the whole value we write in mixed fraction as 4 goes , 4 X 4 is 16, 4
will be our whole number and  (18−16) we have 2 left over. 2 will be our numerator and our denominator
will be the same 4, making as 4 \(\frac{2}{4}\) = 4 + \(\frac{2}{4}\) and
given that each jar can hold 4 cups of water but as we are getting more than 4 it will overflow,
so it is not possible to add another \(\frac{3}{4}\) cup of water to the purple jar.
e. Jars more than half full means each jars should be above 2 cups means to check we need
denominator to be same that is 4 so we multiply denominator and numerator by 4,
we get \(\frac{8}{4}\) to be more than half full means each jars numerators should be more than 8 and all jars denominators being same as 4, now we check with green jar, \(\frac{8}{4}\)  and
\(\frac{3}{4}\) as 3 is not greater than 8 so not green jar, blue jar we have \(\frac{6}{4}\) cups
here also numerator 6 is not more than 8 so even we will not conside blue jar,
orange jar \(\frac{9}{4}\) now here numerator 8 is less than 9 so we will consider orange jar,
Now red jar as red jar has 3 cups which is above 2 cups we consider red jar and we have purple jar as
\(\frac{15}{4}\) here numerator 15 is more than numerator 8 so we consider purple jar,
therefore jars with more than half full are Orange, Red and Purple.

f. Now we will add another \(\frac{3}{4}\) cup of water to the green jar means
\(\frac{3}{4}\) + \(\frac{3}{4}\) as denominators as same we add numerators
we get \(\frac{6}{4}\) which is equal to blue color jar, we know the jar that has the
least amount of water makes the lowest sound before it was the green jar which had least amount of water,
so it had lowest sound, so now after adding \(\frac{3}{4}\) cup of water to green jar it has become equal
to blue jar so now both green and blue jar has least amount of water so they both make the lowest sound.

Multiply Whole Numbers and Fractions Activity

Three In a Row: Fraction Multiplication

Directions:

  1. Players take turns.
  2. On your turn, spin both spinners.
  3. Multiply the whole number and the fraction or mixed number. Cover the product.
  4. If the product is already covered, you lose your turn.
  5. The first player to get three in a row wins!

Big Ideas Math Answer Key Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 117
Answer:

Multiply Whole Numbers and Fractions Chapter Practice

9.1 Understand Multiples of Unit Fractions

Write the fraction as a multiple of a unit fraction.

Question 1.
Big Ideas Math Answer Key Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 118
Answer:
\(\frac{3}{10}\)= \(\frac{1}{10}\) + \(\frac{1}{10}\) + \(\frac{1}{10}\)
= 3 X \(\frac{1}{10}\).

Explanation:
We write \(\frac{3}{10}\) in addition equation as \(\frac{3}{100}\)=
\(\frac{1}{10}\) + \(\frac{1}{10}\) + \(\frac{1}{10}\) and
multiplication as unit fraction \(\frac{3}{10}\) represents 3 parts that are \(\frac{1}{10}\) of the whole, \(\frac{3}{10}\) = 3 X \(\frac{1}{10}\)

Question 2.
Big Ideas Math Answer Key Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 119
Answer:
\(\frac{4}{8}\)= \(\frac{1}{8}\) + \(\frac{1}{8}\) +
\(\frac{1}{8}\) + \(\frac{1}{8}\) = 4 X \(\frac{1}{8}\).

Explanation:
We write \(\frac{4}{8}\) in addition equation as \(\frac{4}{8}\)=
\(\frac{1}{8}\) + \(\frac{1}{8}\) + \(\frac{1}{8}\) +
\(\frac{1}{8}\) and multiplication as unit fraction \(\frac{4}{8}\) represents 4 parts
that are \(\frac{1}{8}\) of the whole, \(\frac{4}{8}\) = 4 X \(\frac{1}{8}\)

Question 3.
\(\frac{7}{2}\)
Answer:
\(\frac{7}{2}\) = \(\frac{1}{2}\) + \(\frac{1}{2}\) +
\(\frac{1}{2}\) + \(\frac{1}{2}\) + \(\frac{1}{2}\) +
\(\frac{1}{2}\) + \(\frac{1}{2}\) = 7 X \(\frac{1}{2}\).

Explanation:
we write \(\frac{7}{2}\) in addition equation as  \(\frac{7}{2}\) =
\(\frac{1}{2}\) + \(\frac{1}{2}\) + \(\frac{1}{2}\) + \(\frac{1}{2}\)
+ \(\frac{1}{2}\) + \(\frac{1}{2}\) + \(\frac{1}{2}\) = 7 X \(\frac{1}{2}\)
and multiplication as unit fraction \(\frac{7}{2}\) represents 7 parts that are
\(\frac{1}{2}\) of the whole, \(\frac{7}{2}\) = 7 X \(\frac{1}{2}\).

Question 4.
\(\frac{56}{100}\)
Answer:
\(\frac{56}{100}\)  = \(\frac{14}{25}\) = \(\frac{1}{25}\) +
\(\frac{1}{25}\) + \(\frac{1}{25}\) + \(\frac{1}{25}\) +
\(\frac{1}{25}\) + \(\frac{1}{25}\) + \(\frac{1}{25}\) +
\(\frac{1}{25}\) + \(\frac{1}{25}\) + \(\frac{1}{25}\) +
\(\frac{1}{25}\) + \(\frac{1}{25}\) + \(\frac{1}{25}\) +
\(\frac{1}{25}\) = 14 X \(\frac{1}{25}\)

Explanation:
\(\frac{56}{100}\) first we simplify as both numerator and denominator can be
divided by 4 we get numerator as 14 and denominatior as 25 =  \(\frac{14}{25}\)
now we write \(\frac{14}{25}\) in addition equation as \(\frac{14}{25}\) =
\(\frac{1}{25}\) + \(\frac{1}{25}\) + \(\frac{1}{25}\) + \(\frac{1}{25}\) +
\(\frac{1}{25}\) + \(\frac{1}{25}\) + \(\frac{1}{25}\) + \(\frac{1}{25}\) +
\(\frac{1}{25}\) + \(\frac{1}{25}\) + \(\frac{1}{25}\) + \(\frac{1}{25}\) +
\(\frac{1}{25}\) + \(\frac{1}{25}\) = 14 X \(\frac{1}{25}\) and multiplication as unit fraction \(\frac{14}{25}\) represents 14 parts that are \(\frac{1}{25}\) of the whole,
\(\frac{14}{25}\) = 14 X \(\frac{1}{25}\) .

Question 5.
Logic
What is Newton’s fraction? Write the fraction as a multiple of a unit fraction.
Big Ideas Math Answer Key Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 120
Answer:
Newton’s fraction is 5 X \(\frac{1}{6}\), The fraction is already a multiple of a unit fraction =
5 X \(\frac{1}{6}\) , 5 is multiple and \(\frac{1}{6}\) is a unit fraction.

Explanation:
Given unit fraction is one sixth of a whole is \(\frac{1}{6}\) and Newton’s fraction is 5 unit fractions
means it is equal to \(\frac{1}{6}\) + \(\frac{1}{6}\) + \(\frac{1}{6}\) +
\(\frac{1}{6}\) + \(\frac{1}{6}\) = 5 X \(\frac{1}{6}\) already the
fraction is a multiple of a unit fraction = 5 X \(\frac{1}{6}\) , 5 is multiple and
\(\frac{1}{6}\) is a unit fraction.

9.2 Understand Multiples of Fractions

Write the product as a multiple of a unit fraction. Then find the product.

Question 6.
Big Ideas Math Answer Key Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 121
Answer:
2 X \(\frac{2}{4}\) = 2 X 2 X \(\frac{1}{4}\) = 4 X \(\frac{1}{4}\) = \(\frac{4}{4}\)= 1

Explanation:
First we write \(\frac{2}{4}\)  as  \(\frac{1}{4}\) + \(\frac{1}{4}\) =
2 X \(\frac{1}{4}\) then multiply it by 2 which is  2 X 2 X \(\frac{1}{4}\) =
4 X \(\frac{1}{4}\) = \(\frac{4}{4}\) as numerator and denominator are same
both get cancelled so we get 1 , \(\frac{4}{4}\) = 1.

Question 7.
Big Ideas Math Answer Key Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 122
Answer:
3 X \(\frac{9}{12}\) = 3 X \(\frac{3}{4}\) = 9 X \(\frac{1}{4}\) =
\(\frac{9}{4}\)

Explanantion:
As \(\frac{9}{12}\) can be simplified as both 9,12 can go with 3 leaving numerator as
3 and denominator as 4= \(\frac{3}{4}\), First we write \(\frac{3}{4}\)  as  \(\frac{1}{4}\) +
\(\frac{1}{4}\) + \(\frac{1}{4}\) = 3 X \(\frac{1}{4}\) then multiply it by 3
which is  3 X 3 X \(\frac{1}{4}\) = 9 X \(\frac{1}{4}\) is multiple of unit fraction,
The product is \(\frac{9}{4}\) .

Question 8.
Big Ideas Math Answer Key Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 123
Answer:
\(\frac{3}{5}\) X 4 = 12 X \(\frac{1}{5}\) = \(\frac{12}{5}\).

Explanation:
First we write \(\frac{3}{5}\)  as  \(\frac{1}{5}\) + \(\frac{1}{5}\) +
\(\frac{1}{5}\) = 3 X \(\frac{1}{5}\) then multiply it by 4 which is
4 X 3 X \(\frac{1}{5}\) = 12 X \(\frac{1}{5}\) = \(\frac{12}{5}\).

Question 9.
Big Ideas Math Answer Key Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 124
Answer:
\(\frac{8}{10}\) X 7= \(\frac{4}{5}\) X 7 = 28 X \(\frac{1}{5}\) is
multiple of unit fraction, The product is \(\frac{28}{5}\).

Explanantion:
As \(\frac{8}{10}\) can be simplified as both 8,10 can go with 2, leaving numerator as 4
and denominator as 5 = \(\frac{4}{5}\), First we write \(\frac{4}{5}\)  as
\(\frac{1}{5}\) + \(\frac{1}{5}\) + \(\frac{1}{5}\) + \(\frac{1}{5}\) =
4 X \(\frac{1}{5}\) then multiply it by 7 which is  7 X 4 X \(\frac{1}{5}\) =
28 X \(\frac{1}{5}\) is multiple of unit fraction, The product is \(\frac{28}{5}\).

Question 10.
Big Ideas Math Answer Key Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 125
Answer:
8 X \(\frac{6}{3}\) = 8 X 2 = 16 as we got \(\frac{6}{3}\) =2,
So we get results as whole not in fraction.

Explanation:
As \(\frac{6}{3}\) can be simplified as both can be divided by 2, leaving numerator as 2 and denominator as 1 ,  \(\frac{6}{3}\) = 2  , now it is 8 X 2 = 16 as we got \(\frac{6}{3}\) =2, so we get results as whole not in fraction.

Question 11.
Big Ideas Math Answer Key Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 126
Answer:
10 X \(\frac{30}{8}\) = 10 X \(\frac{15}{4}\) = 150 X \(\frac{1}{4}\) =
\(\frac{150}{4}\) = \(\frac{75}{2}\).

Explanation:
As \(\frac{30}{8}\) can be simplified as both can be divided by 2 , we get numerator as 15 and denominator as 4, \(\frac{30}{8}\) = \(\frac{15}{4}\) = \(\frac{1}{4}\) + \(\frac{1}{4}\) +
\(\frac{1}{4}\) + \(\frac{1}{4}\) + \(\frac{1}{4}\) + \(\frac{1}{4}\) +
\(\frac{1}{4}\) + \(\frac{1}{4}\) + \(\frac{1}{4}\) + \(\frac{1}{4}\) +
\(\frac{1}{4}\) + \(\frac{1}{4}\) + \(\frac{1}{4}\) + \(\frac{1}{4}\) +
\(\frac{1}{4}\) = 15 X \(\frac{1}{4}\) now multiply with 10 X 15 X \(\frac{1}{4}\)
= 150 X \(\frac{1}{4}\) = \(\frac{150}{4}\)  this can be further simplified as 150 and 4 both
can be further divided by 2 making numerator as 75 and denominator as 2 = \(\frac{75}{2}\) .

9.3 Multiply Whole Numbers and Fractions

Multiply

Question 12.
Big Ideas Math Answer Key Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 127
Answer:
2 X \(\frac{1}{2}\) = \(\frac{2}{2}\) = 1
Explanation:
Already it is in unit fraction so 2 X \(\frac{1}{2}\) = \(\frac{2}{2}\) = 1.

Question 13.
Big Ideas Math Answer Key Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 128
Answer:
4 X \(\frac{5}{8}\) = \(\frac{20}{8}\) = \(\frac{10}{4}\) =
\(\frac{5}{2}\).

Explanation :
First we write \(\frac{5}{8}\) in unit fraction as 5 X \(\frac{1}{8}\)
now we multiply with whole 4 X 5 X \(\frac{1}{8}\) = \(\frac{20}{8}\)
further can be simplifed as 20,8 both can be divided by 4 we get numerator as 5 and denominator 2,
therefore 4 X \(\frac{5}{8}\) = \(\frac{5}{2}\) .

Question 14.
Big Ideas Math Answer Key Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 129
Answer:
3 X \(\frac{9}{6}\) = \(\frac{9}{2}\).

Explanantion:
We can simplify \(\frac{9}{6}\) as both goes by 3 giving numerator as 3 and
denominator as 2 so we get \(\frac{9}{6}\) = \(\frac{3}{2}\) now we write in unit fraction
as 3 X \(\frac{1}{2}\) now we multiply by 3, 3 X 3 X \(\frac{1}{2}\) =
9 X \(\frac{1}{2}\) = \(\frac{9}{2}\).

Question 15.
Big Ideas Math Answer Key Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 130
Answer:
5 X  \(\frac{7}{12}\) = \(\frac{35}{12}\).

Explanation:
first we write \(\frac{7}{12}\) as unit fraction = 7 X \(\frac{1}{12}\),
now we multiply with 5 as 5 X 7 X \(\frac{1}{12}\) = 35 X \(\frac{1}{12}\) =
\(\frac{35}{12}\) .

Question 16.
Big Ideas Math Answer Key Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 131
Answer:
7 X \(\frac{30}{100}\) = 7 X \(\frac{3}{10}\) = \(\frac{21}{10}\).

Explanation:
We can simplify \(\frac{30}{100}\) as both goes by 10 giving numerator as 3 and
denominator as 10 so we get \(\frac{3}{10}\)  now we write in unit fraction as
3 X \(\frac{1}{10}\) and multiply by 7 , 7 X 3 X \(\frac{1}{10}\)  = 21 X \(\frac{1}{10}\)
=  \(\frac{21}{10}\) .

Question 17.
Big Ideas Math Answer Key Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 132
Answer:
\(\frac{8}{4}\) X 9 = 2 X 9 = 18.

Explanantion:
We can simplify \(\frac{8}{4}\) as both goes by 4 giving numerator as 2 and denominator
as 1 so we get 2 as whole, now we multiply with 9 we get 2 X 9 = 18.

9.4 Multiply Whole Numbers and Mixed Numbers

Multiply

Question 18.
Big Ideas Math Answer Key Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 133
Answer:
Big Ideas Math Answer Key Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 133= \(\frac{5}{2}\)= 2 \(\frac{1}{2}\)
Explanation:
First we write the mixed number as a fraction then multiply, so 1 \(\frac{1}{4}\)=
1 + \(\frac{1}{4}\)= (1 X 4 + 1) by 4 = \(\frac{5}{4}\) now we multiply it with the whole,
2 X \(\frac{5}{4}\) now we write in unit fraction and multiply  2 X 5 X \(\frac{1}{4}\)=
10 X \(\frac{1}{4}\) = \(\frac{10}{4}\) further we can simplify as 10,4
both can be divided by 2 we get numerator as 5 and denominator as 2, \(\frac{5}{2}\)
as numerator is greater than denominator we can write in mixed fraction as 2 goes in 2 times.
2 will be our whole number 2 X 2 is 4 and we have 1 left over (54), 1 will be our numerator and our
denominator will stay the same 2, \(\frac{5}{2}\)= 2 \(\frac{1}{2}\).

Question 19.
Big Ideas Math Answer Key Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 134
Answer:
Big Ideas Math Answer Key Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 134= \(\frac{69}{6}\) = \(\frac{23}{2}\) = 11 \(\frac{1}{2}\)
Explanation:
First we write the mixed number as a fraction then multiply, so 3 \(\frac{10}{12}\)=
( 3 X 12 + 10 ) by 12 = \(\frac{46}{12}\) this can be further simplified as both 46,12 goes by 2
we get numerator as 23 and denominator as 6 , \(\frac{23}{6}\) now we multiply by 3 =
3 X \(\frac{23}{6}\) = \(\frac{69}{6}\) it can be more simplified as both 69,6 goes by 3
we get numerator as 23 and denominator as 2 making \(\frac{69}{6}\) as \(\frac{23}{2}\)
as numerator is greater than denominator we can write in mixed fraction as 2 goes in 11 times.
11 will be our whole number 2 X 11 is 22 and we have 1 left over (23−22), 1 will be our numerator
and our denominator will stay the same 2, \(\frac{23}{2}\) = 11 \(\frac{1}{2}\).

Question 20.
Big Ideas Math Answer Key Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 135
Answer:
Big Ideas Math Answer Key Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 135= \(\frac{84}{8}\) = \(\frac{21}{2}\) = 10 \(\frac{1}{2}\).
Explanantion:

First we write the mixed number as a fraction then multiply, so 2 \(\frac{5}{8}\) =
( 2 X 8 + 5 ) by 8 = \(\frac{21}{8}\) now we multiply by 4 we get 4 X \(\frac{21}{8}\) =
4 X 21 X \(\frac{1}{8}\) = 84 X \(\frac{1}{8}\) =\(\frac{84}{8}\)
further can be simplified as both 84,8 goes by 4 we get numerator as 21 and denominator as 2
=\(\frac{21}{2}\) as numerator is greater than denominator we can write in mixed fraction
as 2 goes in 10 times. 10 will be our whole number 2 X 10 is 20 and we have 1 left over (21−20),
1 will be our numerator and our denominator will stay the same 2,
\(\frac{21}{2}\) = 10 \(\frac{1}{2}\).

Question 21.
Big Ideas Math Answer Key Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 136
Answer:
Big Ideas Math Answer Key Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 136= \(\frac{40}{3}\) = 13 \(\frac{1}{3}\)
Explanation:
First we write the mixed number as a fraction then multiply, so 2 \(\frac{4}{6}\) =
( 2 X 6 + 4 ) by 6 = \(\frac{16}{6}\) this can be simplified as 16,6 both goes by 2
we get numerator as 8 and denominator as 3 we get  \(\frac{8}{3}\)
now we multiply by 5 we get 5 X \(\frac{8}{3}\) = 5 X 8 X \(\frac{1}{3}\) =
40 X \(\frac{1}{3}\) = \(\frac{40}{3}\) as numerator is greater than denominator
we can write in mixed fraction as 3 goes in 13 times. 13 will be our whole number 3 X 13 is 39 and
we have 1 left over (40−39), 1 will be our numerator and our denominator will stay the same 3,
we get mixed fraction as 13 \(\frac{1}{3}\) .

Question 22.
Big Ideas Math Answer Key Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 137
Answer:
Big Ideas Math Answer Key Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 137= \(\frac{140}{3}\)  = 46 \(\frac{2}{3}\)
Explanation:
First we write the mixed number as a fraction then multiply, so 6 \(\frac{2}{3}\) =
( 6 X 3 + 2 ) by 3 = \(\frac{20}{3}\)  now we multiply by 7  X \(\frac{20}{3}\) =
7 X 20 x \(\frac{1}{3}\)  = 140 X \(\frac{1}{3}\) =\(\frac{140}{3}\)
as numerator is greater than denominator we can write in mixed fraction as 3 goes in 46 times.
46 will be our whole number 3 X 46 is 138 and we have 2 left over (140−138), 2 will be our numerator
and our denominator will stay the same 3, we get mixed fraction as 46 \(\frac{2}{3}\) .

Question 23.
Big Ideas Math Answer Key Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 138
Answer:
Big Ideas Math Answer Key Grade 4 Chapter 9 Multiply Whole Numbers and Fractions 138= \(\frac{470}{5}\) = 94
Explanation:
First we write the mixed number as a fraction then multiply, so 9 \(\frac{2}{5}\) =
( 9 X 5 + 2 ) by 5 = \(\frac{47}{5}\) now we multiply by 10 = 10 X \(\frac{47}{5}\) =
\(\frac{470}{5}\) this can be further simplified as both can be divided by 5 we get numerator as
94 and denominator as 1 so we get only whole number 94.

9.5 Problem Solving: Fraction Operations

Question 24.
Modeling Real Life
Your friend rides her bike \(\frac{2}{6}\) mile to your house and \(\frac{2}{6}\)
mile back home 4 times in 1 week. How far does she bike in all?
Answer:
My friend rides her bike by \(\frac{8}{3}\) miles or 2 \(\frac{2}{3}\) miles.

Explanation:
Given friend rides her bike \(\frac{2}{6}\) mile to your house and \(\frac{2}{6}\) mile
back home, first we calculate for 1 time, so first we will add \(\frac{2}{6}\) + \(\frac{2}{6}\)
as both have same denominators we will add numerators as ( 2 + 2 = 4 ).
So we get \(\frac{4}{6}\) miles for 1 time , now she rides 4 times means 4 X \(\frac{4}{6}\) =
\(\frac{16}{6}\) this can be further simplified as 16,6 both goes by 2 we get numerator as 8 and
denominator as 3 = \(\frac{8}{3}\) as numerator is greater than denominator we can write in
mixed fraction as 3 goes in 2 times. 2 will be our whole number 3 X 2 is 6 and
we have 2 left over (8−6), 2 will be our numerator and our denominator will stay the same 3,
we get mixed fraction as 2 \(\frac{2}{3}\) miles.

Conclusion:

Access the links given in the Big Ideas Math Book 4th Grade Answer Key Chapter 9 Multiply Whole Numbers and Fractions and practice well for the exams. We have solved all the questions in an easy and simple manner. We hope the information provided in this article has brought a smile to your face. Share Big Ideas Math Grade 4 Answer Key Chapter 9 Multiply Whole Numbers and Fractions.

Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation

Big Ideas Math Answers Grade 8 Chapter 8

Do you face any difficulties in the concepts of Big Ideas Math Grade 8 Exponents and Scientific Notation Chapter? If So, you have come the right way and we will provide you the complete Big Ideas Math Grade 8 Answers Chapter 8 Exponents and Scientific Notation Solutions all in one place. Look no further and begin your preparation taking the help of the quick resources and stand out from the rest of the crowd. Consistent practice is the only key to achieve success in your journey of learning math.

Big Ideas Math Book 8th Grade Answer Key Chapter 8 Exponents and Scientific Notation

Middle School Students are advised to go through the Topicwise BIM Grade 8 Chapter 8 Exponents and Scientific Notation Answers provided to resolve their queries at the soonest. All the Concepts here are given by subject experts in a simple and easy-to-understand language. Big Ideas Math Textbook Solution Key is given via direct links and you can download the respective topic and prepare anywhere and anytime.

Performance Task

Lesson: 1 Exponents

Lesson: 2 Product of Powers Property

Lesson: 3 Quotient of Powers Property

Lesson: 4 Zero and Negative Exponents

Lesson: 5 Estimating Quantities

Lesson: 6 Scientific Notation

Lesson: 7 Operations in Scientific Notation

Chapter: 8 – Exponents and Scientific Notation

Exponents and Scientific Notation STEAM Video/ Performance Task

STEAM Video

Carbon Atoms
Carbon is one of the four main elements of life. The number of carbon atoms in a compound can be represented using exponents. In what other real-life situations are exponents used?
Big Ideas Math Answer Key Grade 8 Chapter 8 Exponents and Scientific Notation 1
Watch the STEAM Video “Carbon Atoms.” Then answer the following questions.
1. The table shows the percent carbon by weight for humans and plants. How many pounds of carbon are in a 130-pound person? a 25-pound plant?
Big Ideas Math Answer Key Grade 8 Chapter 8 Exponents and Scientific Notation 2
a. Pounds of Carbon in  130-pound person is 2.57 X 1021.
b. Pounds of Carbon in 25 -pound plant is 1.23975 X 1021.

Explanation:
Given 1 carbon atoms consists of 5 X 1022,
a. So 1 person has 18% of carbon means 18 ÷ 100 X 5 × 1022 =
18 X 5 X 1022-2 = 90 X 1020 = 9 X 10 X 1020 = 9 X 1021 now,
We know 1 gram is equal to 0.00220462 pound
So 0.00220462 X 9 X 1021 as 0.00220462 approximately equal to ≈
2.204 X 10-3 X 9 X 1021 = 19.836 X 1021-3 = 19.836 X 1018 = 1.9836 X 1019  now
in 130-pound person is 130 X 1.9836 X 1019 = 257.868 X 1019 = 2.57 X 1021.
b. So 1 plant  has 45% of carbon means 45 ÷ 100 X 5 × 1022 =
45 X 5 X 1022-2 = 225 X 1020 = 2.25 X 100 X 1020 = 2.25 X 1022 now,
We know 1 gram is equal to 0.00220462 pound
So 0.00220462 X 2.25 X 1022 as 0.00220462 approximately equal to ≈
2.204 X 10-3 X 2.25 X 1022 = 4.959 X 1022-3 = 4.959 X 1019 = 4.959 X 1019  now
in 25-pound plant is 25 X 4.959 X 1019 =123.975 X 1019 = 1.23975 X 1021.

2. Steven says 5 × 1022, carbon atoms are in 1 gram of carbon. How many carbon atoms are in 3 grams of carbon?

In 3 grams of carbon = 3 X (5 x1022) =  15 x 1022 carbon atoms are available

Explanation:
Given Steven says 5 x1022 ,carbon atoms are in 1 gram of carbon,
in 3 grams of carbon it will be (
5 x 1022) X 3 = 5 X 3 X (1022)=
15 x 1022 carbon atoms are available.

Performance Task

Elements in the Universe
After completing this chapter, you will be able to use the concepts you learned to answer the questions in the STEAM Video Performance Task. You will be given information about the atomic masses of the four most common elements in the universe: oxygen, hydrogen, helium, and carbon.
Big Ideas Math Answer Key Grade 8 Chapter 8 Exponents and Scientific Notation 3
You will be asked to solve problems about the amounts of carbon dioxide in Earth’s atmosphere for several years. What might cause the amount of carbon dioxide in the atmosphere to increase over time?

Exponents and Scientific Notation Getting Ready for Chapter 8

Chapter Exploration
1. Work with a partner. Write each distance as a whole number. Which numbers do you know how to write in words? For instance, in words, 102 is equal to one hundred.
Big Ideas Math Answer Key Grade 8 Chapter 8 Exponents and Scientific Notation 4

a. 1027 meters diameter of the observable universe is
10000000 0000000000 0000000000 is equal to
octillion ( quadrilliard )
b. 1021 meters diameter of the Milky Way galaxy is
10 0000000000 0000000000 is equal to
sextillion (trilliard)
c. 1016 meters diameter of the solar system is
10 0000000000 00000 is equal to
10 quadrillion or 10 thousand trillion
d. 107 meters diameter of Earth is 10000000
is equal to ten million (crore (India))
e. 104 meters diameter of Halley’s Comet is 10000
is equal to ten thousand
f. 103 meters diameter of a meteor crater is
1000 is equal to thousand.

Explanation:
a. 1027 meters diameter of the observable universe,
we call the number 10 is called the base
and the number 27 is called the exponent, we multiply 10 by 27 times,
we write 10 to the 27th power as  1,000,000,000,000,000,000,000,000,000
is equal to octillion ( quadrilliard )
b. 1021 meters diameter of the Milky Way galaxy,
we call the number 10 is called the base
and the number 21 is called the exponent, we multiply 10 by 21 times,
we write 10 to the 21th power as 1,000,000,000,000,000,000,000
is equal to sextillion (trilliard)
c. 1016 meters diameter of the solar system
we call the number 10 is called the base
and the number 16 is called the exponent, we multiply 10 by 16 times,
we write 10 to the 16th power as 10,000,000,000,000,000 is equal to
10 quadrillion or 10 thousand trillion
d. 107 meters diameter of Earth
we call the number 10 is called the base
and the number 7 is called the exponent, we multiply 10 by 7 times,
we write 10 to the 7th power as 10,000,000 is equal to ten million (crore (India))
e. 104 meters diameter of Halley’s Comet we call the number 10 is called the base
and the number 4 is called the exponent, we multiply 10 by 4 times,
we write 10 to the 4th power as 10,000 is equal to ten thousand
f. 103 meters diameter of a meteor crater we call the number 10 is called the base
and the number 27 is called the exponent, we multiply 10 by 3 times
we write 10 to the 3th power as 1,000 is equal to thousand.

2. Work with a partner. Write the numbers of wives, sacks, cats, and kits as powers.
Big Ideas Math Answer Key Grade 8 Chapter 8 Exponents and Scientific Notation 5

Man ,wives are  70  + 71 = 1 + 7 = 8,
Man, wives, sacks are  70 + 7= 1 + 49 = 50
Man, wives, sacks, cats are 70 + 7= 1 + 343 = 344,
Man, wives, sacks, cats , kits are 70 + 7= 1 + 2401 = 2402
Total 2402 are going to St. Ives.
Explanation:
Given I met a man with seven wives so
writing as powers man and seven wives means 1 + 71 = 8,
man, each wives had seven sacks means 1 + 7= 49,
man, wives, each sack had seven cats are 1+ 7= 1 + 343 = 344,
man, wives, sacks , Each cat had seven kits are 1+ 7= 1 + 2401 = 2402,
So in total kits, cats, sacks, wives and man are 2402 are going to St. Ives.

Vocabulary

The following vocabulary terms are defined in this chapter. Think about what each term might mean and record your thoughts.
power
exponent of a power
base of a power
scientific notation

Lesson 8.1 Exponents

The expression 35 is called a power. The base is 3. The exponent is 5.
Big Ideas Math Answer Key Grade 8 Chapter 8 Exponents and Scientific Notation 8.1 1

EXPLORATION 1

Using Exponent Notation
Work with a partner.
a. Copy and complete the table.
Big Ideas Math Answer Key Grade 8 Chapter 8 Exponents and Scientific Notation 8.1 2

Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation-1

Explanation:
Given Powers wrote repeated multiplication form and value as
(-3)3= (-3).(-3).(-3)= -27, multiplied -3 three times as power is 3
(-3)4= (-3).(-3).(-3).(-3)= 81,multiplied -3 four times as power is 4
(-3)5= (-3).(-3).(-3).(-3).(-3)= -243,multiplied -3 five times as power is 5
(-3)6= (-3).(-3).(-3).(-3).(-3).(-3)= 729,multiplied -3 five times as power is 6
(-3)7= (-3).(-3).(-3).(-3).(-3).(-3).(-3)= 2187,multiplied -3 five times as power is 7

b. Describe what is meant by the expression (- 3)n. How can you find the value of (- 3)n?
Answer:
The expression (-3)n is called a power. The base is -3. The exponent is n.
We find the value of (- 3)n we multiply -3 with n number of times.

Explanation:
An expression that represents repeated multiplication
of the same factor is called a power. Here the expression
(-3)n is called a power of n and the number -3 is called the base,
and the number n is called the exponent. The exponent corresponds
to the number of times the base is used as a factor.

EXPLORATION 2

Using Exponent Notation
Work with a partner. On a game show, each small cube is worth $3. The small cubes are arranged to form a large cube. Show how you can use a power to find the total value of the large cube. Then write an explanation to convince a friend that your answer is correct.
Big Ideas Math Answer Key Grade 8 Chapter 8 Exponents and Scientific Notation 8.1 3
Answer:

Big Ideas Math Answer Key Grade 8 Chapter 8 Exponents and Scientific Notation 8.1 4
The total value of the large cube is =$3 X (320)320
My answer is correct as given base as 320 and exponent is 320.

Explanation:
Given small cubes are arranged to form large cube,
and power is 320, base is 320 and exponent is 320,
each small cube is worth $3  so the total value of the large
cube is $3 multiplied by 320 and multiplied $3 with 320 by 320 times is
$3 X (320)320 as my answer is correct I say my friend because
given base is 320 and exponent is 320  we write as (320)320  and
multiply by $3 to get the value of the large cube.

Try It

Write the product using exponents.
Question 1.
\(\frac{1}{4} \cdot \frac{1}{4} \cdot \frac{1}{4} \cdot \frac{1}{4} \cdot \frac{1}{4}\)
Answer:
\(\frac{1}{4} \cdot \frac{1}{4} \cdot \frac{1}{4} \cdot \frac{1}{4} \cdot \frac{1}{4}\) = (\(\frac{1}{4}\))5

Explanation:
As \(\frac{1}{4}\) is multiplied by 5 times,
the expression is (\(\frac{1}{4}\))5.
The base is \(\frac{1}{4}\) and the exponent is 5.

Question 2.
0.3 • 0.3 • 0.3 • 0.3 • x • x
Answer:
0.3 • 0.3 • 0.3 • 0.3 • x • x = (0.3)X (x)

Explanation:
As 0.3 is multiplied 4 times we write as (0.3)and x
is multiplied twice so (x)2
so the expression is (0.3)X (x)2.

Evaluate the expression.

Question 3.
122
Answer:
122 = 144

Explanation:
Given 122 means 12 is multiplied twice
as 12 X 12 we get 144.
here base is 12 and exponent is 2.

Question 4.
(- 2)6
Answer:
(- 2)6 = 64

Explanation:
Given (- 2)6 we multiply -2 by 6 times as
-2  X -2 X -2 X -2 X -2 X -2 we get 64
here base is -2 and exponent is 6.

Question 5.
– 54
Answer:
– 54 =625

Explanation:
Given – 5we multiply -5 by 4 times as
-5 X -5 X -5 X -5 = 625,
here base is -5 and 4 is exponent.

Question 6.
\(\left(-\frac{1}{6}\right)^{3}\)
Answer:
\(\left(-\frac{1}{6}\right)^{3}\) = –\(\frac{1}{216}\)

Explanation:
Given \(\left(-\frac{1}{6}\right)^{3}\)  we multiply
– \(\frac{1}{6}\) by 3 times as –\(\frac{1}{6}\) X
– \(\frac{1}{6}\) X –\(\frac{1}{6}\) we get
– \(\frac{1}{216}\) here base is –\(\frac{1}{6}\)
and 3 is exponent.

Evaluate the expression.
Question 7.
9 – 25 . 0.5
Answer:
9 – 25 . 0.5 = -7

Explanation:
Given 9 – 25 . 0.5 = First we solve  25 . 0.5
25 X 0.5 as 0.5 can be written as \(\frac{1}{2}\),
= 32 X \(\frac{1}{2}\) = 16 now we subtract 16 from 9
9-16 = -7

Question 8.
|- 33 ÷ 27|
Answer:
|- 33 ÷ 27| = -1

Explanation:
First we calculate – 33 we multiply -3 by 3 times as
-3 X -3 X -3 = -27 now we divide -27 by 27 we get -1.

Question 9.
(7 . 4 – 43) ÷ 6
Answer:
(7 . 4 – 43) ÷ 6 = -6

Explanation:
First we calculate the value of 43
we multiply 4 by 3 times as 4 X 4 X 4 = 64,
Now we multiply 7 X 4 = 28 now we subtract 64 from 28
we get (7 . 4 – 43) =(28 – 64) = -36 now we divide this by 6 we get
-36 ÷ 6 = -6, So (7 . 4 – 43) ÷ 6 = -6.

Self-Assessment for Concepts & Skills
Solve each exercise. Then rate your understanding of the success criteria in your journal.

WRITING EXPRESSIONS USING EXPONENTS Write the product using exponents.
Question 10.
(- 0.9) • (- 0.9) • (- 0.9)
Answer:
(- 0.9) • (- 0.9) • (- 0.9 )= (-0.9)3

Explanation:
As -0.9 is multiplied by 3 times we write as
(-0.9)3 here -0.9 is base and 3 is exponent.

Question 11.
\(\frac{1}{8}\) • \(\frac{1}{8}\) • y • y • y
Answer:
\(\frac{1}{8}\) • \(\frac{1}{8}\) • y • y • y =
(\(\frac{1}{8}\))2 X (y)3

Explanation:
As \(\frac{1}{8}\) is multiplied by 2 times we write as
(\(\frac{1}{8}\))2
and y is multiplied 3 times we write as (y)3,
So \(\frac{1}{8}\) • \(\frac{1}{8}\) • y • y • y =
(\(\frac{1}{8}\))2 X (y)3.

EVALUATING EXPRESSIONS Evaluate the expression.
Question 12.
112
Answer:
112 = 121

Explanation:
Given 112 means 11 is multiplied twice
as 11 X 11 = 121.

Question 13.
– 63
Answer:
– 63 = – 216

Explanation:
Given – 6 is multiplied by 3 times as
– 6 X -6 X -6 = -216

Question 14.
(- 0.3)4
Answer:
(- 0.3)4 = 0.0081

Explanation:
As (-0.3) is multiplied by 4 times we get
-0.3 X -0.3 X -0.3 X -0.3 = 0.0081

USING ORDER OF OPERATIONS Evaluate the expression.
Question 15.
|- 24 ÷ 22|
Answer:
|- 24 ÷ 22|= -6

Explanation:
First we calculate 22 we get 4,
now we divide -24 by 4 we get -6.

Question 16.
(33 – 6 • 8) ÷ 7
Answer:
(33 – 6 • 8) ÷ 7 = -3

Explanation:
First we calculate 6 X 8 we get 48 Now we subtract
48 from 33 as 33  is 3 X 3 X 3 = 27 we get (27 – 48 )= -21
now we divide -21 by 7 we get -3 therefore
(33 – 6 • 8) ÷ 7 = -3 .

Question 17.
WHICH ONE DOESN’T BELONG?
Which expression does not belong with the other three? Explain your reasoning.
Big Ideas Math Answer Key Grade 8 Chapter 8 Exponents and Scientific Notation 8.1 5
Answer:
The expression – 82 does not belongs to other three.

Explanation:
Given expressions (-2)6,- 82 ,-82 and 2
the values are  (-2)6= -2 X -2 X -2  X -2 X -2 X -2 = 64,
– 82 =- ( 8 X 8) = -64,
82 = 8 X 8 = 64 and 26 = 2 X 2 X 2 X 2 X 2 X 2 = 64
as (-2)6,-82 and 26 have same value 64 only – 82 =-64 is different,
so the expression – 82 does not belongs to other three.

Self-Assessment for Problem Solving
Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 18.
DIG DEEPER!
Consider the diameters of three planets.
Planet A: 109m
Planet B: 107n
Planet C: 108m
a. Write each diameter as a whole number.
b. A dwarf planet is discovered with a radius that is \(\frac{1}{100}\) the radius Planet C. Write the diameter of the dwarf planet as a power.
Answer:
a. Planet A: 109m = 1000000000m
Planet B: 107n = 10000000n
Planet C: 108m = 100000000m
b. diameter = 2 X 106m

Explanation:
The diameters of three planets are given as
Planet A: 109m, Planet B: 107n, Planet C: 108m
in part a,we write whole for Planet A as 10 is multiplied by 9 times
so 10 X 10 X 10 X 10 X 10 X 10 X 10 X 10 X 10 =1000000000m
Planet B as 10 is multiplied by 7 times therefore it is
10 X 10 X 10 X 10 X 10 X 10 X 10 =10000000n and Planet C
as 10 X 10 X 10 X 10 X 10 X 10 X 10 X 10 =100000000m and
in part b, Given a dwarf planet is discovered with a radius that is
\(\frac{1}{100}\) the radius Planet C and
we know diameter = 2 X radius so diameter is
2 X 100000000 X \(\frac{1}{100}\) = 2 X 1000000,
therfore the diameter of the dwarf planet as a power is 2 X 106m.

Question 19.
A fish jumps out of the water at a speed of 12 feet per second. The height y (in feet) of the fish above the surface of the water is represented by the equation y = – 16x2 + 12x, where x is the time (in seconds) since the jump began. The fish reaches its highest point above the surface of the water after 0.375 second. How far above the surface is the fish at this time?
Big Ideas Math Answer Key Grade 8 Chapter 8 Exponents and Scientific Notation 8.1 6
Answer:
The fish is 2.25 feet above the surface at 0.375 second.

Explanation:
The height y (in feet) of the fish above the surface of the water
is represented by the equation y = – 16x2 + 12x,
where x is the time (in seconds) since the jump began.
The fish reaches its highest point above the surface
of the water after 0.375 second, So X = 0.375 second
we substitute In the equation as
y=- 16 (0.375 X 0.375 ) + 12 (0.375)= – 16 ( 0.140625) + 4.5
= -2.25 + 4.5 = 2.25
The fish is 2.25 feet above the surface at 0.375 second.

Exponents Homework & Practice 8.1

Review & Refresh

Sketch a graph that represents the situation.
Question 1.
A trading card becomes more valuable over time. The value increases at a constant rate, and then at a faster and faster rate.
Answer:
Sketch is
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation-2
Explanation:
We represent the graph Trading card as
on X axis Time and on Y axis Value,
Given the value increases at a constant rate,
so first we draw a straight Line with positive slope,
and increase rate is represented by exponential growth as
shown in the figure above.

Question 2.
The water level of a river remains constant, and then decreases at a constant rate.
Answer:
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation-3
Explanation:
Given the water level of a river remains constant
and then decreases at a constant rate so in the graph
we draw a straight Line with positive slope as constant,
and decrease rate is represented as decay shown in the figure above.

The vertices of a figure are given. Rotate the figure as described. Find the coordinates of the image.
Question 3.
A(0, – 4), B(0, – 1), C(2, – 1)
90° clockwise about the origin
Answer:
A/(-4,0)
B/(-1,0)
C/(-1,-2)

Explanation:
When we rotate a figure of 90 degrees clockwise about the origin,
each point of the given figure has to be changed from (x, y) to (y, -x).
So A(0,-4) becomes A/(-4,-0)= A/(-4,0) ,
B(0,-1) becomes B/(-1,-0)= B/(-1,0) and
C(2, – 1) becomes C/(-1,-2)

Question 4.
E(1, 2), F(1, 3), G(4, 3), H(4, 2)
180° about the origin
Answer:
E/(-1,-2)
F/(-1,-3)
G/(-4,-3)
H/(-4,-2)
Explanation:
Rotation of a point through 180°, about the origin when a point A (x, y)
is rotated about the origin O through 180° in anticlockwise
or clockwise direction, it takes the new position A’ (-x, -y)
So E(1,2) becomes E/(-1,-2), F(1,3) becomes F/(-1,-3), G(4,3)
becomes G/(-4,-3) and H(4,2) becomes H/(-4,-2).

Concepts, Skills, & Problem Solving

USING EXPONENT NOTATION Write the power in repeated multiplication form. Then find the value of the power.(See Exploration 1, p. 319.)
Question 5.
44
Answer:
4 X 4 X 4 X 4, 256

Explanation:
First we write 44 in repeated multiplication form as
power is 4 times 4 X 4 X 4 X 4 and the value is 256.

Question 6.
(- 8)2
Answer:
-8 X -8 , 64

Explanation:
First we write (- 8)2 in repeated multiplication form as
– 8 X -8 as power is 2 times and the value is 64.

Question 7.
(- 2)3
Answer:
– 2 X -2 X -2 , -8

Explanation:
First we write (- 2)3  in repeated multiplication form as
– 2 X -2 X -2  as power is 3 times and the value is -8.

WRITING EXPRESSIONS USING EXPONENTS Write the product using exponents.
Question 8.
3 • 3 • 3 • 3
Answer:
3 • 3 • 3 • 3 = (3)4  

Explanation:
We write the product 3 • 3 • 3 • 3 in exponents as (3)4
because 3 is multiplied by 4 times.

Question 9.
(- 6) • (- 6)
Answer:
(- 6) • (- 6) = (-6)2  

Explanation:
We write the product (-6) X (-6)  in exponents as (-6)2
because -6 is multiplied by 2 times.

Question 10.
(- \(\frac{1}{2}\)) • (- \(\frac{1}{2}\)) • (- \(\frac{1}{2}\))
Answer:
(- \(\frac{1}{2}\)) • (- \(\frac{1}{2}\)) • (- \(\frac{1}{2}\))=
(- \(\frac{1}{2}\))3  

Explanation:
We write the product (- \(\frac{1}{2}\)) X(- \(\frac{1}{2}\))
X (- \(\frac{1}{2}\)) in exponent as (- \(\frac{1}{2}\))3
here – \(\frac{1}{2}\) is multiplied by 3 times.

Question 11.
\(\frac{1}{3}\) • \(\frac{1}{3}\) • \(\frac{1}{3}\)
Answer:
\(\frac{1}{3}\) • \(\frac{1}{3}\) • \(\frac{1}{3}\)=
(\(\frac{1}{3}\))3

Explanation:
We write the product ( \(\frac{1}{3}\)) X ( \(\frac{1}{3}\)) X
(\(\frac{1}{3}\)) in exponent as (\(\frac{1}{3}\))3
here \(\frac{1}{3}\) is multiplied by 3 times.

Question 12.
π • π • π • x • x • x • x
Answer:
π • π • π • x • x • x • x = (π)3 X (x)4

Explanation:
We write the product π • π • π • x • x • x • x
in exponent as (π)3 X (x)4 here π is multiplied by
3 times and X is multiplied 4 times.

Question 13.
(- 4) • (- 4) • (- 4) • y • y
Answer:
(- 4) • (- 4) • (- 4) • y • y = (-4)3 X (y)2

Explanation:
Here we write the product (- 4) • (- 4) • (- 4) • y • y
as (-4)3 X (y)2,-4 is multiplied three times and y is
multiplied by 2 times

Question 14.
6.4 • 6.4 • 6.4 • 6.4 • b • b • b
Answer:
6.4 • 6.4 • 6.4 • 6.4 • b • b • b = (6.4)4 X (b)3

Explanation:
Here we write the product 6.4 • 6.4 • 6.4 • 6.4 • b • b • b
as (6.4)4 X (b)3,6.4  is multiplied four times and b is
multiplied by 3 times.

Question 15.
(- t) • (- t) • (- t) • (- t) • (- t)
Answer:
(- t) • (- t) • (- t) • (- t) • (- t ) = (-t)5

Explanation:
Here we write the given product (- t) • (- t) • (- t) • (- t) • (- t)
as (-t)5 because – t is multiplied by 5 times.

Question 16.
– (7 • 7 • 7 • 7 • 7)
Answer:
– (7 • 7 • 7 • 7 • 7) = -(7)5

Explanation:
We write the given product as – (7 • 7 • 7 • 7 • 7)
= -(7)5 here -(7) is multiplied by 5 times.

Question 17.
\(-\left(\frac{1}{4} \cdot \frac{1}{4} \cdot \frac{1}{4} \cdot \frac{1}{4}\right)\)
Answer:
\(-\left(\frac{1}{4} \cdot \frac{1}{4} \cdot \frac{1}{4} \cdot \frac{1}{4}\right)\)= – (\(\frac{1}{4}\))4

Explanation:
We write the given product as
\(-\left(\frac{1}{4} \cdot \frac{1}{4} \cdot \frac{1}{4} \cdot \frac{1}{4}\right)\)
as – (\(\frac{1}{4}\))4 because \(\frac{1}{4}\)
is multiplied by 4 times.

EVALUATING EXPRESSIONS Evaluate the expression.
Question 18.
52
Answer:
52 = 5 X 5 = 25

Explanation:
Given 52 we write the expression as 5 X 5
and the value is 25

Question 19.
– 113
Answer:
– 113 = (-11 X -11 X -11) = -1331

Explanation:
Given – 113  we write it as (-11 X -11 X -11)
we get -1331.

Question 20.
(- 1)6
Answer:
(- 1)6= (-1 X -1 X  -1 X -1 X -1 X -1) = 1

Explanation:
Given (- 1)6 as power is 6 we multiply -1 by
6 times as  (-1 X -1 X  -1 X -1 X -1 X -1) we get 1.

Question 21.
(\(\frac{1}{6}\))6
Answer:
(\(\frac{1}{6}\))6
= \(\frac{1}{6}\)  X \(\frac{1}{6}\) X
\(\frac{1}{6}\) X \(\frac{1}{6}\) X
\(\frac{1}{6}\) X \(\frac{1}{6}\) = \(\frac{1}{46656}\)

Explanation:
Given (\(\frac{1}{6}\))here power is 6 we multiply
\(\frac{1}{6}\) by 6 times as \(\frac{1}{6}\)  X \(\frac{1}{6}\) X
\(\frac{1}{6}\) X \(\frac{1}{6}\) X
\(\frac{1}{6}\) X \(\frac{1}{6}\) we get \(\frac{1}{46656}\)

Question 22.
(- \(\frac{1}{12}\))2
Answer:
(- \(\frac{1}{12}\))2 = –\(\frac{1}{12}\) X –\(\frac{1}{12}\) = \(\frac{1}{144}\)

Explanation:
Given (- \(\frac{1}{12}\))2
the expression has power 2 we write it as –\(\frac{1}{12}\) X –\(\frac{1}{12}\)
we get \(\frac{1}{144}\)
Question 23.
– (\(\frac{1}{9}\))3
Answer:
– (\(\frac{1}{9}\))3 = – \(\frac{1}{9}\) X – \(\frac{1}{9}\) X –\(\frac{1}{9}\)=
– \(\frac{1}{729}\)

Explanation:
Given – (\(\frac{1}{9}\))3 as in the expression we have power 3 we multiply
– \(\frac{1}{9}\) by three times as – \(\frac{1}{9}\) X – \(\frac{1}{9}\) X –\(\frac{1}{9}\) we get –\(\frac{1}{729}\)

Question 24.
YOU BE THE TEACHER
Your friend evaluates the power – 62. Is your friend correct? Explain your reasoning.
Big Ideas Math Answer Key Grade 8 Chapter 8 Exponents and Scientific Notation 8.1 7
Answer:
Yes friend is correct,

Explanation:
Given friend evaluates the power – 62 as -6 X -6 = 36,
as given -6 has power 2 we multiply -6 twice so we get 36,
which is similar to what friend has evaluated so friend is correct.

STRUCTURE Write the prime factorization of the number using exponents.
Question 25.
675
Answer:
675 = 33 X 52

Explanation:
The number 675 is a composite number so, it is possible to factorize it.
In other words, 675 can be divided by 1, by itself and at least by 3 and 5.
A composite number is a positive integer that has at least one positive divisor
other than one or the number itself.
In other words, a composite number is any integer greater than one that is not a prime number.
The prime factorization of 675 = 33 X 52.
The prime factors of 675 are 3 and 5.
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation-4
Question 26.
280
Answer:
280 = 23 X 5 X 7

Explanation:
The number 280 is a composite number so, it is possible to factorize it.
In other words, 280 can be divided by 1, by itself and at least by 2, 5 and 7.
A composite number is a positive integer that has at least one
positive divisor other than one or the number itself.
In other words, a composite number is any integer greater than
one that is not a prime number.
The prime factorization of 280 = 23 X 5 X 7.
The prime factors of 280 are 2, 5 and 7.
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation-5

Question 27.
363
Answer:
363 = 3 X 112

Explanation:
The number 363 is a composite number so, it is possible to factorize it.
In other words, 363 can be divided by 1, by itself and at least by 3 and 11.
A composite number is a positive integer that has at least one
positive divisor other than one or the number itself.
In other words, a composite number is any integer greater than
one that is not a prime number.
The prime factorization of 363 = 3 X 112.
The prime factors of 363 are 3 and 11.
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation-6

Question 28.
PATTERNS
The largest doll is 12 inches tall. The height of each of the other dolls is \(\frac{7}{10}\) the height of the next larger doll. Write an expression involving a power that represents the height of the smallest doll. What is the height of the smallest doll?
Big Ideas Math Answer Key Grade 8 Chapter 8 Exponents and Scientific Notation 8.1 8.1
Answer:
The height of the smallest doll is 4.116 inches.

Explanation:
Given the largest doll is 12 inches tall, The height of each of the other dolls is
\(\frac{7}{10}\) the height of the next larger doll, There are 4 dolls
So  12 X \(\frac{7}{10}\) X \(\frac{7}{10}\)  X \(\frac{7}{10}\) =
\(\frac{4116}{1000}\) = 4.116 inches, therefore the height of the smallest doll is 4.116 inches.

USING ORDER OF OPERATIONS Evaluate the expression.
Question 29.
5 + 2 • 23
Answer:
5 + 2 • 23 = 21

Explanation:
Given 5 + 2 • 23 we first simplify 2and multiply with 2
we get 2 X 2 X 2 X 2 = 16 and add 5, 16 + 5 = 21,
therefore 5 + 2 • 23 = 21

Question 30.
2 + 7 • (- 3)2
Answer:
2 + 7 • (- 3)2 = 65

Explanation:
Given expression as 2 + 7 • (- 3)we first simplify  7 • (- 3)2
we multiply – 3 twice as – 3 X -3 = 9 and multiply with 7 we get
7 X 9 = 63 now we add 2 to 63 now we get 2 + 63 = 65,
therefore 2 + 7 • (- 3)2 = 65.

Question 31.
(132 – 122) ÷ 5
Answer:
(132 – 122) ÷ 5= 5.

Explanation:
We have expression as (132 – 122) ÷ 5 first
we calculate (132 – 122) so 13 X 13 = 169 and
12 X 12 = 144 we subtract 144 from 169 we get
169 – 144 = 25 now we divide 25 by 5  we get 5,
therefore (132 – 122) ÷ 5= 5.

Question 32.
\(\frac{1}{2}\)(43 – 6 • 32)
Answer:
\(\frac{1}{2}\)(43 – 6 • 32) = 5

Explanation:
Given expression as \(\frac{1}{2}\)(43 – 6 • 32) First we evaluate
(43 – 6 • 32) as 32) as 3 X 3 = 9 now multiply by 6 we get 6 X 9 = 54,
43 = 4 x 4 X 4 = 64, So 64 – 54 =10 Now we multiply 10 with \(\frac{1}{2}\)
we get 5, therefore \(\frac{1}{2}\)(43 – 6 • 32) = 5.

Question 33.
|\(\frac{1}{2}\)(7 + 53)|
Answer:
|\(\frac{1}{2}\)(7 + 53)| = 66

Explanation:
given expression is |\(\frac{1}{2}\)(7 + 53)| so first we evaluate
(7 + 53) = 7 + 5 X 5 X 5 = 7 + 125 = 132 now we multiply 132 with \(\frac{1}{2}\)
we get \(\frac{1}{2}\) X 132 = 66. So |\(\frac{1}{2}\)(7 + 53)| = 66.

Question 34.
|(- \(\frac{1}{2}\))3 ÷ (\(\frac{1}{4}\))2|
Answer:
|(- \(\frac{1}{2}\))3 ÷ (\(\frac{1}{4}\))2| = -2

Explanation:
Given expression as |(- \(\frac{1}{2}\))3 ÷ (\(\frac{1}{4}\))2|
first we evaluate (- \(\frac{1}{2}\))3= – \(\frac{1}{2}\) X – \(\frac{1}{2}\) X
– \(\frac{1}{2}\) = – \(\frac{1}{8}\), Now (\(\frac{1}{4}\))2
= \(\frac{1}{4}\) X \(\frac{1}{4}\) = \(\frac{1}{16}\),
now we multiply – \(\frac{1}{8}\) with \(\frac{1}{16}\) = -2.

Question 35.
(92 – 15 • 2) ÷ 17
Answer:
(92 – 15 • 2) ÷ 17 = 3

Explanation:
The expression is (92 – 15 • 2) ÷ 17 we evaluate first (92 – 15 • 2) as
9 X 9 = 81 and 15 X 2 = 30 so 81 – 30 = 51 now we divide 51 by 17
we get 3 as 17 x 3 = 51 therefore (92 – 15 • 2) ÷ 17 = 3.

Question 36.
– 6 • (- 52 + 20)
Answer:
– 6 • (- 52 + 20) = 30

Explanation:
The given expression  is – 6 • (- 52 + 20) we first find
(- 52 + 20) = – 5 X -5 = -25 + 20 = -5 now we multiply -5 with -6
we get – 5 X -6 = 30, So – 6 • (- 52 + 20) = 30.

Question 37.
(- 4 + 12 – 62) ÷ 7
Answer:
(- 4 + 12 – 62) ÷ 7 = – 4

Explanation:
Given expression is (- 4 + 12 – 62) ÷ 7 we calculate  first (- 4 + 12 – 62) as
-4 +12 – (6 X 6)= -4 +12 -36 = -40 + 12 = – 28 now we divide -28 by 7 we
get -4, therefore (- 4 + 12 – 62) ÷ 7 = – 4

Question 38.
STRUCTURE
Copy and complete the table. Compare the values of 2h – 1 with the values of 2h – 1. When are the values the same?
Big Ideas Math Answer Key Grade 8 Chapter 8 Exponents and Scientific Notation 8.1 8
Answer:
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation-7
Comparing the values of 2h – 1 with the values of 2h – 1 both do not
have any same values.

Explanation:
First we calculate 2h -1 we substitute h as 1,2,3,4,5
we get If h is 1 , 21 -1= 2 – 1 = 2, if h is 2 we get 22 -1= 4 -1 = 3,
now h = 3,  23 -1 = 8 – 1 = 7, if h =4 , 24 -1= 16 – 1 = 15 and
h=5, 25 -1= 32 – 1 = 31. So for h =1,2,3,4,5 we get 2h -1 = 2,3,7,15,31 respectively
Now we substitute for h= 1,  2h-1 = 21-1= 20= 1, for h = 2 it is 22-1 = 21 = 2, now h = 3
23-1 = 22 = 2 X 2 = 4 , if h is 4 we get 24-1 = 23 = 2 X 2 X = 8 and if  h is 5
we get 25-1 = 24 = 2 X 2 X 2 X 2 = 16, So for h = 1,2,3,4,5,  we get 2h-1 = 1,
2,4,8,16 respectively. As comparing the values of 2h – 1 with the values of 2h – 1
both do not have any same values. Hence no common values.

Question 39.
MODELING REAL LIFE
Scientists use carbon-14 dating to determine the age of a sample of organic material.
a. The amount C(in grams) of carbon-14 remaining after t years of a
sample of organic material is represented by the equation C = 100(0.99988)t. Find the amount of carbon-14 remaining after 4 years.
b. What percent of the carbon-14 remains after 4 years?
Answer:
a. The amount of carbon – 14 remaining after 4 years is 99.95 grams.
b. The percent of the carbon – 14 remains after 4 years is 99.95%.

Explanation:
a. Given the amount C(in grams) of carbon-14 remaining
after t years of a sample of organic material is represented
by the equation C = 100(0.99988)t we the amount of carbon-14 remaining
after 4 years as  t = 4  we substitute C = 100(0.99988)4 we get C =
100 X 0.99988 X 0.99988 X 0.99988 X 0.99988 = 99.95 grams,
therefore The amount of carbon – 14 remaining after 4 years is 99.95 grams.
b. Now the percent of the carbon – 14 remains after 4 years , we have the amount
of carbon after 4 years is 99.95 grams,
So Percentage is 100 X 99.95 by 100 = 99.95 %.


Question 40.

DIG DEEPER!
The frequency (in vibrations per second) of a note on a piano is represented by the equation F = 440(1.0595)n, where n is the number of notes above A440. Each black or white key represents one note.
Big Ideas Math Answer Key Grade 8 Chapter 8 Exponents and Scientific Notation 8.1 9
a. How many notes do you take to travel from A440 to A?
b. What is the frequency of A?
c. Compare the frequency of A to the frequency of A440.
Answer:
a. There are 12 notes to travel from A440 to A.
b. The frequency of A is 880 vibrations.
c. The frequency of A is twice to the frequency of A440.

Explanation:
a. As each black or white key represents one note, to reach
from A 440 to A if we count there are 12 notes to travel.
b. The frequency (in vibrations per second) of a note on a piano
is represented by the equation F = 440(1.0595)n, For note A, n =12
F = 440(1.0595)12, F = 440 X 1.0595 X 1.0595 X 1.0595 X 1.0595 X
1.0595 X 1.0595 X 1.0595 X 1.0595 X 1.0595 X 1.0595 X 1.0595 X 1.0595 =
F=880.37 therefore the frequency of A is 880 vibrations.
c. The frequency of A to the frequency of A440 is 880 by 440(1.0595)0,
we get approximately 2, So the frequency of A is twice to the frequency of A440.

Lesson 8.2 Product of Powers Property

EXPLORATION 1

Finding Products of Powers
Work with a partner.
a. Copy and complete the table. Use your results to write a general rule for finding am • an, a product of two powers with the same base.
Big Ideas Math Answers 8th Grade Chapter 8 Exponents and Scientific Notation 8.2 1.1
b. Show how to use your rule in part(a) to write each expression below as a single power. Then write a general rule for finding (am)n, a power of a power.
Big Ideas Math Answers 8th Grade Chapter 8 Exponents and Scientific Notation 8.2 1
Answer:
a.Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation-8

General rule for am • an = am+n a product of two powers with the same base
then powers are added.
b.
(73)2 = 73×2 = 76,
(62)2 = 62×2 = 64,
(32)3 = 32×3 = 36,
(22 )4= 22×4 = 28,
((\(\frac{1}{2}\))2)5 = (\(\frac{1}{2}\))2 x 5 = (\(\frac{1}{2}\))10
General rule for finding (am)power of a power,
If two powers have the same base then
we can multiply the powers as (am)n = am x n .

Explanation:
a. Completed the table as shown above as Product,
Repeated Multiplication Form and Power as
(22  X 24) = 22+4  = 2 X 2 X 2 X 2 X 2 X 2 =  26
(-3)2  X (-3)4  = (-3)2+4  = -3 X -3 X -3 X -3 X -3 X -3 = (-3)6
73  X 72  = (7)3+2  = 7 X 7 X 7 X 7 X 7 = 75
5.11  X 5.16  = (5.1)1+6  = 5.1 X 5.1 X 5.1 X 5.1 X 5.1 X 5.1 X5.1 =(5.1)7
(-4)2  X (-4)2 = (-4)2+2  = -4 X -4 X -4 X -4 = (-4)4
10X 105= 103+5 = 10 X 10 X 10 X 10 X 10 X 10 X 10 X 10 = 108
(\(\frac{1}{2}\))5 X (\(\frac{1}{2}\))5=(\(\frac{1}{2}\))5+5
=\(\frac{1}{2}\) X \(\frac{1}{2}\) X \(\frac{1}{2}\) X \(\frac{1}{2}\) X \(\frac{1}{2}\) X \(\frac{1}{2}\) X \(\frac{1}{2}\) X \(\frac{1}{2}\) X \(\frac{1}{2}\) X \(\frac{1}{2}\) = (\(\frac{1}{2}\))10
General rule for am • an = am+n a product of two powers with the same base,
powers are added. If two powers have the same base
then we can multiply the powers.
When we multiply two powers we add their exponents.
b. We write (73)as  73×2 = 76,
(62)2 as  62×2 = 64,
(32)3 as 32×3 = 36,
(22 )4 as 22×4 = 28,
((\(\frac{1}{2}\))2)5 = (\(\frac{1}{2}\))2 x 5 = (\(\frac{1}{2}\))10
wrote each expression as a single power above,
General rule for finding (am)n a power of a power,
If two powers have the same base then
we can multiply the powers as (am)n = am x n .

EXPLORATION 2

Finding Powers of Products
Work with a partner. Copy and complete the table. Use your results to general rule write a for finding (ab)m, a power of a product.
Big Ideas Math Answers 8th Grade Chapter 8 Exponents and Scientific Notation 8.2 2
Answer:
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation-9

Explanation:
Completed the table as shown above as first
Repeated Multiplication Form and Product of Powers as
(2 X 3)3 = 2 X 2 X 2 X 3 X 3 X 3 = 23 X 33
(2 X 5)2 = 2 X 2 X 5 X 5 = 22 X 52
(5 X 4)3 = 5 X 5 X 5 X 4 X 4 X 4 = 53 X 43
(-2 X 4)2 = -2 X -2 X 4 X 4 = -22 X 42
(-3 X 2)4 = -3 X -3 X -3 X -3 X 2 X 2 X 2 X 2 = -34 X 24
We know general rule to write (ab)m  power of a product is am X bm

Try It
Simplify the expression. Write your answer as a power.
Question 1.
62 • 64
Answer:
62 • 64 = 66

Explanation:
Given 62 • 6we have general rule for am • an = am+n
If product of two powers with the same base then
powers are added. As both base are 6 so 62 • 64 = 62+4 = 66

Question 2.
(- \(\frac{1}{2}\))3 • (- \(\frac{1}{2}\))6
Answer:
(- \(\frac{1}{2}\))3 • (- \(\frac{1}{2}\))6 =(- \(\frac{1}{2}\))9
Given (- \(\frac{1}{2}\))3 • (- \(\frac{1}{2}\))6 we have general rule for
am • an = am+n  If product of two powers with the same base
then powers are added as both bases  are same so (- \(\frac{1}{2}\))3 • (- \(\frac{1}{2}\))6 =
(- \(\frac{1}{2}\))3+6 =(- \(\frac{1}{2}\))9

Question 3.
z • z12
Answer:
z • z12 = z13

Explanation:
Given z • z12 we have general rule for am • an = am+n
If product of two powers with the same base then
powers are added, here  both bases are z so z • z12 = z1+12= z13

Simplify the expression. Write your answer as a power.
Question 4.
(43)5
Answer:
(43)=(4)15

Explanation:
Given (43)we have general rule for finding (am)n a power of a power,
If two powers have the same base then we can multiply the
powers as (am)n = am x n . Here both base is 4 so (43)=(43x5) = (4)15

Question 5.
(y2)4
Answer:
(y2)4 =(y)8

Explanation:
Given (y2)4 we have general rule for finding (am)n a power of a power,
If two powers have the same base then we can multiply the
powers as (am)n = am x n . Here both base is y so (y2)=(y2x4) = (y)8

Question 6.
((- 4)3)2
Answer:
((- 4)3)2 = (-4)6

Explanation:
Given ((- 4)3)we use general rule for finding (am)n a power of a power,
If two powers have the same base then we can multiply the
powers as (am)n = am x n . Here both base is 4 so ((- 4)3)2 =(- 4)3x2 = (-4)6

Simplify the expression.
Question 7.
(5y)4
Answer:
(5y)4 = 54 X y4

Explanation:
Given (5y)4 to simplify the expression we use general rule to
write (ab)m  power of a product as am X bm ,So (5y)4 = 54 X y4

Question 8.
(ab)5
Answer:
(ab)5= a5 X b5

Explanation:
Given (ab)5 to simplify the expression we use general rule to
write (ab)m  power of a product as am X bm ,So (ab)5= a5 X b5

Question 9.
(0.5 mn)2
Answer:
(0.5 mn)2= 0.5X m2 X n2

Explanation:
Given (0.5 mn)2 to simplify the expression we use general rule to
write (abc)m  power of a product as am X bm X cm , So (0.5 mn)2= 0.5X m2 X n2

Self-Assessment for Concepts & Skills

Solve each exercise. Then rate your understanding of the success criteria in your journal.

FINDING POWERS Simplify the expression. Write your answer as a power.
Question 10.
47 • 44
Answer:
47 • 44= 411

Explanation:
Given 47 • 4we have general rule for am • an = am+n
If product of two powers with the same base then
powers are added. So 47 • 44 = 47+4 = 411

Question 11.
(g6)3
Answer:
(g6)3=g18

Explanation:
Given (g6)we have general rule for finding (am)n a power of a power,
If two powers have the same base then we can multiply the
powers as (am)n = am x n . So (g6)3= g6x3= g18

Question 12.
(- \(\frac{1}{3}\))5 • (- \(\frac{1}{3}\))7
Answer:
(- \(\frac{1}{3}\))5 • (- \(\frac{1}{3}\))7= (- \(\frac{1}{3}\))12

Explanation:
Given (- \(\frac{1}{3}\))5 • (- \(\frac{1}{3}\))7
we have general rule for am • an = am+n
If product of two powers with the same base then
powers are added, here  both bases are – \(\frac{1}{3}\) so
(- \(\frac{1}{3}\))5 • (- \(\frac{1}{3}\))7= (- \(\frac{1}{3}\))5+7 = (- \(\frac{1}{3}\))12

FINDING A POWER OF A PRODUCT Simplify the expression.
Question 13.
(8t)4
Answer:
(8t)4 = 84 X t4

Explanation:
Given (8t)4 we have general rule to write (ab)m
power of a product as am X bm therefore (8t)4 = 84 X t4

Question 14.
(yz)6
Answer:
(yz)6= y6 X z6

Explanation:
Given (yz)we have general rule to write (ab)m
power of a product as am X bm therefore (yz)6= y6 X z6

Question 15.
(\(\frac{1}{4}\)gh)3
Answer:
(\(\frac{1}{4}\)gh)3= (\(\frac{1}{4}\))3 X g3 X h3

Explanation:
Given (\(\frac{1}{4}\)gh)3 to simplify the expression we use general rule to
write (abc)m  power of a product as am X bm X cm , So (\(\frac{1}{4}\)gh)3= (\(\frac{1}{4}\))3 X g3 X h3

Question 16.
CRITICAL THINKING
Can you use the Product of Powers Property to simplify 52 • 64? Explain.
Answer:
No, we can not use the Product of Powers Property to simplify 52 • 64

Explanation:
Given to simplify 52 • 6as both bases are different
and Product of Powers Property is in general,
for all real numbers  we  multiply two powers having
the same base we add the exponents but here the bases are different
5,6 so no, we can not use the Product of Powers Property to simplify 52 • 64

Question 17.
OPEN-ENDED
Write an expression that simplifies to x12 using the Product of Powers Property.
Answer:
The expression is x2 X x10 simplifies to  x12

Explanation:
Given an expression that simplifies to x12 by
using the Product of Powers Property we write as x12 = x2 X x10 
as we know Product of Powers Property is for am • an = am+n
If product of two powers with the same base then
powers are added. As base is x here we have m + n  = 12,
so we can take powers (m , n) as (2,10) or (6,6) or (1,11) or (3,9) or
(4,8) or (5,7) or (7,5) or (8,4) or (9,3) or (10,2) or (11,1). So I took (2,10)
as m, n and wrote expression as  x2 X x10 = x12 .

Self-Assessment for Problem Solving

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 18.
A newborn blue whale weighs 37 kilograms. An adult blue whale weighs 81 times the weight of the newborn. How many kilograms does the adult blue whale weigh?
Big Ideas Math Answers 8th Grade Chapter 8 Exponents and Scientific Notation 8.2 3
Answer:
The adult blue whale weighs 81 X  37 kilograms = 311 kilograms

Explanation:
Given A newborn blue whale weighs 37 kilograms.
An adult blue whale weighs 81 times the weight of the newborn.
So the adult blue whale weighs 81 X  37 kilograms , we write 81 as
multiple of 3 we get 81 = 3 X 3 X 3 X 3, 81 =   34 therefore
81 X  37 kilograms =  34 X  37 kilograms,
when bases are same powers are added
34 X  37 kilograms = 34+7 = 311 kilograms,
therefore the adult blue whale weighs
81 X  37 kilograms = 311 kilograms.

Question 19.
One megabyte of cell phone storage space is 220 bytes. An app uses 44 megabytes of storage space. How many bytes of storage space does the app use?
Answer:
The app used 228 bytes of storage space.

Explanation:
Given One megabyte of cell phone storage space is 220 bytes.
and an app uses 44 megabytes of storage space.
the number of bytes of storage space does the app use is
44 X 220 bytes as 4 we can write as 2 X 2 = 22 (22)4 X 220
44 X 220 = (22)4 X 220  = 28 X 220= 28+20 = 228 bytes.

Question 19.
DIG DEEPER!
The diagram shows the area of a small circular rug. The radius of a large circular rug is 3 times the radius of the small rug. Write an expression for the area of the large rug in terms of x. Justify your answer.
Big Ideas Math Answers 8th Grade Chapter 8 Exponents and Scientific Notation 8.2 4
Answer:
The area of the large rug in terms of x is 9πx2 X \(\frac{1}{4}\)

Explanation:
We have area of small circular rug as
A= \(\frac{1}{4}\) X πx2
A = π X (\(\frac{x}{2}\))2
hence the radius of small rug is \(\frac{x}{2}\) and radius
of large rug is \(\frac{3x}{2}\), therefore the area of the
large rug is π X (\(\frac{3x}{2}\))2= 9πx2 X \(\frac{1}{4}\),
therefore the area of the large rug in terms of x is 9πx2 X \(\frac{1}{4}\).

Product of Powers Property Homework & Practice 8.2

Review & Refresh

Write the product using exponents.
Question 1.
11 • 11 • 11 • 11 • 11
Answer:
11 • 11 • 11 • 11 • 11 =  115

Explanation:
Given 11 • 11 • 11 • 11 • 11 as 11 is multiplied by 5 times
we write as 11 • 11 • 11 • 11 • 11 =  115

Question 2.
(- 6) • (- 6) • (- 6) • z • z
Answer:
(- 6) • (- 6) • (- 6) • z • z = (-6)3 X z2

Explanation:
Given (- 6) • (- 6) • (- 6) • z • z we have -6 multiplied
by 3 times and z twice so write the product as
(- 6) • (- 6) • (- 6) • z • z = (-6)3 X z2

Find the value of for the given value of x.
Question 3.
y = – 4x; x = 7
Answer:
y= -28

Explanation:
Given  y = -4x and value of x as 7,
we substitute x as y = -4 X 7, So y = -28.

Question 4.
y = 5x + 6; x = – 2
Answer:
y= -4

Explanation:
Given y= 5x + 6 and value of x as -2,
we substitute x as y= 5 X -2 + 6 = -10 + 6 = -4,
therefore y = -4.

Question 5.
y = 10 – 3x ; x = 3
Answer:
y=1

Explanation:
Given y= 10 – 3x and value of x as 3,
we substitute x as  y= 10 – 3 X 3 = 10 – 9 = 1,
therefore y =1.

Question 6.
What is the measure of each interior angle of the regular polygon?
Big Ideas Math Answers 8th Grade Chapter 8 Exponents and Scientific Notation 8.2 5
A. 45°
B. 135°
C. 1080°
D. 1440°
Answer:
The measure of each interior angle of the
regular polygon given is B.135°.

Explanation:
We know each Angle (of a Regular Polygon) =
(n−2) × 180° / n where n is number of sides
in the given figure we 8 sides so n is 8,
the measure of each interior angle is (8-2) X 180° / 8=
6 X 180° / 8 = 135° . So the measure of each
interior angle of the regular polygon given is B.135°.

Concepts, Skills, &Problem Solving

FINDING PRODUCTS OF POWERS Write the expression in repeated multiplication form. Then write the expression as a power. (See Exploration 1, p. 325.)
Question 7.
56 • 53
Answer:
56 • 53 = 5 X 5 X 5 X 5 X 5 X 5 X 5 X 5 X 5 = 59

Explanation:
Given 56 • 53 the expression in repeated multiplication form is
5 X 5 X 5 X 5 X 5 X 5 X 5 X 5 X 5 and expression as power
is 56 • 53= 56+3 = 59

Question 8.
(64)2
Answer:
(64)2 = 6 X 6 X 6 X 6 X 6 X 6 X 6X 6 = 68

Explanation:
Given (64)2 the expression in repeated multiplication form is
6 X 6 X 6 X 6 X 6 X 6 X 6 X 6  and expression as power
is  (64)= 64×2 = 68

Question 9.
(- 8)3 • (- 8)4
Answer:
(- 8)3 • (- 8)4= (- 8)7

Explanation:
Given (- 8)3 • (- 8)4 the expression in repeated multiplication form is
-8 X -8 X -8 X -8 X -8 X -8 X -8 and expression as power
is  (- 8)3 • (- 8)4= (- 8)3+4 = (- 8)7

FINDING POWERS Simplify the expression. Write your answer as a power.
Question 10.
32 • 32
Answer:
32 • 32 = 34

Explanation:
We write the given 32 • 3expression as a power,
so 32 • 32 as bases are same 3 powers are added 32+2 = 34

Question 11.
810 • 84
Answer:
810 • 84 = 814

Explanation:
We write the given 810 • 84 expression as a power,
here 810 • 84  has same bases 8 so same powers are
added as 810+4 = 814

Question 12.
(54)3
Answer:
(54)3 = (5)12

Explanation:
We write the given expression (54)3 as a power,
so (54)3 has powers of powers therefore powers
are multiplied as (5)4 x 3 = 512

Question 13.
((- 3)2)4
Answer:
((- 3)2)4 = (- 3)8

Explanation:
We write the given expression ((- 3)2)4 as a power,
so ((- 32) has powers of powers therefore powers
are multiplied as (-3)2 x 4 = (-3)8

Question 14.
(- 4)5 • (- 4)7
Answer:
(- 4)5 • (- 4)7= (- 4)12

Explanation:
We write the given expression (- 4)5 • (- 4)7 as a power,
here (- 4)5 • (- 4)7 has same bases -4 so same powers are
added as  (- 4)5+7 = (- 4)12.

Question 15.
h6 • h
Answer:
h6 • h = h7

Explanation:
We write the given expression h6 • h as a power,
here h6 • h has same bases h so same powers are
added as  (h)6+1 = (h)7.

Question 16.
(b12)3
Answer:
(b12)3 = (b)36

Explanation:
We write the given expression (b12)3 as a power,
so (b12)3  has powers of powers therefore powers
are multiplied as (b)12 x 3 = (b)36

Question 17.
(\(\frac{2}{3}\))2 • (\(\frac{2}{3}\))6
Answer:
(\(\frac{2}{3}\))2 • (\(\frac{2}{3}\))6= (\(\frac{2}{3}\))8

Explanation:
We write the given expression (\(\frac{2}{3}\))2 • (\(\frac{2}{3}\))6
as a power, here (\(\frac{2}{3}\))2 • (\(\frac{2}{3}\))6
has same bases so same powers are added as (\(\frac{2}{3}\))2+6
= (\(\frac{2}{3}\))8 

Question 18.
(3.83)4
Answer:
(3.83)4= (3.8)12

Explanation:
We write the given expression (3.83)4 as a power,
so (3.83)4  has powers of powers therefore powers
are multiplied as (3.8)3 x 4 = (3.8)12

Question 19.
(n3)5
Answer:
(n3)5 = (n)15

Explanation:
We write the given expression (n3)5 as a power,
so (n3)5 has powers of powers therefore powers
are multiplied as (n)3 X 5= (n)15

Question 20.
((- \(\frac{3}{4}\))5)2
Answer:
((- \(\frac{3}{4}\))5)2= (- \(\frac{3}{4}\))10

Explanation:
We write the given expression ((- \(\frac{3}{4}\))5)2 as a power,
so ((- \(\frac{3}{4}\))5)2 has powers of powers therefore powers
are multiplied as (- \(\frac{3}{4}\))5 x 2= (- \(\frac{3}{4}\))10

Question 21.
(- \(\frac{1}{2}\))8 • (- \(\frac{1}{2}\))9
Answer:
(- \(\frac{1}{2}\))8 • (- \(\frac{1}{2}\))9= (- \(\frac{1}{2}\))17

Explanation:
We write the given expression (-\(\frac{1}{2}\))8 • (- \(\frac{1}{2}\))9
as a power, here (-\(\frac{1}{2}\))8 • (- \(\frac{1}{2}\))9
has same bases so same powers are added as (-\(\frac{1}{2}\))8+9
= (-\(\frac{1}{2}\))17 

YOU BE THE TEACHER Your friend simplifies the expression.
Is your friend correct? Explain your reasoning.
Question 22.
Big Ideas Math Answers 8th Grade Chapter 8 Exponents and Scientific Notation 8.2 6
Answer:
No, Friend is in correct as 52 X 59 = 511 not 2511

Explanation:
Given expression is 52 X 52
here as base is same 5 we add powers
as 52+9  we get 511 not 2511

Question 23.
Big Ideas Math Answers 8th Grade Chapter 8 Exponents and Scientific Notation 8.2 7
Answer:
No, friend is In correct as (r6)4= (r)6 x 4 = (r)24 not r10

Explanation:
Given expression is (r6)4 here we have
powers of powers therefore powers
are multiplied not added so (r6)4= (r)6 x 4 = (r)24
not r10 therefore friend is incorrect.

FINDING A POWER OF A PRODUCT
Simplify the expression.
Question 24.
(6g)3
Answer:
(6g)3 = 216 g3

Explanation:
Given (6g)3 we simplify as 63 X g3= 6 X 6 X 6 X g X g X g
= 216 g3 here we multiplied 6 by 3 times and g by
3 times as both has power 3.

Question 25.
(- 3v)5
Answer:
(- 3v)5 = -243v5

Explanation:
Given (-3v)5 we simplify as
– 3 X -3 X -3 X -3 X -3 X v X v X v X v X v = -243v5,
here we multiplied -3 and v by
5 times as power for both is 5.

Question 26.
(\(\frac{1}{5}\)k)2
Answer:
(\(\frac{1}{5}\)k)2 = \(\frac{1}{25}\)k2

Explanation:
Given expression (\(\frac{1}{5}\)k)2 we simplify as
\(\frac{1}{5}\) X \(\frac{1}{5}\) X k X k = \(\frac{1}{25}\)k2
here we multiply \(\frac{1}{5}\)  and k
by 2 times as power for both is 2.

Question 27.
(1.2 m)4
Answer:
(1.2 m)4 = 2.0736m4

Explanation:
Given expression as (1.2 m)4 we simplify as
1.2 X 1.2 X 1.2 X m X m X m X m is 2.0736m4
here we multiply 1.2 and m by 4 times as
power for both is 4.

Question 28.
(rt)12
Answer:
(rt)12 =  r12 X t12

Explanation:
Given expression as (rt)12 we simplify as
r X r X r X r X r X r X r X r X r X r X r X r X
t X t X t X t X t X t X t X t X t X t X t X t is r12 X t12
here we multiply r and t by 12 times as
power for both is 12.

Question 29.
(- \(\frac{3}{4}\)p)3
Answer:
(- \(\frac{3}{4}\)p)3 = –\(\frac{27}{64}\)p3

Explanation:
Given expression as (- \(\frac{3}{4}\)p)3 we simplify as
– \(\frac{3}{4}\) X – \(\frac{3}{4}\) X – \(\frac{3}{4}\) X p X p X p is
–\(\frac{27}{64}\)p3
here we multiply –\(\frac{3}{4}\) and p by 3 times as
power for both is 3.

Question 30.
PRECISION
Is 32 + 33 equal to 35? Explain.
Answer:
No,32 + 33 is not equal to 35

Explanation:
Given expression is 32 + 33 first we solve
32 as 3 X 3 = 9 and 33 = 3 X 3 X 3 = 27 so
32 + 33= 9 + 27 = 36 now we have 35 which is
equal to 3 X 3 X 3 X 3 X 3 = 243 as 36 ≠ 243 so
no, 32 + 3≠ 35

Question 31.
PROBLEM SOLVING
A display case for the artifact shown is in the shape of a cube. Each side of the display case is three times longer than the width w of the artifact.
Big Ideas Math Answers 8th Grade Chapter 8 Exponents and Scientific Notation 8.2 8
a. Write a power that represents the volume of the case.
b. Simplify your expression in part(a).
Answer:
a. The volume of the case is (3w)3.
b. Simplified form is 27w3.

Explanation:
Given a display case for the artifact shown
is in the shape of a cube. Each side of the display
case is three times longer than the width w of the artifact.
a. we have volume of cube as (edge)3
So the volume of the case = 3w X 3w X 3w = (3w)3.

b. Simplified form of (3w)3= 3 X 3 X 3 X w X w X w = 27w3.

Question 32.
LOGIC
Show that (3 • 8 • x)7 = 67 • 47 • x7.
Answer:
(3 • 8 • x)7 = ( 37 X 27) X 47 X x7 = 67 x 47 x x7

Explanation:
Given expression is (3 • 8 • x)7 we write 8 as
multiple of 2 X 4 so (3 X 2 X 4 X x)7
and now we multiply first 3 and 2 as (6 X 4 X x)7
we simplify we get 67 x 47 x x7
So (3 • 8 • x)7 = 67 • 47 • x7.

Question 33.
MODELING REAL LIFE
The lowest altitude of an altocumulus cloud is about 38 feet. The highest altitude of an altocumulus cloud is about 3 times the lowest altitude. What is the highest altitude of an altocumulus cloud? Write your answer as a power.
Answer:
The highest altitude of an altocumulus cloud is
3 X 38= 39 feet.

Explanation:
Given The lowest altitude of an altocumulus cloud is
about 38 feet. The highest altitude of an altocumulus cloud is
about 3 times the lowest altitude. So the highest altitude
of an altocumulus cloud is 3 X 38   as bases are same
powers are added  31+8 =  39 feet.

Question 34.
GEOMETRY
A square pyramid has a height h and a base with side lengths. The side lengths of the base increase by 50%. Write a formula for the volume of the new pyramid in terms of s and h.
Answer:
The formula for the volume of the
new pyramid is terms of s and h is 3s2h by 4.

Explanation:
Given a square pyramid has a height h and
a base with side lengths. The side lengths
of the base increase by 50%. So volume is
s2\(\frac{h}{3}\) and side length s
increases by 50%, so s + 50%s = s + \(\frac{50}{100}\)s
= s + \(\frac{1}{2}\)s = \(\frac{3s}{2}\).
The new volume is (\(\frac{3s}{2}\))2 X \(\frac{h}{3}\)
= 9s2 X \(\frac{h}{12}\) = 3s2h by 4
therefore the formula for the volume of the
new pyramid is terms of s and h is 3s2h by 4.

Question 35.
MODELING REAL LIFE
The United States Postal Service delivers about 24 • 3 • 53 pieces of mail each second. There are 28 • 34 • 52 seconds in 6 days. How many pieces of mail does the United States Postal Service deliver in 6 days? Write your answer as an expression involving three powers.
Big Ideas Math Answers 8th Grade Chapter 8 Exponents and Scientific Notation 8.2 9
Answer:
Number of pieces of mails does the United States Postal Service
deliver in 6 days is 212 X 35 X 55 .

Explanation:
Given the United States Postal Service delivers about
24 • 3 • 53 pieces of mail each second. There are 28 • 34 • 52
seconds in 6 days. so number of  pieces of mail does
the United States Postal Service deliver in 6 days is
24 X 3 X 53 X  28 X 34 X 5we add the same bases power we get
24 X  28 X 3 X 34 X 53 X 52 as 24+8 X 31+4 X 53+2
= 212 X 35 X 55 therefore number of pieces of mails
does the United States Postal Service
deliver in 6 days is 212 X 35 X 55 .

Question 36.
REASONING
The row numbers y and column numbers x of a chessboard are shown. Each position on the chessboard has a stack of pennies. (Only the first row is shown.) The number of pennies in each stack is 2x • 2y.
Big Ideas Math Answers 8th Grade Chapter 8 Exponents and Scientific Notation 8.2 10
a. Which locations have 32 pennies in their stacks?
b. How much money (in dollars) is in the location with the tallest stack?
c. A penny is about 0.06 inch thick. About how tall is the tallest stack?
Answer:
a. The locations are the combinations of (x,y)=
(1,4),(2,3),(3,2),(4,1).
b. $655.36 is the money in the location with the tallest stack.
c. The tallest stack is 3932.16 inches.

Explanation:
The row numbers y and column numbers x,
The number of pennies in each stack is 2x • 2y
a. The locations that have 32 pennies in their stacks is
2x • 2y = 32 we write 32 as power of 2 we get
2x • 2y = 25 we write 2x • 2y (bases are same powers are added)
as 2x+y = 25 therefore x + y = 5, So the locations
are the combinations of (x, y)= (1,4),(2,3),(3,2),(4,1).
b. The tallest stack will be in the location of (8,8)
so the maximum combination is 28 X 28 = 28+8 = 216
= 65536 pennies. Now converting pennies into dollar
1 penny = 0.01 dollar so 65536 X 0.01 = $655.36,
therefore $655.36 is the money in the location
with the tallest stack.
c. Given each penny is about 0.06 inch thick
the tallest stack is 65536 X 0.06 = 3932.16 inches.

Question 37.
CRITICAL THINKING
Find the value of x in the equation without evaluating the power.
Big Ideas Math Answers 8th Grade Chapter 8 Exponents and Scientific Notation 8.2 11
Answer:
a. x = 3
b. x = 4

Explanation:
a. Given 25 X 2x = 256, we write 256 as 28,
so 25 X 2x = 25 + x = 28 , 5 + x = 8 so x = 8 – 5,
therefore x = 3.
b. Given (\(\frac{1}{3}\))2 X (\(\frac{1}{3}\))x = \(\frac{1}{729}\)
\(\frac{1}{729}\) as multiple of \(\frac{1}{3}\) we get
\(\frac{1}{729}\) as (\(\frac{1}{3}\)))6  so (\(\frac{1}{3}\))2 X (\(\frac{1}{3}\))x = (\(\frac{1}{3}\)))6  as bases are same
we equate powers as 2 + x = 6 therefore x = 6 – 2 = 4, So X = 4.

Lesson 8.3 Quotient of Powers Property

EXPLORATION 1

Finding Quotients of Powers
Work with a partner.
a. Copy and complete the table. Use your results to write a general rule for finding \(\frac{a^{m}}{a^{n}}\), a quotient of two powers with the same base.
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation 8.3 1
b. Use your rule in part(a) to simplify the quotients in the first column of the table above. Does your rule give the results in the third column?
Answer:
a. Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation-10

The general rule for finding \(\frac{a^{m}}{a^{n}}\), a quotient of two powers with the same base is am-n
b. Yes my rule gives the result in third column.

Explanation:
a. Completed the given table as shown above.
We use the general rule for finding \(\frac{a^{m}}{a^{n}}\),
a quotient of two powers with the same base is am-n .
b. We use part(a) rule to simplify the quotients in
the first column of the table above
and rule gives the results in the third column as
\(\frac{2^{4}}{2^{2}}\) =24-2 = 22
\(\frac{-4^{5}}{-4^{2}}\) = (-4)5-2  = (-4)3
\(\frac{7^{7}}{7^{3}}\) = 77-3 = 74
\(\frac{8.5^{9}}{8.5^{6}}\) = (8.5)9-6 = (8.5)3
\(\frac{10^{8}}{10^{5}}\) = 108-5 = 103
\(\frac{3^{12}}{3^{4}}\) = 312-4 = 38
\(\frac{-5^{7}}{-5^{5}}\) = (-5)7-2 =(-5)2
\(\frac{11^{4}}{11^{1}}\) = 114-1 =113
\(\frac{x^{6}}{x^{2}}\) = x6 – 2= x4

Try It

Simplify the expression. Write your answer as a power.
Question 1.
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation 8.3 2
Answer:
\(\frac{9^{7}}{9^{4}}\) = 97-4 = 93

Explanation:
Given \(\frac{9^{7}}{9^{4}}\) we use rule for finding \(\frac{a^{m}}{a^{n}}\),
a quotient of two powers with the same base as am-n
so \(\frac{9^{7}}{9^{4}}\) = 97-4 = 93

Question 2.
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation 8.3 3
Answer:
\(\frac{4.2^{6}}{4.2^{5}}\) = 4.26-5 = 4.21 = 4.2

Explanation:
Given \(\frac{4.2^{6}}{4.2^{5}}\) we use rule for finding \(\frac{a^{m}}{a^{n}}\), a quotient of two powers with the same base as am-n
so \(\frac{4.2^{6}}{4.2^{5}}\) = 4.26-5 = 4.21 = 4.2

Question 3.
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation 8.3 4
Answer:
\(\frac{-8^{8}}{-8^{4}}\) = (-8)8-4 = (-8)

Explanation:
Given \(\frac{-8^{8}}{-8^{4}}\) we use rule for finding \(\frac{a^{m}}{a^{n}}\), a quotient of two powers with the same base as am-n
so \(\frac{-8^{8}}{-8^{4}}\) = (-8)8-4 = (-8)

Question 4.
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation 8.3 5
Answer:
\(\frac{x^{8}}{x^{3}}\) = (x)8-3 = (x)

Explanation:
Given \(\frac{x^{8}}{x^{3}}\) we use rule for finding \(\frac{a^{m}}{a^{n}}\),
a quotient of two powers with the same base as am-n
so \(\frac{x^{8}}{x^{3}}\) = (x)8-3 = (x)

Simplify the expression. Write your answer as a power.
Question 5.
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation 8.3 6
Answer:
\(\frac{6^{7}}{6^{5}}\) X 63 = 65

Explanation:
Given \(\frac{6^{7}}{6^{5}}\) X 6we use rule for finding \(\frac{a^{m}}{a^{n}}\), a quotient of two powers with the same base as am-n
first we solve \(\frac{6^{7}}{6^{5}}\) = = 67-5 = 62
and now we multiply by = 62 X  63 as bases are same 6 now
we add powers as = 62+3 = 6
therefore \(\frac{6^{7}}{6^{5}}\) X 63 = 65

Question 6.
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation 8.3 7
Answer:
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation 8.3 7= 27

Explanation:
First we multiply denominators as bases are same
we add powers so 23  X 25 = 23+5 = 28
now we solve \(\frac{2^{15}}{2^{8}}\) we use rule for finding
\(\frac{a^{m}}{a^{n}}\), a quotient of two powers with the same base as am-n
\(\frac{2^{15}}{2^{8}}\) = 215-8 = 27
therefore Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation 8.3 7= 27

Question 7.
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation 8.3 8
Answer:
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation 8.3 8= m9

Explanation:
First we multiply numerators as bases are same
we add powers so m8  X m6 = m8+6 = m14
now we solve \(\frac{m^{14}}{m^{5}}\) we use rule for finding
\(\frac{a^{m}}{a^{n}}\),  a quotient of two powers with the same base as am-n
\(\frac{m^{14}}{m^{5}}\)  = m14-5 = m9

Simplify the expression. Write your answer as a power.
Question 8.
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation 8.3 9
Answer:
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation 8.3 9= (-5)6

Explanation:
First we calculate separately values of numerators
and denominators then divide , we have numerator (-5)7 X (-5)6
we have same bases as -5 so we add powers as (-5)7+6 = (-5)13
we have denominator (-5)5 X (-5)2
we have same base as -5 so we add powers as (-5)5+2 = (-5)7
Now we have \(\frac{-5^{13}}{-5^{7}}\)  so we use rule
for finding \(\frac{a^{m}}{a^{n}}\), a quotient of
two powers with the same base as am-n
\(\frac{-5^{13}}{-5^{7}}\)  = (-5)13-7 = (-5)6

Question 9.
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation 8.3 10
Answer:
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation 8.3 10= d5

Explanation:
First we calculate separately values of numerators
and denominators then divide , we have numerator (d)5 X (d)9
we have same bases as d so we add powers as (d)5+9 = (d)14
we have denominator (d) X (d)8
we have same base as d so we add powers as (d)1+8 = (d)9
Now we have \(\frac{d^{14}}{d^{9}}\)  so we use rule
for finding \(\frac{a^{m}}{a^{n}}\), a quotient of
two powers with the same base as am-n
\(\frac{d^{14}}{d^{9}}\)  = (d)14-9 = d5

Question 10.
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation 8.3 11
Answer:
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation 8.3 11= p10

Explanation:
First we calculate separately values of numerators
and denominators then divide , we have numerator (p)3 X (p)6 X (p)4
we have same bases as p so we add powers as (p)3+6+4 = (d)13
we have denominator (p)2 X (p)
we have same bases as p so we add powers as (p)2+1 = (p)3
Now we have \(\frac{p^{13}}{p^{3}}\)  so we use rule
for finding \(\frac{a^{m}}{a^{n}}\) a quotient of
two powers with the same base as am-n
\(\frac{p^{13}}{p^{3}}\)  = (p)13-3 = p10

Self-Assessment for Concepts & Skills
Solve each exercise. Then rate your understanding of the success criteria in your journal.

SIMPLIFYING EXPRESSIONS Simplify the expression. Write your answer as a power.
Question 11.
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation 8.3 12
Answer:
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation 8.3 12= (-3)

Explanation:
Given \(\frac{-3^{9}}{-3^{2}}\) we use rule for
finding \(\frac{a^{m}}{a^{n}}\), a quotient of two
powers with the same base as am-n
so \(\frac{-3^{9}}{-3^{2}}\) = (-3)9-2 = (-3)

Question 12.
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation 8.3 13
Answer:
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation 8.3 13= (8)3
Explanation:
First we calculate separately values of numerators
and denominators then divide , we have numerator (8)6 X (8)2
we have same bases as d so we add powers as (8)6+2 = (8)8
\(\frac{8^{8}}{8^{5}}\) we use rule for
finding \(\frac{a^{m}}{a^{n}}\), a quotient of two
powers with the same base as am-n
so \(\frac{8^{8}}{8^{5}}\) = (8)8-5 = (8)3

Question 13.
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation 8.3 14
Answer:
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation 8.3 14= x

Explanation:
First we multiply denominators as bases are same
we add powers so x4  X x6 = x4+6 = x10
now we solve \(\frac{x^{11}}{x^{10}}\) we use rule for finding
\(\frac{a^{m}}{a^{n}}\), a quotient of two powers with the same base as am-n
\(\frac{x^{11}}{x^{10}}\) = x11-10 = x

Question 14.
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation 8.3 15
Answer:
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation 8.3 15= 56
Explanation:
First we calculate separately values of numerators
and denominators then divide , we have numerator (5)6 X (5)3 
we have same bases as 5 so we add powers as (5)6+3 = (5)9
we have denominator 5 X (5)
we have same bases as 5 so we add powers as (5)1+2 = (5)3
Now we have \(\frac{5^{9}}{5^{3}}\)  so we use rule
for finding \(\frac{a^{m}}{a^{n}}\) a quotient of
two powers with the same base as am-n
\(\frac{5^{9}}{5^{3}}\)  = (5)9-3 = 56

Question 15.
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation 8.3 16
Answer:
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation 8.3 16= (-2)5

Explanation:
First we calculate separately values of numerators
and denominators then divide , we have numerator (-2)9 X (-2)4
we have same bases as -2 so we add powers as (-2)9+4 = (-2)13
we have denominator (-2)4 X (-2)4
we have same base as -2 so we add powers as (-2)4+4 = (-2)8
Now we have \(\frac{-2^{13}}{-2^{8}}\)  so we use rule
for finding \(\frac{a^{m}}{a^{n}}\), a quotient of two
powers with the same base as am-n
\(\frac{-2^{13}}{-2^{8}}\)  = (-2)13-8 = (-2)5

Question 16.
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation 8.3 17
Answer:
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation 8.3 17 = b13

Explanation:
First we calculate separately values of numerators
and denominators then divide, we have numerator (b)10 X (b)3 X (b)5
we have same bases as b so we add powers as (b)10+3+5 = (b)18
we have denominator (b)2 X (b)3
we have same base as b so we add powers as (b)2+3 = (b)5
Now we have \(\frac{b^{18}}{b^{5}}\)  so we use rule
for finding \(\frac{a^{m}}{a^{n}}\) a quotient of
two powers with the same base as am-n
\(\frac{b^{18}}{b^{5}}\)  = (b)18-5 = b13

Question 17.
WHICH ONE DOESN’T BELONG?
Which quotient does not belong with the other three? Explain your reasoning.
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation 8.3 18
Answer:
\(\frac{-4^{8}}{-3^{4}}\)  does not belongs with the other three.

Explanation:
Given \(\frac{-10^{7}}{-10^{2}}\)  have same base -10,
\(\frac{6^{3}}{6^{2}}\)  have same base 6,
\(\frac{5^{6}}{5^{3}}\) have same base 5 but
\(\frac{-4^{8}}{-3^{4}}\) has different bases -4 and -3,
So, \(\frac{-4^{8}}{-3^{4}}\) does not belongs with the other three.

Self-Assessment for Problem Solving
Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 18.
You want to purchase a cat tracker. Tracker A detects your cat within a radius of 4 • 102 feet of your home. Tracker B detects your cat within a radius of 104 feet of your home. Which tracker has a greater radius? How many times greater?
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation 8.3 19.1
Answer:
Tracker B has greater radius than Tracker A and
Tracker B is greater by radius of 9,600 feet.

Explanation:
Given Tracker A detects your cat within a radius of 4 • 102 feet of your home.
Tracker B detects your cat within a radius of 104 feet of your home,
Tracker A = 4 X 102 = 4 X 100 = 400 feet
Tracker B = 104 feet =  10 X 10 X 10 X 10 = 10,000 feet as
comparing Tracker A and Tracker B,Tracker B is greater than Tracker A,
by 10,000 – 400 = 9,600 feet, therfore Tracker B
has greater radius than Tracker A and
Tracker B is greater by radius of 9,600 feet.

Question 19.
DIG DEEPER!
An earthquake of magnitude 3.0 is 102 times stronger than an earthquake of magnitude 1.0. An earthquake of magnitude 8.0 is 107 times stronger than an earthquake of magnitude 1.0. How many times stronger is an earthquake of magnitude 8.0 than an earthquake of magnitude 3.0?
Answer:
105 times stronger is an earthquake of magnitude 8.0
more than an earthquake of magnitude 3.0.

Explanation:
Given an earthquake of magnitude 3.0 is 102 times
stronger than an earthquake of magnitude 1.0 and
an earthquake of magnitude 8.0 is 107 times stronger
than an earthquake of magnitude 1.0. So how many times
stronger is an earthquake of magnitude 8.0 than an
earthquake of magnitude 3.0 is \(\frac{10^{7}}{10^{2}}\) ,
so we use rule for finding \(\frac{a^{m}}{a^{n}}\) a quotient of
two powers with the same base as am-n
as \(\frac{10^{7}}{10^{2}}\) = 107-2 = 105
therefore 105 times stronger is an earthquake of magnitude 8.0
more than an earthquake of magnitude 3.0.

Question 20.
The edge length of a cube-shaped crate is the square of the edge length of a cube-shaped box. Write an expression for the number of boxes that can fit in the crate. Justify your answer.
Answer:
The number of boxes that fit in the crate is x3

Explanation:
Let the edge be x, So the volume of box is V = x3
Given the edge length of a cube-shaped crate is the square of the
edge length of a cube-shaped box so
the volume crate with side equal to the side of the box x2 = ( x2)
we use general rule for finding (am)n a power of a power,
If two powers have the same base then
we can multiply the powers as (am)n = am x n so ( x2)3 = x2 x 3=( x)6
The number of boxes that fit in crate is
\(\frac{x^{6}}{x^{3}}\) now we use
rule for finding \(\frac{a^{m}}{a^{n}}\) a quotient of
two powers with the same base as am-n
therefore \(\frac{x^{6}}{x^{3}}\)  = x6-3 = x3
so the number of boxes that fit in the crate is x3

Quotient of Powers Property Homework & Practice 8.3

Review & Refresh

Simplify the expression. Write your answer as a power.
Question 1.
42 • 43
Answer:
42 • 43 = 45

Explanation:
Given expression as 42 • 43
we use general rule for am • an = am+n a product of
two powers with the same base then powers are added.
So 42 • 43 = 42+3 = 45

Question 2.
(a5)5
Answer:
(a5)5 = a25

Explanation:
Given expression is (a5)5 we have general rule for
finding (am)n a power of a power, If two powers have the
same base then we can multiply the
powers as (am)n = am x n . So (a5)5 = a5X5 = a25 .

Question 3.
(xy)7
Answer:
(xy)7= x7 X y7 = x7 y7

Explanation:
Given expression as (xy)7as both as same
power 7 we write as x7 X y7 = x7 y7

The red figure is similar to the blue figure. Describe a similarity transformation between the figures.
Question 4.
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation 8.3 19
Answer:
Similarity between two figures is
dilate the red figure using a scale factor of 2:3
and then reflect the image in the x- axis.

Explanation:
By comparing the side lengths, we can see that
the blue figure is 2:3 the size of red figure,
Similarity between two figures is dilate the red figure
using a scale factor of 2:3 and then reflect the image in the x- axis.

Question 5.
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation 8.3 20
Answer:
Similarity between two figures is
dilate the red figure using a scale factor of 1/2
and then reflect the image in the x- axis.

Explanation:
By comparing the side lengths, we can see that
the blue figure is one-half the size of red figure,
Similarity between two figures is dilate the red figure
using a scale factor of 1/2 and then reflect the image in the x- axis.

Concepts, Skills, & Problem Solving

FINDING QUOTIENTS OF POWERS Write the quotient as repeated multiplication. Then write the quotient as a power.(See Exploration 1, p. 331.)
Question 6.
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation 8.3 21
Answer:
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation 8.3 21= \(\frac{7 X 7 X 7 X 7 X 7 X 7 X 7 X 7 X 7}{7 X 7 X 7 X 7 X 7 X 7}\) = 73

Explanation:
Given \(\frac{7^{9}}{7^{6}}\) the repeated multiplication is
\(\frac{7 X 7 X 7 X 7 X 7 X 7 X 7 X 7 X 7}{7 X 7 X 7 X 7 X 7 X 7}\) and
the quotient as a power we use rule
for finding \(\frac{a^{m}}{a^{n}}\) a quotient of
two powers with the same base as am-n,so \(\frac{7^{9}}{7^{6}}\) is 79-6 = 73

Question 7.
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation 8.3 22
Answer:
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation 8.3 22= \(\frac{-4.5 X -4.5 X -4.5 X -4.5 X -4.5 X -4.5}{-4.5 X -4.5}\) = (-4.5)4.
Explanation:
Given \(\frac{-4.5^{6}}{-4.5^{2}}\) the repeated multiplication is
\(\frac{-4.5 X -4.5 X -4.5 X -4.5 X -4.5 X -4.5}{-4.5 X -4.5}\) and
the quotient as a power we use rule
for finding \(\frac{a^{m}}{a^{n}}\) a quotient of
two powers with the same base as am-n ,so \(\frac{-4.5^{6}}{-4.5^{2}}\)
is (-4.5)6-2 = (-4.5)4.

Question 8.
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation 8.3 23
Answer:
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation 8.3 23= \(\frac{m X m X m X m X m X m X m X m X m X m}{m X m X m X m X m }\) = m5

Explanation:
Given \(\frac{m^{10}}{m^{5}}\) the repeated multiplication is
and \(\frac{m X m X m X m X m X m X m X m X m X m}{m X m X m X m X m }\)
the quotient as a power we use rule
for finding \(\frac{a^{m}}{a^{n}}\) a quotient of
two powers with the same base as am-n so \(\frac{m^{10}}{m^{5}}\)
is (m)10-5  = m5.

DIVIDING POWERS WITH THE SAME BASE Simplify the expression. Write your answer as a power.
Question 9.
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation 8.3 24
Answer:
\(\frac{6^{10}}{6^{4}}\) = 66.

Explanation:
As given expression is \(\frac{6^{10}}{6^{4}}\) we use
rule for finding \(\frac{a^{m}}{a^{n}}\) a quotient of
two powers with the same base as am-n so \(\frac{6^{10}}{6^{4}}\) =
(6)10-4 = 66.

Question 10.
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation 8.3 25
Answer:
\(\frac{8^{9}}{8^{7}}\) = 82.

Explanation:
As given expression is \(\frac{8^{9}}{8^{7}}\) we use
rule for finding \(\frac{a^{m}}{a^{n}}\) a quotient of
two powers with the same base as am-n so \(\frac{8^{9}}{8^{7}}\) =
(8)9-7 = 82.

Question 11.
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation 8.3 26
Answer:
\(\frac{-3^{4}}{-3^{1}}\) = -33.

Explanation:
As given expression is \(\frac{-3^{4}}{-3^{1}}\) we use
rule for finding \(\frac{a^{m}}{a^{n}}\) a quotient of
two powers with the same base as am-n so \(\frac{-3^{4}}{-3^{1}}\) =
(-3)4-1 = -33.

Question 12.
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation 8.3 27
Answer:
\(\frac{4.5^{5}}{4.5^{3}}\) = 4.52.

Explanation:
As given expression is \(\frac{4.5^{5}}{4.5^{3}}\) we use
rule for finding \(\frac{a^{m}}{a^{n}}\) a quotient of
two powers with the same base as am-n so \(\frac{4.5^{5}}{4.5^{3}}\) =
(4.5)5-3 = 4.52.

Question 13.
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation 8.3 28
Answer:
\(\frac{64^{4}}{64^{3}}\) = 64.

Explanation:
As given expression is \(\frac{64^{4}}{64^{3}}\) we use
rule for finding \(\frac{a^{m}}{a^{n}}\) a quotient of
two powers with the same base as am-n so \(\frac{64^{4}}{64^{3}}\) =
(64)4-3 = 64.

Question 14.
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation 8.3 29
Answer:
\(\frac{-17^{5}}{-17^{2}}\) = (-17)3.

Explanation:
As given expression is \(\frac{-17^{5}}{-17^{2}}\) we use
rule for finding \(\frac{a^{m}}{a^{n}}\) a quotient of
two powers with the same base as am-n so \(\frac{-17^{5}}{-17^{2}}\) =
(-17)5-2 = (-17)3.

Question 15.
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation 8.3 30
Answer:
\(\frac{-6.4^{8}}{-6.4^{6}}\) = (-6.4)2.

Explanation:
As given expression is \(\frac{-6.4^{8}}{-6.4^{6}}\) we use
rule for finding \(\frac{a^{m}}{a^{n}}\) a quotient of
two powers with the same base as am-n so \(\frac{-6.4^{8}}{-6.4^{6}}\) =
(-6.4)8-6 = (-6.4)2.

Question 16.
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation 8.3 31
Answer:
\(\frac{π^{11}}{π^{7}}\) = π4.
Explanation:
As given expression is \(\frac{π^{11}}{π^{7}}\) we use
rule for finding \(\frac{a^{m}}{a^{n}}\) a quotient of
two powers with the same base as am-n so \(\frac{π^{11}}{π^{7}}\)=
(π)11-7 = π4.

Question 17.
YOU BE THE TEACHER
Your friend simplifies the quotient. Is your friend correct? Explain your reasoning.
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation 8.3 32
Answer:
No, friend is incorrect as \(\frac{6^{15}}{6^{5}}\) = 610 ≠ 63

Explanation:
Given \(\frac{6^{15}}{6^{5}}\) as per rule finding
\(\frac{a^{m}}{a^{n}}\) a quotient of
two powers with the same base as am-n so \(\frac{6^{15}}{6^{5}}\) = 615-5
= 610 but friend says \(\frac{6^{15}}{6^{5}}\) = 615/5= 63
which is incorrect therefore \(\frac{6^{15}}{6^{5}}\) = 610 ≠ 63
exponents should be subtracted not divided.

SIMPLIFYING AN EXPRESSION Simplify the expression. Write your answer as a power.

Question 18.
\(\frac{7^{5} \cdot 7^{3}}{7^{2}}\)
Answer:
\(\frac{7^{5} \cdot 7^{3}}{7^{2}}\) = 76.

Explanation:
Given Expression as \(\frac{7^{5} \cdot 7^{3}}{7^{2}}\)
first we calculate separately values of numerators
then divide with denominator, we have numerator (7)5 X (7)
we have same bases as 7 so we add powers as (7)5+3 = (7)8
Now we have \(\frac{7^{8}}{7^{2}}\)  so we use rule
for finding \(\frac{a^{m}}{a^{n}}\) a quotient of
two powers with the same base as am-n
So \(\frac{7^{8}}{7^{2}}\)  = (7)8-2 = 76

Question 19.
\(\frac{6^{13}}{6^{4} \cdot 6^{2}}\)
Answer:
\(\frac{6^{13}}{6^{4} \cdot 6^{2}}\) = 67

Explanation:
Given expression \(\frac{6^{13}}{6^{4} \cdot 6^{2}}\)
first we calculate separately values of denominators
then divide with numerator, we have denominator  (6)4 X (6)
we have same bases as 6 so we add powers as (6)4+2 = (6)6
Now we have \(\frac{6^{13}}{6^{6}}\)  so we use rule
for finding \(\frac{a^{m}}{a^{n}}\) a quotient of
two powers with the same base as am-n
So \(\frac{6^{13}}{6^{6}}\) = (6)13-6 = 67

Question 20.
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation 8.3 33
Answer:
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation 8.3 33= = (-6.1)2

Explanation:
Given expression \(\frac{-6.1^{11}}{-6.1^{7} \cdot -6.1^{2}}\)
first we calculate separately values of denominators
then divide with numerator, we have denominator  (-6.1)7 X (-6.1)
we have same bases as -6.1 so we add powers as (-6.1)7+2 = (-6.1)9
Now we have \(\frac{-6.1^{11}}{-6.1^{9}}\)  so we use rule
for finding \(\frac{a^{m}}{a^{n}}\) a quotient of
two powers with the same base as am-n
So \(\frac{-6.1^{11}}{-6.1^{9}}\) = (-6.1)11-9 = (-6.1)2

Question 21.
\(\frac{\pi^{30}}{\pi^{18} \cdot \pi^{4}}\)
Answer:
\(\frac{\pi^{30}}{\pi^{18} \cdot \pi^{4}}\) = (π)8

Explanation:
Given expression is \(\frac{\pi^{30}}{\pi^{18} \cdot \pi^{4}}\)
first we calculate separately values of denominators
then divide with numerator, we have denominator  (π)18 X (π)
we have same bases as π so we add powers as (π)18+4 = (π)22
Now we have \(\frac{\pi^{30}}{\pi^{22} \)  so we use rule
for finding \(\frac{a^{m}}{a^{n}}\) a quotient of
two powers with the same base as am-n
So \(\frac{\pi^{30}}{\pi^{22} \) = (π)30-22 = π8

Question 22.
\(\frac{c^{22}}{c^{8} \cdot c^{9}}\)
Answer:
\(\frac{c^{22}}{c^{8} \cdot c^{9}}\) = c5

Explanation:
Given expression is \(\frac{c^{22}}{c^{8} \cdot c^{9}}\)
first we calculate separately values of denominators
then divide with numerator, we have denominator  c8 X c
we have same bases as c so we add powers as c8+9 = c17
Now we have \(\frac{c^{22}} {c^{17}} \)  so we use rule
for finding \(\frac{a^{m}}{a^{n}}\) a quotient of
two powers with the same base as am-n
So \(\frac{c^{22}} {c^{17}} \) = c22-17 = c5

Question 23.
\(\frac{z^{8} \cdot z^{6}}{z^{8}}\)
Answer:
\(\frac{z^{8} \cdot z^{6}}{z^{8}}\) = z6

Explanation:
Given expression is \(\frac{z^{8} \cdot z^{6}}{z^{8}}\)
first we calculate separately values of numerators
then divide with denominator, we have numerator z6 X z
we have same bases as z so we add powers as z6+8 = z14
Now we have \(\frac{z^{14}}{z^{8}} \)  so we use rule
for finding \(\frac{a^{m}}{a^{n}}\) a quotient of
two powers with the same base as am-n
So \(\frac{z^{14}}{z^{8}} \) = z14-8 = z6

Question 24.
MODELING REAL LIFE
The sound intensity of a normal conversation is 106 times greater than the quietest noise a person can hear. The sound intensity of a jet at takeoff is 1014 times greater than the quietest noise a person can hear. How many times more intense is the sound of a jet at takeoff than the sound of a normal conversation?
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation 8.3 34
Answer:
108 times more intense is the sound of a jet at takeoff than
the sound of a normal conversation.

Explanation:
Given the sound intensity of a normal conversation is 106 times
greater than the quietest noise a person can hear.
The sound intensity of a jet at takeoff is 1014 times
greater than the quietest noise a person can hear.
Therefore more intense is the sound of a jet at takeoff than
the sound of a normal conversation is \(\frac{10^{14}}{10^{6}} \)
so we use rule
for finding \(\frac{a^{m}}{a^{n}}\) a quotient of
two powers with the same base as am-n
So \(\frac{10^{14}}{10^{6}} \) = 1014-6= 108
therefore 108 times more intense is the sound of a jet at
takeoff than the sound of a normal conversation.

SIMPLIFYING AN EXPRESSION Simplify the expression. Write your answer as a power.
Question 25.
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation 8.3 35
Answer:
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation 8.3 35= (-4)5

Explanation:
First we calculate separately values of numerators
and denominators then divide, we have numerator (-4)8 X (-4)
we have same bases as -4 so we add powers as (-4)8+3 = (-4)11
we have denominator (-4)4 X (-4)2
we have same base as -4 so we add powers as (-4)4+2 = (-4)6
Now we have \(\frac{-4^{11}}{-4^{6}}\)  so we use rule
for finding \(\frac{a^{m}}{a^{n}}\) a quotient of
two powers with the same base as am-n
\(\frac{-4^{11}}{-4^{6}}\)  = (-4)11-6 = (-4)5

Question 26.
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation 8.3 36
Answer:
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation 8.3 36= 65

Explanation:
First we calculate separately values of numerators
and denominators then divide, we have numerator (6)2 X (6)12 
we have same bases as 6 so we add powers as (6)2+12 = (6)14
we have denominator (6)1 X (6)8
we have same base as 6 so we add powers as (6)1+8 = (6)9
Now we have \(\frac{6^{14}}{6^{9}}\)  so we use rule
for finding \(\frac{a^{m}}{a^{n}}\) a quotient of
two powers with the same base as am-n
\(\frac{6^{14}}{6^{9}}\)  = (6)14-9 = 65

Question 27.
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation 8.3 37
Answer:
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation 8.3 37= 310

Explanation:
First we calculate separately values of numerators
and denominators then divide, we have numerator (3)2 X (3)6 X (3)5
we have same bases as 3 so we add powers as (3)2+6+5 = (3)13
we have denominator (3)2 X (3)1
we have same base as 3 so we add powers as (3)2+1 = (3)3
Now we have \(\frac{3^{13}}{3^{3}}\)  so we use rule
for finding \(\frac{a^{m}}{a^{n}}\) a quotient of
two powers with the same base as am-n
\(\frac{3^{13}}{3^{3}}\)  = (3)13-3 = 310

Question 28.
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation 8.3 38
Answer:
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation 8.3 38= z10
Explanation:
First we calculate separately values of numerators
and denominators then divide, we have numerator (z)7 X (z)
we have same bases as z so we add powers as z7+6 = z13
we have denominator (z)1 X (z)2
we have same base as z so we add powers as (z)1+2 = (z)3
Now we have \(\frac{3^{13}}{3^{3}}\)  so we use rule
for finding \(\frac{a^{m}}{a^{n}}\) a quotient of
two powers with the same base as am-n
\(\frac{z^{13}}{z^{3}}\)  = (z)13-3 = z10

Question 29.
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation 8.3 39
Answer:
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation 8.3 39= x6

Explanation:
First we calculate separately values of numerators
and denominators then divide, we have numerator (x)5 X (x)13 
we have same bases as x so we add powers as x5+13 = x18
we have denominator (x)4 X (x)8
we have same base as x so we add powers as (x)4+8 = (x)12
Now we have \(\frac{x^{18}}{x^{12}}\)  so we use rule
for finding \(\frac{a^{m}}{a^{n}}\) a quotient of
two powers with the same base as am-n
\(\frac{x^{18}}{x^{12}}\)  = (x)18-12 = x6

Question 30.
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation 8.3 40
Answer:
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation 8.3 40= y11
Explanation:
First we calculate separately all multiple  values of numerators
and denominators then divide, we have numerator (y)8 X (y)2 X (y)4 X (y)7
we have same bases as y so we add powers as y8+2+4+7 = y21
we have denominator (y)7 X (y)1 X (y)we have same base
as y so we add powers as (y)7+1+2 = (y)10
Now we have \(\frac{y^{21}}{y^{10}}\)  so we use rule
for finding \(\frac{a^{m}}{a^{n}}\) a quotient of
two powers with the same base as am-n
\(\frac{y^{21}}{y^{10}}\)  = (y)21-10 = y11

Question 31.
REASONING
The storage capacities and prices of five devices are shown in the table.
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation 8.3 41
a. How many times more storage does Device D have than Device B?
b. Do storage and price have a linear relationship? Explain.
Answer:
a. 4 times more storage capacity Device D have than Device B.
b. No, as the price increases by $20 storage capacity doubles.

Explanation:
Given the storage capacities and prices of five devices are
as shown above in the table,
a. Storage Device D has capacity of 28 GB and Device B has
capacity of 28 GB as Device D has more capacity by
\(\frac{2^{8}}{2^{6}}\) = (2)8-6 = 22= 4,
therefore 4 times more storage capacity Device D
have than Device B.
b. As seen the price increases by $20 storage capacity doubles so
there is no linear relationship between storage and price.

Question 32.
DIG DEEPER!
Consider the equation \(\frac{9^{m}}{9^{n}}\) = 92
a. Find two numbers m and n that satisfy the equation.
b. Describe the number of solutions that satisfy the equation. Explain your reasoning.
Answer:
a. The two numbers m and n are 3,1.
b. We have more number of solutions that satisfy the equation as
explained below

Explanation:
Given the equation \(\frac{9^{m}}{9^{n}}\) = 92
a. We have rule for finding \(\frac{a^{m}}{a^{n}}\) a quotient of
two powers with the same base as am-n so \(\frac{9^{m}}{9^{n}}\)= 9m-n= 92 so
m-n=2, m=2+n we can take n=1 we get m= 2 + 1 = 3 so m,n is 3,1.
the equation is \(\frac{9^{3}}{9^{1}}\)= 93-1= 92
b. As m=2 + n we can take n any natural number from 1 to infinity,
as if n=1 m will be 2+1=3, if n=2 , m will be 2+2 = 4 and so on.
therefore (m, n)=(n+2,n) or (m,m-2)=(3,1)….(5,3),(∞,∞-2) are the
(infinity, infinity minus 2) numbers of solutions that satisfy the equation.

Question 33.
MODELING REAL LIFE
A scientist estimates that there are about 1024 stars in the universe and that each galaxy has, on average, approximately the same number of stars as the Milky Way galaxy. About how many galaxies are in the universe?
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation 8.3 42
Answer:
1013 galaxies are there in the universe.

Explanation:
A scientist estimates that there are about 1024 stars in the
universe and that each galaxy has, on average, approximately
the same number of stars as the Milky Way galaxy.
so number of galaxies in the universe are \(\frac{10^{24}}{10^{1} \cdot 10^{10}}\)
first we calculate separately values of denominators
then divide with numerator, we have denominator  (10)1 X (10)10 
we have same bases as 10 so we add powers as (10)1+10 = (10)11
Now we have \(\frac{10^{24}}{10^{11}} \)  so we use rule
for finding \(\frac{a^{m}}{a^{n}}\) a quotient of
two powers with the same base as am-n
\(\frac{10^{24}}{10^{11}} \)  = (10)24-11 = 1013
therefore 1013 galaxies are there in the universe.

Question 34.
NUMBER SENSE
Find the value of x that makes c = 89 true. Explain how you found your answer.
Answer:
The value of x is 10,

Explanation:
Given \(\frac{8^{3 x}}{8^{2 x}+1}\) = 89 is true,
so we use rule for finding \(\frac{a^{m}}{a^{n}}\) a quotient of
two powers with the same base as am-n
as bases are same 8 so 83x-(2x+1) = 89 now we equate powers we have
3x-(2x+1)= 9, means x-1=9 therefore x = 9 + 1 = 10.

Lesson 8.4 Zero and Negative Exponents

EXPLORATION 1

Understanding Zero Exponents
Work with a partner.
a. Copy and complete the table.
Big Ideas Math Solutions Grade 8 Chapter 8 Exponents and Scientific Notation 8.4 1
b. Evaluate each expression in the first column of the table in part(a). How can you use these results to define a0, where a ≠ 0?
Answer:
a.
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation-11

b. We define  a0

Explanation:
a. We completed the table by using quotient of powers of property,
as  \(\frac{a^{m}}{a^{n}}\) = am-n
then  by the quotient rule for exponents  we can write this as
an-n  =\(\frac{a^{n}}{a^{n}}\) Then this becomes a problem about
dividing fractions. Since the numerator and denominator
are both the same this becomes .

EXPLORATION 2

Understanding Negative Exponents
Work with a partner.
a. Copy and complete the table.
Big Ideas Math Solutions Grade 8 Chapter 8 Exponents and Scientific Notation 8.4 2
b. How can you use the Multiplicative Inverse Property to rewrite the powers containing negative exponents in the first column of the table?
c. Use your results in parts (a) and (b) to define a-n, where a ≠ 0 and n is an integer.
Answer:
a.
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation-12
b. We rewrite the powers containing negative exponents in the first column of the table as
5-3 X 53 as 62X 6-2 as 4 X 3-4 as -4-5 X -45 as =

c. a-n, where a ≠ 0 we write a-n as \(\frac{1}{a^{n}}\)

Explanation:
a. To complete the table first we write Product of Powers Property
then write their power and value.
As here we use Product of Powers Property
for am • an = am+n  If product of two powers with the same base then
powers are added. So 5-3 X 53 = 5-3+3 = 50 = 1, 62X 6-2 = 62-2 = 60 = 1,
4 X 3-4 = 4-4 =0 = 1 and -4-5 X -45 = -4-5+5 =-40 = 1.
b. The inverse property of multiplication states that if you
multiply a number by its reciprocal, also called the multiplicative inverse,
the product will be 1. (a/b)*(b/a)=1,so we rewrite the powers
containing negative exponents in the first column of the table as
5-3 X 53 as 62X 6-2 as 4 X 3-4 as -4-5 X -45 as =

Try It

Evaluate the expression.
Question 1.
4-2
Answer:
4-2 =

Explanation:
Given expression as 4-2 so we write as

Question 2.
(- 2)– 5
Answer:
(- 2)– 5=

Explanation:
Given expression as (-2)-5 so we write as

Question 3.
6-8 • 68
Answer:
6-8 • 68 = 1

Explanation:
we write the given expression 6-8 X 68 as Product of Powers Property
for am • an = am+n  If product of two powers with the same base then
powers are added. So 6-8 X 68 = 6-8+8 = 60 = 1.

Question 4.
\(\frac{(-3)^{5}}{(-3)^{6}}\)
Answer:
\(\frac{(-3)^{5}}{(-3)^{6}}\) = – \(\frac{1}{3}\) or -3-1

Explanation:
Given expression as \(\frac{(-3)^{5}}{(-3)^{6}}\) we use
the quotient rule for exponents  we can write this as \(\frac{a^{n}}{a^{n}}\) = am-n ,
so (-3)5-6 ,= (-3)-1 or – \(\frac{1}{3}\).

Question 5.
\(\frac{1}{5^{7}} \cdot \frac{1}{5^{4}}\)
Answer:
\(\frac{1}{5^{7}} \cdot \frac{1}{5^{4}}\) = \(\frac{1}{5^{11}}\)

Explanation:
Given expression is \(\frac{1}{5^{7}} \cdot \frac{1}{5^{4}}\)
first we calculate separately values of denominators
then divide with numerator, we have denominator  57 X 54
we have same bases as 5 so we add powers as 57+4 = 511
as numerator is 1 we write as \(\frac{1}{5^{11}}\).

Question 6.
\(\frac{4^{5} \cdot 4^{3}}{4^{2}}\)
Answer:
\(\frac{4^{5} \cdot 4^{3}}{4^{2}}\) = 46

Explanation:
Given Expression as \(\frac{4^{5} \cdot 4^{3}}{4^{2}}\)
first we calculate separately values of numerators
then divide with denominator, we have numerator (4)5 X (4)
we have same bases as 4 so we add powers as (4)5+3 = (4)8
Now we have \(\frac{4^{8}}{4^{2}}\)  so we use rule
for finding \(\frac{a^{m}}{a^{n}}\) a quotient of
two powers with the same base as am-n
So \(\frac{4^{8}}{4^{2}}\)  = (4)8-2 = 46

Simplify. Write the expression using only positive exponents.
Question 7.
8x-2
Answer:
8x-2 = \(\frac{8}{x^{2}}\)

Explanation:
Given 8x-2  we write the expression as positive exponents by using

so 8 X \(\frac{1}{x^{2}}\) or \(\frac{8}{x^{2}}\)

Question 8.
b0 • b-10
Answer:
b0 • b-10 = \(\frac{1}{b^{10}}\)

Explanation:
Given b0 • b-10 we write the expression as positive exponents by using

so b0 X \(\frac{1}{b^{10}}\) as  we know b0 =1,
b0 X \(\frac{1}{b^{10}}\)=1 X \(\frac{1}{b^{10}}\) or \(\frac{1}{b^{10}}\).

Question 9.
\(\frac{z^{6}}{15 z^{9}}\)
Answer:
\(\frac{z^{6}}{15 z^{9}}\) = \(\frac{1}{15 z^{3}}\)

Explanation:
Given expression as \(\frac{z^{6}}{15 z^{9}}\) first we write it as
\(\frac{1}{15}\) X \(\frac{z^{6}}{z^{9}}\) now we use rule
for finding \(\frac{a^{m}}{a^{n}}\) a quotient of
two powers with the same base as am-n
as \(\frac{1}{15}\) X z6-9 we get \(\frac{1}{15}\) X z-3
= \(\frac{1}{15 z^{3}}\).

Self-Assessment for Concepts & Skills
Solve each exercise. Then rate your understanding of the success criteria in your journal.

EVALUATING EXPRESSIONS Evaluate the expression.
Question 10.
7-2
Answer:
7-2= \(\frac{1}{49}\)

Explanation:
Given 7-2  we write the expression as positive exponents by using

so \(\frac{1}{7^{2}}\)= \(\frac{1}{49}\)

Question 11.
4-3 • 40
Answer:
4-3 • 40 = \(\frac{1}{64}\)

Explanation:
Given  4-3 X 40 we write the expression as positive exponents by using

so  \(\frac{1}{4^{3}}\) X 40 as  we know 40 =1,
\(\frac{1}{4^{3}}\) X 40 =\(\frac{1}{4^{3}}\) x 1 = \(\frac{1}{64}\).

Question 12.
\(\frac{(-9)^{5}}{(-9)^{7}}\)
Answer:
\(\frac{(-9)^{5}}{(-9)^{7}}\) = \(\frac{1}{-9^{2}}\) = \(\frac{1}{81}\)

Explanation:
Given expression as \(\frac{-9^{5}}{-9^{7}}\) now we use rule
for finding \(\frac{a^{m}}{a^{n}}\) a quotient of
two powers with the same base as am-n
as -95-7 we get -9-2 we write as \(\frac{1}{-9^{2}}\) = \(\frac{1}{81}\).

SIMPLIFYING EXPRESSIONS Simplify. Write the expression using only positive exponents.
Question 13.
10t-5
Answer:
10t-5 = \(\frac{10}{t^{5}}\)

Explanation:
Given 10t-5 we write as 10 X t-5 = now we write t-5 as tSo 10 X t-5 = 10 X

Question 14.
w3 • w-9
Answer:
w3 • w-9 =

Explanation:
Given expression is w3 • w-9  first we have general rule for am • an = am+n
If product of two powers with the same base then
powers are added, so we add powers as w3-9 = w-6,
So w-6 =

Question 15.
\(\frac{r^{8} \cdot r^{8}}{4}\)
Answer:
\(\frac{r^{8} \cdot r^{8}}{4}\) = \(\frac{r^{16}}{4}\)

Explanation:
Given expression is \(\frac{r^{8} \cdot r^{8}}{4}\) first we
solve numerator by using general rule for am • an = am+n
If product of two powers with the same base then
powers are added, so we add powers as r8+8 = r16,
now we write as r16 X \(\frac{1}{4}\) = \(\frac{r^{16}}{4}\).

Question 16.
DIFFERENT WORDS, SAME QUESTION
Which is different? Find “both” answers.
Big Ideas Math Solutions Grade 8 Chapter 8 Exponents and Scientific Notation 8.4 3
Answer:
d bit  is different because writing (-3) X (-3) X (-3) as a power
with an integer base is (-3)3 no negative exponent.
one more answer is all a,b,c bits have fractions but only d bit
is not in fraction form.

Explanation:
a. Writing \(\frac{1}{3 X 3 X 3}\) using negative exponent is 3-3.
b. Writing 3 to the negative third is 3-3.
c. Writing \(\frac{1^{3}}{3^{3}}\) is 3-3.
d. Writing (-3) X (-3) X (-3) as a power  with an integer base is (-3)3.
As a, b, c has value 3-3only bit d has (-3)3.
Therefore d bit  is different because writing (-3) X (-3) X (-3) as a power
with an integer base is (-3)3 we don not have a negative exponent.
one more answer is all a,b,c bits have fractions but only d bit
is not in fraction form.

Self-Assessment for Problem Solving

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 17.
The mass of a grain of sand is about 10-3 gram. About how many grains of sand are in a 10-kilogram bag of sand?
Answer:
There are about 10,000,000 grains of sand.

Explanation:
Given the mass of a grain of sand is about 10-3 gram.
We know 1 kg = 1000 grams ,10 kg = 10 X 1000 = 10,000 grams,
therefore one grain of sand is 10,000 X 1000 = 10,000,000 grains of sand.

Question 18.
A one-celled, aquatic organism called a dinoflagellate is 1000 micrometers long. A microscope magnifies the dinoflagellate 100 times. What is the magnified length of the dinoflagellate in meters? (1 micrometer is 10-6; meter.)
Big Ideas Math Solutions Grade 8 Chapter 8 Exponents and Scientific Notation 8.4 4
Answer:
The magnified length of the dinoflagellate in meters10-1 meters

Explanation:
Given one-celled, aquatic organism called a dinoflagellate is
1000 micrometers long. A microscope magnifies the
dinoflagellate 100 times, the magnified length in meters is
1 micrometer is 10-6 meter now dinoflagellate is 1000 X 10-6 X 100 =
103 X 10-6 X 102 we use general rule for am • an = am+n
If product of two powers with the same base then
powers are added so all have base 10 we add powers as
103-6+2 = 10-1 meters therefore the magnified length of
the dinoflagellate in meters10-1 meters.

Question 19.
DIG DEEPER!
A garden is 12 yards long. Assuming the snail moves at a constant speed, how many minutes does it take the snail to travel the length of the garden? Justify your answer.
Big Ideas Math Solutions Grade 8 Chapter 8 Exponents and Scientific Notation 8.4 5
Answer:
Snail takes 15 minutes to travel the length of the garden

Explanation:
Given a garden is 12 yards long and the constant speed of snail is
5-2  foot per second. We know 1 yard is equal to 3 foot and
speed = distance by time ,So 5-2  = 12 X 3 by time
therefore time = 12 X 3 X 5seconds = 36 X 25 seconds = 900 seconds,
converting seconds in minutes 900 divide by 60 or \(\frac{900}{60}\)= 15 minutes, therefore snail takes 15 minutes to travel the length of the garden.

Zero and Negative Exponents Homework & Practice 8.4

Review & Refresh

Simplify the expression. Write your answer as a power.
Question 1.
\(\frac{10^{8}}{10^{4}}\)
Answer:
\(\frac{10^{8}}{10^{4}}\) = 104

Explanation:
Given expression \(\frac{10^{8}}{10^{4}}\) so
now we use rule for finding \(\frac{a^{m}}{a^{n}}\) a quotient of
two powers with the same base has am-n
So \(\frac{10^{8}}{10^{4}}\) = 108-4 = 104

Question 2.
\(\frac{y^{9}}{y^{7}}\)
Answer:
\(\frac{y^{9}}{y^{7}}\) = y2

Explanation:
Given expression \(\frac{y^{9}}{y^{7}}\) so
now we use rule for finding \(\frac{a^{m}}{a^{n}}\) a quotient of
two powers with the same base has am-n
So \(\frac{y^{9}}{y^{7}}\) = y9-7 = y2

Question 3.
\(\frac{(-3)^{8} \cdot(-3)^{3}}{(-3)^{2}}\)
Answer:
\(\frac{(-3)^{8} \cdot(-3)^{3}}{(-3)^{2}}\) = (-3)9

Explanation:
Given Expression as \(\frac{-3^{8} \cdot -3^{3}}{-3^{2}}\)
first we calculate separately values of numerators
then divide with denominator, we have numerator (-3)8 X (-3)
we have same bases as -3 so we add powers as (-3)8+3 = (-3)11
Now we have \(\frac{-3^{11}}{-3^{2}}\)  so we use rule
for finding \(\frac{a^{m}}{a^{n}}\) a quotient of
two powers with the same base as am-n
So \(\frac{-3^{11}}{-3^{2}}\)  = (-3)11-2 = (-3)9

Tell whether the triangles are similar. Explain.
Question 4.
Big Ideas Math Solutions Grade 8 Chapter 8 Exponents and Scientific Notation 8.4 6
Answer:
Yes, The triangles are similar.

Explanation:
Given two triangles in figure to know if they are similar we have
two triangles are said to be similar if their corresponding angles
are congruent and the corresponding sides are in proportion.
In the above two figures the triangles have the same angle measures,
and the corresponding sides are in proportion. So triangles are similar.

Question 5.
Big Ideas Math Solutions Grade 8 Chapter 8 Exponents and Scientific Notation 8.4 7
Answer:
Yes, The triangles are similar.

Explanation:
Given two triangles in figure to know if they are similar we have
two triangles are said to be similar if their corresponding angles
are congruent and the corresponding sides are in proportion.
In the above two figures the triangles do not have the same angle measures,
So triangles are not similar.

Question 6.
Which data display best orders numerical data and shows how they are distributed?
A. bar graph
B. line graph
C. scatter plot
D. stem-and-leaf plot
Answer:
D. stem-and-leaf plot

Explanation:
A stem-and-leaf plot best orders numerical data and shows
how the data is distributed since it orders the values from
least to greatest and shows how many values lie under each “stem” in the stem-and leaf plot so you can see how the data is distributed.

Concepts, Skills, &Problem Solving

UNDERSTANDING NEGATIVE EXPONENTS Copy and complete the table. (See Exploration 2, p. 337.)
Big Ideas Math Solutions Grade 8 Chapter 8 Exponents and Scientific Notation 8.4 8
Answer:
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation-13
Explanation:
To complete the table first we write Product of Powers Property
then write their power and value.
As here we use Product of Powers Property
for am • an = am+n  If product of two powers with the same base then
powers are added. So 7. 7-4 X 74 = 7-4+4 = 70 = 1,
8. (-2)5X (-2)-5 = (-2)5-5 = (-2)0 = 1.

EVALUATING EXPRESSIONS Evaluate the expression.
Question 9.
\(\frac{8^{7}}{8^{7}}\)
Answer:
\(\frac{8^{7}}{8^{7}}\) = 1

Explanation:
Given expression is \(\frac{8^{7}}{8^{7}}\)= 87 X 8-7 ,
Here we use Product of Powers Property for am • an = am+n
if we have product of two powers with the same base then
powers are added. So 87 X 8-7 = 87-7 = 80 = 1.

Question 10.
50 • 53
Answer:
50 • 53 =  53 = 125.

Explanation:
Given expression is 50 X 53 ,
Here we use Product of Powers Property for am • an = am+n
if we have product of two powers with the same base then
powers are added. So 50 X 53 = 50+3 = 53 = 125.

Question 11.
(- 2)-8 • (- 2)8
Answer:
(- 2)-8 • (- 2)8 = 1

Explanation:
Given expression is -2-8 X -28 ,
Here we use Product of Powers Property for am • an = am+n
if we have product of two powers with the same base then
powers are added. So (-2)-8 X (-2)8 = (-2)-8+8 = (-2)0= 1.

Question 12.
94 • 9-4
Answer:
94 • 9-4 = 1

Explanation:
Given expression is 94 X 9-4 ,
Here we use Product of Powers Property for am • an = am+n
if we have product of two powers with the same base then
powers are added. So 94 X 9-4 = 94-4 = 90= 1.

Question 13.
6-2
Answer:
6-2= \(\frac{1}{36}\)

Explanation:
Given expression as 6-2 we write the expression as positive exponents by using

so \(\frac{1}{6^{2}}\) = \(\frac{1}{36}\).

Question 14.
1580
Answer:
1580 = 1

Explanation:
Given expression is 158it is proven that any number or
expression raised to the power of zero is always equal to 1.
In other words, if the exponent is zero then the result is 1.
So 1580 = 1.

Question 15.
\(\frac{4^{3}}{4^{5}}\)
Answer:
\(\frac{4^{3}}{4^{5}}\) = \(\frac{1}{4^{2}}\) = \(\frac{1}{16}\)

Explanation:
Given expression is \(\frac{4^{3}}{4^{5}}\)
now we use rule for finding \(\frac{a^{m}}{a^{n}}\) a quotient of
two powers with the same base has am-n So \(\frac{4^{3}}{4^{5}}\) =
43-5 = 4-2 = \(\frac{1}{4^{2}}\) = \(\frac{1}{16}\).

Question 16.
\(\frac{-3}{(-3)^{2}}\)
Answer:
\(\frac{-3}{(-3)^{2}}\) = –\(\frac{1}{3}\)

Explanation:
Given expression is \(\frac{-3}{(-3)^{2}}\)
now we use rule for finding \(\frac{a^{m}}{a^{n}}\) a quotient of
two powers with the same base has am-n ,so \(\frac{-3}{(-3)^{2}}\) =
(-3)1-2 = (-3)-1 = –\(\frac{1}{3}\).

Question 17.
22 • 2-4
Answer:
22 • 2-4 = \(\frac{1}{4}\)

Explanation:
Given expression is 22 X 2-4 ,
Here we use Product of Powers Property for am • an = am+n
if we have product of two powers with the same base then
powers are added so 22-4  = 22 X 2-4 = 2-2 = \(\frac{1}{2^{2}}\) = \(\frac{1}{4}\).

Question 18.
3-3 • 3-2
Answer:
3-3 • 3-2 = \(\frac{1}{243}\)

Explanation:
Given expression is 3-3 • 3-2 ,
Here we use Product of Powers Property for am • an = am+n
if we have product of two powers with the same base then
powers are added so 3-3-2  = 3-5 = \(\frac{1}{3^{5}}\) = \(\frac{1}{243}\).

Question 19.
\(\frac{1}{5^{3}} \cdot \frac{1}{5^{6}}\)
Answer:
\(\frac{1}{5^{3}} \cdot \frac{1}{5^{6}}\) = \(\frac{1}{5^{9}}\) =
\(\frac{1}{1953125}\)

Explanation:
Given expression is \(\frac{1}{5^{3}} \cdot \frac{1}{5^{6}}\)
now we use rule for finding \(\frac{a^{m}}{a^{n}}\) a quotient of
two powers with the same base has am-n ,so \(\frac{1}{5^{3}} \cdot \frac{1}{5^{6}}\) =
(5)-3-6 = (5)-9 = \(\frac{1}{5^{9}}\) =
\(\frac{1}{1953125}\).

Question 20.
\(\frac{(1.5)^{2}}{(1.5)^{2} \cdot(1.5)^{4}}\)
Answer:
\(\frac{(1.5)^{2}}{(1.5)^{2} \cdot(1.5)^{4}}\) = \(\frac{1}{5.0625}\)

Explanation:
Given expression is \(\frac{(1.5)^{2}}{(1.5)^{2} \cdot(1.5)^{4}}\)
First we multiply denominators as bases are same
we add powers so (1.5)2  X (1.5)4 = (1.5)2+4 = (1.5)6
now we solve \(\frac{1.5^{2}}{1.5^{6}}\) we use rule for finding
\(\frac{a^{m}}{a^{n}}\), a quotient of two powers with the same base as am-n
\(\frac{1.5^{2}}{1.5^{6}}\) = 1.52-6 = 1.5-4 now we use =\(\frac{1}{1.5^{4}}\) =
\(\frac{1}{5.0625}\).

Question 21.
YOU BE THE TEACHER
Your friend evaluates 4-3. Is your friend correct? Explain your reasoning.
Big Ideas Math Solutions Grade 8 Chapter 8 Exponents and Scientific Notation 8.4 9
Answer:

No, friend is incorrect as 4-3 = \(\frac{1}{64}\) ≠ -64

Explanation:
Given expression is 4-3 we have = \(\frac{1}{4^{3}}\) = \(\frac{1}{64}\),
but friend says = -64 which is in correct as 4-3 = \(\frac{1}{64}\) ≠ -64.

Question 22.
CRITICAL THINKING
How can you write the number 1 as a power with base 2? a power with base 10?
Answer:
21 and 101

Explanation:
Given to write the number 1 as a power with base 2 is 21 and
a power with base 10 is 101.

Question 23.
NUMBER SENSE
Without evaluating, order 50, 54, and 5-5 from least to greatest. Explain your reasoning.
Answer:
Order 50, 54, and 5-5 from least to greatest is 5-5, 50 and 54

Explanation:
Given order 50, 54, and 5-5 from least to greatest as each has base 5,
we take consideration of orders as -5 < 0 < 4 so Order 50, 54, and 5-5
from least to greatest is 5-5, 50 and 54.

SIMPLIFYING EXPRESSIONS
Simplify. Write the expression using only positive exponents.
Question 24.
6y-4
Answer:
6y-4 = \(\frac{6}{y^{4}}\)

Explanation:
Given expression as 6y-4 we write the expression using only
positive exponents for 6 X y-4  as we write first y-4 we have
=
\(\frac{1}{y^{4}}\) now 6 X \(\frac{1}{y^{4}}\) = \(\frac{6}{y^{4}}\).

Question 25.
8-2 • a7
Answer:
8-2 • a7 = \(\frac{a^{7}}{64}\)

Explanation:
Given expression as 8-2 X a7 we write the expression using only
positive exponents for 8-2 X a7 as we write first 8-2 we have
=
\(\frac{1}{8^{2}}\), So a7 X \(\frac{1}{8^{2}}\) = \(\frac{a^{7}}{64}\).

Question 26.
\(\frac{9 c^{3}}{c^{4}}\)
Answer:
\(\frac{9 c^{3}}{c^{4}}\) = \(\frac{9}{c}\)

Explanation:
Given expression is \(\frac{9 c^{3}}{c^{4}}\) = 9 X \(\frac{c^{3}}{c^{4}}\)
First we solve \(\frac{c^{3}}{c^{4}}\) we use rule for finding
\(\frac{a^{m}}{a^{n}}\), a quotient of two powers with the same base as am-n
\(\frac{c^{3}}{c^{4}}\) = c3-4 = c-1
now we use .

Question 27.
\(\frac{5 b^{2}}{b^{3}}\)
Answer:
\(\frac{5 b^{2}}{b^{3}}\) = \(\frac{5}{b}\)

Explanation:
Given expression is \(\frac{5 b^{2}}{b^{3}}\) = 5 X \(\frac{b^{2}}{b^{3}}\)
First we solve \(\frac{b^{2}}{b^{3}}\) we use rule for finding
\(\frac{a^{m}}{a^{n}}\), a quotient of two powers with the same base as am-n
\(\frac{b^{2}}{b^{3}}\) = b2-3 = b-1
now we use .

Question 28.
\(\frac{8 x^{3}}{2 x^{9}}\)
Answer:
\(\frac{8 x^{3}}{2 x^{9}}\) = \(\frac{4}{x^{6}}\)

Explanation:
Given expression as \(\frac{8 x^{3}}{2 x^{9}}\) we use rule for finding
\(\frac{a^{m}}{a^{n}}\), a quotient of two powers with the same base as am-n
\(\frac{x^{3}}{x^{9}}\) = x3-9 = x-6 for x-6 we use

Question 29.
3d-4 • 4d4
Answer:
3d-4 • 4d4 = 12

Explanation:
Given expression 3d-4 • 4d4 first we multiply d-4 X d4 ,
here we use Product of Powers Property for am • an = am+n
if we have product of two powers with the same base then
powers are added, we have same base d so d-4+4 = d0= 1,
Now we multiply 1 with 3,4 as 3 X 4 X 1 we get 12.

Question 30.
m-2 • n3
Answer:
m-2 • n3 =

Explanation:
Given expression is m-2 • n3 so we write m-2 we use

Question 31.
\(\frac{3^{2} \cdot k^{0} \cdot w^{0}}{w^{6}}\)
Answer:
\(\frac{3^{2} \cdot k^{0} \cdot w^{0}}{w^{6}}\) =

Explanation:
Given expression \(\frac{3^{2} \cdot k^{0} \cdot w^{0}}{w^{6}}\) we know
k0 and w0 is 1 now we have \(\frac{3^{2}}{w^{6}}\) X 1 X 1 = \(\frac{9}{w^{6}}\).

Question 32.

OPEN-ENDED
Write two different powers with negative exponents that have the same value. Justify your answer.
Answer:
We  write two different powers with negative exponents that
have the same value are 2and the value will be

Explanation:
Let us take two different powers with negative exponents that
have the same value are 2we have  .

REASONING
In Exercises 33–36, use the table.
Big Ideas Math Solutions Grade 8 Chapter 8 Exponents and Scientific Notation 8.4 10
Question 33.
How many millimeters are in a decimeter?
Answer:
100 millimeters are in a decimeter.

Explanation:
Given to find millimeter are in a decimeter we have as per table
millimeters are in a decimeter we write as 10so 100 millimeters are in a decimeter.

Question 34.
How many micrometers are in a centimeter?
Answer:
10000 micrometers are in a centimeter

Explanation:
Given to find micrometers are in a centimeter we have as per table
micrometers are in a centimeter we write as 10so 10000 micrometers are in a centimeter.

Question 35.
How many nanometers are in a millimeter?
Answer:
1,000,000 nanometers are in a millimeter

Explanation:
Given to find nanometers are in a millimeter  we have as per table
nanometers are in a millimeter we write as 10so 1,000,000 nanometers are in a millimeter.

Question 36.
How many micrometers are in a meter?
Answer:
1,000,000 micrometers are in a meter

Explanation:
Given to find micrometers are in a meter we have as per table
micrometers are in a meter we write as 10so 1,000,000 micrometers are in a meter.

Question 37.
MODELING REAL LIFE
A bacterium is 100 micrometers long. A virus is 1000 times smaller than the bacterium.
a. Using the table above, find the length of the virus in meters.
b. Is the answer to part (a) less than, greater than, or equal to 1 micrometer?
Answer:
a. The length of the virus in meters is

Explanation:
Given a bacterium is 100 micrometers long. A virus is 1000 times
smaller than the bacterium.
a. The length of the virus in meters is 
1 micrometer = 10-6 meters
= 100 X 10-6 by 1000 = 10 X 10-6 X 10-3
we have Product of Powers Property for am • an = am+n
if we have product of two powers with the same base then
powers are added so = therefore the length of the virus in meters is b. We have 1 micrometer as length of the virus it is So t

Question 38.
DIG DEEPER!
Every 2 seconds, someone in the United States needs blood. A sample blood donation is shown.
Big Ideas Math Solutions Grade 8 Chapter 8 Exponents and Scientific Notation 8.4 11
a. One cubic millimeter of blood contains about 104 white blood cells. How many white blood cells are in the donation? (1 mm3 = 10-3 mL)
b. One cubic millimeter of blood contains about 5 × 106 red blood cells. How many red blood cells are in the donation?
c. Compare your answers for parts (a) and (b).
Answer:
a. The white blood cells in the donation are 107 white blood cells
and the blood donation contains 500 ml are  5 X 109 white blood cells.
b. The red blood cells in the donation are 5 X 109  red blood cells
and the blood donation contains 500 ml are  25 X 1011 red blood cells.
c. The white blood cells are 500 times more than the red blood cells.

Explanation:
a. One cubic millimeter of blood contains about 104 white blood cells.
(1 mm3 = 10-3 mL),The white blood cells  in the donation are
1 mm3 = 10-3 mL,
104 = 10-3 mL,
1 mL= 104+3, therefore 1 mL is 107 white blood cells,
Now the blood donation contains 500 ml = 5 X 102 X 10 7 =
5 X 109 white blood cells.
b.One cubic millimeter of blood contains about 5 × 106 red blood cells,
The red blood cells in the donation are
1 mm3 = 10-3 mL so
5 X 106 = 10-3 mL therefore 1 mL= 5 X 106 X 103
= 5 X 106+3 = 5 X 109 red blood cells
Now the blood donation contains 500 ml = 5 X 102 X 5 X 109 = 25 X 102+9
= 25 X  1011 red blood cells.
c. Now comparison of the white blood cells and the red blood cells are
25 X 1011 divide by  5 X 109 = 5 X 1011-9 = 5 X 102 = 5 X 100 = 500 ,
So the white blood cells are 500 times more than the red blood cells.

Question 39.
PRECISION
Describe how to rewrite a power with a positive exponent as a fraction with a power in the denominator. Use the definition of negative exponents to justify your reasoning.
Answer:

Explanation:
To rewrite a power with a positive exponent as a fraction
with a power in the denominator by using the definition of
negative exponents we write the power as 1 divided by a power
with the same base and a negative exponents , Example
an = a-(-n) =

Question 40.
REASONING
The definition of a negative exponent states that a-n = \(\frac{1}{a^{n^{*}}}\).
Explain why this rule does not apply when a = 0.
Answer:

Explanation:
we have negative exponent states that a-n = \(\frac{1}{a^{n^{*}}}\) but this
rule does not apply when a= 0 as 0-n = 0 or
if we use negative exponent we get =
we could get  0 or 1 so it is undefined.

Lesson 8.5 Estimating Quantities

EXPLORATION 1

Work with a partner. Match each picture with the most appropriate distance. Explain your reasoning.
Big Ideas Math Answer Key Grade 8 Chapter 8 Exponents and Scientific Notation 8.5 1
Answer:
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation-14
Explanation:
Here we match each picture with the most appropriate distance
a. If we look at the picture it is very away so we take the
least distance in all as 6 X 10-2 m .
b. If we look at the picture it is very far so we take the
far distance in all as 6 X 103 m .
c. If we look at the picture it can jump very little distance so we take
as 2 X 10-1 m.
d. If we look at the picture it is at near distance so we take
as 1 X 101 m.

EXPLORATION 2

Approximating Numbers
Work with a partner. Match each number in List 1 with its closest approximation in List 2. Explain your method.
Big Ideas Math Answer Key Grade 8 Chapter 8 Exponents and Scientific Notation 8.5 2
Answer:
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation-15
Explanation:
We matched each number in List 1 with its closest
approximation in List 2 as
a. 180,000,000,000,000 is nearly or approximately equal to ≈ 2,00,000,000,000,000,
we have after 2 followed by 14 zeros so we take as 2 X 1014 matches with C.
therefore we match a. to C.
b. 0.0000000011 is nearly or approximately equal to ≈ 1 X 10-9 matches with C,
we have divided 1 by 10 followed by 9 zeros.
therefore we match b. to F.
c. 302,000,000,000 is nearly or approximately equal to ≈ 300,000,000,000,
we have after 3 followed by 11 zeros so we take as 3 X 1011 matches with A.
therefore match c. to A.
d. 0.00000028 is nearly or approximately equal to ≈ 0.0000003,
so 3 X 10-7 matches with E, we have divided 3 by 10 followed by 7 zeros,
therefore we match d. to E.
e. 0.0000097 is nearly or approximately equal to ≈ 0.00001,
so 1 X 10-5 matches with B, we have divided 1 by 10 followed by 5 zeros,
therefore we match e. to B.
f. 330,000,000,000,000 is nearly or approximately equal to ≈ 3,00,000,000,000,000
we have after 3 followed by 14 zeros so we take as 3 X 1014 matches with H.
therefore we match f. to H.
g. 26,000,000,000,000 is nearly or approximately equal to ≈ 30,000,000,000,000
we have after 3 followed by 13 zeros so we take as 3 X 1013 matches with D.
therefore we match g. to D.
h. 0.000023 is nearly or approximately equal to ≈ 0.00002,
so 2 X 10-5 matches with G, we have divided 2 by 10 followed by 5 zeros,
therefore we match h. to G.

Try It

Round the number. Write the result as the product of a single digit and a power of 10.
Question 1.
8,031,426,100
Answer:
8,031,426,100 = 8 X 109

Explanation:
Given number is 8,031,426,100 is nearly or approximately
equal to ≈ 8,000,000,000 so we have 8 followed by 9 zeros,
so we write as 8 X 109.

Question 2.
98,247,836,218
Answer:
98,247,836,218 = 1 X 1011

Explanation:
Given number is 98,247,836,218 is nearly or approximately
equal to ≈ 100,000,000,000 so we have 1 followed by 11 zeros,
so we write as 1 X 1011.

Round the number. Write the result as the product of a single digit and a power of 10.
Question 3.
0.000384509
Answer:
0.000384509 = 4 X 10-4

Explanation:
Given number is 0.000384509 is nearly or approximately
equal to ≈ 0.0004, we have divided 4 by 10 followed
by 4 zeros so we write as 4 X 10-4.

Question 4.
0.00000726
Answer:
0.00000726 = 7 X 10-6

Explanation:
Given number is 0.00000726 is nearly or approximately
equal to ≈ 0.000007, we have divided 7 by 10
followed by  zeros so we write as 7 X 10-6.

Question 5.
The distance from Mercury to Mars is about 105,651,744 miles. The distance from Saturn to Jupiter is about 4 times this distance. What is the approximate distance from Saturn to Jupiter?
Answer:
The distance from Saturn to Jupiter is (105,651,744)4 miles

Explanation:
Given the distance from Mercury to Mars is about 105,651,744 miles,
and the distance from Saturn to Jupiter is about 4 times this distance,
so 4 times of 105,651,744 is 105,651,744  X 105,651,744 X
105,651,744  X 105,651,744  = (105,651,744)4 miles.

Self-Assessment for Concepts & Skills

Solve each exercise. Then rate your understanding of the success criteria in your journal.

APPROXIMATING A NUMBER Round the number. Write the result as the product of a single digit and a power of 10.
Question 6.
899,032,878,300
Answer:
899,032,878,300 = 9 X 1011

Explanation:
Given number is 899,032,878,300 is nearly or approximately
equal to ≈ 900,000,000,000 so we have 9 followed by 11 zeros,
so we write as 9 X 1011.

Question 7.
62,322,118,987
Answer:
62,322,118,987= 6 X 1010

Explanation:
Given number is 62,322,118,987 is nearly or approximately
equal to ≈ 60,000,000,000 so we have 6 followed by 10 zeros,
so we write as 6 X 1010.

Question 8.
0.00000278101
Answer:
0.00000278101 = 3 X 10-6

Explanation:
Given number is 0.00000278101 is nearly or approximately equal to ≈ 0.000003,
we have divided 3 by 10 followed by 6 zeros so we write as
3 X 10-6.

Question 9.
0.000013094
Answer:
0.000013094 = 1 X 10-5.

Explanation:
Given number is 0.000013094 is nearly or approximately
equal to ≈ 0.00001, we have divided 1 by 10
followed by 5 zeros so we write as 1 X 10-5.

Question 10.
APPROXIMATING A QUANTITY
Lake A has a volume of 21,150,427,000 cubic meters. Lake B has a volume that is 2.5 times the volume of Lake A. What is the approximate volume of Lake B?
Answer:

Explanation:
Given Lake A has a volume of 21,150,427,000 cubic meters.
Lake B has a volume that is 2.5 times the volume of Lake A.
The approximate volume of Lake B is 2.5 X 21,150,427,000,
we take 21,150,427,000 is nearly or approximately
equal to ≈ 20,000,000,000 so 2.5 X 20,000,000,000
= 5 X 10,000,000,000 = 5 X 1010.

Self-Assessment for Problem Solving
Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 11.
On average, a small dog’s heart beats about 530,000,000 times
during its lifetime, and a large dog’s heart beats about 1.4 times this amount.
What is the approximate number of heartbeats in the lifetime of a large dog?
Answer:
The approximate number of heartbeats in the lifetime of a
large dog is 7 X 108 times.

Explanation:
Given on average, a small dog’s heart beats about
530,000,000 times during its lifetime and a large dog’s
heart beats about 1.4 times this amount.
So number of heartbeats in the lifetime of a large dog
= 1.5 X 530,000,000 we take 530,000,000 is nearly or approximately
equal to ≈ 500,000,000 so 1.5 X 500,000,000 = 7.5 X 100,000,000
= 7 X 108 times, therefore the approximate number
of heartbeats in the lifetime of a large dog is 7 X 108 times.

Question 12.
DIG DEEPER!
A physicist observes a gamma ray with a wavelength of 0.00000000135 millimeter and an X-ray with a wavelength of 0.00000012 millimeter. (a) About how many times shorter is the wavelength of the gamma ray than the wavelength of the X-ray? (b) The diagram shows wavelengths of visible light. Which ray has a wavelength closer to the wavelength of dark blue light?
Big Ideas Math Answer Key Grade 8 Chapter 8 Exponents and Scientific Notation 8.5 3
Answer:
a.10 or 100 times shorter is the wavelength of the
gamma ray than the wavelength of the X-ray.
b. X-ray has a wave length closer to the wavelength of dark blue light. 

Explanation:
Given a physicist observes a gamma ray with a wavelength
of 0.00000000135 millimeter is nearly or approximately
equal to ≈ 0.000000001 millimeter = 1 X 10-9  millimeters,
and an X-ray with a wavelength 0.00000012 millimeter
is nearly or approximately equal to ≈ 0.0000001 millimeter=
1 X 10-7  millimeters so now comparing wavelength
of the gamma ray with the wavelength of the X-ray,
1 X 10-9  millimeters with 1 X 10-7  millimeters so
we have 10 or 100 times shorter is the wavelength of the
gamma ray than the wavelength of the X-ray.
b. We have dark blue wavelength as 4 X 10-4  millimeters so
it is close to an X-ray with a wavelength 0.00000012 millimeter
is nearly or approximately equal to ≈ 0.0000001 millimeter =
1 X 10-7  millimeters. So X-ray has a wave length closer to the
wavelength of dark blue light.  

Estimating Quantities Homework & Practice 8.5

Review & Refresh

Simplify. Write the expression using only positive exponents.
Question 1.
3x-5
Answer:
3x-5= \(\frac{3}{x^{5}}\)

Explanation:
Given expression as 3x-5 we write the expression using only
positive exponents for 3 X x-5  as we write first x-5 we have
=
\(\frac{1}{x^{5}}\) now 3 X \(\frac{1}{x^{5}}\) = \(\frac{3}{x^{5}}\).

Question 2.
d0 • d-4
Answer:
d0 • d-4 = \(\frac{1}{d^{4}}\)

Explanation:
Given d0 • d-4 we write the expression as positive exponents by using

so d0 X \(\frac{1}{d^{4}}\) as  we know d0 =1,
d0 X \(\frac{1}{d^{4}}\)=1 X \(\frac{1}{d^{4}}\) or \(\frac{1}{d^{4}}\).

Question 3.
\(\frac{a^{6}}{2 a^{11}}\)
Answer:
\(\frac{a^{6}}{2 a^{11}}\) = \(\frac{1}{2 a^{5}}\)

Explanation:
Given expression as \(\frac{a^{6}}{2 a^{11}}\) we use rule for finding
\(\frac{a^{m}}{a^{n}}\), a quotient of two powers with the same base as am-n
\(\frac{a^{6}}{a^{11}}\) = a6-11 = a-5 for a-5 we use

Write an equation in point-slope form of the line that passes through the given point and has the given slope.
Question 4.
(- 1, 2); m = – \(\frac{1}{3}\)
Answer:
The equation in point-slope form is (y-2)= – \(\frac{1}{3}\) (x+1)

Explanation:
Given (- 1, 2); m = – \(\frac{1}{3}\) we know
for straight-line equations  the “point-slope” form with
(x1, y1) and a slope m we have formula y – y1 = m(x – x1)
here we have  (x1, y1) are (-1,2) and a slope m is – \(\frac{1}{3}\),
So the equation in point-slope form is (y-2)= – \(\frac{1}{3}\) (x+1).

Question 5.
(3, 4); m = \(\frac{3}{4}\)
Answer:
The equation in point-slope form is (y-4)= \(\frac{3}{4}\) (x-3)

Explanation:
Given (3, 4); m = \(\frac{3}{4}\) we know
for straight-line equations  the “point-slope” form with
(x1, y1) and a slope m we have formula y – y1 = m(x – x1)
here we have  (x1, y1) are (3,4) and a slope m is \(\frac{3}{4}\),
So the equation in point-slope form is (y-4)= \(\frac{3}{4}\) (x-3).

Question 6.
(1, – 4); m = – 2
Answer:
The equation in point-slope form is (y+4)=-2 (x-1)

Explanation:
Given (1, -4); m = -2 we know
for straight-line equations  the “point-slope” form with
(x1, y1) and a slope m we have formula y – y1 = m(x – x1)
here we have  (x1, y1) are (1,4) and a slope m is – 2,
So the equation in point-slope form is (y+4)= -2(x-1).

Concepts, Skills, & Problem Solving

APPROXIMATING NUMBERS Match the number with its closest approximation. (See Exploration 2, p. 343.)
Question 7.
0.000618
Answer:
0.000618 ≈ 0.0006 ≈ 6 X 10-4
So we match with B with its closest approximation

Explanation:
0.000618 is nearly or approximately equal to ≈ 0.0006 ≈ 6 X 10-4 ,
we have divided 6 by 10 followed by 4 zeros,
therefore we match B with its closest approximation.

Question 8.
7,257,993,201
Answer:
7,257,993,201 ≈ 7,000,000,000 ≈ 7 X 109
therefore we match D with its closest approximation.

Explanation:
Given number is 7,257,993,201 nearly or approximately
equal to ≈ 7,000,000,000 so we have 7 followed by 9 zeros,
so we write as 7 X 109 therefore we match D with its closest approximation.

Question 9.
0.0006781004
Answer:
0.0006781004 ≈ 0.0007 ≈ 7 X 10-4
So we match with C with its closest approximation

Explanation:
0.000618 is nearly or approximately equal to ≈ 0.0007 ≈ 7 X 10-4 ,
we have divided 7 by 10 followed by 4 zeros,
therefore we match C with its closest approximation.

Question 10.
782,309,441
Answer:
782,309,441≈ 800,000,000 ≈ 8 X 108
therefore we match A with its closest approximation.

Explanation:
Given number is 782,309,441 nearly or approximately
equal to ≈ 800,000,000 so we have 8 followed by 8 zeros,
so we write as 8 X 108 therefore we match A with its closest approximation.

A. 8 × 108
B. 6 × 10-4
C. 7 × 10-4
D. 7 × 109

APPROXIMATING A LARGE NUMBER Round the number. Write the result as a product of a single digit and a power of 10.
Question 11.
414,148,636,008
Answer:
414,148,636,008 ≈ 4 X 1011

Explanation:
Given number is 414,148,636,008 nearly or approximately
equal to ≈ 400,000,000,000 so we have 4 followed by 11 zeros,
so we write as 4 X 1011.

Question 12.
231,210
Answer:
231,210 ≈ 2 X 105

Explanation:
Given number is 231,210 nearly or approximately
equal to ≈ 200,000 so we have 2 followed by 5 zeros,
so we write as 2 X 105.

Question 13.
28,007,806,203
Answer:
28,007,806,203 ≈ 3 X 1010

Explanation:
Given number is  28,007,806,203 nearly or approximately
equal to ≈ 30,000,000,000 so we have 3 followed by 10 zeros,
so we write as 3 X 1010.

Question 14.
38,108,996,999
Answer:
38,108,996,999 ≈ 4 X 1010

Explanation:
Given number is  38,108,996,999 nearly or approximately
equal to ≈ 40,000,000,000 so we have 4 followed by 10 zeros,
so we write as 4 X 1010.

Question 15.
1,003,111,391,008
Answer:
1,003,111,391,008 ≈ 1 X 1012

Explanation:
Given number is 1,003,111,391,008 nearly or approximately
equal to ≈ 1,000,000,000,000 so we have 1 followed by 12 zeros,
so we write as 1 X 1012.

Question 16.
627,638,538
Answer:
627,638,538 ≈ 6 X 108

Explanation:
Given number is 627,638,538 nearly or approximately
equal to ≈ 6,00,000,000 so we have 6 followed by 8 zeros,
so we write as 6 X 108.

Question 17.
APPROXIMATING A LARGE NUMBER
A company earns $518,204,500. Round the number. Write the result as a product of a single digit and a power of 10.
Big Ideas Math Answer Key Grade 8 Chapter 8 Exponents and Scientific Notation 8.5 4
Answer:
$518,204,500 ≈ 5 X 108 dollars

Explanation:
Given number is $518,204,500 nearly or approximately
equal to ≈ 500,000,000 so we have 5 followed by 8 zeros,
so we write as 5 X 108 dollars.

APPROXIMATING A SMALL NUMBER Round the number.
Write the result as a product of a single digit and a power of 10.
Question 18.
0.00000124
Answer:
0.00000124 ≈ 1 X 10-6

Explanation:
As 0.00000124 is nearly or approximately equal to ≈ 0.000001 ≈
1 X 0.000001 = 1 X 10-6 or we have divided 1 by 10 followed by 6 zeros.

Question 19.
0.00003946
Answer:
0.00003946 ≈ 4 X 10-5

Explanation:
As 0.00003946 is nearly or approximately equal to ≈ 0.00004 ≈
4 X 0.00001= 4 X 10-5 or we have divided 4 by 10 followed by 5 zeros.

Question 20.
0.00001726
Answer:
0.00001726 ≈ 2 X 10-5

Explanation:
As 0.00001726 is nearly or approximately equal to ≈ 0.00002 ≈ 2 X 10-5,
or we have divided 2 by 10 followed by 5 zeros.

Question 21.
0.00063718
Answer:
0.00063718 ≈ 6 X 10-4

Explanation:
As 0.00063718 is nearly or approximately equal to ≈ 0.0006 ≈ 6 X 10-4,
or we have divided 6 by 10 followed by 4 zeros.

Question 22.
0.00000000305
Answer:
0.00000000305 ≈ 3 X 10-9

Explanation:
As 0.00000000305 is nearly or approximately
equal to ≈ 0.000000003 ≈ 3 X 10-9,
or we have divided 3 by 10 followed by 9 zeros.

Question 23.
0.000000000994
Answer:
0.000000000994 ≈ 1 X 10-9

Explanation:
As 0.000000000994 is nearly or approximately
equal to ≈ 0.000000001 ≈ 1 X 10-9,
or we have divided 1 by 10 followed by 9 zeros.

Question 24.
YOU BE THE TEACHER
Your friend rounds 0.000468 to the nearest ten thousandth and writes the result as a product of a single digit and a power of 10. Is your friend correct? Explain your reasoning.
Big Ideas Math Answer Key Grade 8 Chapter 8 Exponents and Scientific Notation 8.5 5
Answer:
Yes, Friend is correct as 0.000468 to the nearest ten thousandth is 5 X 10-4

Explanation:
As 0.000468 is nearly or approximately to the nearest
ten thousandth is equal to ≈ 0.0005 ≈ 5 X 0.0001= 5 X 10-4,
which is equal to the value of friend, So Yes, Friend is correct
as 0.000468 to the nearest ten thousandth is 5 X 10-4.

Question 25.
APPROXIMATING A QUANTITY
A series of mystery books contains 2,029,242 words. A series of science fiction books contains about 3.5 times the number of words as the mystery book series. What is the approximate number of words in the science fiction book series?
Answer:
The approximate number of words in the
science fiction book series are 7 X 106 words.

Explanation:
Given a series of mystery books contains 2,029,242 words,
a series of science fiction books contains about 3.5 times the
number of words as the mystery book series.
therefore approximate number of words in the science
fiction book series are 3.5 X 2,029,242 as 2,029,242 is
approximately equal to ≈ 2 X 1000,000 so 3.5 X 2 X 1,000,000
= 7 X 106 words.

Question 26.
APPROXIMATING A QUANTITY
A volcanic eruption ejects about 43,600,000,000 cubic feet of volcanic rock. A smaller volcanic eruption ejects about 75% of this amount. What is the approximate amount of volcanic rock that the smaller volcanic eruption ejects?
Big Ideas Math Answer Key Grade 8 Chapter 8 Exponents and Scientific Notation 8.5 6
Answer:
The approximate amount of volcanic rock that the
smaller volcanic eruption ejects is 3 X 1010 cubic feet.

Explanation:
Given a volcanic eruption ejects about 43,600,000,000 cubic feet
of volcanic rock. A smaller volcanic eruption ejects about 75% of
this amount, So the smaller volcanic eruption is 75% X 43,600,000,000
we get \(\frac{75}{100}\) X 43,600,000,000 = 75 X 43,6000000 =
32,700,000,000 is approximately equal to ≈ 3 X 10,000,000,000 = 3 X 1010 cubic feet.

Question 27.
STRUCTURE
Find a number that is approximately 1.5 times 61,040,000,100.
Write the result as the product of a single digit and a power of 10.
Answer:
The number approximate value of 1.5 times 61,040,000,100 is 9 X 1010

Explanation:
The number approximate value of 1.5 times 61,040,000,100 is
1.5 X 61,040,000,100 = 91,560,000,150 is approximately
equal to ≈ 9 X 10,000,000,000 = 9 X 1010 .

Question 28.
APPROXIMATING A QUANTITY
A mitochondrion has a diameter of about 0.00000031 meter. The diameter of a chloroplast is about 3 times that of the mitochondrion. What is the approximate diameter of the chloroplast?
Big Ideas Math Answer Key Grade 8 Chapter 8 Exponents and Scientific Notation 8.5 7
Answer:
The approximate diameter of the chloroplast is 9 X 10-7 meters.

Explanation:
A mitochondrion has a diameter of about 0.00000031 meter.
The diameter of a chloroplast is about 3 times that of the
mitochondrion. So the approximate diameter of the chloroplast is
3 X 0.00000031 = 0.00000093 is approximately
equal to ≈ 9 X 0.0000001 = 9 X 10-7 meters.

Question 29.
MODELING REAL LIFE
A photo taken with a smartphone has 1,227,104 pixels. A photo taken with a camera has 11,943,936 pixels. Approximately how many times more pixels are in the photo taken with the camera?
Answer:
10 times more pixels are in the photo taken with the camera.

Explanation:
Given a photo taken with a smartphone has 1,227,104 pixels and
A photo taken with a camera has 11,943,936 pixels.
Number of times the pixels are in the photo taken with the camera is
\(\frac{11943936}{1227104}\) = 9.7334341669 is approximately
equal to ≈ 10, So 10 times more pixels are in the photo taken with the camera.

Question 30.
MODELING REAL LIFE
A star has a core temperature of about 115,000,000°F. The temperature of a lightning strike is about 10,300°F. Approximately how many times hotter is the core temperature of the star than the temperature of the lightning strike?
Answer:
11165 times hotter is the core temperature of the star than the
temperature of the lightning strike

Explanation:
Given a star has a core temperature of about 115,000,000°F and
the temperature of a lightning strike is about 10,300°F.
Number of times hotter is the core temperature of the star
more than the temperature of the lightning strike is
\(\frac{115000000}{10300}\) = 11,165.048543689
is approximately equal to ≈ 11165, So 11165 times hotter is the
core temperature of the star than the temperature of the lightning strike.

Question 31.
REASONING
The table shows the diameters of five types of animal hairs.
Big Ideas Math Answer Key Grade 8 Chapter 8 Exponents and Scientific Notation 8.5 8
a. Order the hair types from greatest to least diameter.
b. What unit should be used to represent these data? Explain your reasoning.
Answer:
a. The hair types from greatest to least diameter are
Cow, Buffalo, Camel, Donkey, Rat.
b. Millimeters unit should be used to represent these data.

Explanation:
a. Given the table shows the diameters of five types of animal hairs as
Buffalo = 0.00011 is approximately equal to ≈ 1 X 10-4
Rat       = 0.00004 is approximately equal to ≈ 4 X 10-5
Camel  = 0.00008 is approximately equal to ≈ 8 X 10-5
Cow    = 0.00016 is approximately equal to ≈ 1 X 10-4
Donkey = 0.00005 is approximately equal to ≈ 5 X 10-5,
Now from greatest to least  diameters are
1 X 10-4 > 1 X 10-4 > 8 X 10-5 > 5 X 10-5> 4 X 10-5,
therefore the hair types from greatest to least diameter are
Cow, Buffalo, Camel, Donkey, Rat.
b. As the smallest unit of length is millimeter,
Millimeter is the smallest common unit of length that is represented as
‘mm’. The relation between mm and m is 1 meter = 1000 millimeter.
So millimeters unit should be used to represent these data.

Question 32.
PROBLEM SOLVING
The distance between New York City and Princeton is about 68,500 meters. The distance between New York City and San Antonio is about 40 times this distance. What is the approximate distance between New York City and San Antonio? Write the result as the product of a single digit and a power of 10.
Answer:
The approximate distance between New York City and San Antonio is
3 X 106 meters.

Explanation:
Given the distance between New York City and Princeton
is about 68,500 meters. The distance between New York City
and San Antonio is about 40 times this distance.
So the approximate distance between New York City and San Antonio is
40 X 68500 =  2,740,000 is approximately equal to ≈ 3 X 1,000,000 = 3 X 106 meters.

Question 33.
REASONING
Is 5 × 106 a better approximation of 5,447,040 or 5,305,004? Explain.
Answer:
5 × 106 a better approximation of 5,305,004

Explanation:
We have 5,447,040 or 5,305,004 we take approximate values ,
now 5 × 106 is approximately equal to ≈ 5,000,000 now if
we see for 5,000,000 the more near value is 5,305,004 than the
5,447,040. So 5 × 106 a better approximation of 5,305,004.

Question 34.
DIG DEEPER!
A proton weighs 0.00000000000167 nanogram. About how much do 8 protons weigh? Write the result as the product of a single digit and a power of 10. Is your answer an overestimate or an underestimate?
Answer:
The weight of 8 protons is 1 × 10-11
My answer is overestimate.

Explanation:
A proton weighs 0.00000000000167 nanogram.
So 8 protons weighs is 8 X 0.00000000000167 =1.336 X 10-11
is approximately equal to ≈ 1 X 10-11.
My answer is overestimate given proton weighs
0.00000000000167 nanogram is approximately equal to ≈ 1 X 10-12
and the weight of 8 protons is 1 × 10-11 if we see 1 X 10-11 > 1 X 10-12 ,
So my answer is overestimate.

Lesson 8.6 Scientific Notation

EXPLORATION 1

Work with a partner. Use a graphing calculator.
a. Experiment with multiplying very large numbers until your calculator displays an answer that is not in standard form. What do you think the answer means?
b. Enter the function y = 10x into your graphing calculator. Use the table feature to evaluate the function for positive integer values of x until the calculator displays a y-value that is not in standard form. Do the results support your answer in part(a)? Explain.
Big Ideas Math Answers 8th Grade Chapter 8 Exponents and Scientific Notation 8.6 1
c. Repeat part(a) with very small numbers.
d. Enter the function y = (\(\frac{1}{10}\))x into your graphing calculator. Use the table feature to evaluate the function for positive integer values of x until the calculator displays a y-value that is not in standard form. Do the results support your answer in part(c)? Explain.
Big Ideas Math Answers 8th Grade Chapter 8 Exponents and Scientific Notation 8.6 2

a. Means the answer is big so answer is not in standard form it
will be in scientific notation.
b. Until at x =6 the calculator displays a y-value that is not in
standard form. Yes the results support my answer in part(a),
c. Means the answer is small so answer is not in standard form it
will be in scientific notation.
d. Until at x =6 the calculator displays a y-value that is not in
standard form. Yes the results support my answer in part(c).

Explanation:
a. If multiplying very large numbers until my calculator
displays an answer that is not in standard form. We use
scientific notation it is just a shorthand way of expressing
gigantic numbers also  known as an exponential form,
scientific notation has been one of the oldest mathematical approaches.
If numbers are too big  to be simply calculated we refer to
scientific notation to handle these circumstances.
For example 4.5 X 109 years or (on a calculator)
4.5E9 years (1 billion in scientific notation means
10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10)
b. Entering the function y = 10x into graphing calculator.
Using the table feature we evaluate the function for positive integer
values of until x =6 the calculator displays a y-value that is not in standard form.
Yes the results support my answer in part(a) as we multiplied
very large number, calculator displays an answer that is not in standard form.
c. If multiplying very small numbers until my calculator
displays an answer that is not in standard form. We use
scientific notation it is just a shorthand way of expressing
numbers also known as an exponential form,
For example 0.000003426
Step 1: Move the decimal so that there is only one digit in front of the decimal.
0.000003.426
Step 2: Count the number of moves from the original decimal to the new position.
0.000003.426, There are 6 moves
Step 3: Write the new number as a product with a power of ten.
3.426 x 10-6 , 3E-6 the number of moves becomes the exponent.
d. Entering the function y = (\(\frac{1}{10}\))x  into graphing calculator.
Using the table feature we evaluate the function for positive integer
values of until x =6 the calculator displays a y-value that is not in standard form.
Yes the results support my answer in part(c) as we multiplied
very small number, calculator displays an answer that is not in standard form.

Big Ideas Math Answers 8th Grade Chapter 8 Exponents and Scientific Notation 8.6 3

Try It

Write the number in scientific notation.
Question 1.
50,000
Answer:
50,000 = 5 X 104

Explanation:
Given number is 50,000 in scientific notation
we write a number so that it has single digit to
the left of decimal sign and is multiplied by an integer power of 10.
So 50,000 = 5 X 10000 = 5 X 104.

Question 2.
25,000,000
Answer:
25,000,000 = 25 X 106

Explanation:
Given number is 25,000,000 in scientific notation
we write a number so that it has single digit to
the left of decimal sign and is multiplied by an integer power of 10.
So 25,000,000 = 25 X 1000000 = 25 X 106.

Question 3.
683
Answer:
683 = 6.83 X 102

Explanation:
Given number is 683 in scientific notation
we write a number so that it has single digit to
the left of decimal sign and is multiplied by an integer power of 10.
So 683 = 6.83 X 100 = 6.83 X 102 .

Question 4.
0.005
Answer:
0.005 = 5 X 10-3

Explanation:
Given number is 0.005 in scientific notation
we write a number so that it has single digit to
the left of decimal sign and is multiplied by an integer power of 10.
So 0.005 = 5 X 0.001 = 5 X 10-3.

Question 5.
0.00000033
Answer:
0.00000033 = 3.3 X 10-7

Explanation:
Given number is 0.00000033 in scientific notation
we write a number so that it has single digit to
the left of decimal sign and is multiplied by an integer power of 10.
So 0.00000033 = 3.3 X 0.0000001 = 3.3 X 10-7.

Question 6.
0.000506
Answer:
0.000506 = 5.06 X 10-4

Explanation:
Given number is 0.000506 in scientific notation
we write a number so that it has single digit to
the left of decimal sign and is multiplied by an integer power of 10.
So 0.000506 = 5.06 X 0.0001 = 5.06 X 10-4.

Write the number in standard form.
Question 7.
6 × 107
Answer:
6 × 107 = 60,000,000

Explanation:
Given 6 × 10the number in standard form is
6 X 10,000,000 , So 6 × 107 = 60,000,000.

Question 8.
9.9 × 10-5
Answer:
9.9 × 10-5 = 0.000099

Explanation:
Given 9.9 × 10-5  the number in standard form is
1 × 0.000099 = 0.000099.

Question 9.
1.285 × 104
Answer:
1.285 × 10= 12850

Explanation:
Given 1.285 × 104 the number in standard form is
1.285 X 10,000 , So 1.285 × 104 = 12850.

Self-Assessment for Concepts & Skills
Solve each exercise. Then rate your understanding of the success criteria in your journal.

WRITING NUMBERS IN SCIENTIFIC NOTATION Write the number in scientific notation.
Question 10.
675,000,000
Answer:
675,000,000 = 6.75 X 108

Explanation:
Given number is 675,000,000 in scientific notation
we write a number so that it has single digit to
the left of decimal sign and is multiplied by an integer power of 10.
So 675,000,000 = 6.75 X 1,00,000,000 = 6.75 X 108 .

Question 11.
0.000000084
Answer:
0.000000084 = 8.4 X 10-8

Explanation:
Given number is 0.000000084 in scientific notation
we write a number so that it has single digit to
the left of decimal sign and is multiplied by an integer power of 10.
So 0.000000084 = 8.4 X 0.00000001 = 8.4 X 10-8.

Question 12.
0.000012001
Answer:
0.000012001 = 1 X 10-5

Explanation:
Given number is 0.000012001 first we write
in approximately equal to ≈0.00001 now in scientific notation
we write a number so that it has single digit to
the left of decimal sign and is multiplied by an integer power of 10.
So 0.00001= 1 X 0.00001 = 1 X 10-5.

WRITING NUMBERS IN STANDARD FORM Write the number in standard form.
Question 13.
8 × 10-7
Answer:
8 × 10-7 = 0.0000008

Explanation:
Given 8 × 10-7  the number in standard form is
8 × 0.0000001 = 0.0000008.

Question 14.
3.876 × 107
Answer:
3.876 × 107 = 38,760,000

Explanation:
Given 3.876 × 10the number in standard form is
3.876 X 10,000,000 = 38,760,000 so 3.876 × 107 = 38,760,000.

Question 15.
1.11 × 10-5
Answer:
1.11 × 10-5 = 0.0000111

Explanation:
Given 1.11 × 10-5  the number in standard form is
1.11 × 0.00001 = 0.0000111.

Question 16.
WHICH ONE DOESN’T BELONG?
Which number does not belong with the other three? Explain.
Big Ideas Math Answers 8th Grade Chapter 8 Exponents and Scientific Notation 8.6 4
Answer:
d. 10 X 9.2-13  does not belong with the other three

Explanation:
Given 2.8 X 1015 , 4.3 X 10-30 , 1.05 X 1028 all are single digit to
the left of decimal sign and is multiplied by an integer power of 10,
but only 10 X  9.2-13  has power for 9.213 ,so 10 X  9.2-13
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 17.
A series of movies is about 3.285 × 104 seconds long.
How long does it take to watch the series twice?
Express your answer using more-appropriate units.
Answer:
The time it takes to watch the series twice is 18.25 hours.

Explanation:
Given a series of movies is about 3.285 × 104 seconds long
and the time it takes to watch the series twice is
2 X 3.285 × 104 seconds = 6.57 X 104 seconds,
now we convert seconds into hours as 1 hour = 3600 seconds,
so \(\frac{6.57}{3600}\) X 104 = 18.25 hours, therefore
the time it takes to watch the series twice is 18.25 hours.

Question 18.
The total power of a space shuttle during launch is the sum of the power from its solid rocket boosters and the power from its main engines. e power from the solid rocket boosters is 9,750,000,000 watts. What is the power from the main engines?
Big Ideas Math Answers 8th Grade Chapter 8 Exponents and Scientific Notation 8.6 5
Answer:
The power from the main engines is 1.99 X 109 watts.

Explanation:
Given the power from the solid rocket boosters is
9,750,000,000 watts is approximately equal to ≈9.75 X109 watts
Let the main engine power be x, Total Power =
power from the solid rocket boosters + power from the main engines,
1.174 X 1010 watts = 9.75 X 109 watts + x, So x = (11.74 – 1.99) X 109 watts
= 9.75 X 109 watts, therefore the power from the main engines is 1.99 X 109 watts.

Question 19.
The area of a trampoline is about 1.8 × 104 square inches.
Write this number in standard form. Then represent the area of the
trampoline using more-appropriate units.
Answer:
The area of the trampoline is 125 square feet

Explanation:
Given the area of a trampoline is about 1.8 × 104 square inches,
so we get 1.8 × 104 as 18000 square inches, now we
represent the area of the trampoline using more-appropriate units
in square feet, We know 1 square feet = 144 square inches
= \(\frac{18000}{144}\) = 125 square feet,
so the area of the trampoline is 125 square feet.

Question 20.
DIG DEEPER!
The epidermis, dermis, and hypodermis are layers of your skin. The dermis is about 3.5 millimeters thick. The epidermis is about 1.25 × 10-3 meter thick. The hypodermis is about 0.15 centimeter thick. What is the difference in thickness of the thickest layer and the thinnest layer? Justify your answer.
Answer:
The difference in thickness of the thickest layer and
the thinnest layer is 0.002 meters

Explanation:
Given the epidermis, dermis, and hypodermis are layers of your skin.
The dermis is about 3.5 millimeters thick. The epidermis is about
1.25 × 10-3 meter thick. The hypodermis is about 0.15 centimeter thick.
So the difference in thickness of the thickest layer and the thinnest layer is
1 millimeter = 0.001 meters,
dermis = 3.5 X 0.001 = 0.0035 meters,
epidermis = 1.25 X 10-3 = 0.00125 meters,
hypodermis = 0.15 centimeter, we know 1 centimeter = 0.01 meter,
so 0.15 X 0.01 = 0.0015 meters, we have the thickest as 0.0035 meters and
thinnest is 0.00125 meters , So the difference in thickness of
the thickest layer and the thinnest layer is 0.0035 – 0.00125 =
0.00225 is approximately equal to ≈ 0.002 meters.

Scientific Notation Homework & Practice 8.6

Review & Refresh

Round the number. Write the result as the product of a single digit and a power of 10.
Question 1.
0.00000129
Answer:
0.00000129 ≈ 1 X 10-6

Explanation:
As 0.00000129 is nearly or approximately equal to ≈ 0.000001 ≈
1 X 0.000001 = 1 X 10-6 or we have divided 1 by 10 followed by 6 zeros.

Question 2.
4,241,933,200
Answer:
4,241,933,200 ≈ 4 X 109

Explanation:
Given number is 4,241,933,200 nearly or approximately
equal to ≈ 2,000,000,000 so we have 4 followed by 9 zeros,
so we write as 4 X 109.

Question 3.
0.0000001801
Answer:
0.0000001801≈ 2 X 10-7

Explanation:
As 0.0000001801 is nearly or approximately equal to ≈ 0.0000002 ≈
2 X 0.0000001 = 2 X 10-7 or we have divided 2 by 10 followed by 7 zeros.

Question 4.
879,679,466
Answer:
879,679,466 ≈ 9 X 108

Explanation:
Given number is 879,679,466 nearly or approximately
equal to ≈ 9,00,000,000 so we have 9 followed by 8 zeros,
so we write as 9 X 108.

Write the product using exponents.
Question 5.
4 • 4 • 4 • 4 • 4
Answer:
4 • 4 • 4 • 4 • 4 = (4)5  

Explanation:
We write the product 4 • 4 • 4 • 4 • 4 in exponents as (4)5
because 4 is multiplied by 5 times.

Question 6.
3 • 3 • 3 • y • y • y
Answer:
3 • 3 • 3 • y • y • y = 33y3= (3y)3

Explanation:
We write the product 3 • 3 • 3• y • y • y in exponents as (3y)3
because 3 is multiplied by 3 times and y is also multiplied by 3 times.

Question 7.
(- 2) • (- 2) • (- 2)
Answer:
(- 2) • (- 2) • (- 2) = (-2)3

Explanation:
We write the product (-2) • (-2) • (-2) in exponents as (-2)3
because -2 is multiplied by 3 times.

Concepts, Skills, &Problem Solving

USING A GRAPHING CALCULATOR Use a graphing calculator to evaluate x the function when 10. Write the number in standard form. (See Exploration 1, p. 349.)
Question 8.
y = (\(\frac{1}{10}\))x
Answer:
The number in standard form is 0.0000000001

Explanation:
Given y = (\(\frac{1}{10}\))x and x = 10 so
y = (\(\frac{1}{10}\))10 so 1 X 10-10 = 1X 0.0000000001
= 0.0000000001 therefore the number in standard form is 0.0000000001.

Question 9.
y = 20x
Answer:
The number in standard form is 10,240,000,000,000

Explanation:
Given y = 20x and x = 10 so y= (20)10 = 10,240,000,000,000,
therefore the number in standard form is 10,240,000,000,000.

WRITING NUMBERS IN SCIENTIFIC NOTATION Write the number in scientific notation.
Question 10.
0.0021
Answer:
0.0021 = 2.1 X 10-3

Explanation:
Given number is 0.0021 in scientific notation
we write a number so that it has single digit to
the left of decimal sign and is multiplied by an integer power of 10.
So 0.0021 = 2.1 X 0.001 = 2.1 X 10-3.

Question 11.
5,430,000
Answer:
5,430,000 = 5.43 X 106

Explanation:
Given number is 5,430,000 in scientific notation
we write a number so that it has single digit to
the left of decimal sign and is multiplied by an integer power of 10.
So 5,430,000 = 5.43 X 1,000,000 = 5.43 X 106.

Question 12.
321,000,000
Answer:
321,000,000 = 3.21 X 108

Explanation:
Given number is 321,000,000 in scientific notation
we write a number so that it has single digit to
the left of decimal sign and is multiplied by an integer power of 10.
So 321,000,000 = 3.21 X 1,00,000,000 = 3.21 X 108.

Question 13.
0.00000625
Answer:
0.00000625 = 6.25 X 10-6

Explanation:
Given number is 0.00000625 in scientific notation
we write a number so that it has single digit to
the left of decimal sign and is multiplied by an integer power of 10.
So 0.00000625 = 6.25 X 0.000001 = 6.25 X 10-6.

Question 14.
0.00004
Answer:
0.00004 = 4 X 10-5

Explanation:
Given number is 0.00004 in scientific notation
we write a number so that it has single digit to
the left of decimal sign and is multiplied by an integer power of 10.
So 0.00004 = 4 X 0.00001 = 4 X 10-5.

Question 15.
10,700,000
Answer:
10,700,000 = 1.07 X 107

Explanation:
Given number is 10,700,000 in scientific notation
we write a number so that it has single digit to
the left of decimal sign and is multiplied by an integer power of 10.
So 10,700,000 = 1.07 X 10,000,000 = 1.07 X 107.

Question 16.
45,600,000,000
Answer:
45,600,000,000 = 4.56 X 1010

Explanation:
Given number is 45,600,000,000 in scientific notation
we write a number so that it has single digit to
the left of decimal sign and is multiplied by an integer power of 10.
So 45,600,000,000 = 4.56 X 10,000,000,000 = 4.56 X 1010.

Question 17.
0.000000000009256
Answer:
0.000000000009256 = 9.256 X 10-12

Explanation:
Given number 0.000000000009256 is in scientific notation
we write a number so that it has single digit to
the left of decimal sign and is multiplied by an integer power of 10.
So 0.000000000009256 = 9.256 X 0.000000000001 = 9.256 X 10-12.

Question 18.
840,000
Answer:
840,000 = 8.4 X 105

Explanation:
Given number is 840,000 in scientific notation
we write a number so that it has single digit to
the left of decimal sign and is multiplied by an integer power of 10.
So 840,000 = 8.4 X 100,000 = 8.4 X 105.

WRITING NUMBERS IN STANDARD FORM Write the number in standard form.
Question 19
7 × 107
Answer:
7 × 107 = 70,000,000

Explanation:
Given 7 × 10the number in standard form is
7 X 10,000,000 =70,000,000.

Question 20.
8 × 10-3
Answer:
8 X 10-3 = 0.008

Explanation:
Given 8 × 10-3  the number in standard form is
8 × 0.001 = 0.008.

Question 21.
5 × 102
Answer:
5 × 102 = 500

Explanation:
Given 5 × 102 the number in standard form is
7 X 100 = 500.

Question 22.
2.7 × 10-4
Answer:
2.7 × 10-4 = 0.00027

Explanation:
Given 2.7 × 10-4  the number in standard form is
2.7 × 0.0001 = 0.00027.

Question 23.
4.4 × 10-5
Answer:
4.4 × 10-5 = 0.000044

Explanation:
Given 4.4 × 10-5  the number in standard form is
4.4 × 0.00001 = 0.000044.

Question 24.
2.1 × 103
Answer:
2.1 × 103= 2,100

Explanation:
Given 2.1 × 103 the number in standard form is
2.1 X 1000 = 2,100.

Question 25.
1.66 × 109
Answer:
1.66 × 109 = 1,660,000,000

Explanation:
Given 1.66 × 109 the number in standard form is
1.66 X 1,000,000,000 = 1,660,000,000.

Question 26.
3.85 × 10-8
Answer:
3.85 × 10-8 = 0.0000000385

Explanation:
Given 3.85 × 10-8  the number in standard form is
3.85 × 0.00000001 = 0.0000000385.

Question 27.
9.725 × 106
Answer:
9.725 × 106 = 9,725,000

Explanation:
Given 9.725 × 106 the number in standard form is
9.725 X 1,000,000 = 9,725,000.

Question 28.
MODELING REAL LIFE
The U.S.Brig Niagara, a warship from the Battle of Lake Erie in 1813, uses about 28,300 feet of rope to operate its sails and spars. Write this number in scientific notation.
Big Ideas Math Answers 8th Grade Chapter 8 Exponents and Scientific Notation 8.6 6.1
Answer:
28,300 feet = 2.83 X 104

Explanation:
Given the U.S.Brig Niagara, a warship from the Battle of
Lake Erie in 1813, uses about 28,300 feet of rope to
operate its sails and spars this number in scientific notation as
we write a number in scientific notation has single digit to
the left of decimal sign and is multiplied by an integer power of 10.
So 28,300 = 2.83 X 10,000 = 2.83 X 104.

Question 29.
MODELING REAL LIFE
The radius of a fishing line is 2.5 × 10-4 feet. Write this ×number in standard form. Then write your answer using inches.
Answer:
The number in standard form is 0.00025 feet,
In inches it is 0.003 inches.

Explanation:
The radius of a fishing line is 2.5 × 10-4 feet,
The number in standard form is 2.5 X 0.0001 = 0.00025 feet,
Now to convert feet into inches we know 1 foot is 12 inch so
0.00025 X 12 = 0.003 inches.

Question 30.
MODELING REAL LIFE
Platelets are cell-like particles in the blood that help form blood clots.
Big Ideas Math Answers 8th Grade Chapter 8 Exponents and Scientific Notation 8.6 6
a. How many platelets are in 3 milliliters of blood? Write your answer in standard form.
b. An adult human body contains about 5 liters of blood. How many platelets are in an adult human body?
Answer:
a. There are 810,000,000 milliliters platelets in 3 milliliters of blood.
b. There are 1,350,000,000,000 milliliters platelets are in an adult human body.

Explanation:
a. Given blood has 2.7 X 108 platelets per milliliters,
So the number of platelets in 3 milliliters of blood is
3 X 2.7 X 108 = 8.1 X 108 = 810,000,000 milliliters.
b. An adult human body contains about 5 liters of blood,
So the number of platelets in an human body are, we know
1 liter = 1000 milliliters so 5 liters is equal to 5,000 milliliters,
5000 X 2.7 X 108 = 1,350,000,000,000 milliliters ,therefore
There are 1,350,000,000,000 milliliters platelets are in an adult human body.

CHOOSING APPROPRIATE UNITS Match each value with the most appropriate unit of measurement.
Big Ideas Math Answers 8th Grade Chapter 8 Exponents and Scientific Notation 8.6 7
Answer:
We match each value with the most appropriate unit of measurement
as 31. D, 32. C, 33. A , 34. B

Explanation:
Given
a. Height of a skyscraper is : 2.6 X 102 = 2.6  X 100 = 260,
so the most appropriate unit of measurement is meters as
it is used to measure big lengths.
b. Distance between two asteroids : 2.5 X 105 = 2.5 X 100,000 = 250,000
so the most appropriate unit of measurement is miles as
it is used to measure long distances.
c. Depth of bathtub: 1.6 X 101 = 16,
so the most appropriate unit of measurement is inches as
it is used to measure the length of small objects.
d. Length of memory chip : 7.8 X 100 = 7.8 X 1 = 7.8
so the most appropriate unit of measurement is millimeters as
it is used to measure very short lengths or thicknesses.

Question 35.
NUMBER SENSE
Describe how the value of a number written in scientific notation changes when you increase the exponent by 1.
Answer:
The number increases by 10 times

Explanation:
Whenever we increase the exponent by 1
in scientific notation the number becomes 10 times of
previous number. Example we have number 280 we write
number in scientific notation as single digit to
the left of decimal sign and is multiplied by an integer power of 10.
So 280 = 2.8 X 100 = 2.8 X 102 now if we increase exponent by 1 we get
2.8 X 103 as 2.8 X 1000 = 2800 here the value has increase by 10 times from
280 to 2800.

Question 36.
PROBLEM SOLVING
The area of the Florida Keys National Marine Sanctuary is about 9600 square kilometers. The area of the Florida Reef Tract is about 16.2% of the area of the sanctuary. What is the area of the Florida Reef Tract? Write your answer in scientific notation.
Big Ideas Math Answers 8th Grade Chapter 8 Exponents and Scientific Notation 8.6 8
Answer:
The area of the Florida Reef Tract is 1.5552 X 103 square kilometers

Explanation:
Given the area of the Florida Keys National Marine Sanctuary
is about 9600 square kilometers. The area of the Florida Reef
Tract is about 16.2% of the area of the sanctuary.
So the area of the Florida Reef Tract is \(\frac{16.2}{100}\) X 9600 =
1555.2 square kilometers, Now we convert 1555.2 in scientific notation as
single digit to the left of decimal sign and is multiplied
by an integer power of 10, So 1.552 X 1000 = 1.552 X 103 square kilometers,
therefore the area of the Florida Reef Tract is 1.5552 X 103 square kilometers.

Question 37.
REASONING
A gigameter is 1.0 × 106 kilometers. How many square kilometers
are in 5 square gigameters?
Answer:
There are  5 X 1012 kilometers2 are there in 5 square gigameters

Explanation:
Given a gigameter is 1.0 × 106 kilometers, So 1 Square kilometer =
(1.0 X 106)2 kilometers2 = 1.0 X 1012 kilometers2, So
5 gigameter2= 5 X 1012 kilometerstherefore there are
5 X 1012 kilometers2 are there in 5 square gigameter.

Question 38.
PROBLEM SOLVING
There are about 1.4 × 109 cubic kilometers of water on Earth.
About 2.5% of the water is freshwater. How much freshwater is on Earth?
Answer:
Fresh water available on earth is 3.5 X 107 kilometers3

Explanation:
Given there are about 1.4 × 109 cubic kilometers of water on Earth
and about 2.5% of the water is freshwater so the amount of
fresh water available is 2.5 % of 1.4 × 109 cubic kilometers
= \(\frac{2.5}{100}\) X 1.4 × 109 = 3.5 X 107 kilometers3 ,
therefore fresh water available on earth is 3.5 X 107 kilometers3 .

Question 39.
CRITICAL THINKING
The table shows the speed of light through each of five media. Determine in which media light travels the fastest and the slowest.
Big Ideas Math Answers 8th Grade Chapter 8 Exponents and Scientific Notation 8.6 9
Answer:
Fastest is Vacuum 3.0 X 108 m/sec and slowest is Glass 2.01 X 108 m/sec,

Explanation:
Given the table shows the speed of light through each of five media
Air is 6.7 X 108 mi/sec now convert into meters per second
6.7 X 108 X \(\frac{1609}{3600}\) = 2.99 X 108 m/sec,
Now speed in glass is 6.6 X 108 ft/sec now convert into meters per second
6.6 X 108 X 0.3048 = 2.01 X 108 m/sec,
Now speed in Ice is 2.3 X 105 km/sec now convert into meters per second
2.3 X 105 X 1000 = 2.3 X 108 m/sec, already Vacuum is in m/sec,
so now speed in Water is 2.3 X 1010 cm/sec now
convert into meters per second 2.3 X 1010 X \(\frac{1}{100}\) =
2.3 X 108 m/sec, Now
Medium            Speed in m/sec
Air                    2.99 X 108 m/sec
Glass                2.01 X 108 m/sec
Ice                    2.3 X 108 m/sec
Vacuum            3.0 X 108 m/sec
Water                2.3 X 108 m/sec , So now if we see at speeds the
fastest is Vacuum 3.0 X 108 m/sec and slowest is Glass  2.01 X 108 m/sec.

Question 40.

STRUCTURE
The mass of an atom or molecule is measured in atomic mass units. Which is greater, a carat or a milligram? Explain.
Big Ideas Math Answers 8th Grade Chapter 8 Exponents and Scientific Notation 8.6 10
Answer:
A carat is greater unit than milligram,

Explanation:
The mass of an atom or molecule is measured in atomic mass units.
One atomic mass unit(amu) is equivalent to
1 amu = 8.3 X 10-24 carat
1 amu = 1.66 X 10-21 milligram
now 8.3 X 10-24 carat = 1.66 X 10-21 milligram
8.3 carat = 1.66 X 103 milligram
So 1 carat = \(\frac{1660}{8.3}\) = 200 milligrams,
therefore A carat is greater unit than milligram.

Lesson 8.7 Operations in Scientific Notation

EXPLORATION 1

Adding and Subtracting in Scientific Notation
Work with a partner.
a. Complete the table by finding the sum and the difference of Expression 1 and Expression 2. Write your answers in scientific notation. Explain your method.
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation 8.7 1
b. Use your results in part(a) to explain how to find (a × 10n) + (b × 10n) and (a × 10n) – (b × 10n).
Answer:
a.
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation-16

b. (a × 10n) + (b × 10n) = (a + b ) X 10n
(a × 10n) – (b × 10n) = (a-b) X 10n

Explanation:
a. One of the properties of quantities with exponents is that
numbers with exponents can be added and subtracted
only when they have the same base and exponent.
Since all numbers in scientific notation have the same base (10),
we need only worry about the exponents.
To be added or subtracted, two numbers in scientific notation they
must be manipulated so that their bases have the same exponent,
this will ensure that corresponding digits in their coefficients
have the same place value.
So completed the table by finding
the sum and the difference of Expression 1 and Expression 2.
1. Sum = ( 3 X 104)  + ( 1 X 104) = ( 3 + 1) X 104 = 4 X 104,
Difference = ( 3 X 104)  – ( 1 X 104) = ( 3 – 1) X 104 = 2 X 104,
2. Sum = ( 4 X 10-3)  + ( 2 X 10-3) = ( 4 + 2) X 10-3 = 6 X 10-3,
Difference = ( 4 X 10-3)  – ( 2 X 10-3) = ( 4 – 2) X 10-3 = 2 X 10-3,
3. Sum=( 4.1 X 10-7)  + ( 1.5 X 10-7) = ( 4.1 + 1.5) X 10-7 = 5.6 X 10-7,
Difference = ( 4.1 X 10-7)  – ( 1.5 X 10-7) = ( 4.1 – 1.5) X 10-7 = 2.6 X 10-7,
4. Sum = ( 8.3 X 106)  + ( 1.5 X 106) = ( 8.3 + 1.5) X 106 = 9.8 X 106,
Difference = ( 8.3 X 106)  – ( 1.5 X 106) = ( 8.3 – 1.5) X 106 = 6.8 X 106.
b. (a × 10n) + (b × 10n) can be completed using the distributive property
of multiplication over addition, i.e., factor out the common factor 10n
as (a × 10n) + (b × 10n) = (a + b ) X 10n and
(a × 10n) – (b × 10n) can be completed using the distributive property
of multiplication over subtraction, i.e., factor out the common factor 10n
as (a × 10n) – (b × 10n) = (a-b) X 10n.

EXPLORATION 2

Multiplying and Dividing in Scientific Notation
Work with a partner.
a. Complete the table by finding the product and the quotient of
Expression 1 and Expression 2. Write your answers in scientific notation.
Explain your method.
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation 8.7 2
b. Use your results in part(a) to explain how to find (a × 10n) × (b × 10m) and
(a × 10n) ÷ (b ÷ 10m). Describe any properties that you use.
Answer:
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation-17
b. (a × 10n) × (b × 10m)  = (a × b) X (10n × 10m) = (a × b) X (10n+m) ,
Separately we multiply the coefficients and exponents and
we use Product of Powers Property for an • am = an+m
If product of two powers with the same base then
powers are added and (a × 10n) ÷ (b ÷ 10m) = a ÷ b X (10nm),
Separately we divide the coefficients and exponents.
Noted that when we are dividing exponential terms,
always subtract the denominator from the numerator.
We use the general rule for finding \(\frac{a^{m}}{a^{n}}\),
a quotient of two powers with the same base is am-n .

Explanation:
a. To multiply numbers in scientific notation these are the steps:
If the numbers are not in scientific notation, convert them.
Regroup the numbers using the commutative and
associative properties of exponents.
Now multiply the two
numbers written in scientific notation, we work out
the coefficients and the exponents separately.
we use the product rule; bm x b n = b(m + n) to multiply the bases.
Join the new coefficient to the new power of 10 to get the answer.
If the product of the coefficients is greater than 9, convert it to
scientific notation and multiply by the new power of 10.

To divide two numbers written in scientific notation, follow the steps below:
Separately we divide the coefficients and exponents.
For the division of bases, use the division rule of exponents,
where the exponents are subtracted.
Combine the result of coefficients by the new power of 10.
If the quotient from division of coefficients is not
less than 10 and greater than 1, convert it to scientific notation
and multiply it by the new power of 10.
Noted that when you dividing exponential terms,
always subtract the denominator from the numerator.
So completed the table by finding the product and the
quotient of Expression 1 and Expression 2.
Now first we do write Products as
1.(3 X 104)  X  ( 1 X 104) = (3 X 1) X ( 104+4) = 3 X 108 .
2.(4 X 103)  X  (2 X 102) = (4 X 2 )  X  ( 103+2) = 8 X 105.
3.(7.7 X 10-2)  X  (1.1 X 10-3) = (7.7 X 1.1 ) X (10-2+(-3)) = 8.47 X 10-5.
4.(4.5 X 105)  X  (3 X 10-1) = (4.5 X 3
)  X  ( 105+(-1)) = 13.5 X 104.
Now we write Quotients as separately and divide the
coefficients and exponents.
For the division of bases, use the division rule of exponents,
where the exponents are subtracted.
1. (3 X 104) ÷ ( 1 X 104) = (3 ÷ 1) X ( 104-4) = 3 X 100 = 3 X 1 = 3.
2. (4 X 103) ÷ (2 X 102) = (4 ÷ 2 )  X  ( 103-2) = 2 X 101 = 2 X 10 = 20.
3. (7.7 X 10-2) ÷ (1.1 X 10-3) = (7.7 ÷ 1.1 ) X (10-2-(-3)) =7 X 101 =7 X 10 = 70
4. (4.5 X 105)  ÷  (3 X 10-1) = (4.5 ÷ 3)  X  ( 105-(-1)) = 1.5 X 106.
b. Here we multiply the coefficients separately  and
use Product of Powers Property for an • am = an+m
we have rule of product of two powers with the same base then
we add powers, so (a × 10n) × (b × 10m)  =
(a × b) X (10n × 10m) = (a × b) X (10n+m).
Separately we divide the coefficients and exponents.
Noted that when we are dividing exponential terms,
always subtract the denominator from the numerator.
We use the general rule for finding \(\frac{a^{m}}{a^{n}}\),
a quotient of two powers with the same base is am-n .
So first we divide divide a ÷ b and then divide exponential terms
therefore (a × 10n) ÷ (b ÷ 10m) = a ÷ b X (10n-m).

Try It

Find the sum or difference.
Question 1.
(8.2 × 102) + (3.41 × 10-1)
Answer:
(8.2 × 102) + (3.41 × 10-1) =8203.41 X 10-1= 820.341

Explanation:
Given Expressions as (8.2 × 102) + (3.41 × 10-1)
as the powers of 10 differ we  need to modify before we factor.
We work around this by using our exponent property bm · bn = b (m+n)
to rewrite the 102 as 10-1 ·So we take 103 and then grouping the 103 with 8.2.
So 8.2 X 102 becomes  8200 X 10-1 therefore
(8.2 × 102) + (3.41 × 10-1) =(8.2 X 103 × 10-1) + (3.41 × 10-1) =
(8200 X  10-1) + (3.41 × 10-1)
using the distributive property of multiplication over addition,
we get (8200+3.41) X 10-1 =8203.41 X 10-1= 820.341

Question 2.
(7.8 × 10-5) – (4.5 × 10-5)
Answer:
(7.8 × 10-5) – (4.5 × 10-5) = 3.3 X 10-5

Explanation:
Given expressions as (7.8 × 10-5) – (4.5 × 10-5)
using the distributive property of multiplication over subtraction,
we get (7.8 – 4.5 ) X 10-5  = 3.3 X 10-5.

Find the product.
Question 3.
6 × (8 × 10-5)
Answer:
6 × (8 × 10-5) = 48 X 10-5

Explanation:
As given expression is 6 × (8 × 10-5) so we use
associative law of multiplication as a X (b X c) = (a X b ) X c,
So 6 × (8 × 10-5) = (6 × 8) × 10-5 = 48 X 10-5.

Question 4.
(7 × 102) × (3 × 105)
Answer:
(7 × 102) × (3 × 105) = 21 X 107

Explanation:
Given expressions as (7 × 102) × (3 × 105)
we work out the coefficients and the exponents separately.
and use the product rule; bm x b n = b(m + n) to multiply the bases.
So (7 × 102) × (3 × 105)  = (7 X 3) X (102 × 105) = 21 X (102+5) = 21 X 107 .

Question 5.
(2 × 104) × (6 × 10-7)
Answer:
(2 × 104) × (6 × 10-7) = 12 X 10-3

Explanation:
Given expressions as (2 × 104) × (6 × 10-7)
we work out the coefficients and the exponents separately.
and use the product rule; bm x b n = b(m + n) to multiply the bases.
So (2 × 104) × (6 × 10-7)  = (2 X 6) X (104 × 10-7) = 12 X (104-7) = 12 X 10-3 .

Question 6.
(3 × 108) × (9 × 103)
Answer:
(3 × 108) × (9 × 103) = 27 X 1011

Explanation:
Given expressions as (3 × 108) × (9 × 103)
we work out the coefficients and the exponents separately.
and use the product rule; bm x b n = b(m + n) to multiply the bases.
So (3 × 108) × (9 × 103)  = (3 X 9) X (108 × 103) = 27 X (108+3) = 27 X 1011 .

Find the quotient.

Question 7.
(9.2 × 1012) ÷ 4.6
Answer:
(9.2 × 1012) ÷ 4.6 = 2 X 1012

Explanation:
Given expression as (9.2 × 1012) ÷ 4.6,
Separately we divide the coefficients
and multiply exponent with base
so (9.2 ÷ 4.6) X 1012
= 2 X 1012.

Question 8.
(1.5 × 10-3) ÷ (7.5 × 102)
Answer:
(1.5 × 10-3) ÷ (7.5 × 102) = 0.2 X 10-5

Explanation:
Given expressions as (1.5 × 10-3) ÷ (7.5 × 102),
Separately we divide the coefficients and exponents.
We use the division rule of exponents,
where the exponents are subtracted.
Noted that when we are dividing exponential terms,
always subtract the denominator from the numerator.
So (1.5 × 10-3) ÷ (7.5 × 102) = (1.5 ÷ 7.5 ) X ( 10-3-2) = 0.2 X 10-5.

Question 9.
(3.75 × 10-8) ÷ (1.25 × 10-7)
Answer:
(3.75 × 10-8) ÷ (1.25 × 10-7) = 3 X 10-1 or 0.3

Explanation:
Given expressions as (3.75 × 10-8) ÷ (1.25 × 10-7),
Separately we divide the coefficients and exponents.
We use the division rule of exponents,
where the exponents are subtracted.
Noted that when we are dividing exponential terms,
always subtract the denominator from the numerator.
So (3.75 × 10-8) ÷ (1.25 × 10-7) =
(3.75 ÷ 1.25 ) X ( 10-8-(-7)) = 3 X 10-1= 0.3.

Question 10.
(9.2 × 106) ÷ (2.3 × 1012)
Answer:
(9.2 × 106) ÷ (2.3 × 1012) = 4 X 10-6

Explanation:
Given expressions as (9.2 × 106) ÷ (2.3 × 1012) ,
Separately we divide the coefficients and exponents.
We use the division rule of exponents,
where the exponents are subtracted.
Noted that when we are dividing exponential terms,
always subtract the denominator from the numerator.
So (9.2 × 106) ÷ (2.3 × 1012)  =
(9.2 ÷ 2.3 ) X ( 106-12) = 4 X 10-6.

Self-Assessment for Concepts & Skills
Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 11.
WRITING
Describe how to add or subtract two numbers written in scientific notation with
the same power of 10.
Answer:
Example (a × 10m) + (b × 10m) can be completed using the distributive property
of multiplication over addition, i.e., factor out the common factor 10m
as (a × 10m) + (b × 10m) = (a + b ) X 10m , Here we have the same power 10.

Explanation:
One of the properties of quantities with exponents is that
numbers with exponents can be added and subtracted
only when they have the same base and exponent.
Since all numbers in scientific notation have the same base (10),
we need only worry about the exponents.
To be added or subtracted, two numbers in scientific notation they
must be manipulated so that their bases have the same exponent,
this will ensure that corresponding digits in their coefficients
have the same place value.

Question 12.
NUMBER SENSE
Two numbers written in scientific notation have different powers of 10. Do you have to rewrite the numbers so they have the same power of 10 before multiplying or dividing? Explain.
Answer:
No need to rewrite the numbers so that they have
the same power of 10 before multiplying or dividing.

Explanation:
The steps to multiply two numbers in scientific notation is
multiply the coefficients round to the number of
significant figures in the coefficient with the smallest number
of significant figures and we a
dd the exponents.
The steps to divide two numbers in scientific notation is
divide the coefficients–round to the number of significant
figures in the coefficient with the smallest number of significant figures
and we subtract the exponents.
No need to rewrite the numbers so that they have
the same power of 10 before multiplying or dividing.
In multiplication we add exponents with same base and
in division we subtract exponents.
Noted that when we are dividing exponential terms,
always subtract the denominator from the numerator.

OPERATIONS IN SCIENTIFIC NOTATION Evaluate the expression. Write your answer in scientific notation.
Question 13.
(7.26 × 104) + (3.4 × 104)
Answer:
(7.26 × 104) + (3.4 × 104) = 10.66 X 104

Explanation:
Given expressions as (7.26 × 104) + (3.4 × 104) as it is
(a × 10n) + (b × 10n) can be completed using the distributive property
of multiplication over addition, i.e., factor out the common factor 10n
as (a × 10n) + (b × 10n) = (a + b ) X 10n so (7.26 × 104) + (3.4 × 104) =
(7.26 + 3.4 ) X 104 =10.66 X 104 .

Question 14.
(2.8 × 10-5) – (1.6 × 10-6)
Answer:
(2.8 × 10-5) – (1.6 × 10-6) = 26.4 X 10-6

Explanation:
Given Expressions as (2.8 × 10-5) – (1.6 × 10-6)
as the powers of 10 differ we need to modify before we factor.
We work around this by using our exponent property bm · bn = b (m+n)
to rewrite the 10-5 as 10-6 · So we take 10-5 and then grouping the 10-5 with 2.8.
So 2.8 X 10-5 becomes  28 X 10-6 therefore (2.8 × 10-5) – (1.6 × 10-6)=
(2.8 ×10 X 10-6) – (1.6 × 10-6) = (28 X 10-6) – (1.6 × 10-6)
using the distributive property of multiplication over subtraction,
we get (28-1.6) X 10-6 = 26.4 X 10-6.

Question 15.
(2.4 × 104) × (3.8 × 10-6)
Answer:
(2.4 × 104) × (3.8 × 10-6) = 9.12 X 10-2

Explanation:
Given expressions as (2.4 × 104) × (3.8 × 10-6)
we work out the coefficients and the exponents separately.
and use the product rule; bm x b n = b(m + n) to multiply the bases.
So (2.4 × 104) × (3.8 × 10-6) = (2.4 X 3.8) X (104 × 10-6) =
9.12 X (104-6) = 9.12 X 10-2 .

Question 16.
(5.2 × 10-3) ÷ (1.3 × 10-12)
Answer:
(5.2 × 10-3) ÷ (1.3 × 10-12) = 4 X 109

Explanation:
Given expressions as (5.2 × 10-3) ÷ (1.3 × 10-12) ,
Separately we divide the coefficients and exponents.
We use the division rule of exponents,
where the exponents are subtracted.
Noted that when we are dividing exponential terms,
always subtract the denominator from the numerator.
So (5.2 × 10-3) ÷ (1.3 × 10-12)  =
(5.2 ÷ 1.3 ) X ( 10-3+12) = 4 X 109.

Self-Assessment for Problem Solving
Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 17.
It takes the Sun about 2.3 × 108 years to orbit the center of the Milky Way. It takes Pluto about 2.5 × 102 years to orbit the Sun. How ×many times does Pluto orbit the Sun while the Sun completes one orbit around the Milky Way?
Answer:
9.2 X 105 times

Explanation:
Given It takes the Sun about 2.3 × 108 years to orbit the
center of the Milky Way. It takes Pluto about 2.5 × 102 years
to orbit the Sun. So number of times does Pluto orbit the Sun
while the Sun completes one orbit around the Milky Way is
dividing the number of years the sun takes by
the number of years Pluto takes. So we divide by dividing the factors and
dividing the powers of 10 then rewrite in scientific notation
as  (2.3 × 108 ) ÷ (2.5 × 102) = \(\frac{2.3}{2.5}\) X (108-2) =
0.92 X 106 = 0.92 X 10 X 105 = 9.2 X105 times.

Question 18.
A person typically breathes about 8.64 × 108 liters of air per day.
The life expectancy of a person in the United States at birth is about 29,200 days.
Estimate the total amount of air a person born in the United States breathes over a lifetime.
Answer:
The total amount of air a person born in the United States
breathes over a lifetime is 2.52288 X 1013 liters

Explanation:
A person typically breathes about 8.64 × 108 liters of air per day.
The life expectancy of a person in the United States at birth is
about 29,200 days. The total amount of air a person born in the
United States breathes over a lifetime is 29,200 X 8.64 × 108 =
292 X 102 X 8.64 X 108 =(when bases are same powers are added)
2.92 X 8.64 X 1010 =2522.88 X1010 = 2.522 X 1013 liters.
therefore the total amount of air a person born in the United States
breathes over a lifetime is 2.52288 X 1013 liters.

Question 19.
In one week, about 4100 movie theaters each sold an average of 2200 tickets for Movie A. About 3.6 × 107 total tickets were sold at the theaters during the week. An article claims that about 25% of all tickets sold during the week were for Movie A. Is this claim accurate? Justify your answer.
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation 8.7 3
Answer:
Yes, the claim is accurate.

Explanation:
Given in one week, about 4100 movie theaters each sold
an average of 2200 tickets for Movie A. About 3.6 × 107 total
tickets were sold at the theaters during the week.
Number of tickets  for movie A =
4100 X 2200 = 9,020,000= 9.02 X 106 =
is approximately equal to ≈ 9 X 106 tickets.
Now 25% of total movie tickets is \(\frac{25}{100}\) X  3.6 × 107 =
25 X 3.6 X 105 = 90 X 105 = 9 X 10 X 105 =9 X 106 tickets.
Therefore, the claim of 25% of movie tickets sold for movie A
is approximately accurate.

Operations in Scientific Notation Homework & Practice 8.7

Review & Refresh

Write the number in scientific notation.
Question 1.
0.0038
Answer:
0.0038 = 3.8 X 10-3.

Explanation:
Given number is 0.0038 in scientific notation
we write a number so that it has single digit to
the left of decimal sign and is multiplied by an integer power of 10.
So 0.0038 = 3.8 X 0.001 = 3.8 X 10-3.

Question 2.
74,000,000
Answer:
74,000,000 = 7.4 X 107

Explanation:
Given number is 74,000,000 in scientific notation
we write a number so that it has single digit to
the left of decimal sign and is multiplied by an integer power of 10.
So 74,000,000 = 7.4 X 10,000,000 = 7.4 X 107 .

Question 3.
0.0000475
Answer:
0.0000475 = 4.75 X 10-5

Explanation:
Given number is 0.0000475 in scientific notation
we write a number so that it has single digit to
the left of decimal sign and is multiplied by an integer power of 10.
So 0.0000475 = 4.75 X 0.00001 = 4.75 X 10-5.

Find the values of the ratios (red to blue) of the perimeters and areas of the similar figures.
Question 4.
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation 8.7 4
Answer:
Ratios of perimeter is \(\frac{3}{4}\),
Ratios of area is \(\frac{9}{16}\)

Explanation:
The ratios of corresponding sides are 9/12. These all reduce to 3/4.
It is then said that the scale factor of these two similar squares is 3 : 4.

The perimeter of red square is 4 X 9
the perimeter of blue square is 4 X 12
When we compare the ratios of the perimeters of these similar squares,
we get 4 X 9 : 4 X 12= 9 : 12 = 3 : 4

Now Area of square is side square
so the area of red square is 92 = 81
the area of blue square is 122= 144
So the ratio of their areas is 81 : 144 = 9 :16.
Therefore Ratios of perimeter is \(\frac{3}{4}\),
Ratios of area is \(\frac{9}{16}\)

Question 5.
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation 8.7 5
Answer:
Perimeter is \(\frac{3}{2}\),
Area is \(\frac{9}{4}\)

Explanation:
The ratios of corresponding sides are 6/4. These all reduce to 3/2.
It is then said that the scale factor of these two similar triangles is 3 : 2.

The perimeter of red triangle is 6 + 6 + 6 = 18 and
the perimeter of blue triangle is 4 + 4 + 4=12.
When we compare the ratios of the perimeters of these similar triangles,
we get 18 : 12 = 3:2.
We know if two similar triangles have a scale factor of a : b,
then the ratio of their areas is a2 : b2.
We have scale factor as 3 : 2 so the ratio of their areas is 32 : 22= 9 : 4.
Therefore the Perimeter is \(\frac{3}{2}\) and Area is \(\frac{9}{4}\).

Concepts, Skills, & Problem Solving

OPERATIONS IN SCIENTIFIC NOTATION Find the sum, difference, product, and quotient of Expression 1 and Expression 2. Write your answers in scientific notation. (See Explorations 1 and 2, p. 355.)
Question 6.
3 × 103 Expression 1
2 × 103 Expression 2
Answer:
Sum = (3 X 103) + (2 × 103) = 5 X 103,
Difference = (3 X 103) – (2 × 103) = 1 X 103 = 103,
Product = (3 X 103) X  (2 × 103) = 6 X 106,
Quotient = (3 X 103) ÷ (2 × 103) = \(\frac{3}{2}\) = 1.5

Explanation:
For sum:
Given expressions as (3 × 103) + (2 × 103) as it is
(a × 10n) + (b × 10n) can be completed using the distributive property
of multiplication over addition, i.e., factor out the common factor 10n
as (a × 10n) + (b × 10n) = (a + b ) X 10n  so (3× 103) + (2 × 103) =
(3 + 2) X 103 =5 X 103.
For  difference:
Given expressions as (3 × 103) – (2 × 103)
using the distributive property of multiplication over subtraction,
we get (3-2) X 103  = 1 X 103 = 103
For product:
Given expressions as (3 × 103) × (2 × 103)
we work out the coefficients and the exponents separately.
and use the product rule; bm x b n = b(m + n) to multiply the bases.
So (3 × 103) × (2 × 103) = (3 X 2) X (103 × 103) = 6 X (103+3) = 6 X 106.
For quotient:
Given expressions as (3 × 103) ÷ (2 × 103) ,
Separately we divide the coefficients and exponents.
We use the division rule of exponents,
where the exponents are subtracted.
Noted that when we are dividing exponential terms,
always subtract the denominator from the numerator.
So (3×103) ÷ (2 × 103)  =
(3 ÷ 2 ) X ( 103-3) = (3 ÷ 2 ) X 100 = (3 ÷ 2 ) X 1 = (3 ÷ 2 ) =  \(\frac{3}{2}\) =1.5.

Question 7.
6 × 10-4 Expression 1
1.5 × 10-4 Expression 2
Answer:
Sum = (6 X 10-4) + (1.5 × 10-4) = 7.5 X 10-4,
Difference = (6 X 10-4) – (1.5 × 10-4) = 4.5 X 10-4 ,
Product = (6 X 10-4) X  (1.5 × 10-4) = 9 X 10-8,
Quotient = (6 X 10-4) ÷ (1.5 × 10-4) = 4 X 100,

Explanation:
For sum:
Given expressions as (6 × 10-4) + (1.5 × 10-4) as it is
(a × 10n) + (b × 10n) can be completed using the distributive property
of multiplication over addition, i.e., factor out the common factor 10n
as (a × 10n) + (b × 10n) = (a + b ) X 10n  so (3× 103) + (2 × 103) =
(6 + 1.5) X 10-4 =7.5 X 10-4.
For  difference:
Given expressions as (6 × 10-4) – (1.5 × 10-4)
using the distributive property of multiplication over subtraction,
we get (6-1.5) X 10-4  = 4.5 X 10-4.
For product:
Given expressions as (6 × 10-4) × (1.5 × 10-4)
we work out the coefficients and the exponents separately.
and use the product rule; bm x b n = b(m + n) to multiply the bases.
So (6 × 10-4) × (1.5 × 10-4) = (6 X 1.5) X (10-4 × 10-4) = 6 X (10-4-4) = 9 X 10-8.
For quotient:
Given expressions as (6 × 10-4) ÷ (1.5 × 10-4) ,
Separately we divide the coefficients and exponents.
We use the division rule of exponents,
where the exponents are subtracted.
Noted that when we are dividing exponential terms,
always subtract the denominator from the numerator.
So (6×10-4) ÷ (1.5 × 10-4)  =
(6 ÷ 1.5 ) X ( 10-4-(-4)) = (6 ÷ 1.5 ) X 100 = (4) X 100 or 4.

ADDING AND SUBTRACTING IN SCIENTIFIC NOTATION Find the sum or difference. Write your answer in scientific notation.
Question 8.
(2 × 105) + (3.8 × 105)
Answer:
(2 × 105) + (3.8 × 105) = 5.8 X 105.

Explanation:
Given expressions as (2 × 105) + (3.8 × 105) as it is
(a × 10n) + (b × 10n) can be completed using the distributive property
of multiplication over addition, i.e., factor out the common factor 10n
as (a × 10n) + (b × 10n) = (a + b ) X 10n  so (2× 105) + (3.8 × 105) =
(2 + 3.8) X 105 =5.8 X 105.

Question 9.
(6.33 × 10-9) – (4.5 × 10-9)
Answer:
(6.33 × 10-9) – (4.5 × 10-9) = 1.83 X 10-9.

Explanation:
Given expressions as (6.33 × 10-9) – (4.5 × 10-9)
using the distributive property of multiplication over subtraction,
we get (6.33 – 4.5) X 10-9  = 1.83 X 10-9.

Question 10.
(9.2 × 108) – (4 × 108)
Answer:
(9.2 × 108) – (4 × 108) = 5.2 X 108.

Explanation:
Given expressions as (9.2 × 108) – (4 × 108)
using the distributive property of multiplication over subtraction,
we get (9.2 – 4) X 108  = 5.2 X 108.

Question 11.
(7.2 × 10-6) + (5.44 × 10-6)
Answer:
(7.2 × 10-6) + (5.44 × 10-6) = 12.64 X 10-6 = 1.264 X 10-5

Explanation:
Given expressions as (7.2 × 10-6) + (5.44 × 10-6) as it is
(a × 10n) + (b × 10n) can be completed using the distributive property
of multiplication over addition, i.e., factor out the common factor 10n
as (a × 10n) + (b × 10n) = (a + b ) X 10n  so (2× 105) + (3.8 × 105) =
(7.2 + 5.44) X 10-6 =12.64 X 10-6 = 1.264 X 10-5.

Question 12.
(7.8 × 107) – (2.45 × 106)
Answer:
(7.8 × 107) – (2.45 × 106) = 7.55 X 107

Explanation:
Given expressions as (7.8 × 107) – (2.45 × 106)
as the powers of 10 differ we need to modify before we factor.
We work around this by using our exponent property bm · bn = b (m+n)
to rewrite the 106 as 107 · So we take 107 and then grouping the 107 with 2.45.
So 2.45 X 106 becomes  0.245 X 107 therefore (7.8 × 107) – (0.245 × 107)=
(7.8 X 107) – (2.45 X 10-1X 101 X 106) = (7.8 X 107) – (0.245 × 107)
using the distributive property of multiplication over subtraction,
we get (7.8 – 0.245) X 107 = 7.55 X 107.

Question 13.
(5 × 10-5) + (2.46 × 10-3)
Answer:
(5 × 10-5) + (2.46 × 10-3) = 2.51 X 10-5.

Explanation:
Given expressions as (5 × 10-5) + (2.46 × 10-3)
as the powers of 10 differ we need to modify before we factor.
We work around this by using our exponent property bm · bn = b (m+n)
to rewrite the 10-5 as 10-3 · So we take 10-3 and then grouping the 10-3 with 5.
So 5 X 10-5 becomes  0.05 X 10-3 therefore (5 × 10-5) + (2.46 × 10-3)=
(5 X 10-2 X 10-3) + (2.46 X 10-3) = (0.05 X 10-3) + (2.46 × 10-3)
using the distributive property of multiplication over addition,
we get (0.05 + 2.46) X 10-5 = 2.51 X 10-5.

Question 14.
(9.7 × 106) + (6.7 × 105)
Answer:
(9.7 × 106) + (6.7 × 105) = 103.7 X 105 or 1.037 X 107

Explanation:
Given expressions as (9.7 × 106) + (6.7 × 105)
as the powers of 10 differ we need to modify before we factor.
We work around this by using our exponent property bm · bn = b (m+n)
to rewrite the 106 as 105 · So we take 105 and then grouping the 105 with 9.7.
So 9.7 X 106 becomes  97 X 105 therefore (9.7 × 106) + (6.7 × 105)=
(9.7 X 101 X 105) + (6.7 X 105) = (97 X 105) + (6.7 × 105)
using the distributive property of multiplication over addition,
we get (97 + 6.7) X 105 = 103.7 X 105 = 1 .037 X 102 X 105 = 1.037 X 107 .

Question 15.
(2.4 × 10-1) – (5.5 × 10-2)
Answer:
(2.4 × 10-1) – (5.5 × 10-2) = 1.85 X 10-1

Explanation:
Given expressions as (2.4 × 10-1) – (5.5 × 10-2)
as the powers of 10 differ we need to modify before we factor.
We work around this by using our exponent property bm · bn = b (m+n)
to rewrite the 10-2 as 10-1 · So we take 10-1 and then grouping the 10-1 with 5.5.
So 5.5 X 10-2 becomes  0.55 X 10-1 therefore (2.4 × 10-1) – (5.5 × 10-2) =
(2.4 X 10-1) – (5.5 X 10-1X 10-1) = (2.4 X 10-1) – (0.55 × 10-1)
using the distributive property of multiplication over subtraction,
we get (2.4 – 0.55) X 10-1 = 1.85 X 10-1.

Question 16.
YOU BE THE TEACHER
Your friend adds 2.5 × 109 and 5.3 × 108. Is your friend correct? Explain your reasoning.
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation 8.7 6
Answer:
Yes, Friend is correct

Explanation:
Given expressions as 2.5 × 109 and 5.3 × 108
we do sum as the powers of 10 differ we need to modify before we factor.
We work around this by using our exponent property bm · bn = b (m+n)
to rewrite the 108 as 109 · So we take 109 and then grouping the 109 with 5.3.
So 5.3 X 109 becomes  0.53 X 109 therefore (2.5 × 109) + (5.3 × 108) =
(2.5 X 109) + (5.3 X 10-1 X 10 X 108) = (2.5 X 109) + (0.53 × 109)
using the distributive property of multiplication over addition,
we get (2.5 + 0.53 ) X 109 = 3.03 X 109.
As the results are same friend is correct.

MULTIPLYING AND DIVIDING IN SCIENTIFIC NOTATION
Find the product or quotient. Write your answer in scientific notation.
Question 17.
5 × (7 × 107)
Answer:
5 × (7 × 107) = 3.5 X 108

Explanation:
As given expression is 5 × (7 × 107) so we use
associative law of multiplication as a X (b X c) = (a X b ) X c,
So 5 × (7 × 107) = (5 X 7) × 107 = 35 X 107.= 3.7 X 10 X 107 = 3.7 X 108.

Question 18.
(5.8 × 10-6) ÷ (2 × 10-3)
Answer:
(5.8 × 10-6) ÷ (2 × 10-3) = 2.9 X 10-3.

Explanation:
Given expressions as (5.8 × 10-6) ÷ (2 × 10-3) ,
Separately we divide the coefficients and exponents.
We use the division rule of exponents,
where the exponents are subtracted.
Noted that when we are dividing exponential terms,
always subtract the denominator from the numerator.
So (5.8 × 10-6) ÷ (2 × 10-3)  = (5.8 ÷ 2 ) X ( 10-6+3) = 2.9 X 10-3.

Question 19.
(1.2 × 10-5) ÷ 4
Answer:
(1.2 × 10-5) ÷ 4 = 3 X 10-6

Explanation:
Given expression is (1.2 × 10-5) ÷ 4 so (1.2 ÷ 4) X 10-5= 0.3 X 10-5=
3 X 10-1X 10-5= 3 X 10-6.

Question 20.
(5 × 10-7) × (3 × 106)
Answer:
(5 × 10-7) × (3 × 106) = 15 X 10-1 = 1.5

Explanation:
Given expressions as (5 × 10-7) × (3 × 106)
we work out the coefficients and the exponents separately.
and use the product rule; bm x b n = b(m + n) to multiply the bases.
So (5 × 10-7) × (3 × 106)  = (5 X 3) X (10-7 × 106) = 15 X (10-7+6) = 15 X 10-1 = 1.5.

Question 21.
(3.6 × 107) ÷ (7.2 × 107)
Answer:
(3.6 × 107) ÷ (7.2 × 107) = 5 X 10-1

Explanation:
Given expressions as (5.8 × 10-6) ÷ (2 × 10-3) ,
Separately we divide the coefficients and exponents.
We use the division rule of exponents,
where the exponents are subtracted.
Noted that when we are dividing exponential terms,
always subtract the denominator from the numerator.
So (3.6 × 107) ÷ (7.2 × 107)  = (3.6 ÷ 7.2 ) X ( 10-7+7) = 0.5 X 100 = 5 X 10-1.

Question 22.
(7.2 × 10-1) × (4 × 10-7)
Answer:
(7.2 × 10-1) × (4 × 10-7) = 2.88 X 10-7

Explanation:
Given expressions as (7.2 × 10-1) × (4 × 10-7)
we work out the coefficients and the exponents separately.
and use the product rule; bm x b n = b(m + n) to multiply the bases.
So (7.2 × 10-1) × (4 × 10-7)  = (7.2 X 4) X (10-1 × 10-7) =
28.8 X (10-1-7) = 28.8  X 10-8 = 2.88 X 10-7.

Question 23.
(6.5 × 108) × (1.4 × 10-5)
Answer:
(6.5 × 108) × (1.4 × 10-5) = 9.1  X 103

Explanation:
Given expressions as (6.5 × 108) × (1.4 × 10-5)
we work out the coefficients and the exponents separately.
and use the product rule; bm x b n = b(m + n) to multiply the bases.
So (6.5 × 108) × (1.4 × 10-5)  = (6.5 X 1.4) X (108 × 10-5) =
9.1 X (108-5) = 9.1  X 103.

Question 24.
(2.8 × 104) ÷ (2.5 × 106)
Answer:
(2.8 × 104) ÷ (2.5 × 106) = 1.12 X 10-2

Explanation:
Given expressions as (2.8 × 104) ÷ (2.5 × 106) ,
Separately we divide the coefficients and exponents.
We use the division rule of exponents,
where the exponents are subtracted.
Noted that when we are dividing exponential terms,
always subtract the denominator from the numerator.
So (2.8 × 104) ÷ (2.5 × 106)  = (2.8 ÷ 2.5 ) X ( 104-6) = 1.12 X 10-2.

MATCHING You use technology to find four sums.
Match the sum with its standard form.
Question 25.
4.3E8
Answer:
4.3E8 matches with C. 430,000,000

Explanation:
Given number 4.3E8 here  nearly or approximately equals to ≈ 4.3 X 108,
The exponent “8” says to use the 10 by eight times in a multiplication as
4.3 E 8 = 4.3 X 10 X 10 X 10 X 10 X 10 X 10 X 10 X 10 = 430,000,000,
So 4.3E8 we will match with C.

Question 26.
4.3E – 8
Answer:
4.3E – 8 matches with B. 0.000000043

Explanation:
Given number 4.3E – 8 nearly or approximately equals to ≈ 4.3 X 10-8,
The exponent “-8” says to use the 10-1 by eight times in a multiplication as
4.3 E 8 = 4.3 X 0.1 X 0.1 X 0.1 X 0.1 X 0.1 X 0.1 X 0.1 X 0.1 = 0.000000043,
so 4.3E -8 we will match with B.

Question 27.
4.3E10
Answer:
4.3E10 matches with D. 43,000,000,000

Explanation:
Given number 4.3E10  nearly or approximately equals to ≈ 4.3 X 1010,
The exponent “10” says to use the 10 by ten times in a multiplication as
4.3 E 10 = 4.3 X 10 X 10 X 10 X 10 X 10 X 10 X 10 X 10 X 10 X 10 =
43,000,000,000 so 4.3E10 we will match with D.

Question 28.
4.3E – 10
Answer:
4.3E – 10 matches with A.0.00000000043

Explanation:
Given number 4.3E – 10 nearly or approximately equals to ≈ 4.3 X 10-10,
The exponent “-10” says to use the 10-1 by eight times in a multiplication as
4.3 E 8 = 4.3 X 0.1 X 0.1 X 0.1 X 0.1 X 0.1 X 0.1 X 0.1 X 0.1 X 0.1 X 0.1
= 0.00000000043 so 4.3E -10 we will match with A.

A. 0.00000000043
B. 0.000000043
C. 430,000,000
D. 43,000,000,000

Question 29.
MODELING REAL LIFE
How many times greater is the thickness of a dime than the thickness of a dollar bill?
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation 8.7 7
Answer:
The dime is about 12 times thicker than the dollar bill.

Explanation:
Given the thickness of dime is 0.135 = 1.35 X 10-1 and
the thickness of a dollar is 1.0922 X 10-2 divide the thickness of
dime with the thickness of dollar as (1.35 × 10-1) ÷ (1.0922 × 10-2)
now we separately divide the coefficients and exponents.
We use the division rule of exponents,
where the exponents are subtracted.
Noted that when we are dividing exponential terms,
always subtract the denominator from the numerator.
So (1.35 × 10-1) ÷ (1.0922 × 10-2)  = (1.35 ÷ 1.0922 ) X ( 10-1+2) =
1.2360 X 101= 12.36 approximately equals to ≈ 12.
Therefore the dime is about 12 times thicker than the dollar bill.

Question 30.
MULTIPLE CHOICE
On a social media website, Celebrity A has about 8.6 × 106 followers and Celebrity B has about 4.1 × 106 followers. Determine which of the following is the best estimate for the number of followers for Celebrity A compared to the number of followers for Celebrity B.
A. more than 2 times greater
B. less than 2 times greater
C. more than 20 times greater
D. less than 20 times greater
Answer:
The best estimate for the number of followers for Celebrity A
compared to the number of followers for Celebrity B is
A. more than 2 times greater.

Explanation:
Given on a social media website, Celebrity A has about
8.6 × 106 followers and Celebrity B has about 4.1 × 106 followers.
To find the best estimate for the number of followers for Celebrity A
compared to the number of followers for Celebrity B we divide them as
the number of followers for Celebrity A to the number of
followers for Celebrity B as (8.6 × 106) ÷ (4.1× 106)
now we separately divide the coefficients and exponents.
We use the division rule of exponents,
where the exponents are subtracted.
Noted that when we are dividing exponential terms,
always subtract the denominator from the numerator.
So (8.6× 106) ÷ (4.1 × 106)  = (8.6 ÷ 4.1) X ( 106-6) =
2.097 X 100= 2.097 X 1 = 2.097 approximately equals to ≈ 2.
The best estimate for the number of followers for Celebrity A
compared to the number of followers for Celebrity B is
A. more than 2 times greater.

REASONING Evaluate the expression. Write your answer in scientific notation.
Question 31.
5,200,000 × (8.3 × 102) – (3.1 × 108)
Answer:
5,200,000 × (8.3 × 102) – (3.1 × 108) = 4.006 X 109

Explanation:
Given expressions as 5,200,000 × (8.3 × 102) – (3.1 × 108),
So first we solve 5,200,000 × (8.3 × 102) as 5.2 X 106 X (8.3 × 102),
we work out the coefficients and the exponents separately.
and use the product rule; bm x b n = b(m + n) to multiply the bases.
(5.2 X 8.3) X (106 X 102
) = (43.16) X (106+2) = 43.16 X 108,
now we subtract expressions as (43.16 X 108) – (3.1 × 108) =
using the distributive property of multiplication over subtraction,
we get (43.16 – 3.1) X 108  = 40.06 X 108 = 4.006 X 10 X 108 = 4.006 X 109.

Question 32.
(9 × 10-3) + (2.4 × 10-5) ÷ 0.0012
Answer:
(9 × 10-3) + (2.4 × 10-5) ÷ 0.0012 = 7.52

Explanation:
Given (9 × 10-3) + (2.4 × 10-5) ÷ 0.0012 first we solve
(9 × 10-3) + (2.4 × 10-5) given expressions as 9 × 10-3 and 2.4 × 10-5
we do sum as the powers of 10 differ we need to modify before we factor.
We work around this by using our exponent property bm · bn = b (m+n)
to rewrite the 10-5 as 103 · So we take 10-5 and then grouping the 10-5 with 2.4.
So 2.4 X 10-5 becomes  0.024 X 10-3 therefore (9 × 10-3) + (2.4 × 10-5)=
(9X 10-3) + (2.4 X 10-2 X 10-3) = (9 X 10-3) + (0.024 × 10-3)
using the distributive property of multiplication over addition,
we get (9 + 0.024 ) X 10-3 = 9.024 X 10-3.
So (9 × 10-3) + (2.4 × 10-5) = 9.024 X 10-3 now we write 0.0012 as
1.2 X 10-3 therefore we divide (9.024 X 10-3) ÷ (1.2 X 10-3),
now we separately divide the coefficients and exponents.
We use the division rule of exponents,
where the exponents are subtracted.
Noted that when we are dividing exponential terms,
always subtract the denominator from the numerator.

So (9.024 ÷ 1.2)  X (10-3 ÷ 10-3) so 7.52 X 10-3+3 = 7.52 X 100 =
7.52 X 1 = 7.52.

Question 33.
GEOMETRY
Find the perimeter of the rectangle at the right.
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation 8.7 8
Answer:
The perimeter of the rectangle at the right is 1.962 X 108 cm.

Explanation:
Given area of rectangle as 5.612 X 1014cm2 and breadth as
9.2 X 107 cm , We will find the length first we take length as l,
we know area of rectangle is length X breadth so
5.612 X 1014 cm2 = l X 9.2 X 107
so l= (5.612 X 1014 ) ÷ (9.2 X 107) now we separately divide
the coefficients and exponents. We use the division rule of exponents,
where the exponents are subtracted.
Noted that when we are dividing exponential terms,
always subtract the denominator from the numerator.
(5.612 X 9.2)  X (1014 ÷ 107) = 0.61 X (1014-7) = 0.61X 107=0.61 X 107 cm,
therefore length of rectangle is 0.61 X 107 cm now we will calculate
perimeter of rectangle as we know perimeter of rectangle =
2 x ( length + breadth)= 2 X ((0.61 X 107)+(9.2 X 107))cm,
now we calculate first (0.61 X 107)+(9.2 X 107) using the
distributive property of multiplication over addition,
we get (0.61 + 9.2) X 107 = 9.81 X 107  now 2 X (9.81X 107)=
19.62 X 107= 1.962 X 108 cm. Therefore the perimeter of the
rectangle at the right is 1.962 X 108 cm.

Question 34.
DIG DEEPER!
A human heart pumps about 7 × 10-2 liter of blood per heartbeat.
The average human heart beats about 72 times per minute.
How many liters of blood does a heart pump in 1 year? 70 years?
Answer:
2.65 X 106 liters of blood a heart pump in 1 year.
1.855 X 108 liters of blood a heart pump for 70 years.

Explanation:
Given a human heart pumps about 7 × 10-2 liter of blood per heartbeat.
The average human heart beats about 72 times per minute.
The the number of liters per minute multiply the number of
heartbeats per minute by the number of liters per heartbeat as
72 X 7 × 10-2 = 504 × 10-2 =5.04 liters per minute,
now we multiply by 60 to get liters per hour since
there are 60 minutes in 1 hour so 60 X 5.04 = 302.4 liters per hour,
now we multiply by 24 to get liters per day since
there are 24 hours in 1 day so 302.4 X 24 = 7257.6 liters per day,
now we multiply by 365 to get liters per year since
there are 365 days in 1 year and write in scientific notation
and round to two decimal places as 7257.6 X 365 = 2649024 =
2.649024 X 106 approximately equal to ≈ 2.65 X 106 liters per year
Now for 70 years it is 70 X 2.65 X 106 = 185.5 X 106 = 1.855 X 100 X 106 =
1.855 X 102 X 106 =1.855 X 108 liters for 70 years.

Question 35.
MODELING REAL LIFE
Use the Internet or another reference to find the populations and areas (in square miles) of India, China, Argentina, the United States, and Egypt. Round each population to the nearest million and each area to the nearest thousand square miles.
a. Write each population and area in scientific notation.
b. Use your answers to part(a) to find and order the population densities (people per square mile) of each country from least to greatest.
Answer:
a. In India : Population = 1.311 X 109 , Area = 1.269 X 106
In China : Population = 1.371 X 109 , Area = 3.705 X 10
In Argentina : Population = 4.3 X 107 , Area = 1.074 X 106
In United States : Population = 3.21 X 108, Area = 3.797 X 106
In Egypt: Population = 9.2 X 107 , Area = 3.9 X 106

b. The population densities are
India = 1.033 X 103
China = 3.7 X 102
Argentina =4.0 X 101
United States = 8.45 X 101
and in Egypt = 2.36 X 101
Each country from least to greatest is
Egypt, Argentina, United States, China, India.

Explanation:
The following were found by doing a google search :
we write each population and area in scientific notation,
We  know 1 billion = 109 and 1 million = 106 ,
a. In India : Population = 1.311 billion= 1.311 X 10
Area = 1.268 million = 1.269 X 106
In China : Population = 1.371 billion = 1.371 X 10
Area =3.705 million= 3.705 X 10
In Argentina : Population= 43 million = 4.3 X 10
Area = 1.074 million = 1.074 X 106
In United States : Population=321 million = 3.21 X 10
Area =3.797 million= 3.797 X 106
In Egypt: Population = 92 million= 9.2 X 10
Area =390,000= 3.9 X 106

b. Divide each country’s population by its area to
get its population density, and we separately divide
the coefficients and exponents. We use the division rule of exponents,
where the exponents are subtracted.
Noted that when we are dividing exponential terms,
always subtract the denominator from the numerator.

India = ( 1.311 X 109 ) ÷ ( 1.269 X 106) = 1.033 X 103
(1.311 ÷ 1.269) X (109-6) = 1.033 X 103 ,
China = ( 1.371 X 109 ) ÷ ( 3.705 X 106) =
(1.371 ÷ 3.705) X (109-6) = 0.370 X 103 = 3.7 X 102,
Argentina = ( 4.3 X 107 ) ÷ ( 1.074 X 106) =
(4.3 ÷ 1.074) X (107-6) = 4 X 101,
United States = ( 3.21 X 108 ) ÷ ( 3.797 X 106) =
(3.21 ÷ 3.797) X (108-6) = 0.845 X 102 = 8.45X 101,
Egypt = (9.2 X 107 ) ÷ ( 3.9 X 106) =
(9.2 ÷ 3.9) X (107-6) = 2.36 X 101,
Now we write each country from least to greatest density as
2.36 X 101 < 4 X 101 < 8.45X 101< 3.7 X 102,1.033 X 103 so
Egypt, Argentina, United States, China, India.

Exponents and Scientific Notation Connecting Concepts

Using the Problem-Solving Plan
Question 1.
Atoms are made of protons, neutrons, and electrons. The table shows the numbers of protons and the masses of several atoms. Use a line of best fit to estimate the mass (in grams) of an atom that has 29 protons.
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation cc 1

Understand the problem.
You know the numbers of protons and the masses of several atoms. You are asked to use the line of best fit to estimate the mass of an atom that has 29 protons.

Make a plan.
Use a graphing calculator to find an equation of the line of best fit.
Then evaluate the equation when x = 29.

Solve and check.
Use the plan to solve the problem. Then check your solution.
Answer:
Mass of an atom that has 29 protons is 4.843 X 10-23 grams

Explanation:
Given 1 proton has 1.67 X 10-24 grams of mass,
So 29 protons has 29 X 1.67 X 10-24 grams,
48.43 X 10-24 = 4.843 X 10-23 grams.

Question 2.
Modoc Country, California, is 74.9 miles long and 56.2 miles wide. 263A map of the county is drawn using a scale factor of 2.11 × 10-6. What is the perimeter of the county on the map? Express your answer using more appropriate units.
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation cc 2
Answer:
The perimeter of the country on the map is 6.66631 X 1011 inches

Explanation:
Given Modoc Country, California, is 74.9 miles long and 56.2 miles wide.
First we calculate perimeter it is 2 X ( 74.9 + 56.2) mile = 2 X 131.1 =
262.2 miles, So perimeter of Modoc Country, California is 262.2 miles.
Now on map we will use in inches , we know 1 mile = 5280 feet and
1 feet is equal to 12 inches so 262.2 we will convert into inches.
262.2 X 5280 X 12 = 1406592 inches. Now we are given with
scaling factor on the map as 2.11 × 10-6 equals to 1 inch on the map,
So now perimeter of the country on the map is
(1406592 ) ÷ 2.11 × 10-6 = 666631.27 X 106 = 6.66631 X 1011 inches.

Question 3.
A research company estimates that in the United States, 7733 about 8.37 × 107 adult males and 6.59 × 107 adult females watch NFL football, while 3.13 × 107 adult males and 5.41 × 107 adult females do not watch NFL football. Organize the results in a two-way table. Include the marginal frequencies.
Answer:
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation-18

Explanation:
Given A research company estimates that in the United States, 7733 about
8.37 × 107 adult males and 6.59 × 107 adult females watch NFL football,
while 3.13 × 107 adult males and 5.41 × 107 adult females do not watch NFL football.
Organized the results in two-way table above,
Entries in the “Total” row and “Total” column are called
marginal frequencies or the marginal distribution.
Entries in the body of the table are called joint frequencies.
So total adult watched NFL football match are
(8.37 X 107) + (6.59 X 107)=14.96 X 107 and do not
watch NFL football are(3.13 X 107) + (5.41 X 107) = 8.54 X 107.

Performance Task

Elements in the Universe
At the beginning of this chapter, you watched a STEAM Video called “Carbon Atoms.” 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 Answers Grade 8 Chapter 8 Exponents and Scientific Notation cc 3
Answer:

Exponents and Scientific Notation Chapter Review

Review Vocabulary

Write the definition and give an example of each vocabulary term.
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation cr 1

Let us take an example : 52 ,
An expression that represents repeated multiplication
of the same factor is called a power.
Here the number 5 is called the base,
and the number 2 is called the exponent.
The exponent corresponds to the number of times
the base is used as a factor.
A method for expressing very large or very small numbers
as a product of decimal less than 10 and multiplied
by a power of 10.
Example: 8  X 107 or 5 X 10-2.
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation-23
Graphic Organizers

You can use a Definition and Example Chart to organize information about a concept.
Here is an example of a Definition and Example Chart for the vocabulary term power.
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation cr 2

Choose and complete a graphic organizer to help you study the concept.
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation cr 3
1. Product of Powers Property
2. Power of a Power Property
3. Power of a Product Property
4. Quotient of Powers Property
5. negative exponents
6. scientific notation
7. adding and subtracting numbers in scientific notation
8. multiplying and dividing numbers in scientific notation

Graphic Organizer:
1. Product of Powers Property
2. Power of a Power Property
3. Power of a Product Property
4. Quotient of Powers Property
5. negative exponents & power of a quotient
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation-19
6. scientific notation
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation-20
7. adding and subtracting numbers in scientific notation
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation-21
8. multiplying and dividing numbers in scientific notation
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation-22

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 Answers Grade 8 Chapter 8 Exponents and Scientific Notation cr 4

8.1 Exponents (pp. 319–324)
Learning Target: Use exponents to write and evaluate expressions. Write the product using exponents.

Question 1.
(- 9) • (- 9) • (- 9) • (- 9) • (- 9)
Answer:
(- 9) • (- 9) • (- 9) • (- 9) • (- 9) =(-9)5

Explanation:
As -9 is multiplied by 5 times we write as
(-9)5 here -9 is base and 5 is exponent.

Question 2.
2 • 2 • 2 • n • n
Answer:
2 • 2 • 2 • n • n = (2)X (n)

Explanation:
As 2 is multiplied 3 times we write as (2)and n
is multiplied twice so (n)2
so the expression is (2)X (n)2 .

Evaluate the expression.
Question 3.
113
Answer:
113 = 1331

Explanation:
Given 113 means 11 is multiplied thrice
as 11 X 11 X 11 we get 1331.
here base is 11 and exponent is 3.

Question 4.
-(\(\frac{1}{2}\))4
Answer:
-(\(\frac{1}{2}\))4 = – \(\frac{1}{16}\)

Explanation:
Given -(\(\frac{1}{2}\))4 means \(\frac{1}{2}\) is multiplied four times
as- (\(\frac{1}{2}\) X \(\frac{1}{2}\) X \(\frac{1}{2}\) X \(\frac{1}{2}\) ) = – \(\frac{1}{16}\).

Question 5.
|\(\frac{1}{2}\left(16-6^{3}\right)\)|
Answer:
|\(\frac{1}{2}\left(16-6^{3}\right)\)| = 100

Explanation:
Given expression as |\(\frac{1}{2}\left(16-6^{3}\right)\)|=
first we solve 63 =6 X 6 X 6 = 216 now (16-216) = -200
so |\(\frac{1}{2}\) X -200|= |-100|= 100.

Question 6.
The profit P(in dollars) earned by a local merchant selling x items is 3 represented by the equation P = 0.2x3 – 10. How much more profit does he earn selling 15 items than 5 items?
Answer:
More profits  he earn by selling 15 items than 5 items is $650

Explanation:
Given P = 0.2x3 – 10
Profit by selling 5 items is
0.2 X (5)3 – 10 = 0.2 X 125 – 10 = 25 – 10 = $15.
Profit by selling 15 items is
0.2 X (15)3 – 10 = 0.2 X 3375 – 10 = 675 – 10 = $665.
More profits  he earn by selling 15 items than 5 items is
$665 -15 = $650.

8.2 Product of Powers Property (pp. 325–330)
Learning Target: Generate equivalent expressions involving products of powers.

Simplify the expression. Write your answer as a power.
Question 7.
p5 • p2
Answer:
p5 • p2 = p7

Explanation:
We write the given p5 • p7 expression as a power,
we have general rule for am • an = am+n
If product of two powers with the same base then
powers are added,
so p5 • p2 as bases are same p powers are added p5+2 = p7.

Question 8.
(n11)2
Answer:
(n11)2 = n22

Explanation:
We write the given expression (n11)2 as a power,
General rule for finding (am)power of a power,
If two powers have the same base then
we can multiply the powers as (am)n = am x n .
so (n11)2 has powers of powers therefore powers
are multiplied as (n)11 X 2= (n)22

Question 9.
\(\left(-\frac{2}{5}\right)^{3} \cdot\left(-\frac{2}{5}\right)^{2}\)
Answer:
\(\left(-\frac{2}{5}\right)^{3} \cdot\left(-\frac{2}{5}\right)^{2}\) = –

Explanation:
Given expression is \(\left(-\frac{2}{5}\right)^{3} \cdot\left(-\frac{2}{5}\right)^{2}\)
we have general rule for am • an = am+n
If product of two powers with the same base then
powers are added, so we add powers as (-\(\frac{2}{5}\))3+2 =
(-\(\frac{2}{5}\))5 So \(\left(-\frac{2}{5}\right)^{3} \cdot\left(-\frac{2}{5}\right)^{2}\) =

Question 10.
Simplify (- 2k)4.
Answer:
(- 2k)4 = (-2)4 X k4

Explanation:
Given to simplify the expression (- 2k)we use general rule to
write (ab)m  power of a product as am X bm ,So (- 2k)4 = (-2)4 X k4

Question 11.
Write an expression that simplifies to x24 using the Power of a Power Property.
Answer:
x24 = (x12)2

Explanation :
To simplify  x24 using the Power of a Power Property,
we use general rule for finding (am)power of a power,
If two powers have the same base then
we can multiply the powers as (am)n = am x n ,here m X n = 24,
lets take m, n as 12,2 so that 12 X 2 = 24, therefore (x12)2 =  x12 x 2 =  x24.
therefore x24 = (x12)2 .

Question 12.
You send an email with a file size of 4 kilobytes. One kilobyte is 210 bytes. What is the file size of your email in bytes?
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation cr 12
Answer:
The file size of my email in bytes is

Explanation:
Given I send an email with a file size of 4 kilobytes. One kilobyte is 210 bytes.
The file size of my email in bytes is 4 X 210 bytes as 4 can be written as 2 X 2 =22,
So 22 X 210 we have general rule for am • an = am+n
If product of two powers with the same base then
powers are added, 22 X 210 = 22+10 = 212 bytes.

Question 13.
Explain how to use properties of powers 2 to simplify the expression 27 • 32.
Answer:
27 • 32 = 33 X 32 = 35

Explanation:
Given expression as 27 • 32 first we write 27 as power of 3,
so 27 = 3 X 3 X 3 = 33 now 27 • 32 = 33 X 32 ,
we have general rule for am • an = am+n
If product of two powers with the same base then
powers are added, 33 X 32 = 33+2= 35.

8.3 Quotient of Powers Property (pp. 331–336)

Learning Target: Generate equivalent expressions involving quotients of powers.

Simplify the expression. Write your answer as a power.
Question 14.
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation cr 14
Answer:
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation cr 14= (8)8-3 = (8)

Explanation:
Given \(\frac{8^{8}}{8^{3}}\) we use rule for finding \(\frac{a^{m}}{a^{n}}\),
a quotient of two powers with the same base as am-n
so \(\frac{8^{8}}{8^{3}}\) = (8)8-3 = (8)

Question 15.
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation cr 15
Answer:
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation cr 15= 510
Explanation:
First we calculate separately values of numerators
and denominators then divide, we have numerator (5)2 X (5)
we have same bases as 5 so we add powers as (5)2+9 = (5)11
we have denominator  (5)1
Now we have \(\frac{5^{11}}{5^{1}}\)  so we use rule
for finding \(\frac{a^{m}}{a^{n}}\) a quotient of
two powers with the same base as am-n
\(\frac{5^{11}}{5^{1}}\)  = (5)11-1 = 510.

Question 16.
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation cr 16
Answer:
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation cr 16= w4
Explanation:
First we calculate separately values of numerators
and denominators then divide, we have numerator (w)8 X (w)
we have same bases as w so we add powers as (w)8+5 = (w)13
we have denominator (w)7 X (w)2
we have same base as w so we add powers as (w)7+2 = (w)9
Now we have \(\frac{w^{13}}{w^{9}}\)  so we use rule
for finding \(\frac{a^{m}}{a^{n}}\) a quotient of
two powers with the same base as am-n
\(\frac{w^{13}}{w^{9}}\)  = (w)13-9 = w4.

Question 17.
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation cr 17
Answer:
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation cr 17= m5

Explanation:
First we calculate separately values of numerators
and denominators then divide, we have numerator (m)8 X (m)10 X (m)2
we have same bases as m so we add powers as (m)8+10+2 = (m)20
we have denominator (m)6 X (m)9
we have same base as m so we add powers as (m)6+9 = (m)15
Now we have \(\frac{m^{20}}{m^{15}}\)  so we use rule
for finding \(\frac{a^{m}}{a^{n}}\) a quotient of
two powers with the same base as am-n
\(\frac{m^{20}}{m^{15}}\)  = (m)20-15 = m5.

Question 18.
Write an expression that simplifies to x3 using the Quotient of Powers Property.
Answer:
x3 = \(\frac{x^{6}}{x^{3}}\)

Explanation:
To simplify x3 using the Quotient of Powers Property,
we use rule for finding \(\frac{a^{m}}{a^{n}}\) a quotient of
two powers with the same base as am-n here we have m-n = 3,So lets take
m, n as 6,3 so that 6-3 = 3 therefore \(\frac{x^{6}}{x^{3}}\)= x6-3 = x3.
therefore x3 = \(\frac{x^{6}}{x^{3}}\).

Question 19.
At the end of a fiscal year, a company has made 1.62 × 77 dollars in profit. The company employs 73 people. How much will each person receive if the company divides the profit equally among its employees?
Answer:
Each person will receive $3889.62 if the company divides
the profit equally among its employees.

Explanation:
Given at the end of a fiscal year, a company has made 1.62 × 77 dollars in profit,
The company employs 73 people. To find how much each person
receive if the company divides the profit equally among its employees is
1.62 × 77 ÷ 73 we use the division rule of exponents,
where the exponents are subtracted. 1.62 X ( 77 ÷ 73 ) = 1.62 X ( 77-3)
=
1.62 X 74 = 1.62 X 7 X 7 X 7 X 7 = 3889.62, Therefore each person
will receive $3889.62 if the company divides the profit equally among its employees.

8.4 Zero and Negative Exponents (pp. 337–342)
Learning Target: Understand the concepts of zero and negative exponents.

Evaluate the expression.
Question 20.
2-4
Answer:
4-2

Explanation:
Given expression as 2-4 so we write as

Question 21.
950
Answer:
950 = 1

Explanation:
Given expression is 95it is proven that any number or
expression raised to the power of zero is always equal to 1.
In other words, if the exponent is zero then the result is 1.
So 950 = 1.

Question 22.
\(\frac{8^{2}}{8^{4}}\)
Answer:
\(\frac{8^{2}}{8^{4}}\) = \(\frac{1}{8^{2}}\) = \(\frac{1}{64}\).

Explanation:
Given expression as \(\frac{8^{2}}{8^{4}}\) now we use rule
for finding \(\frac{a^{m}}{a^{n}}\) a quotient of
two powers with the same base as am-n
as 82-4 we get 8-2 we write as \(\frac{1}{8^{2}}\) = \(\frac{1}{64}\).

Question 23.
(- 12)-7 • (- 12)7
Answer:
(- 12)-7 • (- 12)7= 1

Explanation:
Given expression is -12-7 X -127 ,
Here we use Product of Powers Property for am • an = am+n
if we have product of two powers with the same base then
powers are added. So (12)-7 X (12)7 = (12)-7+7 = (12)0= 1.

Question 24.
\(\frac{1}{7^{9}} \cdot \frac{1}{7^{6}}\)
Answer:
\(\frac{1}{7^{9}} \cdot \frac{1}{7^{6}}\) = \(\frac{1}{7^{15}}\)

Explanation:
Given expression is \(\frac{1}{7^{9}} \cdot \frac{1}{7^{6}}\)
first we calculate separately values of denominators
then divide with numerator, we have denominator  79 X 76
we have same bases as 7 so we add powers as 79+6 = 715
as numerator is 1 we write as \(\frac{1}{7^{15}}\).

Question 25.
\(\frac{9^{4} \cdot 9^{2}}{9^{2}}\)
Answer:
\(\frac{9^{4} \cdot 9^{2}}{9^{2}}\) = 9

Explanation:
Given Expression as \(\frac{9^{4} \cdot 9^{2}}{9^{2}}\)
first we calculate separately values of numerators
then divide with denominator, we have numerator (9)4 X (9)
we have same bases as 9 so we add powers as (9)4+2 = (9)6
Now we have \(\frac{9^{6}}{9^{2}}\)  so we use rule
for finding \(\frac{a^{m}}{a^{n}}\) a quotient of
two powers with the same base as am-n,
So \(\frac{9^{6}}{9^{2}}\)  = (9)6-2 = (9)4.

Simplify. Write the expression using only positive exponents.
Question 26.
x-2 • x0
Answer:
x-2 • x0 = \(\frac{1}{x^{2}}\)

Explanation:
Given expression is x-2 • x= \(\frac{1}{x^{2}}\).

Question 27.
y-8y3
Answer:
y-8y3 = \(\frac{1}{y^{5}}\)

Explanation:
Given expression is y-8 • y= \(\frac{1}{y^{5}}\).

Question 28.
\(\frac{3^{-1} \cdot z^{5}}{z^{-2}}\)
Answer:
\(\frac{3^{-1} \cdot z^{5}}{z^{-2}}\) =

Explanation:
Given expression is \(\frac{3^{-1} \cdot z^{5}}{z^{-2}}\)
we use rule for finding \(\frac{a^{m}}{a^{n}}\),
a quotient of two powers with the same base as am-n
for \(\frac{z^{5}}{z^{-2}}\) as z= z7
now we have 3-1 X z7

Question 29.
Write an expression that simplifies to x-4.
Answer:
x-4 =

Explanation:
To simplify  x-4 we use rule for finding \(\frac{a^{m}}{a^{n}}\),
a quotient of two powers with the same base as am-n,So m-n = -4 we take
m, n as 4,8 which makes (4-8) = -4 therefore x-4 =

Question 30.
Water flows from a shower head at a rate of 24-1 gallon per second. How many gallons do you use when taking a 15-minute shower? a 20-minute shower?
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation cr 30
Answer:
In 15 minutes we use in 20 minutes shower we use 50

Explanation:
Given water flows from a shower head at a rate of 24-1 gallon per second,
to know how many gallons we used in 15-minutes first
we convert minutes to seconds as 1 minute is equal to 60 seconds,
so 15 minutes is equal to 15 X 60 = 900 seconds now number of gallons we
used in 15 minutes are 900 X 24-1 =

Question 31.
Explain two different methods for simplifying w-2 • w5.
Answer:
One is product of powers = w-2 • w5 = w-2+5 = w3 or
other is quotient of two powers with the same base = w5-2 = w3

Explanation:
To simplify w-2 • w5 first we use
other method is we write as .

8.5 Estimating Quantities (pp. 343–348)

Learning Target: Round numbers and write the results as the product of a single digit and a power of 10.

Round the number. Write the result as a product of a single digit and a power of 10.
Question 32.
29,197,543
Answer:
29,197,543 = 3 X 107

Explanation:
Given number is 29,197,543 is nearly or approximately
equal to ≈ 30,000,000 so we have 3 followed by 11 zeros,
so we write as 3 X 107.

Question 33.
0.000000647
Answer:
0.000000647 = 6 X 10-7

Explanation:
Given number 0.000000647 is nearly or approximately
equal to ≈ 0.0000006, we have divided 6 by 10 followed
by 7 zeros so we write as 6 X 10-7.

Question 34.
The speed of light is 299,792,458 meters per second. About how far can a light beam travel in 3 seconds? Write your answer as a product of a single digit and a power of 10.
Answer:
A light beam can travel 9 X 108 meters in 3 seconds.

Explanation:
Given the speed of light is 299,792,458 meters per second.
to know about how far can a light beam travel in 3 seconds is
3 X 299,792,458 now we write 299,792,458 is nearly or approximately
equal to ≈ 300,000,000 so 3 X 300,000,000 = 900,000,000
so we have 9 followed by 8 zeros so we write as 9 X 108 meters.
Therefore a light beam can travel 9 X 108 meters in 3 seconds.

Question 35.
The population of Albany, New York is about 98,989 and the population of Moscow, Russia is about 12,235,448. Approximately how many times greater is the population of Moscow than the population of Albany?
Answer:
100 times greater is the population of Moscow than the population of Albany.

Explanation:
Given the population of Albany, New York is about 98,989 and
the population of Moscow, Russia is about 12,235,448 if we see the
the population of Moscow is greater than the population of Albany.
So 12,235,448

8.6 Scientific Notation (pp. 349–354)
Learning Target: Understand the concept of scientific notation.

Write the number in scientific notation.
Question 36.
0.00036
Answer:
0.00036= 3.6 X 10-4

Explanation:
Given number is 0.00036 in scientific notation
we write a number so that it has single digit to
the left of decimal sign and is multiplied by an integer power of 10.
So 0.00036 = 3.6 X 0.0001 = 3.6 X 10-4.

Question 37.
800,000
Answer:
800,000 = 8 X 105

Explanation:
Given number is 800,000 in scientific notation
we write a number so that it has single digit to
the left of decimal sign and is multiplied by an integer power of 10.
So 800,000 = 8 X 100000 = 8 X 105.

Question 38.
79,200,000
Answer:
79,200,000 = 7.92 X 107

Explanation:
Given number is 79,200,000 in scientific notation
we write a number so that it has single digit to
the left of decimal sign and is multiplied by an integer power of 10.
So 79,200,000 = 7.29 X 10000000 = 7.29 X 107.

Write the number in standard form.
Question 39.
2 × 107
Answer:

2 X 107 = 20,000,000

Explanation:
Given 2 × 10the number in standard form is
2 X 10,000,000 = 20,000,000.
So 2 X 107 = 20,000,000.

Question 40.
4.8 × 10-3
Answer:
4.8 × 10-3  = 0.0048

Explanation:
Given 4.8 × 10-3  the number in standard form is
4.8 × 0.001 = 0.0048.

Question 41.
6.25 × 105
Answer:
6.25 × 105 = 6,25,000

Explanation:
Given 6.25 105  the number in standard form is
6.25 X 100,000 = 6,25,000.
So 6.25 X 105 = 6,25,000.

Question 42.
The mass of a single dust particle is 7.52 × 10-10 kilogram. What is the mass of a dust ball made of 100 dust particles? Express your answer using more-appropriate units.
Answer:
The mass of a dust ball made of 100 dust particles is 7.52 10-8 kilogram

Explanation:
Given the mass of a single dust particle is 7.52 × 10-10 kilogram.
So the mass of a dust ball made of 100 dust particles is
7.52 × 10-10 X 100 = 7.52 X 10-10 X 102 = 7.52 × 10-10+2 = 7.52 × 10-8.
Therefore the mass of a dust ball made of 100 dust
particles is 7.52 10-8 kilogram.

8.7 Operations in Scientific Notation (pp. 355–360)
Learning Target: Perform operations with numbers written in scientific notation.

Evaluate the expression. Write your answer in scientific notation.
Question 43.
(4.2 × 108) + (5.9 × 109)
Answer:
(4.2 × 108) + (5.9 × 109) = 6.32 X 109

Explanation:
Given Expressions as (4.2 × 108) + (5.9 × 109)
as the powers of 10 differ we  need to modify before we factor.
We work around this by using our exponent property bm · bn = b (m+n)
to rewrite the 108 as 109 ·So we take 109 and then grouping the 109 with 4.2.
So 4.2 X 108 becomes  0.42 X 109 therefore
(4.2 × 108) + (5.9 × 109) =(0.42 X 101X 108) + (5.9 X 109) =
(0.42 X  109) + (5.9 × 109)
using the distributive property of multiplication over addition,
we get (0.42+5.9) X 109 = 6.32 X 109.

Question 44.
(5.9 × 10-4) – (1.8 × 10-4)
Answer:
(5.9 × 10-4) – (1.8 × 10-4) = 4.1 X 10-4

Explanation:
Given expressions as (5.9 × 10-4) – (1.8 × 10-4)
using the distributive property of multiplication over subtraction,
we get (5.9 – 1.8 ) X 10-4  = 4.1 X 10-4.

Question 45.
(7.7 × 108) × (4.9 × 10-5)
Answer:
(7.7 × 108) × (4.9 × 10-5) =3.773 X 104

Explanation:
Given expressions as (7.7 × 108) × (4.9 × 10-5)
we work out the coefficients and the exponents separately.
and use the product rule; bm x b n = b(m + n) to multiply the bases.
So (7.7 × 108) × (4.9 × 10-5)  = (7.7 X 4.9) X (108 × 10-5) =
37.73 X (108-5) = 3.773 X 10 X 103 = 3.773 X 104.

Question 46.
(3.6 × 105) ÷ (1.8 × 109)
Answer:
(3.6 × 105) ÷ (1.8 × 109) = 2 X 10-4

Explanation:
Given expressions as (3.6 × 105) ÷ (1.8 × 109) ,
Separately we divide the coefficients and exponents.
We use the division rule of exponents,
where the exponents are subtracted.
Noted that when we are dividing exponential terms,
always subtract the denominator from the numerator.
So (3.6 × 105) ÷ (1.8 × 109)  = (3.6 ÷ 1.8 ) X ( 105-9) = 2 X 10-4.

Question 47.
A white blood cell has a diameter of about 0.000012 meter. How many times greater is the diameter of a white blood cell than the diameter of a red blood cell?
Big Ideas Math Answers Grade 8 Chapter 8 Exponents and Scientific Notation cr 47
Answer:
1.5 times greater is the diameter of a white blood cell
than the diameter of a red blood cell

Explanation:
Given a white blood cell has a diameter of about 0.000012 meter and
red blood cell has a diameter of about 8 X 10-6 meter,
the number of times greater is the diameter of a white blood cell
than the diameter of a red blood cell is 0.000012 ÷ 8 X 10-6 we write
0.000012 approximately equal to ≈ 1.2 X 10-5 now separately we
divide the coefficients and exponents.
We use the division rule of exponents,
where the exponents are subtracted.
Noted that when we are dividing exponential terms,
always subtract the denominator from the numerator.

so (1.2 X 10-5) ÷ 8 X 10-6 = (1.2 ÷ 8)  X (10-5 ÷ 10-6) =
0.15 X (10-5+6) = 0.15 X 10 = 1.5 therefore 1.5 times greater is
the diameter of a white blood cell than the diameter of a red blood cell.

Exponents and Scientific Notation Practice Test

Write the product using exponents.
Question 1.
(- 15) • (- 15) • (- 15)
Answer:
(- 15) • (- 15) • (- 15) = (-15)3

Explanation:
As -15 is multiplied by 3 times we write as
(-15)3 here -15 is base and 3 is exponent.

Question 2.
4 • 4 • x • x • x
Answer:
4 • 4 • x • x • x = (4)X (x)

Explanation:
As 4 is multiplied 2 times we write as (4)and x
is multiplied thrice so (x)3
so the expression is 4 • 4 • x • x • x = (4)X (x)

Evaluate the expression.
Question 3.
10 + 33 ÷ 9
Answer:
10 + 33 ÷ 9 = 13

Explanation:
Given expression 10 + 33 ÷ 9 first we write 9 as
3 X 3 = 32 ÷now we divide 33 ÷ 32 so we use rule
for finding \(\frac{a^{m}}{a^{n}}\) a quotient of
two powers with the same base as am-n, 33 ÷ 32 = ( 33-2) = 31= 3,
now we add 10 so 10 + 3 = 13.

Question 4.
Big Ideas Math Solutions Grade 8 Chapter 8 Exponents and Scientific Notation pt 4
Answer:
Big Ideas Math Solutions Grade 8 Chapter 8 Exponents and Scientific Notation pt 4= (-2)-1=
Explanation:
First we calculate separately values of numerators
and denominators then divide, we have numerator (-2)1 X (-2)-4 
we have same bases as -2 so we add powers as (-2)1-4 = (-2)-3
we have denominator (-2)-2
Now we have \(\frac{-2^{-3}}{-2^{-2}}\) so we use rule
for finding \(\frac{a^{m}}{a^{n}}\) a quotient of
two powers with the same base as am-n
\(\frac{-2^{-3}}{-2^{-2}}\) = (-2)-3+2 = (-2)-1n

Simplify the expression. Write your answer as a power.
Question 5.
910 • 9
Answer:
910 • 9 = 911

Explanation:
We write the given 910 • 91 expression as a power,
we have general rule for am • an = am+n
If product of two powers with the same base then
powers are added, so 910 • 91 as bases are same 9
powers are added 910+1 = 911.

Question 6.
(66)5
Answer:
(66)= 630

Explanation:
We write the given expression (66)5 as a power,
General rule for finding (am)power of a power,
If two powers have the same base then
we can multiply the powers as (am)n = am x n .
so (66)5 has powers of powers therefore powers
are multiplied as (6)6 X 5= (6)30

Question 7.
Big Ideas Math Solutions Grade 8 Chapter 8 Exponents and Scientific Notation pt 7
Answer:
Big Ideas Math Solutions Grade 8 Chapter 8 Exponents and Scientific Notation pt 7= (-3.5)6
Explanation:
First we calculate separately values of numerators
and denominators then divide, we have numerator (-3.5)13 X (-3.5)
we have same bases as -3.5 so we add powers as (-3.5)13+2 = (-3.5)15
we have denominator (-3.5)9
Now we have \(\frac{-3.5^{15}}{-3.5^{9}}\) so we use rule
for finding \(\frac{a^{m}}{a^{n}}\) a quotient of
two powers with the same base as am-n
\(\frac{-3.5^{15}}{-3.5^{9}}\) = (-3.5)15-9 = (-3.5)6.

Question 8.
Simplify (2y)7.
Answer:
(2y)7 =(2)7 X y7

Explanation:
Given to simplify the expression (2y)we use general rule to
write (ab)m  power of a product as am X bm ,So (2y)7 = (2)7 X y7

Round the number. Write the result as a product of a single digit and a power of 10.
Question 9.
4,610,428,970
Answer:
4,610,428,970 = 5 X 109

Explanation:
Given number is 4,610,428,970 is nearly or approximately
equal to ≈ 5,000,000,000 so we have 5 followed by 9 zeros,
so we write as 4,610,428,970 = 5 X 109 .

Question 10.
0.00000572
Answer:
0.00000572 = 5.72 X 10-6

Explanation:
Given number 0.00000572 is nearly or approximately
equal to ≈ 0.000006, we have divided 6 by 10 followed
by 6 zeros so we write as 6 X 10-6.

Write the number in standard form.
Question 11.
3 × 107
Answer:

3 X 107 = 30,000,000

Explanation:
Given 3 × 10the number in standard form is
3 X 10,000,000 = 30,000,000.
So 3 X 107 = 30,000,000.

Question 12.
9.05 × 10-3
Answer:
9.05 × 10-3 = 0.00905

Explanation:
Given 9.05 × 10-3  the number in standard form is
9.05 × 0.001 = 0.00905.

Evaluate the expression. Write your answer in scientific notation.
Question 13.
(7.8 × 107) + (9.9 × 107)
Answer:
(7.8 × 107) + (9.9 × 107) = 1.77 X 108

Explanation:
Given expressions as (7.8 × 107) + (9.9 × 107)
using the distributive property of multiplication over addition,
we get (7.8 × 107) + (9.9 × 107)  = (7.8 + 9.9 ) X 107 = 17.7 X 107 =
1.77 X 10 X 107 =1.77 X 108.

Question 14.
(6.4 × 105) – (5.4 × 104)
Answer:
(6.4 × 105) – (5.4 × 104) = 5.86 X 105

Explanation:
Given expressions as (6.4 × 105) – (5.4 × 104)
as the powers of 10 differ we need to modify before we factor.
We work around this by using our exponent property bm · bn = b (m+n)
to rewrite the 104 as 105 · So we take 104 and then grouping the 104 with 5.4.
So 5.4 X 104 becomes  0.54 X 105 therefore = (6.4 × 105) – (5.4 × 104) =
(6.4 X 105) – (5.4 X 10-1X 10 X 104) = (6.4 X 105) – (0.54 × 105)
using the distributive property of multiplication over subtraction,
we get (6.4 – 0.54) X 105 = 5.86 X 105.

Question 15.
(3.1 × 106) × (2.7 × 10-2)
Ans;
(3.1 × 106) × (2.7 × 10-2) = 8.37 X 104.

Explanation:
Given expressions as (3.1 × 106) × (2.7 × 10-2)
we work out the coefficients and the exponents separately.
and use the product rule; bm x b n = b(m + n) to multiply the bases.
So (3.1 × 106) × (2.7 × 10-2) = (3.1 X 2.7) X (106 × 10-2) =
8.37 X (106-2) = 8.37 X 104.

Question 16.
(9.6 × 107) ÷ (1.2 × 10-4)
Answer:
(9.6 × 107) ÷ (1.2 × 10-4) = 8 X 1011

Explanation:
Given expressions as (9.6 × 107) ÷ (1.2 × 10-4),
Separately we divide the coefficients and exponents.
Noted that when we are dividing exponential terms,
always subtract the denominator from the numerator.
So (9.6 × 107) ÷ (1.2 × 10-4) = (9.6 ÷ 1.2 ) X ( 107-(-4)) = 8 X 1011.

Question 17.
Is (xy2)3 the same as (xy3)2? Explain.
Answer:
(xy2)3 is not same as (xy3)2

Explanation:
Given expressions (xy2)3 the same as (xy3)
first we solve the expressions with general rule for
finding (am)power of a power,
If two powers have the same base then
we can multiply the powers as (am)n = am x n as
(xy2)3 = (x)3 X (y2)3 = (x)3 X (y2x3) = x3 X y6.
Now (xy3)2 = (x)2 X (y3)2 = (x)2 X (y3×2) = x2 X y6.
Now comparing  x3 X y6 with x2 X y6 as x powers are not same so
(xy2)3 is not same as (xy3)2.

Question 18.
One scoop of rice weighs about 39 milligrams.
a. Write a linear function that relates the weight of rice to the number of scoops. What is the weight of 5 scoops of rice?
b. A grain of rice weighs about 33 milligrams. About how many grains of rice are in 1 scoop?
Answer:
a. The linear function that relates the weight of rice to the
number of scoops is y= 39x ,The weight of 5 scoops of rice
is 98415 grams,
b. There are 729 grains in 1 scoop of rice

Explanation:
Given One scoop of rice weighs about 39 milligrams.
Let us take x as number of scoops and y represent
the weight of rice,
a.The linear function that relates the
weight of rice to the number of scoops is y= 39x ,
so the weight of 5 scoops of rice = y= 39 X 5 =
19683 X 5 = 98415 grams,
The weight of 5 scoops of rice is 98415 grams,
b. A grain of rice weighs about 33 milligrams the number of
grains of rice are in 1 scoop is 39 ÷ 33,We use the division rule
of exponents, where the exponents are subtracted.
So (39 ÷ 33) = 39-3 = 36 = 3 X 3 X 3 X 3 X 3 X 3 = 729,
therefore there are 729 grains in 1 scoop of rice.

Question 19.
There are about 10,000 taste buds on a human tongue.
Write this number in scientific notation.
Answer:
10,000 = 1 x 104 taste buds are there on a human tongue.

Question 20.
From 1978 to 2008, the amount of lead allowed in the air in the
United States was 1.5 × 10-6 gram per cubic meter.
In 2008, the amount allowed was reduced by 90%.
What is the new amount of lead allowed in the air?
Big Ideas Math Solutions Grade 8 Chapter 8 Exponents and Scientific Notation pt 20
Answer:
1.5 X 10-7 gram per cubic meter is the new amount of lead allowed in the air.

Explanation:
Given from 1978 to 2008, the amount of lead allowed in the air in the
United States was 1.5 × 10-6 gram per cubic meter.
In 2008, the amount allowed was reduced by 90%.
to know the new amount of lead allowed in the air, we see
as it was reduced by 90% then 10% is still allowed, to find 10% of
original amount multiply by 0.1 so 0.1 X 1.5 × 10-6 = 0.15 X 10-6 =
= 1.5 X 10-7 gram per cubic meter, Therefore 1.5 X 10-7 gram per cubic meter
is the new amount of lead allowed in the air.

Exponents and Scientific Notation Cumulative Practice

Big Ideas Math Solutions Grade 8 Chapter 8 Exponents and Scientific Notation cp 1
Answer :
Bit B, 3072 years.

Explanation:
Cats were tamed 3 X 210 years ago in Egypt, So it was
3 X 210 = 3 X 2 X 2 X 2 X 2 X 2 X 2 X 2 X 2 X 2 X 2 = 3072 Years,
So bit b is matched.

Question 1.
Mercury’s distance from the Sun is approximately 5.79 × 107 kilometers. What is this distance in standard form?
A. 5,790,000 km
B. 57,900,000 km
C. 579,000,000 km
D. 5,790,000,000 km
Answer:
The distance of 5.79 × 107 kilometers in standard form is approximately
equal to ≈ 57,900,000 km so matches with  B bit.

Explanation:
Given Mercury’s distance from the Sun is approximately 5.79 × 107 kilometers,
So the distance in standard form is the number in standard form is
5.79 X 10,000,000 = 57,900,000 km. So 5.79 X 107 km matches with B bit
57,900,000 km.

Question 2.
Your friend solves the problem. What should your friend change to correctly answer the question?
Big Ideas Math Solutions Grade 8 Chapter 8 Exponents and Scientific Notation cp 2
F. The left side of the equation should equal 360° instead of 180°.
G. The sum of the acute angles should equal 90.°
H. Evaluate the smallest angle when x = 15.
I. Evaluate the largest angle when x = 15.
Answer:
I. Evaluate the largest angle when x = 15.

Explanation:
Friend already solved x = 15°,So  the largest angle in the triangle
is 8x = 8 X 15°= 120°,So friend needs to
Evaluate the largest angle when x = 15. So bit I is the correct answer.

Question 3.
Which expression is equivalent to the expression 2423?
A. 212
B. 47
C. 48
D. 128
Answer:
2423 = 128, D bit

Explanation:
Given expression is 2423 we know when bases are same
product of powers property says add the powers.
So 2423 = 24 X 23 = 24+3= 27 =  2 X 2 X 2 X 2 X 2 X 2 X 2 = 128,
therefore 2423 =128, so it matches with bit D.

Question 4.
You randomly survey students in your school about whether they have a pet. You display your results in the two-way table. How many female students took the survey?
Big Ideas Math Solutions Grade 8 Chapter 8 Exponents and Scientific Notation cp 4
Answer:
Female 46 students took the survey.

Explanation:
As per two-way table it displays number of students in school
about whether they have a pet or not , So yes are 35 and no are 11,
Therefore total number of female students are 35 + 11 = 46,
So 46 female students took the survey.

Question 5.
A bank account pays interest so that the amount in the account doubles every 10 years. The account started with $5,000 in 1940. Which expression represents the amount (in dollars) in the account n decades later?
F. 2n • 5000
G. 5000 (n + 1)
H. 5000n
I. 2n + 5000
Answer:
F. 2n • 5000

Explanation:
Given a bank account pays interest so that the amount in the account
doubles every 10 years. The account started with $5,000 in 1940,
The expression that represents the amount (in dollars) in
the account n decades later is since the amount is getting doubled
every decade, So it will be increasing in multiples of  2,
so the correct answer is 2n X 5000 so bit  F.

Question 6.
The formula for the volume V of a pyramid is V = \(\frac{1}{3}\)Bh.
Which equation represents a formula for the height h of the pyramid?
A. h= \(\frac{1}{3}\) VB
B. h = \(\frac{3 V}{B}\)
C. h = \(\frac{V}{3B}\)
D. h = V – \(\frac{1}{3}\)B
Answer:
B. h = \(\frac{3 V}{B}\)

Explanation:
Given the formula for the volume V of a pyramid is V = \(\frac{1}{3}\)Bh,
If we cross multiply 3 with V divide by B we get height,
So the formula for the height h of the pyramid is h = 3 V X \(\frac{1}{B}\).
Therefore the correct answer is bit B, h = \(\frac{3 V}{B}\).

Question 7.
The gross domestic product (GDP) is a way to measure how much a country produces economically in a year. The table below shows the approximate population and GDP for the United States.
Big Ideas Math Solutions Grade 8 Chapter 8 Exponents and Scientific Notation cp 7
Part A Write the population and the GDP using scientific notation.
Part B Find the GDP per person for the United States using your answers from PartA. Write your answer in scientific notation. Show your work and explain your reasoning.
Answer:
Part A : Population is 3.24 X 108,
GDP is $1.86 X 1013
Part B :
The GDP per person for the United States is 5.74 X 105

Explanation:
Part A:
Given the population of United States, 2016 is 324,000,000 the scientific
notation is 3.24 X 100,000,000 = 3.24 X 108 and GDP is
$18,600,000,000,000 in scientific notation is $1.86 X 10,000,000,000,000 =
$1.86 X 1013.
Part B :
The GDP per person is dividing the GDP of a country by its population,
So ($1.86 X 1013) ÷ (3.24 X 108) ,Separately we divide the coefficients and exponents.
Noted that when we are dividing exponential terms,
always subtract the denominator from the numerator,
Therefore (1.86 ÷ 3.24) X (1013 ÷ 108)  = 0.574 X (1013-8) =
0.574 X 105 = 0.574  X 10 X 104 = 5.74 X 105.

Question 8.
What is the equation of the line shown in the graph?
Big Ideas Math Solutions Grade 8 Chapter 8 Exponents and Scientific Notation cp 8
Answer:
F. y = –
Explanation:
The equation of line is y = mx + b, Where m is slope and
b is y – intercept So m= (y2-y1) ÷ (x2 -x1)
from graph we have (x1,y1)= (3,2) and (x2,y2) = (-3,4)
m= (4-2) ÷ (-3-3) = 2 ÷ (-6) = (-1 ÷ 3) = now y-intercept is y value of the point where the line intersects the y- axis.
So here it is 3, Now the equation of the line is y =-y = -(1 ÷

Question 9.
Which graph represents a linear function?
Big Ideas Math Solutions Grade 8 Chapter 8 Exponents and Scientific Notation cp 9
Answer:
Graph In B bit represents a linear function

Explanation:
Linear function is one where if there is a constant rate of change,
If we see all the graphs only in graph B there is constant rate of change,
in graph A it is increasing, decreasing and again increasing,
in graph C it is higher and at a time decreased
in graph D it decreased and again increased,
Therefore only in graph B , therefore only in graph B
it represents a linear function.

Question 10.
Find (- 2.5)-2.
Big Ideas Math Solutions Grade 8 Chapter 8 Exponents and Scientific Notation cp 10
Answer:
(- 2.5)-2 =

Explanation:
Given expression as (- 2.5)-2 we write the expression as
positive exponents by using So (- 2.5)-2  is

Question 11.
Two lines have the same y-intercept. The slope of one line is 1, and
the slope of the other line is – 1. What can you conclude?
F. The lines are parallel.
G. The lines meet at exactly one point.
H. The lines meet at more than one point.
I. The situation described is impossible.
Answer:
G. The lines meet at exactly one point.

Explanation:
If two lines have different slopes, they cannot be the same line.
However, if they share a y-intercept, that means they cross the
y-axis at the same y value. Since the x value is constant on the
y-axis (0), they also share an x-value here. If they share an x value
and corresponding y value, they intersect at that point,
in this case their y-intercept and two lines can’t intersect
at more than one point, So we conclude the lines meet at exactly one point.

Question 12.
Which list of ordered pairs represents the mapping diagram?
Big Ideas Math Solutions Grade 8 Chapter 8 Exponents and Scientific Notation cp 12
A. (1, 2), (2, 0), (3, – 2)
B. (1, 0), (2, 2), (3, – 2)
C. (1, 0), (2, 2), (2, – 2), (3, – 2)
D. (0, 1), (2, 2), (- 2, 2), (- 2, 3)
Answer:
C. (1, 0), (2, 2), (2, – 2), (3, – 2)

Explanation:
A mapping shows how the elements are paired.
Its like a flow chart for a function, showing the input and output values.
A mapping diagram consists of two parallel columns..
Lines or arrows are drawn from domain to range,
to represent the relation between any two elements.
So by seeing as 1 is related 0, 2 is related to 2, -2 and 3 is related to -2,
therefore the ordered pairs are (1,0),(2,2),(2,-2),(3,-2) matches with
bit C.

Eureka Math Grade 3 Module 1 End of Module Assessment Answer Key

Engage NY Eureka Math 3rd Grade Module 1 End of Module Assessment Answer Key

Eureka Math Grade 3 Module 1 Answer Key

Eureka Math Grade 3 Module 1 End of Module Assessment Task Answer Key

Question 1.
Mr. Lewis arranges all the desks in his classroom into 6 equal groups of 4.
How many desks are in his classroom? Show a picture and
multiplication sentence in your work.
a. What does the product in your multiplication sentence represent?

24 desks are there in Mr. Lewis classroom,
Multiplication Sentence: 4 X 6 = 24,
the product in your multiplication sentence represent
there are 24 desks.
Eureka Math Grade 3 Module 1 End of Module Assessment Task Answer Key-1

Explanation:
Given Mr. Lewis arranges all the desks in his classroom into
6 equal groups of 4.
So total desks his classroom are 24,
Shown a picture and multiplication sentence is
4 X 6 = 24.

b. Fill in the blanks below to complete a related division sentence.
_24_ ÷ 4 = ___6___,

Explanation:
Related division sentence is 24 ÷ 4 = 6,
when 24 is divided into 4 times we get 6 parts each.

c. What does the quotient in Part (b) represent?
Quotient represents 6 parts,

Explanation:
The quotient in the division sentence 24 ÷ 4 represents
24 is divided by 4 then we get quotient as 6 equal parts.

Question 2.
a. Draw an array that shows 9 rows of 2.
Write a multiplication sentence to represent the array, and
circle the factor that represents the number of rows.
b. Draw another array that shows 2 rows of 9.
Write a different multiplication sentence, and circle the factor
that represents the size of the row.
c. Explain the relationship between the two arrays using
number sentences and words.

a.
Eureka Math Grade 3 Module 1 End of Module Assessment Task Answer Key-2
Explanation:
Drawn an array that shows 9 rows of 2,
Wrote a multiplication sentence to represent the array as
9 X 2 = 18 and circled the factor 9 that represents
the number of rows.

b.
Eureka Math Grade 3 Module 1 End of Module Assessment Task Answer Key-3
Explanation:
Drawn another array that shows 2 rows of 9.
Wrote a different multiplication sentence as 2 X 9 = 18
and circled the factor that 2 represents the size of the row.

c. The relationship between the two arrays is the results is same,

Explanation:
The relationship between the two arrays is same as
number sentences is 9 X 2 = 18, 2 X 9 = 18, means
if 9 rows are multiplied by 2 times we get 18 and if
2 is multiplied by 9 times also we get result as 18 only.
therefore if x rows and y columns are there then x X y =
y X x the product result will be the same.

Question 3.
Ms. Park buys a tray of apples for a class party.
There are 5 rows of 4 red apples. There is 1 row of 4 green apples.
a. The picture below shows Ms. Park’s apples. Fill in the blanks to
complete the expressions.
Eureka Math Grade 3 Module 1 End of Module Assessment Task Answer Key 1
Eureka Math Grade 3 Module 1 End of Module Assessment Task Answer Key-4
Explanation:
Given Ms. Park buys a tray of apples for a class party.
There are 5 rows of 4 red apples.
There is 1 row of 4 green apples,
The expression is Total apples : 6 X 4,
Red apples : 5 X 4, Green apples : 1 X 4.

b. Fill in the unknowns in the equation below to match
the picture of the apples in Part (a). Use the break apart and
distribute strategy to find the total number of apples Ms. Park bought.
_____ × 4 = _____ × 4 + _____ × 4
Ms. Park bought ________ apples.

Equation is 6 X 4 = ((5 X 4) + (1 X 4)),
Total number of apples Ms. Park bought 24 apples.

Explanation:
The equation below to match the picture of the apples in
Part (a). Using the break apart and distribute strategy is
Equation is 6 X 4 = ((5 X 4) + (1 X 4)),
Total number of apples Ms. Park bought are 24 apples.

c. Lilly brings 8 green apples for the class party.
Show Lilly’s green apples on the picture in Part (a).
Then, fill in the unknowns in the equation below to match
the new picture. Solve to find the total number of apples.
_____ × 4 = _____ × 4 + _____ × 4
There are ________ apples in all.
Eureka Math Grade 3 Module 1 End of Module Assessment Task Answer Key-5

Equation is 8 X 4 = ((5 X 4) + (3 X 4)),
Total number of apples are 32.

Explanation:
Given Lilly brings 8 green apples for the class party.
Shown Lilly’s green apples in the picture in Part (a)
as above, Then, filled the unknowns in the equation below to match
the new picture as 8 X 4 = ((5 X 4) + (3 X 4)), on
solving to find the total number of apples are 20 + 12 =
32 apples.

Question 4.
Mr. Myer’s class plays a game. The class earns 5 points
each time they answer a question correctly. The class earns
50 points playing the game on Monday.
a. How many questions did the class answer correctly?
Show a picture and division sentence in your work.

The class answered 10 questions correctly,
Eureka Math Grade 3 Module 1 End of Module Assessment Task Answer Key-6
The division sentence is 50 ÷ 5= 10,

Explanation:
Given Mr. Myer’s class plays a game, The class earns 5 points
each time they answer a question correctly. The class earns
50 points playing the game on Monday.
So the number of questions the class answered correctly
is (50 ÷ 5= 10) 10 questions.
Shown a picture and division sentence in the picture above,
Division Sentence : 50 ÷ 5= 10.

b. Mr. Myer uses the equation 5 × _____ = 50 to find
how many questions the class answered correctly.
Is his method correct? Why or why not?

His method is correct only but it is time consuming,

Explanation:
Given Mr. Myer uses the equation 5 × _____ = 50 to find
how many questions the class answered correctly.
His method is correct, Why because he goes on
substituting numbers till he gets the answer,
Mr. Myer gets the answer as 5 X 10 = 50, But his method
is time consuming and little long.

c. The class answered 7 questions correctly on Tuesday.
What is the total number of points the class
earned on both days?

The total number of points the class
earned on both days is 50 + 35 = 85 points,

Explanation:
Given the class answered 7 questions correctly on Tuesday.
So the total number of points the class earned on both days
is Monday – 50 points, Tuesday – 5 X 7 = 35 points,
In total 50 + 35 = 85 points.

Question 5.
Complete as many problems as you can in 100 seconds.
Your teacher will time you and tell you when to stop.
Eureka Math Grade 3 Module 1 End of Module Assessment Task Answer Key 5

Eureka Math Grade 3 Module 1 End of Module Assessment Task Answer Key 6

All most completed the given problems in 100 seconds,

Explanation:
Completed as many problems as I can in 100 seconds,
till my teacher says when to stop as shown below,
Eureka Math Grade 3 Module 1 End of Module Assessment Task Answer Key-7

Eureka Math Grade 3 Module 1 End of Module Assessment Task Answer Key-8

Eureka Math Grade 3 Module 1 Mid Module Assessment Answer Key

Engage NY Eureka Math 3rd Grade Module 1 Mid Module Assessment Answer Key

Eureka Math Grade 3 Module 1 Answer Key

Eureka Math Grade 3 Module 1 Mid Module Assessment Task Answer Key

Question 1.
Mrs. Tran plants 2 rows of 5 carrots in her garden.
a. Draw an array that represents Mrs. Tran’s carrots. Use an X to show each carrot.
Eureka Math Grade 3 Module 1 Mid Module Assessment Task Answer Key-1
Explanation:
Given Mrs. Tran plants 2 rows of 5 carrots in her garden,
Drawn an array that represents Mrs. Tran’s carrots,
Using  an X to show each carrot as shown above.

b. Mrs. Tran adds 3 more rows of 5 carrots to her garden.
• Use circles to show her new carrots on the array in Part (a).
• Fill in the blanks below to show how she added the five rows.
__ fives + __ fives = __ fives
• Write a sentence to explain your thinking.

Eureka Math Grade 3 Module 1 Mid Module Assessment Task Answer Key-2

(_2) X_ fives + (3)_X_ fives = _(5)_X fives,
2 times 5 and 3 times 5 makes 5 times fives,

Explanation:
Given Mrs. Tran adds 3 more rows of 5 carrots to her garden.
Using circles shown her new carrots on the array in Part (a),
To show how she added the five rows as 2 X 5 + 3 X 5 = 5 X 5,
Sentence is 2 times by five and 3 times by five makes 5 by five times.

c. Find the total number of carrots Mrs. Tran planted.

Total number of carrots Mrs. Tran planted are 25,

Explanation:
Total number of carrots Mrs. Tran planted are ( 2 X 5 + 3 X 5) =
10 + 15 = 25 carrots.

d. Write a multiplication sentence to describe the array
representing the total number of carrots Mrs. Tran planted.

Multiplication Sentence : (2 + 3) X 5 = 5 X 5,

Explanation:
Multiplication sentence to describe the array
representing the total number of carrots Mrs. Tran planted is
(2 + 3) X 5 = 5 X 5 = 25.

Question 2.
Mrs. Tran picks 15 tomatoes from her garden.
She puts 5 tomatoes in each bag.
a. Draw Mrs. Tran’s bags of tomatoes.
b. Write a multiplication sentence that describes your drawing in Part (a).

a. Eureka Math Grade 3 Module 1 Mid Module Assessment Task Answer Key-3
Explanation:
Drawn to show Mrs. Tran’s bags of tomatoes,
5 tomatoes in each bag.

b. Multiplication Sentence : 5 X 3 = 15,

Explanation:
Multiplication sentence that describes my drawing in Part (a)
is we have 5 tomatoes in each bag so 5 X 3 = 15 and Mrs. Tran’s
picked 15 tomatoes from her garden.

Question 3.
Mrs. Tran plants 12 sunflowers in her garden. She plants them in 3 rows.
a. Fill in the blanks below to make a true division sentence.
What does the answer represent?
__ ÷ __ = __

a. Division sentence : 12 ÷ 3 = 4,
4 sunflowers in a row,

Explanation:
Given Mrs. Tran plants 12 sunflowers in her garden.
She plants them in 3 rows.
Division sentence is 12 ÷ 3 = 4,
The 4 represents there are 4 sunflower plants in a row.

b. Mrs. Tran adds 2 more identical rows of sunflowers to
her 3 original rows. Draw an array to show how many flowers she has now.

Eureka Math Grade 3 Module 1 Mid Module Assessment Task Answer Key-4
Explanation:
Drawn an array to show the number of sun flowers
Mrs. Tran has 24 now.

c. Mrs. Tran figured out how many flowers she planted.
Her work is shown in the box below. Would Mrs. Tran get
the same result if she multiplied 5 × 4? Explain why or why not.
Eureka Math Grade 3 Module 1 Mid Module Assessment Task Answer Key 1
Yes, Mrs. Tran get the same result if she
multiplied 5 X 4,

Explanation:
Mrs. Tran worked as (3 X 4) + (2 X 4) = 12 + 8 = 20,
and Mrs. Tran got the same result if she multiply 5 X 4 = 20,
because in the equation both 3 and 2 are multiplied by 4,
So we add 3 and 2 first we get 5 and multiply by 4, So
the result remains same 20 for 5 X 4 also.

Eureka Math Grade 3 Module 1 Lesson 21 Answer Key

Engage NY Eureka Math 3rd Grade Module 1 Lesson 21 Answer Key

Eureka Math Grade 3 Module 1 Answer Key

Engage NY Math Grade 3 Module 1 Lesson 21 Pattern Sheet Answer Key

Multiply.
EngageNY Math Grade 3 Module 1 Lesson 21 Pattern Sheet Answer Key 1
EngageNY Math Grade 3 Module 1 Lesson 21 Pattern Sheet Answer Key 2

multiply by 5 (1–5)
Eureka Math Grade 3 Module 1 Lesson 21 Answer Key-1
Eureka Math Grade 3 Module 1 Lesson 21 Answer Key-2
Explanation :
Multiplied the given numbers and wrote respectively.

Eureka Math Grade 3 Module 1 Lesson 21 Problem Set Answer Key

Question 1.
Jason earns $6 per week for doing all his chores.
On the fifth week, he forgets to take out the trash,
so he only earns $4. Write and solve an equation to show
how much Jason earns in 5 weeks.
Eureka Math Grade 3 Module 1 Lesson 21 Problem Set Answer Key 3
Jason earns ___$28__________.
Eureka Math Grade 3 Module 1 Lesson 21 Answer Key-3
Jason earns $28 in 5 weeks,

Explanation:
Given Jason earns $6 per week for doing all his chores.
On the fifth week, he forgets to take out the trash,
so he only earns $4. Equation of Jason’s earnings is
$6 +$6 + $6 +$6 + $4 = $28 is Jason earnings in 5 weeks.

Question 2.
Miss Lianto orders 4 packs of 7 markers. After passing out
1 marker to each student in her class, she has 6 left.
Label the tape diagram to find how many students are in Miss Lianto’s class.
Eureka Math Grade 3 Module 1 Lesson 21 Problem Set Answer Key 3.1
There are __22__ students in Miss Lianto’s class.
Eureka Math Grade 3 Module 1 Lesson 21 Answer Key-4
There are __22__ students in Miss Lianto’s class.

Explanation:
Given Miss Lianto orders 4 packs of 7 markers. After passing out
1 marker to each student in her class, she has 6 left.
Labeled the tape diagram to show how many students
are in Miss Lianto’s class as
4 packs of 7 markers means number of markers are
7 X 4 = 28 or 7 + 7 + 7 + 7 = 28,
Each student got 1 marker in her class and 6 left,
means number of students in Miss Lianto’s class are
28 – 6 = 22 students as labeled in the tape diagram.
Therefore, there are  22 students in Miss Lianto’s class.

Question 3.
Orlando buys a box of 18 fruit snacks. Each box comes with
an equal number of strawberry-, cherry-, and grape-flavored snacks.
He eats all of the grape-flavored snacks. Draw and label a tape
diagram to find how many fruit snacks he has left.
Eureka Math Grade 3 Module 1 Lesson 21 Answer Key-5
Fruit snacks Orlando left are 12,

Explanation:
Given Orlando buys a box of 18 fruit snacks. Each box comes with
an equal number of strawberry-, cherry-, and grape-flavored snacks,
Orlando eats all of the grape-flavored snacks.
Drawn and labeled a tape
diagram to show how many fruit snacks Orlando has left are
Total fruit snacks 18, equal number of 3 fruit flavored snacks,
So each flavor is 18 ÷ 3 = 6 each, Now Orlando ate all of the grape –
flavored snacks, So fruit snacks Orlando has left are 18 – 6 = 12.
Therefore, Orlando has left 12 fruit snacks.

Question 4.
Eudora buys 21 meters of ribbon. She cuts the ribbon so that each
piece measures 3 meters in length.
a. How many pieces of ribbon does she have?
b. If Eudora needs a total of 12 pieces of the shorter ribbon,
how many more pieces of the shorter ribbon does she need?

a. Eudora has 7 pieces of ribbon,

Explanation:
Given Eudora buys 21 meters of ribbon. She cuts the ribbon
so that each piece measures 3 meters in length,
So number of pieces of ribbon she has is 21 ÷ 3 = 7 pieces of
ribbon.

b. Eudora needs more 5 pieces of the shorter ribbon,

Explanation:
If Eudora needs a total of 12 pieces of the shorter ribbon,
She needs more pieces of the shorter ribbon as 12 – 7 = 5 pieces.

Engage NY Math 3rd Grade Module 1 Lesson 21 Exit Ticket Answer Key

Ms. Egeregor buys 27 books for her classroom library.
She buys an equal number of fiction, nonfiction, and poetry books.
She shelves all of the poetry books first. Draw and label a tape diagram
to show how many books Ms. Egeregor has left to shelve.

Ms. Egeregor has left with 18 more books to shelve,
Eureka Math Grade 3 Module 1 Lesson 21 Answer Key-6

Explanation:
Given Ms. Egeregor buys 27 books for her classroom library.
She buys an equal number of fiction, nonfiction, and poetry books.
So each numbers of books are 27 ÷ 3 = 9,
9 books of fiction, 9 books of nonfiction and 9 books of poetry.
She shelves all of the poetry books first.
Drawn and labeled a tape diagram
to show number of books Ms. Egeregor has left to shelve is
27 – 9 = 18 more books.

Eureka Math 3rd Grade Module 1 Lesson 21 Homework Answer Key

Question 1.
Tina eats 8 crackers for a snack each day at school.
On Friday, she drops 3 and only eats 5. Write and solve
an equation to show the total number of crackers Tina eats during the week.
Eureka Math 3rd Grade Module 1 Lesson 21 Homework Answer Key 4
Tina eats ____37______ crackers.

Tina eats 37 crackers during the week,
Eureka Math Grade 3 Module 1 Lesson 21 Answer Key-7

Explanation:
Given Tina eats 8 crackers for a snack each day at school.
On Friday, she drops 3 and only eats 5. So Wrote and solved
an equation to show the total number of crackers Tina
ate during the week is 8 + 8 + 8 + 8 + 5 = 37 crackers.

Question 2.
Ballio has a reading goal. He checks 3 boxes of 9 books
out from the library. After finishing them, he realizes that
he beat his goal by 4 books! Label the tape diagrams to find
Ballio’s reading goal.
Eureka Math 3rd Grade Module 1 Lesson 21 Homework Answer Key 5
Ballio’s goal is to read ___23____ books.
Eureka Math Grade 3 Module 1 Lesson 21 Answer Key-8
Explanation :
Given Ballio has a reading goal. He checks 3 boxes of 9 books
out from the library. 3 X 9 = 27 books,
After finishing them, he realizes that he beat his goal by 4 books!
So his goal was 27 – 4 = 23 books,
Labeled the tape diagrams to show Ballio’s reading goal as 23 books.

Question 3.
Mr. Nguyen plants 24 trees around the neighborhood pond.
He plants equal numbers of maple, pine, spruce, and birch trees.
He waters the spruce and birch trees before it gets dark.
How many trees does Mr. Nguyen still need to water?
Draw and label a tape diagram.

Mr. Nguyen still needs to water 12 trees,

Explanation:
Given Mr. Nguyen plants 24 trees around the neighborhood pond.
He plants equal numbers of maple, pine, spruce, and birch trees.
So each numbers of trees are 24 ÷ 4 = 6 trees,
He waters the spruce and birch trees before it gets dark.
So number of more trees Mr. Nguyen still need to water is
24 – 12 = 12 trees.
Eureka Math Grade 3 Module 1 Lesson 21 Answer Key-9
Drawn and labeled a tape diagram to show Mr. Nguyen
still needs to water 12 trees.

Question 4.
Anna buys 24 seeds and plants 3 in each pot. She has 5 pots.
How many more pots does Anna need to plant all of her seeds?

Anna needs 3 more pots to plant all of her seeds.

Explanation:
Given Anna buys 24 seeds and plants 3 in each pot. She has 5 pots.
So she planted 5 X 3 = 15 seeds, Seeds left are 24 – 15 = 9 seeds,
As each pot she can plant 3 seeds she needs pots for 9 seeds as 9 ÷ 3 = 3,
Therefore, number of more pots Anna need to plant all of her seeds
are 3 pots.

Eureka Math Grade 3 Module 1 Lesson 20 Answer Key

Engage NY Eureka Math 3rd Grade Module 1 Lesson 20 Answer Key

Eureka Math Grade 3 Module 1 Answer Key

Eureka Math Grade 3 Module 1 Lesson 20 Sprint Answer Key

A
Skip-Count by 5
Eureka Math Grade 3 Module 1 Lesson 20 Sprint Answer Key 21
Eureka Math Grade 3 Module 1 Lesson 20 Sprint Answer Key 22
Eureka Math Grade 3 Module 1 Lesson 20 Sprint Answer Key 23
Eureka Math Grade 3 Module 1 Lesson 20 Sprint Answer Key 24
Eureka Math Grade 3 Module 1 Lesson 20 Answer Key-1
Eureka Math Grade 3 Module 1 Lesson 20 Answer Key-2
Eureka Math Grade 3 Module 1 Lesson 20 Answer Key-3
Eureka Math Grade 3 Module 1 Lesson 20 Answer Key-4

Question 1.
0, 5, __
0, 5, 10,

Explanation:
Given 0, 5, ___ skip count by 5,
0, 0 + 5 = 5, 5 + 5 = 10,
So, 0, 5, 10.

Question 2.
5, 10, __
5, 10, 15,

Explanation:
Given 5, 10, ___ skip count by 5,
5, 5 + 5 = 10, 10 + 5 = 15,
So, 5, 10, 15.

Question 3.
10, 15, __
10, 15, 20,

Explanation:
Given 10, 15, ___ skip count by 5,
10, 10 + 5 = 15, 15 + 5 = 20,
So, 10, 15, 20.

Question 4.
15, 20, __
15, 20, 25,

Explanation:
Given 15, 20, ___ skip count by 5,
15, 15 + 5 = 20, 20 + 5 = 25,
So, 15, 20, 25.

Question 5.
20, 25, __
20, 25, 30,

Explanation:
Given 20, 25, ___ skip count by 5,
20, 20 + 5 = 25, 25 + 5 = 30,
So, 20, 25, 30.

Question 6.
25, 30, __
25, 30, 35,

Explanation:
Given 25, 30, ___ skip count by 5,
25, 25 + 5 = 30, 30 + 5 = 35,
So, 25, 30, 35.

Question 7.
30, 35, __
30, 35, 40,

Explanation:
Given 30, 35, ___ skip count by 5,
30, 30 + 5 = 35, 35 + 5 = 40,
So, 30, 35, 40.

Question 8.
35, 40, __
35, 40, 45,

Explanation:
Given 35, 40, ___ skip count by 5,
35, 35 + 5 = 40, 40 + 5 = 45,
So, 35, 40, 45.

Question 9.
40, 45, __
40, 45, 50,

Explanation:
Given 40, 45, ___ skip count by 5,
40, 40 + 5 = 45, 45 + 5 = 50,
So, 40, 45, 50.

Question 10.
50, 45, __
50, 45, 40,

Explanation:
Given 50, 45, ___ skip count by 5,
50, 50 – 5 = 45, 45 – 5 = 40,
So, 50, 45, 40.

Question 11.
45, 40, __
45, 40, 35,

Explanation:
Given 45, 40, ___ skip count by 5,
45, 45 – 5 = 40, 40 – 5 = 35,
So, 45, 40, 35.

Question 12.
40, 35, __
40, 35, 30,

Explanation:
Given 40, 35, ___ skip count by 5,
40, 40 – 5 = 35, 35 – 5 = 30,
So, 40, 35, 30.

Question 13.
35, 30, __
35, 30, 25,

Explanation:
Given 35, 30, ___ skip count by 5,
35, 35 – 5 = 30, 30 – 5 = 25,
So, 35, 30, 25.

Question 14.
30, 25, __
30, 25, 20

Explanation:
Given 30, 25, ___ skip count by 5,
30, 30 – 5 = 25, 25 – 5 = 20,
So, 30, 25, 20.

Question 15.
25, 20, __
25, 20,15,

Explanation:
Given 25, 20, ___ skip count by 5,
25, 25 – 5 = 20, 20 – 5 = 15,
So, 25, 20, 15.

Question 16.
20, 15, __
20, 15, 10,

Explanation:
Given 20, 15, ___ skip count by 5,
20, 20 – 5 = 15, 15 – 5 = 10,
So, 20, 15, 10.

Question 17.
15, 10, __
15, 10, 5,

Explanation:
Given 15, 10, ___ skip count by 5,
15, 15 – 5 = 10, 10 – 5 = 5,
So, 15, 10, 5.

Question 18.
0, __, 10
0, 5, 10,

Explanation:
Given 0, ___, 10 skip count by 5,
0, 0 + 5 = 5, 5 + 5 = 10,
So, 0, 5, 10.

Question 19.
25, __, 35
25, 30, 35,

Explanation:
Given 25, ___, 35 skip count by 5,
25, 25 + 5 = 30, 30 + 5 = 35,
So, 25, 30, 35.

Question 20.
5, __, 15
5, 10, 15,

Explanation:
Given 5, ___, 15 skip count by 5,
5, 5 + 5 = 10, 10 + 5 = 15,
So, 5, 10, 15.

Question 21.
30, __, 40
30, 35, 40,

Explanation:
Given 30, ___, 40 skip count by 5,
30, 30 + 5 = 35, 35 + 5 = 40,
So, 30, 35, 40.

Question 22.
10, __, 20
10, 15, 20,

Explanation:
Given 10, ___, 20 skip count by 5,
10, 10 + 5 = 15, 15 + 5 = 20,
So, 10, 15, 20.

Question 23.
35, __, 45
35, 40, 45,

Explanation:
Given 35, ___, 45 skip count by 5,
35, 35 + 5 = 40, 40 + 5 = 45,
So, 35, 40, 45.

Question 24.
15, __, 25
15, 20, 25,

Explanation:
Given 15, ___, 25 skip count by 5,
15, 15 + 5 = 20, 20 + 5 = 25,
So, 15, 20, 25.

Question 25.
40, __, 50
40, 45, 50,

Explanation:
Given 40, ___, 50 skip count by 5,
40, 40 + 5 = 45, 45 + 5 = 50,
So, 40, 45, 50.

Question 26.
25, __, 15
25, 20, 15,

Explanation:
Given 25, ___, 15 skip count by 5,
25, 25 – 5 = 20, 20 – 5 = 15,
So, 25, 20, 15.

Question 27.
50, __, 40
50, 45, 40,

Explanation:
Given 50, ___, 40 skip count by 5,
50, 50 – 5 = 45, 45 – 5 = 40,
So, 50, 45, 40.

Question 28.
20, __, 10
20, 15, 10,

Explanation:
Given 20, ___, 10 skip count by 5,
20, 20 – 5 = 15, 15 – 5 = 10,
So, 20, 15, 10.

Question 29.
45, __, 35
45, 40, 35,

Explanation:
Given 45, ___, 35 skip count by 5,
45, 45 – 5 = 40, 40 – 5 = 35,
So, 45, 40, 35.

Question 30.
15, __, 5
15, 10, 5,

Explanation:
Given 15, ___, 5 skip count by 5,
15, 15 – 5 = 10, 10 – 5 = 5,
So, 15, 10, 5.

Question 31.
40, __, 30
40, 35, 30,

Explanation:
Given 40, ___, 30 skip count by 5,
40, 40 – 5 = 35, 35 – 5 = 30,
So, 40, 35, 30.

Question 32.
10, __, 0
10, 5, 0,

Explanation:
Given 10, ___, 0 skip count by 5,
10, 10 – 5 = 5, 5 – 5 = 0,
So, 10, 5, 0.

Question 33.
35, __, 25
35, 30, 25,

Explanation:
Given 35, ___, 25 skip count by 5,
35, 35 – 5 = 30, 30 -5 = 25,
So, 35, 30, 25.

Question 34.
__, 10, 5
15, 10, 5,

Explanation:
Given  ___,10, 5 skip count by 5,
15, 5 + 5 = 10, 5 + 0 = 5,
So, 15, 10, 5.

Question 35.
__, 35, 30
40, 35, 30,

Explanation:
Given  ___,35, 30 skip count by 5,
40, 30 + 5 = 35, 25 + 5 = 30,
So, 40, 35, 30.

Question 36.
__, 15, 10
20, 15, 10,

Explanation:
Given  ___,15, 10 skip count by 5,
20, 10 + 5 = 15, 5 + 5 = 10,
So, 20, 15, 10.

Question 37.
__, 40, 35
45, 40, 35,

Explanation:
Given  ___,40, 35 skip count by 5,
45 = 40 + 5, 40 = 35 + 5, 35 = 30 + 5,
So, 45, 40, 35.

Question 38.
__, 20, 15
25, 20, 15,

Explanation:
Given  ___,20, 15 skip count by 5,
25 = 20 + 5, 20 = 15 + 5, 15 = 10 + 5,
So, 25, 20, 15.

Question 39.
__, 45, 40
50, 45, 40,

Explanation:
Given  ___,45, 40 skip count by 5,
50 = 45 + 5, 45 = 40 + 5, 40 = 35 + 5,
So, 50, 45, 40.

Question 40.
50, 55, __
50, 55, 60,

Explanation:
Given 50, 55, ___ skip count by 5,
50, 50 + 5 = 55, 55 + 5 = 60,
So, 50, 55, 60.

Question 41.
45, 50, __
45, 50, 55,

Explanation:
Given 45, 50, ___ skip count by 5,
45, 45 + 5 = 50, 50 + 5 = 55,
So, 45, 50, 55.

Question 42.
65, __, 55
65, 60, 55

Explanation:
Given 65, ___, 55 skip count by 5,
65, 55 + 5 = 60, 50 + 5 = 55,
So, 65, 60, 55.

Question 43.
55, 60, __
55, 60, 65,

Explanation:
Given 55, 60, ___ skip count by 5,
50 + 5 = 55, 55 + 5 = 60, 60 + 5 = 65,
So, 55, 60, 65.

Question 44.
60, 65, __
60, 65, 70

Explanation:
Given 60, 65, ___ skip count by 5,
55 + 5 = 60, 60 + 5 = 65, 65 + 5 = 70,
So, 60, 65, 70.

B
Skip-Count by 5
Eureka Math Grade 3 Module 1 Lesson 20 Sprint Answer Key 25
Eureka Math Grade 3 Module 1 Lesson 20 Sprint Answer Key 26
Eureka Math Grade 3 Module 1 Lesson 20 Sprint Answer Key 27
Eureka Math Grade 3 Module 1 Lesson 20 Sprint Answer Key 28

Eureka Math Grade 3 Module 1 Lesson 20 Answer Key-5

Eureka Math Grade 3 Module 1 Lesson 20 Answer Key-6
Eureka Math Grade 3 Module 1 Lesson 20 Answer Key-7

Eureka Math Grade 3 Module 1 Lesson 20 Answer Key-8

Question 1.
5, 10, __
5, 10, 15,

Explanation:
Given 5, 10, ___skip count by 5,
0 + 5 = 5, 5 + 5 = 10, 10 + 5 = 15,
So, 5, 10, 15.

Question 2.
10, 15, __
10, 15, 20

Explanation:
Given 10, 15, ___skip count by 5,
5 + 5 = 10, 10 + 5 = 15, 15 + 5 = 20,
So, 10, 15, 20.

Question 3.
15, 20, __
15, 20, 25.

Explanation:
Given 15, 20, ___skip count by 5,
10 + 5 = 15, 15 + 5 = 20, 20 + 5 = 25,
So, 15, 20, 25.

Question 4.
20, 25, __
20, 25, 30.

Explanation:
Given 20 25, ___skip count by 5,
15 + 5 = 20, 20 + 5 = 25, 25 + 5 = 30,
So, 20, 25, 30.

Question 5.
25, 30, __
25, 30, 35.

Explanation:
Given 25, 30, ___skip count by 5,
20 + 5 = 25, 25 + 5 = 30, 30 + 5 = 35,
So, 25, 30, 35.

Question 6.
30, 35, __
30, 35,40.

Explanation:
Given 30, 35, ___skip count by 5,
25 + 5 = 30, 30 + 5 = 35, 35 + 5 = 40,
So, 30, 35, 40.

Question 7.
35, 40, __
35, 40, 45.

Explanation:
Given 35, 40, ___skip count by 5,
30 + 5 = 35, 35 + 5 = 40, 40 + 5 = 45,
So, 35, 40, 45.

Question 8.
40, 45, __
40, 45, 50.

Explanation:
Given 40, 45, ___skip count by 5,
35 + 5 = 40, 40 + 5 = 45, 45 + 5 = 50,
So, 40, 45, 50.

Question 9.
50, 45, __
50, 45, 40.

Explanation:
Given 50, 45, ___skip count by 5,
55 – 5 = 50, 50 – 5 = 45, 45 – 5 = 40,
So, 50, 45, 40.

Question 10.
45, 40, __
45, 40, 35.

Explanation:
Given 45, 40, ___skip count by 5,
50 – 5 = 45, 45 – 5 = 40, 40 – 5 = 35,
So, 45, 40, 35.

Question 11.
40, 35, __
40, 35, 30,

Explanation:
Given 40, 35, ___skip count by 5,
45 – 5 = 40, 40 – 5 = 35, 35 – 5 = 30,
So, 40, 35, 30.

Question 12.
35, 30, __
35, 30, 25,

Explanation:
Given 35, 30, ___skip count by 5,
40 – 5 = 35, 35 – 5 = 30, 30 – 5 = 25,
So, 35, 30, 25.

Question 13.
30, 25, __
30, 25, 20,

Explanation:
Given 30, 25, ___skip count by 5,
35 – 5 = 30, 30 – 5 = 25, 25 – 5 = 20,
So, 30, 25, 20.

Question 14.
25, 20, __
25, 20, 15

Explanation:
Given 25, 20, ___skip count by 5,
30 – 5 = 25, 25 – 5 = 20, 20 – 5 = 15,
So, 25, 20, 15.

Question 15.
20, 15, __
20, 15, 10,

Explanation:
Given 20, 15, ___skip count by 5,
25 – 5 = 20, 20 – 5 = 15, 15 – 5 = 10,
So, 20, 15, 10.

Question 16.
15, 10, __
15, 10, 5,

Explanation:
Given 15, 10, ___skip count by 5,
20 – 5 = 15, 15 – 5 = 10, 10 – 5 = 5,
So, 15, 10, 5.

Question 17.
0, __, 10
0, 5, 10,

Explanation:
Given 0, ___, 10 skip count by 5,
0 + 0 = 0, 0 + 5 = 5, 5 + 5 = 10,
So, 0, 5, 10.

Question 18.
25, __, 35
25, 30, 35,

Explanation:
Given 25, ___, 35 skip count by 5,
20 + 5 = 25, 25 + 5 = 30, 30 + 5 = 35,
So,  25, 30, 35.

Question 19.
5, __, 15
5, 10, 15,

Explanation:
Given 5, ___, 15 skip count by 5,
0 + 5 = 5, 5 + 5 = 10, 10 + 5 = 15,
So, 5, 10,15.

Question 20.
30, __, 40
30, 35, 40,

Explanation:
Given 30, ___, 40 skip count by 5,
25 + 5 = 30, 30 + 5 = 35, 35 + 5 = 40,
So, 30, 35, 40.

Question 21.
10, __, 20
10, 15, 20,

Explanation:
Given 10, ___, 20 skip count by 5,
5 + 5 = 10, 10 + 5 = 15, 15 + 5 = 20,
So, 10, 15, 20.

Question 22.
35, __, 45
35, 40, 45,

Explanation:
Given 35, ___, 45 skip count by 5,
30 + 5 = 35, 35 + 5 = 40, 40 + 5 = 45,
So, 35, 40, 45.

Question 23.
15, __, 25
15, 20, 25,

Explanation:
Given 15, ___, 25 skip count by 5,
10 + 5 = 15, 15 + 5 = 20, 20 + 5 = 25,
So, 15, 20, 25.

Question 24.
35, __, 45
35, 40, 45,

Explanation:
Given 35, ___, 45 skip count by 5,
30 + 5 = 35, 35 + 5 = 40, 40 + 5 = 45,
So, 35, 40, 45.

Question 25.
20, __, 30
20, 25, 30,

Explanation:
Given 20, ___, 30 skip count by 5,
15 + 5 = 20, 20 + 5 = 25, 25 + 5 = 30,
So, 20, 25, 30.

Question 26.
25, __, 15
25, 20, 15,

Explanation:
Given 25, ___, 15 skip count by 5,
30 – 5 = 25, 25 – 5 = 20, 20 – 5 = 15,
So, 25, 20, 15.

Question 27.
50, __, 60
50, 55, 60,

Explanation:
Given 50, ___, 60 skip count by 5,
45 + 5 = 50, 50 + 5 = 55, 55 + 5 = 60,
So, 50, 55, 60.

Question 28.
20, __, 10
20, 15, 10,

Explanation:
Given 20, ___, 10 skip count by 5,
25 – 5 = 20, 20 – 5 = 15, 15 – 5 = 10,
So, 20, 15, 10.

Question 29.
45, __, 35
45, 40, 35,

Explanation:
Given 45, ___, 35 skip count by 5,
50 – 5 = 45, 45 – 5 = 40, 40 – 5 = 35,
So, 45, 40, 35.

Question 30.
15, __, 5
15, 10, 5,

Explanation:
Given 15, ___, 5 skip count by 5,
20 – 5 = 15, 15 – 5 = 10, 10 – 5 = 5,
So, 15, 10, 5.

Question 31.
35, __, 25
35, 30, 25,

Explanation:
Given 35, ___, 25 skip count by 5,
40 + 5 = 35, 35 – 5 = 30, 30 – 5 = 25,
So, 35, 30, 25.

Question 32.
10, __, 0
10, 5, 0,

Explanation:
Given 10, ___, 0 skip count by 5,
15 – 0 = 10, 10 – 5 = 5, 5 – 5 = 0,
So, 10, 5, 0.

Question 33.
35, __, 25
35, 30, 25,

Explanation:
Given 35, ___, 25 skip count by 5,
35, 35 – 5 = 30, 30 – 5 = 25,
So, 35, 30, 25.

Question 34.
__, 15, 10
20, 15, 10,

Explanation:
Given ___, 15, 10 skip count by 5,
20 = 15 + 5, 15 = 10 + 5, 10 = 5 + 5,
So, 20, 15, 10.

Question 35.
__, 40, 35
45, 40, 35,

Explanation:
Given ___, 40, 35 skip count by 5,
45 = 40 + 5, 40 = 35 + 5, 35 = 30 + 5,
So, 45, 40, 35.

Question 36.
__, 20, 15
25, 20, 15,

Explanation:
Given ___, 20, 15 skip count by 5,
25 = 20 + 5, 20 = 15 + 5, 15 = 10 + 5,
So, 25, 20, 15.

Question 37.
__, 45, 40
50, 45, 40,

Explanation:
Given ___, 45, 40 skip count by 5,
50 = 45 + 5, 45 = 40 + 5, 40 = 35 + 5,
So, 50, 45, 40.

Question 38.
__, 10, 5
15, 10, 5

Explanation:
Given ___, 10, 5 skip count by 5,
15 = 10 + 5, 10 = 5 + 5, 5 = 5 + 0,
So, 15, 10, 5.

Question 39.
__, 35, 30
40, 35, 30,

Explanation:
Given ___, 35, 30 skip count by 5,
40 = 35 + 5, 35 = 30 + 5, 30 = 25 + 5,
So, 40, 35, 30.

Question 40.
45, 50, __
45, 50, 55,

Explanation:
Given 45, 50, ___ skip count by 5,
45, 45 + 5 = 50, 50 + 5 = 55,
So, 45, 50, 55.

Question 41.
50, 55, __
50, 55, 60,

Explanation:
Given 55, 55, ___ skip count by 5,
50, 50 + 5 = 55, 55 + 5 = 60,
So, 50, 55, 60.

Question 42.
55, 60, __
55, 60, 65,

Explanation:
Given 55, 60, ___ skip count by 5,
55, 55 + 5 = 60, 60 + 5 = 65,
So, 55, 60, 65.

Question 43.
65, __, 55
65, 60, 55,

Explanation:
Given 65, 60, ___ skip count by 5,
65, 65 – 5 = 60, 60 – 5 = 55,
So, 65, 60, 55.

Question 44.
__, 60, 55
65, 60, 55,

Explanation:
Given 65, 60, ___ skip count by 5,
65 = 60 +5, 60 = 55 + 5, 55 = 50 + 5,
So, 65, 60, 55.

Eureka Math Grade 3 Module 1 Lesson 20 Problem Set Answer Key

Question 1.
Ted buys 3 books and a magazine at the book store.
Each book costs $8. A magazine costs $4.
Eureka Math Grade 3 Module 1 Lesson 20 Problem Set Answer Key 1
a. What is the total cost of the books?
b. How much does Ted spend altogether?

a. Total cost of the books is $24,

Explanation:
Given Ted buys 3 books and each book costs $8,
So, total cost of the books is 3 X $8 = $24.

b. Ted has spend $28 altogether,
Eureka Math Grade 3 Module 1 Lesson 20 Answer Key-9

Explanation:
Given Ted buys 3 books and a magazine at the book store.
Each book costs $8 and  A magazine costs $4, So total
amount spent is Cost of books + Cost of magazine =
(3 X $ 8) + $4  = $24 + $4 = $28.

Question 2.
Seven children share 28 silly bands equally.
Eureka Math Grade 3 Module 1 Lesson 20 Problem Set Answer Key 2
a. How many silly bands does each child get?
b. How many silly bands do 3 children get?
Eureka Math Grade 3 Module 1 Lesson 20 Answer Key-10

a. Each child gets 4 silly bands,

Explanation:
Given seven children share 28 silly bands equally,
so each child get 28 ÷ 7 = 4.

b. 3 children gets 12 silly bands,

Explanation:
We know each child gets 4 silly bands, So 3 children gets
3 X 4 = 12 silly bands.

Question 3.
Eighteen cups are equally packed into 6 boxes.
Two boxes of cups break. How many cups are unbroken?

12 cups are unbroken,

Explanation:
Given eighteen cups are equally packed into 6 boxes
means each box has 18 ÷ 6 = 3 cups.
Two boxes of cups break means 2 X 3 = 6 cups are broken,
Now number of cups unbroken are 18 – 6 = 12 cups.

Question 4.
There are 25 blue balloons and 15 red balloons at a party.
Five children are given an equal number of each color balloon.
How many blue and red balloons does each child get?

Each child gets 5 blue balloons and 3 red balloons in the party,

Explanation:
Given there are 25 blue balloons and 15 red balloons at a party.
Five children are given an equal number of each color balloon.
So each child gets blue balloons as 25 ÷ 5 = 5 and
red balloons as 15 ÷ 5 = 3.

Question 5.
Twenty-seven pears are packed in bags of 3.
Five bags of pears are sold. How many bags of pears are left?

4 bags of pears are left,

Explanation:
Given Twenty-seven pears are packed in bags of 3,
So number of bags of pears are 27 ÷ 3 = 9 bags,
and 5 bags of pears are sold, therefore left bags of pears
are 9 – 5 = 4 bags.

Eureka Math Grade 3 Module 1 Lesson 20 Exit Ticket Answer Key

Question 1.
Thirty-two jelly beans are shared by 8 students.
Engage NY Math 3rd Grade Module 1 Lesson 20 Exit Ticket Answer Key 3
a. How many jelly beans will each student get?
b. How many jelly beans will 4 students get?

a. 4 jelly beans each student will get.
Eureka Math Grade 3 Module 1 Lesson 20 Answer Key-11

Explanation:
Given Thirty-two jelly beans are shared by 8 students,
So each student will get 32 ÷ 8 = 4 jelly beans.

b. 4 students will get 4 X 4 = 16 jelly beans,

Explanation:
As we know each student will get 4 jelly beans,
for 4 students it will be 4 x 4 = 16 jelly beans.

Question 2.
The teacher has 30 apple slices and 20 pear slices.
Five children equally share all of the fruit slices.
How many fruit slices does each child get?

Each child gets 6 apple slices and 4 pear slices,

Explanation:
Given the teacher has 30 apple slices and 20 pear slices,
Five children equally share all of the fruit slices.
So number of fruit slices each child gets is
30 ÷ 5 = 6 apple slices and 20 ÷ 5 = 4 pear slices.

Eureka Math Grade 3 Module 1 Lesson 20 Homework Answer Key

Question 1.
Jerry buys a pack of pencils that costs $3. David buys 4 sets of markers.
Each set of markers also costs $3.
Eureka Math 3rd Grade Module 1 Lesson 20 Homework Answer Key 4
a. What is the total cost of the markers?
b. How much more does David spend on 4 sets of markers than
Jerry spends on a pack of pencils?

a. Total cost of markers are $12,
Eureka Math Grade 3 Module 1 Lesson 20 Answer Key-12

Explanation:
Given David buys 4 sets of markers,
Each set of markers costs $3, So total cost
of markers are 4 X $3 = $12,

b. $9 more David has spend on 4 sets of markers than
Jerry spends on a pack of pencils,

Explanation:
Given Jerry buys a pack of pencils that costs $3.
David buys 4 sets of markers. Each set of markers also costs $3
and we know total cost of markers is $12, Now more David has
than Jerry is $12 – $3 = $9.

Question 2.
Thirty students are eating lunch at 5 tables. Each table has
the same number of students.
a. How many students are sitting at each table?
Eureka Math 3rd Grade Module 1 Lesson 20 Homework Answer Key 5
b. How many students are sitting at 4 tables?
Eureka Math Grade 3 Module 1 Lesson 20 Answer Key-13
a. 6 students are sitting on each table,

Explanation:
Given thirty students are eating lunch at 5 tables.
Each table has the same number of students.
So number of students sitting at each table are
30 ÷ 5 = 6.

b. 24 students are sitting at 4 tables,

Explanation:
We know at each table there are 6 students sitting and
eating lunch, So at 4 tables there are 6 x 4 = 24 students
having lunch.

Question 3.
The teacher has 12 green stickers and 15 purple stickers.
Three students are given an equal number of each color sticker.
How many green and purple stickers does each student get?

Each student will get 4 green stickers and 5 purple stickers,

Explanation:
Given the teacher has 12 green stickers and 15 purple stickers.
Three students are given an equal number of each color sticker.
So number of green and purple stickers each student gets is
12 ÷ 3 = 4 green stickers and 15 ÷ 3 = 5 purple stickers.

Question 4.
Three friends go apple picking. They pick 13 apples on
Saturday and 14 apples on Sunday. They share the apples equally.
How many apples does each person get?

Each person will get 9 apples,

Explanation:
Given three friends go for apple picking. They pick 13 apples on
Saturday and 14 apples on Sunday and they share the apples equally.
Therefore number of apples each person will get is 13 + 14 = 27 apples,
27 ÷ 3 = 9 apples each.

Question 5.
The store has 28 notebooks in packs of 4. Three packs of
notebooks are sold. How many packs of notebooks are left?

4 packs of notebooks are left,

Explanation:
Given the store has 28 notebooks in packs of 4.
Three packs of notebooks are sold. So, number of
packs of notebooks are 28 ÷ 4 = 7 packs,
Now from 7 packs of notebooks 4 packs are sold,
So, left packs of notebooks are 7 – 3 = 4 packs.

Eureka Math Grade 3 Module 1 Lesson 19 Answer Key

Engage NY Eureka Math 3rd Grade Module 1 Lesson 19 Answer Key

Eureka Math Grade 3 Module 1 Answer Key

Eureka Math Grade 3 Module 1 Lesson 19 Problem Set Answer Key

Question 1.
Label the array. Then, fill in the blanks to make true number sentences.
a. 36 ÷ 3 = 12__
Eureka Math Grade 3 Module 1 Lesson 19 Problem Set Answer Key 1
a. 36 ÷ 3 = 12,
Eureka Math Grade 3 Module 1 Lesson 19 Answer Key-1

Explanation:
36 ÷ 3 can be written as (30 ÷ 3) + (6 ÷ 3) = 10 + 2 = 12 or
36 ÷ 3 = 12. Filled the blanks to make true number sentences.

b. 25 ÷ 5 = _5_
Eureka Math Grade 3 Module 1 Lesson 19 Problem Set Answer Key 2
b. 25 ÷ 5 = 5,
Eureka Math Grade 3 Module 1 Lesson 19 Answer Key-2
Explanation:
25 ÷ 5 can be written as (20 ÷ 5) + (5 ÷ 5) = 4  + 1 = 5 or
25 ÷ 5 = 5. Filled the blanks to make true number sentences.

c. 28 ÷ 4 = __7___
Eureka Math Grade 3 Module 1 Lesson 19 Problem Set Answer Key 3

28 ÷ 4 =7,
Eureka Math Grade 3 Module 1 Lesson 19 Answer Key-3
Explanation:
28 ÷ 4 can be written as (20 ÷ 4) + (8 ÷ 4) = 5  + 2 = 7 or
28 ÷ 4 = 7. Filled the blanks to make true number sentences.

d. 32 ÷ 4 = _8_
Eureka Math Grade 3 Module 1 Lesson 19 Problem Set Answer Key 4
32 ÷ 4 = 8,
Eureka Math Grade 3 Module 1 Lesson 19 Answer Key-4
Explanation:
32 ÷ 4 can be written as (20 ÷ 4) + (12 ÷ 4) = 5  + 3 = 8 or
32 ÷ 4 = 8. Filled the blanks to make true number sentences.

Question 2.
Match the equal expressions.
Eureka Math Grade 3 Module 1 Lesson 19 Problem Set Answer Key 5
Matched the equal expressions as
Eureka Math Grade 3 Module 1 Lesson 19 Answer Key-5
Explanation:
Matching the equal expressions as
24 ÷ 2 = (20 ÷ 2) + (4 ÷  2),
24 is written as (20 + 4) ÷ 2.
36 ÷ 3 = (30 ÷ 3) + (6 ÷ 3),
36 is written as (30 + 6) ÷ 3.
39 ÷  3 = (30 ÷  3) + (9 ÷  3),
39 is written as (30 + 9) ÷ 3.
26 ÷ 2 = ( 20 ÷ 2) + (6 ÷ 2),
26 is written as (20 + 6) ÷ 2.

Question 3.
Nell draws the array below to find the answer to 24 ÷ 2. Explain Nell’s strategy.
Eureka Math Grade 3 Module 1 Lesson 19 Problem Set Answer Key 6
Nell’s strategy is 24 ÷ 2 = (12 ÷ 2) + (12 ÷ 2),
6 + 6 = 12, So, 24 ÷ 2 = 12,

Explanation:
To find answer for 24 ÷ 2 Nell draws an array
his strategy is he divided 24 as 12 + 12 then has
divided with 2, So equation for Nell’s drawing is
24 ÷ 2 = (12 ÷ 2) + (12 ÷ 2), 6 + 6 = 12.
Therefore 24 ÷ 2 = 12.

Eureka Math Grade 3 Module 1 Lesson 19 Exit Ticket Answer Key

Complete the equations below to solve 22 ÷ 2 = __11___.

Engage NY Math 3rd Grade Module 1 Lesson 19 Exit Ticket Answer Key 7
22 ÷ 2 = 11,

Explanation:
Eureka Math Grade 3 Module 1 Lesson 19 Answer Key-6
We solve 22 ÷ 2 as below first we write 22 as 20 + 2
and divide by 2, (20 + 2) ÷ 2 = (20 ÷ 2) +(2 ÷ 2) =
10 + 1 = 11, So 22 ÷ 2 = 11.

Eureka Math Grade 3 Module 1 Lesson 19 Homework Answer Key

Question 1.
Label the array. Then, fill in the blanks to make true number sentences.
a. 18 ÷ 3 = _6_
Eureka Math 3rd Grade Module 1 Lesson 19 Homework Answer Key 8
Eureka Math Grade 3 Module 1 Lesson 19 Answer Key-7
18 ÷ 3 = 6,

Explanation:
Labelled the array, filled the blanks to make
true number sentences as 18 ÷ 3 = ( 9 ÷ 3) + (9 ÷ 3) =
3 + 3 = 6 or 18 ÷ 3 = 6.
We wrote 18 ÷ 3 as (9 + 9) ÷ 3.

b. 21 ÷ 3 = _7_
Eureka Math 3rd Grade Module 1 Lesson 19 Homework Answer Key 9
Eureka Math Grade 3 Module 1 Lesson 19 Answer Key-8
21 ÷ 3 = 7,

Explanation:
Labelled the array, filled the blanks to make
true number sentences as 21÷ 3 = (15 ÷ 3) + (6 ÷ 3) =
5 + 2 = 7 or 21 ÷ 3 = 7.
We wrote 21 ÷ 3 as (15 + 6) ÷ 3.

c. 24 ÷ 4 = _6_
Eureka Math 3rd Grade Module 1 Lesson 19 Homework Answer Key 10

Eureka Math Grade 3 Module 1 Lesson 19 Answer Key-9

24 ÷ 4 = 6,

Explanation:
Labelled the array, filled the blanks to make
true number sentences as 24 ÷ 4 = (20÷ 4) + (4 ÷ 4) =
5 + 1 = 6 or 24÷ 4 = 6.
We wrote 24 ÷ 4 as (20 + 4) ÷ 4.

d. 36 ÷ 4 = _9_
Eureka Math 3rd Grade Module 1 Lesson 19 Homework Answer Key 11
Eureka Math Grade 3 Module 1 Lesson 19 Answer Key-10
36 ÷ 4 = 9,

Explanation:
Labelled the array, filled the blanks to make
true number sentences as 36 ÷ 4 = (20÷ 4) + (16 ÷ 4) =
5 + 4 = 9 or 36÷ 4 = 9.
We wrote 36 ÷ 4 as (20 + 16) ÷ 4.

Question 2.
Match equal expressions.
Eureka Math 3rd Grade Module 1 Lesson 19 Homework Answer Key 12

Matched the equal expressions as
Eureka Math Grade 3 Module 1 Lesson 19 Answer Key-11
Explanation:
Matching the equal expressions as
28 ÷ 2 = (20 ÷ 2) + (8 ÷  2),
28 is written as (20 + 8) ÷ 2.
33 ÷ 3 = (30 ÷ 3) + (3 ÷ 3),
33 is written as (30 + 3) ÷ 3.
36 ÷  3 = (30 ÷  3) + (6 ÷  3),
36 is written as (30 + 6) ÷ 3.
26 ÷ 2 = ( 20 ÷ 2) + (6 ÷ 2),
26 is written as (20 + 6) ÷ 2.

Question 3.
Alex draws the array below to find the answer to 35 ÷ 5. Explain Alex’s strategy.
Eureka Math 3rd Grade Module 1 Lesson 19 Homework Answer Key 13

Alex’s strategy is 35 ÷ 5 = (20 ÷ 5) + (15 ÷ 5) =
4 + 3 = 7, So, 35 ÷ 5 = 7,

Explanation:
To find answer for 35 ÷ 5 Alex draws an array
his strategy is he divided 35 as 20 + 15 then has
divided with 5, So equation for Alex’s drawing is
35 ÷ 5 = (20 ÷ 5) + (15 ÷ 5), 4 + 3 = 7,
Therefore 35 ÷ 5 = 7.

Eureka Math Grade 3 Module 1 Lesson 18 Answer Key

Engage NY Eureka Math 3rd Grade Module 1 Lesson 18 Answer Key

Eureka Math Grade 3 Module 1 Answer Key

Eureka Math Grade 3 Module 1 Lesson 18 Sprint Answer Key

A
Add or Subtract Using 5
Eureka Math Grade 3 Module 1 Lesson 18 Sprint Answer Key 21
Eureka Math Grade 3 Module 1 Lesson 18 Sprint Answer Key 22
Eureka Math Grade 3 Module 1 Lesson 18 Sprint Answer Key 23
Eureka Math Grade 3 Module 1 Lesson 18 Sprint Answer Key 24

Eureka Math Grade 3 Module 1 Lesson 18 Answer Key-1
Eureka Math Grade 3 Module 1 Lesson 18 Answer Key-2
Eureka Math Grade 3 Module 1 Lesson 18 Answer Key-3

Eureka Math Grade 3 Module 1 Lesson 18 Answer Key-4

Question 1.
0 + 5 =
0 + 5 = 5,

Explanation:
Given 0 + 5 = adding 5 to 0 we get 5,
So, 0 + 5 = 5.

Question 2.
5 + 5 =
5 + 5 = 10,

Explanation:
Given 5 + 5 = adding 5 to 5 we get 10,
So, 5 + 5 = 10.

Question 3.
10 + 5 =
10 + 5 = 15,

Explanation:
Given 10 + 5 = adding 5 to 10 we get 15,
So, 10 + 5 = 15.

Question 4.
15 + 5 =
15 + 5 = 20,

Explanation:
Given 15 + 5 = adding 5 to 15 we get 20,
So, 15 + 5 = 20.

Question 5.
20 + 5 =
20 + 5 =25,

Explanation:
Given 20 + 5 = adding 5 to 20 we get 25,
So, 20 + 5 = 5.

Question 6.
25 + 5 =
25 + 5 = 30,

Explanation:
Given 25 + 5 = adding 5 to 25 we get 30,
So, 25 + 5 = 30.

Question 7.
30 + 5 =
30 + 5 = 35,

Explanation:
Given 30 + 5 = adding 5 to 30 we get 35,
So, 30 + 5 = 35.

Question 8.
35 + 5 =
35 + 5 = 40,

Explanation:
Given 35 + 5 = adding 5 to 35 we get 40,
So, 35 + 5 = 40.

Question 9.
40 + 5 =
40 + 5 = 45,

Explanation:
Given 40 + 5 = adding 5 to 40 we get 45,
So, 40 + 5 = 45.

Question 10.
45 + 5 =
45 + 5 = 50,

Explanation:
Given 45 + 5 = adding 5 to 45 we get 50,
So, 45 + 5 = 50.

Question 11.
50 – 5 =
50 – 5 = 45,

Explanation:
Given 50 – 5 = subtracting 5 from 50 we get 45,
So, 50 – 5 = 45.

Question 12.
45 – 5 =
45 – 5 = 40,

Explanation:
Given 45 – 5 = subtracting 5 from 45 we get 40,
So, 45 – 5 = 40.

Question 13.
40 – 5 =
40 – 5 = 35,

Explanation:
Given 40 – 5 = subtracting 5 from 40 we get 35,
So, 40 – 5 = 35.

Question 14.
35 – 5 =
35 – 5 = 30,

Explanation:
Given 35 – 5 = subtracting 5 from 35 we get 30,
So, 35 – 5 = 30.

Question 15.
30 – 5 =
30 – 5 = 25,

Explanation:
Given 30 – 5 = subtracting 5 from 30 we get 25,
So, 30 – 5 = 25.

que25 16.
25 – 5 =
25 – 5 = 20,

Explanation:
Given 25 – 5 = subtracting 5 from 25 we get 20,
So, 25 – 5 = 20.

Question 17.
20 – 5 =
20 – 5 = 15,

Explanation:
Given 20 – 5 = subtracting 5 from 20 we get 15,
So, 20 – 5 = 15.

Question 18.
15 – 5 =
15 – 5 = 10,

Explanation:
Given 15 – 5 = subtracting 5 from 15 we get 10,
So, 15 – 5 = 10.

Question 19.
10 – 5 =
10 – 5 = 5,

Explanation:
Given 10 – 5 = subtracting 5 from 10 we get 5,
So, 10 – 5 = 5.

Question 20.
5 – 5 =
5 – 5 = 0,

Explanation:
Given 5 – 5 = subtracting 5 from 5 we get 0,
So, 5 – 5 = 0.

Question 21.
5 + 0 =
5 + 0 = 5,

Explanation:
Given 5 + 0 = adding 5 to 0 we get 5,
So, 5 + 0 = 5.

Question 22.
5 + 5 =
5 + 5 = 10,

Explanation:
Given 5 + 5 = adding 5 to 5 we get 10,
So, 5 + 5 = 10.

Question 23.
10 + 5 =
10 + 5 = 15,

Explanation:
Given 10 + 5 = adding 5 to 10 we get 15,
So, 10 + 5 = 15.

Question 24.
15 + 5 =
15 + 5 = 20,

Explanation:
Given 15 + 5 = adding 5 to 15 we get 20,
So, 15 + 5 = 20.

Question 25.
20 + 5 =
20 + 5 = 25,

Explanation:
Given 20 + 5 = adding 5 to 20 we get 25,
So, 20 + 5 = 25.

Question 26.
25 + 5 =
25 + 5 = 30,

Explanation:
Given 25 + 5 = adding 5 to 25 we get 30,
So, 25 + 5 = 30.

Question 27.
30 + 5 =
30 + 5 = 35,

Explanation:
Given 30 + 5 = adding 5 to 30 we get 35,
So, 30 + 5 = 35.

Question 28.
35 + 5 =
35 + 5 = 40,

Explanation:
Given 35 + 5 = adding 5 to 35 we get 40,
So, 35 + 5 = 40.

Question 29.
40 + 5 =
40 + 5 = 45,

Explanation:
Given 40 + 5 = adding 5 to 40 we get 45,
So, 40 + 5 = 45.

Question 30.
45 + 5 =
45 + 5 = 50,

Explanation:
Given 45 + 5 = adding 5 to 45 we get 50,
So, 45 + 5 = 50.

Question 31.
0 + 50 =
0 + 50 = 50,

Explanation:
Given 0 + 50 = adding 0 to 50 we get 50,
So, 0 + 50 = 50.

Question 32.
50 + 50 =
50 + 50 = 100,

Explanation:
Given 50 + 50 = adding 50 to 50 we get 100,
So, 50 + 50 = 100.

Question 33.
50 + 5 =
50 + 5 = 55,

Explanation:
Given 50 + 5 = adding 5 to 50 we get 55,
So, 50 + 5 = 55.

Question 34.
55 + 5 =
55 + 5 = 60,

Explanation:
Given 55 + 5 = adding 5 to 55 we get 60,
So, 55 + 5 = 55.

Question 35.
60 – 5 =
0 + 50 = 50,

Explanation:
Given 0 + 50 = adding 0 to 50 we get 50,
So, 0 + 50 = 50.

Question 36.
55 – 5 =
55 – 5 = 50,

Explanation:
Given 55 – 5 = subtracting 5 from 55 we get 50,
So, 55 – 5 = 50.

Question 37.
60 + 5 =
60 + 5 = 65,

Explanation:
Given 60 + 5 = adding 5 to 60 we get 65,
So, 60 + 5 = 15.

Question 38.
65 + 5 =
65 + 5 = 70,

Explanation:
Given 65 + 5 = adding 5 to 65 we get 70,
So, 65 + 5 = 70.

Question 39.
70 – 5 =
70 – 5 = 65,

Explanation:
Given 70 – 5 = subtracting 5 from 70 we get 65,
So, 70 – 5 = 65.

Question 40.
65 – 5 =
65 – 5 = 60,

Explanation:
Given 65 – 5 = subtracting 5 from 65 we get 60,
So, 65 – 5 = 60.

Question 41.
100 + 50 =
100 + 50 = 150,

Explanation:
Given 100 + 50 = adding 50 to 100 we get 150,
So, 100 + 50 = 150.

Question 42.
150 + 50 =
150 + 50 = 200,

Explanation:
Given 150 + 50 = adding 50 to 150 we get 200,
So, 150 + 50 = 200.

Question 43.
200 – 50 =
200 – 50 = 150,

Explanation:
Given 200 – 50 = subtracting 50 from 200 we get 150,
So, 200 – 50 = 150.

Question 44.
150 – 50 =
150 – 50 = 100,

Explanation:
Given 150 – 50 = subtracting 50 from 150 we get 100,
So, 150 – 50 = 100.

B
Add or Subtract Using 5
Eureka Math Grade 3 Module 1 Lesson 18 Sprint Answer Key 25
Eureka Math Grade 3 Module 1 Lesson 18 Sprint Answer Key 26
Eureka Math Grade 3 Module 1 Lesson 18 Sprint Answer Key 27
Eureka Math Grade 3 Module 1 Lesson 18 Sprint Answer Key 28

Eureka Math Grade 3 Module 1 Lesson 18 Answer Key-5
Eureka Math Grade 3 Module 1 Lesson 18 Answer Key-6
Eureka Math Grade 3 Module 1 Lesson 18 Answer Key-7
Eureka Math Grade 3 Module 1 Lesson 18 Answer Key-8
Question 1.
5 + 0 =
5 + 0 = 5,

Explanation:
Given 5 + 0 = adding 5 to 0 we get 5,
So, 5 + 0 = 5.

Question 2.
5 + 5 =
5 + 5 = 10,

Explanation:
Given 5 + 5 = adding 5 to 5 we get 10,
So, 5 + 5 = 10.

Question 3.
5 + 10 =
5 + 10 = 15,

Explanation:
Given 5 + 10 = adding 5 to 10 we get 15,
So, 5 + 10 = 15.

Question 4.
5 + 15 =
5 + 15 = 20,

Explanation:
Given 5 + 15 = adding 5 to 15 we get 20,
So, 5 + 15 = 20.

Question 5.
5 + 20 =
5 + 20 = 25,

Explanation:
Given 5 + 20 = adding 5 to 20 we get 25,
So, 5 + 20 = 25.

Question 6.
5 + 25 =
5 + 25 = 30,

Explanation:
Given 5 + 25 = adding 5 to 25 we get 30,
So, 5 + 25 = 30.

Question 7.
5 + 30 =
5 + 30 = 35,

Explanation:
Given 5 + 30 = adding 5 to 30 we get 35,
So, 5 + 30 = 35.

Question 8.
5 + 35 =
5 + 35 = 40,

Explanation:
Given 5 + 35 = adding 5 to 35 we get 40,
So, 5 + 35 = 40.

Question 9.
5 + 40 =
5 + 40 = 45,

Explanation:
Given 5 + 40 = adding 5 to 40 we get 45,
So, 5 + 40 = 45.

Question 10.
5 + 45 =
5 + 45 = 50,

Explanation:
Given 5 + 45 = adding 5 to 45 we get 50,
So, 5 + 45 = 50.

Question 11.
50 – 5 =
50 – 5 = 45,

Explanation:
Given 50 – 5 = subtracting 5 from 50 we get 45,
So, 50 – 5 = 45.

Question 12.
45 – 5 =
45 – 5 = 40,

Explanation:
Given 45 – 5 = subtracting 5 from 45 we get 40,
So, 45 – 5 = 40.

Question 13.
40 – 5 =
40 – 5 = 35,

Explanation:
Given 40 – 5 = subtracting 5 from 40 we get 35,
So, 40 – 5 = 35.

Question 14.
35 – 5 =
35 – 5 = 30,

Explanation:
Given 35 – 5 = subtracting 5 from 35 we get 30,
So, 35 – 5 = 30.

Question 15.
30 – 5 =
30 – 5 = 25,

Explanation:
Given 30 – 5 = subtracting 5 from 30 we get 25,
So, 30 – 5 = 25.

Question 16.
25 – 5 =
25 – 5 = 20,

Explanation:
Given 25 – 5 = subtracting 5 from 25 we get 20,
So, 25 – 5 = 20.

Question 17.
20 – 5 =
20 – 5 = 15,

Explanation:
Given 20 – 5 = subtracting 5 from 20 we get 15,
So, 20 – 5 = 15.

Question 18.
15 – 5 =
15 – 5 = 10,

Explanation:
Given 15 – 5 = subtracting 5 from 15 we get 10,
So, 15 – 5 = 10.

Question 19.
10 – 5 =
10 – 5 = 5,

Explanation:
Given 10 – 5 = subtracting 5 from 10 we get 5,
So, 10 – 5 = 5.

Question 20.
5 – 5 =
5 – 5 = 0,

Explanation:
Given 5 – 5 = subtracting 5 from 5 we get 0,
So, 5 – 5 = 0.

Question 21.
0 + 5 =
0 + 5 = 5,

Explanation:
Given 0 + 5 = adding 5 to 0 we get 5,
So, 0 + 5 = 5.

Question 22.
5 + 5 =
5 + 5 = 10,

Explanation:
Given 5 + 5 = adding 5 to 5 we get 10,
So, 5 + 5 = 10.

Question 23.
10 + 5 =
10 + 5 = 15,

Explanation:
Given 10 + 5 = adding 5 to 10 we get 15,
So, 10 + 5 = 15.

Question 24.
15 + 5 =
15 + 5 = 20,

Explanation:
Given 15 + 5 = adding 5 to 15 we get 20,
So, 15 + 5 = 20.

Question 25.
20 + 5 =
20 + 5 = 25,

Explanation:
Given 20 + 5 = adding 5 to 20 we get 25,
So, 20 + 5 = 25.

Question 26.
25 + 5 =
25 + 5 = 30,

Explanation:
Given 25 + 5 = adding 5 to 25 we get 30,
So, 25 + 5 = 30.

Question 27.
30 + 5 =
30 + 5 = 35,

Explanation:
Given 30 + 5 = adding 5 to 30 we get 35,
So, 30 + 5 = 35.

Question 28.
35 + 5 =
35 + 5 = 40,

Explanation:
Given 35 + 5 = adding 5 to 35 we get 40,
So, 35 + 5 = 40.

Question 29.
40 + 5 =
40 + 5 = 45,

Explanation:
Given 40 + 5 = adding 5 to 40 we get 45,
So, 40 + 5 = 45.

Question 30.
45 + 5 =
45 + 5 = 50,

Explanation:
Given 45 + 5 = adding 5 to 45 we get 50,
So, 45 + 5 = 50.

Question 31.
50 + 0 =
50 + 0 = 50,

Explanation:
Given 50 + 5 = adding 0 to 50 we get 50,
So, 50 + 0 = 50.

Question 32.
50 + 50 =
50 + 50 = 100,

Explanation:
Given 50 + 50 = adding 50 to 50 we get 100,
So, 50 + 50 = 100.

Question 33.
5 + 50 =
5 + 50 = 55,

Explanation:
Given 5 + 50 = adding 5 to 50 we get 55,
So, 5 + 50 = 55.

Question 34.
5 + 55 =
5 + 55 = 60,

Explanation:
Given 5 + 55 = adding 5 to 55 we get 60,
So, 5 + 55 = 60.

Question 35.
60 – 5 =
60 – 5 = 55,

Explanation:
Given 60 – 5 = subtracting 5 from 60 we get 55,
So, 60 – 5 = 55.

Question 36.
55 – 5 =
55 – 5 = 50,

Explanation:
Given 55 – 5 = subtracting 5 from 55 we get 50,
So, 55 – 5 = 50.

Question 37.
5 + 60 =
5 + 60 = 65,

Explanation:
Given 5 + 60 = adding 5 to 60 we get 65,
So, 5 + 60 = 65.

Question 38.
5 + 65 =
5 + 65 = 70,

Explanation:
Given 5 + 65 = adding 5 to 65 we get 70,
So, 5 + 65 = 70.

Question 39.
70 – 5 =
70 – 5 = 65,

Explanation:
Given 70 – 5 = subtracting 5 from 70 we get 65,
So, 65 – 5 = 60.

Question 40.
65 – 5 =
65 – 5 = 60,

Explanation:
Given 65 – 5 = subtracting 5 from 65 we get 60,
So, 65 – 5 = 60.

Question 41.
50 + 100 =
50 + 100 = 150,

Explanation:
Given 50 + 100 = adding 50 to 100 we get 150,
So, 50 + 100 = 150.

Question 42.
50 + 150 =
50 + 150 = 200,

Explanation:
Given 50 + 150 = adding 50 to 150 we get 200,
So, 50 + 150 = 200.

Question 43.
200 – 50 =
200 – 50 = 150,

Explanation:
Given 200 – 50 = subtracting 50 from 200 we get 150,
So, 200 – 50 = 150.

Question 44.
150 – 50 =
150 – 50 = 100,

Explanation:
Given 150 – 50 = subtracting 50 from 150 we get 100,
So, 150 – 50 = 100.

Eureka Math Grade 3 Module 1 Lesson 18 Problem Set Answer Key

Question 1.
8 × 10 = __80____
Eureka Math Grade 3 Module 1 Lesson 18 Problem Set Answer Key 1
5 tens + ______3 tens_________ = 8 tens
(5 × 10) + (___3___ × 10) = 8 × 10
50 + ___30______ = ___80______
8 × 10 = ____80________

8 × 10 = 80
Eureka Math Grade 3 Module 1 Lesson 18 Answer Key-9
5 tens + 3 tens = 8 tens
(5 × 10) + (3 × 10) = 8 × 10
50 + 30 = 80
8 × 10 = 80

Explanation:
Given 8 X 10 =
We can write 8 X 10 as 5 tens + 3 tens,
(5 X 10) + (3 X 10) = 8 X 10,
50 + 30 = 80 or 8 X 10 = 80.

Question 2.
7 × 4 = ___28___
Eureka Math Grade 3 Module 1 Lesson 18 Problem Set Answer Key 2
5 fours + ______2 fours____ = 7 fours
(5 × 4) + (___2___ × 4) = 7 × 4
20 + ____8_____ = ____28_____
7 × 4 = _____28_______

7 x 4 = 28,
Eureka Math Grade 3 Module 1 Lesson 18 Answer Key-10
5 fours + 2 fours = 7 fours
(5 × 4) + (2 × 4) = 7 × 4
20 + 8 = 28
7 × 4 = 28,

Explanation:
Given 7 X 4 =
We can write 7 x 4 as 5 fours + 2 fours,
(5 X 4) + (2 X 4) = 7 X 4,
20 + 8 = 28 or 7 X 4 = 28.

Question 3.
9 × 10 = ___90___
Eureka Math Grade 3 Module 1 Lesson 18 Problem Set Answer Key 3
5 tens + _____4 tens__________ = 9 tens
(5 × 10) + (___4___ × 10) = 9 × 10
___50______ + ___40______ = ___ 90______
9 × 10 = _____90_______

9 X 10 = 90,
Eureka Math Grade 3 Module 1 Lesson 18 Answer Key-11
5 tens + 4 tens = 9 tens
(5 × 10) + (4 × 10) = 9 × 10
50 + 40 =  90
9 × 10 = 90

Explanation:
Given 9 X 10 =
We can write 9 X 10 as 5 tens + 4 tens,
(5 X 10) + (4 X 10) = 9 X 10,
50 + 40 = 90 or 9 X 10 = 90.

Question 4.
10 × 10 = __100____
Eureka Math Grade 3 Module 1 Lesson 18 Problem Set Answer Key 4
____5 tens_________ + ___5 tens_________ = 10 tens
(___5___ × 10) + (__5____ × 10) = 10 × 10
___50_____ + ___50______ = ___100______
10 × 10 = ____100________

10 X 10 = 100
Eureka Math Grade 3 Module 1 Lesson 18 Answer Key-12
5 tens + 5 tens = 10 tens
(5 × 10) + (5 × 10) = 10 × 10
50 + 50 = 100
10 × 10 = 100

Explanation:
Given 10 X 10 =
We can write 10 X 10  as 5 tens + 5 tens,
(5 X 10) + (5 X 10) = 10 X 10,
50 + 50 = 100 or 10 X 10 = 100.

Question 5.
There are 7 teams in the soccer tournament.
Ten children play on each team. How many children are
playing in the tournament? Use the break apart and
distribute strategy, and draw a number bond to solve.
There are ____70______ children playing in the tournament.

There are 70 children playing in the tournament,
7 X 10 =
Eureka Math Grade 3 Module 1 Lesson 18 Answer Key-13
5 tens + 2 tens = 7 tens
(5 × 10) + (2 × 10) = 7 × 10
50 + 20 = 70
7 × 10 = 70

Explanation:
Given there are 7 teams in the soccer tournament.
Ten children play on each team. So number of children
playing in the tournament are 7 X 10 = 70,
Used the break apart and distribute strategy,
drawn a number bond to solve 7 X 10 =
wrote 7 as 5 tens + 2 tens = 7 tens,
(5 × 10) + (2 × 10) = 7 × 10,
50 + 20 = 70,
7 × 10 = 70,
Therefore, there are 70 children playing in the tournament.

Question 6.
What is the total number of sides on 8 triangles?

Total number of 24 sides on 8 triangles,

Explanation:
Given to find the total number of sides on 8 triangles,
as we know a triangle has 3 sides so 8 triangles means
3 sides X  8  = 24 sides,
therefore, total number of 24 sides on 8 triangles.

Question 7.
There are 12 rows of bottled drinks in the vending machine.
Each row has 10 bottles. How many bottles are in the vending machine?

There are 120 bottles in total in the vending machine,

Explanation:
Given there are 12 rows of bottled drinks in the vending machine
and each row has 10 bottles, So total number of bottles in
the vending machine are 12 X 10 = 120 bottles.

Eureka Math Grade 3 Module 1 Lesson 18 Exit Ticket Answer Key

Dylan used the break apart and distribute strategy to
solve a multiplication problem. Look at his work below,
write the multiplication problem Dylan solved,
and complete the number bond.
Engage NY Math 3rd Grade Module 1 Lesson 18 Exit Ticket Answer Key 5

Eureka Math Grade 3 Module 1 Lesson 18 Answer Key-14
Explanation:
Given Dylan used the break apart and distribute strategy to
solve a multiplication problem. Looking at his work,
wrote the multiplication problem Dylan solved,
and completed the number bond as shown above,
as 6 X 4 = (5 X 4) + (1 X 4) = 20 + 4 =24, or 6 X 4 = 24.

Eureka Math Grade 3 Module 1 Lesson 18 Homework Answer Key

Question 1.
Match.
Eureka Math 3rd Grade Module 1 Lesson 18 Homework Answer Key 6
Matched as shown below :
Eureka Math Grade 3 Module 1 Lesson 18 Answer Key-15
Explanation:
Matched expressions as
7 tens = 5 tens + 2 tens = (5 X 10) + (2 X 10) = 70,
8 fours = 5 fours + 3 fours = (5 X 4) + (3 X 4) = 32,
9 tens =  6 tens + 3 tens = (6 X 10) + (3 X 10) = 90,
7 threes = 5 threes + 2 threes = (5 X 3) + (2 X 3) = 21.

Question 2.
9 × 4 = ___36___
Eureka Math 3rd Grade Module 1 Lesson 18 Homework Answer Key 7
(_5_ × 4) + (_4_ × 4) = 9 × 4
_20__ + _16_ = _36_
9 × 4 = _36_.

9 X 4 = 36,
Eureka Math Grade 3 Module 1 Lesson 18 Answer Key-16
(5 X 4 + 4 X 4) = 9 X 4,
20 + 16 = 36,
9 X 4 = 36,

Explanation:
Given 9 X 4 =
We can write 9 X 4 as 5 fours + 4 fours,
(5 X 4) + (4 X 4) = 9 X 4,
20 + 16 = 36 or 9 X 4 = 36.

Question 3.
Lydia makes 10 pancakes. She tops each pancake with
4 blueberries. How many blueberries does Lydia use in all?
Use the break apart and distribute strategy, and
draw a number bond to solve.
Lydia uses __40__ blueberries in all.

Lydia uses 40 blueberries in all,
Eureka Math Grade 3 Module 1 Lesson 18 Answer Key-17
Explanation:
Given Lydia makes 10 pancakes. She tops each pancake with
4 blueberries. So number of  blueberries Lydia use in all are
4 X 10 = 40, Used the break apart and distribute strategy as
(2 X 10) + (2 X 10) = 20 + 20 = 40 and drawn a
number bond to solve 4 x 10 as 40.

Question 4.
Steven solves 7 × 3 using the break apart and distribute strategy.
Show an example of what Steven’s work might look like below.

7 X 3 = 21,
Eureka Math Grade 3 Module 1 Lesson 18 Answer Key-18
Example: Steven has 7 friends and each friend has
3 balloons, how many total balloons Steven will collect
from his friends?

Explanation:
Given Steven has 7 friends and each friend has
3 balloons, how many total balloons Steven will collect
from his friends?
Steven solves 7 × 3 using the
break apart and distribute strategy shown above as
7 X 3 = (4 X 3) + (3 X 3) = 12 + 9 = 21,
or 7 X 3 = 21.
So Steven collects 21 balloons in all.

Question 5.
There are 7 days in 1 week. How many days are there in 10 weeks?

There are 70 days in 10 weeks,

Explanation:
Given there are 7 days in 1 week. So number of days
in 10 weeks are 7 X 10 = 70 days.
therefore there are 70 days in 10 weeks.

Eureka Math Grade 3 Module 1 Lesson 17 Answer Key

Engage NY Eureka Math 3rd Grade Module 1 Lesson 17 Answer Key

Eureka Math Grade 3 Module 1 Answer Key

Eureka Math Grade 3 Module 1 Lesson 17 Sprint Answer Key

A
Multiply or Divide by 4

Eureka Math Grade 3 Module 1 Lesson 17 Sprint Answer Key 1
Eureka Math Grade 3 Module 1 Lesson 17 Sprint Answer Key 2
Eureka Math Grade 3 Module 1 Lesson 17 Sprint Answer Key 3
Eureka Math Grade 3 Module 1 Lesson 17 Sprint Answer Key 4

Eureka Math Grade 3 Module 1 Lesson 17 Answer Key-1
Eureka Math Grade 3 Module 1 Lesson 17 Answer Key-2
Eureka Math Grade 3 Module 1 Lesson 17 Answer Key-3
Eureka Math Grade 3 Module 1 Lesson 17 Answer Key-4
Explanation:
Multiplied or divided the given number by 4,
as shown above.

B
Multiply or Divide by 4

Eureka Math Grade 3 Module 1 Lesson 17 Sprint Answer Key 5
Eureka Math Grade 3 Module 1 Lesson 17 Sprint Answer Key 6
Eureka Math Grade 3 Module 1 Lesson 17 Sprint Answer Key 7
Eureka Math Grade 3 Module 1 Lesson 17 Sprint Answer Key 8

Eureka Math Grade 3 Module 1 Lesson 17 Answer Key-5
Eureka Math Grade 3 Module 1 Lesson 17 Answer Key-6
Eureka Math Grade 3 Module 1 Lesson 17 Answer Key-7
Eureka Math Grade 3 Module 1 Lesson 17 Answer Key-8

Eureka Math Grade 3 Module 1 Lesson 17 Problem Set Answer Key

Question 1.
Use the array to complete the related equations.
Eureka Math Grade 3 Module 1 Lesson 17 Problem Set Answer Key 10

Eureka Math Grade 3 Module 1 Lesson 17 Answer Key-9
Explanation:
Used the array to complete the related equations as shown above and as
1 X 4 = 4,   4 ÷ 4 = 1,
2 X 4 = 8,   8 ÷ 4 = 2,
3 X 4 = 12, 12 ÷ 4 = 3,
4 X 4 = 16, 16 ÷ 4 = 4,
5 X 4 = 20, 20 ÷ 4 = 5,
6 X 4 = 24, 24 ÷ 4 = 6,
7 X 4 = 28, 28 ÷ 4 = 7,
8 X 4 = 32, 32 ÷ 4 = 8,
9 X 4 = 36, 36 ÷ 4 = 9,
10 X 4 = 40, 40 ÷ 4 = 10.

Question 2.
The baker packs 36 bran muffins in boxes of 4.
Draw and label a tape diagram to find the number of boxes he packs.

The baker packs 9 boxes of muffins,
Eureka Math Grade 3 Module 1 Lesson 17 Answer Key-10
Explanation:
The baker packs 36 bran muffins in boxes of 4.
Drawn and labeled a tape diagram,
The number of boxes baker packs is 36 ÷ 4 = 9 boxes.
Therefore, the baker packs 9 boxes of muffins.

Question 3.
The waitress arranges 32 glasses into 4 equal rows.
How many glasses are in each row?

There are 8 glasses in each row,

Explanation:
Given the waitress arranges 32 glasses into 4 equal rows.
Number of glasses in each row are 32 ÷ 4 = 8 (8 X 4 = 32),
Therefore, there are 8 glasses in each row.

Question 4.
Janet paid $28 for 4 notebooks. Each notebook costs the same amount.
What is the cost of 2 notebooks?

The cost of 2 notebooks is $14,

Explanation:
Given Janet paid $28 for 4 notebooks and each
notebook costs the same amount, So cost of each
notebook is $28 ÷ 4 = $7 (4 X 7 = 28),
Now cost for 2 notebooks is 2 X $7 = $14.

Eureka Math Grade 3 Module 1 Lesson 17 Exit Ticket Answer Key

Question 1.
Mr. Thomas organizes 16 binders into stacks of 4.
How many stacks does he make? Draw and label a
number bond to solve.

Eureka Math Grade 3 Module 1 Lesson 17 Answer Key-11
Mr. Thomas makes 4 stacks of 4,

Explanation:
Given Mr. Thomas organizes 16 binders into stacks of 4.
So number of stacks does he make are 16 ÷ 4 = 4 ( 4 X 4 = 16),
Drawn and labeled a number bond to solve number of stacks
Mr. Thomas makes as shown above in the picture.

Question 2.
The chef uses 28 avocados to make 4 batches of guacamole.
How many avocados are in 2 batches of guacamole?
Draw and label a tape diagram to solve.

In 2 batches of guacamole the chef uses 14 avocados,
Eureka Math Grade 3 Module 1 Lesson 17 Answer Key-12
Explanation:
Given the chef uses 28 avocados to make 4 batches of guacamole.
So number of avocados in one batch of guacamole are
28 ÷ 4 =7 (7 X 4 = 28),
Now number of avocados are in 2 batches of guacamole are
2 X 7 = 14 avocados.
Drawn and labeled a tape diagram as shown above.

Eureka Math Grade 3 Module 1 Lesson 17 Homework Answer Key

Question 1.
Use the array to complete the related equations.

Eureka Math Grade 3 Module 1 Lesson 17 Homework Answer Key 11
Eureka Math Grade 3 Module 1 Lesson 17 Answer Key-13
Explanation:
Used the array to complete the related equations as shown above and as
1 X 4 = 4,   4 ÷ 4 = 1,
2 X 4 = 8,   8 ÷ 4 = 2,
3 X 4 = 12, 12 ÷ 4 = 3,
4 X 4 = 16, 16 ÷ 4 = 4,
5 X 4 = 20, 20 ÷ 4 = 5,
6 X 4 = 24, 24 ÷ 4 = 6,
7 X 4 = 28, 28 ÷ 4 = 7,
8 X 4 = 32, 32 ÷ 4 = 8,
9 X 4 = 36, 36 ÷ 4 = 9,
10 X 4 = 40, 40 ÷ 4 = 10.

Question 2.
The teacher puts 32 students into groups of 4.
How many groups does she make? Draw and label a
tape diagram to solve.

Teacher makes 8 groups,
Eureka Math Grade 3 Module 1 Lesson 17 Answer Key-14
Explanation :
Given the teacher puts 32 students into groups of 4.
Teacher makes 32 ÷ 4 = 8 (4 X 8 = 32) groups,
Drawn and labeled tape diagram to solve number of
groups teacher has made.

Question 3.
The store clerk arranges 24 toothbrushes into 4 equal rows.
How many toothbrushes are in each row?

In each row there are 6 toothbrushes,

Explanation:
Given the store clerk arranges 24 toothbrushes into 4 equal rows.
So number of toothbrushes in each row are 24 ÷ 4 = 6 (6 X 4 = 24).

Question 4.
An art teacher has 40 paintbrushes. She divides them equally
among her 4 students. She finds 8 more brushes and divides
these equally among the students, as well. How many brushes
does each student receive?

Each student will receive 12 paintbrushes,

Explanation:
Given An art teacher has 40 paintbrushes. She divides them equally
among her 4 students. She finds 8 more brushes and divides
these equally among the students, as well.
So number of brushes she has 40 + 8 = 48 brushes,
Now number of brushes each student will receive are
48 ÷ 4 = 12 ( 12 X 4 = 48).
Therefore, each student will receive 12 paintbrushes.