Big Ideas Math Answers Grade 1 Chapter 1 Addition and Subtraction Situations

Big Ideas Math Answers Grade 1 Chapter 1

Download Big Ideas Math Book 1st Grade Answer Key Chapter 1 Addition and Subtraction Situations pdf for free of cost. Addition and subtraction are the basics you have to learn in maths to become a master. Make use of Big Ideas Math Book 1st Grade Answer Key Chapter 1 Addition and Subtraction Situations and start your preparation for exams. Quick and easy learning is possible with the help of Big Ideas Math Answers Grade 1 Chapter 1 Addition and Subtraction Situations.

Big Ideas Math Book 1st Grade Answer Key Chapter 1 Addition and Subtraction Situations

The topics covered in this Addition and Subtraction Situations are Vocabulary, Addition: Add To, Solve Add To Problems, etc. If you start practicing the problems from now itself you can score top in the exams. After practicing the problems we suggest you solve the questions given at the end of the chapter. So that you can know in which topic you are lagging and practice number of times. Simply tap the link who wish to practice and solve the problems.

Vocabulary

Lesson: 1 Addition: Add To

Lesson: 2 Solve Add To Problems

Lesson: 3 Solve Put Together Problems

Lesson: 4 Solve Put Together Problems with Both Addends Unknown

Lesson: 5 Solve Take From Problems

Lesson: 6 Solve Compare Problems: More

Lesson: 7 Solve Compare Problems: Fewer

Lesson: 8 Solve Add To Problems with Change Unknown

Lesson: 9 Connect Put Together and Take Apart Problems

Performance Task

Addition and Subtraction Situations Vocabulary

Oraganize it

Use the review words to complete the graphic organizer

Big Ideas Math Answer Key Grade 1 Chapter 1 Addition and Subtraction Situations 1
Answer:
GIVEN:
Big Ideas Math Answer Key Grade 1 Chapter 1 Addition and Subtraction Situations 1
LABELLING:

Define it

Use your vocabulary cards to match.

Big Ideas Math Answer Key Grade 1 Chapter 1 Addition and Subtraction Situations 2
Answer:
GIVEN:
Big Ideas Math Answer Key Grade 1 Chapter 1 Addition and Subtraction Situations 2

EXPLANATION:
MATCHING :

Lesson 1.1 Addition: Add To

Explore and Grow

Use linking cubes to model the story.

Big Ideas Math Answers 1st Grade 1 Chapter 1 Addition and Subtraction Situations 1
Answer:
GIVEN:
Number of bears = 4
Number of more bears added  = 2
EXPLANATION:
ADDITION:
Total number of bears = Number of bears  + Number of more bears added
=4 + 2 =6

Hence, Total number of bears = 6.

Show and Grow

Question 1.
Big Ideas Math Answers 1st Grade 1 Chapter 1 Addition and Subtraction Situations 2
Big Ideas Math Answers 1st Grade 1 Chapter 1 Addition and Subtraction Situations 3
Addition equation: __________ + ___________ = ___________
Answer:
GIVEN:
Number of cats = 5
Number of more cats added = 1
EXPLANATION:
ADDITION:
Total Number of cats = Number of cats  + Number of more cats added
=5 + 1 = 6

Hence, total number of cats = 6.

Apply and Grow: Practice

Question 2.
Big Ideas Math Answers 1st Grade 1 Chapter 1 Addition and Subtraction Situations 4
Big Ideas Math Answers 1st Grade 1 Chapter 1 Addition and Subtraction Situations 5
__________ + ___________ = ___________
Answer:
GIVEN:
Number of crabs = 3
Number of crabs added more = 6
EXPLANATION:
ADDITION:
Total number of crabs = Number of crabs + Number of crabs added more
=3 + 6 = 9

Hence, Total number of crabs = 9.

Question 3.
Big Ideas Math Answers 1st Grade 1 Chapter 1 Addition and Subtraction Situations 6
Big Ideas Math Answers 1st Grade 1 Chapter 1 Addition and Subtraction Situations 7
__________ + ___________ = ___________
Answer:
GIVEN:
Number of pigs = 8
Number of pigs added more = 2
EXPLANATION:
ADDITION:
Total number of pigs = Number of pigs + Number of pigs added more
=8 + 2 = 10

Hence, Total number of pigs = 10.

Question 4.
DIG DEEPER!
Complete the picture and the story.
Big Ideas Math Answers 1st Grade 1 Chapter 1 Addition and Subtraction Situations 8
Big Ideas Math Answers 1st Grade 1 Chapter 1 Addition and Subtraction Situations 9
__________ + ___________ = ___________
Answer:
GIVEN:
Number of owls in the picture = 3
Number of owls added more = 3
EXPLANATION:
ADDITION:
Total number of owls in the picture = Number of owls in the picture  + Number of owls added more
=3 + 3 = 6

Hence,  Total number of owls in the picture = 6.

Think and Grow: Modeling Real Life

Big Ideas Math Answers 1st Grade 1 Chapter 1 Addition and Subtraction Situations 10
Answer:
GIVEN:
Number of fishes = 5
Number of fishes added more = 3
EXPLANATION:
ADDITION:
Total number of fishes = Number of fishes +  Number of fishes added more
=5 + 3 = 8

Hence, Total number of fishes =  8.

Show and Grow

Question 5.
Big Ideas Math Answers 1st Grade 1 Chapter 1 Addition and Subtraction Situations 11
Answer:
GIVEN:
Number of tortoise = 2
Number of tortoise added more = 7
EXPLANATION:
ADDITION:
Total number of tortoise = Number of tortoise + Number of tortoise added more
=2 + 7 = 9

Hence, total number of tortoise = 9.

Addition: Add To Practice 1.1

Question 1.
Big Ideas Math Answers 1st Grade 1 Chapter 1 Addition and Subtraction Situations 12
Big Ideas Math Answers 1st Grade 1 Chapter 1 Addition and Subtraction Situations 13
__________ + ___________ = ___________
Answer:
GIVEN:
Number of foxes = 6
Number of foxes added more = 1
EXPLANATION:
ADDITION:
Total number of foxes = Number of foxes + Number of foxes added more
=6 + 1 = 7
Hence, Total number of foxes = 7.

Question 2.
Big Ideas Math Answers 1st Grade 1 Chapter 1 Addition and Subtraction Situations 14
Big Ideas Math Answers 1st Grade 1 Chapter 1 Addition and Subtraction Situations 15
__________ + ___________ = ___________
Answer:
GIVEN:
Number of bugs = 7
Number of bugs added more = 3
EXPLANATION:
ADDITION:
Total number of bugs = Number of bugs  + Number of bugs added more
=7 + 3 = 10
Hence, the Total number of bugs = 10.

Question 3.
DIG DEEPER!
Big Ideas Math Answers 1st Grade 1 Chapter 1 Addition and Subtraction Situations 16
Big Ideas Math Answers 1st Grade 1 Chapter 1 Addition and Subtraction Situations 17
__________ + ___________ = ___________
Answer:
GIVEN:
Number of parrots = 6
Number of parrots added more = 2
EXPLANATION:
ADDITION:
Total number of parrots = Number of parrots  + Number of parrots added more
=6 + 2 = 8
Hence, Total number of parrots = 8.

Question 4.
Modeling Real Life
There are 2 Big Ideas Math Answers 1st Grade 1 Chapter 1 Addition and Subtraction Situations 18. 7 more Big Ideas Math Answers 1st Grade 1 Chapter 1 Addition and Subtraction Situations 18 join them. How many Big Ideas Math Answers 1st Grade 1 Chapter 1 Addition and Subtraction Situations 18 are there now?

 

___________ Big Ideas Math Answers 1st Grade 1 Chapter 1 Addition and Subtraction Situations 18
Answer:
GIVEN:
Number of ducks = 2
Number of ducks added more = 7
EXPLANATION:
ADDITION:
Total number of ducks = Number of ducks + Number of ducks added more
= 2 + 7 = 9
Hence, the Total number of ducks = 9.

Review and Refresh

Write the number of objects.

Question 5.
Big Ideas Math Answers 1st Grade 1 Chapter 1 Addition and Subtraction Situations 19
Answer:
GIVEN:
Apples in the picture.
COUNT:
Number of apples first row = 4
Number of apples second row = 3
EXPLANATION:
Total number of apples in both rows = Number of apples first row + Number of apples second row
= 4 + 3 = 7
Hence, the Total number of apples in both rows = 7.

Question 5.
Big Ideas Math Answers 1st Grade 1 Chapter 1 Addition and Subtraction Situations 20
Answer:
GIVEN:
Bananas in the picture.
COUNT:
Number of Bananas first row = 4
Number of Bananas second row = 4
Number of Bananas third row = 2
EXPLANATION:
Total number of Bananas in all three rows = Number of Bananas first row + Number of Bananas second row + Number of Bananas third row
= 4 + 4 +2
= 8 + 2
= 10.
Hence, Total number of Bananas in all three rows = 10.

Lesson 1.2 Solve Add To Problems

Explore and Grow

Use linking cubes to model the story.

Big Ideas Math Answers Grade 1 Chapter 1 Addition and Subtraction Situations 1
Answer:
GIVEN:
Number of hats = 5
Number of hats added more = 2
EXPLANATION:
ADDITION:
Total number of hats = Number of hats + Number of hats added more
= 5 + 2 = 7

Hence, Total number of hats= 7.

Show and Grow

Question 1.
Big Ideas Math Answers Grade 1 Chapter 1 Addition and Subtraction Situations 2
Addition equation:
__________ + ___________ = ___________ Big Ideas Math Answers Grade 1 Chapter 1 Addition and Subtraction Situations 3
Answer:
GIVEN:
Number of balls = 4
Number of balls added more = 2
EXPLANATION:
ADDITION:
Total number of balls = Number of balls + Number of balls added more
= 4 + 2 = 6
Hence, number of balls = 6.

Question 2.
Big Ideas Math Answers Grade 1 Chapter 1 Addition and Subtraction Situations 4
Addition equation:
__________ + ___________ = ___________ Big Ideas Math Answers Grade 1 Chapter 1 Addition and Subtraction Situations 5
Answer:
GIVEN:
Number of rabbits = 4
Number of rabbits more added = 4
EXPLANATION:
ADDITION:
Total number of rabbits = Number of rabbits + Number of rabbits more added
= 4 + 4  = 8
Hence, number of rabbits = 8.

Apply and Grow: Practice

Question 3.
Big Ideas Math Answers Grade 1 Chapter 1 Addition and Subtraction Situations 6
__________ + ___________ = ___________ Big Ideas Math Answers Grade 1 Chapter 1 Addition and Subtraction Situations 7
Answer:
GIVEN:
Number of books = 2
Number of books added more = 2
EXPLANATION:
ADDITION:
Total number of books = Number of books + Number of books added more
= 2 + 2 = 4
Hence, Total number of books = 4.

Question 4.
Big Ideas Math Answers Grade 1 Chapter 1 Addition and Subtraction Situations 8
__________ + ___________ = ___________ Big Ideas Math Answers Grade 1 Chapter 1 Addition and Subtraction Situations 9
Answer:
GIVEN:
Number of ants = 3
Number of ants added more = 5
EXPLANATION:
ADDITION:
Total number of ants = Number of ants  + Number of ants added more
= 3 + 5 = 8
Hence, Total number of ants = 8.

Question 5.
Big Ideas Math Answers Grade 1 Chapter 1 Addition and Subtraction Situations 10
__________ + ___________ = ___________ Big Ideas Math Answers Grade 1 Chapter 1 Addition and Subtraction Situations 11
Answer:
GIVEN:
Number of chocolates eaten by me = 9
Number of chocolates eaten by me more = 1
EXPLANATION:
ADDITION:
Total number of chocolates eaten by me = Number of chocolates eaten by me + Number of chocolates eaten by me more
= 9 + 1 = 10.
Hence, Total number of chocolates eaten by me = 10.

Question 6.
DIG DEEPER
Complete the picture and the story
Big Ideas Math Answers Grade 1 Chapter 1 Addition and Subtraction Situations 12
Big Ideas Math Answers Grade 1 Chapter 1 Addition and Subtraction Situations 13
__________ + ___________ = ___________
Answer:
GIVEN:
Number of snails = 4
Number of snails added more = 3
EXPLANATION:
ADDITION:
Total number of snails = Number of snails  + Number of snails added more
= 4 + 3 = 7

Hence, Total number of snails = 7.

Think and Grow: Modeling Real Life

Big Ideas Math Answers Grade 1 Chapter 1 Addition and Subtraction Situations 14
Answer:
GIVEN:
Number of apples with me = 3
Number of apples I purchased = 7
EXPLANATION:
ADDITION:
Total number of apples I am having = Number of apples with me  + Number of apples I purchased
= 3 + 7 = 10.

Hence, Total number of apples I am having = 10.

Show and Grow

Question 7.
Big Ideas Math Answers Grade 1 Chapter 1 Addition and Subtraction Situations 15
Answer:
GIVEN:
Number of cube boxes = 6
Number of cube boxes more added given by your Friend= 3
EXPLANATION:
ADDITION:
Total number of cube boxes = Number of cube boxes + Number of cube boxes more added given by your Friend
= 6 + 3 = 9

Hence, Total number of cube boxes = 9.

Solve Add To Problems Practice 1.2

Question 1.
Big Ideas Math Answers Grade 1 Chapter 1 Addition and Subtraction Situations 16
__________ + ___________ = ___________ Big Ideas Math Answers Grade 1 Chapter 1 Addition and Subtraction Situations 17
Answer:
GIVEN:
Number of pencils = 2
Number of pencils  purchased more = 3
EXPLANATION:
ADDITION:
Total number of pencils  = Number of pencils + Number of pencils purchased more
= 2 + 3 = 5
Hence, Total number of pencils = 5.

Question 2.
Big Ideas Math Answers Grade 1 Chapter 1 Addition and Subtraction Situations 18
__________ + ___________ = ___________ Big Ideas Math Answers Grade 1 Chapter 1 Addition and Subtraction Situations 19
Answer:
GIVEN:
Number of flowers = 6
Number of flowers added more = 4
EXPLANATION:
ADDITION:
Total number of flowers = Number of flowers + Number of flowers added more
= 6 + 4 = 10
Hence, Total number of flowers = 10.

Question 3.
Big Ideas Math Answers Grade 1 Chapter 1 Addition and Subtraction Situations 20
__________ + ___________ = ___________ Big Ideas Math Answers Grade 1 Chapter 1 Addition and Subtraction Situations 21
Answer:
GIVEN:
Number of butterflies = 5
Number of butterflies added more = 4
EXPLANATION:
ADDITION:
Total number of butterflies = Number of butterflies + Number of butterflies added more
= 5 + 4 = 9
Hence, Total number of butterflies = 9.

Question 4.
DIG DEEPER
Complete the picture and the story.
Big Ideas Math Answers Grade 1 Chapter 1 Addition and Subtraction Situations 22
Big Ideas Math Answers Grade 1 Chapter 1 Addition and Subtraction Situations 23
__________ + ___________ = ___________
Answer:
GIVEN:
Number of Frogs = 7
Number of Frogs more added = 3
EXPLANATION:
ADDITION:
Total number of frogs = Number of Frogs + Number of Frogs more added
= 7 + 3 = 10

Hence, Total number of frogs = 10.

Question 5.
Modeling Real Life
You have 2 Big Ideas Math Answers Grade 1 Chapter 1 Addition and Subtraction Situations 24. You buy 4Big Ideas Math Answers Grade 1 Chapter 1 Addition and Subtraction Situations 24. How many Big Ideas Math Answers Grade 1 Chapter 1 Addition and Subtraction Situations 24 do you have now?

_____________ Big Ideas Math Answers Grade 1 Chapter 1 Addition and Subtraction Situations 24

Answer:
GIVEN:
Number of balls = 2
Number of balls purchased more = 4
EXPLANATION:
ADDITION:
Total number of balls = Number of balls + Number of balls purchased more
= 2 + 4 = 6
Hence, Total number of balls = 6.

Review & Refresh

Question 6.
Use the picture to complete the number bond.
Big Ideas Math Answers Grade 1 Chapter 1 Addition and Subtraction Situations 25
Answer:
GIVEN:
Number of lemons = 6
Number of circle bonding given = 3
EXPLANATION:
DIVISION:
The picture of the number bond with lemons = 6/3 = 2 in each circle.
Hence, the picture is below:

Lesson 1.3 Solve Put Together Problems

Explore and Grow

Use counters to model the story.

You have 7 Big Ideas Math Solutions Grade 1 Chapter 1 Addition and Subtraction Situations 1 and 2 Big Ideas Math Solutions Grade 1 Chapter 1 Addition and Subtraction Situations 2. How many toys do you have in all?

 

______ toys
Answer:
GIVEN:
Number of robot toys =7
Number of car toys = 2
EXPLANATION:
ADDITION:
Total number of toys = Number of robot toys + Number of car toys
= 7 + 2 = 9

Hence, total number of toys = 9.

Show and Grow

Question 1.
You have 3 Big Ideas Math Solutions Grade 1 Chapter 1 Addition and Subtraction Situations 3 and 1 Big Ideas Math Solutions Grade 1 Chapter 1 Addition and Subtraction Situations 4 How many balls do you
Big Ideas Math Solutions Grade 1 Chapter 1 Addition and Subtraction Situations 5
Answer:
GIVEN:
Number of volleyballs = 3
Number of footballs = 1
EXPLANATION:
ADDITION:
Total number of balls = Number of volleyballs + Number of footballs
= 3 + 1 = 4
Hence, total number of balls = 4.

Question 2.
There are 2 Big Ideas Math Solutions Grade 1 Chapter 1 Addition and Subtraction Situations 6 and 6 Big Ideas Math Solutions Grade 1 Chapter 1 Addition and Subtraction Situations 7 How many fish are there in all?
Big Ideas Math Solutions Grade 1 Chapter 1 Addition and Subtraction Situations 8
Answer:
GIVEN:
Number of green fishes = 2
Number of yellow fishes = 6
EXPLANATION:
ADDITION:
Total number of fishes = Number of green fishes + Number of yellow fishes
= 2 + 6 =8

Hence, total number of fishes = 8.

Question 3.
You have 1 Big Ideas Math Solutions Grade 1 Chapter 1 Addition and Subtraction Situations 9 and 5 Big Ideas Math Solutions Grade 1 Chapter 1 Addition and Subtraction Situations 10. How many pets do you have in all?
Big Ideas Math Solutions Grade 1 Chapter 1 Addition and Subtraction Situations 11
Answer:
GIVEN:
Number of dogs = 1
Number of cats = 5
EXPLANATION:
ADDITION:
Total number of pets = Number of dogs + Number of cats
= 1 + 5 = 6
Hence, total number of pets = 6.

Question 4.
There are 4 Big Ideas Math Solutions Grade 1 Chapter 1 Addition and Subtraction Situations 12 and 6 Big Ideas Math Solutions Grade 1 Chapter 1 Addition and Subtraction Situations 13. How many animals are there in all?
Big Ideas Math Solutions Grade 1 Chapter 1 Addition and Subtraction Situations 14
Answer:
GIVEN:
Number of Horses = 4
Number of Hens = 6
EXPLANATION:
ADDITION:
Total number of animals = Number of Horses + Number of Hens
= 4 + 6 = 10
Hence, total number of animals = 10.

Question 5.
MP Number Sense
Number Sense Circle two groups of bugs to match the addition problem. Then find the sum.
Big Ideas Math Solutions Grade 1 Chapter 1 Addition and Subtraction Situations 15
3 + 5 = __________
Answer:

Think and Grow: Modeling Real Life

You have 3 Big Ideas Math Solutions Grade 1 Chapter 1 Addition and Subtraction Situations 16 and 4 Big Ideas Math Solutions Grade 1 Chapter 1 Addition and Subtraction Situations 17 . Your friend has 8 flowers. Who has more flowers?
Big Ideas Math Solutions Grade 1 Chapter 1 Addition and Subtraction Situations 18
Model:
Big Ideas Math Solutions Grade 1 Chapter 1 Addition and Subtraction Situations 19
Addition equation:

Who has more flowers?                You                  Friend

Show and Grow
Answer:
GIVEN:
Number of purple flower I am having = 3
Number of yellow flower I am having = 4
EXPLANATION:
ADDITION:
Total number of flowers I am having = Number of purple flower I am having + Number of yellow flower I am having
= 3 + 4 =7
Total Number of flowers my Friend having = 8
COMPARISION:
Number of flowers has more flowers = Total number of flowers I am having  or Total Number of flowers my Friend having
=>
Hence, my Friend has more flowers than me.

Question 6.
You have 4 Big Ideas Math Solutions Grade 1 Chapter 1 Addition and Subtraction Situations 20 and 5 Big Ideas Math Solutions Grade 1 Chapter 1 Addition and Subtraction Situations 21 . Your friend has 7 leaves. Who has more leaves?
Model:
Big Ideas Math Solutions Grade 1 Chapter 1 Addition and Subtraction Situations 22
Addition equation:

Who has more leaves?               You              Friend
Answer:
GIVEN:
Number of  orange leaves with me = 4
Number of  green leaves with me= 5
EXPLANATION:
ADDITION:
Total number of leaves with me= Number of  orange leaves with me + Number of  green leaves with me
= 4 + 5 = 9
Total number of leaves my friend having = 7
COMPARISION:
Number of leaves who is having more = Total number of leaves with me or Total number of leaves my friend having
=
Hence, number of leaves with me are more than my friend.

Solve Put Together Problems Practice 1.3

Question 1.
You buy 1 Big Ideas Math Solutions Grade 1 Chapter 1 Addition and Subtraction Situations 23 and 7 Big Ideas Math Solutions Grade 1 Chapter 1 Addition and Subtraction Situations 24. How many pieces of fruit do you buy in all?
Big Ideas Math Solutions Grade 1 Chapter 1 Addition and Subtraction Situations 25
Answer:
GIVEN:
Number of pineapples purchased = 1
Number of bananas purchased = 7
EXPLANATION:
ADDITION:
Total number of fruits purchased = Number of pineapples purchased + Number of bananas purchased
= 1 + 7 = 8
Hence, Total number of fruits purchased = 8.

Question 2.
There are 5 Big Ideas Math Solutions Grade 1 Chapter 1 Addition and Subtraction Situations 26 and 5 Big Ideas Math Solutions Grade 1 Chapter 1 Addition and Subtraction Situations 27 . How many birds are there in all?
Big Ideas Math Solutions Grade 1 Chapter 1 Addition and Subtraction Situations 28
Answer:
GIVEN:
Number of koala birds = 5
Number of kuku  birds = 5
EXPLANATION:
ADDITION:
Total number of birds = Number of koala birds + Number of kuku  birds
= 5 + 5 = 10
Hence, Total number of birds = 10.

Question 3.
MP Number Sense
Circle two groups of hats to match the addition problem. Then find the sum.
Big Ideas Math Solutions Grade 1 Chapter 1 Addition and Subtraction Situations 29
6 + 1 = 7
Answer:

Question 4.
Modeling Real Life
You have 4 Big Ideas Math Solutions Grade 1 Chapter 1 Addition and Subtraction Situations 30 and 5 Big Ideas Math Solutions Grade 1 Chapter 1 Addition and Subtraction Situations 31. You friend has 8 cars. Who has more cars?
Big Ideas Math Solutions Grade 1 Chapter 1 Addition and Subtraction Situations 32
Who has more cars?                You              Friend
Answer:
GIVEN:
Number of blue cars with me = 4
Number of yellow cars with me = 5
EXPLANATION:
ADDITION:
Total number of cars with me = Number of blue cars with me+ Number of yellow cars with me
= 4 + 5 = 9
Number of cars my friend has = 8
COMPARISION:
Number of cars who has more = Total number of cars with me  or Number of cars my friend has
=
Hence, total number of cars with me are more than my friend.

Review & Refresh

Question 5.
Write the number for each type of pepper. Write the number for the whole.
Big Ideas Math Solutions Grade 1 Chapter 1 Addition and Subtraction Situations 33
Answer:
GIVEN:
Number of yellow pepper =  3 or three
Number of red pepper =  3 or three
EXPLANATION:
ADDITION:
Total number of pepper = Number of yellow pepper + Number of red pepper
=  3 + 3 = 6 or Six.
Hence, Total number of pepper = 6 or Six.

Lesson 1.4 Solve Put Together Problems with Both Addends Unknown

Explore and Grow

Use linking cubes to model the story.

There are 10 Big Ideas Math Answer Key Grade 1 Chapter 1 Addition and Subtraction Situations 34. Some are on the court and some are on the rack Big Ideas Math Answer Key Grade 1 Chapter 1 Addition and Subtraction Situations 34.
Big Ideas Math Answer Key Grade 1 Chapter 1 Addition and Subtraction Situations 35
Answer:
GIVEN:
Number of cubes on the court = 6
Number of cubes on the rack = 4
EXPLANATION:
ADDITION:
Total number of cubes = Number of cubes on the court + Number of cubes on the rack
= 6 + 4 = 10

Hence, total number of cubes = 10.

Show and Grow

Question 1.
There are 7 Big Ideas Math Answer Key Grade 1 Chapter 1 Addition and Subtraction Situations 36 Some are on a tree. Some more are on the ground. Draw the Big Ideas Math Answer Key Grade 1 Chapter 1 Addition and Subtraction Situations 36
Big Ideas Math Answer Key Grade 1 Chapter 1 Addition and Subtraction Situations 37
Addition equation: ________ = ________ + _______
Answer:
GIVEN:
Number of apples on the tree = 4
Number of apples on the ground = 3
EXPLANATION:
ADDITION:
Total number of apples = Number of apples on the tree + Number of apples on the ground
= 4 + 3 = 7.
PICTURE :

Hence, drawn the picture.

Apply and Grow: Practice

Question 2.
There are 5 Big Ideas Math Answer Key Grade 1 Chapter 1 Addition and Subtraction Situations 38 Some are on a couch. Some more are on a rug. Draw the Big Ideas Math Answer Key Grade 1 Chapter 1 Addition and Subtraction Situations 38
Big Ideas Math Answer Key Grade 1 Chapter 1 Addition and Subtraction Situations 39
Answer:
GIVEN:
Number of dogs on a couch = 3
Number of dogs on a rug = 2
EXPLANATION:
ADDITION:
Total number of dogs = Number of dogs on a couch + Number of dogs on a rug
= 3 + 2 = 5
PICTURE:

Hence, picture is drawn.

Question 3.
There are 8 Big Ideas Math Answer Key Grade 1 Chapter 1 Addition and Subtraction Situations 40 Some are in a pond. Some more are in the grass. Draw two different pictures to show the Big Ideas Math Answer Key Grade 1 Chapter 1 Addition and Subtraction Situations 40
Big Ideas Math Answer Key Grade 1 Chapter 1 Addition and Subtraction Situations 41
Answer:
GIVEN:
Total number of snakes = 8
FIGURE:1:
Number of snakes in a pond = 4
Number of snakes in a grass = 4
EXPLANATION:
ADDITION:
Total number of snakes = Number of snakes in a pond + Number of snakes in a grass
= 4 + 4 = 8
FIGURE:2:
Number of snakes in a pond = 3
Number of snakes in a grass = 5
Total number of snakes = Number of snakes in a pond + Number of snakes in a grass
= 3 + 5 = 8
PICTURE:

Hence, Total number of snakes = 8.

Question 4.
DIG DEEPER!
Newton has 4 toy trucks. Some are red. The rest are blue. Write two different addition equations to describe his trucks.
Big Ideas Math Answer Key Grade 1 Chapter 1 Addition and Subtraction Situations 42
Answer:
GIVEN:
EQUATION:1:
Number of red trucks toys Newton has = 3
Number of blue trucks toys Newton has = 1
EXPLANATION:
ADDITION:
Total number of toy trucks Newton has = Number of red trucks toys Newton has + Number of blue trucks toys Newton has
= 3 + 1 =4.
Hence, total number of toy trucks Newton has = 4.
EQUATION:2:
Number of red trucks toys Newton has = 2
Number of blue trucks toys Newton has = 2
EXPLANATION:
ADDITION:
Total number of toy trucks Newton has = Number of red trucks toys Newton has + Number of blue trucks toys Newton has
= 2 + 2 =4.
Hence, total number of toy trucks Newton has = 4.

Think and Grow: Modeling Real Life

You have 7 ribbons. Some are blue. The rest are red. You have more blue ribbons than red ribbons. How many blue and red ribbons can you have?
Big Ideas Math Answer Key Grade 1 Chapter 1 Addition and Subtraction Situations 43
2 blue 5 red           4 blue 3 red
6 blue 1 red           3 blue 4 red
Show how you know:
Answer:
Total number of ribbons = 7.
Some are blue and rest are red.
Blue ribbons are more than red ribbons.
EXPLANATION:
GIVEN:
1. 2 blue 5 red
Blue ribbons are more than red balls.
Number of blue ribbons are less than red.
Hence, this case is not  the correct count of ribbons.
2.  4 blue 3 red
Number of blue ribbons are more than red.
Number of blue ribbons = 4
Number of red ribbons = 3
EXPLANATION:
ADDITION:
Total number of ribbons = Number of blue ribbons + Number of red ribbons
= 4 + 3 = 7.
Hence, this case is the correct count of ribbons.
3.6 blue 1 red
Blue ribbons are more than red balls.
Number of blue ribbons are more than red.
Number of blue ribbons = 6
Number of red ribbons = 1
EXPLANATION:
ADDITION:
Total number of ribbons = Number of blue ribbons + Number of red ribbons
= 6 + 1 = 7.
Hence, this case is the correct count of ribbons.
4.  3 blue 4 red
Number of blue ribbons are less than red.
So, this case is not  the correct count of ribbons.
Hence, Total number of ribbons can be 4 blue 3 red if not 6 blue 1 red ball.

Show and Grow

Question 5.
There are lo kites. Some are green. The rest are yellow. There are more yellow kites than green kites. How many green and yellow kites can there be?
Big Ideas Math Answer Key Grade 1 Chapter 1 Addition and Subtraction Situations 44
4 green 6 yellow               6 green 4 yellow
5 green 5 yellow               3 green 7 yellow
Answer:
GIVEN:
Total number of kites = 10.
Some are green and rest are yellow.
There are more yellow kites than green kites.
EXPLANATION:
1. 4 green 6 yellow
There are more yellow kites than green kites.
Number of green kites = 4
Number of yellow kites = 6
ADDITION:
Total number of kites = Number of green kites + Number of yellow kites
= 4 + 6 = 10.
2.  6 green 4 yellow
Here, number of green kites are more than yellow kites. So, this cannot be the count of kites.
3. 5 green 5 yellow
Here, both the number of green and yellow kites are same. This means the condition given is not fulfilled. So, this cannot be the count of kites.
4.  3 green 7 yellow
There are more yellow kites than green kites.
Number of green kites = 3
Number of yellow kites = 7
ADDITION:
Total number of kites = Number of green kites + Number of yellow kites
= 3 + 7 = 10.
Hence, the total number of kites can be 4 green 6 yellow or 3 green 7 yellow kites.

Solve Put Together Problems with Both Addends Unknown Practice 1.4

Question 1.
There are 6 Big Ideas Math Answer Key Grade 1 Chapter 1 Addition and Subtraction Situations 45 Some are on a hill. Some more are in the grass. Draw the Big Ideas Math Answer Key Grade 1 Chapter 1 Addition and Subtraction Situations 45
Big Ideas Math Answer Key Grade 1 Chapter 1 Addition and Subtraction Situations 46

_____________ = ____________ + ____________
Answer:
Number of ants on hill = 4
Number of ants in the grass = 2
EXPLANATION:
ADDITION:
Total number of ants = Number of ants on hill + Number of ants in the grass
= 4 + 2 = 6
PICTURE:

Hence,  Total number of ants = 6.

Question 2.
There are 8Big Ideas Math Answer Key Grade 1 Chapter 1 Addition and Subtraction Situations 47 Some are in the water. Some more are on the dirt. Draw the Big Ideas Math Answer Key Grade 1 Chapter 1 Addition and Subtraction Situations 47
Big Ideas Math Answer Key Grade 1 Chapter 1 Addition and Subtraction Situations 48
_____________ = ____________ + ____________
Answer:
Number of frogs in the water = 3
Number of frogs on the dirt = 5
EXPLANATION:
ADDITION:
Total number of frogs = Number of frogs in the water+ Number of frogs on the dirt
= 3 + 5 = 8
PICTURE:

Hence, Total number of frogs = 8.
Question 3.
There are 5Big Ideas Math Answer Key Grade 1 Chapter 1 Addition and Subtraction Situations 49 . Some are inside a bin. Some more are outside the bin. Draw two different pictures to show the Big Ideas Math Answer Key Grade 1 Chapter 1 Addition and Subtraction Situations 49 .
Big Ideas Math Answer Key Grade 1 Chapter 1 Addition and Subtraction Situations 50
Answer:
PICTURE:1:
Number of balls are inside a bin = 1
Number of balls are outside the bin = 4
EXPLANATION:
ADDITION:
Total number of balls = Number of balls are inside a bin + Number of balls are outside the bin
= 1 + 4
=5
PICTURE:2:
Number of balls are inside a bin = 2
Number of balls are outside the bin = 3
EXPLANATION:
ADDITION:
Total number of balls = Number of balls are inside a bin + Number of balls are outside the bin
= 2 + 3
= 5

Hence, total number of balls = 5.

Question 4.
Modeling Real Life
You have 9 hats. Some are blue. The rest are red. You have more red hats than blue hats. How many red and blue hats can you have?
4 red 5 blue 6 red 3 blue
3 red 6 blue 7 red 2 blue
Show how you know:
Answer:
GIVEN:
Number of hats = 9
Some are blue. The rest are red.
You have more red hats than blue hats.
EXPLANATION:
1. 4 red 5 blue
Here, red are less than blue hats. So, this are not the count of hats.
2. 6 red 3 blue
Here, red are more than blue hats.
Number of red hats = 6
Number of blue hats = 3
ADDITION:
Total number of hats =  Number of red hats + Number of blue hats
= 6 + 3 = 9
3. 3 red 6 blue
Here, red are less than blue hats. So, this are not the count of hats.
4.  7 red 2 blue
Here, red are more than blue hats.
Number of red hats = 6
Number of blue hats = 3
ADDITION:
Total number of hats =  Number of red hats + Number of blue hats
= 6 + 3 = 9
Hence, Total number of hats can be 6 red 3 blue or 7 red 2 blue.

Review & Refresh

Use the picture to write a subtraction sentence.

Question 5.
Big Ideas Math Answer Key Grade 1 Chapter 1 Addition and Subtraction Situations 51
___________ – ___________ = _____________
Answer:
Number of sea horse in first row = 4
Number of sea horse in second row = 4
EXPLANATION:
SUBTRACTION:
Difference in the rows = Number of sea horse in first row – Number of sea horse in second row
= 4 – 4 = 0
Hence, Difference in the rows = 0.

Question 6.
Big Ideas Math Answer Key Grade 1 Chapter 1 Addition and Subtraction Situations 52
___________ – ___________ = _____________
Answer:
GIVEN:
Number of dolphins left side in the picture = 6
Number of dolphins right side in the picture = 3
EXPLANATION:
SUBTRACTION:
Difference in the dolphins in both the sides of the picture = Number of dolphins left side in the picture – Number of dolphins right side in the picture
= 6 – 3  = 3
Hence, Difference in the dolphins in both the sides of the picture = 3.

Lesson 1.5 Solve Take From Problems

Explore and Grow

Use linking cubes to model the story.

There are 6 Big Ideas Math Answers 1st Grade 1 Chapter 1 Addition and Subtraction Situations 53 3 Big Ideas Math Answers 1st Grade 1 Chapter 1 Addition and Subtraction Situations 53 hop away How many Big Ideas Math Answers 1st Grade 1 Chapter 1 Addition and Subtraction Situations 53 are left.
Big Ideas Math Answers 1st Grade 1 Chapter 1 Addition and Subtraction Situations 54
Answer:
GIVEN:
Total number of rabbits = 6
Number of rabbits left away = 3
EXPLANATION:
SUBTRACTION:
Number of rabbits left = Total number of rabbits – Number of rabbits left away
= 6 – 3 = 3

Hence, Number of rabbits left = 3.

Show and Grow

Question 1.
There are 6 Big Ideas Math Answers 1st Grade 1 Chapter 1 Addition and Subtraction Situations 55.
2 Big Ideas Math Answers 1st Grade 1 Chapter 1 Addition and Subtraction Situations 55 jump away.
How many Big Ideas Math Answers 1st Grade 1 Chapter 1 Addition and Subtraction Situations 55 are left?
Subtraction equation:
___________ – __________ = __________ Big Ideas Math Answers 1st Grade 1 Chapter 1 Addition and Subtraction Situations 55
Answer:
GIVEN:
Total number of monkeys = 6
Number of monkeys jump away = 3
EXPLANATION:
SUBTRACTION:
Number of monkeys left = Total number of monkeys – Number of monkeys jump away
= 6 – 3 = 3
Hence,  Number of monkeys left = 3.

Question 2.
There are 8 Big Ideas Math Answers 1st Grade 1 Chapter 1 Addition and Subtraction Situations 56.
A monkey eats 5 Big Ideas Math Answers 1st Grade 1 Chapter 1 Addition and Subtraction Situations 56
How many Big Ideas Math Answers 1st Grade 1 Chapter 1 Addition and Subtraction Situations 56 are left?
Subtraction equation:
___________ – __________ = __________ Big Ideas Math Answers 1st Grade 1 Chapter 1 Addition and Subtraction Situations 56
Answer:
GIVEN:
Total number of bananas = 8
Number of bananas monkey ate = 5
EXPLANATION:
SUBTRACTION:
Number of bananas left = Total number of bananas – Number of bananas monkey ate
= 8 – 5 = 3
Hence, Number of bananas left = 3.

Apply and Grow: Practice

Question 3.
There are 7 Big Ideas Math Answers 1st Grade 1 Chapter 1 Addition and Subtraction Situations 57.
3 Big Ideas Math Answers 1st Grade 1 Chapter 1 Addition and Subtraction Situations 57 walk away.
How many Big Ideas Math Answers 1st Grade 1 Chapter 1 Addition and Subtraction Situations 57 are left?
___________ – __________ = __________ Big Ideas Math Answers 1st Grade 1 Chapter 1 Addition and Subtraction Situations 57
Answer:
Total number of ants = 7
Number of ants ran away = 3
EXPLANATION:
SUBTRACTION:
Number of ants left =  Total number of ants – Number of ants ran away
= 7 – 3 = 4
Hence, Number of ants left = 4.

Question 4.
There are 8 Big Ideas Math Answers 1st Grade 1 Chapter 1 Addition and Subtraction Situations 58
6 Big Ideas Math Answers 1st Grade 1 Chapter 1 Addition and Subtraction Situations 58 fly away.
How many Big Ideas Math Answers 1st Grade 1 Chapter 1 Addition and Subtraction Situations 58 are left?
___________ – __________ = __________ Big Ideas Math Answers 1st Grade 1 Chapter 1 Addition and Subtraction Situations 58
Answer:
Total number of birds = 8
Number of birds flew = 6
EXPLANATION:
SUBTRACTION:
Number of birds left = Total number of birds – Number of birds flew
= 8 – 6 = 2.
Hence, Number of birds left = 2.

Question 5.
10 Big Ideas Math Answers 1st Grade 1 Chapter 1 Addition and Subtraction Situations 59 play.
5 Big Ideas Math Answers 1st Grade 1 Chapter 1 Addition and Subtraction Situations 59 run away.
How many Big Ideas Math Answers 1st Grade 1 Chapter 1 Addition and Subtraction Situations 59 are left?
___________ – __________ = __________ Big Ideas Math Answers 1st Grade 1 Chapter 1 Addition and Subtraction Situations 59
Answer:
Total number of cats = 10
Number of cats run cats = 5
EXPLANATION:
SUBTRACTION:
Number of cats left = Total number of cats  – Number of cats run cats
= 10 – 5 = 5
Hence, Number of cats left = 5.

Question 6.
MP Precision
Find the difference. Draw a picture to match.
7 – 4 = ______
Answer:
7 – 4 = 3
EXPLANATION:
Out of 7 blocks remove 4 blocks = ????
DIFFERENCE
7 – 4 = 3
PICTURE:

Hence, picture is drawn.

Think and Grow: Modeling Real Life

7 students play tag.
2 of them leave.
How many students are left?
Big Ideas Math Answers 1st Grade 1 Chapter 1 Addition and Subtraction Situations 60
Which equation matches the story?
9 – 7 = 2
7 – 2 = 5
7 – 5 = 2
Show how you know:

 

____ students are left.
Answer:
GIVEN:
Total number of students playing tag = 7
Number of students left away = 2
EXPLANATION:
Number of students left = Total number of students playing tag – Number of students left away
= 7 – 2 = 5
Hence, Number of students left = 2
EQUATION matching the answer = 7 – 2 = 5 not others 9 – 7=2 nor 7 – 5 = 2.

Show and Grow

Question 7.
You have 9 coins. You give 3 coins away. How many coins are left?
Big Ideas Math Answers 1st Grade 1 Chapter 1 Addition and Subtraction Situations 61
Which equation matches the story?
9 – 3 = 6
6 – 3 = 3
Show how you know:

 

____ coins are left.
Answer:
GIVEN:
Total number of coins = 9
Number of coins  given away = 3
EXPLANATION:
SUBTRACTION:
Number of coins left = Total number of coins – Number of coins  given away
= 9 – 3
= 6
Hence, Number of coins left = 6.
EQUATION matching the story is 9 – 3 = 6 not 6 – 3 = 3.

Solve Take From Problems Practice 1.5

Question 1.
There are 5 Big Ideas Math Answers 1st Grade 1 Chapter 1 Addition and Subtraction Situations 62
3 Big Ideas Math Answers 1st Grade 1 Chapter 1 Addition and Subtraction Situations 62 leave.
How many Big Ideas Math Answers 1st Grade 1 Chapter 1 Addition and Subtraction Situations 62 are left?
___________ – __________ = __________ Big Ideas Math Answers 1st Grade 1 Chapter 1 Addition and Subtraction Situations 62
Answer:
GIVEN:
Total number of tigers = 5
Number of tigers left away= 3
EXPLANATION:
SUBTRACTION:
Number of tigers left = Total number of tigers – Number of tigers left away
= 5 – 3
= 2
Hence, Number of tigers left = 2.

Question 2.
There are 7 Big Ideas Math Answers 1st Grade 1 Chapter 1 Addition and Subtraction Situations 63.
A cat eats 5 Big Ideas Math Answers 1st Grade 1 Chapter 1 Addition and Subtraction Situations 63.
How many Big Ideas Math Answers 1st Grade 1 Chapter 1 Addition and Subtraction Situations 63 are left?
___________ – __________ = __________ Big Ideas Math Answers 1st Grade 1 Chapter 1 Addition and Subtraction Situations 63
Answer:
GIVEN:
Total number of fishes = 7
Number of fishes cat ate = 5
EXPLANATION:
SUBTRACTION:
Number of fishes left =  Total number of fishes – Number of fishes cat ate
= 7 – 5
= 2.
Hence, Number of fishes left = 2.

Question 3.
You have 10 Big Ideas Math Answers 1st Grade 1 Chapter 1 Addition and Subtraction Situations 64.
You give away 10 Big Ideas Math Answers 1st Grade 1 Chapter 1 Addition and Subtraction Situations 64.
How many Big Ideas Math Answers 1st Grade 1 Chapter 1 Addition and Subtraction Situations 64 do you have left?
___________ – __________ = __________ Big Ideas Math Answers 1st Grade 1 Chapter 1 Addition and Subtraction Situations 64
Answer:
GIVEN:
Total number of balls = 10
Number of balls given away = 10
EXPLANATION:
Number of balls left = Total number of balls – Number of balls given away
= 10 – 10
= 0
Hence, Number of balls left = 0.

Question 4.
MP Precision
Find the difference. Draw a picture to match.
8 – 2 = _______
Answer:
8 – 2 = 6
EXPLANATION:
Out of 8 blocks remove 2 blocks = ????
Difference:
8 – 2 = 6
PICTURE:

Hence, picture is drawn.

Question 5.
Modeling Real Life
6 students play soccer. 4 of them leave. How many students are left?

Which equation matches the story?
6 – 2 = 4
8 – 6 = 2
6 – 4 = 2
Show how you know:

____ students are left.
Answer:
GIVEN:
Total number of students playing soccer = 6
Number of students left away the soccer = 4
EXPLANATION:
SUBTRACTION:
Number of students left in the game = Total number of students playing soccer – Number of students left away the soccer
=6 – 4
=2
Hence, Equation matches our story is 6 – 4 = 2 not other 6 – 2 = 4, 8 – 6 = 2.

Review & Refresh

Question 6.
Write the numbers of dogs and bones. Circle the number that is greater than the other number.
Big Ideas Math Answers 1st Grade 1 Chapter 1 Addition and Subtraction Situations 65
Answer:

GIVEN:
Number of dogs = 7
Number of dog bones = 10
EXPLANATION:
COMPARISION:
Number of dogs : Number of dog bones
=>
Hence, comparison is done and greater number is also marked circle.

Lesson 1.6 Solve Compare Problems: More

Explore and Grow

Use counters to model the story.

Newton has 4 apples. Descartes has 6 apples. Who has more apples? How many more?
Big Ideas Math Answers Grade 1 Chapter 1 Addition and Subtraction Situations 66
Answer:
Number of apples Descartes has more than Number of apples Newton has.
Therefore, Descartes has 2 apples more than Newton has.
GIVEN:
Number of apples Newton has = 4
Number of apples Descartes has = 6
SUBTRACTION:
Number of apples who has more = Number of apples Descartes has  – Number of apples Newton has
= 6 – 4
=2
Hence, Number of apples Descartes has more than Number of apples Newton has.
Therefore, Descartes has 2 apples more than Newton has.

Show and Grow

Question 1.
You have 7 books. Your friend has 3 books. How many more books do you have?
Big Ideas Math Answers Grade 1 Chapter 1 Addition and Subtraction Situations 67
Subtraction equation: _________ – _________ = __________ more books
Answer:
I am having 4 books more than my friend has.
GIVEN:
Number of books I have = 7
Number of books my friend has = 3
EXPLANATION:
SUBTRACTION:
Number of books who has more = Number of books I have – Number of books my friend has
= 7 – 3
= 4
Hence,  I am having 4 books more than my friend has.

Apply and Grow: Practice

Question 2.
Your friend has 6 crayons. You have 3 crayons. How many more crayons does your friend have?
Big Ideas Math Answers Grade 1 Chapter 1 Addition and Subtraction Situations 68
_________ – _________ = __________ more crayons
Answer:
My friend has 3 crayons more than me.
GIVEN:
Number of crayons my friend has = 6
Number of crayons I am having = 3
EXPLANATION:
SUBTRACTION:
Number of crayons who has more by = Number of crayons my friend has – Number of crayons I am having
= 6 – 3
= 3
Hence, my friend has 3 crayons more than me.

Question 3.
There are 9 purple flowers and 5 yellow flowers. Draw the missing flowers. How many more purple flowers are there?
Big Ideas Math Answers Grade 1 Chapter 1 Addition and Subtraction Situations 69
_________ – _________ = __________ more purple flowers
Answer:
Purples  flowers are 4 more than the yellow flowers.
GIVEN:
Number of purple flowers = 9
Number of yellow flowers = 5
EXPLANATION:
SUBTRACTION:
Difference in the flowers = Number of purple flowers – Number of yellow flowers
= 9 – 5
=4.
PICTURE:
MISSING FLOWERS are 5 yellow flowers. Picture is below:

Hence, purples  flowers are 4 more than the yellow flowers.

Question 4.
You have 2 yellow buckets. Your friend has 6 blue buckets. Draw the missing buckets. How many more buckets does your friend have?
Big Ideas Math Answers Grade 1 Chapter 1 Addition and Subtraction Situations 70
_________ – _________ = __________ more buckets
Answer:
Number of more buckets  my friend having than me are 4 blue buckets.
GIVEN:
Number of yellow buckets I am having = 2
Number of blue buckets my friend having = 6
EXPLANATION:
SUBTRACTION:
Number of more buckets does my friend having = Number of yellow buckets I am having – Number of blue buckets my friend having
= 6 – 2
= 4
PICTURE:
MISSING BUCKETS are 6 blue buckets. Picture is below:

Hence, Number of more buckets  my friend having than me are 4 blue buckets.

Think and Grow: Modeling Real Life

You have 2 dog puppets. You have more cat puppets than dog puppets. How many more cat puppets do you have?
Big Ideas Math Answers Grade 1 Chapter 1 Addition and Subtraction Situations 71
Draw a picture:

 

Subtraction equation:

___________ more cat puppets
Answer:
Cats are 1 more than dog puppets
GIVEN:
Number of dog puppets  = 2
You have more cat puppets than dog puppets.
=> Number of cat puppets  = 3

PICTURE:

 

EXPLANATION:
SUBTRACTION:
Number of cats more than dog puppets = Number of cat puppets  – Number of dog puppets
= 3 – 2
=1
Hence, cats are 1 more than dog puppets.

Show and Grow

Question 5.
You have 9 toy cars. You have more cars than trucks. How many more cars do you have?
Big Ideas Math Answers Grade 1 Chapter 1 Addition and Subtraction Situations 72
Draw a picture:

 

Subtraction equation:

 

__________ more toy cars
Answer:
Number of toy cars more than trucks = 1.
GIVEN:
Number of toy cars = 9
You have more cars than trucks.
=> Number of trucks = 8

PICTURE: MISSING TOYS are 8 trucks.

EXPLANATION:
SUBTRACTION:
Number of toy cars more than trucks =Number of toy cars –  Number of trucks
= 9 – 8
= 1
Hence, Number of toy cars more than trucks = 1.

Solve Compare Problems: More Practice 1.6

Question 1.
There are 4 blue scarves and 3 green scarves. How many more blue scarves are there?
Big Ideas Math Answers Grade 1 Chapter 1 Addition and Subtraction Situations 73
____________ – ____________ = ___________ more blue scarf
Answer:
There are blue scarfs 1 more than red scarfs.
GIVEN:
Number of blue scarves = 4
Number of green scarves = 3
EXPLANATION:
SUBTRACTION:
Difference in the scarves = Number of blue scarves – Number of green scarves
= 4 – 3
= 1
Hence, there are blue scarfs 1 more than red scarfs.

Question 2.
You have 6 strawberries and 2 blueberries. Draw the missing fruit. How many more strawberries do you have?
Big Ideas Math Answers Grade 1 Chapter 1 Addition and Subtraction Situations 74
____________ – ____________ = ___________ more straw berries
Answer:
There are 4 strawberries more than blue berries.
GIVEN:
Number of strawberries = 6
Number of  blueberries = 2
PICTURE:
MISSING FRUITS are 2 blueberries.

EXPLANATION:
SUBTRACTION:
Difference in the berries = Number of strawberries – Number of  blueberries
= 6 – 2
= 4
Hence, there are 4 strawberries more than blue berries.

Question 3.
You have 2 blue towels. Your friend has 7 red towels. Draw the missing towels. How many more towels does your friend have?
Big Ideas Math Answers Grade 1 Chapter 1 Addition and Subtraction Situations 75
____________ – ____________ = ___________ more towels
Answer:
My friend has 5 red towels more than me.
GIVEN:
Number of blue towels I have = 2
Number of red towels my friend has = 7
PICTURE: MISSING TOWELS are 7 red towels.

EXPLANATION:
SUBTRACTION:
Difference in the towels between me and my friend = Number of red towels my friend has – Number of blue towels I have
= 7 – 2
= 5
Hence, my friend has 5 red towels more than me.

Question 4.
Modeling Real Life
You have 3 toy bears. You have more yo-yos than toy bears. How many more yo-yos do you have?
Big Ideas Math Answers Grade 1 Chapter 1 Addition and Subtraction Situations 76
Answer:
There are 2 yo-yos more than toy bears.
GIVEN:
Number of toy bears = 3
You have more yo-yos than toy bears.
Number of yo-yos = 5
PICTURE: MISSING  TOYS are 5 yo-yos. Picture is below:

EXPLANATION:
Difference in the toys = Number of yo-yos – Number of toy bears
= 5 – 3
= 2
Hence, there are 2 yo-yos more than toy bears.

Review & Refresh

Question 5.
Write the number of each type of sea creature. Draw a line through the number that is less than the other number.
Big Ideas Math Answers Grade 1 Chapter 1 Addition and Subtraction Situations 77
Answer:
There are 2 jelly fishes more than star fishes.
GIVEN:
Number of star fish = 3

Read more

Big Ideas Math Answers Grade 2 Chapter 2 Fluency and Strategies within 20

Big Ideas Math Answers Grade 2 Chapter 2

Get Big Ideas Math Book 2rd Grade Answer Key Chapter 2 Fluency and Strategies within 20 on this page. This BIM Grade 2 Chapter 2 Fluency and Strategies within 20 Answer Key is useful for the students while preparing for the exams and preparation tests. You can download Big Ideas Math Answers Grade 2 Chapter 2 Fluency and Strategies within 20 PDF for free of cost.

Big Ideas Math Book 2nd Grade Answer Key Chapter 2 Fluency and Strategies within 20

Understand all the concepts of BIM 2nd Grade Solutions for 2nd Chapter Fluency and Strategies within 20. Most of the students may think that learning fluency and strategies within 20 is not an easy topic. So, we are giving a detailed step by step explanation for all the topics in a simple manner. So go through the following sections and start your preparation with love and joy.

The different topics in Big Ideas Math Grade 2 Chapter 2 Fluency and Strategies within 20 Solution Key are Add in Any Order, Use Doubles, Add Three Numbers, Count On and Count Back to Subtract, Relate Addition, Subtraction, and others.  Just tap the below links and start your preparation. By using these links you can score the highest marks in the exams.

Vocabulary

Lesson: 1 Add in Any Order

Lesson: 2 Use Doubles

Lesson: 3 Add Three Numbers

Lesson: 4 Make a 10 to Add

Lesson: 5 Count On and Count Back to Subtract

Lesson: 6 Relate Addition and Subtraction

Lesson: 7 Get to 10 to Subtract

Lesson: 8 Practice Addition and Subtraction

Lesson: 9 Problem Solving: Addition and Subtraction

Chapter: 2 – Fluency and Strategies within 20

Fluency and Strategies within 20 Vocabulary

Big Ideas Math Answers Grade 2 Chapter 2 Fluency and Strategies within 20 1

Organize It
Use the review words to complete the graphic organizer.
Big Ideas Math Solutions Grade 2 Chapter 2 Fluency and Strategies within 20 2

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-2- Fluency-and-Strategies-within-20-Fluency-and-Strategies-within-20-Vocabulary-Organize It

Define It
Use your vocabulary cards to match.

Question 1.
Big Ideas Math Solutions Grade 2 Chapter 2 Fluency and Strategies within 20 3

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-2- Fluency-and-Strategies-within-20-Fluency-and-Strategies-within-20-Vocabulary-Define-It-Question-1

Big Ideas Math Solutions Grade 2 Chapter 2 Fluency and Strategies within 20 4
Big Ideas Math Solutions Grade 2 Chapter 2 Fluency and Strategies within 20 5

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-2- Fluency-and-Strategies-within-20-Fluency-and-Strategies-within-20-Vocabulary-Define-It-Question-2

Lesson 2.1 Add in Any Order

Explore and Grow

Use the table to find the sum. Change the order of the addends. Write the new equation.
Big Ideas Math Solutions Grade 2 Chapter 2 Fluency and Strategies within 20 6

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-2- Fluency-and-Strategies-within-20-Lesson-2.1-Add-in-Any-Order-Explore-and-GrowBig-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-2- Fluency-and-Strategies-within-20-Lesson-2.1-Add-in-Any-Order-Explore-and-Grow

Explanation:
Changing the order of the addends does not change the sum.
The expressions 5 + 6 and 6 + 5 are both equal to 11. The expressions is represented by red color and sum by green.
The expressions 4 + 9 and 9 + 4 are both equal to 13. The expressions are represented by blue color and sum by yellow.

Show and Grow

Question 1.
Big Ideas Math Solutions Grade 2 Chapter 2 Fluency and Strategies within 20 7
Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-2- Fluency-and-Strategies-within-20-Lesson-2.1-Add-in-Any-Order- Show-and-Grow-question-1

Explanation:
Changing the order of the addends does not change the sum.
The expressions 6 + 3 and 3 + 6 are both equal to 9. The addend 6 is represented by orange color and addend 3 by blue.

Question 2.
Big Ideas Math Solutions Grade 2 Chapter 2 Fluency and Strategies within 20 8

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-2- Fluency-and-Strategies-within-20-Lesson-2.1-Add-in-Any-Order- Show-and-Grow-question-2

Explanation:
Changing the order of the addends does not change the sum.
The expressions 5 + 1 and 1 + 5 are both equal to 6. The addend 5 is represented by yellow color and addend 1 by green.

Find the sum. Then change the order of the addends. Write the new equation.

Question 3.
Big Ideas Math Solutions Grade 2 Chapter 2 Fluency and Strategies within 20 9
Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-2- Fluency-and-Strategies-within-20-Lesson-2.1-Add-in-Any-Order- Show-and-Grow-question-3
Explanation:
Changing the order of the addends does not change the sum.
The expressions 8 +2 and 2 + 8
 are both equal to the sum 10. 

Question 4.
Big Ideas Math Solutions Grade 2 Chapter 2 Fluency and Strategies within 20 10
Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-2- Fluency-and-Strategies-within-20-Lesson-2.1-Add-in-Any-Order- Show-and-Grow-question-4
Explanation:
Changing the order of the addends does not change the sum.
The expressions 4 + 3 and 3 + 4
 are both equal to the sum 7. 

Apply and Grow: Practice

Find the sum. Then change the order of the addends. Write the new equation.

Question 5.
Big Ideas Math Solutions Grade 2 Chapter 2 Fluency and Strategies within 20 11
Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-2- Fluency-and-Strategies-within-20-Lesson-2.1-Add-in-Any-Order-Apply-and-Grow-Practice-Question-5
Explanation:
Changing the order of the addends does not change the sum.
The expressions 4 + 3 and 3 + 4
 are both equal to the sum 7. 

Question 6.
Big Ideas Math Solutions Grade 2 Chapter 2 Fluency and Strategies within 20 12
Answer:
7 + 3 = 10                   3 + 7 = 10
Explanation:
Changing the order of the addends does not change the sum.
The expressions 7 + 3 and 3 +  7
 are both equal to the sum 10. 

Question 7.
Big Ideas Math Solutions Grade 2 Chapter 2 Fluency and Strategies within 20 13
Answer:
6 + 5 = 11                     5 + 6 = 11
Explanation:
Changing the order of the addends does not change the sum.
The expressions 6 + 5 and 5 + 6
 are both equal to the sum 11. 

Question 8.
Big Ideas Math Solutions Grade 2 Chapter 2 Fluency and Strategies within 20 14
Answer:
17 = 9 + 8                         17 = 8 + 9
Explanation:
Changing the order of the addends does not change the sum.
The expressions 9 + 8 and 8 +9
 are both equal to the sum 17. 

Question 9.
Big Ideas Math Solutions Grade 2 Chapter 2 Fluency and Strategies within 20 15
Answer:
2 = 0 + 2                          2 = 2 +0
Explanation:
Changing the order of the addends does not change the sum.
The expressions 0 + 2 and 2 + 0
 are both equal to the sum 2. Adding 0 to any number gives the sum the number itself. 

Question 10.
DIG DEEPER!
Which shape completes the equation?
Big Ideas Math Solutions Grade 2 Chapter 2 Fluency and Strategies within 20 16
Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-2- Fluency-and-Strategies-within-20-Lesson-2.1-Add-in-Any-Order-Apply-and-Grow-Practice-Question-10

Explanation:
Changing the order of the addends does not change the sum.

Think and Grow: Modeling Real Life

You paint 12 shapes. 8 are rectangles. The rest are circles. How many circles do you paint?
Write two equations that describe your shapes:
Big Ideas Math Solutions Grade 2 Chapter 2 Fluency and Strategies within 20 17
Big Ideas Math Solutions Grade 2 Chapter 2 Fluency and Strategies within 20 18

Answer:
Given:
Total number of shapes painted = 12
Number of rectangles painted = 8
Number of circles painted = total shapes painted – number of rectangles painted
12 – 8 = 4
8 + 4 = 12
4 + 8 = 12
Total circles painted = 4.

Show and Grow

Question 11.
There are 13 race cars. 6 of them have numbers. The rest do not. How many race cars do not have numbers?
Write two equations that describe your shapes:
Big Ideas Math Solutions Grade 2 Chapter 2 Fluency and Strategies within 20 19
Big Ideas Math Solutions Grade 2 Chapter 2 Fluency and Strategies within 20 19.1
Explain how you know your answer is correct.
_______________________________
_______________________________
_______________________________
Answer:
Total number of race cars = 13
Number of race cars with numbers = 6.
Number of race cars without number = total number of race cars – Number of race cars with numbers
13 – 6 = 7.
number of race cars with number + number of race cars without number = total number of race cars
6 + 7 = 13
when we reverse the order
7 + 6 = 13
Explanation:
Changing the order of the addends does not change the sum.
Given total race cars = 13
Race cars with number = 6
To find the race cars without number we need to minus race cars with number from the total
13 – 6 = 7

Add in Any Order Homework & Practice 2.1

Question 1.
Big Ideas Math Solutions Grade 2 Chapter 2 Fluency and Strategies within 20 20

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-2- Fluency-and-Strategies-within-20-Add-in-Any-Order-Homework-Practice-2.1-Question-1
Explanation:
Changing the order of the addends does not change the sum.
Given 6 blue linking cubes and 4 red linking cubes
Adding blue and red gives the sum.
6 + 4 = 10.
when we reverse the colors order
adding  red  and  blue gives the sum .
4 + 6 = 10.

Question 2.
Big Ideas Math Solutions Grade 2 Chapter 2 Fluency and Strategies within 20 21
Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-2- Fluency-and-Strategies-within-20-Add-in-Any-Order-Homework-Practice-2.1-Question-2
Explanation:
Changing the order of the addends does not change the sum.
Given 1 yellow linking cubes and 8 orange linking cubes
Adding yellow and orange cube gives the sum
1 + 8 = 9.
when we reverse the colors order
adding orange and yellow cubes gives the sum
8 + 1 = 9.

Find the sum. Then change the order of the addends.
Write the new equation.

Question 3.
Big Ideas Math Solutions Grade 2 Chapter 2 Fluency and Strategies within 20 22
Answer:
7 + 0 = 7                      0 + 7 = 7
Explanation:
Changing the order of the addends does not change the sum.
Adding any number to 0 gives the sum the number itself.

Question 4.
Big Ideas Math Solutions Grade 2 Chapter 2 Fluency and Strategies within 20 23
Answer:
2 + 4 = 6               4 + 2 = 6
Explanation:
Changing the order of the addends does not change the sum.
The expressions 2 + 4 and  4 + 2 gives the same sum 6.

Question 5.
Big Ideas Math Solutions Grade 2 Chapter 2 Fluency and Strategies within 20 24
Answer:
15 = 8 + 7                       7 + 8 = 15
Explanation:
Changing the order of the addends does not change the sum.
The expressions 8 + 7 and 7 +8  gives the same sum 15.

Question 6.
Big Ideas Math Solutions Grade 2 Chapter 2 Fluency and Strategies within 20 25
Answer:
11 = 3 + 8        8 +3 = 11
Explanation:
Changing the order of the addends does not change the sum.
The expressions 3 + 8 and 8 + 3 gives the same sum 11.

Question 7.
Number Sense
Complete each equation.
Big Ideas Math Solutions Grade 2 Chapter 2 Fluency and Strategies within 20 26
Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-2- Fluency-and-Strategies-within-20-Add-in-Any-Order-Homework-Practice-2.1-Question-7
Explanation:
Changing the order of the addends in the expressions.

Question 8.
Modeling Real Life
You have 14 blocks. 8 are cubes. The rest are cylinders. How many cylinders do you have?
Write two equations that describe your blocks:
Big Ideas Math Solutions Grade 2 Chapter 2 Fluency and Strategies within 20 27

Answer:
Total number of Blocks = 14
Number of cubes = 8.
Number of Cylinders = total number Blocks – Number of Cubes
14 – 8 = 6.
Number of cylinders = 6
number of Cubes + number of Cylinders = total number Blocks
8 + 6 = 14
when we reverse the order the expression
6 + 8 = 14
Explanation:
Given total Blocks = 14
Number of cubes = 8
From the above calculation
Number of Cylinder = 6
Blocks = number of cubes + number of cylinders
8 + 6 = 14
Changing the order of the addends does not change the sum.
6 + 8 = 14.

Question 9.
Modeling Real Life
There are 16 kids in a class. 7 have gone camping. The rest have not. How many kids have not gone camping?
Write two equations that describe the kids:
Big Ideas Math Solutions Grade 2 Chapter 2 Fluency and Strategies within 20 28

Answer:
Total number Kids in the class= 16
Number of kids gone camping = 7.
Number of kids who have not gone camping = Total number of kids in the class – Number of kids gone camping
16 – 7 = 9.
Number of kids who have not gone camping = 9
number of kids who have gone camping + number of kids who have not gone camping = total number kids in the class
7 + 9 = 16.
when we reverse the order the expression
9 + 7 = 16.
Explanation:
Given total number kids in the class = 16
Number of kids who have gone camping = 7
From the above calculation
Number of kids who have not gone camping = 9
Total number kids in the class = number of kids who have gone camping + number of kids who have not gone camping
7 + 9 = 16
Changing the order of the addends does not change the sum.
9 + 7 = 16.

Review & Refresh

Question 10.
Which time does not belong with the other three?
Big Ideas Math Solutions Grade 2 Chapter 2 Fluency and Strategies within 20 29

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-2- Fluency-and-Strategies-within-20-Add-in-Any-Order-Homework-Practice-2.1-Review-&-Refresh-Question-10
Explanation:
Half past 5 means 5:30. In the fig first three images indicate 5:30 where as half past 6 means 6:30.

Lesson 2.2 Use Doubles

Explore and Grow

Find each sum. How are the equations alike? How are they different?
Big Ideas Math Solutions Grade 2 Chapter 2 Fluency and Strategies within 20 30

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-2- Fluency-and-Strategies-within-20-Lesson-2.2-Use-Doubles-Explore-and-Grow

Explanation:
First two expressions are double. 6 + 6, 7 + 7
In the next  two expression
one is doubles plus 1 expression 6 + 7.
second is doubles minus 1 expression 7 + 6.

Show and Grow

Find the sum. Write the double you used.

Question 1.
Big Ideas Math Solutions Grade 2 Chapter 2 Fluency and Strategies within 20 31
Answer:
4 + 5 = 9
4 + 4 = 8
Explanation:
Doubles plus 1
4 + 5 is equal to 4 + 4 and 1 more.

Question 2.
Big Ideas Math Solutions Grade 2 Chapter 2 Fluency and Strategies within 20 32
Answer:
7 + 6 = 13
7 + 7 = 14
Explanation:
Doubles minus 1
7 + 6 is equal to 1 less than 7 + 7.

Question 3.
Big Ideas Math Solutions Grade 2 Chapter 2 Fluency and Strategies within 20 33

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-2- Fluency-and-Strategies-within-20-Lesson-2.2-Use-Doubles-Show-and-Grow-Question-3

Explanation:
Doubles minus 1
6 + 5 is equal to 1 less than 6 + 6.

Question 4.
Big Ideas Math Solutions Grade 2 Chapter 2 Fluency and Strategies within 20 34

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-2- Fluency-and-Strategies-within-20-Lesson-2.2-Use-Doubles-Show-and-Grow-Question-4
Explanation:
Doubles plus 1
8 + 9  is equal to 8 + 8 and 1 more.

Apply and Grow: Practice

Question 5.
Big Ideas Math Solutions Grade 2 Chapter 2 Fluency and Strategies within 20 35
Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-2- Fluency-and-Strategies-within-20-Lesson-2.2-Use-Doubles-Apply-and-Grow-Practice-Question-5
Explanation:
Doubles plus 1
3 + 4 is equal to 3 + 3 and 1 more

Question 6.
Big Ideas Math Solutions Grade 2 Chapter 2 Fluency and Strategies within 20 36
Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-2- Fluency-and-Strategies-within-20-Lesson-2.2-Use-Doubles-Apply-and-Grow-Practice-Question-6
Explanation:
Doubles minus 1
8 + 7 is equal to 1 less than 8 + 8

Question 7.
Big Ideas Math Solutions Grade 2 Chapter 2 Fluency and Strategies within 20 37

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-2- Fluency-and-Strategies-within-20-Lesson-2.2-Use-Doubles-Apply-and-Grow-Practice-Question-7
Explanation:
Doubles plus 1
5 + 6 is equal to 1 less than 5 + 5.

Question 8.
Big Ideas Math Solutions Grade 2 Chapter 2 Fluency and Strategies within 20 38
Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-2- Fluency-and-Strategies-within-20-Lesson-2.2-Use-Doubles-Apply-and-Grow-Practice-Question-8
Explanation:
Doubles minus 1
10 +9 is equal to 1 less than 10 + 10.

Question 9.
Big Ideas Math Solutions Grade 2 Chapter 2 Fluency and Strategies within 20 39
Answer:
2 + 3 = 5
2 + 2 = 4
Explanation:
Doubles plus 1
2 + 3 is equal to 2+2 and 1 more.

Question 10.
Big Ideas Math Solutions Grade 2 Chapter 2 Fluency and Strategies within 20 40

Answer:
5 + 4 = 9
5 + 5 = 10
Explanation:
5 + 4 is equal 1 less than 5 + 5.
Doubles minus 1.

Question 11.
Big Ideas Math Solutions Grade 2 Chapter 2 Fluency and Strategies within 20 41

Answer:
6 + 7 = 13
6 + 6 = 12
Explanation:
6 + 7  is equal to 6 + 6 and 1 more.
Doubles plus 1.

Question 12.
Big Ideas Math Solutions Grade 2 Chapter 2 Fluency and Strategies within 20 42
Answer:
9 + 8 = 17
9 + 9 = 18
Explanation:
9 + 8 is equal 1 less than 9 + 9.
Doubles minus 1.

Question 13.
DIG DEEPER!
What double might Newton be thinking?
Big Ideas Math Solutions Grade 2 Chapter 2 Fluency and Strategies within 20 43
Answer:
Given
Sum greater than 4
Addends less than 5
As per the above information the doubles can be 3 + 3, 4 + 4.
Think and Grow: Modeling Real Life

You make 9 mud pies. Your friend makes 1 fewer than you. How many mud pies do you and your friend make in all?
Big Ideas Math Solutions Grade 2 Chapter 2 Fluency and Strategies within 20 44
Addition equation:
What double can you use?
Big Ideas Math Solutions Grade 2 Chapter 2 Fluency and Strategies within 20 45
Answer:
Number of mud pies i make = 9
My friend makes 1 fewer mud pies then me = 9- 1 = 8
Total number of Mud pies we both make = 9 + 8 = 17.
Double
9 + 9 = 18.
9 + 8 is 1 less than 9 + 9
Doubles minus 1.

Show and Grow

Question 14.
You score 6 goals. Your friend scores 1 more than you. How many goals do you and your friend score in all?
Big Ideas Math Solutions Grade 2 Chapter 2 Fluency and Strategies within 20 46
Answer:
Number of goals i scored = 6
Number of goals my friend scored = 6 + 1 = 7
Total number of goals we both scored = 6 + 7 = 13.

Question 15.
DIG DEEPER!
You and your friend jump in 13 puddles in all. You jump in 7. How many puddles does your friend jump in?
Big Ideas Math Solutions Grade 2 Chapter 2 Fluency and Strategies within 20 47
Answer:
Total number of puddles me and my friend jumped = 13
Total number of puddles i jumped = 7
Number of puddles my friend jumped = 13 – 7 = 6.
7 + 6 = 13.

Use Doubles Homework & Practice 2.2

Find the sum. Write the double you used.

Question 1.
Big Ideas Math Solutions Grade 2 Chapter 2 Fluency and Strategies within 20 48
Answer:
4 + 3 = 7
4 + 4 = 8
Explanation:
Doubles minus 1
4 + 3 is equal to 1 less than 4 + 4.
Question 2.
Big Ideas Math Answers Grade 2 Chapter 2 Fluency and Strategies within 20 49
Answer:
5 + 6 = 11
Doubles 5 + 5 = 10
Explanation:
Doubles plus 1
5 + 6 is equal to 5 + 5 and 1 more

Question 3.
Big Ideas Math Answers Grade 2 Chapter 2 Fluency and Strategies within 20 50
Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-2- Fluency-and-Strategies-within-20-Use-Doubles- Homework-&-Practice-2.2-Question-3
Explanation:
Doubles minus 1
9 + 8 is equal to 1 less than 9 + 9

Question 4.
Big Ideas Math Answers Grade 2 Chapter 2 Fluency and Strategies within 20 51
Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-2- Fluency-and-Strategies-within-20-Use-Doubles- Homework-&-Practice-2.2-Question-4
Explanation:
6 + 7 is equal to 6 + 6 and 1 more.
Doubles plus 1.

Question 5.
Big Ideas Math Solutions Grade 2 Chapter 2 Fluency and Strategies within 20 52
Answer:
8 + 7 = 15
8 + 8 = 16
Explanation:
8 + 7 is equal to 1 less than 8 + 8.
Doubles minus 1.

Question 6.
Big Ideas Math Answers Grade 2 Chapter 2 Fluency and Strategies within 20 53

Answer:
9 + 10 = 19
9 + 9 = 18
Explanation:
Doubles plus 1
9 + 10 is equal to 9 + 9 and 1 more.

Question 7.
Number Sense
Which doubles can you use to find 8 + 9?
Big Ideas Math Answers Grade 2 Chapter 2 Fluency and Strategies within 20 54
Answer:
To find the expression 8 + 9
8 + 9 is equal to 8 + 8 and 1 more
8 +9 is equal to 1 less than 9 + 9.

Question 8.
Modeling Real Life
You tell 5 jokes. Your friend tells 1 fewer than you. How many jokes do you and your friend tell in all?
Big Ideas Math Answers Grade 2 Chapter 2 Fluency and Strategies within 20 55
Answer:
Number of jokes i told = 5
Number of jokes my friend told = 5 – 1 = 4
Total number of jokes me and my friend told in together = 5 + 4 = 9
9 jokes together.

Question 9.
Modeling Real Life
Newton paints 3 paw prints. Descartes paints 1 more than Newton. How many paw prints do they paint in all?
Big Ideas Math Answers Grade 2 Chapter 2 Fluency and Strategies within 20 56
Answer:
Number of paw prints by Newton paints = 3
Number of paw prints by Descartes paints is 1 more than newton = 3 + 1 = 4
Total number of paw prints both Newton and Descartes paint =3 + 4 = 7
Total 7 paw prints.

Review & Refresh

Is the equation true or false?

Question 10.
Big Ideas Math Answers Grade 2 Chapter 2 Fluency and Strategies within 20 57
Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Use-Doubles-Homework-&-Practice-2.2-Review-&-Refresh-Question-10

Question 11.
Big Ideas Math Answers Grade 2 Chapter 2 Fluency and Strategies within 20 58
Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Use-Doubles-Homework-&-Practice-2.2-Review-&-Refresh-Question-11

Lesson 2.3 Add Three Numbers

Explore and Grow

Find the sums.
Big Ideas Math Answers Grade 2 Chapter 2 Fluency and Strategies within 20 59
What is the same? What is different?
_________________________
_________________________
_________________________
Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Lesson-2.3-Add-Three-Numbers-Explore-and-Grow
The sum remains same.
The difference is the order of the addends added.
Explanation:
Changing the order of addends does not change the sum.
In all the three expressions the sum is same :14

Show and Grow

Circle two addends to add first. Write their sum. Then find the sum of all three addends.

Question 1.
7 + 3 + 4 = ___
Big Ideas Math Answers Grade 2 Chapter 2 Fluency and Strategies within 20 60
Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Lesson-2.3-Add-Three-Numbers-Show-and-Grow-Question-1
Explanation:
Addends are represented in circle and Sum is represented in square.
First adding two addends 7, 4 give the sum 11
and then adding the another addend 3 to the sum 11.
Gives the sum 14.

Question 2.
1 + 5 + 3 = ___
Big Ideas Math Answers Grade 2 Chapter 2 Fluency and Strategies within 20 61
Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Lesson-2.3-Add-Three-Numbers-Show-and-Grow-Question-2
Explanation:
Addends are represented in circle and Sum is represented in square.
First adding two addends 1 , 5 give the sum 6.
and then adding the another addend 3 to the sum 6.
Gives the sum 9.

Question 3.
9 + 4 + 4 = ___
Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Lesson-2.3-Add-Three-Numbers-Show-and-Grow-Question-3
Explanation:
Addends are represented in circle and Sum is represented in square.
First adding two addends 4, 4 give the sum 8.
and then adding the another addend 9 to the sum 8.
Gives the sum 17.

Question 4.
5 + 2 + 5 = ___
Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Lesson-2.3-Add-Three-Numbers-Show-and-Grow-Question-4
Explanation:
Addends are represented in circle and Sum is represented in square.
First adding two addends 5, 2 give the sum 7.
and then adding the another addend 5 to the sum 7.
Gives the sum 12.

Apply and Grow: Practice

Question 5.
3 + 5 + 5 = ___
Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Lesson-2.3-Add-Three-Numbers-Apply-and-Grow-Practice-Question-5
Explanation:
Addends are represented in circle and Sum is represented in square.
First adding two addends 5, 2 give the sum 7.
and then adding the another addend 5 to the sum 7.
Gives the sum 12.

Question 6.
9 + 0 + 7 = ___
Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Lesson-2.3-Add-Three-Numbers-Apply-and-Grow-Practice-Question-6
Explanation:
Addends are represented in circle and Sum is represented in square.
First adding two addends 9, 0 give the sum 9.
and then adding the another addend 7 to the sum 9.
Gives the sum 16.

Question 7.
4 + 2 + 3 = ___
Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Lesson-2.3-Add-Three-Numbers-Apply-and-Grow-Practice-Question-7
Explanation:
Addends are represented in circle and Sum is represented in square.
First adding two addends 2, 3 give the sum 5.
and then adding the another addend 4 to the sum 5.
Gives the sum 9.

Question 8.
Big Ideas Math Answers Grade 2 Chapter 2 Fluency and Strategies within 20 62
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Lesson-2.3-Add-Three-Numbers-Question-8

Question 9.
Big Ideas Math Answers Grade 2 Chapter 2 Fluency and Strategies within 20 63
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Lesson-2.3-Add-Three-Numbers-Question-9

Question 10.
Big Ideas Math Answers Grade 2 Chapter 2 Fluency and Strategies within 20 64
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Lesson-2.3-Add-Three-Numbers-Question-10

Question 11.
Big Ideas Math Answers Grade 2 Chapter 2 Fluency and Strategies within 20 65
Answer :
9 + 5 + 6 = 20

Question 12.
Big Ideas Math Answers Grade 2 Chapter 2 Fluency and Strategies within 20 66
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Lesson-2.3-Add-Three-Numbers-Question-12

Question 13.
Big Ideas Math Answers Grade 2 Chapter 2 Fluency and Strategies within 20 67
Answer :
3 + 7 + 7 = 17
Explanation:
Addends are represented in circle and Sum is represented in square.
First adding two addends 5, 2 give the sum 7.
and then adding the another addend 5 to the sum 7.
Gives the sum 12.

Question 14.
Structure
Write the missing addends.
Big Ideas Math Answers Grade 2 Chapter 2 Fluency and Strategies within 20 68
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Lesson-2.3-Add-Three-Numbers-Question-14

Think and Grow: Modeling Real Life

You have 6 presents. Newton and Descartes each have 4 presents. How many presents are there in all?
Addition equation:
Big Ideas Math Answers Grade 2 Chapter 2 Fluency and Strategies within 20 69
Answer :
Number of presents with me = 6
Number of presents with Newton = 4
Number of Presents with Descartes = 4
Total Number of presents in all = 6 + 4 + 4 = 14

Show and Grow

Question 15.
You have 8 crayons. Newton and Descartes each have 5 crayons. How many crayons are there in all?
Big Ideas Math Answers Grade 2 Chapter 2 Fluency and Strategies within 20 70
Answer :
Number of Crayons with me = 8
Number of Crayons with Newton = 5
Number of Crayons with Descartes = 5
Total Number of Crayons in all = 8 + 5 + 5 = 18

Question 16.
You, Newton, and Descartes eat 18 apple slices in all. Newton and Descartes each eat 6 slices. How many apple slices do you eat?
Big Ideas Math Answers Grade 2 Chapter 2 Fluency and Strategies within 20 71
Answer :
Number of apples slices eaten by Me , Newton and Descartes = 18
Number of apples eaten by Newton =6
Number of apples eaten by Descartes = 6
Total Number of apples eaten by Newton and Descartes = 6 + 6 = 12
Number of apple slices eaten by me = 18 – 12 = 6

Question 17.
A treasure chest has 20 jewels in all. 7 are blue. 6 are green. The rest are red. How many jewels are red?
Big Ideas Math Answers Grade 2 Chapter 2 Fluency and Strategies within 20 72
Answer :
Number of jewels in Treasure chest = 20
Number of blue jewels = 7
Number of green jewels = 6
Number of blue and green jewels = 7 + 6 = 13
Number of red jewels = 20 – 13 = 7

Add Three Numbers Homework & Practice 2.3

Circle two addends to add first. Write their sum. Then find the sum of all three addends.

Question 1.
Big Ideas Math Answers Grade 2 Chapter 2 Fluency and Strategies within 20 73
Answer :
2 + 5 + 1 = 8
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Add-Three-Numbers-Homework-Practice-2.3-Question-1
Explanation:
Addends are represented in circle and Sum is represented in square.
First adding two addends 5, 1 give the sum 6.
and then adding the another addend 2 to the sum 6.
Gives the sum 8.

Question 2.
Big Ideas Math Answers Grade 2 Chapter 2 Fluency and Strategies within 20 74
Answer
6 + 4 + 4 = 14
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Add-Three-Numbers-Homework-Practice-2.3-Question-2
Explanation:
Addends are represented in circle and Sum is represented in square.
First adding two addends 4, 4 give the sum 8.
and then adding the another addend 6 to the sum 8.
Gives the sum 14.

Question 3.
Big Ideas Math Answers Grade 2 Chapter 2 Fluency and Strategies within 20 75
Answer :
0 + 8 + 7 = 15
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Add-Three-Numbers-Homework-Practice-2.3-Question-3
Explanation:
Addends are represented in circle and Sum is represented in square.
First adding two addends 8, 7 give the sum 15.
and then adding the another addend 0 to the sum 15.
Gives the sum 15.

Question 4.
Big Ideas Math Answers Grade 2 Chapter 2 Fluency and Strategies within 20 76
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Add-Three-Numbers-Homework-Practice-2.3-Question-4
Explanation:
Addends are represented in circle and Sum is represented in square.
First adding two addends 6, 5 give the sum 11.
and then adding the another addend 5 to the sum 11.
Gives the sum 11.

Question 5.
Big Ideas Math Answers Grade 2 Chapter 2 Fluency and Strategies within 20 77
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Add-Three-Numbers-Homework-Practice-2.3-Question-5
Explanation:
Addends are represented in circle and Sum is represented in square.
First adding two addends 4, 9 give the sum 13.
and then adding the another addend 0 to the sum 13.
Gives the sum 13.

Question 6.
Big Ideas Math Answers Grade 2 Chapter 2 Fluency and Strategies within 20 78

Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Add-Three-Numbers-Homework-Practice-2.3-Question-6
Explanation:
Addends are represented in circle and Sum is represented in square.
First adding two addends 3, 3 give the sum 6.
and then adding the another addend 3 to the sum 6.
Gives the sum 9.

Question 7.
2 + 8 + 7 = ___
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Add-Three-Numbers-Homework-Practice-2.3-Question-7
Explanation:
Addends are represented in circle and Sum is represented in square.
First adding two addends 8, 7 give the sum 15.
and then adding the another addend 2 to the sum 15.
Gives the sum 17.

Question 8.
9 + 8 + 1 = ___
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Add-Three-Numbers-Homework-Practice-2.3-Question-8
Explanation:
Addends are represented in circle and Sum is represented in square.
First adding two addends 8, 1 give the sum 9.
and then adding the another addend 9 to the sum 9.
Gives the sum 18.

Question 9.
8 + 7 + 5 = ___
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Add-Three-Numbers-Homework-Practice-2.3-Question-9
Explanation:
Addends are represented in circle and Sum is represented in square.
First adding two addends 7, 5 give the sum 12.
and then adding the another addend 8 to the sum 12.
Gives the sum 20.

Question 10.
DIG DEEPER!
Complete the number puzzle so that each branch has a sum of 20.
Big Ideas Math Answers Grade 2 Chapter 2 Fluency and Strategies within 20 79
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Add-Three-Numbers-Homework-Practice-2.3-Question-10
Explanation :
Sum should be 20
8 + ____ + 7 = 20  so , _____= 20 – 15 = 5
10 + ____ + 10 = 20 so , _______ = 20 – 20 = 0
_____ +  3 + 9 = 20 so , _____ = 20 – 12 = 8
4 + 7 + _____ = 20 so, _____ = 20 – 11 = 9

Question 11.
Modeling Real Life
You, Newton, and Descartes have 17 stampers in all. Newton and Descartes each have 6. How many stampers do you have? Big Ideas Math Answers Grade 2 Chapter 2 Fluency and Strategies within 20 80
Answer :
Number of stampers with me , Newton and Descartes = 17
Number of Stampers with Newton = 6
Number of stampers with Descartes = 6
Total Number of stampers with Newton and Descartes = 6 + 6 = 12
Number of Stampers with me = 17 – 12 = 5 Stampers

Question 12.
Make a quick sketch. Complete the sentences.
Big Ideas Math Answers Grade 2 Chapter 2 Fluency and Strategies within 20 81
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Add-Three-Numbers-Homework-Practice-2.3-Question-12

Lesson 2.4 Make a 10 to Add

Explore and Grow

Use counters and the ten frames to find the sum. How can you make a 10 to solve?
Big Ideas Math Answers Grade 2 Chapter 2 Fluency and Strategies within 20 82
Answer :

Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Lesson-2.4-Make-a-10-to-Add-Explore-Grow
Explanation :
We have 7counters in first table and 6 counters in second table. To make ten counters in first table we should move 3 counters from second table to first table. then,we will have ten counters in first table and 3 in second table. add both 10 + 3 = 13
So, 7 + 6 = 13 .
Show and Grow

Make a 10 to add.

Question 1.
Big Ideas Math Answers Grade 2 Chapter 2 Fluency and Strategies within 20 83
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Lesson-2.4-Make-10-to-Show-Grow-Add-Question-1
Explanation :
To make addend 9 as 10 we should add 1 to the 9 then write addend 7  as  6 + 1 .
then add 9 + 1 we get , 10 and then later add 6 to 10 we get sum , 16 .

Question 2.
Big Ideas Math Answers 2nd Grade Chapter 2 Fluency and Strategies within 20 84
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Lesson-2.4-Make-10-to-Show-Grow-Add-Question-2
Explanation :
To make addend 9 as 10 we should add 1 to the 9 then write addend 5 as  4 + 1 .
then add 9 + 1 we get , 10 and then later add 4 to 10 we get sum , 14 .

Question 3.
6 + 8 = ?
10 + __ = ___
So, 6 + 8 = ___
Answer :
6 + 8 =
4 + 2 + 8 =
10 + 4 = 14
So, 6 + 8 = 14
Explanation :
To make addend 8 as 10 we should add 2 to the 8 then write addend 6  as  4 + 2 .
then add 8 + 2 we get , 10 and then later add 4 to 10 we get sum , 14 .

Question 4.
7 + 4 = ?
10 + __ = ___
So, 7 + 4 = ___
Answer :
7 + 4 =
7 + 3 + 1 =
10 + 1 = 11
So, 7 + 4 = 11
Explanation :
To make addend 7 as 10 we should add 3 to the 7 then write addend 4 as  3 + 1 .
then add 7 + 3 we get , 10 and then later add 1 to 10 we get sum , 11 .

Apply and Grow: Practice

Make a 10 to add.

Question 5.
2 + 9 = ?
10 + __ = ___
So, 2 + 9 = ___
Answer :
2 + 9 =
1 + 1 + 9 =
1 + 10 = 11
So, 2 + 9 = 11
Explanation :
To make addend 9 as 10 we should add 1 to the 9 then write addend 2 as  1 + 1 .
then add 9 + 1 we get , 10 and then later add 1 to 10 we get sum , 11 .

Question 6.
8 + 4 = ?
10 + ___ = __
So, 8 + 4 = ___
Answer :
8 + 4 =
2 + 2 + 8 =
2 + 10 = 12
So, 8 + 4 = 12
Explanation :
To make addend 8 as 10 we should add 2 to the 8 then write addend 4  as  2 + 2 .
then add 2 + 8 we get , 10 and then later add 2 to 10 we get sum , 12 .

Question 7.
9 +6 = ?
10+ ___ = ___
So, 9 + 6 = ___
Answer :
9 + 6 =
9 + 1 + 5 =
10 + 5 = 15
So, 9 + 6 = 15
Explanation :
To make addend 9 as 10 we should add 1 to the 9 then write addend 6  as  5 + 1 .
then add 9 + 1 we get , 10 and then later add 5 to 10 we get sum , 15 .

Question 8.
7 + 7 = ?
10 + __ = ___
So, 7 + 7 = ___
Answer :
7 + 7 =
7 + 3 + 4 =
10 + 4 = 14
So, 7 + 7 = 14
Explanation :
To make addend 7 as 10 we should add 3 to the 7 then write addend 7 as  3 + 4 .
then add 7 + 3 we get , 10 and then later add 4 to 10 we get sum , 14 .

Question 9.
6 + 5 = ___
Answer :
6 + 5 =
6 + 4 + 1 =
10 + 1 = 11
So, 6 + 5 = 11
Explanation :
To make addend 6 as 10 we should add 4 to the 6 then write addend 5  as  4 + 1 .
then add 6 + 4 we get , 10 and then later add 1 to 10 we get sum , 11 .

Question 10.
8 + 9 = ___
Answer :
8 + 9 =
7 + 1 + 9 =
7 + 10 = 17
So, 8 + 9 = 17
Explanation :
To make addend 9 as 10 we should add 1 to the 9 then write addend 8  as  7 + 1 .
then add 9 + 1 we get , 10 and then later add 7 to 10 we get sum , 17 .

Question 11.
7 + 5 = ___
Answer :
7 + 5 =
7 + 3 + 2=
10 + 2 = 12
So, 7 + 5 = 12
Explanation :
To make addend 7 as 10 we should add 3 to the 7 then write addend 5 as  3 + 2 .
then add 7 + 3 we get , 10 and then later add 2 to 10 we get sum , 12 .

Question 12.
6 + 6 = ___
Answer :
6 + 6 =
6 + 4 + 2 =
10 + 2 = 12
So, 6 + 6 = 12
Explanation :
To make addend 6 as 10 we should add 4 to the 6 then write addend 6  as  4 + 2 .
then add 6 + 4 we get , 10 and then later add 2 to 10 we get sum , 12 .

Question 13.
9 + 9 = ___
Answer :
9 + 9 =
9 + 1 + 8 =
10 + 8 = 18
So, 9 + 9 = 18
Explanation :
To make addend 9 as 10 we should add 1 to the 9 then write addend 9 as  1 + 8 .
then add 9 + 1 we get , 10 and then later add 8 to 10 we get sum , 18 .

Question 14.
8 + 7 = __
Answer :
8 + 7 =
5 + 2 + 8 =
10 + 5 = 15
So, 8 + 7 = 15
Explanation :
To make addend 8 as 10 we should add 2 to the 8 then write addend 7  as  5 + 2 .
then add 8 + 2 we get , 10 and then later add 5 to 10 we get sum , 15 .

Question 15.
__ = 4 + 7
Answer :
4 + 7 =
1 + 3 + 7 =
1 + 10 = 11
So, 4 + 7 = 11
Explanation :
To make addend 7 as 10 we should add 3 to the 7 then write addend 4  as  1 + 3 .
then add 3 + 7 we get , 10 and then later add 1 to 10 we get sum , 11 .

Question 16.
__ = 9 + 4
Answer :
9 + 4 =
9 + 1 + 3 =
10 + 3 = 13
So, 9 + 4 = 13
Explanation :
To make addend 9 as 10 we should add 1 to the 9 then write addend 4  as  1 + 3 .
then add 9 + 1 we get , 10 and then later add 3 to 10 we get sum , 13 .

Question 17.
__ = 8 + 8
Answer :
8 + 8 =
6 + 2 + 8 =
6 + 10 = 16
So, 8 + 8 = 16
Explanation :
To make addend 8 as 10 we should add 2 to the 8 then write addend 8 as  6 + 2 .
then add 2 + 8 we get , 10 and then later add 6 to 10 we get sum , 16 .

Question 18.
__ = 6 + 8
Answer :
6 + 8 =
4 + 2 + 8 =
4 + 10 = 14
So, 6 + 8 = 14
Explanation :
To make addend 8 as 10 we should add 2 to the 8 then write addend 6  as  4 + 2 .
then add 2 + 8 we get , 10 and then later add 4 to 10 we get sum , 14 .

Question 19.
Number Sense
How can you make a 10 to find 7 + 6?
Big Ideas Math Answers 2nd Grade Chapter 2 Fluency and Strategies within 20 85
Answer :
7 + 6 =
7 + 3 + 3 =
10 + 3 = 13
So, 7 + 6 = 13

7 + 6 =
3 + 4 + 6 =
3 + 10 = 13
So, 7 + 6 = 13
Explanation :
To make addend 7 as 10 we should add 3 to the 7 then write addend 6  as  3 + 3 .
then add 3 + 7 we get , 10 and then later add 3 to 10 we get sum , 13 .
To make addend 6 as 10 we should add 4 to the 6 then write addend 7  as  3 + 4 .
then add 4 + 6 we get , 10 and then later add 3 to 10 we get sum , 13 .

Think and Grow: Modeling Real Life

There are 13 boys and 4 girls in a line. How many girls must join the line so that the numbers of boys and girls are equal?
Addition equation:
Big Ideas Math Answers 2nd Grade Chapter 2 Fluency and Strategies within 20 86
Answer :
Number of boys = 13
Number of girls = 4
Number of girls required to add more to become 13 = 4 + _____ = 13 = 13 – 4 = 9
Therefore, Number of girls must join the line so that the numbers of boys and girls are equal = 9 girls .

Show and Grow

Question 20.
There are 16 seals and 7 whales. How many whales must join them so that the numbers of seals and whales are equal?
Big Ideas Math Answers 2nd Grade Chapter 2 Fluency and Strategies within 20 87
Answer :
Number of Seals = 16
Number of Whales = 7
Number of whales must join them so that the numbers of seals and whales are equal = 7 + ____ = 16 = 16 – 7 = 9
Therefore, Number of whales must join them so that the numbers of seals and whales are equal = 9

Question 21.
You and your friend count 15 stars in all. You count 8. How many stars does your friend count?
Big Ideas Math Answers 2nd Grade Chapter 2 Fluency and Strategies within 20 88
Answer :
Number of stars Counted by me and my friend = 15
Number of stars counted by me = 8
Number of stars Counted by my friend = 15 – 8 = 7

Question 22.
There are 17 fish in a tank. 9 are green. The rest are blue. How many fish are blue?
Big Ideas Math Answers 2nd Grade Chapter 2 Fluency and Strategies within 20 89
Answer :
Number of Fishes = 17
Number of green Fishes = 9
Number of blue Fishes = 17 – 9 = 8

Make a 10 to Add Homework & Practice 2.4

Make a ten to add.

Question 1.
Big Ideas Math Answers 2nd Grade Chapter 2 Fluency and Strategies within 20 90
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Make-10-Add-Homework-Practice-2.4-Question-1

Question 2.
Big Ideas Math Answers 2nd Grade Chapter 2 Fluency and Strategies within 20 91
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Make-10-Add-Homework-Practice-2.4-Question-2

Question 3.
7 + 6 = ?
10 + __ = ___
So, 7 + 6 = ___
Answer :
7 + 6 =
7 + 3 + 3 =
3 + 10 = 13
So, 7 + 6 = 13
Explanation :
To make addend 7 as 10 we should add 3 to the 7 then write addend 6 as  3 + 3 .
then add 3 + 7 we get , 10 and then later add 3 to 10 we get sum , 13 .

Question 4.
8 + 3 = ___
10 + __ = ___
So, 8 + 3 = __
Answer :
8 + 3=
8 + 2 + 1 =
1 + 10 = 11
So, 8 + 3 = 11
Explanation :
To make addend 8 as 10 we should add 2 to the 8 then write addend 3  as  2 + 1 .
then add 2 + 8 we get , 10 and then later add 1 to 10 we get sum , 11 .

Make a 10 to add.

Question 5.
3 + 9 = ___
Answer :
3 + 9 =
2 + 1 + 9 =
2 + 10 = 12
So, 3 + 9 = 12
Explanation :
To make addend 9 as 10 we should add 1 to the 9 so write addend  3 as  2 + 1 .
then add 1 + 9 we get , 10 and then later add 2 to 10 we get sum , 12 .

Question 6.
8 + 8 = ___
Answer :
8 + 8 =
8 + 2 + 6 =
10 + 6 = 16
So, 8 + 8 = 16
Explanation :
To make addend 8 as 10 we should add 2 to the 8 so write addend  8 as  2 + 6 .
then add 8 + 2 we get , 10 and then later add 6 to 10 we get sum , 16 .

Question 7.
7 +4 = ___
Answer :
7 + 4 =
7 + 3 + 1 =
10 + 1 = 11
So, 7 + 4 = 11
Explanation :
To make addend 7 as 10 we should add 3 to the 7 so write addend  4 as  3 + 1 .
then add 7 + 3 we get , 10 and then later add 1 to 10 we get sum , 11 .

Question 8.
4 + 9 = ___
Answer :
4 + 9 =
3 + 1 + 9 =
10 + 3 = 13
So, 4 + 9 = 13
Explanation :
To make addend 9 as 10 we should add 1 to the 9 so write addend 4 as  3 + 1 .
then add 1 + 9 we get , 10 and then later add 3 to 10 we get sum , 13 .

Question 9.
___ = 6 + 9
Answer :
6 + 9 =
5 + 1 + 9 =
10 + 5 = 15
So, 10 + 5 = 15
Explanation :
To make addend 9 as 10 we should add 1 to the 9 so write addend  6 as  5 + 1 .
then add 1 + 9 we get , 10 and then later add 5 to 10 we get sum , 15 .

Question 10.
___ = 4 + 8
Answer :
4 + 8 =
2 + 2 + 8 =
10 + 2 = 12
So, 4 + 8 = 12
Explanation :
To make addend 8 as 10 we should add 2 to the 8 so write addend  4 as  2 + 2 .
then add 8 + 2 we get , 10 and then later add 2 to 10 we get sum , 12 .

Question 11.
__ = 7 + 8
Answer :
7 + 8 =
5 + 2 + 8 =
10 + 5 = 15
So, 7 + 8 = 15
Explanation :
To make addend 8 as 10 we should add 2 to the 8 so write addend  7 as  5 + 2 .
then add 8 + 2 we get , 10 and then later add 5 to 10 we get sum , 15 .

Question 12.
___ = 9 + 2
9 + 2 =
9 + 1 + 1 =
10 + 1 = 11
So, 9 + 2 = 11
Explanation :
To make addend 9 as 10 we should add 1 to the 9 so write addend 2 as  1 + 1 .
then add 9 + 1 we get , 10 and then later add 1 to 10 we get sum , 11 .

Question 13.
Number Sense
Complete each equation.
Big Ideas Math Answers 2nd Grade Chapter 2 Fluency and Strategies within 20 92

Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Make-10-Add-Homework-Practice-2.4-Question-13

Question 14.
Modeling Real Life
There are 12 giraffes and 5 elephants. How many elephants must join them so that the numbers of giraffes and elephants are equal?
Big Ideas Math Answers 2nd Grade Chapter 2 Fluency and Strategies within 20 93
Answer :
Number of Giraffes = 12
Number of Elephants = 5
Number of more elephants must join them so that the numbers of giraffes and elephants are equal = 12 – 5 = 7
Therefore 7 more elephants must join them so that the numbers of giraffes and elephants will be equal .

Question 15.
Modeling Real Life
There are 14 horses on a farm. 7 are brown. The rest are white. How many are white?
Big Ideas Math Answers 2nd Grade Chapter 2 Fluency and Strategies within 20 94
Answer :
Number of Horses = 14
Number of Brown Horses = 7
Number of White Horses = 14 – 7 = 7

Review & Refresh

Compare.

Question 16.
Big Ideas Math Answers 2nd Grade Chapter 2 Fluency and Strategies within 20 95
Answer :
21 < 22
Explanation :
21 is lesser than 22

Question 17.
Big Ideas Math Answers 2nd Grade Chapter 2 Fluency and Strategies within 20 96
Answer :
43 > 34
Explanation :
43 is greater than 34

Lesson 2.5 Count On and Count Back to Subtract

Explore and Grow

Show two ways to find the difference on the number lines.
Big Ideas Math Answers 2nd Grade Chapter 2 Fluency and Strategies within 20 97
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Lesson 2.5-Count-On-and-Count-Back-to-Subtract-Explore-Grow
Explanation :
Count back to find the difference – start at 13 and count back 9.the arrow ending at 4 shows the difference.
Count on to find the difference – Start at 9 and count on to 13.

Show and Grow

Question 1.
Big Ideas Math Answers 2nd Grade Chapter 2 Fluency and Strategies within 20 98
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Lesson 2.5-Count-On-and-Count-Back-to-Subtract-Show-Grow-Question-1
Explanation :
Count back to find the difference – start at 11 and count back 5.the arrow ending at 6 shows the difference.

Question 2.
Big Ideas Math Answers 2nd Grade Chapter 2 Fluency and Strategies within 20 99
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Lesson 2.5-Count-On-and-Count-Back-to-Subtract-Show-Grow-Question-2
Explanation :
Count back to find the difference – start at 15 and count back 8.the arrow ending at 7 shows the difference.

Question 3.
Big Ideas Math Answers 2nd Grade Chapter 2 Fluency and Strategies within 20 100
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Lesson 2.5-Count-On-and-Count-Back-to-Subtract-Show-Grow-Question-3
Explanation :
Count back to find the difference – start at 9 and count back 6.the arrow ending at 3 shows the difference.

Question 4.
16 – 8 = ___
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Lesson 2.5-Count-On-and-Count-Back-to-Subtract-Show-Grow-Question-4
Explanation :
Count back to find the difference – start at 16 and count back 8.the arrow ending at 8 shows the difference.

Apply and Grow: Practice

Big Ideas Math Answers 2nd Grade Chapter 2 Fluency and Strategies within 20 101

Question 5.
11 – 8 = ___
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Lesson 2.5-Count-On-and-Count-Back-to-Subtract-Apply-and-Grow-Practice-Question-5
Explanation :
Count back to find the difference – start at 11 and count back 8.the arrow ending at 3 shows the difference.

Question 6.
10 – 6 = ___
Answer :

Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Lesson 2.5-Count-On-and-Count-Back-to-Subtract-Apply-and-Grow-Practice-Question-6

Explanation :
Count back to find the difference – start at 10 and count back 6. The arrow ending at 4 shows the difference.

Question 7.
14 – 6 = ___
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Lesson 2.5-Count-On-and-Count-Back-to-Subtract-Apply-and-Grow-Practice-Question-7
Explanation :
Count back to find the difference – start at 14 and count back 6. The arrow ending at 8 shows the difference.

Question 8.
11 – 4 = ___
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Lesson 2.5-Count-On-and-Count-Back-to-Subtract-Apply-and-Grow-Practice-Question-8
Explanation :
Count back to find the difference – start at 11 and count back 4. The arrow ending at 7 shows the difference.

Question 9.
15 – 9 = ___
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Lesson 2.5-Count-On-and-Count-Back-to-Subtract-Apply-and-Grow-Practice-Question-9
Explanation :
Count back to find the difference – start at 15 and count back 9. The arrow ending at 6 shows the difference.

Question 10.
16 – 7 = ___
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Lesson 2.5-Count-On-and-Count-Back-to-Subtract-Apply-and-Grow-Practice-Question-10
Explanation :
Count back to find the difference – start at 16 and count back 7. The arrow ending at 9 shows the difference.

Question 11.
14 – 9 = ___
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Lesson 2.5-Count-On-and-Count-Back-to-Subtract-Apply-and-Grow-Practice-Question-11
Explanation :
Count back to find the difference – start at 14 and count back 9. The arrow ending at 5 shows the difference.

Question 12.
12 – 4 = ___
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Lesson 2.5-Count-On-and-Count-Back-to-Subtract-Apply-and-Grow-Practice-Question-12
Explanation :
Count back to find the difference – start at 12 and count back 4. The arrow ending at 8 shows the difference.

Question 13.
10 – 7 = ___
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Lesson 2.5-Count-On-and-Count-Back-to-Subtract-Apply-and-Grow-Practice-Question-13
Explanation :
Count back to find the difference – start at 10 and count back 7. The arrow ending at 3 shows the difference.

Question 14.
Big Ideas Math Answers 2nd Grade Chapter 2 Fluency and Strategies within 20 102
Answer :
8 – 4 = 4
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Lesson 2.5-Count-On-and-Count-Back-to-Subtract-Apply-and-Grow-Practice-Question-14
Explanation :
Count back to find the difference – start at 8 and count back 4. The arrow ending at 4 shows the difference.

Question 15.
Big Ideas Math Answers 2nd Grade Chapter 2 Fluency and Strategies within 20 103
Answer :
11 – 6 = 5
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Lesson 2.5-Count-On-and-Count-Back-to-Subtract-Apply-and-Grow-Practice-Question-15
Explanation :
Count back to find the difference – start at 11 and count back 6. The arrow ending at 5 shows the difference.

Question 16.
Big Ideas Math Answers 2nd Grade Chapter 2 Fluency and Strategies within 20 104
Answer :
18 – 9 = 9
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Lesson 2.5-Count-On-and-Count-Back-to-Subtract-Apply-and-Grow-Practice-Question-16
Explanation :
Count back to find the difference – start at 18 and count back 9. The arrow ending at 9 shows the difference.

Question 17.
__ = 9 – 2
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Lesson 2.5-Count-On-and-Count-Back-to-Subtract-Apply-and-Grow-Practice-Question-17
Explanation :
Count back to find the difference – start at 9 and count back 2. The arrow ending at 7 shows the difference.

Question 18.
___ = 13 – 5
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Lesson 2.5-Count-On-and-Count-Back-to-Subtract-Apply-and-Grow-Practice-Question-18
Explanation :
Count back to find the difference – start at 13 and count back 5. The arrow ending at 8 shows the difference.

Question 19.
__ = 14 – 7
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Lesson 2.5-Count-On-and-Count-Back-to-Subtract-Apply-and-Grow-Practice-Question-19
Explanation :
Count back to find the difference – start at 14 and count back 7. The arrow ending at 7 shows the difference.

Question 20.
Structure
What strategy did Newton use to solve? How do you know?
Big Ideas Math Answers 2nd Grade Chapter 2 Fluency and Strategies within 20 105
Answer :
13 – 8 = 5
The Method Used here is Count On to find the difference
Explanation :
It started at 8 and count on to 13 .

Think and Grow: Modeling Real Life

There are 13 backpacks in your classroom. 9 are taken. How many backpacks are left?
Big Ideas Math Answers 2nd Grade Chapter 2 Fluency and Strategies within 20 106
Big Ideas Math Answers 2nd Grade Chapter 2 Fluency and Strategies within 20 107
Answer :
Number of backpacks = 13
Number of backpacks taken = 9
Number of backpacks left = 13 – 9 = 4
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Lesson 2.5-Count-On-and-Count-Back-to-Subtract-Think-and-Grow-Modeling-Real-Life
Explanation :
Count back to find the difference – start at 10 and count back 6. The arrow ending at 4 shows the difference.

Show and Grow

Question 21.
There are 17 trays in a stack. 7 are used. How many trays are left?
Big Ideas Math Answers 2nd Grade Chapter 2 Fluency and Strategies within 20 108
Answer :
Number of trays = 17
Number of used trays = 7
Number of trays left  =17 – 7 = 10
Therefore 10 trays are not used .

Question 22.
DIG DEEPER!
Newton wants to download some songs. 8 songs are downloaded. There are 6 songs left. How many songs did Newton want to download?
Big Ideas Math Answers 2nd Grade Chapter 2 Fluency and Strategies within 20 109
Answer :
Number of songs downloaded = 8
Number of songs left for downloading = 7
Total Number of songs = 8 + 7 = 15 .
Therefore , newton want to download 15 songs .

Count On and Count Back to Subtract Homework & Practice 2.5

Question 1.
7 – 5 = __
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Count-On-and-Count-Back-to-Subtract-Homework-&-Practice-2.5-Question-1Explanation :
Count back to find the difference – start at 7 and count back 5. The arrow ending at 2 shows the difference.

Question 2.
11 – 7 = __
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Count-On-and-Count-Back-to-Subtract-Homework-&-Practice-2.5-Question-2
Explanation :
Count back to find the difference – start at 11 and count back 7. The arrow ending at 4 shows the difference.

Question 3.
12 – 5 = __
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Count-On-and-Count-Back-to-Subtract-Homework-&-Practice-2.5-Question-3
Explanation :
Count back to find the difference – start at 12 and count back 5. The arrow ending at 7 shows the difference.

Question 4.
17 – 8 = ___
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Count-On-and-Count-Back-to-Subtract-Homework-&-Practice-2.5-Question-4
Explanation :
Count back to find the difference – start at 17 and count back 8. The arrow ending at 9 shows the difference.

Question 5.
7 – 2 = ___
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Count-On-and-Count-Back-to-Subtract-Homework-&-Practice-2.5-Question-5
Explanation :
Count back to find the difference – start at 7 and count back 2. The arrow ending at 5 shows the difference.

Question 6.
16 – 9 = ___
Answer :

Explanation :
Count back to find the difference – start at 16 and count back 9. The arrow ending at 5 shows the difference.

Question 7.
Big Ideas Math Answers 2nd Grade Chapter 2 Fluency and Strategies within 20 110
Answer :
9 – 3 = 3
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Count-On-and-Count-Back-to-Subtract-Homework-&-Practice-2.5-Question-7
Explanation :
Count back to find the difference – start at 9 and count back 3. The arrow ending at 6 shows the difference.

Question 8.
Big Ideas Math Answers 2nd Grade Chapter 2 Fluency and Strategies within 20 111
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Count-On-and-Count-Back-to-Subtract-Homework-&-Practice-2.5-Question-8
Explanation :
Count back to find the difference – start at 13 and count back 4. The arrow ending at 9 shows the difference.

Question 9.
Big Ideas Math Answers 2nd Grade Chapter 2 Fluency and Strategies within 20 112
Answer :
12 – 6 = 6
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Count-On-and-Count-Back-to-Subtract-Homework-&-Practice-2.5-Question-9
Explanation :
Count back to find the difference – start at 12 and count back 6. The arrow ending at 6 shows the difference.

Question 10.
___ = 10 – 8
Answer :
10 – 8 = 2
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Count-On-and-Count-Back-to-Subtract-Homework-&-Practice-2.5-Question-10
Explanation :
Count back to find the difference – start at 10 and count back 8. The arrow ending at 2 shows the difference.

Question 11.
__ = 8 – 5
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Count-On-and-Count-Back-to-Subtract-Homework-&-Practice-2.5-Question-11
Explanation :
Count back to find the difference – start at 11 and count back 5. The arrow ending at 3 shows the difference.

Question 12.
__ = 12 – 3
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Count-On-and-Count-Back-to-Subtract-Homework-&-Practice-2.5-Question-12
Explanation :
Count back to find the difference – start at 12 and count back 3. The arrow ending at 9 shows the difference.

Question 13.
DIG DEEPER!
Complete the squares so the differences on the outside are correct.
Big Ideas Math Answers 2nd Grade Chapter 2 Fluency and Strategies within 20 113
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Count-On-and-Count-Back-to-Subtract-Homework-&-Practice-2.5-Question-13
Explanation :
9 – 4 = 5
7 – 4 = 3
9 – 7 = 2
4 – 4 = 0
All above equations are correct

Question 14.
Modeling Real Life
15 kids are in line to play four square. 7 of them leave. How many kids are left?
Big Ideas Math Answers 2nd Grade Chapter 2 Fluency and Strategies within 20 114
Answer :
Number of kids = 15
Number of kids left the play = 7
Number of kids left to play = 15 – 7 = 8

Question 15.
Modeling Real Life
Newton has 6 pieces of macaroni. Descartes has 15. How many fewer pieces does Newton have than Descartes?
Big Ideas Math Answers 2nd Grade Chapter 2 Fluency and Strategies within 20 115
Answer :
Number of pieces of macaroni with Newton = 6
Number of pieces of macaroni with Descartes = 15
Number of fewer pieces does Newton have than Descartes = 15 – 6 = 9
Therefore , Newton 9 fewer pieces than Descartes .

Question 16.
Precision
Draw more lines to show fourths.
Big Ideas Math Answers 2nd Grade Chapter 2 Fluency and Strategies within 20 116

Lesson 2.6 Relate Addition and Subtraction

Explore and Grow

Write an addition equation. Complete the part-part-whole model to match. Write a related subtraction equation.
Big Ideas Math Answers 2nd Grade Chapter 2 Fluency and Strategies within 20 117
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Lesson-2.6-Relate-Addition-Subtraction-Explore-Grow
Explanation :
Addition and subtraction are inverse operations because one operation can “undo” the other operation.
Adding 9 and 6 to get 15 is the opposite of 15 minus 9, leaving 6 or the opposite of 15 minus 6, leaving 9

Show and Grow

Question 1.
5 + 3 = ___
8 – 5 = ___
Big Ideas Math Answers 2nd Grade Chapter 2 Fluency and Strategies within 20 118
Answer :
5 + 3 = 8
8 – 5 = 3
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Lesson-2.6-Relate-Addition-Subtraction-Show-Grow-Question-1
Explanation :
Addition and subtraction are inverse operations because one operation can “undo” the other operation.
Adding 5 and 3 to get 8 is the opposite of 8 minus 5, leaving 3 or the opposite of 8 minus 3, leaving 5

Question 2.
4 + 7 = __
11 – 4 = ___
Big Ideas Math Answers 2nd Grade Chapter 2 Fluency and Strategies within 20 119

Answer :
4 + 7 = 11
11 – 4 = 7
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Lesson-2.6-Relate-Addition-Subtraction-Show-Grow-Question-2
Explanation :
Addition and subtraction are inverse operations because one operation can “undo” the other operation.
Adding 4 and 7 to get 11 is the opposite of 11 minus 4, leaving 7 or the opposite of 11 minus 7, leaving 4

Question 3.
3 + 9 = ___
12 – 3 = ___
Answer :
3 + 9 = 12
12 – 3 = 9
Explanation :
Addition and subtraction are inverse operations because one operation can “undo” the other operation.
Adding 3 and 9 to get 12 is the opposite of 12 minus 9, leaving 3 or the opposite of 12 minus 3, leaving 9

Question 4.
8 + 6 = ___
14 – 8 = ___
Answer :
8 + 6 = 14
14 – 8 = 6
Explanation :
Addition and subtraction are inverse operations because one operation can “undo” the other operation.
Adding 8 and 6 to get 14 is the opposite of 14 minus 8, leaving 6 or the opposite of 14 minus 6, leaving 8

Question 5.
7 + 8 = ___
15 – 7 = ___
Answer :
7 + 8 = 15
15 – 7 = 8
Explanation :
Addition and subtraction are inverse operations because one operation can “undo” the other operation.
Adding 7 and 8 to get 15 is the opposite of 15 minus 7, leaving 8 or the opposite of 15 minus 8, leaving 7

Question 6.
9 + 9 = ___
18 – 9 = ___
Answer :
9 + 9 = 18
18 – 9 = 9
Explanation :
Addition and subtraction are inverse operations because one operation can “undo” the other operation.
Adding 9 and 9 to get 18 is the opposite of 18 minus 9, leaving 9

Question 7.
4 + 5 = ___
9 – 4 = ___
Answer :
4 + 5 = 9
9 – 4 = 5
Explanation :
Addition and subtraction are inverse operations because one operation can “undo” the other operation.
Adding 4 and 5 to get 9 is the opposite of 9 minus 4, leaving 5 or the opposite of 9 minus 5, leaving 4

Question 8.
5 + 7 = __
12 – 7 = ___
Answer :
5 + 7 = 12
12 – 7 = 5
Explanation :
Addition and subtraction are inverse operations because one operation can “undo” the other operation.
Adding 5 and 7 to get 12 is the opposite of 12 minus 7, leaving 5 or the opposite of 12 minus 5, leaving 7

Question 9.
8 + 8 = ___
16 – 8 = ___
Answer :
8 + 8 = 16
16 – 8 = 8
Explanation :
Addition and subtraction are inverse operations because one operation can “undo” the other operation.
Adding 8 and 8 to get 16 is the opposite of 16 minus 8, leaving 8

Question 10.
5 + 2 = ___
7 – 2 = ___
Answer :
5 + 2 = 7
7 – 2 = 5
Explanation :
Addition and subtraction are inverse operations because one operation can “undo” the other operation.
Adding 5 and 2 to get 7 is the opposite of 7 minus 2, leaving 5 or the opposite of 7 minus 5, leaving 2

Question 11.
4 + 6 = ___
10 – 4 = ___
Answer :
4 + 6 = 10
10 – 4 = 6
Explanation :
Addition and subtraction are inverse operations because one operation can “undo” the other operation.
Adding 4 and 6 to get 10 is the opposite of 10 minus 4, leaving 6 or the opposite of 10 minus 6, leaving 4

Question 12.
2 + 6 = ___
8 – 2 = ___
Answer :
2 + 6 = 8
8 – 2 = 6
Explanation :
Addition and subtraction are inverse operations because one operation can “undo” the other operation.
Adding 2 and 6 to get 8 is the opposite of 8 minus 2, leaving 6 or the opposite of 8 minus 6, leaving 2

Question 13.
7 + 7 = ___
14 – 7 = ___
Answer :
7 + 7 = 14
14 – 7 = 7
Explanation :
Addition and subtraction are inverse operations because one operation can “undo” the other operation.
Adding 7 and 7 to get 14 is the opposite of 14 minus 7, leaving 7

Question 14.
6 + 5 = ___
11 – 6 = ___
Answer :
6 + 5 = 11
11 – 6 = 5
Explanation :
Addition and subtraction are inverse operations because one operation can “undo” the other operation.
Adding 6 and 5 to get 11 is the opposite of 11 minus 5, leaving 6 or the opposite of 11 minus 6, leaving 5

Question 15.
9 + 6 = ___
15 – 6 = __
Answer :
9 + 6 = 15
15 – 6 = 9
Explanation :
Addition and subtraction are inverse operations because one operation can “undo” the other operation.
Adding 9 and 6 to get 15 is the opposite of 15 minus 9, leaving 6 or the opposite of 15 minus 6, leaving 9

Question 16.
3 + 8 = ___
11 – 8 = ___
11 – 3 = __
Answer :
3 + 8 = 11
11 – 8 = 3
11 – 3 = 8
Explanation :
Addition and subtraction are inverse operations because one operation can “undo” the other operation.
Adding 3 and 8 to get 11 is the opposite of 11 minus 8, leaving 3 or the opposite of 11 minus 3, leaving 8

Question 17.
9 + 7 = ___
16 – 9 = ___
16 – 7 = ___
Answer :
9 + 7 = 16
16 – 9 = 7
16 – 7 = 9
Explanation :
Addition and subtraction are inverse operations because one operation can “undo” the other operation.
Adding 9 and 7 to get 16 is the opposite of 16 minus 9, leaving 7 or the opposite of 16 minus 7, leaving 9

Question 18.
8 + 5 = ___
13 – 5 = ___
13 – 8 = ___
Answer :
8 + 5 = 13
13 – 5 = 8
13 – 8 = 5
Explanation :
Addition and subtraction are inverse operations because one operation can “undo” the other operation.
Adding 8 and 5 to get 13 is the opposite of 13 minus 5, leaving 8 or the opposite of 13 minus 8, leaving 5

Question 19.
Structure
Complete the number bond. Write related addition and subtraction equations.
Big Ideas Math Answers 2nd Grade Chapter 2 Fluency and Strategies within 20 120
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Lesson-2.6-Relate-Addition-Subtraction-Show-Grow-Question-19
Explanation :
Addition and subtraction are inverse operations because one operation can “undo” the other operation.
Adding 7 and 8 to get 15 is the opposite of 15 minus 7, leaving 8 or the opposite of 15 minus 8, leaving 7

Think and Grow: Modeling Real Life

You have 17 glow sticks. You give away 9 of them. How many glow sticks do you have left?
Big Ideas Math Answers 2nd Grade Chapter 2 Fluency and Strategies within 20 121
Which addition fact can help you subtract?
Big Ideas Math Answers 2nd Grade Chapter 2 Fluency and Strategies within 20 123
Answer :
Number of Glow sticks = 17
Number of glow sticks given away = 9
Number of glow sticks left = 17 – 9 = 8

Show and Grow

Question 20.
A box of sidewalk chalk has 20 pieces. You use 10 of them. How many pieces are not used?
Big Ideas Math Answer Key Grade 2 Chapter 2 Fluency and Strategies within 20 124
Answer :
Number of chalks = 20
Number of chalks used = 10
Number of chalks not used = 20 – 10 = 10

Question 21.
There are 7 more butterflies than birds. There are 8 birds. How many butterflies are there?
Big Ideas Math Answers 2nd Grade Chapter 2 Fluency and Strategies within 20 125
Answer :
Number of birds = 8
Number of butterflies = 7 more butterflies than birds = 7 + 8 = 15

Question 22.
A kitchen drawer has 9 fewer forks than spoons. There are 15 spoons. How many forks are there?
Big Ideas Math Answers 2nd Grade Chapter 2 Fluency and Strategies within 20 126
Answer :
Number of spoons = 15
Number of forks = 9 fewer forks than spoons = 15 – 9 = 6

Relate Addition and Subtraction Homework & Practice 2.6

Question 1.
5 + 5 = ___
10 – 5 = ___
Big Ideas Math Answers 2nd Grade Chapter 2 Fluency and Strategies within 20 127
Answer :
5 + 5 = 10
10 – 5 = 5
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Relate-Addition-Subtraction-Homework-Practice-2.6-Question-1
Explanation :
Addition and subtraction are inverse operations because one operation can “undo” the other operation.
Adding 5 and 5 to get 10 is the opposite of 10 minus 5, leaving 5

Question 2.
5 + 8 = ___
13 – 8 = ___
Big Ideas Math Answer Key Grade 2 Chapter 2 Fluency and Strategies within 20 128
Answer :
5 + 8 = 13
13 – 8 = 5
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Relate-Addition-Subtraction-Homework-Practice-2.6-Question-2

Explanation :
Addition and subtraction are inverse operations because one operation can “undo” the other operation.
Adding 5 and 8 to get 13 is the opposite of 13 minus 5, leaving  8 or the opposite of 13 minus 8, leaving 4

Question 3.
3 + 4 = ___
7 – 3 = ___
Answer :
3 + 4 = 7
7 – 3 = 4

Explanation :
Addition and subtraction are inverse operations because one operation can “undo” the other operation.
Adding 3 and 4 to get 7 is the opposite of 7 minus 3, leaving  4 or the opposite of 7 minus 4, leaving 3

Question 4.
8 + 6 = __
14 – 6 = ___
Answer :
8 + 6 = 14
14 – 6 = 8
Explanation :
Addition and subtraction are inverse operations because one operation can “undo” the other operation.
Adding 8 and 6 to get 14 is the opposite of 14 minus 6, leaving 8 or the opposite of 14 minus 8, leaving 6

Question 5.
6 + 6 = ___
12 – 6 = ___
Answer :
6 + 6 = 12
12 – 6 = 6
Explanation :
Addition and subtraction are inverse operations because one operation can “undo” the other operation.
Adding 6 and 6 to get 12 is the opposite of 12 minus 6, leaving 6

Question 6.
5 + 6 = ___
11 – 6 = __
11 – 5 = ___
Answer :
5 + 6 = 11
11 – 6 = 5
11 – 5 = 6
Explanation :
Addition and subtraction are inverse operations because one operation can “undo” the other operation.
Adding 5 and 6 to get 11 is the opposite of 11 minus 6, leaving 5 or the opposite of 11 minus 5, leaving 6

Question 7.
9 + 8 = ___
17 – 9 = ___
17 – 8 = ___
Answer :
9 + 8 = 17
17 – 9 = 8
17 – 8 = 9
Explanation :
Addition and subtraction are inverse operations because one operation can “undo” the other operation.
Adding 9 and 8 to get 17 is the opposite of 17 minus 9, leaving 8 or the opposite of 17 minus 8, leaving 9

Question 8.
7 + 9 = ___
16 – 9 = ___
16 – 7 = ___
Answer :
7 + 9 = 16
16 – 9 = 7
16 – 7 = 9
Explanation :
Addition and subtraction are inverse operations because one operation can “undo” the other operation.
Adding 7 and 9 to get 16 is the opposite of 16 minus 9, leaving 7 or the opposite of 16 minus 7, leaving 9

Question 9.
Number Sense
Find the sums. Write a related subtraction equation for each addition equation.
Big Ideas Math Answer Key Grade 2 Chapter 2 Fluency and Strategies within 20 129
Answer :
4 + 9 = 13 and 13 – 4 = 9
7 + 5 = 12 and 12 – 7 = 5
Explanation :
Addition and subtraction are inverse operations because one operation can “undo” the other operation.
Adding 4 and 9 to get 13 is the opposite of 13 minus 4, leaving 9
Addition and subtraction are inverse operations because one operation can “undo” the other operation.
Adding 7 and 5 to get 12 is the opposite of 12 minus 7, leaving 5

Question 10.
Modeling Real Life
You have 13 colors of modeling clay. You use 7 of them to make a sculpture. How many colors are not used?
Big Ideas Math Answer Key Grade 2 Chapter 2 Fluency and Strategies within 20 130
Answer :
Number of colors with me = 13
Number of colors used = 7
Number of colors not used = 13 – 7 = 6 colors

Question 11.
Modeling Real Life
There are 7 more snails than caterpillars. There are 4 caterpillars. How many snails are there?
Big Ideas Math Answer Key Grade 2 Chapter 2 Fluency and Strategies within 20 131
Answer :
Number of caterpillars = 4
Number of snails = 7 more snails than caterpillars.  = 7 + 4 = 11

Question 12.
You have 6 bears. Some are brown. The rest are black. You have more brown bears than black bears. How many brown and black bears can you have?
Big Ideas Math Answer Key Grade 2 Chapter 2 Fluency and Strategies within 20 132
Show how you know:
Answer :
Option 1 and Option 4
Explanation :
Number of bears = 6
brown bears > black bears
we cant say the exact number of black or brown bears but as per options we can solve
Option 1 – 5 brown and 1 black satisfy the condition.
Option 4 – 4 brown and 2 black satisfy the condition.
brown bears > black bears

Lesson 2.7 Get to 10 to Subtract

Explore and Grow

How can you use the number line to find the missing number?

Big Ideas Math Answer Key Grade 2 Chapter 2 Fluency and Strategies within 20 133
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Lesson-2.7-Get-to-10-to-Subtract-Explore-Grow
Explanation :
Number of linking arrows represent the subtrahend of above equation = 4

Show and Grow

Get to 10 to subtract.

Question 1.
Big Ideas Math Answer Key Grade 2 Chapter 2 Fluency and Strategies within 20 134
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20- Lesson 2.7-Get-to-10-to-Subtract-Show-Grow-Question-1
Explanation :
The subtrahend 6 is written as 3 + 3 so that the Minuend 13 is subtracted with 3 to get 10. then 3 is subtracted from 10 , we get difference as 7.

Question 2.
Big Ideas Math Answer Key Grade 2 Chapter 2 Fluency and Strategies within 20 135
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20- Lesson 2.7-Get-to-10-to-Subtract-Show-Grow-Question-2
Explanation :
The subtrahend 8 is written as 2 + 6 so that the Minuend 12 is subtracted with 2 to get 10. then 6 is subtracted from 10 , we get difference as 4.

Question 3.
14 – 8 = ?
10 – __ = __
So, 14 – 8 = ___.
Answer :
14 – 8
14- (4 + 4)
10 – 4 = 6
So, 14 – 8 = 6
Explanation :
The subtrahend 8 is written as 4 + 4 so that the Minuend 14 is subtracted with 4 to get 10. then 4 is subtracted from 10 , we get difference as 6.

Question 4.
16 – 7 = ?
10 – __ = __
So, 16 – 7 = ___
Answer :
16 – 7
16 – (6 + 1 )
10 – 1= 9
So, 16 – 7 = 9
Explanation :
The subtrahend 7 is written as 6 + 1 so that the Minuend 16 is subtracted with 6 to get 10. then 1 is subtracted from 10 , we get difference as 9.

Apply and Grow: Practice

Get to 10 to subtract.

Question 5.
12 – 6 = ?
10 – __ = __
So, 12 – 6 = __
Answer :
12 – 6 =
12 – (2 + 4 )
10 – 4 = 6
So, 12 – 6 = 6
Explanation :
The subtrahend 6 is written as 2 + 4 so that the Minuend 12 is subtracted with 2 to get 10. then 4 is subtracted from 10 , we get difference as 6.

Question 6.
13 – 7 = __
10 – __ = __
So, 13 – 7 = ___
Answer :
13 – 7 =
13 – (3 + 4)
10 – 4 = 6
So, 13 – 7 = 6
Explanation :
The subtrahend 7 is written as 3 + 4 so that the Minuend 13 is subtracted with 3 to get 10. then 4 is subtracted from 10 , we get difference as 6.

Question 7.
15 – 9 = ?
10 – __ = ___
So, 15 – 9 = ___
Answer :
15 – 9 =
15 – (5 + 4)
10 – 4 = 6
So, 15 – 9 = 6
Explanation :
The subtrahend 9 is written as 5 + 4 so that the Minuend 15 is subtracted with 5 to get 10. then 4 is subtracted from 10 , we get difference as 6.

Question 8.
14 – 6 = ?
10 – __ = __
So, 14 – 6 = __
Answer :
14 – 6 =
14 – ( 4 + 2 )
10 – 2 = 8
So, 14 – 6 = 8
Explanation :
The subtrahend 6 is written as 4 + 2 so that the Minuend 14 is subtracted with 4 to get 10. then 2 is subtracted from 10 , we get difference as 8.

Question 9.
13 – 8 = ___
Answer :
13 – 8 =
13 – ( 3 + 5 )
10 – 5 = 5
So, 13 – 8 = 5
Explanation :
The subtrahend 8 is written as 3 + 5 so that the Minuend 13 is subtracted with 3 to get 10. then 5 is subtracted from 10 , we get difference as 5.

Question 10.
12 – 4= ___
Answer :
12 – 4 =
12 – ( 2 + 2 )
10 – 2 = 8
So, 12 – 4 = 8
Explanation :
The subtrahend 4 is written as 2 + 2 so that the Minuend 12 is subtracted with 2 to get 10. then 2 is subtracted from 10 , we get difference as 8.

Question 11.
17 – 9 = __
Answer :
17 – 9 =
17 – ( 7 + 2 )
10 – 2 = 8
So, 17 – 9 = 8
Explanation :
The subtrahend 9 is written as 7 + 2 so that the Minuend 17 is subtracted with 7 to get 10. then 2 is subtracted from 10 , we get difference as 8.

Question 12.
15 – 8 = __
Answer :
15 – 8 =
15 – ( 5 + 3 )
10 – 3 = 7
So, 15 – 8 = 7
Explanation :
The subtrahend 8 is written as 5 + 23 so that the Minuend 15 is subtracted with 5 to get 10. then 3 is subtracted from 10 , we get difference as 7.

Question 13.
12 – 7 = ___
Answer :
12 – 7 =
12 – ( 2 + 5 )
10 – 5 = 5
So, 12 – 7 = 5
Explanation :
The subtrahend 7 is written as  2 + 5 so that the Minuend 12 is subtracted with 2 to get 10. then 5 is subtracted from 10 , we get difference as 5.

Question 14.
13 – 4 = ___
Answer :
13 – 4 =
13 – ( 3 + 1 )
10 – 1 = 9
So, 13 – 4 = 9
Explanation :
The subtrahend 4 is written as 3 + 1 so that the Minuend 13 is subtracted with 3 to get 10. then 1 is subtracted from 10 , we get difference as 9.

Question 15.
__ = 12 – 3
Answer :
12 – 3 =
12 – ( 2 + 1 )
10 – 1 = 9
So, 12 – 3 = 9
Explanation :
The subtrahend 3 is written as 2 + 1 so that the Minuend 12 is subtracted with 2 to get 10. then 1 is subtracted from 10 , we get difference as 9.

Question 16.
__ = 16 – 7
Answer :
16 – 7 =
16 – ( 6 + 1 )
10 – 1 = 9
So, 16 – 7 = 9
Explanation :
The subtrahend 7 is written as 6 + 1 so that the Minuend 16 is subtracted with 6 to get 10. then 1 is subtracted from 10 , we get difference as 9.

Question 17.
___ = 14 – 9
Answer :
14 – 9 =
14 – ( 4 + 5 )
10 – 5 = 5
So, 14 – 9 = 5
Explanation :
The subtrahend 9 is written as 4 + 5 so that the Minuend 14 is subtracted with 4 to get 10. then 5 is subtracted from 10 , we get difference as 5.

Question 18.
__ = 12 – 9
Answer :
12 – 9 =
12 – ( 2 + 7 )
10 – 7 = 3
So, 12 – 9 = 3
Explanation :
The subtrahend 9 is written as 2 + 7 so that the Minuend 12 is subtracted with 2 to get 10. then 7 is subtracted from 10 , we get difference as 3.

Question 19.
Number Sense
How can you get to 10 to find 13 −5?
Big Ideas Math Answer Key Grade 2 Chapter 2 Fluency and Strategies within 20 136
Answer :
13 – 5 =
13 – ( 3 + 2 )
10 – 2 = 8
So, 13 – 5 = 8
Explanation :
The subtrahend 5 is written as 3+ 2 so that the Minuend 13 is subtracted with 3 to get 10. then 2 is subtracted from 10 , we get difference as 8.

Think and Grow: Modeling Real Life

A first-aid kit has 16 bandages. You use 7 of them. How many bandages are left?
Equation:
Big Ideas Math Answer Key Grade 2 Chapter 2 Fluency and Strategies within 20 137
Answer :
Number of bandages = 16
Number of used bandages = 7
Number of bandages not used = 16 – 7 = 9

Show and Grow

Question 20.
There are 12 elephants at a watering hole. 5 of them leave. How many elephants are left?
Big Ideas Math Answer Key Grade 2 Chapter 2 Fluency and Strategies within 20 138
Answer :
Number of Elephants at watering hole = 12
Number of Elephants left the watering hole  = 5
Number of Elephants at Watering hole now = 12 – 5 = 7

Question 21.
DIG DEEPER!
There are 17 hot-air balloons on the ground. 5 of them take off at 1:00. 4 more take off at 1:30. How many balloons are still on the ground?
Big Ideas Math Answer Key Grade 2 Chapter 2 Fluency and Strategies within 20 139

Explain how you found the number of balloons still on the ground.
______________________________________
______________________________________
Answer :
Number of hot-air balloons on the ground = 17
Number of balloons take off at 1 : 00 = 5
Number of balloons take off at 1 : 30 = 4
Number of balloons left on ground = 17 – 5 – 4 = 8

Get to 10 to Subtract Homework & Practice 2.7

Get to 10 to subtract

Question 1.
Big Ideas Math Answer Key Grade 2 Chapter 2 Fluency and Strategies within 20 140
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Get-to-10-to-Subtract-Homework-Practice-2.7-Question-1
Explanation :
The subtrahend 5 is written as 2+ 3 so that the Minuend 12 is subtracted with 2 to get 10. then 3 is subtracted from 10 , we get difference as 7.

Question 2.
Big Ideas Math Answer Key Grade 2 Chapter 2 Fluency and Strategies within 20 141
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Get-to-10-to-Subtract-Homework-Practice-2.7-Question-2
Explanation :
The subtrahend 6 is written as 5+ 1 so that the Minuend 15 is subtracted with 5 to get 10. then 1 is subtracted from 10 , we get difference as 9.

Question 3.
16 – 8 = ?
10 – __ = __
So, 16 – 8 = __
Answer :
16 – 8
16 – (6 + 2 )
10 – 2 = 8
So, 16 – 8 = 8
Explanation :
The subtrahend 8 is written as 6+ 2 so that the Minuend 16 is subtracted with 6 to get 10. then 2 is subtracted from 10 , we get difference as 8.

Question 4.
13 – 9 = ?
10 – __ = ___
So, 13 – 9 = ___
Answer :
13 – 9
13 – ( 3 + 6 )
10 – 6 = 4
So, 13 – 9 = 4
Explanation :
The subtrahend 9 is written as 3+ 6 so that the Minuend 13 is subtracted with 3 to get 10. then 6 is subtracted from 10 , we get difference as 4.

Question 5.
14 – 5 ___
Answer :
14 – 5
14 – (4 + 1 )
10 – 1 = 9
So, 14 – 5 = 9
Explanation :
The subtrahend 5 is written as 4+ 1 so that the Minuend 14 is subtracted with 4 to get 10. then 1 is subtracted from 10 , we get difference as 9.

Question 6.
16 – 7 = ___
Answer :
16 – 7
16 – (6 + 1 )
10 – 1 = 9
So, 16 – 7 = 9
Explanation :
The subtrahend 7 is written as 6 + 1 so that the Minuend 16 is subtracted with 6 to get 10. then 1 is subtracted from 10 , we get difference as 9.

Question 7.
12 – 8 = __
Answer :
12 – 8
12 – (2 + 6 )
10 – 6 = 4
So, 12 – 8 = 4
Explanation :
The subtrahend 8 is written as 2+ 6 so that the Minuend 12 is subtracted with 2 to get 10. then 6 is subtracted from 10 , we get difference as 4.

Question 8.
13 – 8 = __
Answer :
13 – 8
13 – (3 + 5 )
10 – 5 = 5
So, 13 – 8 = 5
Explanation :
The subtrahend 8 is written as 3+ 5 so that the Minuend 13 is subtracted with 3 to get 10. then 5 is subtracted from 10 , we get difference as 5.

Question 9.
__ = 16 – 9
Answer :
16 – 9
16 – (6 + 3 )
10 – 3 = 7
So, 16 – 9 = 7
Explanation :
The subtrahend 9 is written as 6+ 3 so that the Minuend 16 is subtracted with 6 to get 10. then 3 is subtracted from 10 , we get difference as 7.

Question 10.
___ = 14 – 7
Answer :
14 – 7
14 – (4 + 3 )
10 – 3 = 7
So, 14 – 7 = 7
Explanation :
The subtrahend 7 is written as 4+ 3 so that the Minuend 14 is subtracted with 4 to get 10. then 3 is subtracted from 10 , we get difference as 7.

Question 11.
__ = 12 – 4
Answer :
12 – 4
12 – (2 + 2 )
10 – 2 = 8
So, 12 – 4 = 8
Explanation :
The subtrahend 4 is written as 2+ 2 so that the Minuend 12 is subtracted with 2 to get 10. then 2 is subtracted from 10 , we get difference as 8.

Question 12.
___ = 18 – 9
Answer :
18 – 9
18 – (8 + 1 )
10 – 1 = 9
So, 18 – 9 = 9
Explanation :
The subtrahend 9 is written as 8+ 1 so that the Minuend 18 is subtracted with 8 to get 10. then 1 is subtracted from 10 , we get difference as 9.

Question 13.
Number Sense
Complete each equation.
Big Ideas Math Answer Key Grade 2 Chapter 2 Fluency and Strategies within 20 142
Answer :
14 – 8 = 10 – 6
13 – 6 = 10 – 3
15 – 7 = 10 – 2

Question 14.
Modeling Real Life
Descartes has 18 items to sell at a garage sale. He sells 9 of them. How many items does he have left?
Big Ideas Math Answer Key Grade 2 Chapter 2 Fluency and Strategies within 20 143
Answer :
Number items with Descartes = 18
Number of items sold = 9
Number of items left for sale = 18 – 9 = 9

Question 15.
DIG DEEPER!
There are 12 penguins. 5 of them leave to find food at 5:00. 3 more leave at 5:30. How many penguins are left?
Big Ideas Math Answer Key Grade 2 Chapter 2 Fluency and Strategies within 20 144
Answer :
Number of Penguins = 12
Number of penguins left at 5 : 00 = 5
Number of penguins left at 5 : 30 = 3
Remaining Number of penguins = 12 – 5 – 3 = 4

Review & Refresh

Question 16.
Big Ideas Math Answer Key Grade 2 Chapter 2 Fluency and Strategies within 20 145
Answer :
19 is a Odd Number .
Explanation :
An odd number is a number that cannot be divided into two equal groups.

Question 17.
Big Ideas Math Answer Key Grade 2 Chapter 2 Fluency and Strategies within 20 146
Answer :
12 is a Even number .
Explanation :
An even number is a number that can be divided into two equal groups.  Even numbers end in 2, 4, 6, 8 and 0 regardless of how many digits they have.

Lesson 2.8 Practice Addition and Subtraction

Explore and Grow

You have 6 yellow buttons and 8 red buttons. Write four equations that describe your buttons.
Big Ideas Math Answer Key Grade 2 Chapter 2 Fluency and Strategies within 20 147
Answer :
Number of Yellow buttons = 6
Number of Red buttons = 8
Total Number of buttons = 6 + 8 = 8 + 6 = 14

Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Lesson-2.8-Practice-Addition-Subtraction-Explore-Grow

Show and Grow

Question 1.
3 + 9 = ___
Answer :
3 + 9 = 12
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Lesson-2.8-Practice-Addition-Subtraction-Question-1

Question 2.
8 – 6 = ___
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Lesson-2.8-Practice-Addition-Subtraction-Show-Grow-Question-2
2 + 6 = 8
So, 8 – 6 = 2

Question 3.
11 – 4 = ___
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Lesson-2.8-Practice-Addition-Subtraction-Show-Grow-Question-3
7 + 4 = 11
So, 11 – 4 = 7

Question 4.
7 + 7 = __
Answer :
7 + 7 = 14
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Lesson-2.8-Practice-Addition-Subtraction-Question-4

Question 5.
9 + 8 = ___
Answer :
9 + 8 = 17
8 + 8 = 16
Explanation :
9 + 8 is one more than 8 + 8 or 9 + 8
So, 9 + 8 = 17

Question 6.
10 – 7 = __
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Lesson-2.8-Practice-Addition-Subtraction-Show-Grow-Question-6
3 + 7 = 10
So, 10 – 7 = 3

Apply and Grow: Practice

Question 7.
Big Ideas Math Answer Key Grade 2 Chapter 2 Fluency and Strategies within 20 148
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Lesson-2.8-Practice-Addition-Subtraction-Apply-Grow-Practice-Question-7
Explanation :
Any number added to 0 gives the number itself.

Question 8.
Big Ideas Math Answer Key Grade 2 Chapter 2 Fluency and Strategies within 20 149
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Lesson-2.8-Practice-Addition-Subtraction-Apply-Grow-Practice-Question-8

Question 9.
Big Ideas Math Answer Key Grade 2 Chapter 2 Fluency and Strategies within 20 150
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Lesson-2.8-Practice-Addition-Subtraction-Apply-Grow-Practice-Question-9

Question 10.
Big Ideas Math Answer Key Grade 2 Chapter 2 Fluency and Strategies within 20 151
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Lesson-2.8-Practice-Addition-Subtraction-Apply-Grow-Practice-Question-10

Question 11.
Big Ideas Math Answer Key Grade 2 Chapter 2 Fluency and Strategies within 20 152
Answer : 

Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Lesson-2.8-Practice-Addition-Subtraction-Apply-Grow-Practice-Question-11

Question 12.
Big Ideas Math Answer Key Grade 2 Chapter 2 Fluency and Strategies within 20 153
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Lesson-2.8-Practice-Addition-Subtraction-Apply-Grow-Practice-Question-12
Explanation :
Any number added to 0 gives the number itself.

Question 13.
Big Ideas Math Answer Key Grade 2 Chapter 2 Fluency and Strategies within 20 154
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Lesson-2.8-Practice-Addition-Subtraction-Apply-Grow-Practice-Question-13

Question 14.
Big Ideas Math Answer Key Grade 2 Chapter 2 Fluency and Strategies within 20 155

Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Lesson-2.8-Practice-Addition-Subtraction-Apply-Grow-Practice-Question-14

Question 15.
__ = 9 + 1
Answer :
10 = 9 + 1
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Lesson-2.8-Practice-Addition-Subtraction-Question-15

Question 16.
__ = 4 + 5
Answer :
9 = 4 + 5
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Lesson-2.8-Practice-Addition-Subtraction-Question-16

Question 17.
___ = 11 – 8
Answer :
3 = 11 – 8
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Lesson-2.8-Practice-Addition-Subtraction-Question-17
3 + 8 = 11
So, 11 – 8 = 3

Question 18.
___ = 9 – 5
Answer :
4 = 9 – 5
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Lesson-2.8-Practice-Addition-Subtraction-Question-16
4 + 5 = 9
So, 9 – 5 = 4

Question 19.
__ = 7 + 9
Answer :
16 = 7 + 9
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Lesson-2.8-Practice-Addition-Subtraction-Question-19

Question 20.
___ = 7 – 1
Answer :
6 = 7 – 1
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Lesson-2.8-Practice-Addition-Subtraction-Question-20
1 + 6 = 7
So, 7 – 1 = 6

Question 21.
___ = 8 + 4
Answer :
12 = 8 + 4
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Lesson-2.8-Practice-Addition-Subtraction-Question-21

Question 22.
__ = 14 – 7
Answer :
7 = 14 – 7
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Lesson-2.8-Practice-Addition-Subtraction-Question-22

Question 23.
Repeated Reasoning
Find the missing addend. Explain how you solved.
Big Ideas Math Answer Key Grade 2 Chapter 2 Fluency and Strategies within 20 156
Answer :
8 + _____ = 13
8 + 5 = 13
Explanation :
To get 13 what should be added to 8
if we add 5 to 8 we get sum 13 .

Think and Grow: Modeling Real Life

Show and Grow

How many more students have brown hair than blonde hair?
Big Ideas Math Answer Key Grade 2 Chapter 2 Fluency and Strategies within 20 157
Answer :
Number of students with brown hair = 14
Number of students with black hair = 9
Number of students with blonde hair = 7
Number of more students have brown hair than blonde hair = 14 – 7 = 7
Equation: 14 – 7 = 7
Therefore 7 students more students have brown hair than blonde hair

How many students do not have brown hair?
Equation: 9 + 7
Big Ideas Math Answer Key Grade 2 Chapter 2 Fluency and Strategies within 20 158
Answer :
Number of students with out brown hair = ( number of students with blonde and black hair ) = 9 + 7 = 16

Question 24.
How many fewer students have hazel eyes than brown eyes?
Big Ideas Math Answer Key Grade 2 Chapter 2 Fluency and Strategies within 20 159
Answer :
Number of students with Blue Color= 8
Number of students with Brown color = 12
Number of students with Hazel Color = 4
Number of students fewer have hazel eyes than brown eyes = 12 – 4 = 8

DIG DEEPER!
How many students do not have brown eyes?
Big Ideas Math Answer Key Grade 2 Chapter 2 Fluency and Strategies within 20 160
Answer :
Number of students do not have brown eyes = students with blue color and Hazel color = 8 + 4 = 12

Practice Addition and Subtraction Homework & Practice 2.8

Question 1.
3 + 5 = ___
Answer :
3 + 5 = 8

Question 2.
5 + 7 = ___
Answer :
5 + 7
2 + 3 + 7
= 2 + 10 = 12

Explanation :
To make addend 7 as 10 we should add 3 to the 7 so write addend 5 as  2 + 3 .
then add 3 + 7 we get , 10 and then later add 2 to 10 we get sum , 12 .

Question 3.
13 – 6 = ___
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Practice-Addition-Subtraction-Homework-Practice-2.8-Question-3
Explanation :
The subtrahend 6 is written as 3+ 3 so that the Minuend 13 is subtracted with 3 to get 10. then 3 is subtracted from 10 , we get difference as 7.

Question 4.
12 – 8 = ___
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Practice-Addition-Subtraction-Homework-Practice-2.8-Question-4
Explanation :
The subtrahend 8 is written as 2+ 6 so that the Minuend 12 is subtracted with 2 to get 10. then 6 is subtracted from 10 , we get difference as 4.

Question 5.
6 + 0 = __
Answer :
6 + 0 = 6
Explanation :
Any number added to 0 gives the sum the number itself .

Question 6.
15 – 8 = ___
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Practice-Addition-Subtraction-Homework-Practice-2.8-Question-6
Explanation :
The subtrahend 8 is written as 5+ 3 so that the Minuend 15 is subtracted with 5 to get 10. then 3 is subtracted from 10 , we get difference as 7.

Question 7.
Big Ideas Math Answer Key Grade 2 Chapter 2 Fluency and Strategies within 20 161
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Practice-Addition-Subtraction-Homework-Practice-2.8-Question-7

Question 8.
Big Ideas Math Answer Key Grade 2 Chapter 2 Fluency and Strategies within 20 162
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Practice-Addition-Subtraction-Homework-Practice-2.8-Question-8

Question 9.
Big Ideas Math Answer Key Grade 2 Chapter 2 Fluency and Strategies within 20 163
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Practice-Addition-Subtraction-Homework-Practice-2.8-Question-9

Question 10.
Big Ideas Math Answer Key Grade 2 Chapter 2 Fluency and Strategies within 20 164
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Practice-Addition-Subtraction-Homework-Practice-2.8-Question-10
Explanation :
Any number subtracted with 1 gives the difference of previous number .

Question 11.
___ = 5 + 5
Answer :
5 + 5 = 10

Question 12.
___ = 3 – 0
Answer :
3 – 0 = 3
Explanation :
Any number subtracted with 0 gives the difference the number itself .

Question 13.
Number Sense
Which expressions have a sum or difference that is greater than 10?
Big Ideas Math Answer Key Grade 2 Chapter 2 Fluency and Strategies within 20 165

Question 14.
Modeling Real Life
How many fewer students choose a sandwich than pizza?
Big Ideas Math Answer Key Grade 2 Chapter 2 Fluency and Strategies within 20 166
Answer :
Number of students chooses sandwich = 9
Number of students chooses pasta = 4
Number of students chooses pizza = 15
Number of fewer students chooses a sandwich than pizza = 15 – 9 = 4
Therefore 4 fewer students chooses a sandwich than pizza .

DIG DEEPER!
How many students do not choose pizza?
Big Ideas Math Answer Key Grade 2 Chapter 2 Fluency and Strategies within 20 167

Answer :
Number of students do not choose pizza = number of students who chooses sandwich and pasta
= 9 + 4 = 13
Therefore, Number of students do not choose pizza = 13

Question 15.
Circle groups of 4. Write a repeated addition equation to match.
Big Ideas Math Answer Key Grade 2 Chapter 2 Fluency and Strategies within 20 168
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Practice-Addition-Subtraction-Homework-Practice-2.8-Question-15
Explanation :
Number of birds = 16
Number of groups = 4
Number of birds in each group = 16 ÷ 4 = 4
Therefore, 4 + 4 + 4 + 4 = 16 each group contains 4 birds .

Lesson 2.9 Problem Solving: Addition and Subtraction

Explore and Grow

Model the story.

You have 5 puzzle pieces. Your friend has 9. How many more pieces does your friend have than you?
Big Ideas Math Answer Key Grade 2 Chapter 2 Fluency and Strategies within 20 169
Answer :
Number of puzzles with me = 5
Number of puzzles with my friend = 9
Number of more pieces does my friend have than me = 9 – 5 = 4

Show and Grow

Question 1.
You have 7 more keychains than your friend. You have 15 keychains. How many keychains does your friend have?
Big Ideas Math Answer Key Grade 2 Chapter 2 Fluency and Strategies within 20 170
Answer :
Number of key chains with me = 15
Number of key chains with my friend = You have 7 more keychains than your friend = 15 -7 = 8
Explanation :
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Lesson-2.9-Problem-Solving-Addition-Subtraction-Show-Grow-Question-1

Apply and Grow: Practice

Question 2.
Some friends are at a community pool. 7 more join them. Now there are 12. How many friends were there to start?
Big Ideas Math Answer Key Grade 2 Chapter 2 Fluency and Strategies within 20 172
Answer :
Number of friends at community pool = x
Number of friends joined  = 7
Number of friends now at pool = 15 = x + 7
x + 7 = 15
x = 15 – 7 = 8
Therefore, Number of friends at community pool = 8

Question 3.
Some friends are playing miniature golf. 8 of them leave. There are 6 left. How many friends were there to start?
Big Ideas Math Answer Key Grade 2 Chapter 2 Fluency and Strategies within 20 173
Answer :
Number of friends playing golf = x
Number of friends left = 8
Number of friends now playing golf = 6
Number of friends at the start = x- 8 = 6
x= 8+ 6 = 14
Therefore, 14 friends were there to start.

Question 4.
YOU BE THE TEACHER
Newton has 5 fewer fish than Descartes. Newton has 8 fish. Your friend uses a bar model to find how many fish Descartes has. Is your friend correct? Explain.
Big Ideas Math Answer Key Grade 2 Chapter 2 Fluency and Strategies within 20 174
Big Ideas Math Answer Key Grade 2 Chapter 2 Fluency and Strategies within 20 174.1
Answer :
Yes he is correct
Explanation :
Number of fishes with Newton = 8
Number of fishes with Descartes = Newton has 5 fewer fish than Descartes. that means Descartes 5 more fishes than Newton = 8 + 5 = 13

Think and Grow: Modeling Real Life

15 kids play at a park. 6 of them leave. 9 more kids come to the park. How many kids are at the park now?
Big Ideas Math Answer Key Grade 2 Chapter 2 Fluency and Strategies within 20 175
Big Ideas Math Answer Key Grade 2 Chapter 2 Fluency and Strategies within 20 176
Answer :
Number of kids playing in the park = 15
Number of kids left the park = 6
Number of kids remaining in the park = 15 – 6 = 9
Number of kids joined the park = 9
Total number kids now in the park = 9 + 9 = 18

Question 5.
You have 13 baseball cards. You give 5 away. Then you get 4 more. How many baseball cards do you have now?
Big Ideas Math Answer Key Grade 2 Chapter 2 Fluency and Strategies within 20 157.1
Answer :
Number of base balls cards with me = 13
Number of base balls cards given away = 5
Number of base balls cards remaining with me now = 13 – 5 = 8
Number of base balls cards added = 4
Total Number of base balls cards now = 8 + 4 = 12

Question 6.
DIG DEEPER!
There are 7 adult lions and some cubs in a pride. The pride has 13 lions in all. 3 more cubs are born. How many cubs are in the pride now?
Big Ideas Math Answer Key Grade 2 Chapter 2 Fluency and Strategies within 20 178
Explain how you solved this problem.
Big Ideas Math Answer Key Grade 2 Chapter 2 Fluency and Strategies within 20 179
Answer :
Number of Adults lions = 7
Number of Lions = 13
Number of cubs are more than 3
Number of cubs = 13 – 7 = 6
Therefore, Number of cubs in the pride = 6

Problem Solving: Addition and Subtraction Homework & Practice 2.9

Question 1.
You have 3 more movies than your friend. You have 10 movies. Your friend has 5 video games. How many movies does your friend have?
Big Ideas Math Answer Key Grade 2 Chapter 2 Fluency and Strategies within 20 180
Big Ideas Math Answer Key Grade 2 Chapter 2 Fluency and Strategies within 20 181
Answer :
Number of movies with me = 10
You have 3 more movies than your friend
Number of movies with my friend = 10 – 3 = 7
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Problem-Solving-Addition-Subtraction-Homework-Practice-2.9-Question-1

Question 2.
Some friends are at a skating rink. 6 more join them. Now there are 11. How many friends were there to start?
Big Ideas Math Answer Key Grade 2 Chapter 2 Fluency and Strategies within 20 182
Big Ideas Math Answer Key Grade 2 Chapter 2 Fluency and Strategies within 20 183
Answer :
Number of Friends at skating rink = x
Number of friends joined = 6
Number of Friends now at skating rink = 11
Number of Friends at skating rink = 11 – 6 = 5
Therefore, 5 friends were there to start

Question 3.
Structure
There are 9 bananas in a bunch. A monkey eats some. There are 5 bananas left. Which picture shows how many bananas the monkey ate?
Big Ideas Math Answer Key Grade 2 Chapter 2 Fluency and Strategies within 20 184
Answer :
Number of Bananas = 9
Number of Bananas left = 5
Number of Bananas eaten by Monkey = 9 – 5 = 4
Picture 4 shows 4 Bananas .

Question 4.
Modeling Real Life
You make 4 snow angels and Newton makes 5. Descartes makes 3 more snow angels. How many snow angels are there in all?
Big Ideas Math Answer Key Grade 2 Chapter 2 Fluency and Strategies within 20 185
Answer :
Number of snow angels made by me = 4
Number of Snow angels made by Newton = 5
Number of Snow angels made by Descartes = 3
Total Number of Snow angels = 4 + 5 + 3 = 12

Question 5.
Modeling Real Life
Newton has 9 glitter pens. Descartes has 2 fewer than Newton. How many glitter pens do they have in all?
Big Ideas Math Answer Key Grade 2 Chapter 2 Fluency and Strategies within 20 186
Answer :
Number of Glitter pens with Newton= 9
Number of Glitter pens with Descartes = Descartes has 2 fewer than Newton = 9 – 2 = 7
Number of Glitter pens in all = 9 + 7 = 16

Review & Refresh

Question 6.
Which two shapes combine to make the shape on the left?
Big Ideas Math Answer Key Grade 2 Chapter 2 Fluency and Strategies within 20 186.1
Answer :
Triangle and semi-circle combine to make the above shape .

Fluency and Strategies within 20 Performance Task

Question 1.
You and your friend check out books from the library as shown in the table.
Big Ideas Math Answer Key Grade 2 Chapter 2 Fluency and Strategies within 20 187
a. You have one more adventure book than science books. How many books do you have in all?
Big Ideas Math Answer Key Grade 2 Chapter 2 Fluency and Strategies within 20 188
Answer :
Number of History books with me = 5
Number of science books with me = 7
Number of Adventure books with me = one more adventure books than science books = 7 + 1 = 8
Total Number of Books with me N= science + History + Adventure = 7 + 5 + 8 = 20

b. You and your friend have the same number of books. Your friend has more science books than history books. Your friend has an even number of adventure books. How many science books and adventure books does your friend have?
Big Ideas Math Answer Key Grade 2 Chapter 2 Fluency and Strategies within 20 189
Answer :
Total Number of books with me = Total Number of books with my friend = 20
Science books > History books
Number of history books with my Friend = 8
Number of Science books with my friend let it be = 10
Number of books remaining = 20 – 8 – 10 = 2
Number of Adventure books = 2 even number
So, all the given conditions are satisfied .

c. How many books do you and your friend have in all that are not history books?
Big Ideas Math Answer Key Grade 2 Chapter 2 Fluency and Strategies within 20 190
Answer :
Number of History books with me and my friend = 5 + 8 = 13
Number of Books with me and my friend = 20 + 20 = 40
Number of books in all that are not history books = 40 – 13 = 27 Books.

d. You return 3 books on Monday and 6 books on Thursday. How many books do you have left?
Big Ideas Math Answer Key Grade 2 Chapter 2 Fluency and Strategies within 20 191
Answer :
Number of books with me = 20
Number of books returned on Monday = 3
Number of books returned on Thrusday = 6
Number of books left now = 20 – 3 – 6 = 20 – 9 = 11 Books.

Fluency and Strategies within 20 Activity

To Play:
Place your cube on 0. Roll a die to see how many spaces to move. Write an equation that matches on the Joey Jump Add or Subtract Recording Sheet. Continue until you reach the food and then return to the pouch.
Big Ideas Math Answer Key Grade 2 Chapter 2 Fluency and Strategies within 20 192

Fluency and Strategies within 20 Chapter Practice

2.1 Add in Any Order

Question 1.
Big Ideas Math Answer Key Grade 2 Chapter 2 Fluency and Strategies within 20 193
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Fluency-and-Strategies-within-20-Chapter-Practice-2.1-Add-in-Any-Order-Question-1
Explanation :
Changing the Order of addends doesn’t change the sum. The Expressions 7 + 2 and  2 + 7 have the sum  9 .

Question 2.
Number Sense
Complete each equation.
Big Ideas Math Answer Key Grade 2 Chapter 2 Fluency and Strategies within 20 194
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Fluency-and-Strategies-within-20-Chapter-Practice-2.1-Add-in-Any-Order-Question-2
Explanation :
Changing the Order of addends doesn’t change the sum. The Expression a + b = b +a = c ,the sum .

2.2 Use Doubles

Find the sum. Write the double you used.

Question 3.
6 + 5 = ___
___ + ___ = ___
Answer :
6 + 5 = 11
5 + 5 = 10
Explanation :
Doubles plus 1
6 + 5 is equal to 5 + 5 and 1 more.

Question 4.
8 +9 = ___
___ + ___ = ___
Answer :
8 + 9 = 17
8 + 8 = 16
Explanation :
Doubles plus 1
8 + 9 is equal to 8 + 8 and 1 more.

Question 5.
Big Ideas Math Answer Key Grade 2 Chapter 2 Fluency and Strategies within 20 195
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Fluency-and-Strategies-within-20-Chapter-Practice-2.2-Use-Doubles-Question-5
Explanation :
Changing the Order of addends doesn’t change the sum. The Expressions 3 + 2 and  2 + 3 have the sum  5 .

Question 6.
Big Ideas Math Answer Key Grade 2 Chapter 2 Fluency and Strategies within 20 196
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Fluency-and-Strategies-within-20-Chapter-Practice-2.2-Use-Doubles-Question-6
Explanation :
Changing the Order of addends doesn’t change the sum. The Expressions 7 + 8 and  8 + 7 have the sum  15 .

2.3 Add Three Numbers

Question 7.
Big Ideas Math Answer Key Grade 2 Chapter 2 Fluency and Strategies within 20 197
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Fluency-and-Strategies-within-20-Chapter-Practice-2.3-Add-Three-Numbers-Question-7
Explanation :
Explanation:
Addends are represented in circle and Sum is represented in square.
First adding two addends 8, 0 give the sum 8
and then adding the another addend 6 to the sum 8.
Gives the sum 14.

Question 8.
Big Ideas Math Answer Key Grade 2 Chapter 2 Fluency and Strategies within 20 198
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Fluency-and-Strategies-within-20-Chapter-Practice-2.3-Add-Three-Numbers-Question-8
Explanation:
Addends are represented in circle and Sum is represented in square.
First adding two addends 1, 3 give the sum 4
and then adding the another addend 9 to the sum 4.
Gives the sum 13.

Question 9.
Big Ideas Math Answer Key Grade 2 Chapter 2 Fluency and Strategies within 20 199
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Fluency-and-Strategies-within-20-Chapter-Practice-2.3-Add-Three-Numbers-Question-9
Explanation:
Addends are represented in circle and Sum is represented in square.
First adding two addends 8, 8 give the sum 16
and then adding the another addend 4 to the sum 16.
Gives the sum 20.

Question 10.
4 + 8 + 8 = ___
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Fluency-and-Strategies-within-20-Chapter-Practice-2.3-Add-Three-Numbers-Question-9
Explanation:
Addends are represented in circle and Sum is represented in square.
First adding two addends 8, 8 give the sum 16
and then adding the another addend 4 to the sum 16.
Gives the sum 20.

Question 10.
Structure
Write the missing addends.
Big Ideas Math Answer Key Grade 2 Chapter 2 Fluency and Strategies within 20 200
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Fluency-and-Strategies-within-20-Chapter-Practice-2.3-Add-Three-Numbers-Question-10

2.4 Make a ten to Add

Make a 10 to add

Question 11.
Big Ideas Math Answer Key Grade 2 Chapter 2 Fluency and Strategies within 20 201
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Fluency-and-Strategies-within-20-Chapter-Practice-2.4-Make-a-ten-to-Add-Question-11

Explanation :
To make addend 9 as 10 we should add 1 to the 9 so write addend 8 as  1 + 7 .
then add 9 + 1 we get , 10 and then later add 7 to 10 we get sum , 17 .

Question 12.
Big Ideas Math Answer Key Grade 2 Chapter 2 Fluency and Strategies within 20 202

Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Fluency-and-Strategies-within-20-Chapter-Practice-2.4-Make-a-ten-to-Add-Question-12

Explanation :
To make addend 7 as 10 we should add 3 to the 7 so write addend 6 as  3 + 3 .
then add 7 + 3 we get , 10 and then later add 3 to 10 we get sum , 13 .

Question 13.
9 + 6 = ___
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Fluency-and-Strategies-within-20-Chapter-Practice-2.4-Make-a-ten-to-Add-Question-13

Explanation :
To make addend 9 as 10 we should add 1 to the 9 so write addend 6 as  1 + 5 .
then add 9 + 1 we get , 10 and then later add 5 to 10 we get sum , 15 .

Question 14.
4 + 7 = ___
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Fluency-and-Strategies-within-20-Chapter-Practice-2.4-Make-a-ten-to-Add-Question-14

Explanation :
To make addend 4 as 10 we should add 6 to the 4 so write addend 7 as  6 + 1 .
then add 4 + 6 we get , 10 and then later add 1 to 10 we get sum , 11 .

2.5 Count On and Count Back to Subtract

Big Ideas Math Answer Key Grade 2 Chapter 2 Fluency and Strategies within 20 203

Question 15.
12 – 8 = __
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Fluency-and-Strategies-within-20-Chapter-Practice-2.5-Count-On-and-Count-Back-to-Subtract-Question-15
Start at 12 and count back to 8 . The arrow ends at 4 showing the difference . So, 12 – 8 = 4
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Fluency-and-Strategies-within-20-Chapter-Practice-2.5-Count-On-and-Count-Back-to-Subtract-Question-15-b
Start at 8 count on to 12. So, 12 – 8 = 4

Question 16.
14 – 5 = ___
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Fluency-and-Strategies-within-20-Chapter-Practice-2.5-Count-On-and-Count-Back-to-Subtract-Question-16-a
Start at 14 and count back to 5 . The arrow ends at 9 showing the difference . So, 14 – 5 = 9
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Fluency-and-Strategies-within-20-Chapter-Practice-2.5-Count-On-and-Count-Back-to-Subtract-Question-16-b
Start at 5 count on to 14. So, 14 – 5 = 9

Question 17.
17 – 9 = ___
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Fluency-and-Strategies-within-20-Chapter-Practice-2.5-Count-On-and-Count-Back-to-Subtract-Question-17-a
Start at 17 and count back to 9 . The arrow ends at 8 showing the difference . So, 17 – 9 = 8
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Fluency-and-Strategies-within-20-Chapter-Practice-2.5-Count-On-and-Count-Back-to-Subtract-Question-17-b
Start at 9 count on to 17. So, 17 – 9 = 8

2.6 Relate Addition and Subtraction

Question 18.
7 + 6 = __
13 – 6 = ___
13 – 7 = ___
Big Ideas Math Answer Key Grade 2 Chapter 2 Fluency and Strategies within 20 204
Answer :
7 + 6 = 13
13 – 6 = 7
13 – 7 = 6

Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Fluency-and-Strategies-within-20-Chapter-Practice-2.6-Relate-Addition-and-Subtraction-Question-18
Explanation :
Addition and subtraction are inverse operations because one operation can “undo” the other operation.
Adding 7 and 6 to get 13 is the opposite of 13 minus 7, leaving 6 or the opposite of 13 minus 6, leaving 7.

2.7 Get to 10 to Subtract

Get to 10 to subtract

Question 19.
Big Ideas Math Answer Key Grade 2 Chapter 2 Fluency and Strategies within 20 205
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Fluency-and-Strategies-within-20-Chapter-Practice-2.7-Get-to-10-to-Subtract-Question-19
Explanation :
The subtrahend 4 is written as 2+ 2 so that the Minuend 12 is subtracted with 2 to get 10. then 2 is subtracted from 10 , we get difference as 8.

Question 20.
Big Ideas Math Answer Key Grade 2 Chapter 2 Fluency and Strategies within 20 206
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Fluency-and-Strategies-within-20-Chapter-Practice-2.7-Get-to-10-to-Subtract-Question-20
Explanation :
The subtrahend 8 is written as 5 + 3 so that the Minuend 15 is subtracted with 5 to get 10. then 3 is subtracted from 10 , we get difference as 7.

Question 21.
16 – 9 = ___
Answer :
16 – 9 =
16 – ( 6 + 3 )
10 – 3 = 7
So, 16 – 9 = 7
Explanation :
The subtrahend 9 is written as 6+ 3 so that the Minuend 16 is subtracted with 6 to get 10. then 3 is subtracted from 10 , we get difference as 7.

Question 22.
14 – 7 = ___
Answer :
14 – 7 =
14 – ( 4 + 3 )
10 – 3 = 7
So, 14 – 7 = 7
Explanation :
The subtrahend 7 is written as 4 + 3 so that the Minuend 14 is subtracted with 4 to get 10. then 3 is subtracted from 10 , we get difference as 7.

2.8 Practice Addition and Subtraction

Question 23.
8 + 3 = ___
Answer :
8 + 3 = 11

Question 24.
13 – 7 = ___
Answer :
12 – 4 =
12 – ( 2 + 2 )
10 – 2 = 8
So, 12 – 4 = 8
Explanation :
The subtrahend 5 is written as 2+ 3 so that the Minuend 12 is subtracted with 2 to get 10. then 3 is subtracted from 10 , we get difference as 7.

Question 25.
Big Ideas Math Answer Key Grade 2 Chapter 2 Fluency and Strategies within 20 207
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Fluency-and-Strategies-within-20-Chapter-Practice-2.8-Practice-Addition-and-Subtraction-Question-25

Question 26.
Big Ideas Math Answer Key Grade 2 Chapter 2 Fluency and Strategies within 20 208
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Fluency-and-Strategies-within-20-Chapter-Practice-2.8-Practice-Addition-and-Subtraction-Question-26

Question 27.
Big Ideas Math Answer Key Grade 2 Chapter 2 Fluency and Strategies within 20 209
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Fluency-and-Strategies-within-20-Chapter-Practice-2.8-Practice-Addition-and-Subtraction-Question-27

Question 28.
Big Ideas Math Answer Key Grade 2 Chapter 2 Fluency and Strategies within 20 210
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Fluency-and-Strategies-within-20-Chapter-Practice-2.8-Practice-Addition-and-Subtraction-Question-28

Question 29.
__ = 5 + 8
Answer :
5 + 8
3 + 2 + 8
3 + 10 = 13
So, 5 + 8 = 13
Explanation :
To make addend 8 as 10 we should add 2 to the 8 so write addend 5 as  3 + 2 .
then add 2 + 8 we get , 10 and then later add 3 to 10 we get sum , 13 .

Question 30.
__ = 12 – 5
Answer :
12 – 5 =
12 – ( 2 + 3 )
10 – 3 = 7
So, 12 – 5 = 7
Explanation :
The subtrahend 5 is written as 2+ 3 so that the Minuend 12 is subtracted with 2 to get 10. then 3 is subtracted from 10 , we get difference as 7.

2.9 Problem Solving: Addition and Subtraction

Question 31.
Some friends are playing laser tag. 4 of them leave. There are 8 left. How many friends were there to start?
Big Ideas Math Answer Key Grade 2 Chapter 2 Fluency and Strategies within 20 211
Big Ideas Math Answer Key Grade 2 Chapter 2 Fluency and Strategies within 20 212
Answer :
Number of friends playing = x
Number of friends left = 4
Number of friends left = 8 = x-4
x= 8 + 4 = 12
Big-Ideas-Math-Book-2nd-Grade-Answer-key-Chapter-2-Fluency-and-Strategies-within-20-Fluency-and-Strategies-within-20-Chapter-Practice-2.9-Problem-Solving-Addition-and-Subtraction-Question-31

Conclusion:

I hope that you can resolve all your queries and access your preparation standard using the Big Ideas Math Book Grade 2 Chapter 2 Fluency and Strategies within 20 Solution Key. Step by step explanation for all questions are available at Big Ideas Math 2nd Grade 2nd Chapter Fluency and Strategies within 20 Answer Key. Bookmark our site to find the solution key of all Big Ideas Math Grade 2 Chapters.

Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions

Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions

Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions are aligned based on the latest syllabus and common core curriculum guidelines. Also, you can get a step-by-step explanation by the subject experts for all questions covered in the Big Ideas Math Book Algebra 2 Ch 4 Polynomials Functions Solution Key. Polynomials are a very important concept for high school students to calculate various complex calculations in further chapters. So, it is very crucial to download and practice the questions covered in BIM Textbook solutions of Algebra 2 Ch 4 Polynomial Functions.

Big Ideas Math Book Algebra 2 Answer Key Chapter 4 Polynomial Functions

Improving your Math skills is the best way to score high marks and also become a pro in math concepts. Access the available links below and download free pdf formatted Big Ideas Math Algebra 2 Solutions of Ch 4 Polynomial functions. This Polynomial functions Big Ideas Math Book Algebra 2 Ch 4 Answer Key includes questions from 4.1 to 4.9 lessons exercises, assignment tests, practice tests, chapter tests, quizzes, etc. Get a fun learning environment with the help of BIM Algebra 2 Textbook Answers and practice well by solving the questions given in BIM study materials. 

Polynomial Functions Maintaining Mathematical Proficiency

Simplify the expression.

Question 1.
6x − 4x=2x

Explanation:

Terms can be combined only if they have the exact same variable portion and combining like terms.

Question 2.
12m − m − 7m + 3

Answer:

12m-m-7m+3

=4m+3

Question 3.
3(y + 2) − 4y

Answer:

3(y+2)-4y

3y+6-4y

-y+6

Question 4.
9x − 4(2x − 1)

Answer:

9x-4(2x-1)

=9x-8x+4

=x+4

Question 5.
−(z + 2) − 2(1 − z)

Answer:-

= -z-2-2+z

=-4

Question 6.
−x2 + 5x + x2

Answer:

5x

Find the volume of the solid.

Question 7.
cube with side length 4 inches

solution: 

Volume of cube=sidexsidexside

=4x4x4

=64inches

sphere with radius 2 feet

Question 9.
rectangular prism with length 4 feet, width 2 feet, and height 6 feet

Solution: Given that

length= 4 feet

width=2 feet

height= 6 feet

volume of rectangular prism= length x width x height

See the source image

substitute the values in  formula

volume of rectangular prism= 4 x 2 x 6= 48 cubic feet

Question 10.

right cylinder with radius 3 centimeters and height 5 centimeters

Solution:

Given that

radius=3centimeters

height =5centimeter

Volume of a cylinder = area of base × height = π × r2× h and you can use 3.14 for π.

substitute the values in formula

3.14 x 3 x 3 x 5=141.3cubic centimeter

Question 11.
ABSTRACT REASONING Does doubling the volume of a cube have the same effect on the side length? Explain your reasoning

Solution:

If the side of the cube id doubled. the volume is 8 times larger  .

example: let’s use a cube that has side lenghts of 2 as an  example. the volume of that cube would be lenght x and width x height

2 x 2 x  2 =8. if we double the edges to 4 . it would be 4 x 4 x 4, which is 64. to find how many times increases take the 64 and divide by 8 . 64/8=8 The volume increases by 8 times.

 

Polynomial Functions Mathematical Practices

Monitoring Progress

Use a graphing calculator to determine whether the function is continuous. Explain your reasoning.

Question 1.
f(x) = \(\frac{x^{2}-x}{x}\)

Question 2.
f(x) = x3 − 3

Question 3.
f(x) = \(\sqrt{x^{2}+1}\)

Question 4.
f(x) = | x + 2 |

Question 5.
f(x) = \(\frac{1}{x}\)

Question 6.
f(x) = \(\frac{1}{\sqrt{x^{2}-1}}\)

Question 7.
f(x) = x

Question 8.
f(x) = 2x − 3

Question 9.
f(x) = \(\frac{x}{x}\)

Lesson 4.1 Graphing Polynomial Functions

Essential Question
What are some common characteristics of the graphs of cubic and quartic polynomial functions?
A polynomial function of the form
f(x) = anxn + an – 1xn– 1 +. . .+ a1x + a0
where an ≠ 0, is cubic when n = 3 and quartic when n = 4.

EXPLORATION 1
Identifying Graphs of Polynomial Functions
Work with a partner.
Match each polynomial function with its graph. Explain your reasoning. Use a graphing calculator to verify your answers.
a. f(x) = x3 − x
b. f(x) = −x3 + x
c. f(x) = −x4 + 1
d. f(x) = x4
e. f(x) = x3
f. f(x) = x4 − x2

EXPLORATION 2
Identifying x-Intercepts of Polynomial Graphs
Work with a partner.
Each of the polynomial graphs in Exploration 1 has x-intercept(s) of −1, 0, or 1. Identify the x-intercept(s) of each graph. Explain how you can verify your answers.

Communicate Your Answer

Question 3.
What are some common characteristics of the graphs of cubic and quartic polynomial functions?

Question 4.
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 1
Determine whether each statement is true or false. Justify your answer.
a. When the graph of a cubic polynomial function rises to the left, it falls to the right.
b. When the graph of a quartic polynomial function falls to the left, it rises to the right.

4.1 Lesson

Monitoring Progress

Decide whether the function is a polynomial function. If so, write it in standard form and state its degree, type, and leading coefficient.

Question 1.
f(x) = 7 − 1.6x2 − 5x

Answer:

The standard form of a polynomials has the exponents of the terms arranged in descending order.

The  degree of a polynomial is the highest exponent of a term.

The type of a polynomial depends on the degree.

The leading coefficient is the numerical factor of the terms with highest degree.

Hence the given function f(x) = 7 − 1.6x2 − 5x has the  following characteristic

Standard form f (x) = -1.6x2 -5x +7

Degree: 2

Type: quadratic

Leading coefficient: -1.6

Question 2.
p(x) =x+ 2x-2 + 9.5

Answer:

The function is not a polynomial function because the term 2x-2 has an exponent  that is not a whole number

Question 3.
q(x) = x3 − 6x + 3x4

Answer:

The standard form of a polynomials has the exponents of the terms arranged in descending order.

The  degree of a polynomial is the highest exponent of a term.

The type of a polynomial depends on the degree.

The leading coefficient is the numerical factor of the terms with highest degree.

Hence the given function q(x) = x3 − 6x + 3x4  has the  following characteristic

Standard form : q(x) = 3x4 + x3 – 6x

Degree: 4

Type: quadratic

Leading coefficient: 3

 

Evaluate the function for the given value of x.

Question 4.
f(x) = −x3 + 3x2 + 9; x = 4

Answer:

f(x) = −x3 + 3x2 + 9                         Write Original equation

f(4) =−x3 + 3x2 + 9                        substitute for x

=-(4)3+ 3(4)2 +9                  Evaluate powers and multiply

=-(64)+ 3(16) +9                           Simplify

=-64 + 48 + 9

=-7

Question 5.
f(x) = 3x5 − x4 − 6x + 10; x = −2

Answer:

f(x) = 3x5 − x4 − 6x + 10                Write Original equation

f(-2)=3(-2)5-(-2)4-6(-2)+10            substitute for x

=3(-32)-16+12+10              Evaluate powers and multiply

= -96 -16 +12 +10                Simplify

=-90

Question 6.
Describe the end behavior of the graph of f(x) = 0.25x3 − x2 − 1.

Graph the polynomial Function

Question 7.
f(x) = x4 + x2 − 3

Question 8.
f(x) = 4 − x3

Question 9.
f(x) = x3 − x2 + x − 1

Question 10.
Sketch a graph of the polynomial function f having these characteristics.

  • f is decreasing when x < −1.5 and x > 2.5; f is increasing when −1.5 < x < 2.5.
  • f(x) > 0 when x < −3 and 1 < x < 4; f(x) < 0 when −3 < x < 1 and x > 4.

Use the graph to describe the degree and leading coefficient of f.

Question 11.
WHAT IF? Repeat Example 6 using the alternative model for electric vehicles of
V(t) = −0.0290900t4 + 0.791260t3 − 7.96583t2 + 36.5561t − 12.025.

Graphing Polynomial Functions 4.1 Exercises

Vocabulary and Core Concept Check

Question 1.
WRITING Explain what is meant by the end behavior of a polynomial function.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.1 Question 1

Question 2.
WHICH ONE DOESN’T BELONG? Which function does not belong with the other three? Explain your reasoning.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 2

Monitoring Progress and Modeling with Mathematics

In Exercises 3–8, decide whether the function is a polynomial function. If so, write it in standard form and state its degree, type, and leading coefficient.

Question 3.
f(x) = −3x + 5x3 − 6x2 + 2
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.1 Question 3

Question 4.
p(x) = \(\frac{1}{2}\)x2 + 3x − 4x3 + 6x4 − 1
Answer:

The function is a polynomial function . Written in standard form , the function is p(x)=1/2 x2 + 3x − 4x3 + 6x4 − 1   .

It has degree 4(quartic)   and leading coefficient of 6.

 

Question 5.
f(x) = 9x4 + 8x3 − 6x-2 + 2x
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.1 Question 5

Question 6.
g(x) = \(\sqrt{3}\) − 12x + 13x2
Answer:

Question 7.
h(x) = \(\frac{5}{3}\)x2 − \(\sqrt{7}\)x4 + 8x3 − \(\frac{1}{2}\) + x
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.1 Question 7

Question 8.
h(x) = 3x4 + 2x − \(\frac{5}{x}\) + 9x3 − 7
Answer:

ERROR ANALYSIS In Exercises 9 and 10, describe and correct the error in analyzing the function.

Question 9.
f(x) = 8x3 − 7x4 − 9x − 3x2 + 11
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 3
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.1 Question 9

Question 10.
f(x) = 2x4 + 4x – 9\(\sqrt{x}\) + 3x2 – 8
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4
Answer:

In Exercises 11–16, evaluate the function for the given value of x.

Question 11.
h(x) = −3x4 + 2x3 − 12x − 6; x = −2
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.1 Question 11

Question 12.
f(x) = 7x4 − 10x2 + 14x − 26; x = −7

Answer:

f(x) = 7x4 − 10x2 + 14x − 26            Write Original equation

f(-7)     =  7(-7)4– 10(-7)2+14(-7)-26                   substitute for x

= 7(2401)- 10(49)- 98-26                          Evaluate powers and multiply

= 16807- 490- 98- 26                                           Simplify

= 16,193

Question 13.
g(x) = x6 − 64x4 + x2 − 7x − 51; x = 8
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.1 Question 13

Question 14.
g(x) = −x3 + 3x2 + 5x + 1; x = −12

Answer:

g(x) = −x3 + 3x2 + 5x + 1          Write Original equation

g(-12) =(-12 )3+ 3(-12) 2+ 5(-12 )+1          substitute for x

=1728 + 3(144) – 60 +1                         Evaluate powers and multiply

=1728 + 432 – 60 +1                                    Simplify

=2101

Question 15.
p(x) = 2x3 + 4x2 + 6x + 7; x = \(\frac{1}{2}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.1 Question 15

Question 16.

h(x) = 5x3 − 3x2 + 2x + 4; x = −12

Answer:
h(x) = 5x3 − 3x2 + 2x + 4                      Write Original equation

h(-12)    =5(-12)3-3(-12)2+2(-12)+4          substitute for x

=5(-1728)-3(144)-24+4         Evaluate powers and multiply

=-8640-432-24+4                            Simplify

=-9092

In Exercises 17–20, describe the end behavior of the graph of the function.

Question 17.
h(x) = −5x4 + 7x3 − 6x2 + 9x + 2
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.1 Question 17

Question 18.
g(x) = 7x7 + 12x5 − 6x3 − 2x − 18

Answer:

The function as degree 7 and leading coefficient 7 . Because the  degree is odd and leading coefficient  is positive g(x)->-∞ as x ->-∞  and  g(x)->+∞ as x->+∞

Question 19.
f(x) = −2x4 + 12x8 + 17 + 15x2
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.1 Question 19

Question 20.
f(x) = 11 − 18x2 − 5x5 − 12x4 − 2x
Answer:

f(x)= − 5x5 − 12x4− 18x2− 2x + 11

with the highest exponent is equal to 5, then the polynomial is of odd degree.

Hence the end behaviors are opposite. With the negative leading coefficient

Then   f(x) ─> ∞ as x ─>-∞

f(x)─>-∞ as  x ─>+∞

 

In Exercises 21 and 22, describe the degree and leading coefficient of the polynomial function using the graph.

Question 21.
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 5
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.1 Question 21

Question 22.
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 6
Answer:

The degree of the polynomial is even since one side goes up and other goes up; the leading coefficient is positive  since the left side goes up and the right side goes up.

Question 23.
USING STRUCTURE Determine whether the function is a polynomial function. If so, write it in standard form and state its degree, type, and leading coefficient.
f(x) = 5x3x + \(\frac{5}{2}\)x3 − 9x4 + \(\sqrt{2}\)x2 + 4x − 1 − x-5x5 − 4
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.1 Question 23

Question 24.
WRITING Let f(x) = 13. State the degree, type, and leading coefficient. Describe the end behavior of the function. Explain your reasoning.
Answer:

In Exercises 25–32, graph the polynomial function.

Question 25.
p(x) = 3 − x4
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.1 Question 25

Question 26.
g(x) = x3 + x + 3
Answer:

Question 27.
f(x) = 4x − 9 − x3
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.1 Question 27

Question 28.
p(x) = x5 − 3x3 + 2
Answer:

Question 29.
h(x) = x4 − 2x3 + 3x
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.1 Question 29

Question 30.
h(x) = 5 + 3x2 − x4
Answer:

Question 31.
g(x) = x5 − 3x4 + 2x − 4
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.1 Question 31

Question 32.
p(x) = x6 − 2x5 − 2x3 + x + 5
Answer:

ANALYZING RELATIONSHIPS In Exercises 33–36, describe the x-values for which
(a) f is increasing or decreasing,
(b) f(x) > 0, and
(c) f(x) < 0.

Answer:

Question 33.
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 7
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.1 Question 33

Question 34.
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 8
Answer:

a. The function is increases  when x< 4 and decreases when x>4.

b.The function is positive when x<

Question 35.
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 9
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.1 Question 35

Question 36.
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 10
Answer:

In Exercises 37–40, sketch a graph of the polynomial function f having the given characteristics. Use the graph to describe the degree and leading coefficient of the function f.

Question 37.

  • f is increasing when x > 0.5; f is decreasing when x < 0.5.
  • f(x) > 0 when x < −2 and x > 3; f(x) < 0 when −2 < x < 3.

Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.1 Question 37

Question 38.

  • f is increasing when −2 < x < 3;f is decreasing when x < −2 and x > 3.
  • f(x) > 0 when x < −4 and 1 < x < 5; f(x) < 0 when −4 < x < 1 and x > 5.

Answer:
Question 39.

  • f is increasing when −2 < x < 0 and x > 2;f is decreasing when x < −2 and 0 < x < 2.
  • f(x) > 0 when x < −3, −1 < x < 1, and x > 3; f(x) < 0 when −3 < x < −1 and 1 < x < 3.

Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.1 Question 39

Question 40.

  • f is increasing when x < −1 and x > 1;f is decreasing when −1 < x < 1.
  • f(x) > 0 when −1.5 < x < 0 and x > 1.5; f(x) < 0 when x < −1.5 and 0 < x < 1.5.

Answer:
Question 41.
MODELING WITH MATHEMATICS From 1980 to 2007 the number of drive-in theaters in the United States can be modeled by the function
d(t) = −0.141t3 + 9.64t2 − 232.5t + 2421
where d(t) is the number of open theaters and t is the number of years after 1980.
a. Use a graphing calculator to graph the function for the interval 0 ≤ t ≤ 27. Describe the behavior of the graph on this interval.
b. What is the average rate of change in the number of drive-in movie theaters from 1980 to 1995 and from 1995 to 2007? Interpret the average rates of change.
c. Do you think this model can be used for years before 1980 or after 2007? Explain.
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 11
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.1 Question 41.1
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.1 Question 41.2

Question 42.
PROBLEM SOLVING The weight of an ideal round-cut diamond can be modeled by
w = 0.00583d3 − 0.0125d2 + 0.022d − 0.01
where w is the weight of the diamond (in carats) and d is the diameter (in millimeters). According to the model, what is the weight of a diamond with a diameter of 12 millimeters?
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 12
Answer:

Question 43.
ABSTRACT REASONING Suppose f(x) → ∞ as x → −∞ and f(x) →−∞ as x → ∞. Describe the end behavior of g(x) = −f(x). Justify your answer.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.1 Question 43

Question 44.
THOUGHT PROVOKING Write an even degree polynomial function such that the end behavior of f is given by f(x) → −∞ as x → −∞ and f(x) → −∞ as x → ∞. Justify your answer by drawing the graph of your function.
Answer:

Question 45.
USING TOOLS When using a graphing calculator to graph a polynomial function, explain how you know when the viewing window is appropriate.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.1 Question 45

Question 46.
MAKING AN ARGUMENT Your friend uses the table to speculate that the function f is an even degree polynomial and the function g is an odd degree polynomial. Is your friend correct? Explain your reasoning.
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 13
Answer:

Question 47.
DRAWING CONCLUSIONS The graph of a function is symmetric with respect to the y-axis if for each point (a, b) on the graph, (−a, b) is also a point on the graph. The graph of a function is symmetric with respect to the origin if for each point (a, b) on the graph, (−a, −b) is also a point on the graph.
a. Use a graphing calculator to graph the function y = xn when n = 1, 2, 3, 4, 5, and 6. In each case, identify the symmetry of the graph.
b. Predict what symmetry the graphs of y = x10 and y = x11 each have. Explain your reasoning and then confirm your predictions by graphing.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.1 Question 47.1
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.1 Question 47.2

Question 48.
HOW DO YOU SEE IT? The graph of a polynomial function is shown.
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 14
a. Describe the degree and leading coefficient of f.
b. Describe the intervals where the function is increasing and decreasing.
c. What is the constant term of the polynomial function?
Answer:

Question 49.
REASONING A cubic polynomial function f has a leading coefficient of 2 and a constant term of −5. When f(1) = 0 and f(2) = 3, what is f(−5)? Explain your reasoning.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.1 Question 49

Question 50.
CRITICAL THINKING The weight y (in pounds) of a rainbow trout can be modeled by y = 0.000304x3, where x is the length (in inches) of the trout.
a. Write a function that relates the weight y and length x of a rainbow trout when y is measured in kilograms and x is measured in centimeters. Use the fact that 1 kilogram ≈ 2.20 pounds and 1 centimeter ≈ 0.394 inch.
b. Graph the original function and the function from part (a) in the same coordinate plane. What type of transformation can you apply to the graph of y = 0.000304x3 to produce the graph from part (a)?
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 15
Answer:

Maintaining Mathematical Proficiency

Simplify the expression. (Skills Review Handbook)

Question 51.
xy + x2 + 2xy + y2 − 3x2
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.1 Question 51

Question 52.
2h3g + 3hg3 + 7h2g2 + 5h3g+ 2hg3
Answer:

Question 53.
−wk + 3kz − 2kw + 9zk − kw
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.1 Question 53

Question 54.
a2(m − 7a3) − m(a2 − 10)
Answer:

Question 55.
3x(xy − 4) + 3(4xy + 3) − xy(x2y − 1)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.1 Question 55

Question 56.
cv(9 − 3c) + 2c(v − 4c) + 6c
Answer:

Lesson 4.2 Adding, Subtracting, and Multiplying Polynomials

Essential Question
How can you cube a binomial?

EXPLORATION 1
Cubing Binomials
Work with a partner.
Find each product. Show your steps.
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 16

EXPLORATION 2
Generalizing Patterns for Cubing a Binomial
Work with a partner.

Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 17
a. Use the results of Exploration 1 to describe a pattern for the coefficients of the terms when you expand the cube of a binomial. How is your pattern related to Pascal’s Triangle, shown at the right?
b. Use the results of Exploration 1 to describe a pattern for the exponents of the terms in the expansion of a cube of a binomial.
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 18
c. Explain how you can use the patterns you described in parts (a) and (b) to find the product (2x − 3)3. Then find this product.

Communicate Your Answer

Question 3.
How can you cube a binomial?

Question 4.
Find each product.
a. (x + 2)3
b. (x − 2)3
c. (2x − 3)3
d. (x − 3)3
e. (−2x + 3)3
f. (3x − 5)3

4.2 Lesson

Monitoring Progress

Find the sum or difference.

Question 1.
(2x2 − 6x + 5) + (7x2 − x − 9)

Answer:

Removing the  grouping symbols and combining the like terms, the given expression is

(2x2 − 6x + 5) + (7x2 − x − 9), is equivalent to

=      2x2 − 6x + 5 + 7x2 − x − 9

=   (2x2+7x2)+( − 6x− x)+( 5-9)

=   9x2-7x-4

Question 2.
(3t3 + 8t2 − t − 4) − (5t3 − t2 + 17)

Answer:

(3t3 + 8t2 − t − 4) − (5t3 − t2 + 17)

=3t3 +8t2 -t- 4- 5t3 + t2 -17     (write the opposite of the subtracted polynomial, then add like terms)

=  -2t3+ 9t2 -t -21

Find the product.

Question 3.
(4x2 + x − 5)(2x + 1)

Answer:

(4x2 + x − 5)(2x + 1)

=4x2(2x+1) + x(2x+1) – 5(2x+1)   (we multiply the polynomials in horizontal format)

=8x3+ 4x2+ 2x2+ x- 10x -5

=8x3+ 6x2– 9x -x

Question 4.
(y − 2)(5y2 + 3y − 1)

Answer:

(y − 2)(5y2 + 3y − 1)

=(y-2)5y2+ (y-2)3y – (y-2)    (we multiply the polynomials in horizontal format)

=5y3-10y2+3y3-6y-y+2

=5y3-7y2-7y+2

Question 5.
(m − 2)(m − 1)(m + 3)

Answer:

(m − 2)(m − 1)(m + 3)

=(m2-m-2m+2)(m+3)         (we multiply the first two polynomial in horizontal format , then the result

=(m2-3m+2)(m+3)                       by the last polynomial )

=(m2-3m+2)m+(m2-3m+2)3

=m3-3m2+2m+3m2-9m+6

=m3-7m+6

Question 6.
(3t − 2)(3t + 2)

Answer:

(3t − 2)(3t + 2)

we can use the formula: a2-b2= (a+ b) (a-b)

=(3t)2-22

=9t2-4

Question 7.
(5a + 2)2

Answer:

(5a + 2)2

=(5a)2+2. 5a .2+22       ( we can use the formula : (a+ b)2 = a2+2ab+b2)

=25a2+20a+4

Question 8.

(xy − 3)3

Answer:

(xy-3)3

=(x y)3-3(x y)2.3+3. xy.32-33                ( we can use the formula:(a-b)3=a3-3a2 b +3a b 2-b3

=x3y3– 9x 2y2+27xy-27

Question 9.
(a) Prove the polynomial identity for the cube of a binomial representing a difference: (a − b)3 = a3 − 3a2b + 3ab2 − b3.
(b) Use the cube of a binomial in part (a) to calculate 93.

Answer:

(a) . (a-b)3 =a3– 3a 2b+3ab2-b3               We have to prove the identity:

=   (a-b)3 =(a-b)(a-b)(a-b)                               We start from left side:

=     (a2-2ab+b2)(a-b)

=    a3-a 2b-2a 2b+2ab2 +ab 2-b3

=    a 3-3a 2b+3ab 2– b3

(b). 93=(10-1)3=103-3.102.1+3.10.12-13      

= 1000-300+30-1                         We use the above identity to compute 93

= 729

Question 10.
Use Pascal’s Triangle to expand
(a) (z + 3)4 and
(b) (2t − 1)5

Adding, Subtracting, and Multiplying Polynomials 4.2 Exercises

Vocabulary and Core Concept Check

Question 1.
WRITING Describe three different methods to expand (x + 3)3.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.2 Question 1

Question 2.
WRITING Is (a + b)(a − b) = a2 − b2 an identity? Explain your reasoning.
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3–8, find the sum.

Question 3.
(3x2 + 4x − 1) + (−2x2 − 3x + 2)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.2 Question 3

Question 4.
(−5x2 + 4x − 2) + (−8x2 + 2x + 1)
Answer:

Question 5.
(12x5 − 3x4 + 2x − 5) + (8x4 − 3x3 + 4x + 1)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.2 Question 5

Question 6.
(8x4 + 2x2 − 1) + (3x3 − 5x2 + 7x + 1)
Answer:

Question 7.
(7x6 + 2x5 − 3x2 + 9x) + (5x5 + 8x3 − 6x2 + 2x − 5)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.2 Question 7

Question 8.
(9x4 − 3x3 + 4x2 + 5x + 7) + (11x4 − 4x2 − 11x − 9)
Answer:

In Exercises 9–14, find the difference.

Question 9.
(3x3 − 2x2 + 4x − 8) − (5x3 + 12x2 − 3x − 4)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.2 Question 9

Question 10.
(7x4 − 9x3 − 4x2 + 5x + 6) − (2x4 + 3x3 − x2 + x − 4)
Answer:

Question 11.
(5x6 − 2x4 + 9x3 + 2x − 4) − (7x5 − 8x4 + 2x− 11)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.2 Question 11

Question 12.
(4x5 − 7x3 − 9x2 + 18) − (14x5 − 8x4 + 11x2 + x)
Answer:

Question 13.
(8x5 + 6x3 − 2x2 + 10x) − (9x5 − x3 − 13x2 + 4)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.2 Question 13

Question 14.
(11x4 − 9x2 + 3x + 11) − (2x4 + 6x3 + 2x − 9)
Answer:

Question 15.
MODELING WITH MATHEMATICS During a recent period of time, the numbers (in thousands) of males Mand females F that attend degree-granting institutions in the United States can be modeled by
M = 19.7t2 + 310.5t + 7539.6
F = 28t2 + 368t + 10127.8
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 19
where t is time in years. Write a polynomial to model the total number of people attending degree-granting institutions. Interpret its constant term.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.2 Question 15

Question 16.
MODELING WITH MATHEMATICS A farmer plants a garden that contains corn and pumpkins. The total area (in square feet) of the garden is modeled by the expression 2x2 + 5x + 4. The area of the corn is modeled by the expression x2 − 3x + 2. Write an expression that models the area of the pumpkins.
Answer:

In Exercises 17–24, find the product.

Question 17.
7x3(5x2 + 3x + 1)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.2 Question 17

Question 18.
−4x5(11x3 + 2x2 + 9x + 1)
Answer:

Question 19.
(5x2 − 4x + 6)(−2x + 3)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.2 Question 19

Question 20.
(−x − 3)(2x2 + 5x + 8)
Answer:

Question 21.
(x2 − 2x − 4)(x2 − 3x − 5)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.2 Question 21

Question 22.
(3x2 + x − 2)(−4x2 − 2x − 1)
Answer:

Question 23.
(3x3 − 9x + 7)(x2 − 2x + 1)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.2 Question 23

Question 24.
(4x2 − 8x − 2)(x4 + 3x2 + 4x)
Answer:

ERROR ANALYSIS In Exercises 25 and 26, describe and correct the error in performing the operation.

Question 25.
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 20
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.2 Question 25

Question 26.
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 21
Answer:

In Exercises 27–32, find the product of the binomials.

Question 27.
(x − 3)(x + 2)(x + 4)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.2 Question 27

Question 28.
(x − 5)(x + 2)(x − 6)
Answer:

Question 29.
(x − 2)(3x + 1)(4x − 3)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.2 Question 29

Question 30.
(2x + 5)(x − 2)(3x + 4)
Answer:

Question 31.
(3x − 4)(5 − 2x)(4x + 1)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.2 Question 31

Question 32.
(4 − 5x)(1 − 2x)(3x + 2)
Answer:

Question 33.
REASONING Prove the polynomial identity (a + b)(a − b) = a2 − b2. Then give an example of two whole numbers greater than 10 that can be multiplied using mental math and the given identity. Justify your answers.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.2 Question 33

Question 34.
NUMBER SENSE You have been asked to order textbooks for your class. You need to order 29 textbooks that cost $31 each. Explain how you can use the polynomial identity (a + b)(a − b) = a2 − b2 and mental math to find the total cost of the textbooks.
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 22
Answer:

In Exercises 35–42, find the product.

Question 35.
(x − 9)(x + 9)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.2 Question 35

Question 36.
(m + 6)2
Answer:

Question 37.
(3c − 5)2
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.2 Question 37

Question 38.
(2y − 5)(2y + 5)
Answer:

Question 39.
(7h + 4)2
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.2 Question 39

Question 40.
(9g − 4)2
Answer:

Question 41.
(2k + 6)3
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.2 Question 41

Question 42.
(4n − 3)3
Answer:

In Exercises 43–48, use Pascal’s Triangle to expand the binomial.

Question 43.
(2t + 4)3
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.2 Question 43

Question 44.
(6m + 2)2
Answer:

Question 45.
(2q − 3)4
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.2 Question 45

Question 46.
(g + 2)5
Answer:

Question 47.
(yz + 1)5
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.2 Question 47

Question 48.
(np − 1)4
Answer:

Question 49.
COMPARING METHODS Find the product of the expression (a2 + 4b2)2(3a2 − b2)2 using two different methods. Which method do you prefer? Explain.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.2 Question 49

Question 50.
THOUGHT PROVOKING Adjoin one or more polygons to the rectangle to form a single new polygon whose perimeter is double that of the rectangle. Find the perimeter of the new polygon.
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 23
Answer:

MATHEMATICAL CONNECTIONS In Exercises 51 and 52, write an expression for the volume of the figure as a polynomial in standard form.

Question 51.
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 24
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.2 Question 51

Question 52.
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 25
Answer:

Question 53.
MODELING WITH MATHEMATICS Two people make three deposits into their bank accounts earning the same simple interest rate r.
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 26
Person A’s account is worth 2000(1 + r)3 + 3000(1 + r)2 + 1000(1 + r) on January 1, 2015.
a. Write a polynomial for the value of Person B’s account on January 1, 2015.
b. Write the total value of the two accounts as a polynomial in standard form. Then interpret the coefficients of the polynomial.
c. Suppose their interest rate is 0.05. What is the total value of the two accounts on January 1, 2015?
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.2 Question 53.1
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.2 Question 53.2

Question 54.
PROBLEM SOLVING
The sphere is centered in the cube. Find an expression for the volume of the cube outside the sphere.
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 27
Answer:

Question 55.
MAKING AN ARGUMENT Your friend claims the sum of two binomials is always a binomial and the product of two binomials is always a trinomial. Is your friend correct? Explain your reasoning.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.2 Question 55

Question 56.
HOW DO YOU SEE IT? You make a tin box by cutting x-inch-by-x-inch pieces of tin off the corners of a rectangle and folding up each side. The plan for your box is shown.
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 28
a. What are the dimensions of the original piece of tin?
b. Write a function that represents the volume of the box. Without multiplying, determine its degree.
Answer:

USING TOOLS In Exercises 57–60, use a graphing calculator to make a conjecture about whether the two functions are equivalent. Explain your reasoning.

Question 57.
f(x) = (2x − 3)3; g(x) = 8x3 − 36x2 + 54x − 27
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.2 Question 57

Question 58.
h(x) = (x + 2)5;
k(x) = x5 + 10x4 + 40x3 + 80x2 + 64x
Answer:

Question 59.
f(x) = (−x − 3)4;
g(x) = x4 + 12x3 + 54x2 + 108x + 80
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.2 Question 59

Question 60.
f(x) = (−x + 5)3; g(x) = −x3 + 15x2 − 75x + 125
Answer:

Question 61.
REASONING Copy Pascal’s Triangle and add rows for n = 6, 7, 8, 9, and 10. Use the new rows to expand (x + 3)7 and (x − 5)9.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.2 Question 61.1
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.2 Question 61.2

Question 62.
ABSTRACT REASONING You are given the function f(x) = (x + a)(x + b)(x + c)(x + d). When f(x) is written in standard form, show that the coefficient of x3 is the sum of a, b, c, and d, and the constant term is the product of a, b, c, and d.
Answer:

Question 63.
DRAWING CONCLUSIONS Let g(x) = 12x4 + 8x + 9 and h(x) = 3x5 + 2x3 − 7x + 4.
a. What is the degree of the polynomial g(x) + h(x)?
b. What is the degree of the polynomial g(x) − h(x)?
c. What is the degree of the polynomial g(x) • h(x)?
d. In general, if g(x) and h(x) are polynomials such that g(x) has degree m and h(x) has degree n, and m > n, what are the degrees of g(x) + h(x), g(x) − h(x), and g(x) • h(x)?
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.2 Question 63.1

Question 64.
FINDING A PATTERN In this exercise, you will explore the sequence of square numbers. The first four square numbers are represented below.
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 29
a. Find the differences between consecutive square numbers. Explain what you notice.
b. Show how the polynomial identity (n + 1)2 − n2 = 2n + 1 models the differences between square numbers.
c. Prove the polynomial identity in part (b).
Answer:

Question 65.
CRITICAL THINKING Recall that a Pythagorean triple is a set of positive integers a, b, and c such that a2 + b2 = c2. The numbers 3, 4, and 5 form a Pythagorean triple because 32 + 42 = 52. You can use the polynomial identity (x2 − y2)2 + (2xy)2 = (x2 + y2)2 to generate other Pythagorean triples.
a. Prove the polynomial identity is true by showing that the simplified expressions for the left and right sides are the same. b. Use the identity to generate the Pythagorean triple when x = 6 and y = 5.
c. Verify that your answer in part (b) satisfies a2 + b2 = c2.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.2 Question 65.1
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.2 Question 65.2

Maintaining Mathematical Proficiency

Perform the operation. Write the answer in standard form. (Section 3.2)

Question 66.
(3 − 2i) + (5 + 9i)
Answer:

Question 67.
(12 + 3i) − (7 − 8i)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.2 Question 67

Question 68.
(7i)(−3i)
Answer:

Question 69.
(4 + i)(2 − i)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.2 Question 69

Lesson 4.3 Dividing Polynomials

Essential Question
How can you use the factors of a cubic polynomial to solve a division problem involving the polynomial?

EXPLORATION 1
Dividing Polynomials
Work with a partner.
Match each division statement with the graph of the related cubic polynomial f(x). Explain your reasoning. Use a graphing calculator to verify your answers.
a. \(\frac{f(x)}{x}\) = (x − 1)(x + 2)
b. \(\frac{f(x)}{x-1}\) = (x − 1)(x + 2)
c. \(\frac{f(x)}{x+1}\) = (x − 1)(x + 2)
d. \(\frac{f(x)}{x-2}\) = (x− 1)(x+ 2)
e. \(\frac{f(x)}{x+2}\) = (x− 1)(x+ 2)
f. \(\frac{f(x)}{x-3}\) = (x− 1)(x+ 2)
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 29.1

EXPLORATION 2
Dividing Polynomials

Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 30
Work with a partner. Use the results of Exploration 1 to find each quotient. Write your answers in standard form. Check your answers by multiplying.
a. (x3 + x2 − 2x) ÷ x
b. (x3 − 3x + 2) ÷ (x − 1)
c. (x3 + 2x2 − x − 2) ÷ (x + 1)
d. (x3 − x2 − 4x + 4) ÷ (x − 2)
e. (x3 + 3x2 − 4) ÷ (x + 2)
f. (x3 − 2x2 − 5x + 6) ÷ (x − 3)

Communicate Your Answer

Question 3.
How can you use the factors of a cubic polynomial to solve a division problem involving the polynomial?

4.3 Lesson

Monitoring Progress

Divide using polynomial long division.

Question 1.
(x3 − x2 − 2x + 8) ÷ (x − 1)

Question 2.
(x4 + 2x2 − x + 5) ÷ (x2 − x + 1)

Divide using synthetic division.

Question 3.
(x3 − 3x2 − 7x + 6) ÷ (x − 2)

Question 4.
(2x3 − x − 7) ÷ (x + 3)

Use synthetic division to evaluate the function for the indicated value of x.

Question 5.
f(x) = 4x2 − 10x − 21; x = 5

Question 6.
f(x) = 5x4 + 2x3 − 20x − 6; x = 2

Dividing Polynomials 4.3 Exercises

Monitoring Progress and Modeling with Mathematics

Question 1.
WRITING Explain the Remainder Theorem in your own words. Use an example in your explanation.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.3 Question 1

Question 2.
VOCABULARY What form must the divisor have to make synthetic division an appropriate method for dividing a polynomial? Provide examples to support your claim.
Answer:

Question 3.
VOCABULARY Write the polynomial divisor, dividend, and quotient functions represented by the synthetic division shown at the right.
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 31
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.3 Question 3

Question 4.
WRITING Explain what the colored numbers represent in the synthetic division in Exercise 3.
Answer:

Vocabulary and Core Concept Check

In Exercises 5–10, divide using polynomial long division.

Question 5.
(x2 + x − 17 ) ÷ (x − 4)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.3 Question 5

Question 6.
(3x2 − 14x − 5) ÷ (x − 5)
Answer:

Question 7.
(x3 + x2 + x + 2 ) ÷ (x2 − 1)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.3 Question 7

Question 8.
(7x3 + x2 + x ) ÷ (x2 + 1)

Answer:

Question 9.
(5x4 − 2x3 − 7x2 − 39) ÷ (x2 + 2x − 4)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.3 Question 9

Question 10.
(4x4 + 5x − 4) ÷ (x2 − 3x − 2)
Answer:

In Exercises 11–18, divide using synthetic division.

Question 11.
(x2 + 8x + 1) ÷ (x − 4)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.3 Question 11

Question 12.
(4x2 − 13x − 5) ÷ (x − 2)
Answer:

(4x2 − 13x − 5) ÷ (x − 2)

use synthetic division. Because the divisor is x-2 , k=2

 

 

 

Question 13.
(2x2 − x + 7) ÷ (x + 5)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.3 Question 13

Question 14.
(x3 − 4x + 6) ÷ (x + 3)
Answer:

Question 15.
(x2 + 9) ÷ (x − 3)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.3 Question 15

Question 16.
(3x3 − 5x2 − 2) ÷ (x − 1)
Answer:

Question 17.
(x4 − 5x3 − 8x2 + 13x − 12) ÷ (x − 6)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.3 Question 17

Question 18.
(x4 + 4x3 + 16x − 35) ÷ (x + 5 )
Answer:

ANALYZING RELATIONSHIPS In Exercises 19–22, match the equivalent expressions. Justify your answers.

Question 19.
(x2 + x − 3) ÷ (x − 2)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.3 Question 19

Question 20.
(x2 − x − 3) ÷ (x − 2)
Answer:

Question 21.
(x2 − x + 3) ÷ (x − 2)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.3 Question 21

Question 22.
(x2 + x + 3) ÷ (x − 2)
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 32
Answer:

ERROR ANALYSIS In Exercises 23 and 24, describe and correct the error in using synthetic division to divide x3 – 5x + 3 by x – 2.

Question 23.
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 33
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.3 Question 23

Question 24.
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 34
Answer:

In Exercises 25–32, use synthetic division to evaluate the function for the indicated value of x.

Question 25.
f(x) = −x2 − 8x + 30; x = −1
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.3 Question 25

Question 26.
f(x) = 3x2 + 2x − 20; x = 3
Answer:

Question 27.
f(x) = x3 − 2x2 + 4x + 3; x = 2
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.3 Question 27

Question 28.
f(x) = x3 + x2 − 3x + 9; x = −4
Answer:

Question 29.
f(x) = x3 − 6x + 1; x = 6
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.3 Question 29

Question 30.
f(x) = x3 − 9x − 7; x = 10
Answer:

Question 31.
f(x) = x4 + 6x2 − 7x + 1; x = 3
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.3 Question 31

Question 32.
f(x) = −x4 − x3 − 2; x = 5
Answer:

Question 33.
MAKING AN ARGUMENT You use synthetic division to divide f(x) by (x − a) and find that the remainder equals 15. Your friend concludes that f(15) = a. Is your friend correct? Explain your reasoning.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.3 Question 33

Question 34.
THOUGHT PROVOKING A polygon has an area represented by A = 4x2 + 8x + 4. The figure has at least one dimension equal to 2x + 2. Draw the figure and label its dimensions.
Answer:

Question 35.
USING TOOLS The total attendance A (in thousands) at NCAA women’s basketball games and the number T of NCAA women’s basketball teams over a period of time can be modeled by
A = −1.95x3 + 70.1x2 − 188x + 2150
T = 14.8x + 725
where x is in years and 0 < x < 18. Write a function for the average attendance per team over this period of time.
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 35
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.3 Question 35

Question 36.
COMPARING METHODS The profit P (in millions of dollars) for a DVD manufacturer can be modeled by P = −6x3 + 72x, where x is the number (in millions) of DVDs produced. Use synthetic division to show that the company yields a profit of $96 million when 2 million DVDs are produced. Is there an easier method? Explain.
Answer:

Question 37.
CRITICAL THINKING What is the value of k such that (x3 − x2 + kx − 30) ÷ (x − 5) has a remainder of zero?
A. −14
B. −2
C. 26
D. 32
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.3 Question 37

Question 38.
HOW DO YOU SEE IT? The graph represents the polynomial function f(x) = x3 + 3x2 − x − 3.
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 36
a. The expression f(x) ÷ (x − k) has a remainder of −15. What is the value of k?
b. Use the graph to compare the remainders of (x3 + 3x2 − x − 3) ÷ (x + 3) and (x3 + 3x2 − x − 3) ÷ (x + 1).
Answer:

Question 39.
MATHEMATICAL CONNECTIONS The volume V of the rectangular prism is given by V = 2x3 + 17x2 + 46x + 40. Find an expression for the missing dimension.
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 37
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.3 Question 39

Question 40.
USING STRUCTURE You divide two polynomials and obtain the result 5x2 − 13x + 47 − \(\frac{102}{x+2}\). What is the dividend? How did you find it?
Answer:

Maintaining Mathematical Proficiency

Find the zero(s) of the function. (Sections 3.1and 3.2)

Question 41.
f(x) = x2 − 6x + 9
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.3 Question 41

Question 42.
g(x) = 3(x + 6)(x − 2)
Answer:

Question 43.
g(x) = x2 + 14x + 49
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.3 Question 43

Question 44.
h(x) = 4x2 + 36

Lesson 4.4 Factoring Polynomials

Essential Question
How can you factor a polynomial?

EXPLORATION 1
Factoring Polynomials
Work with a partner.
Match each polynomial equation with the graph of its related polynomial function. Use the x-intercepts of the graph to write each polynomial in factored form. Explain your reasoning.
a. x2 + 5x + 4 = 0
b. x3 − 2x2 − x + 2 = 0
c. x3 + x2 − 2x = 0
d. x3 − x = 0
e. x4 − 5x2 + 4 = 0
f. x4 − 2x3 − x2 + 2x = 0
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 38

EXPLORATION 2
Factoring Polynomials

Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 39
Work with a partner. Use the x-intercepts of the graph of the polynomial function to write each polynomial in factored form. Explain your reasoning. Check your answers by multiplying.
a. f(x) = x2 − x − 2
b. f(x) = x3 − x2 − 2x
c. f(x) = x3 − 2x2 − 3x
d. f(x) = x3 − 3x2 − x + 3
e. f(x) = x4 + 2x3 − x2 − 2x
f. f(x) = x4 − 10x2 + 9

Communicate Your Answer

Question 3.
How can you factor a polynomial?

Question 4.
What information can you obtain about the graph of a polynomial function written in factored form?

4.4 Lesson

Monitoring Progress

Question 1.
x3 − 7x2 + 10x

Question 2.
3n7 − 75n5

Question 3.
8m5 − 16m4 + 8m3

Factor the polynomial completely

Question 4.
a3 + 27

Question 5.
6z5 − 750z2

Question 6.
x3 + 4x2 − x − 4

Question 7.
3y3 + y2 + 9y + 3

Question 8.
−16n4 + 625

Question 9.
5w6 − 25w4 + 30w2

Question 10.
Determine whether x− 4 is a factor of f(x) = 2x2 + 5x − 12.

Question 11.
Show that x − 6 is a factor of f(x) = x3 − 5x2 − 6x. Then factor f(x) completely.

Question 12.
In Example 7, does your answer change when you first determine whether 2 is a zero and then whether −1 is a zero? Justify your answer.

Factoring Polynomials 4.4 Exercises

Vocabulary and Core Concept Check

Question 1.
COMPLETE THE SENTENCE The expression 9x4 − 49 is in _________ form because it can be written as u2 − 49 where u = _____.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.4 Question 1

Question 2.
VOCABULARY Explain when you should try factoring a polynomial by grouping.
Answer:

Question 3.
WRITING How do you know when a polynomial is factored completely?
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.4 Question 3

Question 4.
WRITING Explain the Factor Theorem and why it is useful.
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 5–12, factor the polynomial completely.

Question 5.
x3 − 2x2 − 24x
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.4 Question 5

Question 6.
4k5 − 100k3
Answer:

Question 7.
3p5 − 192p3
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.4 Question 7

Question 8.
2m6 − 24m5 + 64m4
Answer:

Question 9.
2q4 + 9q3 − 18q2
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.4 Question 9

Question 10.
3r6 − 11r5 − 20r4
Answer:

Question 11.
10w10 − 19w9 + 6w8
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.4 Question 11

Question 12.
18v9 + 33v8 + 14v7
Answer:

In Exercises 13–20, factor the polynomial completely.

Question 13.
x3 + 64
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.4 Question 13

Question 14.
y3 + 512
Answer:

Question 15.
g3 − 343
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.4 Question 15

Question 16.
c3 − 27
Answer:

Question 17.
3h9 − 192h6
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.4 Question 17

Question 18.
9n6 − 6561n3
Answer:

Question 19.
16t7 + 250t4
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.4 Question 19

Question 20.
135z11 − 1080z8
Answer:

ERROR ANALYSIS In Exercises 21 and 22, describe and correct the error in factoring the polynomial.

Question 21.
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 40
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.4 Question 21

Question 22.
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 41
Answer:

In Exercises 23–30, factor the polynomial completely.

Question 23.
y3 − 5y2 + 6y − 30
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.4 Question 23

Question 24.
m3 − m2 + 7m − 72
Answer:

Question 25.
3a3 + 18a2 + 8a + 48
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.4 Question 25

Question 26.
2k3 − 20k2 + 5k − 50
Answer:

Question 27.
x3 − 8x2 − 4x + 32
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.4 Question 27

Question 28.
z3 − 5z2 − 9z + 45
Answer:

Question 29.
4q3 − 16q7 − 9q + 36
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.4 Question 29

Question 30.
16n3 + 32n7 − n − 2
Answer:

In Exercises 31–38, factor the polynomial completely.

Question 31.
49k4 − 9
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.4 Question 31

Question 32.
4m4 − 25
Answer:

Question 33.
c4 + 9c2 + 20
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.4 Question 33

Question 34.
y4 − 3y2 − 28
Answer:

Question 35.
16z4 − 81
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.4 Question 35

Question 36.
81a4 − 256
Answer:

Question 37.
3r8 + 3r5 − 60r2
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.4 Question 37

Question 38.
4n12 − 32n7 + 48n2
Answer:

In Exercises 39–44, determine whether the binomial is a factor of the polynomial.

Question 39.
f(x) = 2x3 + 5x2 − 37x − 60; x − 4
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.4 Question 39

Question 40.
g(x) = 3x3 − 28x2 + 29x + 140; x + 7
Answer:

Question 41.
h(x) = 6x5 − 15x4 − 9x3; x + 3
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.4 Question 41

Question 42.
g(x) = 8x5 − 58x4 + 60x3 + 140; x − 6
Answer:

Question 43.
h(x) = 6x4 − 6x3 − 84x2 + 144x; x + 4
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.4 Question 43

Question 44.
t(x) = 48x4 + 36x3 − 138x2 − 36x; x + 2
Answer:

In Exercises 45–50, show that the binomial is a factor of the polynomial. Then factor the polynomial completely.

Question 45.
g(x) = x3 − x2 − 20x; x + 4
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.4 Question 45

Question 46.
t(x) = x3 − 5x2 − 9x + 45; x − 5
Answer:

Question 47.
f(x) = x4 − 6x3 − 8x + 48; x − 6
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.4 Question 47

Question 48.
s(x) = x4 + 4x3 − 64x − 256; x + 4
Answer:

Question 49.
r(x) = x3 − 37x + 84; x + 7
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.4 Question 49

Question 50.
h(x) = x3 − x2 − 24x − 36; x + 2
Answer:

ANALYZING RELATIONSHIPS In Exercises 51–54, match the function with the correct graph. Explain your reasoning.

Question 51.
f(x) = (x − 2)(x − 3)(x + 1)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.4 Question 51

Question 52.
g(x) = x(x + 2)(x + 1)(x − 2)
Answer:

Question 53.
h(x) = (x + 2)(x + 3)(x − 1)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.4 Question 53

Question 54.
k(x) = x(x − 2)(x − 1)(x + 2)
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 42
Answer:

Question 55.
MODELING WITH MATHEMATICS The volume (in cubic inches) of a shipping box is modeled by V = 2x3 − 19x2 + 39x, where x is the length (in inches). Determine the values of x for which the model makes sense. Explain your reasoning.
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 43
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.4 Question 55

Question 56.
MODELING WITH MATHEMATICS The volume (in cubic inches) of a rectangular birdcage can be modeled by V = 3x3 − 17x2 + 29x − 15, where x is the length (in inches). Determine the values of x for which the model makes sense. Explain your reasoning.
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 44
Answer:

USING STRUCTURE In Exercises 57–64, use the method of your choice to factor the polynomial completely. Explain your reasoning.

Question 57.
a6 + a5 − 30a4
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.4 Question 57

Question 58.
8m3 − 343
Answer:

Question 59.
z3 − 7z2 − 9z + 63
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.4 Question 59

Question 60.
2p8 − 12p5 + 16p2
Answer:

Question 61.
64r3 + 729
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.4 Question 61

Question 62.
5x5 − 10x4 − 40x3
Answer:

Question 63.
16n4 − 1
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.4 Question 63

Question 64.
9k3 − 24k2 + 3k − 8
Answer:

Question 65.
REASONING Determine whether each polynomial is factored completely. If not, factor completely.
a. 7z4(2z2 − z − 6)
b. (2 − n)(n2 + 6n)(3n − 11)
c. 3(4y − 5)(9y2 − 6y − 4)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.4 Question 65

Question 66.
PROBLEM SOLVING The profit P (in millions of dollars) for a T-shirt manufacturer can be modeled by P = −x3 + 4x2 + x, where x is the number (in millions) of T-shirts produced. Currently the company produces 4 million T-shirts and makes a profit of $4 million. What lesser number of T-shirts could the company produce and still make the same profit?
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 44.1
Answer:

Question 67.
PROBLEM SOLVING The profit P (in millions of dollars) for a shoe manufacturer can be modeled by P = −21x3 + 46x, where x is the number (in millions) of shoes produced. The company now produces 1 million shoes and makes a profit of $25 million, but it would like to cut back production. What lesser number of shoes could the company produce and still make the same profit?
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.4 Question 67

Question 68.
THOUGHT PROVOKING Find a value of k such that \(\frac{f(x)}{x-k}\) has a remainder of 0. Justify your answer.
Answer:

Question 69.
COMPARING METHODS You are taking a test where calculators are not permitted. One question asks you to evaluate g(7) for the function g(x) = x3 − 7x2 − 4x + 28. You use the Factor Theorem and synthetic division and your friend uses direct substitution. Whose method do you prefer? Explain your reasoning.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.4 Question 69

Question 70.
MAKING AN ARGUMENT You divide f(x) by (x−a) and find that the remainder does not equal 0. Your friend concludes that f(x) cannot be factored. Is your friend correct? Explain your reasoning.
Answer:

Question 71.
CRITICAL THINKING What is the value of k such that x− 7 is a factor of h(x) = 2x3 − 13x2 − kx + 105? Justify your answer.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.4 Question 71

Question 72.
HOW DO YOU SEE IT? Use the graph to write an equation of the cubic function in factored form. Explain your reasoning.
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 45
Answer:

Question 73.
ABSTRACT REASONING Factor each polynomial completely.
a. 7ac2 + bc2 − 7ad2 − bd2
b. x2n − 2xn + 1
c. a5b2 − a2b4 + 2a4b − 2ab3 + a3 − b2
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.4 Question 73

Question 74.
REASONING The graph of the function f(x) = x4 + 3x3 + 2x2 + x + 3 is shown. Can you use the Factor Theorem to factor f(x)? Explain.
Answer:

Question 75.
MATHEMATICAL CONNECTIONS The standard equation of a circle with radius r and center (h, k) is (x − h)2 + (y − k)2 = r2. Rewrite each equation of a circle in standard form. Identify the center and radius of the circle. Then graph the circle.
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 46
a. x2 + 6x + 9 + y2 = 25
b. x2 − 4x + 4 + y2 = 9
c. x2 − 8x + 16 + y2 + 2y + 1 = 36
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.4 Question 75.1
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.4 Question 75.2
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.4 Question 75.3

Question 76.
CRITICAL THINKING Use the diagram to complete parts (a)–(c).
a. Explain why a3 − b3 is equal to the sum of the volumes of the solids I, II, and III.
b. Write an algebraic expression for the volume of each of the three solids. Leave your expressions in factored form.
c. Use the results from part (a) and part (b) to derive the factoring pattern a3 − b3.
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 47
Answer:

Maintaining Mathematical Proficiency

Solve the quadratic equation by factoring.(Section 3.1)

Question 77.
x2 − x − 30 = 0
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.4 Question 77

Question 78.
2x2 − 10x − 72 = 0
Answer:

Question 79.
3x2 − 11x + 10 = 0
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.4 Question 79

Question 80.
9x2 − 28x + 3 = 0
Answer:

Solve the quadratic equation by completing the square.(Section 3.3)

Question 81.
x2 − 12x + 36 = 144
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.4 Question 81

Question 82.
x2 − 8x − 11 = 0
Answer:

Question 83.
3x2 + 30x + 63 = 0
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.4 Question 83

Question 84.
4x2 + 36x − 4 = 0
Answer:

Polynomial Functions Study Skills : Keeping Your Mind Focused

4.1–4.4 What Did You Learn?

Core Vocabulary
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 48

Core Concepts
Section 4.1
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 49

Section 4.2
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 50

Section 4.3
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 51

Section 4.4
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 52

Mathematical Practices

Question 1.
Describe the entry points you used to analyze the function in Exercise 43 on page 164.

Question 2.
Describe how you maintained oversight in the process of factoring the polynomial in Exercise 49 on page 185.

Study Skills
Keeping Your Mind Focused

  • When you sit down at your desk, review your notes from the last class.
  • Repeat in your mind what you are writing in your notes.
  • When a mathematical concept is particularly difficult, ask your teacher for another example.

Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 52.1

Polynomial Functions 4.1 – 4.4 Quiz

Decide whether the function is a polynomial function. If so, write it in standard form and state its degree, type, and leading coefficient.(Section 4.1)

Question 1.
f(x) = 5 + 2x2 − 3x4 − 2x − x3

Question 2.
g(x) = \(\frac{1}{4}\)x3 + 2x − 3x2 + 1

Question 3.
h(x) = 3 − 6x3 + 4x − 2 + 6x

Question 4.
Describe the x-values for which
(a) f is increasing or decreasing,
(b) f(x) > 0, and
(c) f(x) < 0. (Section 4.1)
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 53

Question 5.
Write an expression for the area and perimeter for the figure shown.(Section 4.2)
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 54

Perform the indicated operation.(Section 4.2)

Question 6.
(7x2 − 4) − (3x2 − 5x + 1)

Question 7.
(x2 − 3x + 2)(3x − 1)

Question 8.
(x − 1)(x + 3)(x − 4)

Question 9.
Use Pascal’s Triangle to expand (x + 2)5. (Section 4.2)

Question 10.
Divide 4x4 − 2x3 + x2 − 5x + 8 by x2 − 2x − 1. (Section 4.3)

Factor the polynomial completely.(Section 4.4)

Question 11.
a3 − 2a2 − 8a

Question 12.
8m3 + 27

Question 13.
z3 + z2 − 4z − 4

Question 14.
49b4 − 64

Question 15.
Show that x + 5 is a factor of f(x) = x3 − 2x2 − 23x + 60. Then factor f(x) completely. (Section 4.4)

Question 16.
The estimated price P (in cents) of stamps in the United States can be modeled by the polynomial function P(t)= 0.007t3 − 0.16t2 + 1t + 17, where t represents the number of years since 1990. (Section 4.1)
a. Use a graphing calculator to graph the function for the interval 0 ≤ t ≤ 20. Describe the behavior of the graph on this interval.
b. What was the average rate of change in the price of stamps from 1990 to 2010?

Question 17.
The volume V (in cubic feet) of a rectangular wooden crate is modeled by the function V(x) = 2x3 − 11x2 + 12x, where x is the width (in feet) of the crate. Determine the values of x for which the model makes sense. Explain your reasoning. (Section 4.4)
Big Ideas Math Answers Algebra 2 Chapter 4 Polynomial Functions 55

Lesson 4.5 Solving Polynomial Equations

Essential Question
How can you determine whether a polynomial equation has a repeated solution?

EXPLORATION 1
Cubic Equations and Repeated Solutions
Work with a partner.
Some cubic equations have three distinct solutions. Others have repeated solutions. Match each cubic polynomial equation with the graph of its related polynomial function. Then solve each equation. For those equations that have repeated solutions, describe the behavior of the related function near the repeated zero using the graph or a table of values.
Big Ideas Math Answers Algebra 2 Chapter 4 Polynomial Functions 56
a. x3 − 6x2 + 12x − 8 = 0
b. x3 + 3x2 + 3x + 1 = 0
c. x3− 3x + 2 = 0
d. x3 + x2 − 2x = 0
e. x3 − 3x − 2 = 0
f. x3 − 3x2 + 2x = 0
Big Ideas Math Answers Algebra 2 Chapter 4 Polynomial Functions 57

EXPLORATION 2
Quartic Equations and Repeated Solutions
Work with a partner.
Determine whether each quartic equation has repeated solutions using the graph of the related quartic function or a table of values. Explain your reasoning. Then solve each equation.
a. x4 − 4x3 + 5x2 − 2x = 0
b. x4 − 2x3 − x2 + 2x = 0
c. x4 − 4x3 + 4x2 = 0
d. x4 + 3x3 = 0

Communicate Your Answer

Question 3.
How can you determine whether a polynomial equation has a repeated solution?

Question 4.
Write a cubic or a quartic polynomial equation that is different from the equations in Explorations 1 and 2 and has a repeated solution.

4.5 Lesson

Monitoring Progress

Solve the equation.

Question 1.
4x4 − 40x2 + 36 = 0

Question 2.
2x5 + 24x = 14x3

Find the zeros of the function. Then sketch a graph of the function.

Question 3.
f(x) = 3x4 − 6x2 + 3

Question 4.
f(x) = x3 + x2 − 6x

Question 5.
Find all real solutions of x3 − 5x2 − 2x + 24 = 0.

Question 6.
Find all real zeros of f(x) = 3x4 − 2x3 − 37x2 + 24x + 12.

Question 7.
Write a polynomial function f of least degree that has rational coefficients, aleading coefficient of 1, and the zeros 4 and 1 − \(\sqrt{5}\).

Solving Polynomial Equations 4.5 Exercises

Vocabulary and Core Concept Check

Question 1.
COMPLETE THE SENTENCE If a polynomial function f has integer coefficients, then every rational solution of f(x) = 0 has the form \(\frac{p}{q}\), where p is a factor of the _____________ and q is a factor of the _____________.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.5 Question 1

Question 2.
DIFFERENT WORDS, SAME QUESTION Which is different? Find “both” answers.
Big Ideas Math Answers Algebra 2 Chapter 4 Polynomial Functions 58
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3–12, solve the equation.

Question 3.
z3 − z2 − 12z = 0
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.5 Question 3

Question 4.
a3 − 4a2 + 4a = 0
Answer:

Question 5.
2x4 − 4x3 = −2x2
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.5 Question 5

Question 6.
v3 − 2v2 − 16v = − 32
Answer:

Question 7.
5w3 = 50w
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.5 Question 7

Question 8.
9m5 = 27m3
Answer:

Question 9.
2c4 − 6c3 = 12c2 − 36c
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.5 Question 9

Question 10.
p4 + 40 = 14p2
Answer:

Question 11.
12n2 + 48n = −n3 − 64
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.5 Question 11

Question 12.
y3 − 27 = 9y2 − 27y
Answer:

In Exercises 13–20, find the zeros of the function. Then sketch a graph of the function.

Question 13.
h(x) = x4 + x3 − 6x2
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.5 Question 13

Question 14.
f(x) = x4 − 18x2 + 81
Answer:

Question 15.
p(x) = x6 − 11x5 + 30x4
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.5 Question 15

Question 16.
g(x) = −2x5 + 2x4 + 40x3
Answer:

Question 17.
g(x) = −4x4 + 8x3 + 60x2
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.5 Question 17

Question 18.
h(x) = −x3 − 2x2 + 15x
Answer:

Question 19.
h(x) = −x3 − x2 + 9x + 9
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.5 Question 19

Question 20.
p(x) = x3 − 5x2 − 4x + 20
Answer:

Question 21.
USING EQUATIONS According to the Rational Root Theorem, which is not a possible solution of the equation 2x4 − 5x3 + 10x2 − 9 = 0?
A. −9
B. −\([\frac{1}{2}/latex]
C. [latex][\frac{5}{2}/latex]
D. 3
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.5 Question 21

Question 22.
USING EQUATIONS According to the Rational Root Theorem, which is not a possible zero of the function f(x) = 40x5 − 42x4 − 107x3 + 107x2 + 33x − 36?
A. −[latex][\frac{2}{3}/latex]
B. − [latex][\frac{3}{8}/latex]
C. [latex][\frac{3}{4}/latex]
D. [latex][\frac{4}{5}/latex]
Answer:

ERROR ANALYSIS In Exercises 23 and 24, describe and correct the error in listing the possible rational zeros of the function.

Question 23.
Big Ideas Math Answers Algebra 2 Chapter 4 Polynomial Functions 59
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.5 Question 23

Question 24.
Big Ideas Math Answers Algebra 2 Chapter 4 Polynomial Functions 60
Answer:

In Exercises 25–32, find all the real solutions of the equation.

Question 25.
x3 + x2 − 17x + 15 = 0
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.5 Question 25

Question 26.
x3 − 2x2 − 5x + 6 = 0
Answer:

Question 27.
x3 − 10x2 + 19x + 30 = 0
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.5 Question 27

Question 28.
x3 + 4x2 − 11x − 30 = 0
Answer:

Question 29.
x3 − 6x2 − 7x + 60 = 0
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.5 Question 29

Question 30.
x3 − 16x2 + 55x + 72 = 0
Answer:

Question 31.
2x3 − 3x2 − 50x − 24 = 0
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.5 Question 31

Question 32.
3x3 + x2 − 38x + 24 = 0
Answer:

In Exercises 33–38, find all the real zeros of the function.

Question 33.
f(x) = x3 − 2x2 − 23x + 60
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.5 Question 33

Question 34.
g(x) = x3 − 28x − 48
Answer:

Question 35.
h(x) = x3 + 10x2 + 31x + 30
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.5 Question 35

Question 36.
f(x) = x3 − 14x2 + 55x − 42
Answer:

Question 37.
p(x) = 2x3 −x2 − 27x + 36
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.5 Question 37

Question 38.
g(x) = 3x3 − 25x2 + 58x − 40
Answer:

USING TOOLS In Exercises 39 and 40, use the graph to shorten the list of possible rational zeros of the function. Then find all real zeros of the function.

Question 39.
f(x) = 4x3 − 20x + 16
Big Ideas Math Answers Algebra 2 Chapter 4 Polynomial Functions 61
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.5 Question 39

Question 40.
f(x) = 4x3 − 49x − 60
Big Ideas Math Answers Algebra 2 Chapter 4 Polynomial Functions 62
Answer:

In Exercises 41–46, write a polynomial function f of least degree that has a leading coefficient of 1 and the given zeros.

Question 41.
−2, 3, 6
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.5 Question 41

Question 42.
−4, −2, 5
Answer:

Question 43.
−2, 1 + [latex]\sqrt{7}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.5 Question 43

Question 44.
4, 6 − \(\sqrt{7}\)
Answer:

Question 45.
−6, 0, 3 −\(\sqrt{5}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.5 Question 45

Question 46.
0, 5, −5 + \(\sqrt{8}\)
Answer:

Question 47.
COMPARING METHODS Solve the equation x3 − 4x2 − 9x + 36 = 0 using two different methods. Which method do you prefer? Explain your reasoning.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.5 Question 47.1
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.5 Question 47.2

Question 48.
REASONING Is it possible for a cubic function to have more than three real zeros? Explain.
Answer:

Question 49.
PROBLEM SOLVING At a factory, molten glass is poured into molds to make paperweights. Each mold is a rectangular prism with a height 3 centimeters greater than the length of each side of its square base. Each mold holds 112 cubic centimeters of glass. What are the dimensions of the mold?
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.5 Question 49

Question 50.
MATHEMATICAL CONNECTIONS The volume of the cube shown is 8 cubic centimeters.
a. Write a polynomial equation that you can use to find the value of x.
b. Identify the possible rational solutions of the equation in part (a).
c. Use synthetic division to find a rational solution of the equation. Show that no other real solutions exist.
d. What are the dimensions of the cube?
Big Ideas Math Answers Algebra 2 Chapter 4 Polynomial Functions 63
Answer:

Question 51.
PROBLEM SOLVING Archaeologists discovered a huge hydraulic concrete block at the ruins of Caesarea with a volume of 945 cubic meters. The block is x meters high by 12x − 15 meters long by 12x − 21 meters wide. What are the dimensions of the block?
Big Ideas Math Answers Algebra 2 Chapter 4 Polynomial Functions 64
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.5 Question 51

Question 52.
MAKING AN ARGUMENT Your friend claims that when a polynomial function has a leading coefficient of 1 and the coefficients are all integers, every possible rational zero is an integer. Is your friend correct? Explain your reasoning.
Answer:

Question 53.
MODELING WITH MATHEMATICS During a 10-year period, the amount (in millions of dollars) of athletic equipment Esold domestically can be modeled by E(t) = −20t3 + 252t2 − 280t + 21,614, where t is in years.
a. Write a polynomial equation to find the year when about $24,014,000,000 of athletic equipment is sold.
b. List the possible whole-number solutions of the equation in part (a). Consider the domain when making your list of possible solutions.
c. Use synthetic division to find when $24,014,000,000 of athletic equipment is sold.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.5 Question 53

Question 54.
THOUGHT PROVOKING Write a third or fourth degree polynomial function that has zeros at ± \(\frac{3}{4}\). Justify your answer.
Answer:

Question 55.
MODELING WITH MATHEMATICS You are designing a marble basin that will hold a fountain for a city park. The sides and bottom of the basin should be 1 foot thick. Its outer length should be twice its outer width and outer height. What should the outer dimensions of the basin be if it is to hold 36 cubic feet of water?
Big Ideas Math Answers Algebra 2 Chapter 4 Polynomial Functions 65
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.5 Question 55

Question 56.
HOW DO YOU SEE IT? Use the information in the graph to answer the questions.
Big Ideas Math Answers Algebra 2 Chapter 4 Polynomial Functions 66
a. What are the real zeros of the function f ?
b. Write an equation of the quartic function in factored form.
Answer:

Question 57.
REASONING Determine the value of k for each equation so that the given x-value is a solution.
a. x3 − 6x2 − 7x + k = 0; x = 4
b. 2x3 + 7x2 − kx − 18 = 0; x = −6
c. kx3 − 35x2 + 19x + 30 = 0; x = 5
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.5 Question 57

Question 58.
WRITING EQUATIONS Write a polynomial function gof least degree that has rational coefficients, a leading coefficient of 1, and the zeros −2 + \(\sqrt{7}\) and 3 + \(\sqrt{2}\).
Answer:

In Exercises 59–62, solve f(x) = g(x) by graphing and algebraic methods.

Question 59.
f(x) = x3 + x2 − x − 1; g(x) = −x + 1
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.5 Question 59

Question 60.
f(x) = x4 − 5x3 + 2x2 + 8x; g(x) = −x2 + 6x − 8
Answer:

Question 61.
f(x) = x3 − 4x2 + 4x; g(x) = −2x + 4
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.5 Question 61

Question 62.
f(x) = x4 + 2x3 − 11x2 − 12x + 36;
g(x) = −x2 − 6x − 9
Answer:

Question 63.
MODELING WITH MATHEMATICS You are building a pair of ramps for a loading platform. The left ramp is twice as long as the right ramp. If 150 cubic feet of concrete are used to build the ramps, what are the dimensions of each ramp?
Big Ideas Math Answers Algebra 2 Chapter 4 Polynomial Functions 67
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.5 Question 63

Question 64.
MODELING WITH MATHEMATICS Some ice sculptures are made by filling a mold and then freezing it. You are making an ice mold for a school dance. It is to be shaped like a pyramid with a height 1 foot greater than the length of each side of its square base. The volume of the ice sculpture is 4 cubic feet. What are the dimensions of the mold?
Big Ideas Math Answers Algebra 2 Chapter 4 Polynomial Functions 68
Answer:

Question 65.
ABSTRACT REASONING Let an be the leading coefficient of a polynomial function f and a0 be the constant term. If an has r factors and a0 has s factors, what is the greatest number of possible rational zeros of f that can be generated by the Rational Zero Theorem? Explain your reasoning.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.5 Question 65

Maintaining Mathematical Proficiency

Decide whether the function is a polynomial function. If so, write it in standard form and state its degree, type, and leading coefficient.(Section 4.1)

Question 66.
h(x) = −3x2 + 2x − 9 + \(\sqrt{4}\)x3
Answer:

Question 67.
g(x) = 2x3 − 7x2 − 3x − 1 + x
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.5 Question 67

Question 68.
f(x) = \(\frac{1}{3}\)x2 + 2x3 − 4x4 − \(\sqrt{3}\)
Answer:

Question 69.
p(x) = 2x − 5x3 + 9x2 + \(\sqrt[4]{x}\) + 1
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.5 Question 69

Find the zeros of the function.(Section 3.2)

Question 70.
f(x) = 7x2 + 42
Answer:

Question 71.
g(x) = 9x2 + 81
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.5 Question 71

Question 72.
h(x) = 5x2 + 40
Answer:

Question 73.
f(x) = 8x2 − 1
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.5 Question 73

Lesson 4.6 The Fundamental Theorem of Algebra

Essential Question

How can you determine whether a polynomial equation has imaginary solutions?

EXPLORATION 1
Cubic Equations and Imaginary Solutions
Work with a partner.
Match each cubic polynomial equation with the graph of its related polynomial function. Then find all solutions. Make a conjecture about how you can use a graph or table of values to determine the number and types of solutions of a cubic polynomial equation.
a. x3 − 3x2 + x + 5 = 0
b. x3 − 2x2 − x + 2 = 0
c. x3 − x2 − 4x + 4 = 0
d. x3 + 5x2 + 8x + 6 = 0
e. x3 − 3x2 + x − 3 = 0
f. x3 − 3x2 + 2x = 0
Big Ideas Math Answers Algebra 2 Chapter 4 Polynomial Functions 69

EXPLORATION 2
Quartic Equations and Imaginary Solutions
Work with a partner.
Use the graph of the related quartic function, or a table of values, to determine whether each quartic equation has imaginary solutions. Explain your reasoning. Then find all solutions.
Big Ideas Math Answers Algebra 2 Chapter 4 Polynomial Functions 70
a. x4 − 2x3 − x2 + 2x = 0
b. x4 − 1 = 0
c. x4 + x3 − x − 1 = 0
d. x4 − 3x3 + x2 + 3x − 2 = 0

Communicate Your Answer

Question 3.
How can you determine whether a polynomial equation has imaginary solutions?

Question 4.
Is it possible for a cubic equation to have three imaginary solutions? Explain your reasoning.

4.6 Lesson

Monitoring Progress

Question 1.
How many solutions does the equation x4 + 7x2 − 144 = 0 have?

Question 2.
How many zeros does the function f(x) = x3 − 5x2 − 8x + 48 have? Find all zeros of the polynomial function.

Question 3.
f(x) = x3 + 7x2 + 16x + 12

Question 4.
f(x) = x5 − 3x4 + 5x3 − x2 − 6x + 4

Write a polynomial function f of least degree that has rational coefficients, a leading coefficient of 1, and the given zeros.

Question 5.
−1, 4i

Question 6.
3, 1 + i\(\sqrt{5}\)

Question 7.
\(\sqrt{2}\), 1 − 3i

Question 8.
2, 2i, 4 − \(\sqrt{6}\)

Determine the possible numbers of positive real zeros, negative real zeros, and imaginary zeros for the function.

Question 9.
f(x) = x3 + 9x − 25

Question 10.
f(x) = 3x4 − 7x3 + x2 − 13x + 8

Question 11.
WHAT IF? In Example 5, what is the tachometer reading when the boat travels 20 miles per hour?

The Fundamental Theorem of Algebra 4.6 Exercises

Vocabulary and Core Concept Check

Question 1.
COMPLETE THE SENTENCE The expressions 5 + i and 5 − i are _____________.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.6 Question 1

Question 2.
WRITING How many solutions does the polynomial equation (x + 8)3(x − 1) = 0 have? Explain.
Answer:

Monitoring Progress and Modeling with Mathematics

Question 3.
x4 + 2x3 − 4x2 + x = 0
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.6 Question 3

Question 4.
5y3 − 3y2 + 8y = 0
Answer:

Question 5.
9t6 − 14t3 + 4t − 1 = 0
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.6 Question 5

Question 6.
f(z) = −7z4 + z2 − 25
Answer:

Question 7.
g(s) = 4s5 − s3 + 2s7 − 2
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.6 Question 7

Question 8.
h(x) = 5x4 + 7x8 − x12
Answer:

In Exercises 9–16, find all zeros of the polynomial function.

Question 9.
f(x) = x4 − 6x3 + 7x2 + 6x − 8
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.6 Question 9

Question 10.
f(x) = x4 + 5x3 − 7x2 − 29x + 30
Answer:

Question 11.
g(x) = x4 − 9x2 − 4x + 12
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.6 Question 11

Question 12.
h(x) = x3 + 5x2 − 4x − 20
Answer:

Question 13.
g(x) = x4 + 4x3 + 7x2 + 16x + 12
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.6 Question 13

Question 14.
h(x) = x4 − x3 + 7x2 − 9x − 18
Answer:

Question 15.
g(x) = x5 + 3x4 − 4x3 − 2x2 − 12x − 16
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.6 Question 15

Question 16.
f(x) = x5 − 20x3 + 20x2 − 21x + 20
Answer:

ANALYZING RELATIONSHIPS In Exercises 17–20, determine the number of imaginary zeros for the function with the given degree and graph. Explain your reasoning.

Question 17.
Degree: 4
Big Ideas Math Answers Algebra 2 Chapter 4 Polynomial Functions 70.1
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.6 Question 17

Question 18.
Degree: 5
Big Ideas Math Answers Algebra 2 Chapter 4 Polynomial Functions 71
Answer:

Question 19.
Degree: 2
Big Ideas Math Answers Algebra 2 Chapter 4 Polynomial Functions 72
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.6 Question 19

Question 20.
Degree: 3
Big Ideas Math Answers Algebra 2 Chapter 4 Polynomial Functions 73
Answer:

In Exercises 21–28, write a polynomial function f of least degree that has rational coefficients, a leading coefficient of 1, and the given zeros.

Question 21.
−5, −1, 2
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.6 Question 21

Question 22.
−2, 1, 3
Answer:

Question 23.
3, 4 + i
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.6 Question 23

Question 24.
2, 5 − i
Answer:

Question 25.
4, −\(\sqrt{5}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.6 Question 25

Question 26.
3i, 2 − i
Answer:

Question 27.
2, 1 + i, 2 −\(\sqrt{3}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.6 Question 27

Question 28.
3, 4 + 2i, 1 + \(\sqrt{7}\)
Answer:

ERROR ANALYSIS In Exercises 29 and 30, describe and correct the error in writing a polynomial function with rational coefficients and the given zero(s).

Question 29.
Zeros: 2, 1 + i
Big Ideas Math Answers Algebra 2 Chapter 4 Polynomial Functions 74
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.6 Question 29

Question 30.
Zero: 2 +i
Big Ideas Math Answers Algebra 2 Chapter 4 Polynomial Functions 75
Answer:

Question 31.
OPEN-ENDED Write a polynomial function of degree 6 with zeros 1, 2, and −i. Justify your answer.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.6 Question 31

Question 32.
REASONING Two zeros of f(x) = x3 − 6x2 − 16x + 96 are 4 and −4. Explain why the third zero must also be a real number.
Answer:

In Exercises 33–40, determine the possible numbers of positive real zeros, negative real zeros, and imaginary zeros for the function.

Question 33.
g(x) = x4 − x2 − 6
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.6 Question 33

Question 34.
g(x) = −x3 + 5x2 + 12
Answer:

Question 35.
g(x) = x3 − 4x2 + 8x + 7
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.6 Question 35

Question 36.
g(x) = x5 − 2x3 − x2 + 6
Answer:

Question 37.
g(x) = x5 − 3x3 + 8x − 10
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.6 Question 37

Question 38.
g(x) = x5 + 7x4 − 4x3 − 3x2 + 9x − 15
Answer:

Question 39.
g(x) = x6 + x5 − 3x4 + x3 + 5x2 + 9x − 18
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.6 Question 39

Question 40.
g(x) = x7 + 4x4 − 10x + 25
Answer:

Question 41.
REASONING Which is not a possible classification of zeros for f(x) = x5 − 4x3 + 6x2 + 2x − 6? Explain.
A. three positive real zeros, two negative real zeros, and zero imaginary zeros
B. three positive real zeros, zero negative real zeros, and two imaginary zeros
C. one positive real zero, four negative real zeros, and zero imaginary zeros
D. one positive real zero, two negative real zeros, and two imaginary zeros
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.6 Question 41

Question 42.
USING STRUCTURE Use Descartes’s Rule of Signs to determine which function has at least 1 positive real zero.
A. f(x) = x4 + 2x3 − 9x2 − 2x – 8
B. f(x) = x4 + 4x3 + 8x2 + 16x + 16
C. f(x) = −x4 − 5x2 − 4
D. f(x) = x4 + 4x3 + 7x2 + 12x + 12
Answer:

Question 43.
MODELING WITH MATHEMATICS From 1890 to 2000, the American Indian, Eskimo, and Aleut population P (in thousands) can be modeled by the function P = 0.004t3 − 0.24t2 + 4.9t + 243, where t is the number of years since 1890. In which year did the population first reach 722,000?
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.6 Question 43

Question 44.
MODELING WITH MATHEMATICS Over a period of 14 years, the number N of inland lakes infested with zebra mussels in a certain state can be modeled by
N = −0.0284t4 + 0.5937t3 − 2.464t2 + 8.33t − 2.5
where t is time (in years). In which year did the number of infested inland lakes first reach 120?
Big Ideas Math Answers Algebra 2 Chapter 4 Polynomial Functions 76
Answer:

Question 45.
MODELING WITH MATHEMATICS For the 12 years that a grocery store has been open, its annual revenue R (in millions of dollars) can be modeled by the function
R = 0.0001(−t4 + 12t3 − 77t2 + 600t + 13,650)where t is the number of years since the store opened. In which year(s) was the revenue $1.5 million?
Big Ideas Math Answers Algebra 2 Chapter 4 Polynomial Functions 77
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.6 Question 45

Question 46.
MAKING AN ARGUMENT Your friend claims that 2 − i is a complex zero of the polynomial function f(x) = x3 − 2x2 + 2x + 5i, but that its conjugate is not a zero. You claim that both 2 − i and its conjugate must be zeros by the Complex Conjugates Theorem. Who is correct? Justify your answer.
Answer:

Question 47.
MATHEMATICAL CONNECTIONS A solid monument with the dimensions shown is to be built using 1000 cubic feet of marble. What is the value of x?
Big Ideas Math Answers Algebra 2 Chapter 4 Polynomial Functions 78
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.6 Question 47

Question 48.
THOUGHT PROVOKING Write and graph a polynomial function of degree 5 that has all positive or negative real zeros. Label each x-intercept. Then write the function in standard form.
Answer:

Question 49.
WRITING The graph of the constant polynomial function f(x) = 2 is a line that does not have any x-intercepts. Does the function contradict the Fundamental Theorem of Algebra? Explain.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.6 Question 49

Question 50.
HOW DO YOU SEE IT? The graph represents a polynomial function of degree 6.
Big Ideas Math Answers Algebra 2 Chapter 4 Polynomial Functions 79
a. How many positive real zeros does the function have? negative real zeros? imaginary zeros?
b. Use Descartes’s Rule of Signs and your answers in part (a) to describe the possible sign changes in the coefficients of f(x).
Answer:

Question 51.
FINDING A PATTERN Use a graphing calculator to graph the function f(x) = (x + 3)n for n = 2, 3, 4, 5, 6, and 7.
a. Compare the graphs when n is even and n is odd.
b. Describe the behavior of the graph near the zero x = −3 as n increases.
c. Use your results from parts (a) and (b) to describe the behavior of the graph of g(x) = (x − 4)20 near x = 4.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.6 Question 51.1
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.6 Question 51.2

Question 52.
DRAWING CONCLUSIONS Find the zeros of each function.
f(x) = x5 − 5x + 6
g(x) = x3 − 7x + 6
h(x) = x4 + 2x3 + x2 + 8x − 12
k(x) = x5 − 3x4 − 9x3 + 25x2 − 6x
a. Describe the relationship between the sum of the zeros of a polynomial function and the coefficients of the polynomial function.
b. Describe the relationship between the product of the zeros of a polynomial function and the coefficients of the polynomial function.
Answer:

Question 53.
PROBLEM SOLVING You want to save money so you can buy a used car in four years. At the end of each summer, you deposit $1000 earned from summer jobs into your bank account. The table shows the value of your deposits over the four-year period. In the table, g is the growth factor 1 + r, where r is the annual interest rate expressed as a decimal.
Big Ideas Math Answers Algebra 2 Chapter 4 Polynomial Functions 80
a. Copy and complete the table.
b. Write a polynomial function that gives the value v of your account at the end of the fourth summer in terms of g.
c. You want to buy a car that costs about $4300. What growth factor do you need to obtain this amount? What annual interest rate do you need?
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.6 Question 53.1

Maintaining Mathematical Proficiency

Describe the transformation of f(x) = x2 represented by g. Then graph each function. (Section 2.1)

Question 54.
g(x) = −3x2
Answer:

Question 55.
g(x) = (x − 4)2 + 6
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.6 Question 55

Question 56.
g(x) = −(x − 1)2
Answer:

Question 57.
g(x) = 5(x + 4)2
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.6 Question 57

Write a function g whose graph represents the indicated transformation of the graph of f.(Sections 1.2and 2.1)

Question 58.
f(x) = x; vertical shrink by a factor of \(\frac{1}{3}\) and a reflection in the y-axis
Answer:

Question 59.
f(x) = | x + 1 ∣− 3; horizontal stretch by a factor of 9
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.6 Question 59

Question 60.
f(x) = x2; reflection in the x-axis, followed by a translation 2 units right and 7 units up
Answer:

Lesson 4.7 Tranformations of Polynomial Functions

Essential Question
How can you transform the graph of a polynomial function?

EXPLORATION 1
Transforming the Graph of a Cubic Function
Work with a partner.
The graph of the cubic function f(x) = x3 is shown. The graph of each cubic function g represents a transformation of the graph of f. Write a rule for g. Use a graphing calculator to verify your answers.
Big Ideas Math Answers Algebra 2 Chapter 4 Polynomial Functions 81
Big Ideas Math Answers Algebra 2 Chapter 4 Polynomial Functions 81.1

EXPLORATION 2
Transforming the Graph of a Quartic Function
Work with a partner.
The graph of the quartic function f(x) = x4 is shown. The graph of each quartic function g represents a transformation of the graph of f. Write a rule for g. Use a graphing calculator to verify your answers.
Big Ideas Math Answers Algebra 2 Chapter 4 Polynomial Functions 82
Big Ideas Math Answers Algebra 2 Chapter 4 Polynomial Functions 82.1

Communicate Your Answer

Question 3.
How can you transform the graph of a polynomial function?

Big Ideas Math Answers Algebra 2 Chapter 4 Polynomial Functions 83
Question 4.

Describe the transformation of f(x) = x4 represented by g(x) = (x + 1)4 + 3. Then graph g.

4.7 Lesson

Monitoring Progress

Question 1.
Describe the transformation of f(x) = x4 represented by g(x) = (x − 3)4 − 1. Then graph each function.

Question 2.
Describe the transformation of f(x) = x3 represented by g(x) = 4(x + 2)3. Then graph each function.

Question 3.
Let f(x) = x5 − 4x + 6 and g(x) = −f(x). Write a rule for g and then graph each function. Describe the graph of g as a transformation of the graph of f.

Question 4.
Let the graph of g be a horizontal stretch by a factor of 2, followed by a translation 3 units to the right of the graph of f(x) = 8x3 + 3. Write a rule for g.

Question 5.
WHAT IF? In Example 5, the height of the pyramid is 6x, and the volume (in cubic feet) is represented by V(x) = 2x3. Write a rule for W. Find and interpret W(7).

Tranformations of Polynomial Functions 4.7 Exercises

Question 1.
COMPLETE THE SENTENCE The graph of f(x) = (x + 2)3 is a ____________ translation of the graph of f(x) = x3.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.7 Question 1

Question 2.
VOCABULARY Describe how the vertex form of quadratic functions is similar to the form f(x) = a(x − h)3 + k for cubic functions.
Answer:

In Exercises 3–6, describe the transformation of f represented by g. Then graph each function.

Question 3.
f(x) = x4, g(x) = x4 + 3
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.7 Question 3

Question 4.
f(x) = x4, g(x) = (x − 5)4
Answer:

Question 5.
f(x) = x5, g(x) = (x − 2)5 − 1
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.7 Question 5

Question 6.
f(x) = x6, g(x) = (x + 1)6 − 4
Answer:

ANALYZING RELATIONSHIPS In Exercises 7–10, match the function with the correct transformation of the graph of f. Explain your reasoning.
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 84

Question 7.
y = f(x − 2)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.7 Question 7

Question 8.
y = f(x + 2) + 2
Answer:

Question 9.
y = f(x − 2) + 2
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.7 Question 9

Question 10.
y = f(x) − 2
Answer:

Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 85

In Exercises 11–16, describe the transformation of f represented by g. Then graph each function.

Question 11.
f(x) = x4, g(x) = −2x4
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.7 Question 11

Question 12.
f(x) = x6, g(x) = −3x6
Answer:

Question 13.
f(x) = x3, g(x) = 5x3 + 1
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.7 Question 13

Question 14.
f(x) = x4, g(x) = \(\frac{1}{2}\)x4 + 1
Answer:

Question 15.
f(x) = x5, g(x) = \(\frac{3}{4}\)(x + 4)5
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.7 Question 15

Question 16.
f(x) = x4, g(x) = (2x)4− 3
Answer:

In Exercises 17–20, write a rule for g and then graph each function. Describe the graph of g as a transformation of the graph of f.

Question 17.
f(x) = x4 + 1, g(x) = f(x + 2)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.7 Question 17

Question 18.
f(x) = x5 − 2x + 3, g(x) = 3f(x)
Answer:

Question 19.
f(x) = 2x3 − 2x2 + 6, g(x) = − \(\frac{1}{2}\)f(x)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.7 Question 19

Question 20.
f(x) = x4 + x3 − 1, g(x) = f(−x) − 5
Answer:

Question 21.
ERROR ANALYSIS Describe and correct the error in graphing the function g(x) = (x + 2)4 − 6
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 86
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.7 Question 21

Question 22.
ERROR ANALYSIS Describe and correct the error in describing the transformation of the graph of f(x) = x5 represented by the graph of g(x) = (3x)5− 4.
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 87
Answer:

Question 23.
f(x) = x3 − 6; translation 3 units left, followed by a reflection in the y-axis
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.7 Question 23

Question 24.
f(x) = x4 + 2x + 6; vertical stretch by a factor of 2, followed by a translation 4 units right
Answer:

Question 25.
f(x) = x3 + 2x2 − 9; horizontal shrink by a factor of \(\frac{1}{3}\) and a translation 2 units up, followed by a reflection in the x-axis
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.7 Question 25

Question 26.
f(x) = 2x5 − x3 + x2 + 4; reflection in the y-axis and a vertical stretch by a factor of 3, followed by a translation 1 unit down
Answer:

Question 27.
MODELING WITH MATHEMATICS The volume V(in cubic feet) of the pyramid is given by V(x) = x3 − 4x. The function W(x) = V(3x) gives the volume (in cubic feet) of the pyramid when x is measured in yards. Write a rule for W. Find and interpret W(5).
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 88
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.7 Question 27

Question 28.
MAKING AN ARGUMENT The volume of a cube with side length x is given by V(x) = x3. Your friend claims that when you divide the volume in half, the volume decreases by a greater amount than when you divide each side length in half. Is your friend correct? Justify your answer.
Answer:

Question 29.
OPEN-ENDED Describe two transformations of the graph of f(x) = x5 where the order in which the transformations are performed is important. Then describe two transformations where the order is not important. Explain your reasoning.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.7 Question 29

Question 30.
THOUGHT PROVOKING Write and graph a transformation of the graph of f(x) = x5 − 3x4 + 2x − 4 that results in a graph with a y-intercept of −2.
Answer:

Question 31.
PROBLEM SOLVING A portion of the path that a hummingbird flies while feeding can be modeled by the function
f(x) = −\(\frac{1}{5}\)x(x − 4)2(x − 7), 0 ≤ x ≤ 7 w
here x is the horizontal distance (in meters) and f(x) is the height (in meters). The hummingbird feeds each time it is at ground level.
a. At what distances does the hummingbird feed?
b. A second hummingbird feeds 2 meters farther away than the first hummingbird and flies twice as high. Write a function to model the path of the second hummingbird.
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 89
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.7 Question 31

Question 32.
HOW DO YOU SEE IT? Determine the real zeros of each function. Then describe the transformation of the graph of f that results in the graph of g.
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 90
Answer:

Question 33.
MATHEMATICAL CONNECTIONS Write a function V for the volume (in cubic yards) of the right circular cone shown. Then write a function W that gives the volume (in cubic yards) of the cone when x is measured in feet. Find and interpret W(3).
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 91
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.7 Question 33

Maintaining Mathematical Proficiency

Find the minimum value or maximum value of the function. Describe the domain and range of the function, and where the function is increasing and decreasing.(Section 2.2)

Question 34.
h(x) = (x + 5)2 − 7
Answer:

Question 35.
f(x) = 4 − x2
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.7 Question 35

Question 36.
f(x) = 3(x − 10)(x + 4)
Answer:

Question 37.
g(x) = −(x + 2)(x + 8)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.7 Question 37

Question 38.
h(x) = \(\frac{1}{2}\)(x − 1)2 − 3
Answer:

Question 39.
f(x) = −2x2 + 4x − 1
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.7 Question 39

Lesson 4.8 Analyzing Graphs of Polynomial Functions

Essential Question
How many turning points can the graph of a polynomial function have?
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 92
A turning point of the graph of a polynomial function is a point on the graph at which the function changes from

  • increasing to decreasing, or
  • decreasing to increasing.

EXPLORATION 1
Approximating Turning Points
Work with a partner.
Match each polynomial function with its graph. Explain your reasoning. Then use a graphing calculator to approximate the coordinates of the turning points of the graph of the function. Round your answers to the nearest hundredth.
a. f(x) = 2x2 + 3x − 4
b. f(x) = x2 + 3x + 2
c. f(x) = x3 − 2x2 − x + 1
d. f(x) = −x3 + 5x − 2
e. f(x) = x4 − 3x2 + 2x − 1
f. f(x) = −2x5 − x2 + 5x + 3
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 93
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 94

Communicate Your Answer

Question 2.
How many turning points can the graph of a polynomial function have?

Question 3.
Is it possible to sketch the graph of a cubic polynomial function that has no turning points? Justify your answer.

Monitoring Progress

Graph the function.

Question 1.
f(x) = \(\frac{1}{2}\)(x + 1)(x − 4)

Question 2.
f(x) = \(\frac{1}{4}\)(x + 2)(x − 1)(x − 3)

Question 3.
Find all real zeros of f(x) = 18x3 + 21x2 − 13x − 6.

Question 4.
Graph f(x) = 0.5x3 + x2 − x + 2. Identify the x-intercepts and the points where the local maximums and local minimums occur. Determine the intervals for which the function is increasing or decreasing.

Determine whether the function is even, odd, or neither.

Question 5.
f(x) = −x2 + 5

Question 6.
f(x) = x4 − 5x3

Question 7.
f(x) = 2x5

Analyzing Graphs of Polynomial Functions 4.8 Exercises

Vocabulary and Core Concept Check

Question 1.
COMPLETE THE SENTENCE A local maximum or local minimum of a polynomial function occurs at a ______________ point of the graph of the function.
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 4.8 a 1

Question 2.
WRITING Explain what a local maximum of a function is and how it may be different from the maximum value of the function.
Answer:

ANALYZING RELATIONSHIPS In Exercises 3–6, match the function with its graph.

Question 3.
f(x) = (x − 1)(x − 2)(x + 2)
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 4.8 a 3

Question 4.
h(x) = (x + 2)2(x + 1)
Answer:

Question 5.
g(x) = (x + 1)(x − 1)(x + 2)
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 4.8 a 5

Question 6.
f(x) = (x − 1)2(x + 2)
Answer:

Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 95

In Exercises 7–14, graph the function.

Question 7.
f(x) = (x − 2)2(x + 1)
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 4.8 a 7

Question 8.
f(x) = (x + 2)2(x + 4)2
Answer:

Question 9.
h(x) = (x + 1)2(x − 1)(x − 3)
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 4.8 a 9

Question 10.
g(x) = 4(x + 1)(x + 2)(x − 1)
Answer:

Question 11.
h(x) = \(\frac{1}{3}\)(x − 5)(x + 2)(x − 3)
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 4.8 a 11

Question 12.
g(x) = \(\frac{1}{12}\)(x + 4)(x + 8)(x − 1)
Answer:

Question 13.
h(x) = (x− 3)(x2 + x + 1)
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 4.8 a 13

Question 14.
f(x) = (x − 4)(2x2 − 2x + 1)
Answer:

ERROR ANALYSIS In Exercises 15 and 16, describe and correct the error in using factors to graph f.

Question 15.
f(x) = (x + 2)(x − 1)2
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 96
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 4.8 a 15

Question 16.
f(x) = x2(x − 3)
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 97
Answer:

In Exercises 17–22, find all real zeros of the function.

Question 17.
f(x) = x3 − 4x2 − x + 4
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 4.8 a 17

Question 18.
f(x) = x3 − 3x2 − 4x + 12
Answer:

Question 19.
h(x) = 2x3 + 7x2 − 5x − 4
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 4.8 a 19

Question 20.
h(x) = 4x3 − 2x2 − 24x − 18
Answer:

Question 21.
g(x) = 4x3 + x2 − 51x + 36
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 4.8 a 21

Question 22.
f(x) = 2x3 − 3x2 − 32x − 15
Answer:

In Exercises 23–30, graph the function. Identify the x-intercepts and the points where the local maximums and local minimums occur. Determine the intervals for which the function is increasing or decreasing.

Question 23.
g(x) = 2x3 + 8x2 − 3
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 4.8 a 23

Question 24.
g(x) = −x4 + 3x
Answer:

Question 25.
h(x) = x4 − 3x2 + x
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 4.8 a 25

Question 26.
f(x) = x5 − 4x3 + x2 + 2
Answer:

Question 27.
f(x) = 0.5x3 − 2x + 2.5
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 4.8 a 27

Question 28.
f(x) = 0.7x4 − 3x3 + 5x
Answer:

Question 29.
h(x) = x5 + 2x2 − 17x − 4
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 4.8 a 29

Question 30.
g(x) = x4 − 5x3 + 2x2 + x − 3
Answer:

In Exercises 31–36, estimate the coordinates of each turning point. State whether each corresponds to a local maximum or a local minimum. Then estimate the real zeros and find the least possible degree of the function.

Question 31.
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 98
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 4.8 a 31

Question 32.
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 99
Answer:

Question 33.
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 100
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 4.8 a 33

Question 34.
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 101
Answer:

Question 35.
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 102
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 4.8 a 35

Question 36.
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 103
Answer:

OPEN-ENDED In Exercises 37 and 38, sketch a graph of a polynomial function f having the given characteristics.

Question 37.

  • The graph of f has x-intercepts at x = −4, x = 0, and x = 2.
  • f has a local maximum value when x = 1.
  • f has a local minimum value when x = −2.

Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 4.8 a 37

Question 38.

  • The graph of f has x-intercepts at x = −3, x = 1, and x = 5.
  • f has a local maximum value when x = 1.
  • f has a local minimum value when x = −2 and when x = 4.

Answer:

In Exercises 39–46, determine whether the function is even, odd, or neither.

Question 39.
h(x) = 4x7
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 4.8 a 39

Question 40.
g(x) = −2x6 + x2
Answer:

Question 41.
f(x) = x4 + 3x2 − 2
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 4.8 a 41

Question 42.
f(x) = x5 + 3x3 − x
Answer:

Question 43.
g(x) = x2 + 5x + 1
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 4.8 a 43

Question 44.
f(x) = −x3 + 2x − 9
Answer:

Question 45.
f(x) = x4 − 12x2
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 4.8 a 45

Question 46.
h(x) = x5 + 3x4
Answer:

Question 47.
USING TOOLS When a swimmer does the breaststroke, the function
S = −241t7 + 1060t6 − 1870t5 + 1650t4 − 737t3 + 144t2 − 2.43t
models the speed S (in meters per second) of the swimmer during one complete stroke, where t is the number of seconds since the start of the stroke and 0 ≤ t ≤ 1.22. Use a graphing calculator to graph the function. At what time during the stroke is the swimmer traveling the fastest?
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 104
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 4.8 a 47

Question 48.
USING TOOLS During a recent period of time, the number S (in thousands) of students enrolled in public schools in a certain country can be modeled by S = 1.64x3 − 102x2 + 1710x + 36,300, where x is time (in years). Use a graphing calculator to graph the function for the interval 0 ≤ x ≤ 41. Then describe how the public school enrollment changes over this period of time.
Answer:

Question 49.
WRITING Why is the adjective local, used to describe the maximums and minimums of cubic functions, sometimes not required for quadratic functions?
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 4.8 a 49

Question 50.
HOW DO YOU SEE IT? The graph of a polynomial function is shown.
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 105
a. Find the zeros, local maximum, and local minimum values of the function.
b. Compare the x-intercepts of the graphs of y = f(x) and y = −f(x).
c. Compare the maximum and minimum values of the functions y = f(x) and y = −f(x).
Answer:

Question 51.
MAKING AN ARGUMENT Your friend claims that the product of two odd functions is an odd function. Is your friend correct? Explain your reasoning.
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 4.8 a 51

Question 52.
MODELING WITH MATHEMATICS You are making a rectangular box out of a 16-inch-by-20-inch piece of cardboard. The box will be formed by making the cuts shown in the diagram and folding up the sides. You want the box to have the greatest volume possible.
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 106
a. How long should you make the cuts?
b. What is the maximum volume?
c. What are the dimensions of the finished box?
Answer:

Question 53.
PROBLEM SOLVING Quonset huts are temporary, all-purpose structures shaped like half-cylinders. You have 1100 square feet of material to build a quonset hut.
a. The surface area S of a quonset hut is given by S = πr2 + πrℓ. Substitute 1100 for S and then write an expression for ℓ in terms of r.
b. The volume V of a quonset hut is given by V = 1 — 2πr2ℓ. Write an equation that gives V as a function in terms of r only.c. Find the value of r that maximizes the volume of the hut.
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 107
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 4.8 a 53

Question 54.
THOUGHT PROVOKING Write and graph a polynomial function that has one real zero in each of the intervals −2 < x < −1, 0 < x < 1, and 4 < x < 5. Is there a maximum degree that such a polynomial function can have? Justify your answer.
Answer:

Question 55.
MATHEMATICAL CONNECTIONS A cylinder is inscribed in a sphere of radius 8 inches. Write an equation for the volume of the cylinder as a function of h. Find the value of h that maximizes the volume of the inscribed cylinder. What is the maximum volume of the cylinder?
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 108
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 4.8 a 55.1
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 4.8 a 55.2

Maintaining Mathematical Proficiency

State whether the table displays linear data, quadratic data, or neither. Explain

Question 56.
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 109
Answer:

Question 57.
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 110
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 4.8 a 57

Lesson 4.9 Modeling with Polynomial Functions

Essential Question
How can you find a polynomial model for real-life data?

EXPLORATION 1
Modeling Real-Life Data
Work with a partner.
The distance a baseball travels after it is hit depends on the angle at which it was hit and the initial speed. The table shows the distances a baseball hit at an angle of 35° travels at various initial speeds.
Big Ideas Math Algebra 2 Solutions Chapter 4 Polynomial Functions 111
a. Recall that when data have equally-spaced x-values, you can analyze patterns in the differences of the y-values to determine what type of function can be used to model the data. If the first differences are constant, then the set of data fits a linear model. If the second differences are constant, then the set of data fits a quadratic model.Find the first and second differences of the data. Are the data linear or quadratic? Explain your reasoning.
Big Ideas Math Algebra 2 Solutions Chapter 4 Polynomial Functions 112
Big Ideas Math Algebra 2 Solutions Chapter 4 Polynomial Functions 113
b. Use a graphing calculator to draw a scatter plot of the data. Do the data appear linear or quadratic? Use the regression feature of the graphing calculator to find a linear or quadratic model that best fits the data.12019075400
Big Ideas Math Algebra 2 Solutions Chapter 4 Polynomial Functions 113.1
c. Use the model you found in part (b) to find the distance a baseball travels when it is hit at an angle of 35° and travels at an initial speed of 120 miles per hour.
d. According to the Baseball Almanac, “Any drive over400 feet is noteworthy. A blow of 450 feet shows exceptional power, as the majority of major league players are unable to hit a ball that far. Anything in the 500-foot range is genuinely historic.” Estimate the initial speed of a baseball that travels a distance of 500 feet.

Communicate Your Answer

Question 2.
How can you find a polynomial model for real-life data?

Question 3.
How well does the model you found in Exploration 1(b) fit the data? Do you think the model is valid for any initial speed? Explain your reasoning.

4.9 Lesson

Monitoring Progress

write a cubic function whose graph passes through the given points. 

Question 1.
(−4, 0), (0, 10), (2, 0), (5, 0)

Question 2.
(−1, 0), (0, −12), (2, 0), (3, 0)

Question 3.
Use finite differences to determine the degree of the polynomial function that fits the data. Then use technology to find the polynomial function.
Big Ideas Math Algebra 2 Solutions Chapter 4 Polynomial Functions 118

Use a graphing calculator to find a polynomial function that fits the data.

Question 4.
Big Ideas Math Algebra 2 Solutions Chapter 4 Polynomial Functions 119

Question 5.
Big Ideas Math Algebra 2 Solutions Chapter 4 Polynomial Functions 120

Modeling with polynomial Functions 4.9 Exercises

Vocabulary and Core Concept Check

Question 1.
COMPLETE THE SENTENCE When the x-values in a set of data are equally spaced, the differences of consecutive y-values are called ________________.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.9 a 1

Question 2.
WRITING Explain how you know when a set of data could be modeled by a cubic function.
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3–6, write a cubic function whose graph is shown.

Question 3.
Big Ideas Math Algebra 2 Solutions Chapter 4 Polynomial Functions 121
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.9 a 3

Question 4.
Big Ideas Math Algebra 2 Solutions Chapter 4 Polynomial Functions 122
Answer:

Question 5.
Big Ideas Math Algebra 2 Solutions Chapter 4 Polynomial Functions 123
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.9 a 5

Question 6.
Big Ideas Math Algebra 2 Solutions Chapter 4 Polynomial Functions 124
Answer:

In Exercises 7–12, use finite differences to determine the degree of the polynomial function that fits the data. Then use technology to find the polynomial function.

Question 7.
Big Ideas Math Algebra 2 Solutions Chapter 4 Polynomial Functions 125
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.9 a 7

Question 8.
Big Ideas Math Algebra 2 Solutions Chapter 4 Polynomial Functions 126
Answer:

Question 9.
(−4, −317), (−3, −37), (−2, 21), (−1, 7), (0, −1), (1, 3), (2, −47), (3, −289), (4, −933)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.9 a 9

Question 10.
(−6, 744), (−4, 154), (−2, 4), (0, −6), (2, 16), (4, 154), (6, 684), (8, 2074), (10, 4984)
Answer:

Question 11.
(−2, 968), (−1, 422), (0, 142), (1, 26), (2, −4), (3, −2), (4, 2), (5, 2), (6, 16)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.9 a 11

Question 12.
(1, 0), (2, 6), (3, 2), (4, 6), (5, 12), (6, −10), (7, −114), (8, −378), (9, −904)
Answer:

Question 13.
ERROR ANALYSIS Describe and correct the error in writing a cubic function whose graph passes through the given points.
Big Ideas Math Algebra 2 Solutions Chapter 4 Polynomial Functions 127
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.9 a 13

Question 14.
MODELING WITH MATHEMATICS The dot patterns show pentagonal numbers. The number of dots in the nth pentagonal number is given by f(n) = \(\frac{1}{2}\)n(3n − 1). Show that this function has constant second-order differences.
Big Ideas Math Algebra 2 Solutions Chapter 4 Polynomial Functions 128
Answer:

Question 15.
OPEN-ENDED Write three different cubic functions that pass through the points (3, 0), (4, 0), and (2, 6). Justify your answers.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.9 a 15

Question 16.
MODELING WITH MATHEMATICS The table shows the ages of cats and their corresponding ages in human years. Find a polynomial model for the data for the first 8 years of a cat’s life. Use the model to estimate the age (in human years) of a cat that is 3 years old.
Big Ideas Math Algebra 2 Solutions Chapter 4 Polynomial Functions 129
Answer:

Question 17.
MODELING WITH MATHEMATICS The data in the table show the average speeds y (in miles per hour) of a pontoon boat for several different engine speeds x (in hundreds of revolutions per minute, or RPMs). Find a polynomial model for the data. Estimate the average speed of the pontoon boat when the engine speed is 2800 RPMs.
Big Ideas Math Algebra 2 Solutions Chapter 4 Polynomial Functions 130
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.9 a 17

Question 18.
HOW DO YOU SEE IT? The graph shows typical speeds y (in feet per second) of a space shuttle x seconds after it is launched.
Big Ideas Math Algebra 2 Solutions Chapter 4 Polynomial Functions 131
a. What type of polynomial function models the data? Explain.
b. Which nth-order finite difference should be constant for the function in part (a)? Explain.
Answer:

Question 19.
MATHEMATICAL CONNECTIONS The table shows the number of diagonals for polygons with n sides. Find a polynomial function that fits the data. Determine the total number of diagonals in the decagon shown.
Big Ideas Math Algebra 2 Solutions Chapter 4 Polynomial Functions 132
Big Ideas Math Algebra 2 Solutions Chapter 4 Polynomial Functions 133
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.9 a 19

Question 20.
MAKING AN ARGUMENT Your friend states that it is not possible to determine the degree of a function given the first-order differences. Is your friend correct? Explain your reasoning.
Answer:

Question 21.
WRITING Explain why you cannot always use finite differences to find a model for real-life data sets.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.9 a 21

Question 22.
THOUGHT PROVOKING A, B, and C are zeros of a cubic polynomial function. Choose values for A, B, and C such that the distance from A to B is less than or equal to the distance from A to C. Then write the function using the A, B, and C values you chose.
Answer:

Question 23.
MULTIPLE REPRESENTATIONS Order the polynomial functions according to their degree, from least to greatest.
A.f(x) = −3x + 2x2 + 1
Big Ideas Math Algebra 2 Solutions Chapter 4 Polynomial Functions 134
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.9 a 23

Question 24.
ABSTRACT REASONING Substitute the expressions z, z + 1, z + 2, …….. , z + 5 for x in the function f(x) = ax3 + bx2 + cx + d to generate six equally-spaced ordered pairs. Then show that the third-order differences are constant.
Answer:

Maintaining Mathematical Proficiency

Solve the equation using square roots.(Section 3.1)

Question 25.
x2 − 6 = 30
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.9 a 25

Question 26.
5x2 − 38 = 187
Answer:

Question 27.
2(x − 3)2 = 24
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.9 a 27

Question 28.
\(\frac{4}{3}\)(x + 5)2 = 4
Answer:

Solve the equation using the Quadratic Formula.(Section 3.4)

Question 29.
2x2 + 3x = 5
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.9 a 29

Question 30.
2x2 + \(\frac{1}{2}\) = 2x
Answer:

Question 31.
2x2 + 3x =−3x2 + 1
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.9 a 31

Question 32.
4x − 20 = x2
Answer:

Polynomial Functions Performance Task: For the Birds-Wildlife Management

Core Vocabulary
Big Ideas Math Algebra 2 Solutions Chapter 4 Polynomial Functions 135

Core Concepts
Section 4.5
Big Ideas Math Answer Key Algebra 2 Chapter 4 Polynomial Functions 136

Section 4.6
Big Ideas Math Answer Key Algebra 2 Chapter 4 Polynomial Functions 137

Section 4.7

Big Ideas Math Answer Key Algebra 2 Chapter 4 Polynomial Functions 138

Section 4.8
Big Ideas Math Answer Key Algebra 2 Chapter 4 Polynomial Functions 139

Section 4.9
Big Ideas Math Answer Key Algebra 2 Chapter 4 Polynomial Functions 140

Mathematical Practices

Question 1.
Explain how understanding the Complex Conjugates Theorem allows you to construct your argument in Exercise 46 on page 203.

Question 2.
Describe how you use structure to accurately match each graph with its transformation in Exercises 7–10 on page 209.

Performance Task
For the Birds -Wildlife Management
How does the presence of humans affect the population of sparrows in a park? Do more humans mean fewer sparrows? Or does the presence of humans increase the number of sparrows up to a point? Are there a minimum number of sparrows that can be found in a park, regardless of how many humans there are? What can a mathematical model tell you?
Big Ideas Math Algebra 2 Solutions Chapter 4 Polynomial Functions 204
To explore the answers to these questions and more, go to BigIdeasMath.com.

Polynomial Functions Chapter Review

Decide whether the function is a polynomial function. If so, write it in standard form and state its degree, type, and leading coefficient.

Question 1.
h(x) = −x3 + 2x2 − 15x7

Question 2.
p(x) = x3 − 5x0.5 + 13x2 + 8

Graph the polynomial function.

Question 3.
h(x) = x2 + 6x5 − 5

Question 4.
f(x) = 3x4 − 5x2 + 1

Question 5.
g(x) = −x4 + x + 2

Find the sum or difference.

Question 6.
(4x3 − 12x2 − 5) − (−8x2 + 4x + 3)

Question 7.
(x4 + 3x3 − x2 + 6) + (2x4 − 3x + 9)

Question 8.
(3x2 + 9x + 13) − (x2 − 2x + 12)

Find the product.

Question 9.
(2y2 + 4y − 7)(y + 3)

Question 10.
(2m + n)3

Question 11.
(s + 2)(s + 4)(s − 3)

Use Pascal’s Triangle to expand the binomial.

Question 12.
(m + 4)4

Question 13.
(3s + 2)5

Question 14.
(z + 1)6

Divide using polynomial long division or synthetic division.

Question 15.
(x3 + x2 + 3x − 4) ÷ (x2 + 2x + 1)

Question 16.
(x4 + 3x3 − 4x2 + 5x + 3) ÷ (x2 + x + 4)

Question 17.
(x4 − x2 − 7) ÷ (x + 4)

Question 18.
Use synthetic division to evaluate g(x) = 4x3 + 2x2 − 4 when x = 5.

Factor the polynomial completely.

Question 19.
64x3 − 8

Question 20.
2z5 − 12z3 + 10z

Question 21.
2a3 − 7a2 − 8a + 28

Question 22.
Show that x + 2 is a factor of f(x) = x4 + 2x3 − 27x − 54. Then factor f(x) completely.

Find all real solutions of the equation.

Question 23.
x3 + 3x2 − 10x − 24 = 0

Question 24.
x3 + 5x2 − 2x − 24 = 0

Write a polynomial function f of least degree that has rational coefficients, a leading coefficient of 1, and the given zeros.

Question 25.
1, 2 − \(\sqrt{3}\)

Question 26.
2, 3, \(\sqrt{5}\)

Question 27.
−2, 5, 3 + \(\sqrt{6}\)

Question 28.
You use 240 cubic inches of clay to make a sculpture shaped as a rectangular prism. The width is 4 inches less than the length and the height is 2 inches more than three times the length. What are the dimensions of the sculpture? Justify your answer.

Write a polynomial function f of least degree that has rational coefficients, a leading coefficient of 1, and the given zeros.

Question 29.
3, 1 + 2i

Question 30.
−1, 2, 4i

Question 31.
−5, −4, 1 −i\(\sqrt{3}\)

Determine the possible numbers of positive real zeros, negative real zeros, and imaginary zeros for the function.

Question 32.
f(x) = x4 − 10x + 8

Question 33.
f(x) = −6x4 − x3 + 3x2 + 2x + 18

Describe the transformation of f represented by g. Then graph each function.

Question 34.
f(x) = x3, g(x) = (−x)3 + 2

Question 35.
f(x) = x4, g(x) = −(x + 9)4

Write a rule for g.

Question 36.
Let the graph of g be a horizontal stretch by a factor of 4, followed by a translation 3 units right and 5 units down of the graph of f(x) = x5 + 3x.

Question 37.
Let the graph of g be a translation 5 units up, followed by a reflection in the y-axis of the graph of f(x) = x4 − 2x3 − 12.

Graph the function. Identify the x-intercepts and the points where the local maximums and local minimums occur. Determine the intervals for which the function is increasing or decreasing.

Question 38.
f(x) = −2x3 − 3x2 − 1

Question 39.
f(x) = x4 + 3x3 − x2 − 8x + 2

Determine whether the function is even, odd, or neither.

Question 40.
f(x) = 2x3 + 3x

Question 41.
g(x) = 3x2 − 7

Question 42.
h(x) = x6 + 3x5

Question 43.
Write a cubic function whose graph passes through the points (−4, 0), (4, 0), (0, 6), and (2, 0).

Question 44.
Use finite differences to determine the degree of the polynomial function that fits the data. Then use technology to find the polynomial function.
Big Ideas Math Algebra 2 Solutions Chapter 4 Polynomial Functions 206

Polynomial Functions Chapter Test

Write a polynomial function f of least degree that has rational coefficients, a leading coefficient of 1, and the given zeros.

Question 1.
3, 1 − \(\sqrt{2}\)

Question 2.
−2, 4, 3i

Find the product or quotient.

Question 3.
(x6 − 4)(x2 − 7x + 5)

Question 4.
(3x4 − 2x3 − x − 1) ÷ (x2 − 2x + 1)

Question 5.
(2x3 − 3x2 + 5x − 1) ÷ (x + 2)

Question 6.
(2x + 3)3

Question 7.
The graphs of f(x) = x4 and g(x) = (x − 3)4 are shown.
a. How many zeros does each function have? Explain.
b. Describe the transformation of f represented by g.
c. Determine the intervals for which the function g is increasing or decreasing.
Big Ideas Math Algebra 2 Solutions Chapter 4 Polynomial Functions 207

Question 8.
The volume V (in cubic feet) of an aquarium is modeled by the polynomial function V(x) = x3 + 2x2 − 13x + 10, where x is the length of the tank.
a. Explain how you know x = 4 is not a possible rational zero.
b. Show that x − 1 is a factor of V(x). Then factor V(x) completely.
c. Find the dimensions of the aquarium shown.
Big Ideas Math Algebra 2 Solutions Chapter 4 Polynomial Functions 208

Question 9.
One special product pattern is (a − b)2 = a2 − 2ab + b2. Using Pascal’s Triangle to expand (a − b)2 gives 1a2 + 2a(−b) + 1(−b)2. Are the two expressions equivalent? Explain.

Question 10.
Can you use the synthetic division procedure that you learned in this chapter to divide any two polynomials? Explain.

Question 11.
Let T be the number (in thousands) of new truck sales. Let C be the number (in thousands) of new car sales. During a 10-year period, T and C can be modeled by the following equations where t is time (in years).
T = 23t4 − 330t3 + 3500t2 − 7500t + 9000
C = 14t4 − 330t3 + 2400t2 − 5900t + 8900
a. Find a new model S for the total number of new vehicle sales.
b. Is the function S even, odd, or neither? Explain your reasoning.

Question 12.
Your friend has started a golf caddy business. The table shows the profits p (in dollars) of the business in the first 5 months. Use finite differences to find a polynomial model for the data. Then use the model to predict the profit after 7 months.
Big Ideas Math Algebra 2 Solutions Chapter 4 Polynomial Functions 209

Polynomial Functions Cumulative Assessment

Question 1.
The synthetic division below represents f(x) ÷ (x− 3). Choose a value for m so that x − 3 is a factor of f(x). Justify your answer.
Big Ideas Math Algebra 2 Solutions Chapter 4 Polynomial Functions 210

Question 2.
Analyze the graph of the polynomial function to determine the sign of the leading coefficient, the degree of the function, and the number of real zeros. Explain.
Big Ideas Math Algebra 2 Solutions Chapter 4 Polynomial Functions 211

Question 3.
Which statement about the graph of the equation 12(x− 6) = −( y + 4)2 is not true?
A. The vertex is (6, −4).
B. The axis of symmetry is y = −4.
C. The focus is (3, −4).
D. The graph represents a function.

Question 4.
A parabola passes through the point shown in the graph. The equation of the axis of symmetry is x = −a. Which of the given points could lie on the parabola? If the axis of symmetry was x = a, then which points could lie on the parabola? Explain your reasoning.
Big Ideas Math Algebra 2 Solutions Chapter 4 Polynomial Functions 212

Question 5.
Select values for the function to model each transformation of the graph of f(x) = x.
Big Ideas Math Algebra 2 Solutions Chapter 4 Polynomial Functions 213
a. The graph is a translation 2 units up and 3 units left.
b. The graph is a translation 2 units right and 3 units down.
c. The graph is a vertical stretch by a factor of 2, followed by a translation 2 units up.
d. The graph is a translation 3 units right and a vertical shrink by a factor of \(\frac{1}{2}\), followed by a translation 4 units down.

Question 6.
The diagram shows a circle inscribed in a square. The area of the shaded region is21.5 square meters. To the nearest tenth of a meter, how long is each side of the square?
Big Ideas Math Algebra 2 Solutions Chapter 4 Polynomial Functions 214
A. 4.6 meters
B. 8.7 meters
C. 9.7 meters
D. 10.0 meters

Question 7.
Classify each function as even, odd, or neither. Justify your answer.
a. f(x) = 3x5
b. f(x) = 4x3 + 8x
c. f(x) = 3x5 + 12x2+ 1
d. f(x) = 2x4
e. f(x) = x11 − x7
f. f(x) = 2x8 + 4x4 + x2 − 5

Question 8.
The volume of the rectangular prism shown is given by V = 2x3 + 7x2 − 18x − 63. Which polynomial represents the area of the base of the prism?
A. 2x2 + x − 21
B. 2x2 + 21 − x
C. 13x + 21 + 2x2
D. 2x2 − 21 − 13x
Big Ideas Math Algebra 2 Solutions Chapter 4 Polynomial Functions 215

Question 9.
The number R (in tens of thousands) of retirees receiving Social Security benefits is represented by the function R = 0.286t3 − 4.68t2 + 8.8t + 403, 0 ≤ t ≤ 10
where t represents the number of years since 2000. Identify any turning points on the given interval. What does a turning point represent in this situation?

Big Ideas Math Answers Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison

Big Ideas Math Answers Grade 3 Chapter 11

Are you been looking for assistance to solve Big Ideas Math Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison questions? If yes, then stay here. On this page, we have given a direct link to download Big Ideas Math Book Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison Answer Key PDF for free of cost. This solution key is useful for the students to complete their homework within time. It contains a detailed explanation solution for every problem of BIM 3rd Grade 11th Chapter Understand Fraction Equivalence and Comparison.

Big Ideas Math Book Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison Solutions

The exact solution and step by step answer for each and every question from BIM 3rd Grade 11th Chapter Understand Fraction Equivalence and Comparison are included below. The various lessons from Big Ideas Math Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison are Equivalent Fractions, Equivalent Fractions on a Number Line, Relate Fractions and Whole Numbers, Compare Fractions with the Same Denominator, Compare Fractions with the Same Numerator, Compare Fractions on a Number Line, Compare Fractions, and Compare and Order Fractions.

By learning these topics, you will be able to solve real-life math questions. Tap on the quick links mentioned here to get questions and answers of every concept. A better preparation plan and study material like Big Ideas Math Answers Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison is helpful to improve your math skills.

Lesson 1 Equivalent Fractions

Lesson 2 Equivalent Fractions on a Number Line

Lesson 3 Relate Fractions and Whole Numbers

Lesson 4 Compare Fractions with the Same Denominator

Lesson 5 Compare Fractions with the Same Numerator

Lesson 6 Compare Fractions on a Number Line

Lesson 7 Compare Fractions

Lesson 8 Compare and Order Fractions

Understand Fraction Equivalence and Comparison

Lesson 11.1 Equivalent Fractions

Explore and Grow   

Use the model to write fractions that are the same size as \(\frac{1}{2}\)
Big Ideas Math Answer Key Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.1 1
Answer: fraction \(\frac{4}{8}\) , fraction\(\frac{8}{16}\) are the same size as fraction \(\frac{1}{2}\).

Reasoning
Can you write a fraction with a denominator of that is the same size as \(\frac{3}{4}\)? Explain.
Answer: It is nothing but equivalent fraction \(\frac{3}{4}\) and \(\frac{21}{28}\) are equivalent fraction.

Think and Grow: Model Equivalent Fractions

Two or more numbers that have the same value are equivalent. Two or more fractions that name the same part of a whole are equivalent fractions.

Example
Use the models to find an equivalent fraction for \(\frac{2}{3}\) .
Big Ideas Math Answer Key Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.1 2

Answer:
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.1-2

Show and Grow

Use the models to find an equivalent fraction. Both models show the same whole.
Question 1.
Big Ideas Math Answer Key Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.1 3
Answer:\(\frac{2}{8}\)

Explanation: \(\frac{1}{4}\) and 2/8 are equivalent fractions both gives the same answer.
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.1-3

Question 2.
Big Ideas Math Answer Key Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.1 4
Answer: 1/3

Explanation: \(\frac{2}{6}\) and \(\frac{2}{8}\) are equivalent fractions both gives the same answer.
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.1-4

Question 3.
Shade 1 part of the model. Then divide the model into 4 equal parts. Write the equivalent fraction.
Big Ideas Math Answer Key Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.1 5
Answer: \(\frac{2}{4}\)

Explanation:
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.1-5
\(\frac{1}{4}\) and \(\frac{2}{4}\) are equivalent fractions both gives the same answer.

Apply and Grow: Practice

Use the models to find an equivalent fraction. Both models show the same whole.
Question 4.
Big Ideas Math Answer Key Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.1 6
Answer: \(\frac{4}{8}\)

Explanation:
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.1-6
\(\frac{2}{4}\) and \(\frac{4}{8}\) are equivalent fractions both gives the same answer.

Question 5.
Big Ideas Math Answer Key Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.1 7
Answer: \(\frac{3}{3}\)
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.1-7

Explanation: \(\frac{6}{6}\) and \(\frac{3}{3}\) are equivalent fractions both gives the same answer.

Question 6.
Shade 1 part of the model. Then divide the model into 6 equal parts. Write the equivalent fraction.
Big Ideas Math Answer Key Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.1 8
Answer: \(\frac{2}{6}\)
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.1-8

Explanation: \(\frac{1}{3}\) and \(\frac{2}{6}\) are equivalent fractions both gives the same answer.

Find the equivalent fraction
Question 7.
Big Ideas Math Answer Key Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.1 9
Answer: \(\frac{4}{8}\)
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.1-9

Explanation:
The equivalent fractions can be defined as fractions with different numerators and denominators that represent the same value or proportion of the whole.
Here the numerator of both the fractions is different.

Question 8.
Big Ideas Math Answer Key Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.1 10
Answer: \(\frac{2}{2}\)
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.1-10

Explanation:
The equivalent fractions can be defined as fractions with different numerators and denominators that represent the same value or proportion of the whole.
Here the numerator of both the fractions is different.

Question 9.
Big Ideas Math Answer Key Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.1 11
Answer: \(\frac{4}{6}\)
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.1-11

Explanation:
The equivalent fractions can be defined as fractions with different numerators and denominators that represent the same value or proportion of the whole.
Here the numerator of both the fractions is different.

Question 10.
Big Ideas Math Answer Key Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.1 12
Answer: \(\frac{1}{2}\)
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.1-12

Explanation:
The equivalent fractions can be defined as fractions with different numerators and denominators that represent the same value or proportion of the whole.
Here the numerator of both the fractions is different.

Question 11.
Structure
Descartes shades \(\frac{3}{4}\) of a rectangle. Divide and shade the model to show an equivalent fraction for \(\frac{3}{4}\).
Big Ideas Math Answer Key Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.1 13
Answer: \(\frac{6}{8}\)
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.1-13

Explanation:
The number equivalent to \(\frac{3}{4}\) is \(\frac{6}{8}\).
So you have to divide the rectangle into 8 equal parts and shade 6 parts among them.

Question 12.
Which One Doesn’t Belong?
Which model does not belong with the other three?
Big Ideas Math Answer Key Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.1 14
Answer: 4 figure is different

Explanation: From the above pictures 1 and 2 are equivalent fractions the first rectangle represent \(\frac{4}{8}\) and the second represents \(\frac{2}{4}\) both give the answer as \(\frac{1}{2}\)
The fourth figure has 4 parts and 1 part is shaded. So, the name of the fraction is \(\frac{1}{4}\)

Think and Grow: Modeling Real Life

You, Newton, and Descartes divide your posters for a science fair as shown.You finish 3 parts, Newton finishes 2 parts, and Descartes finishes 4 parts.Who has finished the same amount?
Big Ideas Math Answer Key Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.1 15

Answer:
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.1-15

Show and Grow

Question 13.
You, Newton, and Descartes divide your submarine sandwiches as shown. You eat 1 part, Newton eats 2 parts, and Descartes eats 2 parts. Who eats the same amount?
Big Ideas Math Answer Key Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.1 16
Answer:
Given,
You eat 1 part, Newton eats 2 parts, and Descartes eats 2 parts.
You and Newton eats the same amount

Question 14.
DIG DEEPER!
You and your friend have small pizzas. You cut your pizza sixths. Your friend cuts her pizza into eighths. You eat \(\frac{3}{6}\) of your pizza. Your 6 friend eats the same amount of her pizza. What fraction of her pizza does your friend eat? How many slices does your friend eat? Explain.
Answer:
Given,
You and your friend have small pizzas. You cut your pizza sixths.
Your friend cuts her pizza into eighths. You eat \(\frac{3}{6}\) of your pizza. Your 6 friend eats the same amount of her pizza.
That means \(\frac{6}{8}\) = \(\frac{3}{4}\)
Thus your friend eats \(\frac{3}{4}\) part of the pizza.

Equivalent Fractions Homework & Practice 11.1

Use the models to find an equivalent fraction. Both models show the same whole.
Question 1.
Big Ideas Math Answer Key Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.1 17
Answer: \(\frac{2}{3}\)

Big-Ideas-Math-Answer-Key-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.1-17

Explanation: \(\frac{4}{6}\) and \(\frac{2}{3}\) are equivalent fractions both gives the same answer.

Question 2.
Big Ideas Math Answer Key Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.1 18
Answer: \(\frac{3}{6}\)

Big-Ideas-Math-Answer-Key-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.1-18

Explanation: \(\frac{1}{2}\) and \(\frac{3}{6}\) are equivalent fractions both gives the same answer.

Question 3.
Shade 1 part of the model. Then divide the model into 8 equal parts. Write the equivalent fraction.
Big Ideas Math Answer Key Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.1 19
Answer: \(\frac{2}{8}\)

Big-Ideas-Math-Answer-Key-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.1-19 (1)

Big-Ideas-Math-Answer-Key-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.1-19
Explanation: \(\frac{1}{4}\) and \(\frac{2}{8}\) are equivalent fractions both gives the same answer.

Find the equivalent fraction
Question 4.
Big Ideas Math Answer Key Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.1 20
Answer: \(\frac{8}{8}\)

Big-Ideas-Math-Answer-Key-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.1-20

Explanation: \(\frac{2}{2}\) and \(\frac{8}{8}\) are equivalent fractions both gives the same answer.

Question 5.
Big Ideas Math Answer Key Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.1 21
Answer: \(\frac{3}{4}\)
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.1-21
Explanation: \(\frac{6}{8}\) and \(\frac{3}{4}\) are equivalent fractions both gives the same answer.

Question 6.
Big Ideas Math Answer Key Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.1 22
Answer: \(\frac{2}{6}\)
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.1-22
Explanation: \(\frac{1}{3}\) and \(\frac{2}{6}\) are equivalent fractions both gives the same answer.

Question 7.
Big Ideas Math Answer Key Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.1 23
Answer: \(\frac{1}{2}\)
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.1-23
Explanation: \(\frac{2}{4}\) and \(\frac{1}{2}\) are equivalent fractions both gives the same answer.

Question 8.
Open-Ended
Divide one model into an odd number of equal parts and the other model into an even number of equal parts. Then model and write two equivalent fractions.
Big Ideas Math Answer Key Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.1 24
Answer: \(\frac{1}{2}\) = \(\frac{2}{4}\)
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.1-24
Explanation: \(\frac{1}{2}\) and \(\frac{2}{4}\) are equivalent fractions both gives the same answer.

Question 9.
Modeling Real Life
You, Newton, and Descartes divide your portrait canvases as shown. You paint 2 parts, Newton paints 2 parts, and Descartes paints 8 parts. Who paints the same amount of the portrait canvas?
Big Ideas Math Answer Key Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.1 25
Answer: You and Descartes

Explanation:
You, Newton, and Descartes divide your portrait canvases as shown.
You paint 2 parts, Newton paints 2 parts, and Descartes paints 8 parts.
You completed 2 of 2 parts.
Descartes paints 8 of 8 parts.
\(\frac{2}{2}\) = 1
\(\frac{8}{8}\) = 1
Thus you and Descartes completes the same amount of the portrait canvas.

Question 10.
DIG DEEPER!
You and your friend paint 2 roundabouts for a park. You divide your roundabout into thirds. Your friend divides his roundabout into sixths. You paint \(\frac{1}{3}\) of your roundabout. Your friend paints the same amount of his roundabout. What fraction does your friend paint? Explain.
Big Ideas Math Answer Key Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.1 26
Answer:
Given,
You and your friend paint 2 roundabouts for a park. You divide your roundabout into thirds.
Your friend divides his roundabout into sixths.
You paint \(\frac{1}{3}\) of your roundabout. Your friend paints the same amount of his roundabout.
\(\frac{6}{6}\)

Review & Refresh

Question 11.
Round to the nearest ten to estimate the sum.
Big Ideas Math Answer Key Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.1 27
Answer: 540
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.1-27

Explanation:
The nearest ten estimation for 431 is 430
and for 109 Nearest ten estimation  is 110
hence
430
+ 110
540

Question 12.
Round to the nearest hundred to estimate the sum.
Big Ideas Math Answer Key Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.1 28
Answer: 820

Explanation:
The nearest ten estimation for 551 is 550
and for 268 Nearest ten estimation  is 270
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.1-28

Lesson 11.2 Equivalent Fractions on a Number Line

Explore and Grow

Use Fraction Strips to label thirds on the number line.
Big Ideas Math Answers 3rd Grade Chapter 11 Understand Fraction Equivalence and Comparison 11.2 1
Use Fraction Strips to label sixths on the number line.
Big Ideas Math Answers 3rd Grade Chapter 11 Understand Fraction Equivalence and Comparison 11.2 2
Use the number lines to complete the equivalent fraction.
Big Ideas Math Answers 3rd Grade Chapter 11 Understand Fraction Equivalence and Comparison 11.2 3
Answer: \(\frac{2}{6}\)
Big-Ideas-Math-Answers-3rd-Grade-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.2-3
Explanation: \(\frac{1}{3}\) and \(\frac{2}{6}\) are equivalent fractions both gives the same answer.

Structure
How can you tell whether fractions are equivalent using a number line?
Answer:
Step 1: Locate the original fraction on the number line. We’ll first need to locate our fraction on the number line.
Step 2: Divide each your newly created divisions from step 1 into two even spaces.
Step 3: Find the equivalent fraction.
Step 4: Repeat steps 2 and 3 to find more equivalent fractions.

Think and Grow: Equivalent Fractions on a Number Line

You can use a number line to find equivalent fractions. Equivalent fractions represent the same point on a number line.

Example
Use a number line to find an equivalent fraction for \(\frac{3}{4}\).
Big Ideas Math Answers 3rd Grade Chapter 11 Understand Fraction Equivalence and Comparison 11.2 4
Step 1: Plot \(\frac{3}{4}\) on a number line.
Big Ideas Math Answers 3rd Grade Chapter 11 Understand Fraction Equivalence and Comparison 11.2 5
Step 2: Divide the number line into eighths. Label each tick mark to show eighths.
Big Ideas Math Answers 3rd Grade Chapter 11 Understand Fraction Equivalence and Comparison 11.2 6

Answer:
Big-Ideas-Math-Answers-3rd-Grade-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.2-6

Show and Grow

Question 1.
Use the number line to find an equivalent fraction.
Big Ideas Math Answers 3rd Grade Chapter 11 Understand Fraction Equivalence and Comparison 11.2 7
Answer: \(\frac{4}{6}\)
Big-Ideas-Math-Answers-3rd-Grade-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.2-7
The fractions \(\frac{2}{3}\) and \(\frac{4}{6}\) are on the same point.
So, \(\frac{2}{3}\) = \(\frac{4}{6}\)

Question 2.
Write two fractions that name the point shown.
Big Ideas Math Answers 3rd Grade Chapter 11 Understand Fraction Equivalence and Comparison 11.2 8
Answer: \(\frac{2}{4}\) = \(\frac{1}{2}\)
Big-Ideas-Math-Answers-3rd-Grade-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.2-8
The fractions \(\frac{1}{2}\) and \(\frac{2}{4}\) are on the same point.
So,\(\frac{1}{2}\) = \(\frac{2}{4}\)

Apply and Grow: Practice

Write two fractions that name the point shown.
Question 3.
Big Ideas Math Answers 3rd Grade Chapter 11 Understand Fraction Equivalence and Comparison 11.2 9
Answer: \(\frac{4}{6}\) = \(\frac{1}{2}\)

Big-Ideas-Math-Answers-3rd-Grade-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.2-9

The fractions \(\frac{4}{6}\) and \(\frac{1}{2}\) are on the same point.
So,\(\frac{4}{6}\) = \(\frac{1}{2}\)

Question 4.
Big Ideas Math Answers 3rd Grade Chapter 11 Understand Fraction Equivalence and Comparison 11.2 10
Answer: \(\frac{4}{8}\) = \(\frac{1}{2}\)

Big-Ideas-Math-Answers-3rd-Grade-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.2-10

The fractions \(\frac{4}{8}\) and \(\frac{1}{2}\) are on the same point.
So,\(\frac{4}{8}\) = \(\frac{1}{2}\)

Question 5.
Big Ideas Math Answers 3rd Grade Chapter 11 Understand Fraction Equivalence and Comparison 11.2 11
Answer: \(\frac{1}{4}\) = \(\frac{2}{8}\)

Big-Ideas-Math-Answers-3rd-Grade-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.2-11

The fractions \(\frac{1}{4}\) and \(\frac{2}{8}\) are on the same point.
So,\(\frac{1}{4}\) = \(\frac{2}{8}\)

Question 6.
Big Ideas Math Answers 3rd Grade Chapter 11 Understand Fraction Equivalence and Comparison 11.2 12
Answer:  \(\frac{1}{2}\) = \(\frac{2}{4}\)

Big-Ideas-Math-Answers-3rd-Grade-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.2-12
The fractions \(\frac{1}{2}\) and \(\frac{2}{4}\) are on the same point.
So,\(\frac{1}{2}\) = \(\frac{2}{4}\)

Question 7.
YOU BE THE TEACHER
Your friend says \(\frac{3}{4}\) and \(\frac{6}{8}\) are equivalent because they are not the same distance from 0. Is your friend correct? Explain.
Big Ideas Math Answers 3rd Grade Chapter 11 Understand Fraction Equivalence and Comparison 11.2 13
Answer: No your friend is not correct. \(\frac{3}{4}\) and \(\frac{6}{8}\) are equivalent fractions irrespective of the distance on the number line.

Question 8.
Reasoning
Explain why \(\frac{1}{3}\) is equal to two \(\frac{1}{6}\)s
Answer:
Divide \(\frac{1}{3}\) by 2. You get \(\frac{1}{6}\)

Think and Grow: Modeling Real Life

Newton rests after biking \(\frac{2}{3}\) of a race. Descartes rests after biking \(\frac{2}{6}\) of the same race. Do they rest at the same point along the race path?
Big Ideas Math Answers 3rd Grade Chapter 11 Understand Fraction Equivalence and Comparison 11.2 14
Model:
Big Ideas Math Answers 3rd Grade Chapter 11 Understand Fraction Equivalence and Comparison 11.2 15
They ______ rest at the same point.
Explain: No they do not rest at the same point. Because \(\frac{2}{3}\) is not equal to \(\frac{2}{6}\)
Big-Ideas-Math-Answers-3rd-Grade-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.2-15

Show and Grow

Question 9.
Newton hikes \(\frac{7}{8}\) of a trail. Descartes hikes \(\frac{3}{4}\) of the same trail. Do they hike the same distance along the trail?
Answer:
Newton hikes \(\frac{7}{8}\) of a trail. Descartes hikes \(\frac{3}{4}\) of the same trail.
You have to compare both the fractions.
\(\frac{7}{8}\) is greater than \(\frac{3}{4}\).
By this, we can say that does not hike the same distance along the trail.

Question 10.
Newton chases Descartes for \(\frac{3}{6}\) mile. Descartes turns around and chases Newton an equal distance.Write two equivalent fractions that can describe how far Descartes chases Newton.
Answer:
The equivalent fractions of \(\frac{3}{6}\) are
\(\frac{3}{6}\) × \(\frac{2}{2}\) = \(\frac{6}{12}\)
\(\frac{3}{6}\) = \(\frac{1}{2}\)
Thus the two The equivalent fractions of \(\frac{3}{6}\) are \(\frac{6}{12}\), \(\frac{1}{2}\)

Question 11.
DIG DEEPER!
You cut a quiche into 8 equal slices. Your family eats \(\frac{1}{2}\) of the quiche. How many slices does your family eat? Explain.
Answer:
Given,
You cut a quiche into 8 equal slices. Your family eats \(\frac{1}{2}\) of the quiche.
\(\frac{1}{2}\) × 8 = \(\frac{1}{4}\)
Thus your family eats 4 equal slices.

Equivalent Fractions on a Number Line Homework & Practice 11.2

Question 1.
Use the number line to find an equivalent fraction.
Big Ideas Math Answers 3rd Grade Chapter 11 Understand Fraction Equivalence and Comparison 11.2 16
Answer: \(\frac{1}{4}\)

Big-Ideas-Math-Answers-3rd-Grade-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.2-16
By seeing the above number we can find the equivalent fraction of \(\frac{2}{8}\).
\(\frac{1}{4}\) and \(\frac{2}{8}\) are on the same point.

Question 2.
Write two fractions that name the point shown
Big Ideas Math Answers 3rd Grade Chapter 11 Understand Fraction Equivalence and Comparison 11.2 17
Answer: \(\frac{1}{2}\) = \(\frac{2}{4}\)

Big-Ideas-Math-Answers-3rd-Grade-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.2-17

By seeing the above number we can find the equivalent fraction of \(\frac{1}{2}\).
\(\frac{1}{2}\) = \(\frac{2}{4}\)

Write two fractions that name the same point shown.
Question 3.
Big Ideas Math Answers 3rd Grade Chapter 11 Understand Fraction Equivalence and Comparison 11.2 18
Answer: \(\frac{6}{8}\) = \(\frac{3}{4}\)

Big-Ideas-Math-Answers-3rd-Grade-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.2-18

By seeing the above number we can find the equivalent fraction of \(\frac{6}{8}\).
\(\frac{6}{8}\) = \(\frac{3}{4}\)

Question 4.
Big Ideas Math Answers 3rd Grade Chapter 11 Understand Fraction Equivalence and Comparison 11.2 19
Answer: \(\frac{4}{6}\) = \(\frac{2}{3}\)
Big-Ideas-Math-Answers-3rd-Grade-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.2-19

By seeing the above number we can find the equivalent fraction of \(\frac{4}{6}\).
\(\frac{4}{6}\) = \(\frac{2}{3}\)

Question 5.
Which One Doesn’tBelong?
Which not fraction does belong with the other three? Explain.
Big Ideas Math Answers 3rd Grade Chapter 11 Understand Fraction Equivalence and Comparison 11.2 20
Answer: \(\frac{2}{3}\)

Big-Ideas-Math-Answers-3rd-Grade-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.2-20
The fraction \(\frac{2}{3}\) does not belong to other three fractions.

Question 6.
Reasoning
How do you know that \(\frac{3}{8}\) and \(\frac{3}{4}\) are equivalent when plotting the fractions on a number line?
Answer:
No, they are not equivalent. The numerators are equals, but the denominators are not, they are not equivalent. If the first one would be 6/8, they would be equivalent, because we can simplify the fractions by 2 and we got 3/4 again.

Question 7.
Modeling Real Life
You run \(\frac{6}{8}\) of a race. Your friend runs \(\frac{3}{4}\) of the same race. Do you and your friend run the same distance?
Big Ideas Math Answers 3rd Grade Chapter 11 Understand Fraction Equivalence and Comparison 11.2 21.1
Answer:
Given,
You run \(\frac{6}{8}\) of a race. Your friend runs \(\frac{3}{4}\) of the same race.
The equivalent fraction of \(\frac{3}{4}\) is \(\frac{6}{8}\)
By this we can say that you and your friend run the same distance.

Question 8.
DIG DEEPER!
You have a frame that holds 8 pictures. You fill \(\frac{1}{4}\) of the frame. How many pictures do you put in the frame? Explain.
Answer:
Given,
You have a frame that holds 8 pictures. You fill \(\frac{1}{4}\) of the frame.
\(\frac{1}{4}\) × 8 = 2
Thus you put 2 pictures in the frame.

Review & Refresh

What fraction of the whole is shaded?
Question 9.
Big Ideas Math Answers 3rd Grade Chapter 11 Understand Fraction Equivalence and Comparison 11.2 21
Answer: \(\frac{2}{3}\)
By seeing the figure we can name the fraction of the shaded part. The box is divided into 3 parts among them 2 parts are shaded.

Big-Ideas-Math-Answers-3rd-Grade-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.2-21

Question 10.
Big Ideas Math Answers 3rd Grade Chapter 11 Understand Fraction Equivalence and Comparison 11.2 22
Answer: \(\frac{4}{6}\)
By seeing the figure we can name the fraction of the shaded part. The circle is divided into 6 parts among them 4 parts are shaded.

Big-Ideas-Math-Answers-3rd-Grade-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.2-22

 

Lesson 11.3 Relate Fractions and Whole Numbers

Explore and Grow

Use Fraction Strips to complete the fractions. Draw to show your models.
Big Ideas Math Answers Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.3 1
Answer:

Big-Ideas-Math-Answers-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.3-1

Explanation: The equivalent fractions can be defined as fractions with different numerators and denominators that represent the same value or proportion of the whole.

Repeated Reasoning
How many more fourths did you use to model 2 than you did to model 1? How many more fourths would you need to model 3? Complete the fraction.
Big Ideas Math Answers Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.3 2
Answer: 12
Big-Ideas-Math-Answers-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.3-2

Explanation: The equivalent fractions can be defined as fractions with different numerators and denominators that represent the same value or proportion of the whole.

Think and Grow: Equivalent Fractions and Whole Numbers

Example
Write the numbers 1 and 2 as fractions. The number line shows 2 wholes. Each whole is divided into1 equal part.
Big Ideas Math Answers Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.3 3

Answer:
Big-Ideas-Math-Answers-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.3-3

Example
Write equivalent fractions for the numbers 1 and 2.
Big Ideas Math Answers Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.3 4

Answer:
Big-Ideas-Math-Answers-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.3-4

Show and Grow

Question 1.
Complete the number line. Then complete the statements.
Big Ideas Math Answers Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.3 5
Answer:

Big-Ideas-Math-Answers-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.3-5

Explanation: The equivalent fractions can be defined as fractions with different numerators and denominators that represent the same value or proportion of the whole.

Apply and Grow: Practice

Question 2.
Complete the number line. Then write equivalent fractions for the numbers 3 and 4.
Big Ideas Math Answers Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.3 6
Answer:

Big-Ideas-Math-Answers-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.3-6

Explanation: The equivalent fractions can be defined as fractions with different numerators and denominators that represent the same value or proportion of the whole.

Write two equivalent fractions for the whole number.
Question 3.
Big Ideas Math Answers Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.3 7
Answer:

Big-Ideas-Math-Answers-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.3-7
Explanation: The equivalent fractions can be defined as fractions with different numerators and denominators that represent the same value or proportion of the whole.

Question 4.
Big Ideas Math Answers Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.3 8
Answer:
Big-Ideas-Math-Answers-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.3-8
Explanation: The equivalent fractions can be defined as fractions with different numerators and denominators that represent the same value or proportion of the whole.

Question 5.
Big Ideas Math Answers Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.3 9
Answer:
Big-Ideas-Math-Answers-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.3-9
Explanation: The equivalent fractions can be defined as fractions with different numerators and denominators that represent the same value or proportion of the whole.

Write the equivalent whole number.
Question 6.
Big Ideas Math Answers Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.3 10
Answer:
Big-Ideas-Math-Answers-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.3-10
Explanation: The equivalent fractions can be defined as fractions with different numerators and denominators that represent the same value or proportion of the whole.

Question 7.
Big Ideas Math Answers Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.3 11
Answer:
Big-Ideas-Math-Answers-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.3-11
Explanation: The equivalent fractions can be defined as fractions with different numerators and denominators that represent the same value or proportion of the whole.

Question 8.
Big Ideas Math Answers Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.3 12

Answer:
Big-Ideas-Math-Answers-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.3-12

Explanation: The equivalent fractions can be defined as fractions with different numerators and denominators that represent the same value or proportion of the whole.

Question 9.
You cut a sandwich into 4 pieces. You eat the whole sandwich. What fraction of the sandwich do you eat?
Big Ideas Math Answers Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.3 13
Answer:
Given,
You cut a sandwich into 4 pieces. You eat the whole sandwich.
\(\frac{4}{4}\) = 1
Thus you eat \(\frac{4}{4}\) fraction of the sandwich.

Question 10.
Number Sense
Write three fractions that are equivalent to 6 using the denominators 1, 2, and 6.
Answer:
\(\frac{6}{6}\) = 1
\(\frac{12}{6}\) = 2
\(\frac{36}{6}\) = 6

Question 11.
Use a number line to show \(\frac{5}{5}\) = 1.
Answer:
Big Ideas Math Book 3rd Grade Answer Key Chapter 11

Think and Grow: Modeling Real Life

You run around a track 12 times. Each lap is \(\frac{1}{4}\) mile. How many miles do you run? Write your answer as a whole number and as a fraction.
Model:
Big Ideas Math Answers Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.3 14
You run _____ or ______ miles.

Answer:
Given,
You run around a track 12 times. Each lap is \(\frac{1}{4}\) mile.
12 × \(\frac{1}{4}\) = 3 mile
You run 3 miles.

Show and Grow

Question 12.
You ride your bike around a city block 24 times. Each time around is \(\frac{1}{6}\) mile. How many miles do you ride? Write your answer as a 6 whole number and as a fraction.
Answer:
Given,
You ride your bike around a city block 24 times. Each time around is \(\frac{1}{6}\) mile.
24 × \(\frac{1}{6}\) = 4
Thus you ride 4 miles.

Question 13.
There are 5 Great Lakes: Lake Erie, Lake Huron, Lake Michigan, Lake Ontario, and Lake Superior. You complete a project and use 1 whole poster for each lake. Write a fraction that gives the number of posters you use in all.
Big Ideas Math Answers Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.3 15
Answer:
Lake Michigan is \(\frac{1}{1}\)
Lake Huron is \(\frac{2}{1}\)
Lake Erie is \(\frac{3}{1}\)
Lake Ontario is \(\frac{4}{1}\)
Lake Superior is \(\frac{5}{1}\)

Question 14.
DIG DEEPER!
You and your friends order 3 pizzas. Each pizza is cut into 8 slices. You and your friends eat \(\frac{16}{8}\) of the pizzas. Write a fraction that shows the number of eighths that are left.
Answer:
Given that,
You and your friends order 3 pizzas. Each pizza is cut into 8 slices.
You and your friends eat \(\frac{16}{8}\) of the pizzas.
3 × 8 = 24 slices
\(\frac{24}{8}\)– \(\frac{16}{8}\) = \(\frac{8}{8}\)
Thus 8 slices are left in the pizza.

Equivalent Fractions on a Number Line Homework & Practice 11.3

Question 1.
Complete the number line. Then write equivalent fractions for the numbers 3 and 4.
Big Ideas Math Answers Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.3 16
Answer:
Big-Ideas-Math-Answers-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.3-16
Explanation: The equivalent fractions can be defined as fractions with different numerators and denominators that represent the same value or proportion of the whole.

Write two equivalent fractions for the whole number.
Question 2.
Big Ideas Math Answers Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.3 17
Answer:
Big-Ideas-Math-Answers-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.3-17
Explanation: The equivalent fractions can be defined as fractions with different numerators and denominators that represent the same value or proportion of the whole.

Question 3.
Big Ideas Math Answers Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.3 18
Answer:
Big-Ideas-Math-Answers-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.3-18
Explanation: The equivalent fractions can be defined as fractions with different numerators and denominators that represent the same value or proportion of the whole.

Question 4.
Big Ideas Math Answers Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.3 19
Answer:
Big-Ideas-Math-Answers-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.3-19
Explanation: The equivalent fractions can be defined as fractions with different numerators and denominators that represent the same value or proportion of the whole.

Write the equivalent whole number.
Question 5.
Big Ideas Math Answers Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.3 20
Answer:
Big-Ideas-Math-Answers-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.3-20
Explanation: The equivalent fractions can be defined as fractions with different numerators and denominators that represent the same value or proportion of the whole.

Question 6.
Big Ideas Math Answers Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.3 21
Answer:
Big-Ideas-Math-Answers-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.3-21
Explanation: The equivalent fractions can be defined as fractions with different numerators and denominators that represent the same value or proportion of the whole.

Question 7.
Big Ideas Math Answers Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.3 22
Answer:
Big-Ideas-Math-Answers-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.3-22
Explanation: The equivalent fractions can be defined as fractions with different numerators and denominators that represent the same value or proportion of the whole.

Question 8.
Precision
Match each whole number to its equivalent fractions.
Big Ideas Math Answers Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.3 23
Answer:
Big-Ideas-Math-Answers-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.3-23
Explanation: The equivalent fractions can be defined as fractions with different numerators and denominators that represent the same value or proportion of the whole.

Question 9.
YOU BE THE TEACHER
Descartes says a fraction equivalent to 3 has a 3 in the denominator and a 1 in the numerator. Is Descartes correct? Explain.
Answer:
Descartes says a fraction equivalent to 3 has a 3 in the denominator and a 1 in the numerator.
Yes 3 is equivalent to \(\frac{3}{1}\)
Thus Descartes is correct.

Question 10.
Patterns
Describe and complete the pattern.
Big Ideas Math Answers Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.3 24
Answer:
Big-Ideas-Math-Answers-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.3-24
The fraction is the multiples of 8. So, the pattern is 8/8, 16/8, 24/8, 32/8, 40/8.

Question 11.
Modeling Real Life
You run around a park 18 times. Each lap is \(\frac{1}{6}\) mile. How many miles do you run? Write your answer as a whole number and as a fraction.
Answer:
Given that,
You run around a park 18 times. Each lap is \(\frac{1}{6}\) mile.
18 × \(\frac{1}{6}\) = 3
Thus you run 3 miles or \(\frac{18}{6}\) miles

Question 12.
Modeling Real Life
Christopher Columbus had 3 ships on his first NiñaPintavoyage: the , the , and the Santa Maria. You complete a project and use 1 whole poster for each ship. Write a fraction that gives the number of posters you use in all.
Answer:
Given,
Christopher Columbus had 3 ships on his first NiñaPintavoyage: the , the , and the Santa Maria.
You complete a project and use 1 whole poster for each ship.
\(\frac{3}{3}\) = 1
Thus you use 1 poster for each ship.
Thus you use 3 posters for all the ships.

Question 13.
DIG DEEPER!
A teacher has 4 tables. Each table is divided into 4 sections. \(\frac{12}{4}\) of the tables have 4 supplies on them. Write a fraction that shows the number of fourths not that do have supplies on them.
Answer:
Given,
A teacher has 4 tables. Each table is divided into 4 sections.
4 × 4 = 16 sections
\(\frac{12}{4}\) of the tables have 4 supplies on them.
\(\frac{12}{4}\) = 3
3 of the tables have 4 supplies.
The fraction that shows the number of fourths not that do have supplies on them is \(\frac{3}{4}\)

Review & Refresh

Estimate. Then find the sum. Check whether your answer is reasonable.
Question 14.
Estimate: _____
Big Ideas Math Answers Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.3 25
Answer: 890

Explanation:
The nearest ten estimation for 837 is 840
and for 46 Nearest ten estimation  is 50
840
+  50
890

Question 15.
Estimate: _____
Big Ideas Math Answers Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.3 26
Answer: 510

Explanation:
The nearest ten estimation for 396 is 400
and for 108 Nearest ten estimation  is 110
400
+ 110
510

Question 16.
Estimate: _____
Big Ideas Math Answers Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.3 27
Answer: 830

Explanation:
The nearest ten estimation for 551 is 550
and for 279 Nearest ten estimation  is 280
550
+ 280
830

Lesson 11.4 Compare Fractions with the Same Denominator

Explore and Grow

Color to show each fraction. Circle the greater fraction.
Big Ideas Math Solutions Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.4 1
Answer: \(\frac{5}{6}\) is the greater fraction

Big-Ideas-Math-Solutions-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.4-1

Construct Arguments
Explain to your partner how you can compare fractions with the same denominator.
Answer:
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

Think and Grow: Compare Fractions That have the Same Denominator

Example
Compare \(\frac{3}{8}\) and \(\frac{7}{8}\).
One Way: Use Fraction Strips.
The fractions have the same denominator, 8.
Big Ideas Math Solutions Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.4 2
Big Ideas Math Solutions Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.4 3

Answer:
Big-Ideas-Math-Solutions-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.4-3

Another Way:
When the denominators are the same, the whole is divided into the same number of equal parts. So, look at the numerators 3 and 7 to compare.The fraction with the greater numerator is the greater fraction.
Big Ideas Math Solutions Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.4 4

Answer:
Big-Ideas-Math-Solutions-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.4-4

Show and Grow

Shade to compare the fractions
Question 1.
Big Ideas Math Solutions Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.4 5
Answer:

Big-Ideas-Math-Solutions-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.4-5
Explanation: 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

Question 2.
Big Ideas Math Solutions Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.4 6
Answer:

Big-Ideas-Math-Solutions-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.4-6

Explanation: 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

Apply and Grow: Practice

Shade to compare the fractions.
Question 3.
Big Ideas Math Solutions Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.4 7
Answer:
Big-Ideas-Math-Solutions-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.4-7

Explanation: 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

Question 4.
Big Ideas Math Solutions Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.4 8
Answer:
Big-Ideas-Math-Solutions-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.4-8

Explanation: Here the numerator and denominator of the fractions are the same so

Compare.
Question 5.
Big Ideas Math Solutions Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.4 9
Answer:
Big-Ideas-Math-Solutions-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.4-9

Explanation:
Compare the fractions. The denominators of both the fractions are the same so you have to compare the numerators.
2 is less than 3.
So, \(\frac{2}{3}\) < \(\frac{3}{3}\)

Question 6.
Big Ideas Math Solutions Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.4 10
Answer:
Big-Ideas-Math-Solutions-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.4-10
Compare the fractions. The denominators of both the fractions are the same so you have to compare the numerators.
So, \(\frac{1}{2}\) = \(\frac{1}{2}\)

Question 7.
Big Ideas Math Solutions Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.4 11
Answer:
Big-Ideas-Math-Solutions-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.4-11

Explanation:
Compare the fractions. The denominators of both the fractions are the same so you have to compare the numerators.
3 is greater than 1.
So, \(\frac{3}{8}\) > \(\frac{1}{8}\)

Question 8.
Big Ideas Math Solutions Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.4 12
Answer:
Big-Ideas-Math-Solutions-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.4-12

Explanation:
Compare the fractions. The denominators of both the fractions are the same so you have to compare the numerators.
4 is greater than 2.
So, \(\frac{4}{6}\) > \(\frac{2}{6}\)

Question 9.
Big Ideas Math Solutions Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.4 13
Answer:
Big-Ideas-Math-Solutions-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.4-13

Explanation:
Compare the fractions. The denominators of both the fractions are the same so you have to compare the numerators.
5 is less than 6.
So, \(\frac{5}{8}\) < \(\frac{6}{8}\)

Question 10
Big Ideas Math Solutions Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.4 14
Answer:
Big-Ideas-Math-Solutions-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.4-14

Explanation:
Compare the fractions. The denominators of both the fractions are the same so you have to compare the numerators.
0 is less than 1.
So, \(\frac{0}{4}\) < \(\frac{1}{4}\)

Question 11.
Number Sense
Which statements are true?
Big Ideas Math Solutions Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.4 15
Answer:
Big-Ideas-Math-Solutions-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.4-15

Question 12.
Writing
How can you show that \(\frac{3}{8}\) is less than \(\frac{6}{8}\)?
Answer:
Big-Ideas-Math-Solutions-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.4-7 (1)

Explanation:
Compare the fractions. The denominators of both the fractions are the same so you have to compare the numerators.
3 is less than 6.
So, \(\frac{3}{8}\) < \(\frac{6}{8}\)

Question 13.
Structure
Which statement correctly compares the fraction of circles shaded in Group A to the fraction of circles shaded in Group B?
Big Ideas Math Solutions Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.4 16
Answer:
Big-Ideas-Math-Solutions-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.4-16

Think and Grow: Modeling Real Life

Newton and Descartes have the same number of chores. Newton finishes \(\frac{2}{3}\) of his chores. Descartes finishes \(\frac{1}{3}\) of his chores. Who finishes more of his chores?
Model:
______ finishes more of his chores.
Explain:

Answer: Newton finishes more of his chores

Explanation:
Given,
Newton and Descartes have the same number of chores.
Newton finishes \(\frac{2}{3}\) of his chores. Descartes finishes \(\frac{1}{3}\) of his chores.
Compare both the fractions \(\frac{2}{3}\) and \(\frac{1}{3}\)
\(\frac{2}{3}\) > \(\frac{1}{3}\)
Therefore Newton finishes more of his chores.

Show and Grow

Question 14.
At the pet store, a green lizard is \(\frac{5}{6}\) foot long. A brown lizard is \(\frac{6}{6}\) foot long. You pick up the longer lizard. Which lizard do you pick up?
Answer:
Given,
At the pet store, a green lizard is \(\frac{5}{6}\) foot long.
A brown lizard is \(\frac{6}{6}\) foot long. You pick up the longer lizard.
Compare both the fractions \(\frac{5}{6}\) and \(\frac{6}{6}\)
\(\frac{5}{6}\) < \(\frac{6}{6}\)
Therefore you pick up brown lizard.

Question 15.
You and a friend each have a burrito. The burritos are the same size. You eat \(\frac{3}{4}\) of your burrito. Your friend eats \(\frac{1}{4}\) of his burrito. Who has less burrito left to eat? Explain.
Big Ideas Math Solutions Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.4 17
Answer:
Given,
You and a friend each have a burrito. The burritos are the same size.
You eat \(\frac{3}{4}\) of your burrito. Your friend eats \(\frac{1}{4}\) of his burrito.
\(\frac{3}{4}\) > \(\frac{1}{4}\)
You have less burrito left to eat.

Question 16.
DIG DEEPER!
You walk \(\frac{5}{8}\) of the distance to the library from your home. Your friend walks \(\frac{5}{8}\) of the distance to school from his home. You walk a greater distance than your friend. Explain how this is possible.
Answer:
Given,
You walk \(\frac{5}{8}\) of the distance to the library from your home. Your friend walks \(\frac{5}{8}\) of the distance to school from his home.
\(\frac{5}{8}\) = \(\frac{5}{8}\)

Compare Fractions with the Same Denominator Homework & Practice 11.4

Shade to compare the fractions
Question 1.
Big Ideas Math Solutions Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.4 18
Answer:
Big-Ideas-Math-Solutions-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.4-18

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
3 is greater than 1
So, \(\frac{3}{4}\) > \(\frac{1}{4}\)

Question 2.
Big Ideas Math Solutions Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.4 19
Answer:
Big-Ideas-Math-Solutions-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.4-19
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
\(\frac{1}{2}\) < \(\frac{2}{2}\)

Compare
Question 3.
Big Ideas Math Solutions Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.4 20
Answer:
Big-Ideas-Math-Solutions-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.4-20
Explanation: 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.
5 is less than 7.
So, \(\frac{5}{8}\) < \(\frac{7}{8}\)

Question 4.
Big Ideas Math Solutions Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.4 21
Answer:
Big-Ideas-Math-Solutions-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.4-21

Explanation: 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.
6 is greater than 5.
So, \(\frac{6}{6}\) > \(\frac{5}{6}\)

Question 5.
Big Ideas Math Solutions Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.4 22
Answer:
Big-Ideas-Math-Solutions-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.4-22

Explanation: 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.
1 is less than 2.
So, \(\frac{1}{3}\) < \(\frac{2}{3}\)

Question 6.
Big Ideas Math Solutions Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.4 23
Answer:
Big-Ideas-Math-Solutions-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.4-23

Explanation: 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.
2 is less than 4.
So, \(\frac{2}{4}\) < \(\frac{4}{4}\)

Question 7.
Big Ideas Math Solutions Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.4 24
Answer:
Big-Ideas-Math-Solutions-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.4-24

Explanation: 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.
0 is less than 1.
So, \(\frac{0}{3}\) < \(\frac{1}{3}\)

Question 8.
Big Ideas Math Solutions Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.4 25
Answer:
Big-Ideas-Math-Solutions-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.4-25

Explanation: 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.
3 is less than 4.
So, \(\frac{3}{6}\) < \(\frac{4}{6}\)

Question 9.
YOU BE THE TEACHER
Your friend says \(\frac{3}{4}>\frac{2}{4}\). Is your friend correct? Explain.
Answer: Your friend is correct.
\(\frac{3}{4}\) = 0.75
\(\frac{1}{2}\) = 0.50
\(\frac{3}{4}>\frac{2}{4}\)

Question 10.
Logic
Two fractions are equivalent and have the same denominator. What must be true about the numerators of the fractions?
Answer:
Two fractions are equivalent fractions when they represent the same part of a whole. Since equivalent fractions do not always have the same numerator and denominator, one way to determine if two fractions are equivalent is to find a common denominator and rewrite each fraction with that denominator.

Open-Ended
Write a number to make the statement true.
Question 11.
Big Ideas Math Solutions Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.4 26
Answer: 5

Explanation:
Here the denominators of both the fractions are the same. You have to write the numerator of less than 6.
So, to make the statement true we are writing 5.
Big-Ideas-Math-Solutions-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.4-26

Question 12.
Big Ideas Math Solutions Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.4 27
Answer: 4
Big-Ideas-Math-Solutions-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.4-27

Explanation:
Here the denominators of both the fractions are the same. You have to write the numerator of greater than 2.
So, to make the statement true we are writing 4.

Question 13.
Big Ideas Math Solutions Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.4 28
Answer: 1
Big-Ideas-Math-Solutions-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.4-28

Explanation:
Here the denominators of both the fractions are the same. You have to write the numerator of less than 4.
So, to make the statement true we are writing 1.

Question 14.
Modeling Real Life
You and your friend had the same number of baby teeth. You have lost \(\frac{3}{4}\) of your baby teeth. Your friend has lost \(\frac{2}{4}\) of her baby teeth. Who has lost more teeth?
Big Ideas Math Solutions Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.4 29
Answer: you have lost more teeth

Explanation:
Given,
You and your friend had the same number of baby teeth. You have lost \(\frac{3}{4}\) of your baby teeth. Your friend has lost \(\frac{2}{4}\) of her baby teeth.
\(\frac{3}{4}\) = 0.75
\(\frac{1}{2}\) = 0.50
\(\frac{3}{4}>\frac{2}{4}\)
Thus you lost more teeth.

Question 15.
Modeling Real Life
Two classes each paint a mural. The murals are the same size. The third-grade class paints \(\frac{5}{8}\) of one mural. The fourth-grade class paints \(\frac{7}{8}\) of the other mural. Which class has less left to paint? Explain.
Answer: The fourth-grade class has one part left to paint

Review & Refresh

Find the product.
Question 16.
9 × 80 = _____
Answer: 720

Explanation:
Multiply 9 with 80 we get 720
9 × 80 = 720

Question 17.
5 × 30 = _____
Answer: 150

Explanation:
Multiply 5 with 30 we get 150
5 × 30 = 150

Question 18.
6 × 40 = _____
Answer: 240

Explanation:
Multiply 6 with 40 we get 240
6 × 40 = 240

Lesson 11.5 Compare Fractions with Same Numerator

Explore and Grow

Color to show each fraction. Circle the greater fraction.
Big Ideas Math Answer Key Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.5 1
Answer:
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.5-1

Precision
How can you use Fraction Strips to check your answer? Draw to show.
Answer:
Big Ideas Math Grade 3 Chapter 11 Answer Key

Think and Grow: Compare Fractions That Have the Same Numerator

Example
Compare \(\frac{2}{4}\) and \(\frac{2}{6}\).
One Way: Use Fraction Strips
Big Ideas Math Answer Key Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.5 2

Answer:
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.5-2

Another Way:
When the numerators are the same, look at the denominators 4 and 6 to compare the sizes of the parts.
• The more parts the whole is divided into, the smaller the parts are.
• The fewer parts the whole is divided into, the larger the parts are.
The fraction with the greater denominator is the lesser fraction.
Big Ideas Math Answer Key Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.5 3

Answer:
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.5-3

Show and Grow

Shade to compare the fractions.
Question 1.
Big Ideas Math Answer Key Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.5 4
Answer:
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.5-4
Explanation: if the numerators are the same, denominators are different then we have to compare the Denominator of both the fractions. The fraction with the Greater denominators is the lesser fraction

Question 2.
Big Ideas Math Answer Key Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.5 5
Answer:
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.5-5
Explanation: if the numerators are the same, denominators are different then we have to compare the Denominator of both the fractions. The fraction with the Greater denominators is the lesser fraction

Apply and Grow: Practice

Shade to compare the fractions
Question 3.
Big Ideas Math Answer Key Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.5 6
Answer:
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.5-6
Explanation: if the numerators are the same, denominators are different then we have to compare the Denominator of both the fractions. The fraction with the Greater denominators is the lesser fraction

Question 4.
Big Ideas Math Answer Key Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.5 7
Answer:
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.5-7
Explanation: if the numerators are the same, denominators are different then we have to compare the Denominator of both the fractions. The fraction with the Greater denominators is the lesser fraction

Compare
Question 5.
Big Ideas Math Answer Key Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.5 8
Answer:
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.5-8
Explanation: if the numerators are the same, denominators are different then we have to compare the Denominator of both the fractions. The fraction with the Greater denominators is the lesser fraction

Question 6.
Big Ideas Math Answer Key Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.5 9
Answer:
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.5-9
Explanation: if the numerators are the same, denominators are different then we have to compare the Denominator of both the fractions. The fraction with the Greater denominators is the lesser fraction

Question 7.
Big Ideas Math Answer Key Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.5 10
Answer:
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.5-10
Explanation: Here the numerator and denominator of the fractions are the same so  \(\frac{6}{8}\) = \(\frac{6}{8}\)

Question 8.
Big Ideas Math Answer Key Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.5 11
Answer:
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.5-11
Explanation: if the numerators are the same, denominators are different then we have to compare the Denominator of both the fractions. The fraction with the Greater denominators is the lesser fraction

Question 9.
Big Ideas Math Answer Key Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.5 12
Answer:
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.5-12
Explanation: 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

Question 10.
Big Ideas Math Answer Key Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.5 13
Answer:
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.5-13
Explanation: if the numerators are the same, denominators are different then we have to compare the Denominator of both the fractions. The fraction with the Greater denominators is the lesser fraction

Question 11.
Use the models to compare two fractions that have the same numerators.
Big Ideas Math Answer Key Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.5 14
Answer:
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.5-14
Here the numerators are the same and the denominators are different. The denominator with the highest number will be the lesser fraction.
8 > 4.
Thus 2/8 > 2/4

Question 12.
You spend \(\frac{2}{3}\) of an hour playing soccer on Wednesday. You spend \(\frac{2}{6}\) of an hour playing soccer on Friday. On which day do you spend more time playing soccer? Explain.
Answer:
Given,
You spend \(\frac{2}{3}\) of an hour playing soccer on Wednesday.
You spend \(\frac{2}{6}\) of an hour playing soccer on Friday.
\(\frac{2}{3}\) > \(\frac{2}{6}\)
Thus you spend more time playing soccer on wednesday.

Question 13.
YOU BE THE TEACHER
Your friend says \(\frac{1}{4}>\frac{1}{2}\) because \(\frac{1}{4}\) of one circle is bigger than \(\frac{1}{2}\) of the other circle. Is your friend correct? Explain.
Big Ideas Math Answer Key Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.5 15
Answer: Your friend is incorrect.
The size of the circles is different.
\(\frac{1}{4}>\frac{1}{2}\)

Think and Grow: Modeling Real Life

Newton and Descartes have piggy banks that are the same size. Newton fills \(\frac{3}{4}\) of his bank with pennies. Descartes fills \(\frac{3}{6}\) of his bank with pennies. Whose bank is filled more?
Model:
Big Ideas Math Answer Key Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.5 16
Explain:
______ bank is filled more.
Answer: Newton bank is filled more

Explanation:
Given,
Newton and Descartes have piggy banks that are the same size.
Newton fills \(\frac{3}{4}\) of his bank with pennies. Descartes fills \(\frac{3}{6}\) of his bank with pennies.
\(\frac{3}{4}\) > \(\frac{3}{6}\)
Thus Newton bank is filled more

Show and Grow

Question 14.
You use \(\frac{1}{3}\) of a bottle of glitter for an art project. Your friend uses \(\frac{1}{4}\) of a bottle for his art project. The bottles are the same size. Who uses more glitter?
Answer:
Given,
You use \(\frac{1}{3}\) of a bottle of glitter for an art project.
Your friend uses \(\frac{1}{4}\) of a bottle for his art project. The bottles are the same size.
\(\frac{1}{3}\) > \(\frac{1}{4}\)
Thus you used more glitter.

Question 15.
A train and a plane are traveling the same distance from New York to Pittsburgh. The train completes \(\frac{5}{8}\) of its journey. The plane completes \(\frac{5}{6}\) of its journey. Which is farther away from Pittsburgh? Explain.
Answer:
A train and a plane are traveling the same distance from New York to Pittsburgh.
The train completes \(\frac{5}{8}\) of its journey. The plane completes \(\frac{5}{6}\) of its journey.
\(\frac{5}{8}\) < \(\frac{5}{6}\)

Question 16.
DIG DEEPER!
You cut a piece of lasagna that is \(\frac{1}{8}\) of a tray. Your friend cuts a piece of lasagna that is \(\frac{1}{6}\) of a tray. Your piece is larger than your friend’s piece. Explain how this is possible.
Answer:
You cut a piece of lasagna that is \(\frac{1}{8}\) of a tray.
Your friend cuts a piece of lasagna that is \(\frac{1}{6}\) of a tray.
Your piece is larger than your friend’s piece.
\(\frac{1}{8}\) < \(\frac{1}{6}\)
Your piece is smaller than your friend’s piece.

Compare Fractions with Same Numerator Homework & Practice 11.5

Shade to compare the fractions
Question 1.
Big Ideas Math Answer Key Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.5 17
Answer:
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.5-17
Explanation: if the numerators are the same, denominators are different then we have to compare the Denominator of both the fractions. The fraction with the Greater denominators is the lesser fraction

Question 2.
Big Ideas Math Answer Key Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.5 18
Answer:
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.5-18
Explanation: if the numerators are the same, denominators are different then we have to compare the Denominator of both the fractions. The fraction with the Greater denominators is the lesser fraction

Compare
Question 3.
Big Ideas Math Answer Key Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.5 19
Answer:
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.5-19
Explanation: if the numerators are the same, denominators are different then we have to compare the Denominator of both the fractions. The fraction with the Greater denominators is the lesser fraction

Question 4.
Big Ideas Math Answer Key Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.5 20
Answer:
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.5-20
Explanation: if the numerators are the same, denominators are different then we have to compare the Denominator of both the fractions. The fraction with the Greater denominators is the lesser fraction

Question 5.
Big Ideas Math Answer Key Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.5 21
Answer:
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.5-21
Explanation: if the numerators are the same, denominators are different then we have to compare the Denominator of both the fractions. The fraction with the Greater denominators is the lesser fraction

Question 6.
Big Ideas Math Answer Key Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.5 22
Answer:
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.5-22
Explanation: if the numerators are the same, denominators are different then we have to compare the Denominator of both the fractions. The fraction with the Greater denominators is the lesser fraction

Question 7.
Big Ideas Math Answer Key Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.5 23
Answer:
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.5-23
Explanation: if the numerators are the same, denominators are different then we have to compare the Denominator of both the fractions. The fraction with the Greater denominators is the lesser fraction

Question 8.
Big Ideas Math Answer Key Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.5 24
Answer:
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.5-24
Explanation: if the numerators are the same, denominators are different then we have to compare the Denominator of both the fractions. The fraction with the Greater denominators is the lesser fraction

Question 9.
Use the models to compare two fractions that have the same numerators.
Big Ideas Math Answer Key Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.5 25
Answer:
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.5-25
Explanation: if the Numerator and denominators are different then we have to make the denominator as common and then compare the fractions. So \(\frac{1}{2}\) Equal to \(\frac{4}{8}\).

Question 10.
Newton says that if two fractions have the same numerator, then the fraction with the greater denominator is the greater fraction. Is Newton correct? Explain.
Answer: wrong

Explanation: if the numerators are the same, denominators are different then we have to compare the Denominator of both the fractions. The fraction with the Greater denominators is the lesser fraction.
Example: \(\frac{4}{4}\) Greater than \(\frac{4}{5}\).

Question 11.
Writing
How is comparing fractions with the same numerator similar to comparing fractions with the same denominator? How is it different?
Answer:
if the numerators are the same, denominators are different then we have to compare the Denominator of both the fractions. The fraction with the Greater denominators is the lesser fraction.
Example: \(\frac{4}{4}\) Greater than \(\frac{4}{5}\).
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.
Example: \(\frac{5}{8}\) Greater than \(\frac{4}{8}\).

Question 12.
Modeling Real Life
You play a video game for \(\frac{1}{2}\) of an hour. Your friend plays the video game for \(\frac{1}{6}\) of an hour. Who plays the video 6 game longer? Explain.
Big Ideas Math Answer Key Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.5 26
Answer:
Given,
You play a video game for \(\frac{1}{2}\) of an hour.
Your friend plays the video game for \(\frac{1}{6}\) of an hour.
\(\frac{1}{2}\) > \(\frac{1}{6}\)
Thus you play a video game for a long time.

Question 13.
Modeling Real Life
Two cars have the same-sized gasoline tank. Tank A is \(\frac{4}{6}\) full. Tank B is \(\frac{4}{8}\) full. Which tank is less full? Explain.
Answer:
Given,
Two cars have the same-sized gasoline tank.
Tank A is \(\frac{4}{6}\) full.
Tank B is \(\frac{4}{8}\) full.
\(\frac{4}{6}\) > \(\frac{4}{8}\)
Tank B is less filled.

Review & Refresh

Plot the fraction on the number line.
Question 14.
\(\frac{1}{4}\)
Big Ideas Math Answer Key Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.5 27
Answer:
Plot the given fraction on the below number line..
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.5-27

Question 15.
\(\frac{2}{3}\)
Big Ideas Math Answer Key Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.5 28
Answer:
Plot the given fraction on the below number line..
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.5-28

Lesson 11.6 Compare Fractions on a Number Line

Explore and Grow

Use Fraction Strips to plot each fraction on the number line. Circle the greater fraction.
Big Ideas Math Answers 3rd Grade Chapter 11 Understand Fraction Equivalence and Comparison 11.6 1
Answer:

Explanation: 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

Construct Arguments
Tell your partner how to use a number line to compare the fractions.
Big Ideas Math Answers 3rd Grade Chapter 11 Understand Fraction Equivalence and Comparison 11.6 2
Answer:
Step-1 Draw a number line from 0 to 1
Step-2 plot the fractions from \(\frac{0}{7}\) to \(\frac{7}{7}\)
Step-3 Now put a mark on \(\frac{1}{7}\) \(\frac{3}{7}\)
Step-4 Now compare the distance of both the fractions on the number line
Step-5 Here 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
Step-6 \(\frac{1}{7}\) less than \(\frac{3}{7}\)

Think and Grow: Compare Fractions on a Number Line

On a number line, fractions to the left are less than fractions to the right. Fractions to the right are greater than fractions to the left.
Example
Use a number line to compare \(\frac{3}{4}\) and \(\frac{1}{4}\).
Plot the fractions on a number line. Both have the same denominator.
Big Ideas Math Answers 3rd Grade Chapter 11 Understand Fraction Equivalence and Comparison 11.6 3
Big Ideas Math Answers 3rd Grade Chapter 11 Understand Fraction Equivalence and Comparison 11.6 4

Answer:
Big-Ideas-Math-Answers-3rd-Grade-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.6-3

Example
Use number lines to compare \(\frac{2}{8}\) and \(\frac{2}{3}\).
Plot each fraction on a number line. Both fractions have the same numerator.
Big Ideas Math Answers 3rd Grade Chapter 11 Understand Fraction Equivalence and Comparison 11.6 5

Answer:
Big-Ideas-Math-Answers-3rd-Grade-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.6-5

Show and Grow

Use the number line to compare the fractions.
Question 1.
Big Ideas Math Answers 3rd Grade Chapter 11 Understand Fraction Equivalence and Comparison 11.6 6
Answer:
Big-Ideas-Math-Answers-3rd-Grade-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.6-6
Explanation: 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. So \(\frac{5}{8}\) Greater than \(\frac{4}{6}\).

Question 2.
Big Ideas Math Answers 3rd Grade Chapter 11 Understand Fraction Equivalence and Comparison 11.6 7
Answer:
Big-Ideas-Math-Answers-3rd-Grade-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.6-7
Explanation: if the Numerator and denominators are different then we have to make the denominator as common and then compare the fractions. So \(\frac{1}{2}\) equal to  \(\frac{2}{4}\).

Apply and Grow: Practice

Compare.
Question 3.
Big Ideas Math Answers 3rd Grade Chapter 11 Understand Fraction Equivalence and Comparison 11.6 8
Answer:
Big-Ideas-Math-Answers-3rd-Grade-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.6-8
Explanation: if the Numerator and denominators are different then we have to make the denominator as common and then compare the fractions. So \(\frac{3}{4}\) equal to  \(\frac{6}{8}\).

Question 4.
Big Ideas Math Answers 3rd Grade Chapter 11 Understand Fraction Equivalence and Comparison 11.6 9
Answer:
Big-Ideas-Math-Answers-3rd-Grade-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.6-9
Explanation: if the numerators are the same, denominators are different then we have to compare the Denominator of both the fractions. The fraction with the Greater denominators is the lesser fraction. So \(\frac{4}{6}\) lesser than \(\frac{4}{4}\).

Question 5.
Big Ideas Math Answers 3rd Grade Chapter 11 Understand Fraction Equivalence and Comparison 11.6 10
Answer:
Big-Ideas-Math-Answers-3rd-Grade-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.6-10
Explanation: if the numerators are the same, denominators are different then we have to compare the Denominator of both the fractions. The fraction with the Greater denominators is the lesser fraction. So \(\frac{1}{2}\) Greater than \(\frac{4}{4}\).

Question 6.
Big Ideas Math Answers 3rd Grade Chapter 11 Understand Fraction Equivalence and Comparison 11.6 11
Answer:
Big-Ideas-Math-Answers-3rd-Grade-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.6-11
Explanation: if the numerators are the same, denominators are different then we have to compare the Denominator of both the fractions. The fraction with the Greater denominators is the lesser fraction. So \(\frac{2}{8}\) lesser than \(\frac{2}{6}\).

Question 7.
Big Ideas Math Answers 3rd Grade Chapter 11 Understand Fraction Equivalence and Comparison 11.6 12
Answer:
Big-Ideas-Math-Answers-3rd-Grade-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.6-12
Explanation: if the Numerator and denominators are different then we have to make the denominator as common and then compare the fractions. So \(\frac{4}{8}\) equal to \(\frac{1}{2}\).

Question 8.
Big Ideas Math Answers 3rd Grade Chapter 11 Understand Fraction Equivalence and Comparison 11.6 13
Answer:
Big-Ideas-Math-Answers-3rd-Grade-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.6-13
Explanation: if the numerators are the same, denominators are different then we have to compare the Denominator of both the fractions. The fraction with the Greater denominators is the lesser fraction. So \(\frac{4}{4}\) Greater than \(\frac{4}{6}\).

Question 9.
Big Ideas Math Answers 3rd Grade Chapter 11 Understand Fraction Equivalence and Comparison 11.6 14
Answer:
Big-Ideas-Math-Answers-3rd-Grade-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.6-14
Explanation: 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. So \(\frac{5}{8}\) lesser than \(\frac{7}{8}\).

Question 10.
Big Ideas Math Answers 3rd Grade Chapter 11 Understand Fraction Equivalence and Comparison 11.6 15
Answer:
Big-Ideas-Math-Answers-3rd-Grade-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.6-15
Explanation: 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. So \(\frac{1}{6}\) Lesser than \(\frac{2}{6}\).

Question 11.
Write a fraction that is greater than \(\frac{5}{6}\).
Answer: \(\frac{6}{6}\).

Explanation: 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. So \(\frac{6}{6}\) Greater than \(\frac{5}{6}\).

Question 12.
Write a fraction that is less than \(\frac{3}{4}\).
Answer: \(\frac{1}{4}\).

Explanation: 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. So \(\frac{1}{4}\) lesser than \(\frac{3}{4}\).

Question 13.
Newton buys \(\frac{1}{4}\) pound of cheese. Descartes buys \(\frac{3}{4}\) pound of cheese. Who buys more cheese?
Big Ideas Math Answers 3rd Grade Chapter 11 Understand Fraction Equivalence and Comparison 11.6 16
Answer:  Descartes buys More \(\frac{3}{4}\) pound of cheese.

Explanation: 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. So \(\frac{1}{4}\) lesser than \(\frac{3}{4}\).

Question 14.
Number Sense
Which fractions are greater than \(\frac{2}{4}\)?
Big Ideas Math Answers 3rd Grade Chapter 11 Understand Fraction Equivalence and Comparison 11.6 17
Answer:  \(\frac{4}{4}\) is greater than \(\frac{2}{4}\)

Explanation: 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. So \(\frac{4}{4}\) is greater than \(\frac{2}{4}\)

Think and Grow: Modeling Real Life

Does the recipe call for more salt or more pepper?
Big Ideas Math Answers 3rd Grade Chapter 11 Understand Fraction Equivalence and Comparison 11.6 18
Models:
Big Ideas Math Answers 3rd Grade Chapter 11 Understand Fraction Equivalence and Comparison 11.6 19
The recipe calls for more _______.
Explain:

Answer:
By seeing the above table we can find the recipe call for more salt or more pepper.
The recipe calls for more salt.
\(\frac{1}{4}\) > \(\frac{1}{2}\)

Show and Grow

Question 15.
Does the recipe above call for more Creole seasoning or more lemon juice?
Answer:
By seeing the above table we can find the recipe call for more Creole seasoning or more lemon juice.
The recipe calls for more lemon juice.
\(\frac{3}{4}\) > \(\frac{3}{8}\)

Question 16.
You and your friend each catch a fish. Your fish is \(\frac{5}{3}\) feet long, and your friend’s fish is \(\frac{2}{3}\) foot long. Whose fish is shorter? Explain.
Answer:
Given,
You and your friend each catch a fish.
Your fish is \(\frac{5}{3}\) feet long, and your friend’s fish is \(\frac{2}{3}\) foot long.
\(\frac{5}{3}\) > \(\frac{2}{3}\)
5 is greater than 2.
Thus your friend’s fish is shorter.

Question 17.
DIG DEEPER!
You eat \(\frac{4}{8}\) of a quesadilla. Your friend eats \(\frac{2}{8}\) of a different quesadilla. Can you tell who ate more quesadilla? Explain
Big Ideas Math Answers 3rd Grade Chapter 11 Understand Fraction Equivalence and Comparison 11.6 20
Answer:
Given,
You eat \(\frac{4}{8}\) of a quesadilla. Your friend eats \(\frac{2}{8}\) of a different quesadilla.
\(\frac{4}{8}\) > \(\frac{2}{8}\)
Thus you ate more quesadilla.

Compare Fractions on a Number Line Homework & Practice 11.6

Use the number line to compare the fractions
Question 1.
Big Ideas Math Answers 3rd Grade Chapter 11 Understand Fraction Equivalence and Comparison 11.6 21
Answer:
Big-Ideas-Math-Answers-3rd-Grade-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.6-21
Explanation: 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. So \(\frac{2}{8}\) Lesser than \(\frac{5}{8}\).

Question 2.
Big Ideas Math Answers 3rd Grade Chapter 11 Understand Fraction Equivalence and Comparison 11.6 22
Answer:
Big-Ideas-Math-Answers-3rd-Grade-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.6-22
Explanation: if the Numerator and denominators are different then we have to make the denominator as common and then compare the fractions. So \(\frac{2}{6}\) Equal to \(\frac{1}{3}\).

Compare
Question 3.
Big Ideas Math Answers 3rd Grade Chapter 11 Understand Fraction Equivalence and Comparison 11.6 23
Answer:
Explanation: if the numerators are the same, denominators are different then we have to compare the Denominator of both the fractions. The fraction with the Greater denominators is the lesser fraction. So \(\frac{1}{3}\) Greater than \(\frac{1}{4}\).

Question 4.
Big Ideas Math Answers 3rd Grade Chapter 11 Understand Fraction Equivalence and Comparison 11.6 24
Answer:
Big-Ideas-Math-Answers-3rd-Grade-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.6-24
Explanation: 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. So \(\frac{3}{6}\) lesser than \(\frac{4}{6}\).

Question 5.
Big Ideas Math Answers 3rd Grade Chapter 11 Understand Fraction Equivalence and Comparison 11.6 25
Answer:
Big-Ideas-Math-Answers-3rd-Grade-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.6-25
Explanation: 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. So \(\frac{1}{3}\) lesser than \(\frac{2}{3}\).

Question 6.
Big Ideas Math Answers 3rd Grade Chapter 11 Understand Fraction Equivalence and Comparison 11.6 26
Answer:
Big-Ideas-Math-Answers-3rd-Grade-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.6-26
Explanation: if the Numerator and denominators are different then we have to make the denominator as common and then compare the fractions. So \(\frac{1}{3}\) Equal to \(\frac{2}{6}\).

Question 7.
Big Ideas Math Answers 3rd Grade Chapter 11 Understand Fraction Equivalence and Comparison 11.6 27
Answer:
Big-Ideas-Math-Answers-3rd-Grade-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.6-27
Explanation: if the Numerator and denominators are different then we have to make the denominator as common and then compare the fractions. So \(\frac{2}{4}\) Greater than \(\frac{2}{6}\).

Question 8.
Big Ideas Math Answers 3rd Grade Chapter 11 Understand Fraction Equivalence and Comparison 11.6 28
Answer:
Big-Ideas-Math-Answers-3rd-Grade-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.6-28
Explanation: 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. So \(\frac{1}{6}\) lesser than \(\frac{3}{6}\).

Question 9.
Big Ideas Math Answers 3rd Grade Chapter 11 Understand Fraction Equivalence and Comparison 11.6 29
Answer:
Big-Ideas-Math-Answers-3rd-Grade-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.6-29
Explanation: if the numerators are the same, denominators are different then we have to compare the Denominator of both the fractions. The fraction with the Greater denominators is the lesser fraction. So \(\frac{4}{8}\) lesser than \(\frac{4}{4}\).

Question 10.
Big Ideas Math Answers 3rd Grade Chapter 11 Understand Fraction Equivalence and Comparison 11.6 30
Answer:
Big-Ideas-Math-Answers-3rd-Grade-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.6-30
Explanation: if the numerators are the same, denominators are different then we have to compare the Denominator of both the fractions. The fraction with the Greater denominators is the lesser fraction. So \(\frac{1}{6}\) lesser than \(\frac{1}{3}\).

Question 11.
Big Ideas Math Answers 3rd Grade Chapter 11 Understand Fraction Equivalence and Comparison 11.6 31
Answer:
Big-Ideas-Math-Answers-3rd-Grade-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.6-31
Explanation: 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. So \(\frac{3}{4}\) Greater than \(\frac{1}{4}\).

Question 12.
Write a fraction that is greater than \(\frac{1}{3}\).
Answer: \(\frac{2}{3}\)

Explanation: if the numerators are the same, denominators are different then we have to compare the Denominator of both the fractions. The fraction with the Greater denominators is the lesser fraction. So \(\frac{2}{3}\) Greater than \(\frac{1}{3}\).

Question 13.
Write a fraction that is less than \(\frac{1}{2}\).
Answer: \(\frac{1}{6}\)

Explanation: if the numerators are the same, denominators are different then we have to compare the Denominator of both the fractions. The fraction with the Greater denominators is the lesser fraction. So \(\frac{1}{6}\) lesser than \(\frac{1}{2}\).

Question 14.
Writing
Explain how to compare \(\frac{2}{3}\) and \(\frac{4}{3}\) on a number line.
Answer:
Big Ideas Math Answer Grade 3 Chapter 11
By seeing the number line we can say that \(\frac{2}{3}\) < \(\frac{4}{3}\)

Question 15.
Number Sense
Which fractions are less than \(\frac{4}{6}\)?

Big Ideas Math Answers 3rd Grade Chapter 11 Understand Fraction Equivalence and Comparison 11.6 32
Answer:
\(\frac{2}{6}\) and \(\frac{1}{6}\) are less than \(\frac{4}{6}\).

Modeling Real Life
Use the recipe shown.
Big Ideas Math Answers 3rd Grade Chapter 11 Understand Fraction Equivalence and Comparison 11.6 33
Question 16.
Does the recipe call for more butter or more sugar? Explain.
Answer:
Recipe call for butter is \(\frac{1}{3}\)
Recipe call for sugar is \(\frac{2}{3}\)
\(\frac{1}{3}\) < \(\frac{2}{3}\)
Thus the recipe call for more sugar.

Question 17.
Does the recipe call for less flour or less nuts? Explain.
Answer:
Recipe call for flour is \(\frac{3}{2}\)
Recipe call for nuts is \(\frac{3}{4}\)
\(\frac{3}{2}\) > \(\frac{3}{4}\)
Thus the recipe call for less nuts.

Review & Refresh

Find the missing number.
Question 18.
98 + ____ = 98
Answer: 0

Explanation:
Add 98 with 0 we get 98

Question 19.
52 + (____ + 16) = (52 + 28) + 16
Answer: 28

Explanation:
The associative property allows us to change groupings of addition and keep the same value. (a+b)+c = a+(b+c)
hence the missing number is 28

Question 20.
33 + _____ + 46 = 33 + 46
Answer: 0
Explanation:
Fill the blank with 0 we get 33 + 46

Question 21.
14 + 67 = 67 + ______
Answer: 148

Explanation:
Add 14,67 & 67 we get 148

Lesson 11.7 Compare Fractions

Explore and Grow

Use a strategy to find the greater fraction.
\(\frac{2}{3}\)
\(\frac{2}{8}\)

Use a different strategy to check your answer.
Answer:
if the numerators are the same, denominators are different then we have to compare the Denominator of both the fractions. The fraction with the Greater denominators is the lesser fraction

Construct Arguments
Tell your partner which strategy you prefer. Explain.
Answer:
I will compare the denominators to find which fraction is greater or lesser.
In the given fraction \(\frac{2}{3}\), \(\frac{2}{8}\)
In this case, the numerators are the same and denominators are different.
The denominator with the greatest number will be the smallest one.
\(\frac{2}{3}\) > \(\frac{2}{8}\)

Think and Grow: Compare Fractions

Example
Compare \(\frac{5}{6}\) and \(\frac{2}{3}\).
Big Ideas Math Answers Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.7 1
Use Fraction Strips.
In some cases, it is easier to compare the This is an image333parts that are missing. Notice that each fraction is 1 part away from being equivalent to 1. Compare the missing pieces.
Big Ideas Math Answers Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.7 2

Answer:
Big-Ideas-Math-Answers-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.7-2

Example
Compare \(\frac{3}{8}\) and \(\frac{5}{8}\).
The denominators are the same. Compare the numerators, 3 and 5.
Big Ideas Math Answers Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.7 3

Answer:
Big-Ideas-Math-Answers-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.7-3

Example
Compare \(\frac{4}{6}\) and \(\frac{4}{8}\).
The numerators are the same. Look at the denominators 6 and 8 to compare the sizes of the parts. The fraction with the greater denominator is the lesser fraction.
Big Ideas Math Answers Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.7 4

Answer:
Big-Ideas-Math-Answers-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.7-4

Show and Grow

Compare.
Question 1.
Big Ideas Math Answers Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.7 5
Answer:

Big-Ideas-Math-Answers-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.7-5

Explanation: if the numerators are the same, denominators are different then we have to compare the Denominator of both the fractions. The fraction with the Greater denominators is the lesser fraction. So \(\frac{1}{2}\) Greater than \(\frac{1}{8}\).

Question 2.
Big Ideas Math Answers Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.7 6
Answer:
Big-Ideas-Math-Answers-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.7-6
Explanation: if the Numerator and denominators are different then we have to make the denominator as common and then compare the fractions. So \(\frac{3}{4}\) lesser than \(\frac{5}{6}\).

Question 3.
Big Ideas Math Answers Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.7 7
Answer:
Big-Ideas-Math-Answers-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.7-7
Explanation: if the numerators are the same, denominators are different then we have to compare the Denominator of both the fractions. The fraction with the Greater denominators is the lesser fraction. So \(\frac{0}{6}\) Equal to \(\frac{0}{2}\).

Question 4.
Big Ideas Math Answers Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.7 8
Answer:
Big-Ideas-Math-Answers-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.7-8
Explanation: 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. So \(\frac{8}{8}\) Greater than \(\frac{7}{8}\).

Apply and Grow: Practice

Compare
Question 5.
Big Ideas Math Answers Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.7 9
Answer:
Big-Ideas-Math-Answers-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.7-9
Explanation: 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. So \(\frac{1}{4}\) lesser than \(\frac{2}{4}\).

Question 6.
Big Ideas Math Answers Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.7 10
Answer:
Big-Ideas-Math-Answers-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.7-10
Explanation: if the numerators are the same, denominators are different then we have to compare the Denominator of both the fractions. The fraction with the Greater denominators is the lesser fraction. So \(\frac{3}{4}\) Greater than \(\frac{3}{6}\).

Question 7.
Big Ideas Math Answers Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.7 11
Answer:
Big-Ideas-Math-Answers-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.7-11
Explanation: if the Numerator and denominators are different then we have to make the denominator as common and then compare the fractions. So \(\frac{7}{8}\) Greater than \(\frac{3}{4}\).

Question 8.
Big Ideas Math Answers Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.7 12
Answer:
Big-Ideas-Math-Answers-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.7-12
Explanation: 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. So \(\frac{1}{6}\) lesser than \(\frac{3}{6}\).

Question 9.
Big Ideas Math Answers Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.7 13
Answer:
Big-Ideas-Math-Answers-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.7-13
Explanation: if the Numerator and denominators are different then we have to make the denominator as common and then compare the fractions. So \(\frac{1}{4}\) equal to \(\frac{2}{8}\).

Question 10.
Big Ideas Math Answers Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.7 14
Answer:
Big-Ideas-Math-Answers-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.7-14
Explanation: if the Numerator and denominators are different then we have to make the denominator as common and then compare the fractions. So \(\frac{2}{4}\) equal to \(\frac{2}{8}\).

Question 11.
Big Ideas Math Answers Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.7 15
Answer:
Big-Ideas-Math-Answers-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.7-15
Explanation: if the Numerator and denominators are different then we have to make the denominator as common and then compare the fractions. So \(\frac{1}{2}\) equal to \(\frac{4}{8}\).

Question 12.
Big Ideas Math Answers Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.7 16
Answer:
Big-Ideas-Math-Answers-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.7-16
Explanation: if the Numerator and denominators are different then we have to make the denominator as common and then compare the fractions. So \(\frac{2}{3}\) lesser than \(\frac{5}{6}\).

Question 13.
Big Ideas Math Answers Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.7 17
Answer:
Big-Ideas-Math-Answers-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.7-17
Explanation: 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. So \(\frac{1}{3}\) lesser than \(\frac{2}{3}\).

Complete the statement.
Question 14.
Big Ideas Math Answers Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.7 18
Answer:
Big-Ideas-Math-Answers-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.7-18
Explanation: 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. So \(\frac{3}{6}\) lesser than \(\frac{5}{6}\).

Question 15.
Big Ideas Math Answers Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.7 19
Answer:
Big-Ideas-Math-Answers-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.7-19
Explanation: 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. So \(\frac{2}{8}\) Greater than \(\frac{1}{8}\).

Question 16.
Big Ideas Math Answers Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.7 20
Answer:
Big-Ideas-Math-Answers-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.7-20
Explanation: 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. So \(\frac{1}{4}\) lesser than \(\frac{3}{4}\).

Question 17.
Newton and Descartes have the same-sized lunch. Newton eats \(\frac{2}{3}\) of his lunch. Descartes eats \(\frac{1}{2}\) of his lunch. Who eats more of his lunch?
Big Ideas Math Answers Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.7 21
Answer:
Given,
Newton and Descartes have the same-sized lunch. Newton eats \(\frac{2}{3}\) of his lunch. Descartes eats \(\frac{1}{2}\) of his lunch.
\(\frac{2}{3}\) > \(\frac{1}{2}\)
Thus Newton eats more of his lunch.

Question 18.
DIG DEEPER!
Use the fractions and symbols to make two true statements.
Big Ideas Math Answers Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.7 22
Answer:
Big-Ideas-Math-Answers-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.7-22

Think and Grow: Modeling Real Life

Newton walks \(\frac{3}{8}\) mile to school. Descartes walks \(\frac{5}{6}\) mile to school. Who walks farther to school?
Big Ideas Math Answers Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.7 23
Understand the problem:
Make a plan:
Solve:
_______ walks farther to school.

Answer:
Given,
Newton walks \(\frac{3}{8}\) mile to school.
Descartes walks \(\frac{5}{6}\) mile to school.
\(\frac{5}{6}\) > \(\frac{3}{8}\)
Thus Descartes walks farther to the school.

Show and Grow

Question 19.
Two apartment buildings have the same number of floors. New carpeting is installed on \(\frac{2}{3}\) of the floors in Building A and \(\frac{3}{6}\) of the floors in Building B. Which building has more floors with new carpeting?
Answer:
Given,
Two apartment buildings have the same number of floors.
New carpeting is installed on \(\frac{2}{3}\) of the floors in Building A and \(\frac{3}{6}\) of the floors in Building B.
\(\frac{2}{3}\) > \(\frac{3}{6}\)
Building A has more floors with new carpeting.

Question 20.
Newton buys \(\frac{5}{8}\) pound of peanuts and \(\frac{1}{2}\) pound of cashews.Which weighs less? Explain.
Big Ideas Math Answers Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.7 24
Answer:
Given,
Newton buys \(\frac{5}{8}\) pound of peanuts and \(\frac{1}{2}\) pound of cashews.
Compare both the fractions
\(\frac{5}{8}\) > \(\frac{1}{2}\)
Thus Cashews weights less than peanuts.

Question 21.
DIG DEEPER!
You are painting a backdrop for a school play. You paint \(\frac{1}{3}\) of the backdrop blue. You paint more of the backdrop yellow than blue. What fraction can represent the part of the backdrop that is yellow? Explain.
Answer:
Given,
You are painting a backdrop for a school play.
You paint \(\frac{1}{3}\) of the backdrop blue. You paint more of the backdrop yellow than blue.
\(\frac{3}{3}\) – \(\frac{1}{3}\) = \(\frac{2}{3}\)
Thus the fraction can represent the part of the backdrop that is yellow is \(\frac{2}{3}\)

Compare Fractions Homework & Practice 11.7

Compare
Question 1.
Big Ideas Math Answers Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.7 25
Answer:
Big-Ideas-Math-Answers-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.7-25
Explanation: if the Numerator and denominators are different then we have to make the denominator as common and then compare the fractions. So \(\frac{2}{4}\) Greater than \(\frac{2}{6}\).

Question 2.
Big Ideas Math Answers Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.7 26
Answer:
Big-Ideas-Math-Answers-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.7-26
Explanation: if the Numerator and denominators are different then we have to make the denominator as common and then compare the fractions. So \(\frac{1}{2}\) equal \(\frac{3}{6}\).

Question 3.
Big Ideas Math Answers Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.7 27
Answer:
Big-Ideas-Math-Answers-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.7-27
Explanation: 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. So \(\frac{1}{4}\) lesser than \(\frac{3}{4}\).

Question 4.
Big Ideas Math Answers Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.7 28
Answer:
Big-Ideas-Math-Answers-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.7-28
Explanation: 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. So \(\frac{4}{6}\) lesser than \(\frac{6}{6}\).

Question 5.
Big Ideas Math Answers Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.7 29
Answer:
Big-Ideas-Math-Answers-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.7-29
Explanation: if the Numerator and denominators are different then we have to make the denominator as common and then compare the fractions. So \(\frac{3}{4}\) equal \(\frac{6}{8}\).

Question 6.
Big Ideas Math Answers Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.7 30
Answer:
Big-Ideas-Math-Answers-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.7-30
Explanation: if the Numerator and denominators are different then we have to make the denominator as common and then compare the fractions. So \(\frac{1}{2}\) lesser than\(\frac{7}{8}\).

Question 7.
Big Ideas Math Answers Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.7 31
Answer:
Big-Ideas-Math-Answers-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.7-31
Explanation: if the Numerator and denominators are different then we have to make the denominator as common and then compare the fractions. So \(\frac{3}{6}\) Greater than \(\frac{3}{8}\).

Question 8.
Big Ideas Math Answers Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.7 32
Answer:
Big-Ideas-Math-Answers-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.7-32
Explanation: if the Numerator and denominators are different then we have to make the denominator as common and then compare the fractions. So \(\frac{2}{3}\) lesser than\(\frac{3}{4}\).

Complete the statement
Question 9.
Big Ideas Math Answers Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.7 33
Answer:
Big-Ideas-Math-Answers-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.7-33
Explanation: 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. So \(\frac{5}{6}\) lesser than \(\frac{6}{6}\).

Question 10.
Big Ideas Math Answers Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.7 34
Answer:
Big-Ideas-Math-Answers-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.7-34
Explanation: 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. So \(\frac{7}{8}\) Greater than \(\frac{6}{8}\).

Question 11.
Big Ideas Math Answers Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.7 35
Answer:
Big-Ideas-Math-Answers-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.7-35
Explanation: 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. So \(\frac{2}{4}\) lesser than \(\frac{3}{4}\).

Question 12.
Number Sense
Which statements are true?
Big Ideas Math Answers Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.7 36
Answer:
Big-Ideas-Math-Answers-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.7-36

Question 13.
Writing
Explain how to compare \(\frac{2}{3}\) and \(\frac{7}{8}\) using missing pieces.
Answer:
The least common denominator (LCD) is: 24.
Rewriting as equivalent fractions with the LCD:
2/3 = 16/24
7/8 = 21/24
Comparing the numerators of the equivalent fractions we have:
16/24 < 21/24
\(\frac{2}{3}\) < \(\frac{7}{8}\)

Question 14.
Modeling Real Life
Newton and Descartes each have a phone with the same amount of storage. Newton’s storage is \(\frac{3}{4}\) full. Descartes’s storage is \(\frac{2}{8}\) full. Whose phone has more storage left?
Big Ideas Math Answers Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.7 37
Answer:
Given,
Newton and Descartes each have a phone with the same amount of storage.
Newton’s storage is \(\frac{3}{4}\) full. Descartes’s storage is \(\frac{2}{8}\) full
\(\frac{3}{4}\) > \(\frac{2}{8}\)
Descartes phone has more storage left

Question 15.
DIG DEEPER!
A restaurant wants to serve two different-sized hamburgers. The larger hamburger will be \(\frac{1}{2}\) pound. Name a fraction that could represent the other hamburger’s size.
Big Ideas Math Answers Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.7 38
Answer:
Given,
A restaurant wants to serve two different-sized hamburgers. The larger hamburger will be \(\frac{1}{2}\) pound.
The other size could represent \(\frac{2}{4}\)

Review & Refresh

Question 16.
There are 37 students on a bus. The bus stops 5 times. Seven students get off at each stop. How many students are left on the bus?
Answer:
Given,
There are 37 students on a bus. The bus stops 5 times. Seven students get off at each stop.
5 × 7 = 35
37 – 35 = 2
Therefore 2 students left on the bus

Lesson 11.8 Compare and Order Fractions

Explore and Grow

Plot the fractions on the number line. Order the fractions from least to greatest.
Big Ideas Math Solutions Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.8 1
Answer:
Big-Ideas-Math-Solutions-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.8-1

Explanation: Ordering fraction simply means arranging fraction either from the smallest to the largest (ascending order) or largest to smallest (descending order). There are two common methods of ordering fractions.

Reasoning
How can you use the fraction \(\frac{1}{2}\) to compare and order fractions?
Answer: \(\frac{1}{2}\), \(\frac{2}{2}\), \(\frac{3}{2}\), \(\frac{4}{2}\)

Think and Grow: Compare and Order Fractions

Example
Order the fractions \(\frac{7}{8}, \frac{1}{8}\) and \(\frac{5}{8}\) from least to greatest.
Plot the fractions on the number line. All 3 fractions have the same denominator,8.
Big Ideas Math Solutions Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.8 2
Big Ideas Math Solutions Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.8 3

Answer:
Big-Ideas-Math-Solutions-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.8-2

Example
Order the fractions \(\frac{2}{4}, \frac{2}{3}\) and \(\frac{2}{6}\) and from least to greatest.
Use Fraction Strips. All three fractions have the same numerator, 2.
Big Ideas Math Solutions Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.8 4

Show and Grow

Question 1.
Order the fractions \(\frac{3}{3}, \frac{3}{4}\) and \(\frac{3}{8}\) from least to greatest.
Big Ideas Math Solutions Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.8 5
Answer:
Big-Ideas-Math-Solutions-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.8-5

Apply and Grow: Practice

Order the fractions from least to greatest
Question 2.
Big Ideas Math Solutions Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.8 6
Answer:
Big-Ideas-Math-Solutions-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.8-6
Explanation: 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
by using this statement we can write the order of fractions from least to greatest

Question 3.
Big Ideas Math Solutions Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.8 7
Answer:
Big-Ideas-Math-Solutions-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.8-7

Explanation: if the numerators are the same, denominators are different then we have to compare the Denominator of both the fractions. The fraction with the Greater denominators is the lesser fraction
by using this statement we can write the order of fractions from least to greatest

Question 4.
Big Ideas Math Solutions Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.8 8
Answer:
Big-Ideas-Math-Solutions-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.8-8
Explanation: 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
by using this statement we can write the order of fractions from least to greatest

Question 5.
Big Ideas Math Solutions Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.8 9
Answer:
Big-Ideas-Math-Solutions-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.8-9

Explanation: if the numerators are the same, denominators are different then we have to compare the Denominator of both the fractions. The fraction with the Greater denominators is the lesser fraction
by using this statement we can write the order of fractions from least to greatest

Order the fractions from greatest to least.
Question 6.
Big Ideas Math Solutions Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.8 10
Answer:
Big-Ideas-Math-Solutions-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.8-10
Explanation: 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
by using this statement we can write the order of fractions from Greatest to least

Question 7.
Big Ideas Math Solutions Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.8 11
Answer:
Big-Ideas-Math-Solutions-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.8-11
Explanation: if the numerators are the same, denominators are different then we have to compare the Denominator of both the fractions. The fraction with the Greater denominators is the lesser fraction
by using this statement we can write the order of fractions from Greatest to least

Question 8.
Reasoning
Three fractions have the same denominator. How do you know which fraction is the greatest?
Answer:
Explanation: 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.
Example:
Big-Ideas-Math-Solutions-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.8-10

Question 9.
Open-Ended
Write a fraction that is between \(\frac{2}{8}\) and \(\frac{2}{3}\).
Answer:

Question 10.
Patterns
The fractions below are in order from least to greatest. Describe and complete the pattern.
Big Ideas Math Solutions Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.8 12
Answer:
Big-Ideas-Math-Solutions-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.8-12

Think and Grow: Modeling Real Life

A construction crew replaces \(\frac{1}{8}\) mile of a road on Monday, \(\frac{1}{4}\)mile of the road on Tuesday, and \(\frac{1}{6}\) mile of the road on Wednesday. On which day does the crew replace the longest piece of the road? On which day does the crew replace the shortest piece? Explain.
Big Ideas Math Solutions Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.8 13
Model:
The crew replaces the longest piece of the road on ______ and the shortest piece on ______.
Explain:
The crew replaces the longest piece of the road on Tuesday and the shortest piece on Monday.

Explanation:
A construction crew replaces \(\frac{1}{8}\) mile of a road on Monday, \(\frac{1}{4}\)mile of the road on Tuesday, and \(\frac{1}{6}\) mile of the road on Wednesday
Order of fraction from longest piece to shortest piece is \(\frac{1}{4}\) , \(\frac{1}{6}\) \(\frac{1}{8}\)

Show and Grow

Question 11.
You place three plants in order from shortest to tallest. A cactus is \(\frac{4}{6}\) foot tall, a fern is \(\frac{4}{4}\) foot tall, and an aloe vera plant is \(\frac{4}{8}\) foot tall. Which plant is first? Which plant is last? Explain.
Answer:
Given,
place three plants in order from shortest to tallest. A cactus is \(\frac{4}{6}\) foot tall, a fern is \(\frac{4}{4}\) foot tall, and an aloe vera plant is \(\frac{4}{8}\) foot tall
\(\frac{4}{8}\) , \(\frac{4}{6}\) , \(\frac{4}{4}\)

Question 12.
You measure the lengths of three spiders. The longest one is \(\frac{3}{4}\) inch long. The shortest one is \(\frac{1}{4}\) inch long. What is a possible length for the third spider? Explain.
Big Ideas Math Solutions Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.8 14
Answer: \(\frac{2}{4}\)

Explanation:
Given,
You measure the lengths of three spiders. The longest one is \(\frac{3}{4}\) inch long. The shortest one is \(\frac{1}{4}\) inch long.
so possibility length for the third spider is \(\frac{2}{4}\)

Question 13.
DIG DEEPER!
You are ordering three fractions. You know which fraction is the greatest and which fraction is the least. What do you know about the third fraction?
Answer:
Example:
if You measure the lengths of three spiders. The longest one is \(\frac{3}{4}\) inch long. The shortest one is \(\frac{1}{4}\) inch long.
so possibility length for the third spider is \(\frac{2}{4}\)

Compare and Order Fractions Homework & Practice 11.8

Order the fractions from least to greatest
Question 1.
Big Ideas Math Solutions Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.8 15
Answer:
Big-Ideas-Math-Solutions-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.8-15

Explanation: if the numerators are the same, denominators are different then we have to compare the Denominator of both the fractions. The fraction with the Greater denominators is the lesser fraction
by using this statement we can write the order of fractions from Least to Greatest

Question 2.
Big Ideas Math Solutions Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.8 16
Answer:
Big-Ideas-Math-Solutions-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.8-16
Explanation: 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
by using this statement we can write the order of fractions from Least to Greatest

Question 3.
Big Ideas Math Solutions Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.8 17
Answer:
Big-Ideas-Math-Solutions-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.8-17
Explanation: 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
by using this statement we can write the order of fractions from Least to Greatest

Question 4.
Big Ideas Math Solutions Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.8 18
Answer:
Big-Ideas-Math-Solutions-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.8-18
Explanation: if the numerators are the same, denominators are different then we have to compare the Denominator of both the fractions. The fraction with the Greater denominators is the lesser fraction
by using this statement we can write the order of fractions from Least to Greatest

Order the fractions from greatest to least.
Question 5.
Big Ideas Math Solutions Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.8 19
Answer:
Big-Ideas-Math-Solutions-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.8-19
Explanation: if the numerators are the same, denominators are different then we have to compare the Denominator of both the fractions. The fraction with the Greater denominators is the lesser fraction
by using this statement we can write the order of fractions from Greatest to least

Question 6.
Big Ideas Math Solutions Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.8 20
Answer:
Big-Ideas-Math-Solutions-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.8-20
Explanation: 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
by using this statement we can write the order of fractions from Greatest to least

Question 7.
Big Ideas Math Solutions Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.8 21
Answer:
Big-Ideas-Math-Solutions-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.8-21
Explanation: 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
by using this statement we can write the order of fractions from Greatest to least

Question 8.
Big Ideas Math Solutions Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.8 22
Answer:
Big-Ideas-Math-Solutions-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.8-22
Explanation: if the numerators are the same, denominators are different then we have to compare the Denominator of both the fractions. The fraction with the Greater denominators is the lesser fraction
by using this statement we can write the order of fractions from Greatest to least

Question 9.
Reasoning
Three fractions have the same numerator. How do you know which fraction is the greatest?
Answer:if the numerators are the same, denominators are different then we have to compare the Denominator of both the fractions. The fraction with the Greater denominators is the lesser fraction.
Example:
Big-Ideas-Math-Solutions-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.8-22

Question 10.
Precision
Which set of fractions is ordered from least to greatest?
Big Ideas Math Solutions Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.8 23
Answer:
Big-Ideas-Math-Solutions-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-11.8-23
Explanation: if the numerators are the same, denominators are different then we have to compare the Denominator of both the fractions. The fraction with the Greater denominators is the lesser fraction
by using this statement we can write the order of fractions from least to greatest

Question 11.
Modeling Real Life
You survey your classmates about their favorite subject. \(\frac{5}{8}\) of the students choose math, \(\frac{1}{8}\) choose reading, \(\frac{2}{8}\) and choose science.Which subject receives the most votes? Which8subject receives the least votes? Explain.
Answer:
Given, You survey your classmates about their favorite subject. \(\frac{5}{8}\) of the students choose math, \(\frac{1}{8}\) choose reading, \(\frac{2}{8}\) and choose science
Math subject receives the most votes
\(\frac{5}{8}\) , \(\frac{2}{8}\) , \(\frac{1}{8}\)
Which subject receives the least vote is
reading of  \(\frac{1}{8}\) having least vote

Question 12.
Modeling Real Life
A carpenter has three drill bits. The thickest one is \(\frac{3}{4}\) inch. The thinnest one is \(\frac{3}{8}\) inch. What is a possible width for the third drill bit? Explain.
Big Ideas Math Solutions Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 11.8 24
Answer:  \(\frac{3}{6}\)

Explanation:
Given
A carpenter has three drill bits. The thickest one is \(\frac{3}{4}\) inch.
The thinnest one is \(\frac{3}{8}\) inch
The possibility of Width for the third drill bit is \(\frac{3}{6}\)

Review & Refresh

Question 13.
You clean the lunch tables. There are 6 rows of tables with 7 tables in each row. How many tables do you clean?
Answer: 42

Explanation:
Given clean lunch tables There are 6 rows of tables with 7 tables in each row
we apply the row multiply columns concept
so 6 × 7 we get 42
Therefore the number of tables cleaned is 42

Understand Fraction Equivalence and Comparison Performance Task

Question 1.
You invite five friends to dinner. You start making chili using the recipe shown at 4:00.
Big Ideas Math Answer Key Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 1
a.What time should you tell your friends the chili will be ready?
Answer: by seeing the above image you can tell to your friends that the chili will be ready in 60 minutes

b.Which ingredient do you use the same amount of as crushed garlic?
Answer: chili powder is the ingredient that use the same amount as crushed garlic

c. Do you use more chopped green pepper or onion?
Answer: we use more chopped green pepper than onion

d. How many ounces of beans do you use in all?
Answer: 2 8 ounce  cans of pinto beans and 2 8- an ounce cans of black beans
so we get 32 ounces of beans in all

e. You know there is \(\frac{1}{4}\) cup of diced tomatoes in each serving. How many cups of diced tomatoes do you need?
Answer:
No.of servings are 16
You know there is \(\frac{1}{4}\) cup of diced tomatoes in each serving
\(\frac{1}{4}\) × 16 we get 4
Thus we need 4 cups of diced tomatoes

f. You and your friends each eat 2 servings of chili. How many servings are left? Write an equation to solve. Use letters to represent unknown numbers.
Answer:  12
No.of servings are 16
You and your friends each eat 2 servings of chili we get 2+2= 4
thus 16-4 we get 12 servings are left

Big Ideas Math Answer Key Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 2

Understand Fraction Equivalence and Comparison Activity

Fraction Spin and Compare
Directions:
1.Players take turns.
2.On your turn, spin both spinners. Cover a box that makes the statement true.
3.If there are no fractions left that make the statement true, then you lose your turn.
4.Play until all boxes are covered.
5.The player with the most boxes covered wins!
Big Ideas Math Answer Key Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison 3

Understand Fraction Equivalence and Comparison Chapter Practice

11.1 Equivalent Fractions

Use the models to find an equivalent fraction. Both models show the same whole.
Question 1.
Big Ideas Math Answer Key Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison chp 1
Answer: \(\frac{2}{8}\)
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-chp-1
Explanation: \(\frac{1}{4}\) and \(\frac{2}{8}\) are equivalent fractions both gives the same answer.

Question 2.
Big Ideas Math Answer Key Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison chp 2
Answer:  \(\frac{4}{4}\)
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-chp-2
Explanation: \(\frac{8}{8}\) and \(\frac{4}{4}\) are equivalent fractions both gives the same answer.

Question 3.
Which One Doesn’t Belong?
Which one does belong with the other three?
Big Ideas Math Answer Key Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison chp 3
Answer:
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-chp-3
Explanation: above all the figures has the fraction \(\frac{4}{6}\) except circled figure
so second doest belong to the other three

11.2 Equivalent Fractions on a Number Line

Write two fractions that name the point shown.
Question 4.
Big Ideas Math Answer Key Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison chp 4
Answer:
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-chp-4
Explanation: \(\frac{6}{8}\) and \(\frac{3}{4}\) are equivalent fractions both gives the same answer.

Question 5.
Big Ideas Math Answer Key Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison chp 5
Answer:
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-chp-5
Explanation: \(\frac{2}{4}\) and \(\frac{1}{2}\) are equivalent fractions both gives the same answer.

11.3 Relate Fractions and Whole Numbers

Question 6.
Complete the number line. Then write equivalent fractions for the numbers 1 and 2.
Big Ideas Math Answer Key Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison chp 6
Answer:
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-chp-6

Write two equivalent fractions for the whole number.
Question 7.
Big Ideas Math Answer Key Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison chp 7
Answer:
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-chp-7

Explanation:
2 is equivalent to the fractions \(\frac{8}{4}\) and \(\frac{12}{6}\).

Question 8.
Big Ideas Math Answer Key Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison chp 8
Answer:
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-chp-8

Explanation:
6 is equivalent to the fractions \(\frac{18}{3}\) and \(\frac{24}{4}\).

Question 9.
Big Ideas Math Answer Key Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison chp 9
Answer:
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-chp-9

Explanation:
4 is equivalent to the fractions \(\frac{8}{2}\) and \(\frac{32}{8}\).

Write the equivalent whole number.
Question 10.
Big Ideas Math Answer Key Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison chp 10
Answer:
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-chp-10

The number equivalent to \(\frac{1}{1}\) and \(\frac{4}{4}\) is 1.

Question 11.
Big Ideas Math Answer Key Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison chp 11
Answer:
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-chp-11
The number equivalent to \(\frac{24}{8}\) and \(\frac{12}{4}\) is 3.

Question 12.
Big Ideas Math Answer Key Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison chp 12
Answer:
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-chp-12
The number equivalent to \(\frac{24}{6}\) and \(\frac{16}{4}\) is 4.

11.4 Compare Fractions with the Same Denominator

Shade to compare the fractions.
Question 13.
Big Ideas Math Answer Key Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison chp 13
Answer:
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-chp-13
Explanation: 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. So \(\frac{2}{3}\) Greater than \(\frac{1}{3}\).

Question 14.
Big Ideas Math Answer Key Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison chp 14
Answer:
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-chp-14
Explanation: 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. So \(\frac{1}{2}\) lesser than \(\frac{2}{2}\).

Compare
Question 15.
Big Ideas Math Answer Key Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison chp 15
Answer:
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-chp-15
Explanation: 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. So \(\frac{2}{4}\) lesser than \(\frac{3}{4}\).

Question 16.
Big Ideas Math Answer Key Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison chp 16
Answer:
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-chp-16
By seeing the above number we can find the equivalent fraction of
\(\frac{1}{3}\) = \(\frac{1}{3}\)

Question 17.
Big Ideas Math Answer Key Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison chp 17
Answer:
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-chp-17
Explanation: 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. So \(\frac{1}{6}\) lesser than \(\frac{5}{6}\).

11.5 Compare Fractions with the Same Numerator

Compare.
Question 18.
Big Ideas Math Answers 3rd Grade Chapter 11 Understand Fraction Equivalence and Comparison chp 18
Answer:
Big-Ideas-Math-Answers-3rd-Grade-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-chp-18
Explanation: if the numerators are the same, denominators are different then we have to compare the Denominator of both the fractions. The fraction with the Greater denominators is the lesser fraction. So \(\frac{1}{8}\) lesser than \(\frac{1}{6}\).

Question 19.
Big Ideas Math Answers 3rd Grade Chapter 11 Understand Fraction Equivalence and Comparison chp 19
Answer:
Big-Ideas-Math-Answers-3rd-Grade-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-chp-19
Explanation: if the numerators are the same, denominators are different then we have to compare the Denominator of both the fractions. The fraction with the Greater denominators is the lesser fraction. So \(\frac{4}{6}\) lesser than \(\frac{4}{4}\).

Question 20.
Big Ideas Math Answers 3rd Grade Chapter 11 Understand Fraction Equivalence and Comparison chp 20
Answer:
Big-Ideas-Math-Answers-3rd-Grade-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-chp-20
Explanation: if the numerators are the same, denominators are different then we have to compare the Denominator of both the fractions. The fraction with the Greater denominators is the lesser fraction. So \(\frac{2}{3}\) Greater than \(\frac{2}{4}\).

Question 21.
Big Ideas Math Answers 3rd Grade Chapter 11 Understand Fraction Equivalence and Comparison chp 21
Answer:
Big-Ideas-Math-Answers-3rd-Grade-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-chp-21
Explanation: if the numerators are the same, denominators are different then we have to compare the Denominator of both the fractions. The fraction with the Greater denominators is the lesser fraction. So \(\frac{3}{6}\) Greater than \(\frac{3}{8}\).

Question 22.
Big Ideas Math Answers 3rd Grade Chapter 11 Understand Fraction Equivalence and Comparison chp 22
Answer:
Big-Ideas-Math-Answers-3rd-Grade-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-chp-22
Explanation: if the numerators are the same, denominators are different then we have to compare the Denominator of both the fractions. The fraction with the Greater denominators is the lesser fraction. So \(\frac{6}{8}\) lesser than \(\frac{6}{6}

Question 23.
Big Ideas Math Answers 3rd Grade Chapter 11 Understand Fraction Equivalence and Comparison chp 23
Answer:
Big-Ideas-Math-Answers-3rd-Grade-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-chp-23
Explanation: if the numerators are the same, denominators are different then we have to compare the Denominator of both the fractions. The fraction with the Greater denominators is the lesser fraction. So [latex]\frac{3}{8}\) lesser than \(\frac{3}{4}

Question 24.
Modeling Real Life
Your glass of orange juice is [latex]\frac{1}{2}\) full. Your friend’s glass of orange juice is \(\frac{1}{3}\) full. Your friend has more orange juice. Explain how this is possible.
Big Ideas Math Answers 3rd Grade Chapter 11 Understand Fraction Equivalence and Comparison chp 24
Answer:
Given,
Your glass of orange juice is \(\frac{1}{2}\) full.
Your friend’s glass of orange juice is \(\frac{1}{3}\) full. Your friend has more orange juice.
\(\frac{1}{2}\) > \(\frac{1}{3}\)
Thus your friend has more orange juice.

11.6 Compare Fractions on a Number Line

Compare
Question 25.
Big Ideas Math Answers 3rd Grade Chapter 11 Understand Fraction Equivalence and Comparison chp 25
Answer:
Big-Ideas-Math-Answers-3rd-Grade-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-chp-25
Explanation: 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. So \(\frac{2}{3}\) Greater than \(\frac{1}{3}\).

Question 26.
Big Ideas Math Answers 3rd Grade Chapter 11 Understand Fraction Equivalence and Comparison chp 26
Answer:
Big-Ideas-Math-Answers-3rd-Grade-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-chp-26
Explanation: if the numerators are the same, denominators are different then we have to compare the Denominator of both the fractions. The fraction with the Greater denominators is the lesser fraction. So \(\frac{2}{6}\) lesser than \(\frac{2}{4}

Question 27.
Big Ideas Math Answers 3rd Grade Chapter 11 Understand Fraction Equivalence and Comparison chp 27
Answer:
Big-Ideas-Math-Answers-3rd-Grade-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-chp-27
Explanation: if the numerators are the same, denominators are different then we have to compare the Denominator of both the fractions. The fraction with the Greater denominators is the lesser fraction. So [latex]\frac{4}{8}\) lesser than \(\frac{4}{6}

Question 28.
Big Ideas Math Answers 3rd Grade Chapter 11 Understand Fraction Equivalence and Comparison chp 28
Answer:
Big-Ideas-Math-Answers-3rd-Grade-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-chp-28
Explanation: 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. So [latex]\frac{5}{8}\) Greater than \(\frac{3}{8}\).

Question 29.
Big Ideas Math Answers 3rd Grade Chapter 11 Understand Fraction Equivalence and Comparison chp 29
Answer:
Big-Ideas-Math-Answers-3rd-Grade-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-chp-29
Explanation: 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. So \(\frac{5}{6}\) Greater than \(\frac{2}{6}\).

Question 30.
Write a fraction that is greater than \(\frac{1}{8}\).
Answer: \(\frac{3}{8}\).

Explanation:
The fraction greater than \(\frac{1}{8}\) is \(\frac{3}{8}\).

Question 31.
Write a fraction that is less than \(\frac{2}{3}\).
Answer: \(\frac{1}{3}\).

Explanation:
The fraction less than \(\frac{2}{3}\) is \(\frac{1}{3}\).

11.7 Compare Fractions

Compare.
Question 32.
Big Ideas Math Answers 3rd Grade Chapter 11 Understand Fraction Equivalence and Comparison chp 32
Answer:
Big-Ideas-Math-Answers-3rd-Grade-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-chp-32
Explanation: if the numerators are the same, denominators are different then we have to compare the Denominator of both the fractions. The fraction with the Greater denominators is the lesser fraction. So \(\frac{1}{1}\) Greater than \(\frac{1}{2}\).

Question 33.
Big Ideas Math Answers 3rd Grade Chapter 11 Understand Fraction Equivalence and Comparison chp 33
Answer:
Big-Ideas-Math-Answers-3rd-Grade-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-chp-33
Explanation: 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. So \(\frac{3}{6}\) Greater than \(\frac{5}{6}\).

Question 34.
Big Ideas Math Answers 3rd Grade Chapter 11 Understand Fraction Equivalence and Comparison chp 34
Answer:
Big-Ideas-Math-Answers-3rd-Grade-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-chp-34
Explanation: if the numerators are the same, denominators are different then we have to compare the Denominator of both the fractions. The fraction with the Greater denominators is the lesser fraction. So \(\frac{4}{6}\) Greater than \(\frac{4}{8}\).

Question 35.
Big Ideas Math Answers 3rd Grade Chapter 11 Understand Fraction Equivalence and Comparison chp 35
Answer:
Big-Ideas-Math-Answers-3rd-Grade-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-chp-35
Explanation: 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. So \(\frac{1}{3}\) lesser than \(\frac{2}{3}\).

Question 36.
Big Ideas Math Answers 3rd Grade Chapter 11 Understand Fraction Equivalence and Comparison chp 36
Answer:
Big-Ideas-Math-Answers-3rd-Grade-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-chp-36
Explanation: if the numerators are the same, denominators are different then we have to compare the Denominator of both the fractions. The fraction with the Greater denominators is the lesser fraction. So \(\frac{2}{2}\) Greater than \(\frac{2}{8}\).

Question 37.
Big Ideas Math Answers 3rd Grade Chapter 11 Understand Fraction Equivalence and Comparison chp 37
Answer:
Big-Ideas-Math-Answers-3rd-Grade-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-chp-37
Explanation: if the Numerator and denominators are different then we have to make the denominator as common and then compare the fractions. So \(\frac{3}{4}\) lesser than \(\frac{7}{8}\).

Question 38.
Number Sense
Which statements are true?
Big Ideas Math Answers 3rd Grade Chapter 11 Understand Fraction Equivalence and Comparison chp 38
Answer:
Big-Ideas-Math-Answers-3rd-Grade-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-chp-38
all the statements are true

11.8 Compare and Order Fractions

Order the fractions from least to greatest.
Question 39.
Big Ideas Math Answers 3rd Grade Chapter 11 Understand Fraction Equivalence and Comparison chp 39
Answer:
Big-Ideas-Math-Answers-3rd-Grade-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-chp-39
Explanation: if the numerators are the same, denominators are different then we have to compare the Denominator of both the fractions. The fraction with the Greater denominators is the lesser fraction.

Question 40.
Big Ideas Math Answers 3rd Grade Chapter 11 Understand Fraction Equivalence and Comparison chp 40
Answer:
Big-Ideas-Math-Answers-3rd-Grade-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-chp-40
Explanation: 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.

Question 41.
Big Ideas Math Answers 3rd Grade Chapter 11 Understand Fraction Equivalence and Comparison chp 41
Answer:
Big-Ideas-Math-Answers-3rd-Grade-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-chp-41
Explanation: 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.

Question 42.
Big Ideas Math Answers 3rd Grade Chapter 11 Understand Fraction Equivalence and Comparison chp 42
Answer:
Big-Ideas-Math-Answers-3rd-Grade-Chapter-11-Understand-Fraction-Equivalence-and-Comparison-chp-42
Explanation: if the numerators are the same, denominators are different then we have to compare the Denominator of both the fractions. The fraction with the Greater denominators is the lesser fraction.

Question 43.
Modeling Real Life
You, your friend, and your cousin have the same-sized aquarium. You fill your aquarium \(\frac{2}{3}\) full, your friend fills hers \(\frac{2}{6}\) full, and your cousin fills his \(\frac{2}{4}\) full. Which aquarium is the least full? Which aquarium is the most full? Explain.
Big Ideas Math Answers 3rd Grade Chapter 11 Understand Fraction Equivalence and Comparison chp 43
Answer:
Given,
You, your friend, and your cousin have the same-sized aquarium.
You fill your aquarium \(\frac{2}{3}\) full, your friend fills hers \(\frac{2}{6}\) full, and your cousin fills his \(\frac{2}{4}\) full.
\(\frac{2}{6}\), \(\frac{2}{4}\), \(\frac{2}{3}\)

Conclusion:

I wish the detailed solutions of Big Ideas Math 3rd Grade Book Chapter 11 Understand Fraction Equivalence and Comparison are useful for the students to learn the concept. Please share this BIM Book Grade 3 Chapter 11 Understand Fraction Equivalence and Comparison Answer Key to your friends to solve their doubts and queries. You can contact us through the comment section. Stay in touch with us to get the latest editions of all the Big Ideas Math Grade 3 Chapters solutions.

Big Ideas Math Answers Grade 5 Chapter 3 Add and Subtract Decimals

Big Ideas Math Answers Grade 5 Chapter 3 Add and Subtract Decimals

All topics in Grade 5 Chapter 3 Add and Subtract Decimals are provided below. Big Ideas Math Answers Grade 5 Chapter 3 Add and Subtract Decimals topics are given with problems, answers, explanations, practice tests, quiz, and, etc. Take every option we provided for you to prepare well for the exam. Students can easily become math experts by referring to Big Ideas Math Answers Grade 5 Chapter 3 Add and Subtract Decimals. Download Big Ideas Grade 5 Math Answers Ch 3 Add and Subtract Decimals to refer to in later times of your practice. Free PDF of BIM Grade 5 Chapter 3 Math Answers are given on this page.

Big Ideas 5th Grade Chapter 3 Add and Subtract Decimals Math Book Answer Key

Practice all the problems given here and cross-check the answers to verify your preparation level. As per your preparation, concentrate on the topics. Practice more and gain more knowledge using Big Ideas Math Answers Grade 5 Chapter 3 Add and Subtract Decimals. Big Ideas Math Book 5th Grade Answer Key Chapter 3 Add and Subtract Decimals is the best source to prepare for the exams and grab knowledge.

Lesson: 1 Estimate Sum and Differences

Lesson: 2 Use Models to Add or Subtract Decimals

Lesson: 3 Add Decimals

Lesson: 4 Subtract Decimals

Lesson: 5 Add and Subtract Decimals

Lesson: 6 Use Mental Maths to Add or Subtract Decimals

Lesson: 7 Problem Solving: Money

Chapter: 3 – Add and Subtract Decimals

Lesson 3.1 Estimate Sum and Differences

Explore and Grow

Estimate each sum or difference.
Big Ideas Math Answer Key Grade 5 Chapter 3 Add and Subtract Decimals 3.1 1
Answer:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.1 Estimate Sum and Differences

Construct Arguments
Compare your answers to your partner’s answers. Explain why they are the same or why they are different.

Answer:
My Friend and My answers are the same in the process of addition and subtraction.

Think and Grow: Estimate Sums and Differences

You can use rounding or compatible numbers to estimate sums and differences of decimals.+
Example
Estimate 27.21 + 48.73.
One Way: Use rounding. Round each addend to the nearest whole number. Then find the sum of the rounded numbers.
Big Ideas Math Answer Key Grade 5 Chapter 3 Add and Subtract Decimals 3.1 2
So, 27.21 + 48.73 is about _____.
Answer:
So, 27.21 + 48.73 is about 76.

Explanation:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.1-Think and Grow-Estimate Sums and Differences-example

Another Way: Use compatible numbers.
Big Ideas Math Answer Key Grade 5 Chapter 3 Add and Subtract Decimals 3.1 3
So, 27.21 + 48.73 is about _______.
Answer:
So, 27.21 + 48.73 is about 80.

Explanation:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.1-Think and Grow-Estimate Sums and Differences-example.

Example
Estimate 388.5 + 103.2.
One Way: Use rounding. Round each number to the nearest ten. Then find the difference of the rounded numbers.
Big Ideas Math Answer Key Grade 5 Chapter 3 Add and Subtract Decimals 3.1 4
So, 388.5 – 103.2 is about ______.
Answer:
So, 388.5 – 103.2 is about 492.

Explanation:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.1-Think and Grow-Estimate Sums and Differences-Example..

Another Way: Use compatible numbers.
Big Ideas Math Answer Key Grade 5 Chapter 3 Add and Subtract Decimals 3.1 5
So, 388.5 – 103.2 is about ______.
Answer:
So, 388.5 – 103.2 is about  505.

Explanation:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.1-Think and Grow-Estimate Sums and Differences-Example....

Show and Grow

Estimate the sum or difference.
Question 1.
Big Ideas Math Answer Key Grade 5 Chapter 3 Add and Subtract Decimals 3.1 6
Answer:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.1-Think and Grow-Estimate Sums and Differences-Show and Grow1
Sum of 112.18 + 47.99 = 160.17

Question 2.
Big Ideas Math Answer Key Grade 5 Chapter 3 Add and Subtract Decimals 3.1 7
Answer:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.1-Think and Grow-Estimate Sums and Differences-Show and Grow2

Difference of 68.8 – 29.7 = 39.1.

Apply and Grow: Practice

Estimate the sum or difference.
Question 3.
Big Ideas Math Answer Key Grade 5 Chapter 3 Add and Subtract Decimals 3.1 8
Answer:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.1-Think and Grow-Apply and Grow-Practice1
Difference of 7.95 – 1.72 = 6.23.

Question 4.
Big Ideas Math Answer Key Grade 5 Chapter 3 Add and Subtract Decimals 3.1 9
Answer:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.1-Think and Grow-Apply and Grow-Practice4
Sum of 237.5 + 345.1 = 582.6.

Question 5.
67.5 + 15.1
Answer:
Sum of 67.5 + 15.1 = 82.6
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.1-Think and Grow-Apply and Grow-Practice5

Question 6.
418.9 – 14.7
Answer:
Difference of 418.9 – 14.7 = 404.2.
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.1-Think and Grow-Apply and Grow-Practice6

Question 7.
13.9 – 8.21
Answer:
Difference of 13.9 – 8.21 = 5.69.
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.1-Think and Grow-Apply and Grow-Practice7

Question 8.
208.6 + 56.9
Answer:
Sum of 208.6 + 56.9 = 265.5.
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.1-Think and Grow-Apply and Grow-Practice8

Question 9.
Number Sense
Descartes estimates a sum by rounding each number to the nearest ten. His estimate is 140. Which sums could he have estimated?
83.6 + 59.8
106.92 + 44.68
130.8 + 4.68
73.5 + 67.4
Answer:
Descartes estimates a sum of 73.5 + 67.4 = 140.9 to be 140 by rounding each number to the nearest ten.

Explanation:
First, we need to calculate the sum by rounding each number to the nearest ten:
83.6 + 59.8 = 143.4
106.92 + 44.68 = 151.6
130.8 + 4.68 = 135.48
73.5 + 67.4 = 140.9

Question 10.
Precision
Does rounding 209.11 – 104.53 to the nearest ten or to the nearest hundred give an estimate that is closer to the actual difference? Explain.
Answer:
Estimation to  the difference by rounding to nearest ten ( 209 – 105 = 104 )is closer to the actual difference ( 209.11 – 104.53 = 104.58 ).

Explanation:
First, Estimate the actual difference:
209.11 – 104.53 = 104.58
Second, Estimate the difference by rounding to nearest ten:
209 – 105 = 104
Third, Estimate the difference by rounding to nearest hundred:
200 – 100 = 100

Question 11.
DIG DEEPER!
Newton packs a suitcase to fly on a plane. His suitcase needs to weigh less than 50.0 pounds. Should Newton overestimate or underestimate the weight of his suitcase? Explain.
Answer: Newton should overestimate the weight of his suitcase because the weigh should not be more than 50.0 pounds.

Think and Grow: Modeling Real Life

Example
About how many feet taller is One World Trade Center than the Empire State Building?
Big Ideas Math Answer Key Grade 5 Chapter 3 Add and Subtract Decimals 3.1 10
Round the height of each building to the nearest hundred because you do not need a precise answer.
One World Trade Center: 1,800 feet
Empire State Building: 1,500 feet
Subtract the estimated height of the Empire State Building from the estimated height of One World Trade Center.
Big Ideas Math Answer Key Grade 5 Chapter 3 Add and Subtract Decimals 3.1 11
One World Trade Center is about ______ feet taller than the Empire State Building.
Answer:
One World Trade Center is about  300 feet taller than the Empire State Building.

Explanation:
One World Trade Center: 1,800 feet
Empire State Building: 1,500 feet
Subtract the estimated height of the Empire State Building from the estimated height of One World Trade Center.
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.1-Think and Grow-Modeling Real Life
Difference of 1800 – 1500 = 300 feet.

Show and Grow

Question 12.
About how many more tons does the African elephant weigh than the hippopotamus?
Big Ideas Math Answer Key Grade 5 Chapter 3 Add and Subtract Decimals 3.1 12
Answer:
Weight of African Elephant is 3.4 Tons more than the Weight of Hippopotamus.

Explanation:
Weight of African Elephant = 5.7 Tons
Weight of Hippopotamus = 2.3 Tons
Difference:
Weight of African Elephant – Weight of Hippopotamus = 5.7 – 2.3 = 3.4 Tons

 

Question 13.
One cup of wild rice has 6.5 grams of protein. One cup of shredded chicken has 36.9 more grams of protein than 1 cup of wild rice. About how many grams of protein are in 1 cup of chicken?
Answer:
1 cup of chicken = 43.4 grams of protein.

Explanation:
One cup of wild rice = 6.5 grams
One cup of shredded chicken has 36.9 more grams of protein than 1 cup of wild rice
=> 1 cup of chicken = 36.9 grams + 6.5 grams
=> 1 cup of chicken = 43.4 grams of protein.

Question 14.
The speed Earth orbits the Sun is 18.5 miles per second. This speed is 3.3 miles per second slower than the speed Venus orbits the Sun. About how fast does Venus orbit the Sun?
Answer:
Speed of Venus orbit the Sun is 15.2 miles fast than the Earth orbit per second.

Explanation:
Speed of Earth orbits the Sun = 18.5 miles
This speed is 3.3 miles per second slower than the speed Venus orbits the Sun
=> Speed of  Venus orbit the Sun = 18.5 – 3.3 miles = 15.2 miles

Estimate Sum and Differences Homework & Practice 3.1

Estimate the sum or difference.
Question 1.
Big Ideas Math Answer Key Grade 5 Chapter 3 Add and Subtract Decimals 3.1 13
Answer:
Sum of 76.81 + 2.17 =78.98.
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.1-Estimate Sum and Differences Homework & Practice 3.1.1

Question 2.
Big Ideas Math Answer Key Grade 5 Chapter 3 Add and Subtract Decimals 3.1 14
Answer:
Difference of 79.1 – 32.8 = 46.3
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.1-Estimate Sum and Differences Homework & Practice 3.1.2

Question 3.
418.44 – 23.61
Answer:
Difference of 418.44 – 23.61 = 442.05
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.1-Estimate Sum and Differences Homework & Practice 3.1.3

Question 4.
90.7 + 2.3
Answer:
Sum of 90.7 + 2.3 = 93
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.1-Estimate Sum and Differences Homework & Practice 3.1.4

Question 5.
6.24 – 3.8
Answer:
Difference of 6.24 – 3.8 = 2.44
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.1-Estimate Sum and Differences Homework & Practice 3.1.5...

Question 6.
837.51 + 149.27
Answer:
Sum of 837.51 + 149.27 = 986.78
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.1-Estimate Sum and Differences Homework & Practice 3.1.6

Question 7.
Open-Ended
Write a decimal addition problem that has an estimated sum of 35.
Answer:
Sum of 16.3+18.7 = 35

Explanation:
16.3+18.7 =??
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.1-Estimate Sum and Differences Homework & Practice 3.1.7-Open-Ended

Question 8.
Reasoning
Describe a situation when you would estimate the difference of 25.4 and 19.8 to the nearest whole number.
Answer:
Estimate the difference between 25.4  and  19.8 by to round each number to the nearest whole number and then subtract the rounded numbers.

Explanation:
Rounding number to nearest whole number of 25.4 = 25
Rounding number to nearest whole number of 19.8 = 20
Difference = Rounding number to nearest whole number of 25.4 – Rounding number to nearest whole number of 19.8
= 25 – 20
= 5

Question 9.
Modeling Real Life
Your friend participates in a 3.1-mile race. She walks for 1.4 miles of the race and runs the rest. About how many miles does your friend run?
Answer:
Number of miles my Friend runs = 1.7 miles.

Explanation:
Number of miles the race is my Friend participates  = 3.1 mils
Number of miles my Friend walks = 1.4 miles
Number of miles my Friend runs = Number of miles the race is my Friend participates  – Number of miles my Friend walks
=>  3.1 – 1.4
=> 1.7 miles.

Question 10.
DIG DEEPER!
Your phone has 16,000 megabytes of total storage space. About how much storage space is free?
Big Ideas Math Answer Key Grade 5 Chapter 3 Add and Subtract Decimals 3.1 15
Answer:
Free Storage space left in Phone = 2944.8 megabytes

Explanation:
Total storage space of Phone = 16000 megabytes
Memory Storage for Photos = 3111.1 megabytes
Memory Storage for Applications = 5345.3 megabytes
Memory Storage for Music = 4598.8 megabytes
Free Storage space = Total storage space of Phone – ( Memory Storage for Photos + Memory Storage for Applications +
Memory Storage for Music)
=> 16000 – ( 3111.1 + 5345.3 + 4598.8 )
=> 16000 – 13055.2
=> 2944.8 megabytes

Review & Refresh

Find the sum.
Question 11.
\(\frac{2}{8}+\frac{4}{8}\) = ______
Answer:
\(\frac{2}{8}+\frac{4}{8}\) = \(\frac{3}{4}\) .

Explanation:
\(\frac{2}{8}+\frac{4}{8}\) = \(\frac{2+4}{8}\) = \(\frac{6}{8}\) = \(\frac{3}{4}\)

Question 12.
\(\frac{1}{2}+\frac{5}{2}\) = ______
Answer:
\(\frac{1}{2}+\frac{5}{2}\) = \({3}\) .

Explanation:
\(\frac{1}{2}+\frac{5}{2}\) = \(\frac{1+5}{2}\) = \(\frac{6}{2}\) = \({3}\)

Question 13.
1 + \(\frac{1}{4}\) = ______
Answer:
1 + \(\frac{1}{4}\) = \(\frac{5}{4}\) .

Explanation:
1 + \(\frac{1}{4}\) = \(\frac{4+1}{4}\) = \(\frac{5}{4}\)

Lesson 3.2 Use Models to Add or Subtract Decimals

Explore and Grow

Use base ten blocks to find 1.4 + 0.7.
Big Ideas Math Answers 5th Grade Chapter 3 Add and Subtract Decimals 3.2 1

Use base ten blocks to find 1.16 – 0.89.
Answer:
So, 1.16 – 0.89 = 0.27

Explanation:
1.16 – 0.89 = Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.2 Use Models to Add or Subtract Decimals. – Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.2 Use Models to Add or Subtract Decimals...= Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.2 Use Models to Add or Subtract Decimals.....

Repeated Reasoning
Why can you use base ten blocks to model ones, tenths, and hundredths?
Answer:
Yes, we can use base ten blocks to model ones, tenths, and hundredths because they represent the numeral values in blocks which helps is east calculation and understanding.

Think and Grow: Use Models to Add or Subtract Decimals

You have used models to add or subtract whole numbers. You can use similar models to add or subtract decimals.
Example
Find 1.5 + 2.7.
Make a quick sketch. Add the tenths, then the ones.
Regroup 10 tenths as 1 one.
Big Ideas Math Answers 5th Grade Chapter 3 Add and Subtract Decimals 3.2 2
So, 1.5 + 2.7 = _____.
Answer:
So, 1.5 + 2.7 = 3.2

Explanation:
Big Ideas Math Answers 5th Grade Chapter 3 Add and Subtract Decimals 3.2 2

Example
Find 2.42 – 1.24.

So, 2.42 – 1.24 = _____.
Answer:
So, 2.42 – 1.24 = 1.18

Explanation:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.2-Think and Grow-Use Models to Add or Subtract Decimals-Example

Show and Grow

Make a quick sketch to find the sum or difference.
Question 1.
3.15 + 0.92 = _____
Big Ideas Math Answers 5th Grade Chapter 3 Add and Subtract Decimals 3.2 4
Answer:
3.15 + 0.92 = 4.07

Explanation:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.2-Show and Grow1

Question 2.
1.26 – 0.83 = _____
Big Ideas Math Answers 5th Grade Chapter 3 Add and Subtract Decimals 3.2 5
Answer:
1.26 – 0.83 = 0.43

Explanation:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.2-Show and Grow2

Apply and Grow: Practice

Make a quick sketch to find the sum or difference.
Question 3.
1.61 + 0.22 = _____
Answer:
1.61 + 0.22 = 1.83

Explanation:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.2-Apply and Grow-Practice3

Question 4.
2.4 – 1.3 = _____
Answer:
2.4 – 1.3 = 1.1

Explanation:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.2-Apply and Grow-Practice4

Question 5.
2.2 + 0.9 = _______
Answer:
2.2 + 0.9 = 3.1

Explanation:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.2-Apply and Grow-Practice5

Question 6.
1.56 – 1.08 = ______
Answer:
1.56 – 1.08 = 0.48

Explanation:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.2-Apply and Grow-Practice6

Question 7.
Number Sense
Newton adds two decimals that are each greater than 0.5. What can you conclude about the sum?
Answer:
If Newton’s adds two decimals that are each greater than 0.5 then the Sum of the decimal numbers is going to be more than 0.5.

Explanation:
First Decimal Number= 1.2
Second Decimal Number = 0.8
Sum = First Decimal Number + Second Decimal Number
= 1.2 + 0.8 = 2.0

Question 8.
Structure
Write an equation represented by the quick sketch.
Big Ideas Math Answers 5th Grade Chapter 3 Add and Subtract Decimals 3.2 6
Answer:
The Equation for the given quick Sketch :
3.54 + 1.63 = 5.17

Question 9.
DIG DEEPER!
Use a model to find 1.31 – 0.4.
Answer:
1.31 – 0.4 = 0.91.54

Explanation:
1.31 – 0.4 = ???
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.2-Apply and Grow-Practice6-DIG DEEPER!-9

Think and Grow: Modeling Real Life

Example
How much longer is the duration of the wooden roller-coaster ride than the duration of the steel roller-coaster ride?
Big Ideas Math Answers 5th Grade Chapter 3 Add and Subtract Decimals 3.2 7
Find the difference of the durations: 4.25 – 1.50.
Make a quick sketch of 4.25.
Big Ideas Math Answers 5th Grade Chapter 3 Add and Subtract Decimals 3.2 8
So, 4.25 – 1.50
Answer:
So, 4.25 – 1.50 = 2.75.

Explanation:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.2-Think and Grow-Modeling Real Life.
So, 4.25 – 1.50 = ???
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.2-Think and Grow-Modeling Real Life...

Show and Grow

Question 10.
A football player completes a 40-yard dash in 5.84 seconds. Another player completes it in 4.56 seconds. How much longer does it take the first player to complete the 40-yard dash than the second player?
Answer:
Time taken by the the first player to complete the 40-yard dash 1.28 seconds more than the second player.

Explanation:
Time taken by the First player to complete the 40-yard dash = 5.84 seconds
Time taken by the Second player to complete the 40-yard dash = 4.56 seconds
Time taken by the the first player to complete the 40-yard dash than the second player = Time taken by the First player to complete the 40-yard dash – Time taken by the Second player to complete the 40-yard dash
= 5.84 seconds – 4.56 seconds
= 1.28 seconds.

Question 11.
What is the total length of the narwhal?
Big Ideas Math Answers 5th Grade Chapter 3 Add and Subtract Decimals 3.2 9
Answer:
Total Length of whole body Narwhal = 16.40 feet.

Explanation:
Length of Narwhal tusk = 4.92 feet
Length of Narwhal Body = 11.48 feet
Total Length of whole body Narwhal = Length of Narwhal tusk + Length of Narwhal Body
= 4.92 feet + 11.48 feet
= 16.40 feet.

Question 12.
DIG DEEPER!
An Olympic gymnast scores 6.6 points for skill. She starts with 10.0 points for execution, then loses 0.4 point for execution errors. What is the gymnast’s final score?
Answer:
Final score points of Olympic gymnast = 16.2.

Explanation:
Score points of Olympic gymnast for skill = 6.6
Score points of Olympic gymnast for execution = 10.0
Score points lost of Olympic gymnast for execution errors = 0.4
Final score points of Olympic gymnast = Score points of Olympic gymnast for skill + Score points of Olympic gymnast for execution – Score points lost of Olympic gymnast for execution errors
= 6.6 + 10.0 – 0.4
= 16.6 – 0.4
= 16.2

Use Models to Add or Subtract Decimals Homework & Practice 3.2

Make a quick sketch to find the sum or difference.
Question 1.
2.6 + 1.3 = ______
Big Ideas Math Answers 5th Grade Chapter 3 Add and Subtract Decimals 3.2 10
Answer:
2.6 + 1.3 = 3.9.

Explanation:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.2-Use Models to Add or Subtract Decimals Homework & Practice 3.2.1

Question 2.
1.36 – 0.25 = _____
Big Ideas Math Answers 5th Grade Chapter 3 Add and Subtract Decimals 3.2 11
Answer:
1.36 – 0.25 = 1.11

Explanation:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.2-Use Models to Add or Subtract Decimals Homework & Practice 3.2.2

Question 3.
0.84 – 0.17 = _____
Answer:
0.84 – 0.17 = 0.67.

Explanation:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.2-Use Models to Add or Subtract Decimals Homework & Practice 3.2.3

Question 4.
1.25 + 1.55 = _____
Answer:
1.25 + 1.55 = 2.80.

Explanation:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.2-Use Models to Add or Subtract Decimals Homework & Practice 3.2.4

Question 5.
2.6 – 1.9 = ______
Answer:
2.6 – 1.9 = 0.7.

Explanation:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.2-Use Models to Add or Subtract Decimals Homework & Practice 3.2.5

Question 6.
3.9 + 0.6 = ______
Answer:
3.9 + 0.6 = 4.5.

Explanation:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.2-Use Models to Add or Subtract Decimals Homework & Practice 3.2.6

Question 7.
Structure
Write an equation represented by the quick sketch.
Big Ideas Math Answers 5th Grade Chapter 3 Add and Subtract Decimals 3.2 12
Answer:

Question 8.
Number Sense
What is the greatest possible difference of two decimals to the hundredths place that are between 0 and 1?
Answer:
0.01 is the greatest possible difference of two decimals to the hundredths place that are between 0 and 1.

Question 9.
DIG DEEPER!
Write a decimal addition equation that requires regrouping the tenths. Explain how you know you need to regroup to add.
Answer:
Find the Sum of 1.5 +2.7 = 4.2

Explanation:

Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.2-DIG DEEPER-9.

Question 10.
Modeling Real Life
The fastest 1-mile sack race was completed in 16.68 minutes. The fastest 1-mile piggyback race was completed in 11.18 minutes. How much longer did it take to complete the sack race than the piggyback race?
Answer:
Time taken to complete the sack race than the piggyback race = 5.5 minutes.

Explanation:
Time taken to complete fastest 1-mile sack race = 16.68 minutes
Time taken to complete fastest 1-mile piggyback race  = 11.18 minutes
Time taken to complete the sack race than the piggyback race = Time taken to complete fastest 1-mile sack race – Time taken to complete fastest 1-mile piggyback race
= 16.68  – 11.18
= 5.5 minutes

Question 11
Modeling Real Life
How many pounds of honey do the hives produce in all?
Big Ideas Math Answers 5th Grade Chapter 3 Add and Subtract Decimals 3.2 13
Answer:
Total number of pounds of Honey produced by Hive A  and Hive B = 36.23 lb.

Explanation:
Number of pounds of Honey Hive A  produce = 15.25 lb.
Number of pounds of Honey Hive B  produce = 20.98 lb.
Total number of pounds of Honey produced Hive A and Hive B = Number of pounds of Honey Hive A  produce + Number of pounds of Honey Hive B  produce
= 15.25 + 20.98
= 36.23 lb.

Review & Refresh

Find the perimeter of the rectangle.
Question 12.
Big Ideas Math Answers 5th Grade Chapter 3 Add and Subtract Decimals 3.2 14
Answer:
Perimeter of Rectangle = 102 yards.

Explanation:
Length of Rectangle = 32 yards.
Width of Rectangle = 19 yards
Perimeter of Rectangle = 2 (l + w)
= 2( 32 + 19 )
= 2 × 51
= 102 yards.

Question 13.
Big Ideas Math Answers 5th Grade Chapter 3 Add and Subtract Decimals 3.2 15
Answer:
Perimeter of Rectangle = 114 mm.

Explanation:
Length of Rectangle = 36 mm
Width of Rectangle = 21 mm
Perimeter of Rectangle = 2 (l + w)
= 2 ( 36 + 21 )
= 2 × 57
= 114 mm.

Lesson 3.3 Add Decimals

Explore and Grow

Find the sum.
Big Ideas Math Answers Grade 5 Chapter 3 Add and Subtract Decimals 3.3 1
Big Ideas Math Answers Grade 5 Chapter 3 Add and Subtract Decimals 3.3 2

Use a model to find the sum. Draw your model.
Big Ideas Math Answers Grade 5 Chapter 3 Add and Subtract Decimals 3.3 3
Answer:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.3 Add DecimalsBig-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.3 Add Decimals..
Explanation:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.3 Add Decimals..

Construct Arguments
+-?
Answer:
The adding of decimals is same as to that of adding of whole numbers because here the process we do is adding the numbers in both.

Think and Grow: Add Decimals

To add decimals, add like place values.
Example
Add: 52.76 + 38.18. Estimate ______
Big Ideas Math Answers Grade 5 Chapter 3 Add and Subtract Decimals 3.3 4
Answer:
Sum of 52.76  +38.18 = 90.94
90.94 is close to 91.

Show and Grow

Find the sum. Check whether your answer is reasonable.
Question 1.
Big Ideas Math Answers Grade 5 Chapter 3 Add and Subtract Decimals 3.3 5
Answer:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.3 Add Decimals-Show and Grow1..
5.6 is close to 6.

Explanation:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.3 Add Decimals-Show and Grow1

Question 2.
Big Ideas Math Answers Grade 5 Chapter 3 Add and Subtract Decimals 3.3 6
Answer:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.3 Add Decimals-Show and Grow2
1.76 is close to 2.

Explanation:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.3 Add Decimals-Show and Grow2

Question 3.
Big Ideas Math Answers Grade 5 Chapter 3 Add and Subtract Decimals 3.3 7
Answer:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.3 Add Decimals-Show and Grow3.
93.7 is close to 94.

Explanation:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.3 Add Decimals-Show and Grow3.

Question 4.
Big Ideas Math Answers Grade 5 Chapter 3 Add and Subtract Decimals 3.3 7
Answer:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.3 Add Decimals-Show and Grow4...
16.51 is close to 17.

Explanation:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.3 Add Decimals-Show and Grow4

Apply and Grow: Practice

Find the sum. Check whether your answer is reasonable.
Question 5.
Big Ideas Math Answers Grade 5 Chapter 3 Add and Subtract Decimals 3.3 9
Answer:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.3 Add Decimals-Show and Grow3...
8.89 is close to 9.

Explanation:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.3 Add Decimals-Apply and Grow-Practice1
Check:
8.89 – 1.07
= 7.82.

Question 6.
Big Ideas Math Answers Grade 5 Chapter 3 Add and Subtract Decimals 3.3 10
Answer:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.3 Add Decimals-Show and Grow6...
38.7 is close to 39.

Explanation:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.3 Add Decimals-Apply and Grow-Practice6
Check:
38.7 – 15.1
= 23.6.

Question 7.
Big Ideas Math Answers Grade 5 Chapter 3 Add and Subtract Decimals 3.3 11
Answer:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.3 Add Decimals-Show and Grow7....
640.89 is close to 641.

Explanation:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.3 Add Decimals-Show and Grow7

Check:
640.89 – 78.04
= 562.65.

Question 8.
2.5 + 4.6 = _____
Answer:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.3 Add Decimals-Show and Grow8..
7.1 is close to 7.

Explanation:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.3 Add Decimals-Show and Grow8

Check:
7.1 – 4.6
= 2.5.

Question 9.
107.5 + 63.5 = _____
Answer:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.3 Add Decimals-Show and Grow9..
171.0 is close to 171.

Explanation:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.3 Add Decimals-Show and Grow9
Check:
171.0 – 63.5
= 107.5.

Question 10.
49.87 + 32.53 = ______
Answer:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.3 Add Decimals-Show and Grow10..
82.40 is close to 82.

Explanation:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.3 Add Decimals-Show and Grow10
Check:
82.40 – 32.53
= 49.87.

Question 11.
96.5 + 4.1 = _____
Answer:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.3 Add Decimals-Show and Grow11..
100.6 is close to 101.

Explanation:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.3 Add Decimals-Show and Grow11
Check:
100.6 – 4.1
= 96.5.

Question 12.
Number Sense
Which expressions have a sum greater than 24.7?
20.9 + 3.8
15.96 + 9.51
14.23 + 9.85
10.62 + 14.11
Answer:
The Sum of 15.96 + 9.51= 25.74 which is  greater than 24.7.

Explanation:
First, find the sum:
20.9 + 3.8 = 24.7
15.96 + 9.51 = 25.47
14.23 + 9.85 = 24.08
10.62 + 14.11 = 24.73

Question 13.
YOU BE THE TEACHER
Your friend finds 34.5 + 6.8 and gets a sum of 413. Without adding, explain how you know your friend’s answer is correct.
Answer:
My Friend’s answer is incorrect because he did not add the digits correctly .

Explanation:
Find the Sum :
34.5 + 6.8 = 41.3
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.3 Add Decimals-YOU BE THE TEACHER13

Think and Grow: Modeling Real Life

Example
You have 6.5 megabytes of storage space left on your tablet. Do you have enough space to download both of the ebooks?
Add the ebook file sizes.
Big Ideas Math Answers Grade 5 Chapter 3 Add and Subtract Decimals 3.3 12
Compare the sum of the file sizes to the available space on your tablet.
You ______ have enough room to download both ebooks.
Answer:
I  have enough space to download both ebooks.

Explanation:
Storage space for tale of the Monster = 3.8 megabytes
Storage space for In the Game = 2.5 megabytes
Total storage for tale of the Monster and In the Game = 3.8 + 2.5 = 6.3 megabytes
Storage space left on the Tablet = 6.5 megabytes
Storage space left after downloading the eBooks = Storage space left on the Tablet  – Total storage for tale of the Monster and In the Game
= 6.5 – 6.3
= 0.2 megabytes

Show and Grow

Question 14.
A chef needs 3.75 pounds of fish for a dinner party. She catches a fish that is 2.50 pounds and another fish that is 1.75 pounds. Does she catch enough fish for her dinner party?
Answer:
She catches enough fish for her dinner party in fact 0.5 pounds more weight fish she catches.

Explanation:
Weight of first fish = 2.50 pounds
Weight of Second Fish = 1.75 pounds
Weight of both Fishes = 2.50 + 1.75 = 4.25 pounds
Weight of the fish needed for a dinner party = 3.75 pounds
Difference:
Weight of fishes caught – Weight of the fish needed for a dinner party
= 4.25 – 3.75
= 0.5 pounds

Question 15.
Your school’s long jump record is 5.51 meters. You jump 3.38 meters on your first attempt. Your second attempt is 1.56 meters longer than your first. Do you break your school’s record?
Answer:
I have broke the school record by jumping more 2.81 meters than the record distance.

Explanation:
Distance long jump record in the school = 5.51 meters
Distance jumped on my first attempt = 3.38 meters
Your second attempt is 1.56 meters longer than your first.
=> Distance jumped on my second attempt = 1.56+ 3.38 meters = 4.94 meters
Total distance jumped by me = Distance jumped on my first attempt + Distance jumped on my second attempt
= 3.38 + 4.94
= 8.32 meters
Difference:
Total distance jumped by me – Distance long jump record in the school
= 8.32 – 5.51
= 2.81 meters

Question 16.
The melting point of aluminum is 403.86 degrees Celsius cooler than the melting point of gold. What is the melting point of gold?
Big Ideas Math Answers Grade 5 Chapter 3 Add and Subtract Decimals 3.3 13
Answer:
The melting point of Gold = 1063.32  degrees Celsius.

Explanation:
The melting point of aluminum = 660.32 degrees Celsius
The melting point of aluminum is 403.86 degrees Celsius cooler than the melting point of gold.
=> Let the melting point of Gold be X
=> X- 403.86 = 660.32
=> X = 660.32 + 403
=> X = 1063.32  degrees Celsius.

Add Decimals Homework & Practice 3.3

Find the sum. Check whether your answer is reasonable.
Question 1.
Big Ideas Math Answers Grade 5 Chapter 3 Add and Subtract Decimals 3.3 14
Answer:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.3 Add Decimals-Add Decimals Homework & Practice 3.3.1..
8.8 is close to 9.

Explanation:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.3 Add Decimals-Add Decimals Homework & Practice 3.3.1
Check:
8.8 – 1.3 = 7.5.

Question 2.
Big Ideas Math Answers Grade 5 Chapter 3 Add and Subtract Decimals 3.3 15
Answer:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.3 Add Decimals-Add Decimals Homework & Practice 3.3.2..
683.89 is close to 684.

Explanation:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.3 Add Decimals-Add Decimals Homework & Practice 3.3.2
Check:
683.89 – 82.31 = 601.58.

Question 3.
19.73 + 7.16 = _____
Answer:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.3 Add Decimals-Add Decimals Homework & Practice 3.3.3..
26.89 is close to 27.

Explanation:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.3 Add Decimals-Add Decimals Homework & Practice 3.3.3
Check:
26.89 – 7.16 = 19.73.

Question 4.
84.6 + 44.7 = ______
Answer:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.3 Add Decimals-Add Decimals Homework & Practice 3.3.4..
129.3 is close to 129.

Explanation:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.3-Add Decimals Homework & Practice 3.3.4
Check:
129.3 – 44.7 = 84.6.

Find the sum. Check whether your answer is reasonable.
Question 5.
123.98 + 47.65 = _____
Answer:
123.98 + 47.65 = 171.63.
171.63 is close to 172.

Explanation:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.3 Add Decimals-Add Decimals Homework & Practice 3.3.5
Check:
171.63 – 47.65
= 123.98.

Question 6.
427.8 + 13.2 = _____
Answer:
427.8 + 13.2 = 441.0
441 is close to 441.

Explanation:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.3 Add Decimals-Add Decimals Homework & Practice 3.3.6
Check:
441.0 – 13.2 = 427.8.

Question 7.
Reasoning
Your friend adds 46.7 and 5.4. Without calculating, is the sum greater than or less than 51? Explain.
Answer:
Answer is greater than 51 because the Sum of 46.7 + 5.4 = 52.1.

Explanation:
First Find the Sum:
46.7 + 5.4 = 52.1.
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.3 Add Decimals-Add Decimals Homework & Practice 3.3.7

Question 8.
Writing
Explain why it is important to use place value to line up the numbers when adding decimals.
Answer:
When we add whole numbers,  in the Value (cents) column above, you line up the numbers so that the digits in the ones place-value column are aligned. In order to keep the numbers in the proper place-value column when adding decimals, align the decimal points.

Question 9.
Modeling Real Life
You need 10.5 pounds of fruit for a fruit salad. You buy 3.8 pounds of apples and 6.7 pounds of pears. Do you buy enough fruit?
Big Ideas Math Answers Grade 5 Chapter 3 Add and Subtract Decimals 3.3 16
Answer:
Yes, there are enough fruits for a fruit salad.

Explanation:
Weight of fruits for a fruit salad needed = 10.5 pounds
Weight of Apples purchased = 3.8 pounds
Weight of pears purchased = 6.7 pounds
Total Weight of apples And Pears purchased = Weight of Apples purchased + Weight of pears purchased
= 3.8  + 6.7
= 10.5 pounds
=> Compare:
Weight of fruits for a fruit salad needed : Total Weight of apples And Pears purchased
= 10.5 pounds :10.5 pounds

Question 10.
Modeling Real Life
A scientist has 200.59 grams of mercury. She has 135.23 fewer grams of zinc. How many grams of zinc does she have?
Answer:
Zinc she has is 65.36 grams.

Explanation:
Mercury a Scientist has = 200.59 grams
She has 135.23 fewer grams of zinc
=> Let the Zinc be X
=> X + 135.23 = 200.59
=> X = 200.59 – 135.23
=> X = 65.36 grams.

Review & Refresh

Find the factor pairs for the number.
Question 11.
24
Answer:
Factor pairs of 24 :
1 × 24
2 × 12
3 × 8
4 × 6

Question 12.
36
Answer:
Factor pairs of 36 :
1 × 36
2 × 18
3 × 12
4 × 9
6 × 6

Question 13.
17
Answer:
Factor pairs of 17 :
1 × 17
17 is a prime number so it has 1 and itself as factors.

Lesson 3.4 Subtract Decimals

Explore and Grow

Find the difference.
Big Ideas Math Solutions Grade 5 Chapter 3 Add and Subtract Decimals 3.4 1
Big Ideas Math Solutions Grade 5 Chapter 3 Add and Subtract Decimals 3.4 2
Use a model to find the difference. Draw your model.
Big Ideas Math Solutions Grade 5 Chapter 3 Add and Subtract Decimals 3.4 3
Answer:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.4 Subtract DecimalsBig-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.4 Subtract Decimals..

Explanation:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.4 Subtract Decimals-

Construct Arguments
How is subtracting decimals similar to subtracting whole numbers?
Answer:
Subtracting decimals is similar to subtracting whole numbers because the process is same for both kind of numbers that is subtracting.

Think and Grow: Subtract Decimals

To subtract decimals, subtract like place values.

Example
Subtract: 18.93 – 7.66.
Big Ideas Math Solutions Grade 5 Chapter 3 Add and Subtract Decimals 3.4 4

Show and Grow

Find the difference. Then check your answer.
Question 1.
Big Ideas Math Solutions Grade 5 Chapter 3 Add and Subtract Decimals 3.4 5
Answer:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.4 Subtract Decimals-Show and Grow1..

Explanation:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.4 Subtract Decimals-Show and Grow1

Check:
36.3 + 35.3 = 71.6

Question 2.
Big Ideas Math Solutions Grade 5 Chapter 3 Add and Subtract Decimals 3.4 6
Answer:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.4 Subtract Decimals-Show and Grow2..

Explanation:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.4 Subtract Decimals-Show and Grow2
Check:
52.43 + 10.65 = 63.08.

Question 3.
Big Ideas Math Solutions Grade 5 Chapter 3 Add and Subtract Decimals 3.4 7
Answer:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.4 Subtract Decimals-Show and Grow3..

Explanation:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.4 Subtract Decimals-Show and Grow3
Check:
15.56 + 9.19 = 24.75.

Apply and Grow: Practice

Find the difference. Then check your answer.
Question 4.
Big Ideas Math Solutions Grade 5 Chapter 3 Add and Subtract Decimals 3.4 8
Answer:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.4 Subtract Decimals-Apply and Grow-Practice4..

Explanation:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.4 Subtract Decimals-Apply and Grow-Practice4
Check:
5.7 + 1.8 = 7.5.

Question 5.
Big Ideas Math Solutions Grade 5 Chapter 3 Add and Subtract Decimals 3.4 9
Answer:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.4 Subtract Decimals-Apply and Grow-Practice5..

Explanation:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.4 Subtract Decimals-Apply and Grow-Practice5
Check:
2.81 + 81.11 = 83.92.

Question 6.
Big Ideas Math Solutions Grade 5 Chapter 3 Add and Subtract Decimals 3.4 10
Answer:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.4 Subtract Decimals-Apply and Grow-Practice6..

Explanation:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.4 Subtract Decimals-Apply and Grow-Practice6
Check:
8.9 + 5.7 = 14.6.

Question 7.
856.02 – 48.12 = _______
Answer:
856.02 – 48.12 = 807.90.

Explanation:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.4 Subtract Decimals-Apply and Grow-Practice7
Check:
48.12 + 807.90 = 856.02.

Question 8.
3.76 – 0.53 = _____
Answer:
3.76 – 0.53 = 3.23.

Explanation:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.4 Subtract Decimals-Apply and Grow-Practice8
Check:
0.53 + 3.23 = 3.76.

Question 9.
27.5 – 21.7 = ______
Answer:
27.5 – 21.7 = 5.8.

Explanation:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.4 Subtract Decimals-Apply and Grow-Practice9
Check:
21.7 + 5.8 = 27.5.

Question 10.
737.2 – 2.8 = _____
Answer:
737.2 – 2.8 = 734.4.

Explanation:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.4 Subtract Decimals-Apply and Grow-Practice10
Check:
2.8 + 734.4 = 737.2.

Question 11.
YOU BE THE TEACHER
Your friend finds 534.8 – 23.9. Is your friend correct? Explain.
Big Ideas Math Solutions Grade 5 Chapter 3 Add and Subtract Decimals 3.4 11
Answer:
My Friend’s answer is incorrect because he misplaced the digits and then did their difference.

Explanation:
First find the Difference:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.4 Add Decimals-YOU BE THE TEACHER11

Question 12.
DIG DEEPER!
Find y.
Big Ideas Math Solutions Grade 5 Chapter 3 Add and Subtract Decimals 3.4 12
Answer:
y = 2.2

Explanation:
d + 26.4 = 38.7
=> d = 38.7 – 26.4
=> d = 12.3
d = y + 10.1
Substitute d = 12.3 in d = y + 10.1
=> d = y + 10.1
=> 12.3 = y + 10.1
=> y = 12.3 – 10.1
=> y = 2.2

Think and Grow: Modeling Real Life

Example
The world record for the longest dog tongue is 18.58 centimeters. The previous record was 11.43 centimeters. By how many centimeters was the record broken?
Big Ideas Math Solutions Grade 5 Chapter 3 Add and Subtract Decimals 3.4 13
Subtract the previous record,11.43 centimeters, from the current record,18.58 centimeters.
Big Ideas Math Solutions Grade 5 Chapter 3 Add and Subtract Decimals 3.4 14
So, the record was broken by ________ centimeters.

Answer:
Length of the record was broken by 7.15 centimeters.
Big Ideas Math Solutions Grade 5 Chapter 3 Add and Subtract Decimals 3.4 14

Explanation:
Length of world record for the longest dog tongue = 18.58 centimeters
Length of previous world record for the longest dog tongue = 11.43 centimeters
Length of the record was broken = Length of world record for the longest dog tongue – Length of previous world record for the longest dog tongue
= 18.58 centimeters – 11.43 centimeters
= 7.15 centimeters.

Show and Grow

Question 13.
The average high summer temperature in Death Valley, California, is 46.7 degrees Celsius. The hottest temperature ever recorded is 56.7 degrees Celsius. How much greater is the hottest recorded temperature than the average high temperature?
Answer:
The hottest recorded temperature is greater by 10 degrees Celsius than the average high temperature.

Explanation:
The average high summer temperature in Death Valley, California = 46.7 degrees Celsius
The hottest temperature ever recorded = 56.7 degrees Celsius.
Difference:
=The hottest temperature ever recorded  – The average high summer temperature in Death Valley, California
= 56.7 – 46.7
= 10 degrees Celsius

Question 14.
An athlete’s fitness tracker reads 2.69 miles at noon. The figure shows the athlete’s fitness tracker before going to sleep that night. How many miles were recorded after noon?
Big Ideas Math Solutions Grade 5 Chapter 3 Add and Subtract Decimals 3.4 15
Answer:
Reading of an athlete’s fitness tracker recorded after noon = 3.65 miles.

Explanation:
Reading of an athlete’s fitness tracker  at noon = 2.69 miles
Reading of an athlete’s fitness tracker before going to sleep=6.34 miles
Reading of an athlete’s fitness tracker recorded after noon = Reading of an athlete’s fitness tracker before going to sleep – Reading of an athlete’s fitness tracker  at noon
= 6.34 – 2.69
= 3.65 miles.

Question 15.
DIG DEEPER!
Your friend is using a pattern to knit a 139.7-centimeter-long scarf. He has already knitted 61.8 centimeters. How many centimeters of scarf does he have left to knit? Write your answer as a mixed number.
Answer:
Length of scarf  left to knit = 77.9/100 centimeters

Explanation:
Length of long scarf pattern to knit = 139.7 centimeter
Length of scarf already knitted = 61.8 centimeters
Length of scarf  left to knit = Length of long scarf pattern to knit – Length of scarf already knitted
= 139.7 centimeter – 61.8 centimeters
= 77.9 centimeters

Subtract Decimals Homework & Practice 3.4

Find the difference. Then check your answer.
Question 1.
Big Ideas Math Solutions Grade 5 Chapter 3 Add and Subtract Decimals 3.4 16
Answer:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.4 Subtract Decimals-Subtract Decimals Homework & Practice 3.4.1
Check:
21.5 + 13.0
= 34.5.

Question 2.
Big Ideas Math Solutions Grade 5 Chapter 3 Add and Subtract Decimals 3.4 17
Answer:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.4 Subtract Decimals-Subtract Decimals Homework & Practice 3.4.2
Check:
38.51 + 958.23
= 996.74.

Question 3.
Big Ideas Math Solutions Grade 5 Chapter 3 Add and Subtract Decimals 3.4 18
Answer:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.4 Subtract Decimals-Subtract Decimals Homework & Practice 3.4.3
Check:
3.8 + 19.7
= 23.5.

Find the difference. Then check your answer.
Question 4.
856.02 – 48.12 = ______
Answer:
856.02 – 48.12
= 807.90
Check:
48.12 + 807.90
= 856.02

Question 5.
3.76 – 0.53 = ______
Answer:
3.76 – 0.53
= 3.23
Check:
0.53 + 3.23
= 3.76.

Question 6.
27.5 – 21.7 = ______
Answer:
27.5 – 21.7
= 5.8
Check:
21.7 + 5.8
= 27.5.

Question 7.
737.2 – 2.8 = _______
Answer:
737.2 – 2.8
= 734.4.
Check:
2.8 + 734.4
= 737.2.

Question 8.
Number Sense
Which statements describe the difference of 134.6 and 78.9?
The difference is 55.7.
The difference is greater than 55.
The difference is less than 55.07.
Answer:
The difference is 55.7 is the correct statement.

Explanation:
First find the Difference:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.4 Subtract Decimals-Subtract Decimals Homework & Practice 3.4-Number Sense
The difference is 55.7.
The difference is greater than 55.
The difference is less than 55.07.

Question 9.
DIG DEEPER!
Find the missing digits.
Big Ideas Math Solutions Grade 5 Chapter 3 Add and Subtract Decimals 3.4 19
Answer:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.4 Subtract Decimals-Subtract Decimals Homework & Practice 3.4-DIG DEEPER-9
Explanation:
Missing digit in tenths place:
Add tenths place 3 + 8 =11, => 1
Subtract the tenths digits and ones digits to get the missing digit in tens place.

Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.4 Subtract Decimals-Subtract Decimals Homework & Practice 3.4-DIG DEEPER-9

Question 10.
Modeling Real Life
The world record for the longest fingernails on a pair of female hands is 576.41 centimeters. The world record for the longest fingernails on a single hand is 909.60 centimeters. How much longer is the world record for the longest fingernails on a single hand than the world record for the longest fingernails on a pair of female hands?
Answer:
The world record for the longest fingernails on a single hand is 333.19 centimeters longer than the world record for the longest fingernails on a pair of female hands.

Explanation:
Length of the world record for the longest fingernails on a pair of female hands = 576.41 centimeters.
Length of the world record for the longest fingernails on a single hand = 909.60 centimeters.
Difference:
Length of the world record for the longest fingernails on a single hand – Length of the world record for the longest fingernails on a pair of female hands
= 909.60 –  576.41
= 333.19 centimeters.

Review & Refresh

Find the product.
Question 11.
80 × 10 = ______
Answer:
Multiplication:
80 × 10 = 800.

Question 12.
30 × 90 = _______
Answer:
Multiplication:
30 × 90 = 2700.

Question 13.
60 × 70 = _______
Answer:
Multiplication:
60 × 70 = 4200.

Lesson 3.5 Add and Subtract Decimals

Explore and Grow

Add the decimals.
Big Ideas Math Answer Key Grade 5 Chapter 3 Add and Subtract Decimals 3.5 1
Answer:
1.36 + 2.5 + 2.78
= 3.86 + 2.78
= 6.64.

Explanation:
Count the given values:
1.36 + 2.5 + 2.78 = ???
Add the decimal values:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.5 Add and Subtract Decimals

Reasoning
How did you find the sum? What other ways can you find the sum?
Answer:
First count the figures given place values and later add the decimals values.
Other ways are like counting total figures n doing the addition of them.

Think and Grow: Add and Subtract Decimals

Example
Add: 7.39 + 4.16 + 9. Estimate ________
Use place value to line up the addends. Add the hundredths, then the tenths, and then the ones.
Big Ideas Math Answer Key Grade 5 Chapter 3 Add and Subtract Decimals 3.5 2

Answer:
7.39 + 4.16 + 9.00 = 20.55.
20.55 is close to 21.

Explanation:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.5 Add and Subtract Decimals-Think and Grow-Add and Subtract Decimals

Example
Find 18.46 + 13.31 – 5.6. Estimate _______
Use order of operations.
Big Ideas Math Answer Key Grade 5 Chapter 3 Add and Subtract Decimals 3.5 3
Answer:
26.17 is close to 26.

Explanation:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.5 Add and Subtract Decimals-Think and Grow-Add and Subtract Decimals-example

Show and Grow

Evaluate. Check whether your answer is reasonable.
Question 1.
6.84 + 2.9 + 5.22 = ______
Answer:
6.84 + 2.9 + 5.22 = 14.96
14.96 is close to 15.

Explanation:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.5 Add and Subtract Decimals-Show and Grow1
Check:
14.96 – 5.22 – 2.9
= 9.74 – 2.9
=6.84.

Question 2.
41.55 – 14.09 + 42.4 = ______
Answer:
41.55 – 14.09 + 42.4 = 69.86.
69.86 is close to 70.

Explanation:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.5 Add and Subtract Decimals-Show and Grow2
Check:
69.86 – 42.4 + 14.09
= 27.46 + 14.09
= 41.55.

Apply and Grow: Practice

Evaluate. Check whether your answer is reasonable.
Question 3.
5.3 – 1.7 – 2 = _____
Answer:
5.3 – 1.7 – 2 = 3.6 – 2 = 1.6
1.6 is close to 2.

Explanation:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.5 Add and Subtract Decimals-Apply and Grow-Practice3
Check:
1.6 + 2 + 1.7
= 3.6 + 1.7
= 5.3.

Question 4.
14.08 + 6.15 – 9.3 = ______
Answer:
14.08 + 6.15 – 9.3 = 10.93
10.93 is close to 11.

Explanation:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.5 Add and Subtract Decimals-Apply and Grow-Practice4
Check:
10.93 + 9.3 – 6.15
= 20.23 – 6.15
= 14.08.

Question 5.
24.7 – 5.1 + 3.75 = _____
Answer:
24.7 – 5.1 + 3.75 = 19.6 + 3.75 = 23.35
23.35 is close to 23.

Explanation:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.5 Add and Subtract Decimals-Apply and Grow-Practice5
Check:
23.35 – 3.75 + 5.1
= 19.6 + 5.1
= 24.7.

Question 6.
32.6 + 6 + 15.1 = ______
Answer:
32.6 + 6 + 15.1 = 38.6 + 15.1 = 53.7.
53.7 is close to 54.

Explanation:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.5 Add and Subtract Decimals-Apply and Grow-Practice6
Check:
53.7 – 15.1 – 6
= 38.6 – 6
= 32.6

 

Question 7.
You have $35.50. You buy a movie for $15.89 and music for $12.50. How much money do you have left?
Answer:
Amount left with me = $7.11.

Explanation:
Total money I have = $35.50
Cost of movie purchased = $15.89
Cost of music purchased = $12.50
Total amount spent on music and movie = $15.89 + $12.50 = $28.39
Amount left with me = Total money I have – Total amount spent on music and movie
= $35.50 – $28.39
= $7.11

Question 8.
You ride your bike each day over the weekend. How far do you ride your bike altogether?
Big Ideas Math Answer Key Grade 5 Chapter 3 Add and Subtract Decimals 3.5 4
Answer:
Total Distance covered over all travelled on bike = 7.55 miles.

Explanation:
Distance covered on Friday = 1.6 miles
Distance covered on Saturday = 2.85 miles
Distance covered on Sunday = 3.1 miles
Total Distance covered over all = Distance covered on Friday+ Distance covered on Saturday + Distance covered on Sunday
= 1.6 + 2.85 + 3.1
= 4.45 + 3.1
= 7.55 miles.

Question 9.
Logic
Your friend finds 5.31 + 2.8 – 1.27 and gets the answer 4.32. Is your friend’s answer reasonable? Explain.
Answer:
My friend answer is incorrect because the correct answer is 7 not 4.32.

Explanation:
First do the adding the equation:
5.31 + 2.8 – 1.27
= 8.11 – 1.27
= 6.84
6.84 is close to 7.

Question 10.
DIG DEEPER!
Find the missing digits.
Big Ideas Math Answer Key Grade 5 Chapter 3 Add and Subtract Decimals 3.5 5
Answer:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.5 Add and Subtract Decimals-DIG DEEPER-10
Explanation:
First add hundredths place then tenths values.
Secondly add ones and tens values checking the sum and numbers given for adding.
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.5 Add and Subtract Decimals-DIG DEEPER-10

Think and Grow: Modeling Real Life

Example
A lifeguard walks around the perimeter of the lazy river ride. How far does the lifeguard walk?
Find the perimeter of the lazy river by adding all of the side lengths.
Big Ideas Math Answer Key Grade 5 Chapter 3 Add and Subtract Decimals 3.5 6
The lifeguard walks _______ feet.
Answer:
The lifeguard walks  637.50 feet.

Explanation:
Length of the side of the river = 216.75 feet
Length of the side of the river = 216.75 feet
Length of the base of the river = 204 feet
Perimeter  of the river = Length of the side of the river + Length of the side of the river + Length of the base of the river
= 216.75 + 216.75  + 204
= 433.50 + 204
= 637.50 feet.

Show and Grow

Question 11.
You ride your skateboard around the perimeter of the park. How far do you ride your skateboard?
Big Ideas Math Answer Key Grade 5 Chapter 3 Add and Subtract Decimals 3.5 7
Answer:
Perimeter of the park = 184.8 m

Explanation:
Length of side of the park = 67.2 m
Length of the base of the park = 39.6 m
Length of the Height of the park = 78 m
Perimeter of the park = Length of side of the park + Length of the base of the park + Length of the Height of the park
= 67.2 + 39.6 + 78
= 106.8 + 78
= 184.8 m

Question 12.
A German shepherd weighs 24.75 pounds less than the Bernese mountain dog. A husky weighs 21.67 pounds less than the German shepherd. How much does the husky weigh?
Big Ideas Math Answer Key Grade 5 Chapter 3 Add and Subtract Decimals 3.5 8
Answer:
Weight of Husky dog = 56.91 lb.

Explanation:
Weight of Bernese mountain dog = 103.33 lb.
A German shepherd weighs 24.75 pounds less than the Bernese mountain dog.
=> 103.33 – 24.75
=.> 78.58 lb.
A husky weighs 21.67 pounds less than the German shepherd.
=> 78.58 – 21.67
=> 56.91 lb.

Question 13.
DIG DEEPER!
Your friend launches a water balloon 136.33 yards. Your cousin launches a balloon 15.67 yards farther than your friend and 8.33 yards farther than you. How much farther do you launch a balloon than your friend?
Answer:
I have launched 24 yards a balloon farther than my friend.

Explanation:
Length of my friend a water balloon = 136.33 yards
Your cousin launches a balloon 15.67 yards farther than your friend and 8.33 yards farther than you.
=>  136.33 + 15.67 = 152 + 8.33 = 160.33 yards
Difference:
Your cousin launches a balloon 15.67 yards farther than your friend and 8.33 yards farther than you – Length of my friend a water balloon
= 160.33 – 136.33
= 24 yards

Add and Subtract Decimals Homework & Practice 3.5

Evaluate. Check whether your answer is reasonable.
Question 1.
2.8 + 7 + 5.3 = _______
Answer:
2.8 + 7 + 5.3 = 9.8 + 5.3 = 15.1.
15.1 is close to 15.

Explanation:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.5 Add and Subtract Decimals-Add and Subtract Decimals Homework & Practice 3.5.1
Check:
15.1 – 5.3 – 7
= 9.8 – 7
= 2.8.

Question 2.
14.86 + 5.72 – 3 = ______
Answer:
14.86 + 5.72 – 3 = 20.58 – 3 = 17.58.
17.58 is close to 18.

Explanation:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.5 Add and Subtract Decimals-Add and Subtract Decimals Homework & Practice 3.5.2
Check:
17.58 + 3 – 5.72
= 20.58 – 5.72
= 14.86.

Question 3.
28.95 – 4.12 – 0.7 = ______
Answer:
28.95 – 4.12 – 0.7 = 24.83 – 0.7 = 24.13.
24.13 is close to 24.

Explanation:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.5 Add and Subtract Decimals-Add and Subtract Decimals Homework & Practice 3.5.3
Check:
24.13 + 0.7 + 4.12
= 24.83 + 4.12
= 28.95.

Question 4.
10.6 – 10.35 + 34.27 = _____
Answer:
10.6 – 10.35 + 34.27 = 0.25 + 34.27 = 34.52.
34.52 is close to 35.

Explanation:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.5 Add and Subtract Decimals-Add and Subtract Decimals Homework & Practice 3.5.4
Check:
34.52 – 34.27 + 10.35
= 0.25 + 10.35
= 10.60.

Question 5.
19.36 + 24 + 8.5 = ______
Answer:
19.36 + 24 + 8.5 = 43.36 + 8.5 = 51.86.
51.86 is close to 52.

Explanation:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.5 Add and Subtract Decimals-Add and Subtract Decimals Homework & Practice 3.5.5
check:
51.86 – 8.5 -24
= 43.36 – 24
= 19.36.

Question 6.
5 – 0.2 – 1.37 = ______
Answer:
5 – 0.2 – 1.37 = 4.8 – 1.37 = 3.43.
3.43 is close to 3.

Explanation:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.5 Add and Subtract Decimals-Add and Subtract Decimals Homework & Practice 3.5.6
Check:
3.43 + 1.37 + 0.2
= 4.8 + 0.2
= 5.

Question 7.
Which One Doesn’t Belong?
Which addition problem does not involve adding zeros to line up the place values?
5.39 + 1.76 + 40.9
25 + 1.8 + 6.3
18.5 + 6.2 + 0.7
12.49 + 10.8 + 0.13
Answer:
The equations 25 + 1.8 + 6.3, 18.5 + 6.2 + 0.7 and 12.49 + 10.8 + 0.13 does not involve adding of zeros to line up the place values.

Explanation:
First, find the sum:
5.39 + 1.76 + 40.9 = 7.15 + 40.9 = 48.05
25 + 1.8 + 6.3 = 26.8 + 6.3 =  33.1
18.5 + 6.2 + 0.7 = 24.7 + 0.7 = 25.4
12.49 + 10.8 + 0.13 = 23.29 + 0.13 = 23.42

Question 8.
Logic
Which expressions have an estimate greater than 10?
4.9 + 2.5 + 3.8
8.1 – 4.3 + 2.7
10.9 – 0.2 – 0.1
10.8 + 3.6 – 5.2
Answer:
The Equation 4.9 + 2.5 + 3.8 and 10.9 – 0.2 – 0.1 have an estimate greater than 10.

Explanation:
First, add and subtract the equation:
4.9 + 2.5 + 3.8 = 7.4 + 3.8 = 11.2
8.1 – 4.3 + 2.7 = 3.8 + 2.7 = 6.5
10.9 – 0.2 – 0.1 = 10.7 – 0.1 =10.6
10.8 + 3.6 – 5.2 = 14.4 – 5.2 = 9.2

Question 9.
Modeling Real Life
You run across the length of the soccer field until you reach the middle, and then run back. Then you run across the entire length of the field. How far do you run altogether?
Big Ideas Math Answer Key Grade 5 Chapter 3 Add and Subtract Decimals 3.5 9
Answer:
Length of the total run altogether = 137.16 m.

Explanation:
Length of the entire soccer field = 91.44 m
Length of the half soccer field = 45.72 m
Length of the total run altogether = Length of the entire soccer field + Length of the half soccer field
= 91.44  + 45.72
= 137.16 m

Question 10.
DIG DEEPER!
Three pumpkins are catapulted at a contest. How much farther does Pumpkin C travel than Pumpkin A?
• Pumpkin A travels 0.61 mile.
• Pumpkin B travels 0.28 mile farther than Pumpkin A.
• Pumpkin C travels 0.06 mile farther than Pumpkin B.
Big Ideas Math Answer Key Grade 5 Chapter 3 Add and Subtract Decimals 3.5 10
Answer:
Pumpkin C travel 0.34 miles longer than Pumpkin A.

Explanation:
Pumpkin A travels= 0.61 mile.
Pumpkin B travels 0.28 mile farther than Pumpkin A.
=> 0.61 + 0.28 = 0.89 miles
Pumpkin C travels 0.06 mile farther than Pumpkin B.
=> 0.89 + 0.06 = 0.95 miles
Difference between the Pumpkin C and Pumpkin A= Pumpkin C travels – Pumpkin A travels
= 0.95 – 0.61
= 0.34 miles

Review & Refresh

Determine whether the shape has line symmetry. Draw each line of symmetry.
Question 11.
Big Ideas Math Answer Key Grade 5 Chapter 3 Add and Subtract Decimals 3.5 11
Answer:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.5 Add and Subtract Decimals-Review & Refresh-11

Explanation:
No, there is no line symmetry in the given shape.

Question 12.
Big Ideas Math Answer Key Grade 5 Chapter 3 Add and Subtract Decimals 3.5 12
Answer:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.5 Add and Subtract Decimals-Review & Refresh-12

Explanation:
No, there is no line symmetry in the given shape.

Question 13.
Big Ideas Math Answer Key Grade 5 Chapter 3 Add and Subtract Decimals 3.5 13
Answer:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.5 Add and Subtract Decimals-Review & Refresh-13

Explanation:
No, there is no line symmetry in the given shape.

Lesson 3.6 Use Mental Maths to Add or Subtract Decimals

Explore and Grow

Use mental math to find each sum.
6.5 + 3.1 + 3.5 = ______
10. 79 + 6.25 = ______
Answer:
6.5 + 3.1 + 3.5 = 13.1.
10. 79 + 6.25 = 17.04.

Explanation:
Given 6.5 + 3.1 + 3.5,
6.5 and 3.5 are compatible numbers. Use addition properties to rewrite the problem.
6.5 + 3.1 + 3.5 = 3.1 + (6.5 + 3.5) Associative Property
= 3.1 + 10
= 13.1

10. 79 + 6.25 = 17.04.

Reasoning
Explain how you used mental math to find each.
Answer:
For the first one, I have rearranged the numbers  to add compatibles numbers first. Later used Associative Property to add them.
Second equation directly I have added the given numbers.

Think and Grow: Use mental math to Add or Subtract Decimals

Example
find 12.4 + 3.9 + 7.6.
Notice that 12.4 and 7.6 are compatible numbers. Use addition properties to rewrite the problem.
12.4 + 3.9 + 7.6 = 3.9 + 12.4 + 7.6 Commutative Property
= 3.9 + (12.4 + 7.6) Associative Property
= 3.9 + 20
= _____
So, 12.4 + 3.9 + 7.6 = _______.

Answer:
So, 12.4 + 3.9 + 7.6 = 23.9.

Explanation:
Notice that 12.4 and 7.6 are compatible numbers. Use addition properties to rewrite the problem.
12.4 + 3.9 + 7.6 = 3.9 + 12.4 + 7.6 Commutative Property
= 3.9 + (12.4 + 7.6) Associative Property
= 3.9 + 20
= 23.9.

Example
Use mental math to find 6.75 – 4.46.
Use compensation.
Big Ideas Math Answers 5th Grade Chapter 3 Add and Subtract Decimals 3.6 1
You subtracted 0.04 more than 4.46, so you must add 0.04 to2.25 to find the answer.
Big Ideas Math Answers 5th Grade Chapter 3 Add and Subtract Decimals 3.6 2
So, 6.75 – 4.46 = _____.

Answer:
So, 6.75 – 4.46 = 2.29.

Explanation:
Big Ideas Math Answers 5th Grade Chapter 3 Add and Subtract Decimals 3.6 1
Subtract 0.04 more than 4.46, so you must add 0.04 to 2.25 to find the answer.
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-Lesson 3.5 Add and Subtract Decimals-Think and Grow-Use mental math to Add or Subtract Decimals

Show and Grow

Use mental math to find the sum or difference.
Question 1.
4.25 + 1.36 + 2.75 = ______
Answer:
4.25 + 1.36 + 2.75 = 8.36.

Explanation:
4.25 + 1.36 + 2.75 = ????
Notice that 4.25 and 2.75 are compatible numbers. Use addition properties to rewrite the problem.
4.25 + 1.36 + 2.75 = 1.36 + 4.25 + 2.75 Commutative Property
= 1.36 + (4.25 + 2.75 )Associative Property
= 1.36 + 7.00
= 8.36

Question 2.
12.78 + 5.25 = _____
Answer:
12.78 + 5.25 = 18.03.

Explanation:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-3.6-Show and Grow2
Add 0.05 more than 5.25, so you must subtract 0.05 to 18.08 to find the answer.
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-3.6-Show and Grow2..

Question 3.
17.4 – 13.6 = _____
Answer:
17.4 – 13.6 = 3.8.

Explanation:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-3.6-Show and Grow3
Subtract 0.4 more than 13.6, so you must add 0.4 to 3.4 to find the answer.
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-3.6-Show and Grow3...

Question 4.
29.8 – 2.27 = _____
Answer:
29.8 – 2.27 = 27.53.

Explanation:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-3.6-Show and Grow4
Subtract 0.03 more than 2.27, so you must add 0.03 to 27.50 to find the answer.
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-3.6-Show and Grow4..

Apply and Grow: Practice

Use mental math to find the sum or difference.
Question 5.
1.25 + 2.45 + 1.75 = _____
Answer:
So, 1.25 + 2.45 + 1.75 = 5.45.

Explanation:
1.25 + 2.45 + 1.75 = ???
Notice that 1.25 and 1.75 are compatible numbers. Use addition properties to rewrite the problem.
1.25 + 2.45 + 1.75 =  2.45 + 1.25 + 1.75 Commutative Property
= 2.45 + (1.25 + 1.75) Associative Property
= 2.45 + 3.00
= 5.45.

Question 6.
8.4 + 6.15 + 2.6 = ______
Answer:
So, 8.4 + 6.15 + 2.6 = 17.15.

Explanation:
8.4 + 6.15 + 2.6 = ???
Notice that 8.4 and 2.6 are compatible numbers. Use addition properties to rewrite the problem.
8.4 + 6.15 + 2.6 = 6.15 + 8.4 + 2.6 Commutative Property
= 6.15 + (8.4 + 2.6)  Associative Property
= 6.15 + 11.00
= 17.15

Question 7.
17.82 – 15.61 = _____
Answer:
17.82 – 15.61 = 2.21.

Explanation:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-3.6-Apply and Grow-Practice7
Subtract 0.09 more than 15.61, so you must add 0.09 to 2.12 to find the answer.
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-3.6-Apply and Grow-Practice7..

Question 8.
35.7 – 13.8 = _____
Answer:
35.7 – 13.8 = 21.9.

Explanation:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-3.6-Apply and Grow-Practice8
Subtract 0.2 more than 13.8, so you must add 0.2 to 21.7 to find the answer.
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-3.6-Apply and Grow-Practice8..

Question 9.
Two years ago, your friend was1.25 meters tall. Today, your friend is1.48 meters tall. How many meters did your friend grow?
Answer:
Height my friend grew = 0.23 meters.

Explanation:
Height of my Friend two years ago = 1.25 meters
Height of my Friend today = 1.48 meters
Height my friend grew = Height of my Friend today – Height of my Friend two years ago
= 1.48 – 1.25
= 0.23 meters

Question 10.
Clothing items are donated to an organization. How many pounds of clothing are donated altogether?
Big Ideas Math Answers 5th Grade Chapter 3 Add and Subtract Decimals 3.6 3
Answer:
Total weight of all clothing donated together = 134.8 pounds.

Explanation:
Weight of Shirts donated = 37.4 pounds.
Weight of Pants donated = 54.8 pounds.
Weight of Coats donated = 42.6 pounds.
Total weight of all clothing donated together = Weight of Shirts donated + Weight of Pants donated + Weight of Coats donated
= 37.4 +  54.8 + 42.6
= 134.8 pounds.

Question 11.
Reasoning
Can you use the Commutative Property to find the difference of two decimals? Explain.
Answer:
Yes, we can Commutative Property to find the difference of two decimals because the methodology we doing is subtraction in between the numerals. So, there wont be any difference expect the points.

 

Question 12.
DIG DEEPER!
Use each digit from 1 through 8 once to complete the problem.
Big Ideas Math Answers 5th Grade Chapter 3 Add and Subtract Decimals 3.6 4
Answer:

Explanation:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-3.6-DIG DEEPER-12

Think and Grow: Modeling Real Life

Example
A city receives 75.5 inches of snowfall and 45.27 inches of rainfall in 1 year. How much more snowfall does the city receive than rainfall?
Big Ideas Math Answers 5th Grade Chapter 3 Add and Subtract Decimals 3.6 5
Use compensation to find the difference between the amounts of snowfall and rainfall: 75.5 – 45.27.
Big Ideas Math Answers 5th Grade Chapter 3 Add and Subtract Decimals 3.6 6
So, the city receives _______ inches more snowfall than rainfall.
Answer:
So, the city receives 30.23 inches more snowfall than rainfall.

Explanation:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-3.6-Think and Grow-Modeling Real Life

Show and Grow

Question 13.
On average, a left-handed professional baseball player runs from home plate to first base in 4.2 seconds. The record is 3.1 seconds. How much faster is the record than the average time?
Big Ideas Math Answers 5th Grade Chapter 3 Add and Subtract Decimals 3.6 7
Answer:
1.1 seconds is faster the record than the average time.

Explanation:
Time taken on an average, a left-handed professional baseball player runs from home plate to first base = 4.2 seconds
Time of record = 3.1 seconds
Difference in the time = Time taken on an average, a left-handed professional baseball player runs from home plate to first base  – Time of record
= 4.2 – 3.1
= 1.1 seconds

Question 14.
DIG DEEPER!
The distances of each part of a triathlon are shown in the table. What is the total distance? How many more miles are traveled on land than in water?
Big Ideas Math Answers 5th Grade Chapter 3 Add and Subtract Decimals 3.6 8
Answer:
Total Distance travelled on land than in water = 30.07 miles.

Explanation:
Distance travelled to Swimming = 0.93 miles
Distance travelled on Bike = 24.8 miles
Distance travelled by Running = 6.2 miles
Total Distance travelled on land = Distance travelled on Bike – Distance travelled by Running
= 24.8  + 6.2
= 31 miles
Total Distance travelled on land than in water = Total Distance travelled on land  – Distance travelled to Swimming
= 31 – 0.93
= 30.07 miles

Question 15.
DIG DEEPER!
You paint a meter stick for a craft. You paint 30.7 centimeters blue and 67.9 centimeters green. How many millimeters are not painted?
Answer:
Length of meter stick not painted = 14 millimeters

Explanation:
Length of blue paint painted on meter stick = 30.7 centimeters
Length of green paint painted on meter stick = 67.9 centimeters
Total length of meter stick painted = Length of blue paint painted on meter stick  + Length of green paint painted on meter stick
= 30.7 + 67.9
= 98.6 centimeters
Conversion:
1 meter = 100 centimeters
Length of meter stick not painted = 100 centimeters – 98.6 centimeters = 1.4 centimeters.
1 centimeter = 10 millimeters
1.4 centimeter = ???
cross multiplication:
1.4 × 10 ÷ 1
= 14 ÷ 1
= 14 millimeters.

Use Mental Maths to Add or Subtract Decimals Homework & Practice 3.6

Use mental math to find the sum or difference
Question 1.
2.73 + 2.14 = ______
Answer:
2.73 + 2.14 = 4.87.

Explanation:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-3.6-Use Mental Maths to Add or Subtract Decimals Homework & Practice 3.6.1
Add 0.06 more than 2.14, so you must subtract 0.06 to 4.93 to find the answer.
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-3.6-Use Mental Maths to Add or Subtract Decimals Homework & Practice 3.6.1..

Question 2.
54.3 + 14.1 = ______
Answer:
54.3 + 14.1 = 68.40

Explanation:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-3.6-Use Mental Maths to Add or Subtract Decimals Homework & Practice 3.6.2

Add 0.9 more than 14.1, so you must subtract 0.9 to 69.3 to find the answer.
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-3.6-Use Mental Maths to Add or Subtract Decimals Homework & Practice 3.6.2..

Question 3.
19.86 – 7.2 = ______
Answer:

Explanation:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-3.6-Use Mental Maths to Add or Subtract Decimals Homework & Practice 3.6.3

Subtract 0.8 more than 7.2, so you must add 0.8 to 11.86 to find the answer.
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-3.6-Use Mental Maths to Add or Subtract Decimals Homework & Practice 3.6.3..

Question 4.
41.19 – 0.58 = _______
Answer:
41.19 – 0.58 =

Explanation:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-3.6-Use Mental Maths to Add or Subtract Decimals Homework & Practice 3.6.4
Subtract 0.02 more than 0.58 , so you must add 0.02 to 40.59 to find the answer.
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-3.6-Use Mental Maths to Add or Subtract Decimals Homework & Practice 3.6.4..

Question 5.
6.5 + 2.32 + 1.5 = _____
Answer:
6.5 + 2.32 + 1.5 = 10.32.

Explanation:
Notice that 6.5 and 1.5 are compatible numbers. Use addition properties to rewrite the problem.
6.5 + 2.32 + 1.5  = 2.32 + 6.5 + 1.5 Commutative Property
= 2.32 + (6.5 + 1.5)  Associative Property
= 2.32 + 8.0
= 10.32.

Explanation:

Question 6.
2.25 – 1.15 = ______
Answer:
2.25 – 1.15 = 1.10.

Explanation:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-3.6-Use Mental Maths to Add or Subtract Decimals Homework & Practice 3.6.6
Subtract 0.05 more than 1.15 , so you must add 0.05 to 1.05 to find the answer.
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-3.6-Use Mental Maths to Add or Subtract Decimals Homework & Practice 3.6.6..

Question 7.
YOU BE THE TEACHER
Your friend finds 23.74 – 13.18. Is your friend correct? Explain.
Big Ideas Math Answers 5th Grade Chapter 3 Add and Subtract Decimals 3.6 9
Answer:
My Friend answer is correct because he used mental math to do the sum of the given numbers.

Explanation:
Big Ideas Math Answers 5th Grade Chapter 3 Add and Subtract Decimals 3.6 9

Question 8.
Writing
Describe a situation when you would use the Commutative Property to add decimals using mental math.
Answer:

The Commutative Property of Addition states that the order of the addends does not change the sum.
Let’s test the property using simple whole numbers.

4+5+9=18       4+9+5=18         5+4+9=18        5+9+4=18         9+5+4=18          9+4+5=18

As you can see, we can add the three addends (4, 5, and 9) in many different orders. The Commutative Property of Addition works also works for four, five, six addends, and it works for decimal addends, too.

Question 9.
DIG DEEPER!
Use mental math to find the missing number.
Answer:

Question 10.
Modeling Real Life
How much wider is the wingspan of the albatross than the length of an 8.89-foot-long alligator?
Big Ideas Math Answers 5th Grade Chapter 3 Add and Subtract Decimals 3.6 10
Answer:
2.06 feet wider is the wingspan of the albatross than the length of an 8.89-foot-long alligator.

Explanation:
Length of Albatross wingspan = 10.95 feet
Length of  long Alligator = 8.89 foot
Difference:
Length of Albatross wingspan – Length of  long Alligator
= 10.95 – 8.89
= 2.06 feet

Question 11.
Modeling Real Life
Your friend buys 3.5 meters of blue ribbon, 2.25 meters of red ribbon, and 3.75 meters of white ribbon. How much ribbon does she buy in all?
Answer:
Total Length of all ribbons = 9.50 meters.

Explanation:
Length of blue ribbon = 3.5 meters
Length of red ribbon = 2.25 meters
Length of white ribbon = 3.75 meters
Total Length of all ribbons = Length of blue ribbon + Length of red ribbon + Length of white ribbon
= 3.5 + 2.25 + 3.75
= 5.75 + 3.75
= 9.50 meters

Review & Refresh

Use the figure.
Big Ideas Math Answers 5th Grade Chapter 3 Add and Subtract Decimals 3.6 11
Question 12.
Name a line segment.
Answer:
Line segment = CD or DE or EF

Explanation:
In  geometry, a line segment is a part of a  line that is bounded by two distinct end  points , and contains every point on the line between its endpoints.

Question 13.
Name two different rays.
Answer:
EF and AB are two different Rays.

Explanation:
In geometry, a ray can be defined as a part of a  line that has a fixed starting point but no end point. It can extend infinitely in one direction.

Question 14.
Name two different lines.
Answer:
EA and DF are different lines.

Explanation:
In geometry, a line can be defined as a straight one- dimensional figure that has no thickness and extends endlessly in both directions. It is often described as the shortest distance between any two points.

Lesson 3.7 Problem Solving: Money

Explore and Grow

Use addition or subtraction to make several item Cost conclusions about the prices of the items in the table.
Big Ideas Math Answers Grade 5 Chapter 3 Add and Subtract Decimals 3.7 1
Answer:
Total cost of all items = $180.77.

Explanation:
Cost of Video game = $54.93
Cost of Selfie stick = $15.08
Cost of Power Bank = $27.36
Cost of Speaker = $83.40
Total cost of all items = Cost of Video game + Cost of Selfie stick + Cost of Power Bank + Cost of Speaker
= $54.93 + $15.08 + $27.36 + $83.40
= $70.01 + $27.36 + $83.40
= $97.37 + $83.40
= $180.77.

Precision
Switch papers with your partner. Check your partner’s work for reasonableness.
Answer: Both the answers are reasonableness because the answers procedure is same and answers match.

Think and Grow: Problem Solving: Money

Example
You have$11.83. You earn $13.75 more doing chores. Then you buy headphones for $12.39. How much money do you have left?
Understand the Problem
What do you know?
• You have $11.83.
• You earn $13.75 more.
• You buy headphones for $12.39.

What do you need to find?
• You need to find how much money you have left.

Make a Plan
How will you solve?
Write and solve an equation to find the amount of money you have left.

Solve
Big Ideas Math Answers Grade 5 Chapter 3 Add and Subtract Decimals 3.7 2
Let s represent the amount of money you have left.
s = 11.83 + 13.75 – 12.39
= ______ − 12.39
= ______
You have $ ______ left.
Answer:
Amount left out with me = $13.19.

Explanation:
Amount of money I have = $11.83
Amount of money I have earned = $13.75
Cost of Headphones = $12.39
Amount left out with me = Amount of money I have + Amount of money I have earned + Cost of Headphones
= $11.83 + $13.75 – $12.39
= $25.58 – $12.39
= $13.19.

Show and Grow

Question 1.
You have $32.63. You earn $28.42 more washing windows. Then you buy a video game for $35.95.How much money do you have left?
Big Ideas Math Answers Grade 5 Chapter 3 Add and Subtract Decimals 3.7 3
Answer:
Amount of money left with me = $25.1.

Explanation:
Amount of money I have = $32.63
Amount of money earned = $28.42
Cost of Video game = $35.95
Amount of money left with me = Amount of money I have + Amount of money earned + Cost of Video game
= $32.63 + $28.42 – $35.95
= $61.05 – $35.95
= $25.1.

Apply and Grow: Practice

Understand the problem. What do you know? What do you need to find? Explain.
Question 2.
You buy 3 cell phone games that cost $5 each and an in-game purchase for $1.69. How much money do you spend?
Answer:
Total Cost of money spent = $16.69

Explanation:
Cost of 3 cell phones games = 3 × $5 = $15
Cost of an in-game = $1.69
Total Cost of money spent = Cost of 3 cell phones games + Cost of an in-game
= $15 + $1.69
= $16.69.

Question 3.
Newton buys a movie ticket that costs $4.50, a drink that costs $2.25, and a bag of popcorn that costs $4.75. How much money does Newton spend?
Big Ideas Math Answers Grade 5 Chapter 3 Add and Subtract Decimals 3.7 4
Answer:
Total amount spent = $11.50.

Explanation:
Cost of movie ticket = $4.50
Cost of a drink = $2.25
Cost of a bag of popcorn = $4.75
Total amount spent = Cost of movie ticket + Cost of a drink + Cost of a bag of popcorn
= $4.50 + $2.25 + $4.75
= $6.75 + $4.75
= $11.50.

Understand the problem. Then make a plan. How will you solve? Explain.
Question 4.
Your friend buys a remote control car for$17.89 and some batteries for $5.49. He gives the cashier $25. How much change does he receive?
Answer:
Amount of change he receives = $1.62.

Explanation:
Cost of remote control car = $17.89
Cost of Some batteries = $5.49
Money given to cashier = $25
Amount of change he receives = Money given to cashier- Cost of remote control car + Cost of Some batteries –
= $25 – $17.89 + $5.49
= $25 – $23.38
= $1.62.

Question 5.
Your cousin has $156.34 in his bank account. He withdraws $15.50 one day and $6.75 the next day.  How much money is in his bank account now?
Answer:
Amount left in his bank account = $134.09.

Explanation:
Amount of money my cousin has in bank account = $156.34
Amount of money he withdraws one day = $15.50
Amount of money he withdraws next day = $6.75
Amount left in his bank account = Amount of money my cousin has in bank account – Amount of money he withdraws one day –
Amount of money he withdraws next day
= $156.34 – $15.50 – $6.75
= $156.34 – $22.25
= $134.09.

Question 6.
Adult admission to open skate night costs $5.75. Admission for children under 13 costs $1.50 less than adult admission. What is the total admission cost for two 12-year-olds?
Big Ideas Math Answers Grade 5 Chapter 3 Add and Subtract Decimals 3.7 5
Answer:
Total admission cost for two 12-year-olds = $10.

Explanation:
Cost of Adult admission to open skate night = $5.75
Cost of Admission for children under 13 costs $1.50 less than adult admission
=> $5.75 – $1.50
=> $4.25
Total admission cost for two 12-year-olds = Cost of Adult admission to open skate night  + Cost of Admission for children under 13
= $5.75 + $4.25
= $10.

Question 7.
Descartes buys a book for $14.99 and a bookmark. He pays with a $20 bill and receives $3.96 in change. How much does the bookmark cost?
Answer:
Cost of the book mark = $1.05.

Explanation:
Cost of the book and a bookmark = $14.99
Amount paid by a bill = $20
Amount of money received as change = $3.96
Cost of the book mark = Amount paid by a bill – Cost of the book and a bookmark – Amount of money received as change
= $20 – $14.99 – $3.96
= $5.01 – $3.96
= $1.05.

Think and Grow: Modeling Real Life

Example
You pay for colored pencils, a coloring book, and a pencil sharpener with two $20 bills. Your change is $17.42. How much does the pencil sharpener cost?
Big Ideas Math Answers Grade 5 Chapter 3 Add and Subtract Decimals 3.7 6
Think: What do you know? What do you need to find? How will you solve?
Write and solve an equation to find the cost of the pencil sharpener.
Big Ideas Math Answers Grade 5 Chapter 3 Add and Subtract Decimals 3.7 7
Let c represent the cost of the pencil sharpener.
c = 40 – (8.99 + 5.99) – 17.42
= 40 – ______ – 17.42
= ____ − 17.42
= _____
The pencil sharpener costs $ _____.
Answer:
Cost of the pencil sharpener costs = $7.6.

Explanation:
Cost of the coloring book = $5.99
Cost of the colored pencils = $8.99
Amount received as change = $17.42
Amount paid in two bills = $20 × 2 = $40
Cost of the pencil sharpener costs = Amount paid in two bills  – (Cost of the coloring book + Cost of the colored pencils) + Amount received as change
= $40 – ( $5.99 + $8.99 ) – $17.42
= $40 – $14.98 – $17.42
= $25.02 – $17.42
= $7.6.

Show and Grow

Question 8.
Newton pays for a new collar, a bag of dog food, and a box of bones with four $5 bills. His change is $4.02. How much does the box of bones cost?
Big Ideas Math Answers Grade 5 Chapter 3 Add and Subtract Decimals 3.7 8
Answer:
Cost of the box of bones = $3.39.

Explanation:
Cost of a bag of dog food = $5.49
Cost of a dog’s new collar = $7.10
Amount of change he receives = $4.02
Amount paid in four bills = $5 × 4 = $20.
Cost of the box of bones = Amount paid in four bills – (Cost of a bag of dog food – Cost of a dog’s new collar) – Amount of change he receives
= $20 – ($5.49 + $7.10) – $4.02
= $20 – $12.59 – $4.02
= $7.41 -$4.02
= $3.39.

Question 9.
DIG DEEPER!
Your friend buys the scooter and the helmet. Your friend has two gift cards that are worth the same amount. Your friend uses both gift cards and pays an additional $2.49. What is the amount on each gift card?
Big Ideas Math Answers Grade 5 Chapter 3 Add and Subtract Decimals 3.7 9
Answer:
Cost of the each gift cards = $75.

Explanation:
Cost of the Helmet = $27.50
Cost of the Scooter = $124.99
Amount of money paid additionally = $2.49
Let the gifts cards be X.
= 2Cost of the each gift cards + Amount of money paid additionally = (Cost of the Helmet + Cost of the Scooter)
=> 2X + $2.49 = $27.50 + $124.99
=> 2X + $2.49 = $152.49
=> 2X = $152.49 – $2.49
=> 2X = $150.00
=> X = $150.00 ÷ 2
=> X = $75.

Problem Solving: Money Homework & Practice 3.7

Understand the problem. What do you know? What do you need to find? Explain.
Question 1.
A musician buys an electronic keyboard for $89.94 and 2 books of sheet music for $7.99 each. How much does she spend in all?
Answer:
Total cost of money spent = $105.92.

Explanation:
Cost of an electronic keyboard = $89.94
Cost of two books of sheet music = $7.99 × 2 = $15.98
Total cost of money spent = Cost of an electronic keyboard + Cost of two books of sheet music
= $89.94 +  $15.98
= $105.92

Question 2.
You have$28.49. You spend $14.95 at a bowling alley. You win $35.00 for setting the record score for your age group. How much money do you have now?
Answer:
Amount of money I have now = $48.54.

Explanation:
Amount of money I have = $28.49
Amount spent at a bowling alley = $14.95
Amount of money won = $35.00
Amount of money I have now = Amount of money I have – Amount spent at a bowling alley  + Amount of money won
=> $28.49 – $14.95 + $35.00
=> $13.54 +  $35.00
=> $48.54.

Understand the problem. Then make a plan. How will you solve? Explain.
Question 3.
Newton wins a $50 gift card toa craft store. He uses the card to buy a scrapbook for $19.99 and a set of markers for $10.75. How much money is left on his gift card?
Answer:
Amount of money left on his gift card = $19.26.

Explanation:
Amount of money won for a gift card to a craft store = $50
Amount of money spent to buy a scrapbook by using card = $19.99
Amount of money spent to buy a set of markers by using card = $10.75
Amount of money left on his gift card = Amount of money won for a gift card to a craft store – (Amount of money spent to buy a scrapbook by using card  + Amount of money spent to buy a set of markers by using card)
= $50 – ($19.99 + $10.75)
= $ 50 – $30.74
= $19.26.

Question 4.
A dancer buys a tutu for $48.75 and the ballet shoes. She gives the cashier $80. How much change does she receive?
Big Ideas Math Answers Grade 5 Chapter 3 Add and Subtract Decimals 3.7 10
Answer:
Amount of change received = $4.76.

Explanation:
Cost of the  ballet shoes = $26.49
Cost of the tutu = $48.75
Amount given to cashier = $80
Amount of change received = Amount given to cashier – (Cost of the tutu + Cost of the ballet shoes )
= $80 – ( $ 26.49 + $48.75)
= $80 – $75.24
= $4.76

Question 5.
A theater director buys props for$116.20 and costumes for $494.85. He has $194.27 left. How much money did he have to start?
Answer:
Amount of money he  have started = $805.32.

Explanation:
Cost of props = $116.20
Cost of the costumes = $494.85
Amount left with him = $194.27
Amount of money he  have started = Cost of the costumes + Cost of props + Amount left with him
= $116.20 + $494.85 + $194.27
= $611.05 + $194.27
= $805.32

Question 6.
Descartes uses a metal detector at a beach. He finds 3 quarters, 6 dimes,2 nickels, and 5 pennies. He drops1 quarter and 2 pennies into the water. Then he finds 4 more dimes. How much money does Descartes have now?
Answer:
Total amount of money Descartes have now = 163 pennies.

Explanation:
Number of dimes found = 6
Number of quarters found = 3
Number of pennies found = 5
Number of nickels found = 2
Number of quarters dropped into water= 1
Number of pennies dropped into water = 2
Number of dimes found later = 4
Conversion:

Total number of quarters = Number of quarters found  – Number of quarters dropped into water
=> 3 – 1 = 2
Quarter into pennies
=> 1 Quarter = 25 pennies
=> 2 Quarter  = 25 × 2 = 50 pennies.

Total number of dimes = Number of dimes found  + Number of dimes found later
=> 6 +4 = 10
Dimes into pennies
=> 1 Dime = 10 pennies
=> 10 Dime = 10 × 10 = 100 pennies

Total number of Nickels = Number of nickels found
=> 2
Nickels into pennies
=> 1 Nickel = 5 pennies
=> 2 nickels = 2 × 5 = 10 pennies

Total Number of pennies = Number of pennies found – Number of pennies dropped into water
=> 5 – 2
=>3 pennies
Total amount of money Descartes have now = Total number of quarters  + Total number of dimes + Total number of Nickels + Total Number of pennies
= 50 pennies + 100 pennies + 10 pennies + 3pennies
= 163 pennies.

Question 7.
DIG DEEPER!
Descartes wants to buy a ball launcher for Newton for $12.50. He has one $10 bill, two $5 bills, three $1 bills, and two quarters. Make a table to find four ways Descartes could pay for the ball launcher.
Big Ideas Math Answers Grade 5 Chapter 3 Add and Subtract Decimals 3.7 11
Answer:
Amount of change Descartes receives = $11.00.

Explanation:
Cost of a ball launcher = $12.50
Number of bills he has worth $10 = 1
Number of bills he has worth $5 = 2
Number of bills he has worth $1 = 3
Number of quarters he has = 2
Conversion:
Quarter into Dollars
1 Quarter = $0.25
=>2 Quarter = 0.25 × 2 = $0.50.

Amount given to cashier =  Number of bills he has worth $10 + Number of bills he has worth $5 + Number of bills he has worth $1 +
Number of quarters he has
=> $10 + $5 + $5 + $1 + $1 + $1 + $0.50
=> $23.50

Amount of change Descartes receives = Amount given to cashier – Cost of a ball launcher
=> $23.50 – $12.50
= $11.00
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-3.7-Problem Solving-Money Homework & Practice 3.7.7

Question 8.
Modeling Real Life
Your science teacher pays for an ultraviolet light, a beaker, and a glowing liquid kit with five $5 bills. The ultraviolet light costs $13.29. The beaker costs $7.99. Your teacher’s change is $0.77. What is the cost of the glowing liquid kit?
Big Ideas Math Answers Grade 5 Chapter 3 Add and Subtract Decimals 3.7 12
Answer:
Cost of the glowing liquid kit = $2.95

Explanation:
Cost of the ultraviolet light = $13.29
Cost of the beaker = $7.99
Amount of change teacher receives= $0.77

Your science teacher pays an ultraviolet light, a beaker, and a glowing liquid kit with five $5 bills
=> Cost of the ultraviolet light  + Cost of the beaker  + Cost of the glowing liquid kit + Amount of change teacher receives = 5 × $5
=> $13.29 + $7.99 + Cost of the glowing liquid kit + $0.77 = $25
=> $22.05 + Cost of the glowing liquid kit = $25
=> Cost of the glowing liquid kit = $25 – $22.05
=> Cost of the glowing liquid kit = $2.95

Review & Refresh

Find the equivalent length.
Question 9.
10 ft = _____ in.
Answer:
10 ft = 120 in.

Explanation:
Conversion:
1 feet = 12 inches
=> 10 feet = 10 × 12 =120 inches.

Question 10.
2 mi = ______ ft
Answer:
2 mi = 10560 ft

Explanation:
Conversion:
1 Mile = 5280 feet
2 mi = 5280 × 2 = 10560 ft

Question 11.
\(\frac{1}{3}\) yd = ______ ft
Answer:
\(\frac{1}{3}\) yd = \({1}\) feet.

Explanation:
Conversion:
1 Yard = 3 Feet
\(\frac{1}{3}\) yd = \(\frac{1}{3}\) yd  × \({3}\) = \({1}\) feet

Add and Subtract Decimals Performance Task

The table shows the average amount of rainfall each month Rainfall in Mobile, Alabama. Mobile, Alabama
Big Ideas Math Solutions Grade 5 Chapter 3 Add and Subtract Decimals 1
Question 1.
About how many inches of rainfall does the city get Average Month Rainfall during June, July, and August combined?
Answer:
Average rainfall during June, July, and August = 6.77 inches.

Explanation:
Rainfall rate in June month = 6.10 inches
Rainfall rate in July month = 7.24 inches
Rainfall rate in August month = 6.97 inches
Average rainfall during June, July, and August = Rainfall rate in June month + Rainfall rate in July month + Rainfall rate in August month ÷ 3
= 6.10 + 7.24  + 6.97  ÷ 3
= 20.31 ÷ 3
= 6.77 inches

Question 2.
How many more inches of rainfall does the city usually get in July than in October?
Answer:
3.54 inches. of Rainfall is more in the city in July  month than in October month.

Explanation:
Rainfall in July month = 7.24 inches
Rainfall in October month = 3.70 inches
Comparison:
= Rainfall in July month – Rainfall in October month
= 7.24  – 3.70
= 3.54 inches.

Question 3.
a. Round each rainfall amount to the nearest tenth. About how many inches of rainfall does Mobile get each year?
b. How can you check whether your answer is reasonable?
Answer:
a)  The total Rainfall Mobile gets each year = 66.22 inches.
b) We can check the answer is reasonable or not by rounding it to its nearest 10.

Explanation:
a) Round each rainfall amount to the nearest tenth.

Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-3.7-Add and Subtract Decimals Performance Task-3
5 and above rainfall rate is said to be rounded to its nearest 10 as 10.

Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-3.7-Add and Subtract Decimals Performance Task

b)
We can check the answer is reasonable or not by rounding it to its nearest 10.
63 is close reasonable to 66.22.

 

Question 4.
The average yearly amount of rainfall in Seattle, Washington, is about 3 feet.
a. Write an expression to find about how many more inches of rain Mobile, Alabama, usually gets than Seattle each year.
Big Ideas Math Solutions Grade 5 Chapter 3 Add and Subtract Decimals 2
b. Seattle is known for being one of the rainiest cities in the United States, even though it usually gets less rain than Mobile each year. Why do you think Seattle is known as the rainiest city?
Answer:
a) 30.22 inches of rain Mobile, Alabama, usually gets than Seattle each year.

b) Seattle is known for being one of the rainiest cities in the United States, even though it usually gets less rain than Mobile each year. This means that Seattle has more frequent rain, but the light rain results in low totals than compared to rainfall in Mobile, Alabama.

Explanation:
a)The average yearly amount of rainfall in Seattle, Washington, is about 3 feet.
=> Average Rainfall of Seattle, Washington  = 3 feet.
Total Average of  Rainfall in Mobile, Alabama = 66.22 inches
Difference :
1 feet = 12 inches
=> 3 feet = 12 × 3 = 36 inches
Total Average of  Rainfall in Mobile, Alabama – Average Rainfall of Seattle, Washington
= 66.22 – 36
= 30.22 inches

b)
Seattle is known for being one of the rainiest cities in the United States, even though it usually gets less rain than Mobile each year. This means that Seattle has more frequent rain, but the light rain results in low totals than compared to rainfall in Mobile, Alabama.

Add and Subtract Decimals Activity

Decimal Dots
Directions:
1. Players take turns connecting two dots, each using a different color.
2. On your turn, connect two dots, vertically or horizontally. If you close a square around an expression, then evaluate the expression. If you do not close a square, your turn is over.
3. Continue playing until all expressions are evaluated.
4. The player with the most evaluated expressions wins!
Big Ideas Math Solutions Grade 5 Chapter 3 Add and Subtract Decimals 3

Add and Subtract Decimals Chapter Practice

3.1 Estimate the Sum or Differences.

Estimate the sum or difference.
Question 1.
Big Ideas Math Solutions Grade 5 Chapter 3 Add and Subtract Decimals chp 1
Answer:
741.26 + 103.75 = 845.31.

Explanation:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-3.7-Add and Subtract Decimals Chapter Practice.1

 

Question 2.
Big Ideas Math Solutions Grade 5 Chapter 3 Add and Subtract Decimals chp 2
Answer:
89.5 – 41.8 = 47.7.

Explanation:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-3.7-Add and Subtract Decimals Chapter Practice.2

Question 3.
13.47 – 9.82
Answer:
13.47 – 9.82 = 3.65.

Explanation:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-3.7-Add and Subtract Decimals Chapter Practice.3

Question 4.
64.9 + 17.5
Answer:
64.9 + 17.5 = 82.4.

Explanation:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-3.7-Add and Subtract Decimals Chapter Practice.4

Question 5.
Number Sense
Descartes estimates a difference by rounding each number to the nearest whole number. His estimate is 6. Which problems could he have estimated?
97.5 – 91.8
10.3 – 4.6
23.7 – 18.1
7.2 – 0.8
Answer:
His Problems of estimation to 6 are 97.5 – 91.8 and 7.2 – 0.8.

Explanation:
First  round the numbers to nearest whole number later , find the difference:
97.5 – 91.8 = 98 – 92 = 6
10.3 – 4.6 = 10 – 5 = 5
23.7 – 18.1 = 23 – 18 = 5
7.2 – 0.8 = 7 – 1 = 6

3.2 Use Models to Add or Subtract Decimals

Make a quick sketch to find the sum or difference.
Question 6.
0.27 + 0.19
Answer:
0.27 + 0.19 = 0.46.

Explanation:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-3.7-Add and Subtract Decimals Chapter Practice-3.2 Use Models to Add or Subtract Decimals-6

Question 7.
2.52 – 1.36 = ______
Answer:
2.52 – 1.36 = 1.16.

Explanation:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-3.7-Add and Subtract Decimals Chapter Practice-3.2 Use Models to Add or Subtract Decimals-7

3.3 Add Decimals

Find the sum. Check whether your answer is reasonable.
Question 8.
Big Ideas Math Solutions Grade 5 Chapter 3 Add and Subtract Decimals chp 8
Answer:
10.36 + 0.29 = 10.65.
10.65 is reasonable.

Explanation:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-3.7-Add and Subtract Decimals Chapter Practice-3.3 Add Decimals.4
Check:
10.65 – 0.29
=10.36.

Question 9.
Big Ideas Math Solutions Grade 5 Chapter 3 Add and Subtract Decimals chp 9
Answer:
461.2 + 78.5 = 539.7
539.7 is reasonable.

Explanation:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-3.7-Add and Subtract Decimals Chapter Practice-3.3 Add Decimals.9
Check:
539.7 – 78.5
= 461.2.

Question 10.
97.65 + 23.08 = _____
Answer:
97.65 + 23.08 = 120.73.
120.73 is reasonable.

Explanation:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-3.7-Add and Subtract Decimals Chapter Practice-3.3 Add Decimals-10
Check:
120.73 – 23.08
= 97.65.

Question 11.
63.39 + 4.05 = _____
Answer:
63.39 + 4.05 = 67.44.
67.44 is a reasonable.

Explanation:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-3.7-Add and Subtract Decimals Chapter Practice-3.3 Add Decimals-11
Check:
67.44 – 4.05
=63.39.

Question 12.
Modeling Real Life
You want to conduct an experiment. You need 0.5 cup of water and the same amount of oil. How many cups of liquid do you need for the experiment?
Answer:
Total number of cups required of liquid = 1 cup.

Explanation:
Amount of water required = 0.5 cup
Amount of oil required = 0.5 cup
Total number of cups required of liquid = Amount of water required + Amount of oil required
= 0.5 + 0.5
= 1 cup

3.4 Subtract Decimals

Find the difference. Then check your answer.
Question 13.
Big Ideas Math Solutions Grade 5 Chapter 3 Add and Subtract Decimals chp 13
Answer:
86.5 – 49.4 = 37.1.

Explanation:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-3.7-Add and Subtract Decimals Chapter Practice-3.3 Add Decimals- 3.4 Subtract Decimals - 13
Check:
37.1 + 49.4
= 86.5

Question 14.
Big Ideas Math Solutions Grade 5 Chapter 3 Add and Subtract Decimals chp 14
Answer:
37.82 – 5.07 = 32.75.

Explanation:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-3.7-Add and Subtract Decimals Chapter Practice-3.3 Add Decimals- 3.4 Subtract Decimals - 14
Check :
32.75 + 5.07
= 37.82.

Question 15.
582.58 – 33.21 = _____
Answer:
582.58 – 33.21 = 549.37.

Explanation:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-3.7-Add and Subtract Decimals Chapter Practice-3.3 Add Decimals- 3.4 Subtract Decimals - 15
Check:
549.37 + 33.21
= 582.58.

Question 16.
99.2 – 8.6 = _____
Answer:
99.2 – 8.6 = 90.6.

Explanation:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-3.7-Add and Subtract Decimals Chapter Practice-3.3 Add Decimals- 3.4 Subtract Decimals - 16
Check:
90.6 + 8.6
= 99.2.

3.5 Add and Subtract Decimals

Evaluate. Check whether your answer is reasonable.
Question 17.
9.28 – 5.4 – 2.75 = _____
Answer:
9.28 – 5.4 – 2.75 = 1.13.
1.13 is reasonable.

Explanation:
9.28 – 5.4 – 2.75 =  ???
9.28 – 5.4 – 2.75 = 3.88 – 2.75 = 1.13.
Check:
1.13 + 2.75 + 5.4
= 3.88 + 5.4
= 9.28.

Question 18.
2 + 18.6 – 7.9 = ______
Answer:
2 + 18.6 – 7.9 = 12.7.
12.7 is reasonable.

Explanation:
2 + 18.6 – 7.9 = ???
2 + 18.6 – 7.9
= 20.6 – 7.9
= 12.7
Check:
12.7 + 7.9 – 18.6
= 20.6 – 18.6
= 2

Question 19.
26 – 0.8 + 13.1 = ______
Answer:
26 – 0.8 + 13.1 = 38.3
38.3 is reasonable.

Explanation:
26 – 0.8 + 13.1 = ???
26 – 0.8 + 13.1
= 25.2 + 13.1
= 38.3

Check:
38.3 – 13.1 + 0.8
= 25.2 + 0.8
= 26

Question 20.
1.72 + 4 + 35.68
Answer:
1.72 + 4 + 35.68 = 41.4
41.4 is reasonable.

Explanation:
1.72 + 4 + 35.68 = ???
1.72 + 4 + 35.68
= 5.72 + 35.68
= 41.4

Check:
41.4 – 35.68 -4
= 5.72 – 4
= 1.72

Question 21.
Modeling Real Life
How many inches of snowfall Month Snowfall (inches)are recorded altogether?
Big Ideas Math Solutions Grade 5 Chapter 3 Add and Subtract Decimals chp 21
Answer:
Total Snowfall in all three months = 70.50 inches.

Explanation:
Snowfall in the month of December = 27.5
Snowfall in the month of January = 30.75
Snowfall in the month of February = 12.25
Total Snowfall in all three months = Snowfall in the month of December – Snowfall in the month of January – Snowfall in the month of February
= 27.5 + 30.75 + 12.25
= 58.25 + 12.25
= 70.50 inches.

3.6 Use Mental Maths to Add or Subtract Decimals

Use mental math to find the sum or difference.
Question 22.
6.3 + 2.98 + 5.7 = ______
Answer:
6.3 + 2.98 + 5.7 = 14.98.

Explanation:
Given 6.3 + 2.98 + 5.7 =
6.3 and 5.7 are compatible numbers. Use addition properties to rewrite the problem.
6.3 + 2.98 + 5.7 = 2.98 + (6.3 + 5.7) Associative Property
= 2.98 + 12
= 14.98.

Question 23.
9.1 + 4.57 + 8.03 = _____
Answer:
9.1 + 4.57 + 8.03 = 21.7.

Explanation:
Given 9.1 + 4.57 + 8.03
4.57 and 8.03 are compatible numbers. Use addition properties to rewrite the problem.
9.1 + 4.57 + 8.03 = 9.1 + (4.57 + 8.03) Associative Property
= 9.1 + 12.6
= 21.7

Question 24.
25.7 – 14.9 = _____
Answer:
25.7 – 14.9 = 10.8.

Explanation:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-3.7-Add and Subtract Decimals Chapter Practice-3.6 Use Mental Maths to Add or Subtract Decimals-24
Subtract 0.1 more than 14.9 , so you must Add 0.1 to 10.7 to find the answer.
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-3.7-Add and Subtract Decimals Chapter Practice-3.6 Use Mental Maths to Add or Subtract Decimals-24..

Question 25.
9.24 – 2.67 = _____
Answer:
9.24 – 2.67 = 6.57.

Explanation:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-3.7-Add and Subtract Decimals Chapter Practice-3.6 Use Mental Maths to Add or Subtract Decimals-25
Subtract 0.03 more than 2.67 , so you must Add 0.03 to 6.54 to find the answer.
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-3.7-Add and Subtract Decimals Chapter Practice-3.6 Use Mental Maths to Add or Subtract Decimals-25..

Question 26.
Number Sense
Use mental math to find the missing number.
9.2 + n + 2.5 = 15.8
Answer:

Explanation:
Given 6.5 + 3.1 + 3.5,
6.5 and 3.5 are compatible numbers. Use addition properties to rewrite the problem.
6.5 + 3.1 + 3.5 = 3.1 + (6.5 + 3.5) Associative Property
= 3.1 + 10
= 13.1

3.7 Problem Solving: Money

Question 27.
Newton buys a board game for $9.99 and a pack of trading cards for $2.99. He gives the cashier a $20 bill. How much change does he receive?
Big Ideas Math Solutions Grade 5 Chapter 3 Add and Subtract Decimals chp 27
Answer:
Amount of money change received = $7.02.

Explanation:
Cost of a board game = $9.99
Cost of a pack of trading cards = $2.99
Amount of bills paid to cashier = $20
Amount of money change received = Amount of bills paid to cashier – ( Cost of a pack of trading cards + Cost of a board game  )
= $20 – ( $9.99 + $2.99 )
= $20 – $12.98
= $7.02.

Question 28.
Descartes has $378.97 in his bank account. He withdraws $59.50 one day and then deposits $20.44 the next day. How much money is in his bank account now?
Answer:
Amount of money left in the bank account = $339.91.

Explanation:
Amount of money Descartes has in his bank account= $378.97
Amount of money withdrawn = $59.50
Amount of money he deposited later = $20.44
Amount of money left in the bank account = Amount of money Descartes has in his bank account – Amount of money withdrawn +  Amount of money he deposited later
= $378.97 – $59.50 + $20.44
= $319.47 + $20.44
= $339.91

Question 29.
Modeling Real Life
A bowling alley charges $7.50 for each person to bowl and $2.25 to rent a pair of shoes. How much does it cost you and your friend to bowl and rent shoes?
Answer:
Amount of money charged  for me and my friend to bowl and rent shoes = $9.75.

Explanation:
Amount of money charged for each person to bowl = $7.50
Amount of money charged as rent for a pair of shoes = $2.25
Amount of money charged  for me and my friend to bowl and rent shoes = Amount of money charged for each person to bowl +
Amount of money charged as rent for a pair of shoes
= $7.50 + $2.25
= $9.75.

Question 30.
Reasoning
Your friend wants to buy a magic kit for $18.75. He has two $10 bills, one $5 bill, four $1 bills, and 3 quarters. Make a table to find three ways your friend could pay for the magic kit.
Big Ideas Math Solutions Grade 5 Chapter 3 Add and Subtract Decimals chp 30
Answer:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-3.7-Add and Subtract Decimals Chapter Practice-3.6 Use Mental Maths to Add or Subtract Decimals-Reasoning-30

Explanation:
Cost of the a magic kit = $18.75
Number of $10 bills my friend has = 2 = $10 × 2 = $ 20.
Number of $5 bills my friend has = 1 = $5 × 1 = $5.
Number of $1 bills my friend has = 4 = $ 1 × 4 = $ 4.
Number of Quarters my friend has = 3
Conversion:
1 Quarter = 0.25 dollar
=> 3 Quarter = 3 × 0.25 = $0.75.
Amount of money my friend has = Number of $10 bills my friend has + Number of $5 bills my friend has + Number of $1 bills my friend has + Number of Quarters my friend has
= $20 + $5 + $4 + $0.75
= $ 25 + $4 + $0.75
= $29 + $0.75
= $29.75.

Add and Subtract Decimals Cumulative Practice

Question 1.
Using the order of operations, which operation should you perform last to evaluate the expression?
(4 × 6) + (100 ÷ 10) – (3 + 4)
A. addition
B. subtraction
C. multiplication
D. division
Answer:
A. addition.

Explanation:
Given Equation:
According to DMAS rule:
First Division:
=> (4 × 6) + (100 ÷ 10) – (3 + 4)
=> (4 × 6) + 10 – (3 + 4)
Second Multiplication:
=> (4 × 6) + 10 – (3 + 4)
=> 24 + 10 – (3 + 4)
Third Addition:
=> 24 + 10 – (3 + 4)
=> 24 + 10 + 1
=> 34 +1
=> 35.

Question 2.
Which numbers round to 4 when rounded to the nearest whole number?
Big Ideas Math Solutions Grade 5 Chapter 3 Add and Subtract Decimals cp 2
Answer:
4.03, 3.98, 4.1 are the numbers round to 4 when rounded to the nearest whole number.

Explanation:
Round all options to nearest whole number:
4.03 –  4
4.5  – 5
44.2 – 44
3.98 – 4
3.46 – 3
4.1 – 4

Question 3.
Which bike trail does the figure show?
Big Ideas Math Solutions Grade 5 Chapter 3 Add and Subtract Decimals cp 3
A. River Run
B. Straight and Fast
C. Twist and Turn
D. Wood Way

Answer:
Total distance from A to E = 4.5 km
=> River Run bike trail the figure shows.

Explanation:
First add all the distances given in the map:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-3.7-Add and Subtract Decimals Chapter Practice-Add and Subtract Decimals Cumulative Practice-3
Distance from A to B = 1.75 km
Distance from B to C = 1.0 km
Distance from C to D = 0.5 km
Distance from D to E = 0.75 km
Distance from E to A = 0.5 km
Total distance from A to E = Distance from A to B + Distance from B to C + Distance from C to D + Distance from D to E
Distance from E to A
= 1.75 + 1.0  + 0.5  +  0.75  + 0.5
= 2.75 + + 0.5  +  0.75  + 0.5
= 3.0 +  0.75  + 0.5
= 3.75 + 0.5
= 4.25 km.

Question 4.
What is 0.045 written as a fraction?
A. \(\frac{45}{100}\)
B. \(\frac{1,000}{45}\)
C. \(\frac{45}{1,000}\)
D. \(\frac{450}{1,000}\)
Answer:
C. \(\frac{45}{1,000}\)

Explanation:
0.045 fraction= 45 ÷ 1000.

Question 5.
A recipe calls for 3 eggs. Newton makes 5 batches of the recipe, and Descartes makes 2 batches of the recipe. Which expressions can you use to determine how many eggs Newton and Descartes use in all?
Big Ideas Math Solutions Grade 5 Chapter 3 Add and Subtract Decimals cp 5
Answer:
The Expression used = (5 × 3) + ( 2 × 2 ).

Explanation:
A recipe calls for 3 eggs.
Number of recipe batches Newton makes = 5
=>Number of eggs used by Newton = 5 × 3
Number of recipe batches Descartes makes = 2
=>Number of eggs used by  Descartes = 2 × 2
Number of eggs both used = Number of eggs used by Newton + Number of eggs used by  Descartes
= (5 × 3) + ( 2 × 2 )
= 15 + 4
= 19.

Question 6.
What is the difference of 97.63 and 4.98?
Big Ideas Math Solutions Grade 5 Chapter 3 Add and Subtract Decimals cp 6
Answer:
97.63 -4.98 = 92.65.

Explanation:
Difference:
97.63 -4.98 =???
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-3.7-Add and Subtract Decimals Chapter Practice-Add and Subtract Decimals Cumulative Practice-6

Question 7.
What is 60.25 written in word form?
A. sixty twenty-five
B. sixty and twenty-five tenths
C. sixty and twenty-five
D. sixty and twenty-five hundredths
Answer:
D. sixty and twenty-five hundredths

Explanation:
Number name of 60.25 = ????

Question 8.
Find 55.6 – 3.97.
A. 1.59
B. 51.63
C. 52.37
D. 59.57
Answer:
55.6 – 3.97 = 51.63.

Explanation:
55.6 – 3.97 = ???
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-3.7-Add and Subtract Decimals Chapter Practice-Add and Subtract Decimals Cumulative Practice-8

Question 9.
What is the value of the expression?
[21 ÷ (3 + 4)] – 2
A. 9
B. 4 R1
C. 1
D. 2
Answer:
C. 1

Explanation:
[21 ÷ (3 + 4)] – 2
= [21 ÷ (7)] – 2
= 3 – 2
=1

Question 10.
Which statement correctly compares the decimals?
Big Ideas Math Solutions Grade 5 Chapter 3 Add and Subtract Decimals cp 10
Answer:
A. 0.1 > 0.111.

Explanation:
Big Ideas Math Solutions Grade 5 Chapter 3 Add and Subtract Decimals cp 10

Question 11.
What is the value of n?
n – 9.02 = 3.85
A. 12.87
B. 5.17
C. 3.87
D. 12
Answer:
n = 12.87.

Explanation:
n = ??
n – 9.02 = 3.85
n = 3.85 + 9.02
n = 12.87.

Question 12.
Which property is shown?
35 + (15 + 7) = (35 + 15) + 7
A. Associative Property of Addition
B. Commutative Property
C. Distributive Property
D. Associative Property of Multiplication
Answer:
A. Associative Property of Addition.

Explanation:
The associative property states that you can add or multiply regardless of how the numbers are grouped.
35 + (15 + 7) = (35 + 15) + 7

Question 13.
Newton estimates a sum by rounding each number to the nearest ten. His estimate is 140. Which problems could he have estimated?
Big Ideas Math Solutions Grade 5 Chapter 3 Add and Subtract Decimals cp 13
Answer:
123.5 + 19.6, 87.6 + 50.4, 65.43 + 71.59 problem he have estimated.

Explanation:
First, find all sums of given Equations:
123.5 + 19.6 = 124 + 20 = 144.
87.6 + 50.4  = 88 + 50 = 138.
65.43 + 71.59 = 65 + 72 = 137.
98.03 + 34.56 = 98 + 34 = 132.

Question 14.
What is five hundred four thousand, eleven written in standard form?
A. 54,011
B. 54,110
C. 504,110
D. 504,011
Answer:

Explanation:

Question 15.
Which expressions have a value of 8?
Big Ideas Math Solutions Grade 5 Chapter 3 Add and Subtract Decimals cp 15
Answer:
30 ÷ ( 5 × 2 ) + 5, 4 × ( 5 – 3 ) , (16 ÷ 8) + ( 3 × 2 )  are expressions having a value of 8.

Explanation:
First, solve the equations:
30 ÷ ( 5 × 2 ) + 5 = 30 ÷ 10 + 5 = 3 + 5 = 8.
55 – 7 ÷ 6 = 55  – 1.167 = 53.833.
4 × ( 5 – 3 ) = 4 × 2 = 8.
(16 ÷ 8) + ( 3 × 2 ) = 2 + 6 = 8.

Question 16.
Descartes wins a $50 gift card to a sporting goods store. He uses the card to buy two baseballs for $1.79 each, a baseball glove for $14.99, and a baseball bat for $12.25. About how much money is left on his gift card?
A. $21
B. $19
C. $31
D. $18
Answer:
Total amount left on his gift card= $19.18.

Explanation:
Amount of gift card won for a sporting goods store = $50
Cost of the two baseballs  = $1.79 each = > 2 × 1.79 = $3.58
Cost of a baseball glove = $14.99
Cost of a baseball bat = $12.25
Total amount left on his gift card= Amount of gift card won for a sporting goods store – (Cost of the two baseballs + Cost of a baseball glove + Cost of a baseball bat )
= $50 – ( $3.58 + $14.99 + $12.25)
= $50 – ( $18.57 + $12.25)
= $50 – $30.82
= $19.18.

Question 17.
What is the value of 4× 105?
A. 45
B. 400,000
C. 40,000
D. 200
Answer:
4× 105   = B. 400,000.

Explanation:
4× 105
= 4 × 10 ×10 × 10 × 10 × 10
= 4 × 100000
= 400,000

Question 18.
The dimensions of two rectangles are shown.
Big Ideas Math Solutions Grade 5 Chapter 3 Add and Subtract Decimals cp 18
Part A Write an expression to determine how much greater the area of the green rectangle is than the area of the yellow rectangle.
Part B Evaluate the expression.
Part C Explain how you used order of operations to evaluate the expression.
Answer:
Part A : Area of the green rectangle  > Area of the yellow rectangle
9ft × 8ft >  7ft × 5ft

Part B: Evaluate the expression.
9ft × 8ft >  7ft × 5ft
=> 72ft > 35ft

Part C:
I used the order of DMAS rule.
Division, Multiplication, Addition and Subtraction order of operations to evaluate the expression

Explanation:
Part A:
Length of the green rectangle = 9ft
Width of the green rectangle = 8ft
Area of the green rectangle = Length of the green rectangle × Width of the green rectangle
= 9ft × 8ft
= 72 ft.
Length of the yellow rectangle = 7ft
Width of the yellow rectangle = 5ft
Area of the yellow rectangle = Length of the yellow rectangle × Width of the yellow rectangle
= 7ft × 5ft
= 35ft

Part C: Explain how you used order of operations to evaluate the expression.
I used the order of DMAS rule.
Division, Multiplication, Addition and Subtraction order of operations to evaluate the expression.

Add and Subtract Decimals STEAM Performance Task

Students can buy pencils, notebooks, and other supplies at your school’s student store. The table below shows the money in the student store account.
Big Ideas Math Solutions Grade 5 Chapter 3 Add and Subtract Decimals spt 1
Big Ideas Math Solutions Grade 5 Chapter 3 Add and Subtract Decimals spt 2

Question 1.
Line 1 shows how much money was in the account at the start of the school year. Line 2 shows a payment of $85.89 for supplies. Complete the account balance for Line 2.
Answer:
Amount of money left in the account = $14.11.
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-3.7-Add and Subtract Decimals Chapter Practice- Add and Subtract Decimals STEAM Performance Task-1

Explanation:
Amount in the account at the start of the school year = $100.00
Amount in the account used for supplies = $85.89
Amount of money left in the account = Amount in the account at the start of the school year – Amount in the account used for supplies = $100.00 – $85.89
= $ 14.11

Question 2.
The store has forty-four dollars and thirty cents in sales for the first week and thirty-two dollars and twelve cents in sales for the second week. The sales each week are deposited into the student store account. Fill in the deposits and balances for Lines 3 and 4.
Answer:
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-3.7-Add and Subtract Decimals Chapter Practice- Add and Subtract Decimals STEAM Performance Task-2

Explanation:
Amount of money received in sales for the first week = forty-four dollars and thirty cents = $44.30
Amount of money received in sales for the second week = thirty-two dollars and twelve cents = $ 32.12

Question 3.
The balance after the third week of sales is $115.68. How much money did the store have in sales during the third week? Fill in the deposit and balance for Line 5.
Answer:
Amount of money received in sales for the third week  = $25.15.
Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-3.7-Add and Subtract Decimals Chapter Practice- Add and Subtract Decimals STEAM Performance Task-3

Explanation:
Amount of money balance after the third week of sales = $115.68
Amount of money received in sales for the first week = forty-four dollars and thirty cents = $44.30
Amount of money received in sales for the second week = thirty-two dollars and twelve cents = $ 32.12
Amount of money left in the account = $14.11
Amount of money received in sales for the third week + Amount of money received in sales for the first week + Amount of money received in sales for the second week + Amount of money left in the account = Amount of money balance after the third week of sales
= Amount of money received in sales for the third week  + $44.30 + $32.12 + $14.11 = $115.68
=> Amount of money received in sales for the third week  + $76.42 + $14.11 = $115.68
=> Amount of money received in sales for the third week + $90.53 = $115.68
=> Amount of money received in sales for the third week  = $115.68 – $90.53
=>Amount of money received in sales for the third week  = $25.15.

You are in charge of purchasing supplies for the student store. All of the supplies for the student store are purchased and then sold at a higher price so the store can earn a profit.
Big Ideas Math Solutions Grade 5 Chapter 3 Add and Subtract Decimals spt 3
Question 4.
You want to buy 300 more pencils. You find two different quantities to buy.
a. Which pencils will you buy to get $35.99$19.99the best value? Justify your answer.
b. You pay about $0.24 for each pencil. How much do you think the store should charge for each pencil to earn a profit? Explain.
c. How much profit does the store earn from selling 100 pencils using your recommended price? Explain.
d. You want to buy some notebooks for a total of $25.24 and 300 pencils. You have a $5 coupon. How much do you spend on notebooks and pencils?
e. Write out the check to buy the supplies. What is the account balance now? Fill in the payment and balance for line 6.
Big Ideas Math Solutions Grade 5 Chapter 3 Add and Subtract Decimals spt 4
Answer:
a) I would like to purchase 300 pencils from Pencil assortment A because the cost of each pencil is lesser than Pencil assortment B.

b)Cost the pencil to be sold = $0.30 because 0.06 is the profit range of 4 % out of 0.24.

c) The profit store owns from selling 100 pencils using your recommended price is $30.

d) Cost of the Notebooks and pencils = $115.24

e) Big-Ideas-Math-Answers-Grade-5-Chapter-3-Add-and-Subtract-Decimals-3.7-Add and Subtract Decimals Chapter Practice- Add and Subtract Decimals STEAM Performance Task-4

Explanation:
a) Cost of 150pencil from Pencil assortment A = $ 35.99
=> Cost of each pencil from Pencil assortment A = $35.99 ÷ 150 = $ 0.2399
Cost of 75pencil from Pencil assortment B = $19.99
=> Cost of each pencil from Pencil assortment B = $19.99 ÷ 75 = $0.2665

b) Cost of pencil is purchased by me = $0.24
Cost the pencil to be sold = $0.24 + $0.06 = $0.30
Because 0.06 is the profit range of 4 % out of 0.24.

c) Cost of each pencils to be sold = $0.30
Cost of 100 pencils to be sold = $0.30 × 100 = $30.

d) Total Cost of the Notebooks = $25.24
Cost of 300 pencils = $0.30 ×300 = $90
Amount of coupon used to pay = $5
Cost of the Notebooks and pencils = Total Cost of the Notebooks + Cost of 300 pencils
= $25.24 + $90
= $115.24.

e) Current account balance = $115.68

Conclusion:

Hope the information provided in the above pdf is helpful for all the students in Grade 5. Download Big Ideas Math Answers Grade 5 Chapter 3 Add and Subtract Decimals for free of cost. Keep in touch with us to get the latest updates of all Grade 5 chapters.

Big Ideas Math Answers Grade 2 Chapter 13 Represent and Interpret Data

Big Ideas Math Answers Grade 2 Chapter 13

Do you feel maths is a difficult subject and worried about the test? If yes, then don’t worry we are here to help you out to overcome the difficulties in maths. Get free access to download Big Ideas Math Answers 2nd Grade 13th Chapter Represent and Interpret Data pdf from this page. Students can find various tricks to solve the questions. Learn how to represent data in different ways with the help of BIM Book Grade 2 Chapter 13 Represent and Interpret Data Answers.

Big Ideas Math Book Grade 2 Answer Key Chapter 13 Represent and Interpret Data

It is necessary for the students to go through the topics included in this chapter before starting their preparation. The list of lessons in Chapter 13 Represent and Interpret Data are Sort and Organize Data, Real and Interpret Picture Graphs, Make Picture Graphs, Real and Interpret Bar Graphs, Make Bar Graphs, Make Line Plots, and Measure Objects and Make Line Plots.

You can understand the concepts quickly and easily by referring to Big Ideas Math Answers Grade 2 Chapter 13 Represent and Interpret Data. The quick links are attached at the end for the reference of students.

Vocabulary

Lesson: 1 Sort and Organize Data

Lesson: 2 Real and Interpret Picture Graphs

Lesson: 3 Make Picture Graphs

Lesson: 4 Real and Interpret Bar Graphs

Lesson: 5 Make Bar Graphs

Lesson: 6 Make Line Plots

Lesson: 7 Measure Objects and Make Line Plots

Chapter- 13: Represent and Interpret Data

Represent and Interpret Data Vocabulary

Big Ideas Math Answer Key Grade 2 Chapter 13 Represent and Interpret Data v1
Organize It
Use the review words to complete the graphic organizer.
Big Ideas Math Answer Key Grade 2 Chapter 13 Represent and Interpret Data v2

Answer:
The number of students who choose math = 6.
The number of students who choose science = 5

Explanation:
In the above-given figure,
the number of students who choose math = 6
the number of students who choose science = 5
the more number of students choose maths than science.

.Big-Ideas-Math-Solutions-Grade-2-Chapter-13-Represent and Interpret Data-13-1

Organize It
Use your vocabulary cards to identify the word.
Big Ideas Math Answer Key Grade 2 Chapter 13 Represent and Interpret Data v3

Answer:
1. The pencil lengths of each student = 4, 5, 6, and 7.

Explanation:
The number of pencil lengths of all students = 4, 5, 6, and 7.
the max length of the pencil = 5
In the above-given figure is a line plot.
given that the pencil lengths in inches = 4, 5, 6, and 7.
Big-Ideas-Math-Solutions-Grade-2-Chapter-13-Represent and Interpret Data-13-2

Answer:
2. The more number of students whose favorite hobby is singing = 7.

Explanation:
In the above-given figure,
the total number of students = 8
the number of students whose favorite hobby dancing = 6.
the number of students whose favorite hobby running = 5.
the number of students whose favorite hobby singing = 7.
The more number of students whose favorite hobby is singing = 7.
Big-Ideas-Math-Solutions-Grade-2-Chapter-13-Represent and Interpret Data-13-3

Big Ideas Math Answer Key Grade 2 Chapter 13 Represent and Interpret Data v4.1
Big Ideas Math Answer Key Grade 2 Chapter 13 Represent and Interpret Data v4
Big Ideas Math Answer Key Grade 2 Chapter 13 Represent and Interpret Data v5.1
Big Ideas Math Answer Key Grade 2 Chapter 13 Represent and Interpret Data v5

Lesson 13.1 Sort and Organize Data

Explore and Grow

Look at your color tiles. Complete the tally chart.
Big Ideas Math Answer Key Grade 2 Chapter 13 Represent and Interpret Data 13.1 1
Answer:
The tile colors are blue, green, red, and yellow.

Explanation:
In the above-given figure,
given the tile, colors are blue, green, red, and yellow.
Big-Ideas-Math-Solutions-Grade-2-Chapter-13-Represent and Interpret Data-13.1-1

Write and answer a question about your tally chart.

Answer:
The more number of students they choose yellow = 5.

Explanation:
In the above-given figure,
the number of students who choose blue = 2.
the number of students who choose green = 3.
the number of students who choose red = 4.
the number of students who choose yellow = 5.
the more number of students who choose yellow = 5.

Show and Grow

Question 1.
Use the data to complete the tally chart.
Big Ideas Math Answer Key Grade 2 Chapter 13 Represent and Interpret Data 13.1 2
How many students chose math? ________
Which subject is the most favorite? _________

Answer:
The students who choose math = 7.
The most favorite subject = math.

Explanation:
The students who choose math = 7.
The students who choose science = 4.
The students who choose social studies = 2.
The students who choose language arts = 3.

Big-Ideas-Math-Solutions-Grade-2-Chapter-13-Represent and Interpret Data-13.1-2

Big-Ideas-Math-Solutions-Grade-2-Chapter-13-Represent and Interpret Data-13.1-3

Apply and Grow: Practice

Question 2.
Use the data to complete the tally chart.
Big Ideas Math Answer Key Grade 2 Chapter 13 Represent and Interpret Data 13.1 3
Which animal is the least favorite? ________
How many students did not choose fox? How do you know?

Answer:
The least favorite animal = owl
the students did not choose fox = 12.

Explanation:
In the above-given figure,
the least favorite animal = owl.
the students did not choose fox = 12.
the total number of students = 18
18 – 6 = 12
12 students did not choose fox.
Big-Ideas-Math-Solutions-Grade-2-Chapter-13-Represent and Interpret Data-13.1-4

Did more students choose fox or owl?
_______
How many more? ______ more

Did fewer students choose reindeer or polar bear?
________
How many fewer? ______ fewer

Answer:
The more students choose fox = 7.
the more students = 7
the fewer students choose a polar bear.
fewer = 3

Explanation:
The students who choose fox = 7.
the students who choose more = 7
the students who choose fewer = polar bear
polar bear = 3
Big-Ideas-Math-Solutions-Grade-2-Chapter-13-Represent and Interpret Data-13.1-4

Question 3.
Reasoning
Which sentences are correct?
Big Ideas Math Answer Key Grade 2 Chapter 13 Represent and Interpret Data 13.1 4
You survey 30 students.
11 students chose whale.
4 more students chose seal than penguin.
20 students did not choose seal.

Answer:
4 more students choose seal than penguin is correct.

Explanation:
In the above-given figure,
the penguin = 14
whale = 6
seal = 10

Think and Grow: Modeling Real Life

Newton wants to survey25 students. How many more students does he need to survey?
Big Ideas Math Answer Key Grade 2 Chapter 13 Represent and Interpret Data 13.1 5
Addition equation:

_______ students

Answer:
The more students does he need to survey = 6 students.

Explanation:
In the above-given figure,
the total number of students = 19.
newton wants 25 students.
19 + 6 = 25
Big-Ideas-Math-Solutions-Grade-2-Chapter-13-Represent and Interpret Data-13.1-5

Show and Grow

Question 4.
Descartes wants to survey 20 students. How many more students does he need to survey?
Big Ideas Math Answer Key Grade 2 Chapter 13 Represent and Interpret Data 13.1 6
_______ students

Answer:
The more students do he need to survey = 2 students.

Explanation:
In the above-given figure,
the total number of students = 18.
Descartes wants 20 students.
18 + 2 = 20.
Big-Ideas-Math-Solutions-Grade-2-Chapter-13-Represent and Interpret Data-13.1-6

How many more students need to choose sneakers so that the numbers of students who choose sneakers and sandals are equal?
______ students

Answer:
The students need to choose sneakers so that the number of students who choose sneakers and sandals is equal = 2 students.

Explanation:
In the above-given figure,
the sandals = 6
sneakers = 4
6 – 4 = 2
2 students.

Sort and Organize Data Homework & Practice 13.1

Question 1.
Use the data to complete the tally chart.
Big Ideas Math Answer Key Grade 2 Chapter 13 Represent and Interpret Data 13.1 7
Which activity is the most favorite? _______
Did more students choose catch or sidewalk chalk? _______
How many more?
______ more

Answer:
The activity the most favorite = tag.
The more students choose catch than sidewalk chalk.
0ne more.

Explanation:
The most favorite activity = tag.
Most students choose catch than the sidewalk.
sidewalk = 2
catch = 4
bubbles are more.
Big-Ideas-Math-Solutions-Grade-2-Chapter-13-Represent and Interpret Data-13.1-7

Question 2.
Modeling Real Life
Newton wants to survey 15 friends. How many more friends does he need to survey?
Big Ideas Math Answer Key Grade 2 Chapter 13 Represent and Interpret Data 13.1 8

Answer:
The more friends he needs to survey = 4.

Explanation:
Given that newton wants 15 friends to survey.
but in the figure, he has only 11 friends.
11 + 4 = 15
the more friends he wants to survey = 4
Big-Ideas-Math-Solutions-Grade-2-Chapter-13-Represent and Interpret Data-13.1-8

Question 3.
Writing
In Exercise 2, what question did Newton ask?
__________________
__________________

Answer:
Newton wants to survey 15 friends.

Explanation:
In the above-given question, he wants to survey more friends.
the more friends = 4
in the figure given that 11 friends.
11 + 4 = 15.
Big-Ideas-Math-Solutions-Grade-2-Chapter-13-Represent and Interpret Data-13.1-9

Question 4.
Modeling Real Life
You want to survey 30 students. How many more students do you need to survey?
Big Ideas Math Answer Key Grade 2 Chapter 13 Represent and Interpret Data 13.1 9
______ students
How many more students need to choose coloring so that the numbers of students who choose coloring and puppets are equal?
______ students

Answer:
The more students you need to survey = 10 students.

Explanation:
In the above-given figure,
The total number of students = 20
The more number of students you need to survey = 10.
Big-Ideas-Math-Solutions-Grade-2-Chapter-13-Represent and Interpret Data-13.1-10
Review & Refresh

Question 5.
153 − 10 = ______

Answer:
153 – 10 = 143

Explanation:
153 – 10 = 143
Big-Ideas-Math-Solutions-Grade-2-Chapter-13-Represent and Interpret Data-13.1-12

Question 6.
978 − 10 = ______

Answer:
978 – 10 = 968

Explanation:
978 – 10 = 968
Big-Ideas-Math-Solutions-Grade-2-Chapter-13-Represent and Interpret Data-13.1-13

Question 7.
642 − 100 = ______

Answer:
642 – 100 = 542

Explanation:
642 – 100 = 542
Big-Ideas-Math-Solutions-Grade-2-Chapter-13-Represent and Interpret Data-13.1-14

Question 8.
1,000 −100 = ______

Answer:
1000 – 100 = 900

Explanation:
1000 – 100 = 900
Big-Ideas-Math-Solutions-Grade-2-Chapter-13-Represent and Interpret Data-13.1-15

Lesson 13.2 Real and Interpret Picture Graphs

Explore and Grow

How are the tally chart and the picture graph the same? How are they different?
Big Ideas Math Answers 2nd Grade Chapter 13 Represent and Interpret Data 13.2 1
__________________
___________________

Answer:
Yes, the tally chart and picture graph represents the same.

Explanation:
The favorite bird = 2.
cat = 4
dog = 7.
So in the tally chart and the picture graph represents the same.
Big-Ideas-Math-Solutions-Grade-2-Chapter-13-Represent and Interpret Data-13.2-1

Show and Grow

Question 1.
Big Ideas Math Answers 2nd Grade Chapter 13 Represent and Interpret Data 13.2 2
How many students chose butterfly? _______
Which insect is the most favorite? ______
Which insect is the least favorite? _______

Answer:
The students who choose butterfly = 7
The most favorite insect= 8
The least favorite insect = 2

Explanation:
In the above-given figure,
The insects bumblebee = 2
ladybug = 8
grasshopper = 6
butterfly = 7
The students who choose butterfly = 7
The most favorite insect= 8
The least favorite insect = 2
Big-Ideas-Math-Solutions-Grade-2-Chapter-13-Represent and Interpret Data-13.2-2

Apply and Grow: Practice

Question 2.
Big Ideas Math Answers 2nd Grade Chapter 13 Represent and Interpret Data 13.2 3
How many students chose sporting event? _______
Which school trip is the least favorite? _______
How many more students chose science center than the park? _______

Answer:
The students who choose sporting event = 8
The least favorite school trip = park
The more students who choose science center than the park = 3

Explanation:
In the above-given figure,
the science center = 5
park = 2
sporting event = 8
museum = 3
The students who choose sporting event = 8
The least favorite school trip = park
The more students who choose science center than the park = 3
Big-Ideas-Math-Solutions-Grade-2-Chapter-13-Represent and Interpret Data-13.2-3

Question 3.

Number Sense
Use the numbers to complete the sentences.
Big Ideas Math Answers 2nd Grade Chapter 13 Represent and Interpret Data 13.2 4
Newton has ______ siblings.
Your friend has _______ siblings.
Your friend has ______ more siblings than your cousin.

Answer:
Newton has 2 siblings.
your friend has 4 siblings.
your friend has 3 more siblings than your cousin.

Explanation:
In the above-given figure,
your cousin has 1 sibling.
your friend has 4 siblings.
newton has 2 siblings.
Descartes has 3 siblings.
Big-Ideas-Math-Solutions-Grade-2-Chapter-13-Represent and Interpret Data-13.2-4

Think and Grow: Modeling Real Life

Big Ideas Math Answers 2nd Grade Chapter 13 Represent and Interpret Data 13.2 5
Do more students like crabs and sea turtles or octopuses and jellyfish?
Addition equations:
Big Ideas Math Answers 2nd Grade Chapter 13 Represent and Interpret Data 13.2 6
More students like ______ and _______.

Answer:
The more students like sea turtle = 9
More students like octopus and sea turtle.

Explanation:
In the above-given figure,
the students like crab = 6
the students who like octopus = 8
the students who like jellyfish = 3
the students who like sea turtle = 9
The more students like sea turtle = 9
More students like octopus and sea turtle.
Big-Ideas-Math-Solutions-Grade-2-Chapter-13-Represent and Interpret Data-13.2-5

Show and Grow

Question 4.
Big Ideas Math Answers 2nd Grade Chapter 13 Represent and Interpret Data 13.2 7
Do you see more cars and motorcycles or vans and trucks?
You see more _______ and ______.

Answer:
We see more cars.

Explanation:
In the above-given figure,
the cars = 9
vans = 8
truck  4
motorcycles = 2
we see more cars than other vehicles.

Real and Interpret Picture Graphs Homework & Practice 13.2

Question 1.
Big Ideas Math Answers 2nd Grade Chapter 13 Represent and Interpret Data 13.2 8
Which fruit do exactly 7 students like most? _______
How many students like the fruit with the fewest votes? _______
How many more students chose banana than pear? _____

Answer:
The fruits exactly 7 students like most = banana.
the students who like the fewest = apple.
the more students choose banana than pears = 3

Explanation:
In the above-given figure,
The students who like apple = 4
the students who like pears = 5
The students who like grapes = 7
the students who like banana = 8
The fruits exactly 7 students like most = banana.
the students who like the fewest = apple.
the more students choose banana than pears = 3
Big-Ideas-Math-Solutions-Grade-2-Chapter-13-Represent and Interpret Data-13.2-6

Question 2.
Writing
Use the picture graph.
Big Ideas Math Answers 2nd Grade Chapter 13 Represent and Interpret Data 13.2 9
Write two true statements about the picture graph.
____________________
____________________

Answer:
The students who choose bus = 6
The students who choose walk= 4
The students who choose car = 2
The students who choose subway = 5

Explanation:
In the above- given figure,
the students who choose bus = 6
The students who choose walk= 4
The students who choose car = 2
The students who choose subway = 5
the more students choose the bus.
fewer students choose cars.
Big-Ideas-Math-Solutions-Grade-2-Chapter-13-Represent and Interpret Data-13.2-7

Question 3.
Modeling Real Life
Use the picture graph.
Big Ideas Math Answers 2nd Grade Chapter 13 Represent and Interpret Data 13.2 10
Do more students like to play video games and read or play outside and watch TV?
More students like to ______ and ______.

Answer:
The more students like to play video games.

Explanation:
In the above-given figure,
the students who like to play outside = 6
the students who like to play video games = 9
the students who like to watch tv= 7
the students who like to read = 3
the more students like to play videogames and watching tv
Big-Ideas-Math-Solutions-Grade-2-Chapter-13-Represent and Interpret Data-13.2-8
Review & Refresh

Question 4.
What is the best estimate of the length of a keyboard?
Big Ideas Math Answers 2nd Grade Chapter 13 Represent and Interpret Data 13.2 11
Answer: 18 inches

Lesson 13.3 Make Picture Graphs

Explore and Grow

Look at your color tiles. Complete the tally chart and the picture graph.
Big Ideas Math Answers Grade 2 Chapter 13 Represent and Interpret Data 13.3 1

Answer:
The number of students who like blue = 2
The number of students who like green = 3
The number of students who like red = 4
The number of students who like yellow = 5
Big-Ideas-Math-Solutions-Grade-2-Chapter-13-Represent and Interpret Data-13.3-1

Show and Grow

Question 1.
Complete the picture graph.
Big Ideas Math Answers Grade 2 Chapter 13 Represent and Interpret Data 13.3 2
Which fruit is the most favorite? ______
Which fruit is the least favorite? ________
How many more students chose apple than banana? _________

Answer:
The favorite fruit = apple.
the least favorite = strawberry
The students choose apple than banana = 1

Explanation:
In the above-given figure,
The students who choose orange = 4
The students who choose strawberry = 2
The students who choose apple = 6
The students who choose banana = 5
The favorite fruit = apple.
the least favorite = strawberry
The students choose apple than banana = 1
Big-Ideas-Math-Solutions-Grade-2-Chapter-13-Represent and Interpret Data-13.3-2

Apply and Grow: Practice

Question 2.
Complete the picture graph.
Big Ideas Math Answers Grade 2 Chapter 13 Represent and Interpret Data 13.3 3
How many students chose the most favorite bird? ________
How many students chose flamingo or penguin? _______

Answer:
The students who choose penguin = 7
the students who choose flamingo or penguin = 6 or 7

Explanation:
In the above-given figure,
the students who choose penguin = 7
the students who choose flamingo = 5
the students who choose own = 3
the students who choose parrot = 2
The students who choose penguin = 7
the students who choose flamingo or penguin = 6 or 7

Big-Ideas-Math-Solutions-Grade-2-Chapter-13-Represent and Interpret Data-13.3-3

Question 3.
Complete the picture graph.
Big Ideas Math Answers Grade 2 Chapter 13 Represent and Interpret Data 13.3 4
Newton gives 4 coins to Descartes. How many coins does Newton have now? How do you know?
____________________
____________________

Answer:
Newton has zero coins.

Explanation:
In the above-given figure,
the newton has 4 coins.
Descartes has 6 coins.
I have 6 coins.
Given that newton gives 4 coins to Descartes.
so newton has 0 coins.
Big-Ideas-Math-Solutions-Grade-2-Chapter-13-Represent and Interpret Data-13.3-4

Think and Grow: Modeling Real Life

You ask 20 students to name their eye colors. 9 have brown eyes. 3 more students have brown eyes than blue eyes. The rest have green eyes. Complete the picture graph.
Big Ideas Math Answers Grade 2 Chapter 13 Represent and Interpret Data 13.3 5

Answer:
The students who have green eyes = 8

Explanation:
In the above-given question, given that
9 have brown eyes.
3 students have blue eyes.
the students who have green = 8
20 – 8 = 12
9 + 3 = 12.
Big-Ideas-Math-Solutions-Grade-2-Chapter-13-Represent and Interpret Data-13.3-5

Show and Grow

Question 4.
You ask 15 students which community helper is their favorite. 6 choose police officer. 2 fewer students choose doctor than police officer. The rest choose firefighter. Complete the picture graph.
Big Ideas Math Answers Grade 2 Chapter 13 Represent and Interpret Data 13.3 6
How did you find how many students chose firefighter?

Answer:
The students who choose firefighter = 5

Explanation:
The students who choose police officer = 6
The students who choose doctor = 2
The students who choose fire fighter= 5
8 + 5 = 15
15 – 10 = 5
Big-Ideas-Math-Solutions-Grade-2-Chapter-13-Represent and Interpret Data-13.3-6

Make Picture Graphs Homework & Practice 13.3

Question 1.
Complete the picture graph.
Big Ideas Math Answers Grade 2 Chapter 13 Represent and Interpret Data 13.3 7
Who has the least medals? _______
How many more medals do you have than your friend? _________
How many medals do Newton and Descartes have in all? ________

Answer:
Your friend has the least medals.
the medals I have more than my friend = 2
Newton and Descartes have all the medals = 11

Explanation:
In the above-given figure,
I have 4 medals.
my friend has 2 medals.
Newton has 5 medals.
descartes have 6 medals.
Big-Ideas-Math-Solutions-Grade-2-Chapter-13-Represent and Interpret Data-13.3-10

Question 2.
Complete the picture graph.
Big Ideas Math Answers Grade 2 Chapter 13 Represent and Interpret Data 13.3 8

2 more students chose tomato soup. How many students chose tomato now? How do you know?

Answer:
The students who choose tomato soup = 8

Explanation:
In the above-given figure,
The students who choose alphabet = 7
The students who choose vegetable = 3
the students who choose tomato = 6
given that 2 more students choose tomato soup.
6 + 2 = 8
Big-Ideas-Math-Solutions-Grade-2-Chapter-13-Represent and Interpret Data-13.3-11

Question 3.
Modeling Real Life
You ask 20 students about their favorite way to exercise. 4 like to walk. 6 like to swim. The rest like to bike. Complete the picture graph.
Big Ideas Math Answers Grade 2 Chapter 13 Represent and Interpret Data 13.3 9
How did you find how many students like to bike?

Answer:
The students who like to walk = 4.
The students who like to swim = 6.
Rest of the students who like bike = 10

Explanation:
In the above-given figure,
The students who like to walk = 4.
The students who like to swim = 6.
Rest of the students who like bike = 10
20 – 4 + 6
20 – 10
10
Big-Ideas-Math-Solutions-Grade-2-Chapter-13-Represent and Interpret Data-13.3-12

Review & Refresh

Question 4.
Big Ideas Math Answers Grade 2 Chapter 13 Represent and Interpret Data 13.3 10

Answer:
415 – 273 = 142

Explanation:
415 – 273 = 142
Big-Ideas-Math-Solutions-Grade-2-Chapter-13-Represent and Interpret Data-13.3-7

Question 5.
Big Ideas Math Answers Grade 2 Chapter 13 Represent and Interpret Data 13.3 11

Answer:
583 – 127 = 456

Explanation:
583 – 127 = 456
Big-Ideas-Math-Solutions-Grade-2-Chapter-13-Represent and Interpret Data-13.3-8

Question 6.
Big Ideas Math Answers Grade 2 Chapter 13 Represent and Interpret Data 13.3 12

Answer:
892 – 105 = 787

Explanation:
892 – 105 = 787
Big-Ideas-Math-Solutions-Grade-2-Chapter-13-Represent and Interpret Data-13.3-9

Lesson 13.4 Real and Interpret Bar Graphs

Explore and Grow

How are the tally chart and the bar graph the same? How are they different?
Big Ideas Math Solutions Grade 2 Chapter 13 Represent and Interpret Data 13.4 1
Based on the given values they are the same.
If they give different values they are different.

Answer:

The number of students who choose baseball = 6
The number of students who choose basketball = 5
The number of students who choose soccer = 7

Explanation:
In the above-given figure,
The number of students who choose baseball = 6
The number of students who choose basketball = 5
The number of students who choose soccer = 7
The students whose favorite spot = soccer.

Show and Grow

Question 1.
Big Ideas Math Solutions Grade 2 Chapter 13 Represent and Interpret Data 13.4 2
How many students chose computer? _______
Which activity is the least favorite? ________

Answer:
The students who choose computer = 8.
The least favorite activity = reading.

Explanation:
In the above-given figure,
The students who choose painting = 4
the students who choose computer = 8
the students who choose reading = 3
the students who choose games = 6
The students who choose computer = 8.
The least favorite activity = reading
Big-Ideas-Math-Solutions-Grade-2-Chapter-13-Represent and Interpret Data-13.4-1

Apply and Grow: Practice

Question 2.
Big Ideas Math Solutions Grade 2 Chapter 13 Represent and Interpret Data 13.4 3.1
How many students chose breakfast? _______
Which meal is the most favorite? ________
How many students chose the meal that is least favorite? ________
How many more students chose snack than lunch? _______

Answer:
The students who choose breakfast = 7.
the most favorite meal = Dinner.
The students chose the meal that is least favorite = 4
The students choose snack than lunch = 2

Explanation:
The students who choose breakfast = 7.
The students who choose lunch = 4.
The students who choose dinner = 9.
The students who choose snack = 6.
The students who choose breakfast = 7.
the most favorite meal = Dinner.
The students chose the meal that is least favorite = 4
The students choose snack than lunch = 2
Big-Ideas-Math-Solutions-Grade-2-Chapter-13-Represent and Interpret Data-13.4-2

Question 3.
DIG DEEPER!
Order the meals in Exercise 2 from the least favorite to the most favorite.
______, ______, ________, ________

Answer:
Lunch, snack, breakfast, and dinner.

Explanation:
In the above-given figure
given that order the meals from least favorite to the most favorite.
the least favorite = lunch.
the most favorite = dinner.
lunch, snack, breakfast, dinner.
Big-Ideas-Math-Solutions-Grade-2-Chapter-13-Represent and Interpret Data-13.4-3

Think and Grow: Modeling Real Life

Big Ideas Math Solutions Grade 2 Chapter 13 Represent and Interpret Data 13.4 3
A student chooses a food that has 1 more vote than eggs and toast combined. Which food does the student choose?
The student chooses ________.

Answer:
The students who choose 1 more than eggs = toast.

Explanation:
In the above-given figure,
The students who choose cereal = 9
The students who choose toast = 5
The students who choose pancakes = 8
The students who choose eggs = 3
The students who choose 1 more than eggs = toast.
Big-Ideas-Math-Solutions-Grade-2-Chapter-13-Represent and Interpret Data-13.4-4

Show and Grow

Question 4.
Big Ideas Math Solutions Grade 2 Chapter 13 Represent and Interpret Data 13.4 4
A student chooses an activity that has the same number of votes as crafts and hiking combined. Which activity does the student choose?
The student chooses ________

Answer:
The student who chooses an activity that has the same number of votes as crafts and hiking combined = Archery.

Explanation:
In the above-given figure,
the student who chooses archery = 6
the student who chooses crafts = 4
the student who chooses Hiking = 2
the student who chooses swimming = 7
The student who chooses an activity that has the same number of votes as crafts and hiking combined = Archery.
Big-Ideas-Math-Solutions-Grade-2-Chapter-13-Represent and Interpret Data-13.4-5

Real and Interpret Bar Graphs Homework & Practice 13.4

Question 1.
Big Ideas Math Solutions Grade 2 Chapter 13 Represent and Interpret Data 13.4 5
How many students chose police officer? _________
How many more students chose teacher than nurse? _________

Answer:
The students who choose police officer = 7.
The more students choose teacher than nurse = 1

Explanation:
In the above-given figure,
The students who choose teacher job = 5
The students who choose police officer job = 7
The students who choose sports player job = 8
The students who choose nurse job = 4
The students who choose police officer = 7.
The more students choose teacher than nurse = 1
Big-Ideas-Math-Solutions-Grade-2-Chapter-13-Represent and Interpret Data-13.4-6

Question 2.
Writing
How can you use a bar graph to find how many students were surveyed?
__________________
__________________

Answer:
The number of students surveyed = 24 students.

Explanation;
In the above-given graph
The students who choose teacher job = 5
The students who choose police officer job = 7
The students who choose sports player job = 8
The students who choose nurse job = 4
The total number of students were surveyed = 24 students.

Question 3.
Modeling Real Life
Use the bar graph.
Big Ideas Math Solutions Grade 2 Chapter 13 Represent and Interpret Data 13.4 6
A student chooses a writing tool that has 3 fewer votes than crayon and marker combined. Which writing tool does the student choose?
The student chooses ______.

Answer:
The student chooses a writing tool that has 3 fewer votes than crayon and marker combined = pen.

Explanation:
In the above-given graph,
the student who chooses pencil = 4
the student who chooses pen = 6
the student who chooses crayon = 2
the student who chooses marker = 5
The student chooses a writing tool that has 3 fewer votes than crayon and marker combined = pen.
Big-Ideas-Math-Solutions-Grade-2-Chapter-13-Represent and Interpret Data-13.4-7

Review & Refresh

Find the missing number.
Question 4.
Big Ideas Math Solutions Grade 2 Chapter 13 Represent and Interpret Data 13.4 7
Answer:
576 + 153  = 729

Explanation:
729 – 576 = 153
Big-Ideas-Math-Solutions-Grade-2-Chapter-13-Represent and Interpret Data-13.4-8

Question 5.
Big Ideas Math Solutions Grade 2 Chapter 13 Represent and Interpret Data 13.4 8
Answer:
431 + 389 = 820

Explanation:
820 – 431 = 389
Big-Ideas-Math-Solutions-Grade-2-Chapter-13-Represent and Interpret Data-13.4-9

Question 6.
Big Ideas Math Solutions Grade 2 Chapter 13 Represent and Interpret Data 13.4 9

Answer:
128 + 521 = 649

Explanation:
649 – 128 = 521
Big-Ideas-Math-Solutions-Grade-2-Chapter-13-Represent and Interpret Data-13.4-10

Lesson 13.5 Make Bar Graphs

Explore and Grow

Look at your Instrument Cards. Complete the tally chart and the bar graph.
Big Ideas Math Answer Key Grade 2 Chapter 13 Represent and Interpret Data 13.5 1
Answer:
The students who choose drum = 2
The students who choose trumpet = 3
The students who choose tuba = 4

Explanation:
In the above-given figure,
The students who choose drum = 2
The students who choose trumpet = 3
The students who choose tuba = 4
Big-Ideas-Math-Solutions-Grade-2-Chapter-13-Represent and Interpret Data-13.5-1

Show and Grow

Question 1.
Complete the bar graph.
Big Ideas Math Answer Key Grade 2 Chapter 13 Represent and Interpret Data 13.5 2
Which book type is the least favorite? _________
How many more students chose fiction than history? _______

Answer:
The least favorite book = history
The more students choose fiction than history = 3

Explanation:
In the above-given figure,
The students who choose history = 3
The students who choose fiction = 7
The students who choose science = 5
The students who choose poetry = 6
The least favorite book = history
The more students choose fiction than history = 3
Big-Ideas-Math-Solutions-Grade-2-Chapter-13-Represent and Interpret Data-13.5-2

Apply and Grow: Practice

Question 2.
Complete the bar graph.
Big Ideas Math Answer Key Grade 2 Chapter 13 Represent and Interpret Data 13.5 3
How many raisins are there? _________
How many more almonds are there than dried fruit? _______

Answer:
The raisins are = 6
The more almonds are there than dried fruit = 3

Explanation:
In the above-given figure,
The number of peanuts = 9
the number of raisins = 6
The number of almonds = 7
the number of dried fruit = 4
The raisins are 6
the more almonds are there than dried fruits = 3
Big-Ideas-Math-Solutions-Grade-2-Chapter-13-Represent and Interpret Data-13.5-3

Question 3.
Complete the bar graph.
Big Ideas Math Answer Key Grade 2 Chapter 13 Represent and Interpret Data 13.5 4
Which drink was chosen the most? _______
How many more people chose iced tea than water? _______

Answer:
The number of people who choose the drink most = Lemonade
The people who choose iced tea over water = 3

Explanation:
The number of people who choose water = 5
The number of people who chose fruit punch = 2
The number of people who choose lemonade = 9
The number of people who choose  iced tea 8
The number of people who choose the drink most = Lemonade
The people who choose iced tea over water = 3
Big-Ideas-Math-Solutions-Grade-2-Chapter-13-Represent and Interpret Data-13.5-4

Think and Grow: Modeling Real Life

You classify 30 animals as fish, mammals, or reptiles.11 are fish. 7 are reptiles. The rest are mammals. Complete the bar graph.
Big Ideas Math Answer Key Grade 2 Chapter 13 Represent and Interpret Data 13.5 5

Answer:
The mammals = 12.

Explanation:
In the above-given figure,
Given that the fishes are 11.
reptiles = 7
mammals = 12
11 + 7 = 18
30 – 18 = 12
Big-Ideas-Math-Solutions-Grade-2-Chapter-13-Represent and Interpret Data-13.5-5

Show and Grow

Question 4.
You ask 29 students if they want to collect seashells, fossils, or stickers. 8 said fossils. 12 said stickers. The rest said seashells. Complete the bar graph.
Big Ideas Math Answer Key Grade 2 Chapter 13 Represent and Interpret Data 13.5 6

DIG DEEPER!
You ask 3 more students which object they want to collect. 1 said fossils. 2 said seashells. How many fewer students chose fossils than seashells now?
_______ students

Answer:
The fewer students choose fossils than seashells now = 2.

Explanation:
In the above-given figure,
The number of students who choose seashells = 9
The number of students who choose fossils = 8
The number of students who choose stickers = 12
given that 29 students.
11 + 9 = 20
29 – 20 = 9
Big-Ideas-Math-Solutions-Grade-2-Chapter-13-Represent and Interpret Data-13.5-6
Make Bar Graphs Homework & Practice 13.5

Question 1.
Complete the bar graph.
Big Ideas Math Answer Key Grade 2 Chapter 13 Represent and Interpret Data 13.5 7
How many more votes did Spot receive than Sparkle? ______

Answer:
The more votes did spot receive than sparkle = 7

Explanation:
In the above-given figure,
The number of votes for spot = 9
The number of votes for sparkle = 2
The number of votes for flip = 7
The number of votes for star = 6
The more votes did spot receive than sparkle = 7
the spot receives 7 more votes than sparkle.
Big-Ideas-Math-Solutions-Grade-2-Chapter-13-Represent and Interpret Data-13.5-7

Question 2.
Complete the bar graph.
Big Ideas Math Answer Key Grade 2 Chapter 13 Represent and Interpret Data 13.5 8
How many total animals are in the pet store? _______

Answer:
The total animals in the pet store = 20.

Explanation:
In the above-given figure,
Given that fish = 7
hamster = 5
snake = 4
rabbit = 4
total animals in the pet = 20 animals.
Big-Ideas-Math-Solutions-Grade-2-Chapter-13-Represent and Interpret Data-13.5-8

Question 3.
Modeling Real Life
You, your friend, and your cousin hand out a total of 25 flyers. You hand out 12. Your friend hands out 9. Your cousin hands out the rest. Complete the bar graph.
Big Ideas Math Answer Key Grade 2 Chapter 13 Represent and Interpret Data 13.5 9
DIG DEEPER!
You hand out 1 more flyer. Your friend hands out4 more. How many fewer flyers does your cousin hand out than your friend now?
_______ flyers

Answer:
The many fewer flyers do your cousin hand out than your friend now = 9

Explanation:
In the above-given figure,
The flyers handed out by me = 12.
The flyers handed out by my friend = 9.
The flyers handed out by my cousin = 4.
given that flyers handed out by me = 12
totally there are 25 flyers.
12 + 9 = 25
Big-Ideas-Math-Solutions-Grade-2-Chapter-13-Represent and Interpret Data-13.5-9
Review & Refresh

Question 4.
861 − 410 = ______

Answer:
861 – 410 = 451

Explanation:
861 – 410 = 451
Big-Ideas-Math-Solutions-Grade-2-Chapter-13-Represent and Interpret Data-13.5-10

Question 5.
624 − 320 = ______

Answer:
624 – 320 = 304

Explanation:
624 – 320 = 304
Big-Ideas-Math-Solutions-Grade-2-Chapter-13-Represent and Interpret Data-13.5-11

Lesson 13.6 Make Line Plots

Explore and Grow

How are the thumb lengths shown on the number line?
Big Ideas Math Answers 2nd Grade Chapter 13 Represent and Interpret Data 13.6 1
Answer:
The thumb lengths for thumb 1, thumb 4, thumb 7 = 4 cms.
thumb 2 , thumb 6 = 5 cms.
thumb 3 = 3 cm
thumb 5 = 6cm

Explanation:
In the above-given figure,
given the thumb lengths.
The thumb lengths for thumb 1, thumb 4, thumb 7 = 4 cms.
thumb 2 , thumb 6 = 5 cms.
thumb 3 = 3 cm
thumb 5 = 6cm

Show and Grow

Question 1.
Complete the line plot.
Big Ideas Math Answers 2nd Grade Chapter 13 Represent and Interpret Data 13.6 2
How many long jumps are 43 inches long? _____
What is the most common long jump length? ______ inches

Answer:
One long-jump is 43 inches long = child 3.
the most common long jump length = child 1 and child 4.

Explanation:
In the above-given figure,
given that the long jump lengths.
child 1 and child 4 = 41
child 2 = 44
child 5 = 45
One long-jump is 43 inches long = child 3.
the most common long jump length = child 1 and child 4.
Big-Ideas-Math-Solutions-Grade-2-Chapter-13-Represent and Interpret Data-13.6-1

Apply and Grow: Practice

Question 2.
Complete the line plot.
Big Ideas Math Answers 2nd Grade Chapter 13 Represent and Interpret Data 13.6 3
What is the most common puppy length? ______ inches
How many fewer puppies are 11 inches long than 10 inches long? ______
How many puppies are 8 or 9 inches long? ______
How many puppies are shorter than 11 inches? How do you know?
_____________________
____________________

Answer:
The most common puppy length = puppy 2, 3, 5, 6, and 7.
The fewer puppies are 11 inches longer than 10 inches long = puppy 1.
The puppies which are 8 or 9 inches long = 3 puppies.
The puppies which are shorter than 11 inches = 6 puppies.

Explanation:
In the above-given figure,
given the puppy lengths.
the puppy lengths with puppy 1 = 11 inches.
puppy 2 = 9 inches.
puppy 3 = 10 inches.
puppy 4 = 12 inches.
puppy 5 = 9 inches.
puppy 6 = 10 inches.
puppy 7 = 10 inches.
puppy 8 = 8 inches.
The most common puppy length = puppy 2, 3, 5, 6, and 7.
The fewer puppies are 11 inches longer than 10 inches long = puppy 1.
The puppies which are 8 or 9 inches long = 3 puppies.
The puppies which are shorter than 11 inches = 6 puppies.
Big-Ideas-Math-Solutions-Grade-2-Chapter-13-Represent and Interpret Data-13.6-2

Think and Grow: Modeling Real Life

9 people measure the length of a guitar. The line plot shows the measured lengths. How long do you think the guitar is? Explain.
Big Ideas Math Answers 2nd Grade Chapter 13 Represent and Interpret Data 13.6 4.2
______ inches

Answer:
The length of the guitar = 35 inches.

Explanation:
In the above-given figure,
the number of inches in guitar lengths are:
the guitar length with 34 inches.
the guitar length with 35 inches.
the guitar length with 36 inches.

Show and Grow

Question 3.
8 people measure the length of a playground. The line plot shows the measured lengths.How long do you think the playground is? Explain.
Big Ideas Math Answers 2nd Grade Chapter 13 Represent and Interpret Data 13.6 5
______ meters
DIG DEEPER!
Why are the measurements different?

Answer:
The length o the playground = 51 meters.

Explanation:
In the above-given figure,
the number of meters of playgrounds is:
given that 8 people measure the length of the playground.
6 people choose 50 meters.
1 people choose 49 meters.
1 people choose 48 meters.

Make Line Plots Homework & Practice 13.6

Question 1.
Complete the line plot.
Big Ideas Math Answers 2nd Grade Chapter 13 Represent and Interpret Data 13.6 6
How many leaves are 13 centimeters long? _______
What is the most common leaf length? _______ centimeters
How many more leaves are 14 centimeters long than 12 centimeters long? ______
How many leaves are 14 or 15 centimeters? ________

Answer:
No leaves are 13 centimeters long.
the most common leaf lengths are leaf 2, leaf 3, and leaf 6.
the more leaves are 14 centimeters long than 12 centimeters long = leaf 1 and leaf 4.
The leaves are 14 or 15 centimeters = 5 leaves.

Explanation:
In the above-given figure,
given that, the leaf lengths.
the leaf 1 = 14 centimeters.
the leaf 2 = 15 centimeters
the leaf 3 = 15 centimeters
the leaf 4 = 14 centimeters
the leaf 5 = 12 centimeters
the leaf 6 = 15 centimeters
No leaves are 13 centimeters long.
the most common leaf lengths are leaf 2, leaf 3, and leaf 6.
the more leaves are 14 centimeters long than 12 centimeters long = leaf 1 and leaf 4.
The leaves are 14 or 15 centimeters = 5 leaves.
Big-Ideas-Math-Solutions-Grade-2-Chapter-13-Represent and Interpret Data-13.6-3
Question 2.
Complete the line plot.
Big Ideas Math Answers 2nd Grade Chapter 13 Represent and Interpret Data 13.6 7
How many sharks are longer than 14 feet? How do you know?
____________

Answer:
The sharks which are longer than 14 feet = 3 sharks.

Explanation:
In the above-given figure,
given that the shark lengths in feets.
shark 1 = 14
shark 2 = 15
shark 3 = 16
shark 4 = 14
shark 5 = 12
shark 6 = 16
The sharks which are longer than 14 feet = 3 sharks.
Big-Ideas-Math-Solutions-Grade-2-Chapter-13-Represent and Interpret Data-13.6-4

Question 3.
Modeling Real Life
8 people measure the length of a bus. The line plot shows the measured lengths. How long do you think the school bus is? Explain.
Big Ideas Math Answers 2nd Grade Chapter 13 Represent and Interpret Data 13.6 8
______ meters

Answer:
The length of the school bus = 105 meters.

Explanation:
In the above-given figure,
the lengths of the school bus is given.
given that 8 people measure the length of a bus.
the lone plot shows the measured lengths.
2 people measure the 14 meters.
1 people measure the 12 inches.
Big-Ideas-Math-Solutions-Grade-2-Chapter-13-Represent and Interpret Data-13.6-5

Review & Refresh

Question 4.
Big Ideas Math Answers 2nd Grade Chapter 13 Represent and Interpret Data 13.6 9
Answer:
127.

Explanation:
36 + 4 + 22 + 65
40 + 22 + 65
62 + 65
127
Big-Ideas-Math-Solutions-Grade-2-Chapter-13-Represent and Interpret Data-13.6-5

Question 5.
Big Ideas Math Answers 2nd Grade Chapter 13 Represent and Interpret Data 13.6 10
Answer:
208.

Explanation:
95 + 68 + 45
163 + 45
208
Big-Ideas-Math-Solutions-Grade-2-Chapter-13-Represent and Interpret Data-13.6-6

Question 6.
Big Ideas Math Answers 2nd Grade Chapter 13 Represent and Interpret Data 13.6 11
Answer:
226.

Explanation:
76 + 50 + 18 + 82
126 + 18 + 82
144 + 82
226
Big-Ideas-Math-Solutions-Grade-2-Chapter-13-Represent and Interpret Data-13.6-7

Lesson 13.7 Measure Objects and Make Line Plots

Explore and Grow

Use a ruler to measure the caterpillars on Lengths of Caterpillars. Use the lengths to complete the chart and the line plot.
Big Ideas Math Answers Grade 2 Chapter 13 Represent and Interpret Data 13.7 1

Answer:
The long length of the caterpillar = 9.

Explanation:
We have to assume the lengths of the cater pillars.
cater pillar 1 = 6 cms.
cater pillar 2 = 7 cms.
cater pillar 3 = 9 cms.
cater pillar 4 = 9 cms.
cater pillar 5 = 7 cms.
cater pillar 6 = 10 cms.
cater pillar 7 = 8 cms.
cater pillar 8 = 7 cms.
The long length of the caterpillar = 9
Big-Ideas-Math-Solutions-Grade-2-Chapter-13-Represent and Interpret Data-13.6-8

Show and Grow

Question 1.
Measure the lengths of 4 pencils. Complete the line plot.
Big Ideas Math Answers Grade 2 Chapter 13 Represent and Interpret Data 13.7 2
What is the length of the longest pencil? _______ inches
How much longer is the longest pencil than the shortest pencil? _______ inches

Answer:
The length of the longest  pencil = 10 inches.
The longer is the longest pencil than the shortest penccil = 5 inches.

Explanation:
In the above-given figure,
given that the length of the 4 pencils.
the length of the pencil 1 = 5 inches.
pencil 2 = 6 inches.
pencil 3 = 8 inches.
pencil 4 = 10 inches.
The length of the longest  pencil = 10 inches.
The longer is the longest pencil than the shortest penccil = 5 inches.
Big-Ideas-Math-Solutions-Grade-2-Chapter-13-Represent and Interpret Data-13.6-9

Apply and Grow: Practice

Question 2.
Measure the lengths of 8 shoes. Complete the line plot.
Big Ideas Math Answers Grade 2 Chapter 13 Represent and Interpret Data 13.7 3
What is the length of the longest shoe? ______ inches
What is the length of the shortest shoe? _______ inches
How much shorter is the shortest shoe than the longest shoe? _______ inches
You measure 5 more shoes and they are each6 inches long. How does the line plot change?
_________________________
_________________________

Answer:
The length of the longest shoe = 10 inches.
The length of the shortest shoe = 6 inches.
The shorter is the shortest shoe than the longest shoe = 5 inches.
The line plot changes.

Explanation:
In the above-given figure,
we have to assume  the examples.
length of the shoe 1 = 6 inches.
length of the shoe 2 = 7 inches.
length of the shoe 3 = 6 inches.
length of the shoe 4 = 7 inches.
length of the shoe 5 = 8 inches.
length of the shoe 6 = 9 inches.
length of the shoe 7 = 10 inches.
length of the shoe 8 = 8 inches.
The length of the longest shoe = 10 inches.
The length of the shortest shoe = 6 inches.
The shorter is the shortest shoe than the longest shoe = 5 inches.
The line plot changes.
Big-Ideas-Math-Solutions-Grade-2-Chapter-13-Represent and Interpret Data-13.6-10

Think and Grow: Modeling Real Life

A fire station is building a new garage for emergency vehicles. Complete the sentence. Explain.
Big Ideas Math Answers Grade 2 Chapter 13 Represent and Interpret Data 13.7 4
The garage should be more than _______ meters long.
____________________

Answer:
The garage should be more than 16 meters long.

Explanation:
In the above-given figure,
Given that the fire station is building a new garage for emergency vehicles.
the garage should be more than 6 meters long and the shorter than 16 meters long.
Big-Ideas-Math-Solutions-Grade-2-Chapter-13-Represent and Interpret Data-13.6-11

Show and Grow

Question 3.
You want to put some school supplies in a pencil box. Complete the sentence. Explain.
Big Ideas Math Answers Grade 2 Chapter 13 Represent and Interpret Data 13.7 5
The pencil box should be more than _______ centimeters long.
______________________
Answer:
Big-Ideas-Math-Solutions-Grade-2-Chapter-13-Represent and Interpret Data-13.6-11

Measure Objects and Make Line Plots Homework & Practice 13.7

Question 1.
Big Ideas Math Answers Grade 2 Chapter 13 Represent and Interpret Data 13.7 6
Measure the lengths of 5 socks. Complete the line plot.
What is the length of the longest sock? ______ inches
What is the length of the shortest sock? _______ inches
How much longer is the longest sock than the shortest sock? _______ inches

Answer:
The length of the longest sock = 10 inches.
The length of the shortest sock = 7 inches.
The longer is the longest sock than the shortest sock = 3 inches.

Explanation:
In the above-given figure,
we have to assume the:
sock 1 = 7
sock 2 = 8
sock 3 = 8
sock 4 = 9
sock 5 = 10
The length of the longest sock = 10 inches.
The length of the shortest sock = 7 inches.
The longer is the longest sock than the shortest sock = 3 inches.
Big-Ideas-Math-Solutions-Grade-2-Chapter-13-Represent and Interpret Data-13.6-12

Question 2.
Measure the lengths of 4 hands. Complete the line plot.
Big Ideas Math Answers Grade 2 Chapter 13 Represent and Interpret Data 13.7 7
How many hands are longer than 5 inches? How do you know?
__________________________
___________________________

Answer:
The hands that are longer than 5 inches = 3

Explanation:
We have to assume the hands in inches.
given that hand 1 = 7 inches.
hand 2 = 6 inches.
hand 3 = 5 inches.
hand 4 = 6 inches.
The hands that are longer than 5 inches = 3
Big-Ideas-Math-Solutions-Grade-2-Chapter-13-Represent and Interpret Data-13.6-13

Question 3.
Modeling Real Life
Newton wants to put his dog bones in a box. Complete the sentence. Explain.
Big Ideas Math Answers Grade 2 Chapter 13 Represent and Interpret Data 13.7 8
The box should be more than ______ inches long.
____________________

Answer:
The bo should be more than 11 inches.

Explanation:
In the above-given figure,
Given that newton wants to put his dog bones in a box.
the box should be more than 11 inches.
the box should be less than 4 inches.
Review & Refresh

Question 4.
10 more than 347 is _______.

Answer:
357

Explanation:
Given that,
10 more than 347 = 357.
347 + 10 = 357

Question 5.
100 less than 926 is _________.

Answer:
826

Explanation:
Given that 100 less than 926 = 826
926 – 100 = 826

Represent and Interpret Data Performance Task

You measure the lengths of 8 writing tools. The tools and their lengths are shown.
Big Ideas Math Solutions Grade 2 Chapter 13 Represent and Interpret Data 1
Question 1.
Organize the writing tool lengths on the line plot.
Big Ideas Math Solutions Grade 2 Chapter 13 Represent and Interpret Data 2

Answer:
The maximum length of the pencil = 12cms.

Explanation:
In the above-given figure,
given that
pencil 1 = 9 cm
pencil 2 = 9 cm
pencil 3 = 6 cm
pencil 4 = 9 cm
pencil 5 = 12 cm
pencil 6 = 12 cm
pencil 7 = 12 cm
pencil  8 = 12 cm
the maximum length of the pencil = 12 cms
Big-Ideas-Math-Solutions-Grade-2-Chapter-13-Represent and Interpret Data-13.7-1

Question 2.
Use the line plot to complete the equation. Why is the sum 8?
_____ + ______ + ______ = 8

Answer:
2 + 2 + 4 = 8

Question 3.
Measure your pencil. Add the length of your pencil to the line plot in Exercise 1.

Answer:
My pencil length is = 11 cms.

Explanation:
In the above-given figure,
given that
pencil 1 = 9 cm
pencil 2 = 9 cm
pencil 3 = 6 cm
pencil 4 = 9 cm
pencil 5 = 12 cm
pencil 6 = 12 cm
pencil 7 = 12 cm
pencil  8 = 12 cm
the maximum length of the pencil = 12 cms
Assuming my pencil length = 11 cms
Big-Ideas-Math-Solutions-Grade-2-Chapter-13-Represent and Interpret Data-13.7-2

Question 4.
Compare the lengths of your pencil and one of the writing tools above.

Answer:
The length of my pencil = 11 cms.

Explanation:
In the above-given figure,
given that
pencil 1 = 9 cm
pencil 2 = 9 cm
pencil 3 = 6 cm
pencil 4 = 9 cm
pencil 5 = 12 cm
pencil 6 = 12 cm
pencil 7 = 12 cm
pencil  8 = 12 cm
the maximum length of the pencil = 12 cms
Assuming my pencil length = 11 cms
Big-Ideas-Math-Solutions-Grade-2-Chapter-13-Represent and Interpret Data-13.7-2

Represent and Interpret Data Activity

Spin and Graph
To Play: Spin 10 times. Complete the tally chart. Then complete the bar graph. Answer the Spin and Graph Questions about your graph.
Big Ideas Math Solutions Grade 2 Chapter 13 Represent and Interpret Data 3

Represent and Interpret Data Chapter Practice

13.1 Sort and Organize Data

Question 1.
Use the data to complete the tally chart.
Big Ideas Math Solutions Grade 2 Chapter 13 Represent and Interpret Data chp 1
How many students chose magic show? _______
Which event is the least favorite? __________

Answer:
The students who choose the magic show = 5 students.
The last favorite = dancing.

Explanation:
In the above-given figure,
The students who choose magic show = 5
The students who choose face painting = 4
The students who choose games = 2
The students who choose dancing = 1
The students who choose the magic show = 5 students.
The last favorite = dancing.
Big-Ideas-Math-Solutions-Grade-2-Chapter-13-Represent and Interpret Data-13.7-3

Question 2.
Modeling Real Life
You want to survey30 students. How many more students do you need to ask?
Big Ideas Math Solutions Grade 2 Chapter 13 Represent and Interpret Data chp 2
______ students
How many more students need to have red hair so that the numbers of students with red hair and blonde hair are equal?
______ students

Answer:
The more students need to ask = 3 students.
The more students need to have red hair so that the numbers of students with red hair and blonde hair are equal = 7 students.

Explanation:
In the above-given figure,
the hair color which is in red = 2
the hair color which is in black = 5
the hair color which is brown = 10
the hair color which is in blonde = 9
The more students need to ask = 3 students.
The more students need to have red hair so that the numbers of students with red hair and blonde hair are equal = 7 students.
Big-Ideas-Math-Solutions-Grade-2-Chapter-13-Represent and Interpret Data-13.7-4

13.2 Read and Interpret Picture Graphs

Question 3.
Big Ideas Math Solutions Grade 2 Chapter 13 Represent and Interpret Data chp 3
Which type of book do exactly 7 students like best? _______
How many more students chose magazine than fiction? ________

Answer:
The type of book does exactly 7 students like best = comics.
The more students choose magazine than fiction = 4.

Explanation:
In the above-given figure,
The students who choose comics = 7
The students who choose fiction = 4
The students who choose nonfiction = 5
The students who choose magazine = 8
The type of book does exactly 7 students like best = comics.
The more students choose magazine than fiction = 4.
Big-Ideas-Math-Solutions-Grade-2-Chapter-13-Represent and Interpret Data-13.7-5

13.3 Make Picture Graphs

Question 4.
Complete the picture graph.
Big Ideas Math Solutions Grade 2 Chapter 13 Represent and Interpret Data chp 4
How many fewer students chose tiger than cheetah? ______
How many students chose the least favorite cat? _______
How many students chose panther or lion? __________

Answer:
The fewer students choose tiger than cheetah = 4
The students who choose least favorite cat = tiger
the students who choose panther = 5
the students who choose lion = 4

Explanation:
In the above-given figure,
Given that,
The fewer students choose tiger than cheetah = 4
The students who choose least favorite cat = tiger
the students who choose panther = 5
the students who choose lion = 4
Big-Ideas-Math-Solutions-Grade-2-Chapter-13-Represent and Interpret Data-13.7-6

13.4 Read and Interpret Bar Graphs

Question 5.
Big Ideas Math Solutions Grade 2 Chapter 13 Represent and Interpret Data chp 5
How much rain fell on Tuesday? ______
Which day did it rain the least? _______
How much more rain fell on Monday than Wednesday? _________

Answer:
The rain fell on Tuesday = 3 inches.
The day it rains the least = thursday.
the more rain fell on Monday than Wednesday = 1 inch.

Explanation:
In the above-given graph,
the rain fell on Monday = 5 inches.
the rain fell on Tuesday = 3 inches.
the rain fell on Wednesday = 4 inches.
the rain fell on Thursday = 2 inches.
The rain fell on Tuesday = 3 inches.
The day it rains the least = Thursday.
the more rain fell on Monday than Wednesday = 1 inch.

Big-Ideas-Math-Solutions-Grade-2-Chapter-13-Represent and Interpret Data-13.7-7
13.5 Make Bar Graphs

Question 6.
Complete the bar graph.
Big Ideas Math Solutions Grade 2 Chapter 13 Represent and Interpret Data chp 6
How many hurricanes were there in 2012 and 2013? _______

Answer:
The hurricanes were there in 2012 and 2013 = 12.

Explanation:
In the above-given figure,
given that the hurricanes in the north Atlantic.
in 2012 = 10
2013 = 2
2014 = 5
2015 = 4
The hurricanes were there in 2012 and 2013 = 12.
Big-Ideas-Math-Solutions-Grade-2-Chapter-13-Represent and Interpret Data-13.7-8

13.6 Read and Interpret Line Plots

Question 7.
Complete the line plot.
Big Ideas Math Solutions Grade 2 Chapter 13 Represent and Interpret Data chp 7
How many feathers are longer than 13 centimeters? How do you know?
__________________
__________________

Answer:
The feathers which are longer than 13 centimeters = 2 feathers.

Explanation:
In the above-given figure,
The feathers with the lengths in centimeters.
feather 1 = 14 cms.
feather 2 = 13 cms.
feather 3 = 15 cms.
feather 4 = 12 cms.
The feathers which are longer than 13 centimeters = 2 feathers.
Big-Ideas-Math-Solutions-Grade-2-Chapter-13-Represent and Interpret Data-13.7-9

13.7 Measure Objects and Make Line Plots

Question 8.
Use an inch ruler to measure the lengths of 5 toys to the nearest inch. Then complete the line plot.
Big Ideas Math Solutions Grade 2 Chapter 13 Represent and Interpret Data chp 8
What is the length of the longest toy? _______ inches

Answer:
The length of the longest toy = 9inches.

Explanation:
In the above-given figure,
we have to assume that
the length of the toy 1 = 6 inches.
the length of the toy 2 = 7 inches.
the length of the toy 3 = 6 inches
the length of the toy 4 = 8 inches.
the length of the toy 5 = 9 inches.
The length of the longest toy = 9inches.
Big-Ideas-Math-Solutions-Grade-2-Chapter-13-Represent and Interpret Data-13.7-10

Conclusion:

I hope that the solutions explained here are helpful for the students while preparing for the exams. So, download Big Ideas Math 2nd Grade 13th Chapter Represent and Interpret Data Answer Key pdf for free of cost and begin preparation. Students can clear their doubts by writing a comment in the below comment section. Also, bookmark our site to get the latest edition solutions for Big Ideas Math Book Grade 2 Chapters.

Big Ideas Math Answers Grade 2 Chapter 14 Money and Time

Big Ideas Math Answers Grade 2 Chapter 14

Students can find Big Ideas Math Book Grade 2 Chapter 14 Money and Time Answers in the following sections. You can definitely score good marks on the test with the help of the BIM 2nd Grade Book 14th Chapter Money and Time Answer Key. The students can download and practice questions from Big Ideas Math Answers Grade 2 14th Chapter Money and Time PDF for free of cost. It will help you to complete the homework within time.

Big Ideas Math Book 2nd Grade Answer Key Chapter 14 Money and Time

Check out the BIM Book Grade 2 Chapter 14 Money and Time Solution Key and prepare well for the exam. You will see the step by step detailed explanation for each and every question from Big Ideas Math Textbook Grade 2 Chapter 14 Money and Time.

The different topics covered in this chapter are Find Total Values of Coins, Order to Find Total Values of Coins, Show Money Amounts in Different Ways, Make One Dollar, Make Change from One Dollar, Find Total Values of Bills, Problem Solving: Money, Tell Time to the Nearest Five Minutes, Tell Time Before and After the Hour, and Relate A.M. and P.M.

Vocabulary

Lesson: 1 Find Total Values of Coins

Lesson: 2 Order to Find Total Values of Coins

Lesson: 3 Show Money Amounts in Different Ways

Lesson: 4 Make One Dollar

Lesson: 5 Make Change from One Dollar

Lesson: 6 Find Total Values of Bills

Lesson: 7 Problem Solving: Money

Lesson: 8 Tell Time to the Nearest Five Minutes

Lesson: 9 Tell Time Before and After the Hour

Lesson: 10 Relate A.M. and P.M.

Chapter: 10 – Money and Time

Money and Time Vocabulary

Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 1

Organize It
Use the review words to complete the graphic organizer.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 2

 

Answer:
1. Analog clock .
2. The long hand pointing minutes is the minutes hand.
3. The short hand pointing hours is the hour hand.

Define it
Use your vocabulary cards to complete the puzzle.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 3

Answer:

1. Penny
2.Dime
3. Quarter
4. Nickel

Chapter 14 Vocabulary Cards

Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 4
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 5
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 6
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 7
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 8
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 9

Lesson 14.1 Find Total Values of Coins

Explore and Grow

Sort your coins.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 10
Explain how you sorted.
__________________
__________________
__________________
__________________

 

Answer:

Show and Grow

Count on to find the total value.

Question 1.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 11

 

Answer:
36 cents

Explanation :
1 dime = 10 cents
That means, 3x 10 = 30 cents
1 nickel = 5 cents
That means,  1x 5= 5 cents
and 1 cent
So , 3 dime +1 nickel+1 cent = 30 cents+ 5 cents +1 cent = 36 cents.

Question 2.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 12

 

Answer:
42 cents

Explanation:
1 quarter = 25 cents
That means , 1 x25= 25 cents
1 nickel = 5 cents
That means , 3  x 5  = 15 cents & 2 cents additional
Total : 25 cents + 15 cents + 2 cents = 42 cents
Therefore , total value is 42 cents .

Apply and Grow: Practice

Count on to find the total value.

Question 3.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 13

 

Answer:
55 cents

Explanation :
1 quarter = 25 cents
Which means , 2 x 25 = 50
1 nickel = 5 cents
Total : 50 + 5 =55 cents
Total value of coins is 55 cents.

Question 4.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 14

 

Answer:
48 cents

Explanation:
1 quarter = 25 cents
1 dime = 10 cents

Which means,  2 x 10 = 20 cents
and 3 cents
Total : 25 +20 +3 = 48 cents
Therefore, Total value is 48 cents.

Question 5.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 15

 

Answer:
66 cents

Explanation:
1 quarter = 25 cents
1 dime = 10 cents
Which means , 3 x 10 = 30 cents
1 nickel = 5 cents
Which means , 2 x 5 = 10 cents
& 1 cent
Total : 25 +30+ 10 +1=66
Therefore , the total value of coins is 66 cents.

Question 6.
Reasoning
You have 27¢. Which groups of coins could you have?
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 16

 

Answer:

This set of coins have total of 27 cents

Explanation:
1 dime = 10 cents
Which means, 2 x 10 = 20 cents
1nickel = 5 cents
and 2 cents
Total : 20 + 5 + 2 = 27 cents.

Think and Grow: Modeling Real Life

You have 2 quarters, 1 dime, 4 nickels, and 1 penny. How many cents do you have? Do you have enough money to buy the airplane?
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 17
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 18

Answer:
I have 81 cents.
Yes, I have enough money to buy the airplane.

Explanation:
1 quarter = 25 cents
So 2 quarters = 25+25 =50 cents
1 dime =10 cents
1 nickel = 5 cents
So 4 nickels = 4 x 5 = 20 cents
1 penny =1 cent
Total amount is 50+10+20+1 =81 cents
The airplane costs 80 cents so, I have enough money to buy the plain.

Show and Grow

Question 7.
You have 5 dimes, 3 nickels, and 2 pennies. How many cents do you have? Do you have enough money to buy the coloring book?
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 19
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 19.1

Answer:
I have 67 cents.
No, I do not have enough money to buy the coloring book.

Explanation:
1 dime =10 cents
so, 5 dimes = 5 x 10= 50 cents
1 nickel = 5 cents
So, 3 nickels = 3 x 5 = 15 cents
2 penny =2 cents
Total amount is 50+15+2 =67 cents
The coloring book costs 70 cents
So, I do not  have enough money to buy the book.

Question 8.
You have 4 dimes, 1 nickel, and 3 pennies. How many more cents do you need to buy the whistle? Draw and label the coins you need.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 20

Answer:
I have 48 cents.
No , I  do not have enough money to buy the whistle.

Explanation:
1 dime =10 cents
so, 4 dimes =4 x 10= 40 cents
1 nickel = 5 cents
3 pennies =3 cents
Total amount is 40+5+3 =48 cents
The whistle costs 59 cents so, I  do not have enough money to buy the whistle.

Question 9.
DIG DEEPER!
You have 3 quarters, 2 nickels, and 3 pennies. Your friend has 1 quarter and 5 dimes. Who has more money? How much more?
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 21

Answer:
I have 13 cents money more than my friend .

Explanation:
My money :
1 quarter = 25 cents
So 3 quarters = 3 x 25 = 75 cents
1 nickel = 5 cents
So 2 nickel = 2 x 5 = 10 cents
3 pennies = 3 cents
Total amount: 75+10+3 = 88 cents
So, I have a total of 88 cents .
Friends money:
1 quarter= 25 cents
1 dime = 10 cents
So 5 dimes = 5 x 10 = 50 cents
Total amount = 25 +50= 75 cents
Friends money total is 75 cents
Now, the difference between my money and my friends money is
88 cents – 75 cents = 13 cents
That means, I have 13 cents more than my friend.

Find Total Values of Coins Homework & Practice 14.1

Count on to find the total value.

Question 1.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 22

Answer:
65 cents

Explanation:
1 quarter = 25 cents
1 dime = 10 cents
Which means , 4 x 10 = 40 cents
Total : 25 + 40 = 65 cents
Therefore, the total value of coins is 65 cents.

Question 2.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 23

Answer:
38 cents

Explanation:
1 dime = 10 cents
Which means , 3 dime = 3 x 10 = 30 cents
1 nickel = 5 cents
and 3 cents
Total : 30+ 5 + 3 = 38 cents

Question 3.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 24

Answer:
72 cents

Explanation:
1 quarter = 25 cents
1 dime = 10 cents
Which means, 4 x 10 = 40 cents
1 nickel = 5 cents
and 2 cents
Total : 25 + 40+ 5 + 2 =72 cents

Question 4.
DIG DEEPER!
You had 52¢. You lost a coin. Now you have the 5 coins shown. What coin did you lose?
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 24.1

 

Answer:
1 quarter = 25 cents
1 dime = 10 cents
1 nickel = 5 cents
and 2 cents
Total :  25+10+5+2 = 42 cents
Total amount i have is 52 cents
Now , 52 – 42 = 10
1 dime = 10 cents
So , I lost 1 dime.

Question 5.
Precision
Circle coins to show 80¢.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 24.2

 

Answer:
1 dime = 10 cents
80 cents = 8 dimes

Question 6.
Modeling Real Life
You have 3 quarters, 1 nickel, and 4 pennies. How many cents do you have? Do you have enough money to buy the boomerang?
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 25

 

Answer:
No, I do not have enough money to buy the boomerang.

Explanation:
The cost of boomerang is 94 cents
1 quarter = 25 cents
3 x 25 =75 cents
1 nickel =5 cents
4 pennies = 4 cents
Total amount : 75+5+4 =84
I have 84 cents .
No, I do not have enough money to buy the boomerang

Question 7.
Modeling Real Life
You have 1 quarter, 3 dimes, and 1 nickel. How many more cents do you need to buy the toy bird? Draw the coins you need.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 26

Answer:
The cost of bird is 75 cents.

Explanation:
1 quarter = 25 cents.
1 dime = 10 cents,
3 dimes = 3 x 10 = 30 cents
1 nickel = 5 cents
Total amount : 25 +30+ 5 =60 cents
Now,
75- 60 = 15cents
I need 15 more cents to buy the toy bird .

Review & Refresh

Compare.

Question 8.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 27

Answer:
324 > 317
327 is greater 317

Question 9.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 28

Answer:
426 > 206
426 is greater than 206

Question 10.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 29

Answer:
546 < 564
564 is greater than 546

Question 11.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 30

Answer:
931 > 842
931 is greater than 846

Lesson 14.2 Order to Find Total Values of Coins

Explore and Grow

Order your coins from the greatest value to the least value. Draw and label each coin with its value. What is the total value of all the coins?
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 31
Explain how ordering the coins helped you find the total.
_____________________________________
_____________________________________
_____________________________________
_____________________________________

Answer:
1 quarter = 25 cents
1 dime = 10 cents
1 nickel = 5 cents
and 1 penny = 1 cent
Total value of coins :
25+10+5+1=41 cents

>>>

Show and Grow

Draw and label the coins from the greatest value to the least value. Then find the total value.

Question 1.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 32

 

Answer:
45 cents

Explanation:
1 quarter = 25 cents
1 dime = 10 cents
1 nickel = 5 cents
Which means , 2 x 5 = 10 cents
Total value :
25 +10 + 10 = 45 cents

>

Quarter       >         Dime     >   Nickel

Question 2.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 33

 

Answer:
1 quarter = 25 cents
which means, 2quarters = 2 x 25= 50 cents
1 dime = 10 cents
1 nickel = 5 cents
and 1 penny = 1 cents
Total : 50 +10+5+1 = 66 cents
Therefore total value of coins is 66 cents .

>>>

Question 3.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 34

Answer: 31 cents

Explanation:
1 dime = 10 cents
which means , 2 dimes = 2 x 10 = 20 cents
1 nickel = 5 cents
which means , 2 nickel = 2 x5 = 10 cents
and 1 penny = 1 cent
Total : 20 +10 +1 = 31
Therefore, total value of coins is 31 cents

>>

Apply and Grow: Practice

Draw and label the coins from the greatest value to the least value. Then find the total value.

Question 4.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 35

 

Answer:
41 cents
Explanation :
1 quarter = 25 cents
1 dime = 10 cents
1 nickel = 5 cents
and 1 penny = 1 cent
Total value = 25 +10 +5 +1 =41
Therefore total value of coins is 41 cents

>>>

Question 5.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 36

 

Answer:
27 cents

Explanation :
1 dime = 10 cents
which means , 2 dimes = 2 x 10 = 20 cents
1 nickel = 5 cents
and 2 pennies = 2 cents
total : 20 +5 +2= 27
Therefore total value of coins is 27 cents .

>>

Question 6.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 37

Answer:
85 cents

Explanation :
1 quarter = 25 cents
Which means, 2 quarter = 2 x25 = 50 cents
1 dime = 10 cents
Which means, 3 dimes = 3 x 10 = 30 cents
And 1 nickel = 5 cents
Total :
50 +30 +5 = 85 cents
Therefore total value of coins is 85 cents.

>>

Question 7.
Reasoning
You have a dime, a nickel, and one other coin. The total value is 40¢. What is your third coin?
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 38

Answer:
The total value of  3 coins is 40 cents
1 dime = 10 cents
1 nickel = 5 cents
10 cents + 5 cents = 15 cents
Now , 40 cents – 15 cents =  25 cents
So the other coin is quarter
1 quarter = 25 cents
Therefore, 25 +10+ 5 = 40 cents.

Think and Grow: Modeling Real Life

You have 3 nickels and 4 pennies in one pocket. You have 2 dimes and 2 quarters in your other pocket. How much money do you have in all? Do you have enough money to buy the car?
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 39
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 40

Answer:
The cost of car is 9 cents .
The amount i have in one pocket is 3 nickels and 4 pennies
1 nickel = 5 cents
which means , 3 nickels = 3 x 5 = 15 cents
4 pennies = 4 cents
Total = 15 cents + 4 cents = 19 cents
The amount i have in another pocket is 2 dimes , 2 quarters
1 dime = 10 cents
in which , 2 dimes = 2 x 10 = 20 cents
1 quarter = 25 cents
2 quarters = 2 x 25 = 50 cents
Total : 20 cents + 50 cents = 70 cents
Total amount in both the pockets is 70 cents + 19 cents =89 cents
No, I do not have enough money to  buy the car.

Show and Grow

Question 8.
You have 30¢. You find 2 nickels, 1 dime, and 3 pennies in your room. How much money do you have now? Do you have enough money to buy the yo-yo?
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 41
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 42

Answer:
I have 30 cents
Amount I found in my room =
1 nickel = 5 cents ,
2 nickel x 5 = 10 cents.
1 dime =10 cents.
3 pennies = 3 cents.
Total  : 10 +10 + 3
=23 cents
Now, total amount i have is
30 cents +23 cents = 53 cents
The cost of yo-yo is 50 cent
So , now I have enough money to buy the yo- yo

Question 9.
You have 1 nickel, 1 quarter, and 4 dimes. How many more cents do you need to buy the stuffed animal? Draw and label the coin you need.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 43

Answer:
1 nickel = 5 cents
1 quarter= 25 cents
1 dime =10 cents
Which means 4 dimes = 4 x 10=40 cents
Total amount = 5+25+40 =70 cents
But the cost of stuffed animal is 80 cents
So , I do not have enough cents to buy the stuffed animal.
I need 10 more cents to buy the stuffed animal.

Question 10.
DIG DEEPER!
You have 65¢. You give your friend a dime. You have 3 coins left. Draw and label the coins you have left.

Answer:
Given that,
I have 65 cents
The amount I gave to my friend is 1 dime
1 dime = 10 cents
Now,
65 cents – 10 cents = 55 cents
55 cents can be shown in 3 coins as
2 quarters and 1 nickel

Order to Find Total Values of Coins Homework & Practice 14.2

Draw and label the coins from the greatest value to the least value. Then find the total value.

Question 1.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 44

 

Answer: 35 cents

Explanation :
1 dime = 10 cents
which means, 2 dimes = 2 x 10 = 20 cents
and 1 nickel = 5 cents
which means, 3 nickels = 3 x 5 = 15 cents
Total:  20 +15 = 35 cents
Therefore total value of coins is 35 cents.

>

Question 2.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 45

Answer: 63 cents

Explanation:
1 quarter = 25 cents
Which means , 2 quarters = 2 x 25 = 50 cents
1 dime = 10 cents
and 3 pennies = 3 cents
Total : 50 +10+3 = 63 cents

>>

Question 3.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 46

Answer: 46 cents

Explanation :
1 quarter = 25 cents
1 dime = 10 cents
1 nickel = 5 cents
Which means , 2 nickel = 2 x 5 = 10 cents
and 1 penny = 1 cents
Total :
25 +10 +5 +5+1 = 46 cents

>>>

Question 4.
Open-Ended
Draw and label four coins that have a total value of 40¢

Answer: 40 cents can be written as 4 dimes
1 dime = 10 cents
Which means , 4 x 10 = 40 cents.

Question 5.
Modeling Real Life
You have 46¢. You find 4 pennies and 1 nickel in your room. How much money do you have now? Do you have enough money to buy the app?
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 47
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 48

Answer:
I have 46 Cents

Explanation:
The amount I found is
4 pennies =  4 cents
1 nickel = 5 cents
Total  : 5 + 4 = 9 cents
Now,
Total amount I have is 46 cents +9 cents = 55 cents
The cost of app is 50 cents
yes , I have enough money to buy the app.

Question 7.
DIG DEEPER!
You have some nickels and dimes. You have 1 more nickel than dimes. The total value of your coins is 50¢. How many nickels and dimes do you have?
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 49

Answer:
I have  4 nickels and  3 dimes.

Explanation:
1 nickel =5 cents
That means, 4 nickels= 4 x 5 = 20 cents
1 dime = 10 cents
That means , 3 dimes = 3 x 10 = 30
Total: 20 + 30 = 50 cents
So, I have  4 nickels and  3 dimes.

Review & Refresh

Question 8.
Which fruit is the least favorite?
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 50

Answer:
Cherry is the least favorite.

Explanation :
In the given table ,
Total lines for orange fruit are 5
Total lines for cherry fruit are 2
Total lines for apple fruit are  4
On comparing them , 2 is smaller than 4 and 5
so, Cherry is the least favorite.

Lesson 14.3 Show Money Amounts in Different Ways

Explore and Grow

Use your coins to show 25 cents in two different ways. Draw and label the coins.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 51

Answer :
25 cents can be written as
1 quarter
1 quarter = 25 cents

Another way :
25 cents can be written as
2 dime , 1 nickel
1 dime = 10 cents
Which means , 2 dime = 2 x 10= 20
1 nickel = 5 cents.

Did everyone in your class use the same coins?
_____________________________________
_____________________________________
_____________________________________
_____________________________________

Answer:
No, everyone used different coins to make a total value of 25 cents
For example, my friend used 1 dime, 3 nickels
1 dime = 10 cents
1 nickel = 5 cents
Which means , 3 x 5 = 15 cents
Total : 10 + 15 = 25 cents

Show and Grow

Show the amount in two different ways.

Question 1.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 52

 

Answer:
45 cents can be written as
4 dime + 1 nickel
Which means  one dime = 10 cents
4 dime = 4 x 10 = 40
1 nickel = 5 cents
Total: 40+5 = 45
Another way: 1 quarter +2 dime
Which means , 1 quarter =25 cents
1 dime= 10 cents
2 dime = 2 x 10 = 20
Total=: 25+20= 45
Therefore, 45 can be written as  4 dime 1 nickel and 1 quarter 2 dime

Question 2.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 53

Answer:
27 cents can be written as
2 dime +1 nickel+2 penny
Which means , 1 dime = 10 cents
2 dime = 2 x 10= 20 cents
1 nickel = 5 cents2 penny = 2 cents
Total: 20+5+ 2= 27 cents
Another way:
1 quarter+2 pennies
Which means , 1 quarter = 25 cents
2 penny = 2 cents
Total= 25 + 2 = 27 cents
Therefore 27 cents can be written as 2 dime +1 nickel+2 penny and 1 quarter +2 pennies

Question 3.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 54

Answer:
80 cents can be written as
8 dimes
Which means 1 dime = 10 cents
So 8 dimes = 80 cents
Another way:
3 quarter + 1 nickel
Which means , 1 quarter = 25 cents
3 quarters = 3 x 25 = 75 cents
1 nickel = 5 cents
Total : 75 + 5 = 80 cents
Therefore, 80 cents can e written as 8 dimes and 3quarter and 1 nickel .

Apply and Grow: Practice

Show the amount in two different ways.

Question 4.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 55

Answer:
23 cents can be written as
2 dime  3 pennies
Which means , 1 dime = 10 cents
2 dimes =2 x 10 = 20 cents
3 pennies = 3 cents
Total : 20 + 3 = 23 cents
Another way:
4 nickel ,3 pennies
Which means, 1 nickel = 5 cents
4 nickel = 4 x 5 =20 cents
3 pennies = 3 cents
Total= 20 + 3 =23 cents
Therefore 23 cents can be written as 2 dimes + 3 pennies and 4 nickel + 3 pennies

Question 5.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 56

Answer:
49 cents can be written as
1 quarter ,2 dime ,4 pennies
Which means , 1 quarter = 25 cents
1 dime = 10 cents
So,2 dime = 2 x 10 = 20 cents
4 pennies = 4 cents
Total : 25 +20+4 =49 cents
Another way :
49 Pennies
1 penny = 1 cent
49 pennies = 49 cents
Therefore, 49cents can be written as 1 Quarter 2 dime 4 pennies and 49 pennies.

Question 6.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 57

 

Answer:
75 cents can be written as
3 quarters
1 quarter = 25 cents
Which means , 3 quarters = 3 x 25 = 75
Another way:
1 quarter, 5 dimes
1 quarter = 25 cents
1 dime = 10 cents
Which means , 5 dimes = 5 x 10 = 50 cents
Total :  25 + 50 = 75 cents
Therefore , 75 cents can be written as 3 quarters and 1 quarter , 5 dimes

Question 7.
Structure
You have 55¢. You have no quarters. Draw to show what coins you might have.

Answer:
55 cents be written as 5 dimes , 1 nickel .
1 dime = 10 cents
Which means , 5 dimes = 5 x 10 = 50 cents
1 nickel = 5 cents
Total : 50 cents + 5 cents = 55 cents.

Question 8.
YOU BE THE TEACHER
Newton says he drew the fewest number of coins to show 66¢. Is he correct? Explain.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 58

Answer :
No, he is wrong
Newton used 6 coins to show 66 cents .
66 cents can also be shown as
2 quarters , 1 dime , 1 nickel and 1 cents
Total number of coins is 5
1 quarter = 25 cents
Which means , 2 quarters = 2 x 25 = 50 cents
1 dime = 10 cents
1 nickel = 5 cents
and cent
Total value of coins is 50+ 10 + 5 +1 = 66
Therefore ,66 cents can be shown in 5 coins .
so , Newton is wrong.

Think and Grow: Modeling Real Life

Newton has 2 dimes, 1 nickel, and 1 penny. Descartes uses the fewest number of coins to make the same amount. Draw and label their coins.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 59

 

Answer:

Total number of coins newton have is 4
2 dimes , 1 nickel and 1 penny
1 dime = 10 cents
Which means, 2 dimes = 2 x 10 = 20 cents
1 nickel = 5 cents
and 1 penny = 1 cent

Total value of coins is
20 +5+1 = 26 cents.
Given that ,
Descartes uses fewest number of coins to make same amount
26 can also be shown with 2 coins
1 quarter = 25 cents
1 penny = 1 cents
Total value is
25 + 1 = 26 cents

So, 26 cents can be shown with 2 coins
2 is less than 6 so , Descartes uses few coins than Newton to show the same amount.

Show and Grow

Question 9.
Newton has 3 dimes and 2 pennies. Descartes uses the fewest number of coins to make the same amount. Draw and label their coins.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 60

Answer:
Total number of coins newton have is 5
3 dimes , 2 pennies
1 dime = 10 cents
Which means , 3 dimes = 3 x 10 = 30 cents
2 pennies = 2 cents.
Total value of coins is
30 + 2 = 32 cents

Given that,
Descartes used the fewest number of coins to make the same amount
32 cents can be shown in 4 coins
1 quarter, 1 dime and 2 pennies
1 quarter = 25 cents
1 dime = 10 cents
and 2 pennies = 2 cents
Total : 25 + 10 + 2 = 37 cents

 

So, 37 cents can be shown in 4 coins
4 is less than 5.
Descartes used few coins than newton to show the same amount

Question 10.
You use fewer than 5 coins to buy the pack of gum. Draw and label coins to show how you pay.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 60.1

Answer:
The cost of gum is 70 cents
70 cents can be shown in 4 coins as
2 quarters , 2 dime
1 quarter = 25 cents
Which means , 2 x 25 = 50 cents
1 dime = 10 cents
Which means , 2 x 10 = 20 cents
Total amount = 50 + 20 = 70

Question 11.
DIG DEEPER!
You have 2 quarters. Newton and Descartes each have 5 coins and the same amount of money as you. Their coins are different. Draw and label their coins.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 61

Answer:
1 quarter = 25 cents
Which means, 2 quarters= 2 x 25 = 50 cents
50 cents can be shown in 5 coins as 5 dime
1 dime = 10 cents
Which means 50 cents = 5 dimes

Show Money Amounts in Different Ways Homework & Practice 14.3

Show money amounts in different ways.

Question 1.
Big Ideas Math Answers 2nd Grade Chapter 14 Money and Time 62

Answer:
15 cents can be written as:
1 dime +1 Nickel
which means 1 dime = 1o cents
1 nickel =5 cents
10+5 = 15 cents
Another way:
15 cents can also be written as:
1 dime +5 pennies
1 dime =10 cents
5 pennies = 5 cents
Total: 10 +5 = 15 cents
So, 15 cents can be written as 1 dime 1 nickel and 1 dime 5 pennies

Question 2.
Big Ideas Math Answers 2nd Grade Chapter 14 Money and Time 63

 

Answer:
58 cents can be written as
2 quarter 8 pennies
1 quarter = 25 cents
Which means 2 quarters = 2 x 25 = 50 cents
8 pennies = 8 cents
Total : 50 + 8 = 58 cents
Another way :
5 dime 1 nickel 3 pennies
Which means,
1 dime = 10 cents
=5 dime = 5 x 10 = 50cents
1 nickel = 5 cents
1 pennies = 3 cents
Total : 50+ 5+3= 58
So, 58 cents can be written as 2 quarter 8 pennies and 5 dime 1 nickel 3 pennies.

Question 3.
Big Ideas Math Answers 2nd Grade Chapter 14 Money and Time 64

Answer:
90 cents can be written as
3 quarters 1 nickel 1 dime
Which means,
1quarter =25 cents
3 quarters = 3 x 25 = 75 cents
1 nickel =5 cents
1 dime = 10 cents
Total : 75+5+10=  90 cents
Another way :
9 dimes
1dime = 10 cents
9 dimes = 9 x 10 = 90 cents .
So , 90 cents can be written as 3 quarters1 dime 1 nickel and 9 dimes.

Question 4.
Reasoning
Draw to show 60¢ with only 3 coins.

Answer:
60  cents can be drawn in 3 coins as
2 quarters and 1 dime
1 quarter = 25 cents
Which means , 2 x 25 = 50 cents
1 dime = 10 cents
Total : 50 + 10 = 60.

Question 5.
Structure
Draw to show 42¢ without using dimes.

Answer:
42 cents can be drawn as
1 quarter , 3 nickels and 2 pennies
1 quarter = 25 cents
1 nickel = 5 cents
Which means 3 x 5 = 15 cents
2 pennies = 2 cents
Total : 25 +15 + 2 = 42 cents

Question 6.
Modeling Real Life
Newton has 6 dimes and 1 nickel. Descartes uses the fewest number of coins to make the same amount. Draw and label their coins.
Big Ideas Math Answers 2nd Grade Chapter 14 Money and Time 65

Answer :
6 dimes and 1 nickel
Total number of coins is 7
1 dime = 10 cents
Which means, 6 dimes = 6 x 10 = 60 cents
1 nickel = 5 cents
Total : 60 +5 = 65 cents
65 cents can be shown in fewest number of coins as
2 quarters , 1 dime and 1 nickel
Total number of coins is 4

Question 7.
Modeling Real Life
You use fewer than 5 coins to buy the pen. Draw and label coins to show how you pay.
Big Ideas Math Answers 2nd Grade Chapter 14 Money and Time 66

Answer:
The cost of pen is 52 cents
52 cents be shown in fewer than 5 coins as
2 quarter and 2 pennies
1 quarter = 25 cents
Which means, 2 quarters = 2 x 25 = 50 cents
and 2 pennies = 2 cents
Total 50 +2 = 52 cents.

Review & Refresh

Question 8.
A green scarf is 50 inches long. An orange scarf is 40 inches long. A red scarf is 38 inches long. How much longer is the green scarf than the red scarf?
Big Ideas Math Answers 2nd Grade Chapter 14 Money and Time 67

Answer:
22 inches

Explanation:
The length of green scarf is 50 inches
The length of red scarf is 38 inches
Now ,
50 inches – 38 inches = 22 inches
Therefore , Green scarf is 22 inches longer than red scarf.

Lesson 14.4 Make One Dollar

Explore and Grow

Newton has 4 coins. The total value is 100¢. Draw and label his coins

Answer:
Given ,
The total value of money is 100 cents
100 cents can be shown in 4 coins as 4 quarters
1 quarter = 25 cents
Which means 4 quarters = 4 x 25 = 100 cents
So, Newton has 4 quarters

Big Ideas Math Answers 2nd Grade Chapter 14 Money and Time 68
Descartes has 10 coins. The total value is 100¢. Draw and label his coins.

Answer:
Given ,
Total value of money is 100 cents
100 cents be shown in 10 coins as 10 dimes
1 dime = 10 cents
Which means , 10 dimes = 10 x 10 = 100 cents
So, Descartes have 10 dimes .

Show and Grow

Draw coins to make $1. How many cents do you need?

Question 1.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 69

Answer:
$1 = 100 cents
So, 100 cents – 25 cents = 75 cents
I need 75 cents to make $1.
75 Cents can e shown as 3 quarters
1 quarter = 25 cents
3 quarters = 3 x 25 = 7 cents

Question 2.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 70

Answer:
$1 = 100 cents
So, 100 cents – 80 cents = 20 cents
I need 20 cents to make $1.
20 cents can be shown as 2 dimes
1 dime = 10 cents
2 dimes = 2 x 10 = 20 cents

Apply and Grow: Practice

Draw coins to make $1. How many cents do you need?

Question 3.
35¢
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 70.1

Answer:  $1 = 100 cents
So, 100 cents – 35 cents = 65 cents
I need 65 cents to make $1.
65 cents can be shown as
2 quarters , 1 dime and 1 nickel
1 quarter = 25 cents
Which means , 2 quarters =  2 x 25 = 50 cents
1 dime = 10 cents
and 1 nickel = 5 cents
Total : 50 +10 +5 = 65 cents

Question 4.
72¢
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 70.2

Answer:
$1 = 100 cents
So, 100 cents – 72 cents = 28 cents
I need 28 cents to make $1.
28 cents can be shown as
1 quarter , 3 pennies
1 quarter = 25 cents and
3 pennies = 3 cents
Total : 25 +3 = 28 cents

Question 5.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 71

Answer:
$1 = 100 cents
1 dime = 10 cents
so 2 dimes = 2 x 10 = 20 cents
so, 100 cents – 20 cents = 80 cents
I need 20 cents to make $1.
20 cents can be shown as 2 dimes
1 dime = 10 cents
Which means , 2 dimes = 2 x 10 = 20 cents

Question 6.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 72

Answer:  $1 = 100 cents
1 quarter = 25 cents
Which means , 3 x 25 = 75 cents
1 dime = 10 cents
Total : 75 + 10 = 85 cents
Now, 100 – 85 = 15 cents
So, I need 15 cents to make $1.
15 cents can be shown as
1 dime and 1 nickel
1 dime = 10 cents
1 nickel = 5 cents
Total : 10 +5 = 15 cents

Question 7.
Number Sense
Circle coins to make $1.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 73

 

Answer:

Explanation:
$1 is equals to 100  cents
1 quarter = 25 cents
Which means
2 quarters = 2x 25 = 50 cents
1 dime = 10 cents
5 dimes = 5 x 10 = 50 cents
Total : 50 +50 = 100 cents
So, 2 quarters and 5 dimes make $1.

Think and Grow: Modeling Real Life

You have 1 quarter, 3 pennies, and 1 dime in one pocket. You have 2 pennies, 2 nickels, and 4 dimes in your other pocket. How many more cents do you need to make $1?
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 74

Answer:
The amount i have in one pocket is
1 quarter =25 cents
3 pennies = 3 cents
and 1 dime = 10 cents
Total : 25 +10 +3 = 38 cents
The amount I have in another pocket :
2 pennies = 2 cents
1 nickel = 5 cents
which means, 2 nickels = 2 x5 = 10 cents
1 dime = 10 cents
Which means, 4 dimes = 4 x 10 = 40 cents .
Total : 2 + 10 +40 =52 cents
Now , the total amount i have in both pockets is
38+52 = 90 cents
We know that $1 is equal to 100 cents
100 – 90 = 10
10 cents can be shown as 1 dime

.

Show and Grow

Question 8.
You have 2 dimes and 1 nickel in your desk. You have a quarter and 10 pennies in your backpack. How many more cents do you need to make $1?
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 74.1

Answer:
Total amount I have in my desk is
2 dimes and 1 nickel
1 dime = 10 cents
Which means, 2 dime =2 x 10 = 20 cents
1 nickel = 5 cents
Total = 20 +5 = 25 cents
Total amount I have in my pocket is
a quarter and 10 pennies
1 quarter = 25 cents
and 10 pennies = 10 cents
Total : 25 +10 = 35 cents
Total amount I have is
The amount in my desk + the amount in my pocket
Which means, 25 cents +35 cents
=60 cents
We know that ,
$1 is equal to 100 cents
100 cents – 60 cents
= 40 cents.
So, I need 40 cents more to make $1.

Question 9.
A notebook costs $1. You have 5 dimes and 4 pennies. How much more money do you need to buy the notebook?
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 75

Answer:
1 dime = 10 cents
Which means 5 dimes = 5 x 10 = 50 cents
and 4 pennies = 4 cents
Total : 50 cents +4 cents = 54 cents
we know that,
$1 is equal to 100 cents
Now , 100 – 54 = 46 cents
So ,I need 46 cents more to buy the book.

Question 10.
DIG DEEPER!
You have a $1 bill. You have 33 more cents than your friend. How much money does your friend have?
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 76

Answer:
I have $1 bill
Given that I have 33 cents more than my friend
We know that ,$1 is equal to 100 cents
Now,
100 cents – 33 cents = 67 cents
So , my friend have 67 cents.

Make One Dollar Homework & Practice 14.4

Draw coins to make $1. How many cents do you need?

Question 1.
54¢
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 77

Answer:  $1 = 100 cents
so, 100 cents – 54 cents = 46 cents
I need 46 cents to make $1.
46 cents can be shown as
1 quarter , 2 dime and 1 penny
1 quarter = 25 cents
1 dime = 10 cents
Which means, 2 dimes = 2 x 10 = 20 cents
and 1 penny = 1 cent

Question 2.
38¢
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 78

Answer:
$1 = 100 cents
So, 100 cents – 38 cents = 62 cents
I need 62 cents to make $1.
62 cents can be shown as
2 quarters, 1 dime and 2 pennies.
1 quarter = 25 cents
Which means , 2 quarters = 2 x 25 = 50
1 dime= 10 cents
2 pennies = 2 cents

Question 3.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 79

Answer:
$1 =100 cents
Given coins are 1 quarter and 2 pennies
1 quarter = 25 cents
and 2 pennies = 2cents
Total value : 25 +2 = 27
So, 100 – 27 =73
Therefore , I need 73 cents to make $1
73 cents can be shown as
2 quarters , 2 dime and 3 pennies
1 quarter = 25 cents
Which means 2 quarters = 2 x 25 = 50 cents
1 dime = 10 cents
Which means , 2 dimes = 2 x 10 = 20 cents
and 3 pennies = 3 cents
Total : 50 +20 +3 = 73.

Question 4.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 80

Answer:
$1 = 100 cents
Given coins are 1 dime and 1 nickel
1 dime = 10 cents
1 nickel = 5 cents
Total : 10 +5 = 15 cents
Now , 100 cents – 15 cents =85 cents
So , I need 85 cents to make $1.
85 cents can be written as
3 quarters , 1 dime
1 quarter = 25 cents
Which means , 3 quarters = 3 x 25 = 75 cents
1 dime = 10 cents
Total : 75 +10 = 85 cents

Question 5.
Structure
Show $1 using only nickels and dimes.

Answer:
1 dime = 10 cents  and
1 nickel = 5 cents
$1 is equals to 100 cents
100 cents can be shown as 8 dimes 4 nickels
8dimes = 8 x 10 = 80
4 nickels = 4 x 5 = 20
Total value = 100 cents

 

Question 6.
Structure
How many nickels make $1?
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 81

Answer :
20 nickels
1 nickel = 5 cents
Which means ,20 nickels =
20 x 5 = 100
100 cents =$1
So 20 nickels make $1.

Question 7.
Modeling Real Life
You have 1 dime and 4 nickels in a jar. You have 1 quarter and 3 pennies in your pocket. How many more cents do you need to make $1?
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 81.1

Answer:
The amount I have in jar is
1 dime = 10 cents
and 1 nickel = 5 cents
Which means , 4 nickels = 4 x 5 = 20 cents
Total : 10 + 20 = 30 cents
The amount i found in my pocket is
1 quarter = 25 cents
and 3 pennies = 3 cents
Total : 25+ 3 = 28 cents
Now , the total amount i have is
Amount in jar +amount in my pocket
= 30 +28 = 58cents
We know that, $1 is equal to 100 cents
100 – 58 = 42 cents
So, I need 42 cents more to make $1.

Question 8.
Modeling Real Life
A snack costs $1. You have 2 quarters and 2 dimes. How much more money do you need to buy the snack?

Answer:
The cost of snack is $1.
1 quarter = 25 cents
2 quarters = 2x 25= 50 cents
1 dimes = 10 cents
2 dimes= 2 x 10 = 20 cents
Total : 50 +20= 70 cents
$1 = 100 cents
100 – 70 = 30 cents
So , I need 30 cents more to buy the snack.

Review & Refresh

Question 9.
100 – 54 = ____

Answer : 46

By regrouping, 0 in units place becomes 10 and the 0 in tens place becomes 9 after the second regrouping.

Question 10.
200 – 134 = ____

Answer: 64

By regrouping, 0 in units place becomes 10 and the 0 in tens place becomes 9 after the second regrouping.

Lesson 14.5 Make Change from One Dollar

Explore and Grow

Model the story.
Newton buys a bag of fish crackers for 45¢. He pays with a $1 bill. What is his change?
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 82
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 83

Explain how you solved.
_______________________________
_______________________________
_______________________________
_______________________________

 

Answer: 55 cents

Explanation :
$1 is equals to 100 cents
The cost of fire crackers is 45 cents
Now his change is
$1 bill – cost of fire crackers
100 cents -45 cents
=55 cents
So, Newtons change is 55 cents.

Show and Grow

You buy the item shown. You pay with a $1 bill. What is your change?

Question 1.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 84

 

Answer:
65 Cents

Explanation:
$1 is equal to 100 cents
So, 100 cents – 35cents  = 65
Therefore, 65 cents is the change.

Question 2.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 85

Answer: 51 Cents

Explanation:
$1 is equals to 100 cents
So, 100 cents – 49 cents = 51 cents
Therefore: 51 cents is the change.

Apply and Grow: Practice

You buy the item shown. You pay with a $1 bill. What is your change?

Question 3.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 86

Answer: 85 Cents

Explanation:
$1 = 100 cents
So, 100 cents -15 cents =85 cents
Therefore 85 cents is the change.

Question 4.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 87

Answer: 39 Cents

Explanation:
$1 = 100 cents
So, 100 cents -61 cents =39 cents
Therefore 39 cents is the change.

Question 5.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 88

 

Answer: 13 Cents

Explanation:
$1 = 100 cents
So, 100 cents -87 cents =13 cents
Therefore ,13 cents is the change.

Question 6.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 89

 

Answer: 41 Cents

Explanation:
$1 = 100 cents
So, 100 cents -59 cents =41 cents
Therefore ,41 cents is the change.

Question 7.
Reasoning
Newton buys a notebook for 34¢. Descartes buys one for 52¢. You buy one for 48¢. You each pay with $1. Who gets back the most amount of money? How do you know?
__________________
__________________
__________________

Answer:
Newton gets back the most amount of money.

Explanation:
$1 is equals to 100 cents
Given that ,
The cost of newtons book is 34 cents
So, 100 cents – 32 cents = 68 cents .
The cost of Descartes book is 52 cents.
So , 100 cents – 52 cents = 48 cents.
The cost of my book is 48cents .
So, 100 cents – 48 cents = 52 cents
Therefore, newton gets back the most amount of money.

Think and Grow: Modeling Real Life

You pay for some school supplies with $1. Your change is 17¢. How much money did you spend?
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 90
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 91

Answer: 83 Cents

Explanation:
$1 is equals to 100 cents
My change is 17 cents
Now , 100 cents – 17 cents =83 cents
So, the cost of school supplies is 83 cents.

Show and Grow

Question 8.
You pay for some erasers with $1. Your change is 38¢. How much money did you spend?
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 92

 

Answer: 62 Cents

Explanation:
$1 is equals to 100 cents
My change is 38 cents
Now , 100 cents – 38 cents =62 cents
So the cost of erasers is 62 cents.

Question 9.
You buy a toy ring. You pay with $1. You get back 1 quarter, 2 dimes, 1 nickel, and 1 penny. How much does the toy ring cost?
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 93

Answer: 59 Cents.

Explanation:
$1 is equals to 100 cents
My change is 1 quarter = 25 cents
1dime = 10 cents
Which means , 2 dimes = 2 x 10 = 20 cents
1 nickel = 5 cents
1 penny = 1 cent
Total = 25 +10 + 5 +1 =41 cents
100 cents – 41 cents = 59 cents
Therefore , the cost of toy ring is 59 cents.

Question 10.
You buy a banana for 25¢ and an orange for 45¢. You pay with $1. What is your change?
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 94

Answer:  Change is 30 Cents.

Explanation:
The cost of banana is 25 cents and the cost of orange is 45 cents
Total cost : 25 cents + 45 cents = 70 cents
$1 is equals to 100 cents
Now, 100 – 70 = 30 cents
Therefore ,30 cents is the change.

Make Change from One Dollar Homework & Practice 14.5

You buy the item shown. You pay with a $1 bill. What is your change?

Question 1.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 95

Answer: 55 Cents.

Explanation:
$1 = 100 cents
So, 100 cents -45 cents =55 cents
Therefore 55 cents is the change.

Question 2.
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 96

Answer: 24 Cents.

Explanation:
$1 = 100 cents
So, 100 cents -76 cents =24 cents
Therefore, 24 cents is the change.

Question 3.
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 97

Answer: 87 Cents.

Explanation:
$1 = 100 cents
So, 100 cents -13 cents =87 cents
Therefore, 87 cents is the change.

Question 4.
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 98

Answer: 63 Cents.

Explanation:
$1 = 100 cents
So, 100 cents -37 cents = 63 cents
Therefore, 63 cents is the change.

Question 5.
Reasoning
A puzzle costs 68¢. Newton pays for it with a $1 bill. Draw to show his change in two ways.
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 99

Answer: Change is 32 Cents.

Explanation:
$1 is equals to 100 cents
The cost of puzzle is 68 cents
Now, 100 – 68 = 32 cents
Therefore, 32 is the change.
32 cents can be written as 3 dime, 2 pennies

1 dime = 10 cents
3 dimes = 3 x 10 = 30 cents
2 pennies = 2 cents
Total : 30 + 2 = 32 cents
Another way ,
1 quarter 7 pennies

Which means, 1 quarter = 25 cents
7 pennies = 7 cents
Total: 25 + 7 = 32 cents
Therefore 32 cents can be written as 3 dime 2 pennies and 1 quarter 7 pennies.

Question 6.
Modeling Real Life
You buy a pencil sharpener. You give the cashier $1. You get back 2 quarters, 1 nickel, and 3 pennies. How much does the pencil sharpener cost?
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 100

Answer: 42 Cents.

Explanation:
$1 is equals to 100 cents
My change is 2 quarters 1 nickel 3 pennies
1 quarter = 25 cents
Which means , 2 quarters = 2 x 25 =50 cents
1 nickel = 5 cents
3 pennies = 3 pennies
Total: 50 +5+3=58 cents
Now,
100 – 58 = 42 cents
Therefore, the cost of the pencil sharpener is 42 cents .

Question 7.
Modeling Real Life
You buy an onion for 51¢ and a pepper for 22¢. You pay with a $1 bill. What is your change?

Answer: 27 Cents.

Explanation:
The cost of an onion is 51 cents and the cost of 22 cents
Total cost : 51 + 22 =73 cents
$1 is equals to 100 cents
100 cents – 73cents = 27 cents
Therefore, 27 cents is the change.

Review & Refresh

Question 8.
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 101

Answer: There are three straight sides and three vertices.

Question 9.
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 102

Answer: There are six straight sides and six vertices.

Lesson 14.6 Find Total Values of Bills

Explore and Grow

Model the story.
Descartes has three $5 bills and three $1 bills. How much money does he have in all?
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 103

Answer: $18.

Explanation:
Given that ,
Descartes have Three $5 bills and three $1 bill
Three $5 bills = $5+$5+$5 = $15.
and three $1 bills = $3
Total amount =$15+$3= $18.

Show and Grow

Count on to find the total value.

Question 1.
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 104

Answer: $45.

Explanation:
Three $5 bills =$5+$5+$5 = $15
$10 and $20
Total value :
$15 +$10 +$20= $45
Therefore, total value = $45.

Question 2.
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 105

Answer: $32.

Explanation:
Two $1 bills = $2
Two $5 bills = $5+$5= $10
and two $10 = $10 +$10= $20
Total value = $2+$10+$20=$32
Therefore ,the total value is $32.

Apply and Grow: Practice

Count on to find the total value.

Question 3.
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 106

Answer: $16.

Explanation:
$10 ,$5 and $1
Total value :
$10 +$5 +$1 =$16

Question 4.
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 107

Answer: $36.

Explanation:
Total value of $20 , $10 ,$5 and $1
=$20 +$10 +$5 +$1 =$36 .

Question 5.
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 108

Answer: $75.

Explanation:
Total value of a $5 bill, $10 bill and three$20 bills is
$5+$10+$20+$20+$20= $75
Therefore the total value is $75.

Question 6.
Which One Doesn’t Belong?
Which group of bills does not belong with the other two?
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 109

Answer:

Explanation:
The total value of dollar bills is $61
But the remaining dollar sets total value is $65
So ,this set is different from the other two sets.

Think and Grow: Modeling Real Life

You buy some T-shirts for $39. Draw and label bills to show two different ways to pay for the T-shirts. One way should use the fewest number of bills.
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 110

 

Answer:
$39 can be shown as
A $20 bill ,a $10 bill ,a $5 bill and four $1 bills

Another way:
$39 can be shown as
a $20 bill , three $5 bills and four $1 bills.

Show and Grow

Question 7.
You buy a pair of sneakers for $24. Draw and label bills to show two different ways to pay for the sneakers. One way should use the fewest number of bills.
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 111

Answer:
$24 can be shown as

a $20 bill and four $1 bill

    

$24 can also be shown as
A$20 bill and two $2 bills

Question 8.
Newton has three $20 bills, one $10 bill, one $5 bill, and three $1 bills. Does he have enough money to buy a new dog house that costs $80? Explain.

Answer: No , he do not have enough money to buy a dog house

Explanation:
Three $20 bills = $20+$20+$20= $60
$10 , $5 and three $1
Total value :
$60 +$10 +$5+$1 = $76
But, the cost of dog house is $80
So, he do not have enough money to buy a dog house.

Question 9.
DIG DEEPER!
Explain why you would order a group of bills from the greatest value to the least value to find the total value.
_____________________________________
_____________________________________

Find Total Values of Bills Homework & Practice 14.6

Count on to find the total value.

Question 1.
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 112

Answer:
$10 +$ 1= $11
Therefore the total value is $11

Question 2.
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 113

Answer: $50

Explanation:
Two $10 bills = $10 +$10 = $20
Two $5 bills = $5+$5 = $10
Total value = $20+$20+$10 = $50

Question 3.
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 114

Answer: $91

Explanation:
Four $20 bills = $20 +$20 +$20 +$20 = $80
$10 and $ 1
Total value = $80 +$10 +$1 = $91.

Question 4.
YOU BE THE TEACHER
Newton says he drew the fewest number of bills to show $35. Is he correct? Explain.
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 115

Answer:
No , he is wrong.

Explanation:
$35 can be shown in few bills as
a $20 bill, a $10  and a $5 bill

Question 5.
Modeling Real Life
A pair of headphones costs $88. Draw and label bills to show two different ways to pay for the headphones. One way should use the fewest number of bills.
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 116

Answer:
The cost of head phones is $88 .
$ 88 can be shown as
four $20 bills, a $5 bill ,a $2 bill and a$1 bill

   

Question 6.
Modeling Real Life
Descartes has two $20 bills, three $10 bills, and four $1 bills. Does he have enough money to buy a scratching post that costs $62? Explain.
__________________________

__________________________

Answer:
Yes ,he have enough money to buy a scratching post

Explanation:
Descartes have
Two $20 bills
Which means , 2 x 20 = 40
Three $10 bills
Which means, 3 x 10 = 30
Four $ 1bills
Total : $40 +$30 +$4 =
$74
The cost of scratching post is $62
So ,Descartes have enough money to but the post.

Review & Refresh

Question 7.
A photo album has 3 rows of photos. There are 4 photos in each row. How many photos are there in all?
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 117

Answer: 12 Photos.

Explanation:
Given that,
A photo album has 3 rows of photos
and there are 4 photos in each row
Now, the total number of phone are
4 +4+4 = 12
So , there are 12 photos in all.

Lesson 14.7 Problem Solving: Money

Explore and Grow

Model the story.
You buy a book for 60¢. Your friend buys a book for 33¢. How much do you and your friend spend in all?
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 118
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 119
Explain how you solved.
_______________________________
_______________________________
_______________________________

_______________________________

Answer: 93 Cents.

Explanation:
The cost of my book is 60 cents and
The cost of my friends book is 33 cents
Total value of our money is
60 cents + 33 cents
= 93 cents.
Therefore, me and my friend spent 93 cents in all.

Show and Grow

Question 1.
Descartes has two $10 bills and two $5 bills. He has $21 more than Newton. How much money does Newton have?
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 120

 

Answer: $9

Explanation:
Total amount of Descartes is
Two $10 bills
Which means, 2 x $10 bills = $20
Two $5 bills
Which means , 2 x $5 = $10
Total value = $20 +$10 = $30
Descartes have $30
Given that, Descartes have $21 more than newton
now , Newtons money = $30 – $21 = $9
Therefore ,Newton have $9

Question 2.
Descartes has some coins in a jar. He puts in 4 dimes, 1 nickel, and 1 penny. Now he has $1. How many cents were in the jar to start?
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 121

Answer: 54 Cents.

Explanation:
1 dime = 10 cents
Which means , 4 dimes = 4 x 10 = 40
1 nickel = 5 cents
and 1 penny = 1 cents
total value :
40 +5+1= 46 cents
We know that,
$1 is equals to 100 cents
now , 100 cents -46 cents = 54 cents
Therefore, there are 54 cents in the start.

Apply and Grow: Practice

Question 3.
Newton has some money. He loses a $10 bill and three $1 bills. Now he has $19. How much money did he have to start?

Answer: $32.

Explanation:
The total money newton lost is
$10 bill
and three $ 1 bill which means, $3
Total : $10 +$3 = $13
Given that ,
He have $19 after losing some money
Now, the total money newton have on the beginning is
lost money +leftover money
$13 +$19 = $32
Therefore , Newton have $32 in the starting.

Question 4.
Descartes has one $20 bill, three $10 bills, and three $5 bills. He spends $50. How much money does he have left?

Answer: $15

Explanation:
Total money Descartes have is
$20 bill
Three $10 bills = 3 x 10 = 30  and
Three $5 bills = 3 x 5 = 15
Total : $20 +$30 +$15= $65
Given that,
He spent $50
Therefore , Leftover money =
$65 – $50 = $15
So, Descartes is left with $15 .

Question 5.
A joke book costs $1. You have 2 quarters and 1nickel. How much more money do you need to buy the joke book?

Answer: 45 Cents.

Explanation:
$1 is equal to 100 cents
1 quarter = 25 cents
Which means, 2 quarters = 2 x 25 = 50 cents
1 nickel = 5 cents = 5 cents
Total : 50 cents +5 cents = 55 cents
now ,
100 cents – 55 cents = 45 cents
So , I need 45 cents more to buy the book.

Question 6.
YOU BE THE TEACHER
Your friend says that 3 dimes and 2 nickels is 50¢. Is your friend correct? Explain.
_______________________________
_______________________________

Answer: No, He is wrong.

Explanation:
1 dime = 10 cents
Which means , 3 dimes = 3 x 10 = 30 cents
1 nickel = 5 cents
Which means, 2 nickels = 2 x 5 = 10
Total value = 30 + 10 = 40
So, he is wrong

Think and Grow: Modeling Real Life

You have a $20 bill and a $5 bill. Your friend has $10 less than you. Do you and your friend have enough money to buy a $38 skateboard? Explain.
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 122

Answer: No.

Explanation :
The  amount I have is
$20 and $ 5
Total : $20+$5 = $25
Given that , my friend have $10 less than me
Which means,
$25 – $10= $15
But the cost of skateboard is $38
So , me and my friend do not have enough value to buy the skateboard.

Show and Grow

Question 7.
You have 1 quarter, 2 dimes, and 3 pennies. Your friend has 4 nickels and 2 pennies. Do you and your friend have enough money to buy a 75¢ bottle of orange juice? Explain.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 123
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 124

Answer:
My money:
1 quarter = 25 cents
1 dimes = 10 cents
Which means , 2 x 10 = 20 cents
3 pennies = 3 cents
Total : 25 + 20 + 3 = 48 cents
I have 48 cents
My friends money:
1 nickel = 5 cents
Which means , 4 nickels = 4 x 5 = 20 cents
2 pennies = 2 cents
Total : 20 + 2 = 22 cents
My friend have 22 cents
The cost of orange juice is 75 cents.
Therefore, me and my friend do not have enough money to buy the orange juice.

Question 8.
Descartes buys a board game for $19. He has three $5 bills and two $1 bills left over. How much money did he have before he bought the game?
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 125

Answer: $26.

Explanation:
The cost of board game = $19
Total value of leftover money is
Three $5 bills = 3 x 5 = $15
Two $1 bills = 2 x 1 = $2
Total : $15 +$2
= $17
The total money Descartes have in the beginning is
leftover money + cost of board game
=$17 +$19
= $26
Therefore , Descartes have $26 before he bought the game.

Question 9.
DIG DEEPER!
You have 25¢ in your desk, 18¢ in your backpack, and 50¢ in your pocket. You spend 43¢ and lose a quarter. How much money do you have left?
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 126

Answer: 25 cents.

Explanation:
The amount I have in my desk is 25 cents
The amount have in my backpack is 18 cents
The amount I have in my pocket is 50 cents
Now,  The total amount I have with me is:
25 cents +18 cents + 50 cents = 93 cents.
1 quarter = 25 cents
The amount I spend = 43 cents
total : 25 + 43 = 68 cents
Now, The money I have left is
93 cents – 68 cents = 25 cents
Therefore , I am left with 25 cents  .

Problem Solving: Money Homework & Practice 14.7

Question 1.
Newton has $30. Descartes has a $20 bill, a $10 bill, and two $1 bills. How much more money does Descartes have?
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 127

Answer:
The money newton has is $30 .
Descartes have $20 bill ,$10 bill and two $1 bill
Total money : 20 +10+1+1 =32
So Descartes have $32
Now, $32 – $30 = $2
Therefore, Descartes have $2 more than Newton.

Question 2.
You have some money. You spend 2 quarters and 3dimes at the cafeteria. Now you have 20¢. How much money did you have to start?
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 128

Answer: $1.

Explanation:
1 quarter = 25 cents
Which means , 2 quarters = 2 x 25 = 50 cents
1 dime =10 cents
Which means, 3 dimes = 3 x 10 = 30 cents
Total amount : 50 +30 = 80
After spending 80 cents at cafeteria , I have 20 cents
Now , the money i have at start is
80 cents + 20 = 100 cents
100 cents =  $1
Therefore , I have $ 1 in the start.

Question 3.
YOU BE THE TEACHER
You buy a sandwich for 75¢. You pay with $1. Your friend says your change will be 1 quarter. Is your friend correct? Explain.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 129

Answer:
Yes, he is correct .
The cost of sandwich is 75 cents
$1 is equals to 100 cents
100 cents – 75 cents = 25 cents
As we know that ,
1 quarter = 25 cents
so , he is correct.

Question 4.
Modeling Real Life
You have 12 pennies, 2 dimes, and 1 nickel. Your friend has 20¢ more than you. Do you and your friend have enough money to buy a toy car that costs $1? Explain.
_______________________________
_______________________________

Answer: No , me and my friend do not have enough money to buy a toy car

Explanation :
The amount I have is
12 pennies, 2 dimes and 1 nickel
12 pennies = 12 cents
1 dime = 10 cents
Which means 2 dime = 2 x 10 = 20 cents
and 1 nickel = 5 cents
Total amount =
12 +20 +5 = 37 cents
Therefore I have 37 cents
Friends money :
He has 20 cents more than me
Which means, 37 +20 = 57 cents
But , we know that $1 = 100 cents
So we do not have enough money to buy a toy car

Question 5.
Modeling Real Life
Descartes has $45. He spends a $20 bill and a $1 bill. He earns two $5 bills and a $10 bill. How much money does he have now?
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 130

Answer:
Given that , Descartes has $45
He spends $ 20 and $1 bill
total : 20 + 1 = 21
now, $45 -$21= $24
Then,
He earns 2 $5 bills and a $10 bill
Total : $24 +$5 +$5+$10= $44
Therefore ,Descartes have $44 .

Review & Refresh

Question 6.
Which time does not belong with the other three?
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 131

Answer:

Explanation :
In the above clock the time is 4 hours 30 minutes.
But the time in remaining three is 3 hours 3 minutes.
So, it does not belong with the other three.

Lesson 14.8 Tell Time to the Nearest Five Minutes

Explore and Grow

Label the missing minutes around the clock. Then tell the time.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 132

Answer:

  1. 15
  2. 30
  3. 55.
  4. The time is 3 hours 30 minutes

Show and Grow

Write the time.

Question 1.
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 133

Answer: 1 hours 25 minutes.

Explanation:
As the hours hand is in between 1 and 2 and the minutes hand on 5. So, the time is on 25 minutes.
That means, the time is 1 hours and 25 minutes.

Question 2.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 134

Answer:8 hours 10 minutes.

Explanation:
As the hours hand is on 8 and the minutes hand on 10. So, the time is on 10 minutes.
That means, the time is 8 hours and 10 minutes.

Question 3.
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 135

Answer: 6 hours 45 minutes.

Explanation:
As the hours hand is in between 6 and 7 and the minutes hand on 9. So, the time is on 45 minutes.
That means, the time is 6 hours and 45 minutes.

Question 4.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 136

Answer: 5 hours 55 minutes.

Explanation:
As the hours hand is in between 5 and 6 and the minutes hand on 11. So, the time is on 55 minutes.
That means, the time is 5 hours and 55 minutes.

Apply and Grow: Practice

Write the time.

Question 5.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 137

Answer: 3 hours 50 minutes.

Explanation:
As the hours hand is in between 3 and 4 and the minutes hand on 10. So, the time is on 50 minutes.
That means, the time is 3 hours and 50 minutes.

Question 6.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 138

Answer: 11 hours 20 minutes.

Explanation:
As the hours hand is in between 11 and 12 and the minutes hand on 4. So, the time is on 20 minutes.
That means, the time is 11 hours and 20 minutes.

Question 7.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 139

 

Answer: 5 hours 30 minutes.

Explanation:
As the hours hand is in between 5 and 6 and the minutes hand on 6. So, the time is on 30 minutes.
That means, the time is 5 hours and 30 minutes.

Draw to show the time.

Question 8.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 140

Answer:

Question 9.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 141

Answer:

Question 10.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 142

 

Answer:

Question 11.
Patterns
Write the next time in the pattern.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 143

Answer:
8:30
The sequence is differ  by 5 minutes

Question 12.
Precision
The hour hand points between the 4 and the 5. The minute hand points to the 4. What time is it?
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 144

Answer:
As the hour hand points between 4 and 5 .the minute hand points to the 4
The time is 4 hours 20 minutes

Think and Grow: Modeling Real Life

Baseball practice lasts 40 minutes. Show and write the time practice ends.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 145

Answer: 3 hours 55 minutes

Show and Grow

Question 13.
Recess lasts 25 minutes. Show and write the time recess ends.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 146

Answer: 12 hours 35 minutes

Question 14.
DIG DEEPER!
A train ride starts at 6:40. The ride lasts 45 minutes. What time does the ride end?
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 147
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 148

Answer:  7 hours 25 minutes

Tell Time to the Nearest Five Minutes Homework & Practice 14.8

Write the time.

Question 1.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 149

Answer:
10 hours 35 minutes.
As the hours hand is in between 10 and 11 and the minutes hand is near to 7. So, the time is near to 35 minutes.
That means, the time is 10 hours and 35 minutes.

Question 2.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 150

Answer:
6 hours 10 minutes.
As the hours hand is on 6, the minutes hand is on 2. So, the time is near to 10 minutes.
That means, the time is 6 hours and 10 minutes.

Question 3.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 151

Answer:
7 hours 25 minutes.
As the hours hand is in between 7 and 8 and the minutes hand is near to 5. So, the time is near to 25 minutes.
That means, the time is 7 hours and 25 minutes.

Question 4.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 152

 

Answer: 

Question 5.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 153

Answer:

Question 6.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 154

Answer:

Question 7.
Reasoning
The minute hand points to the 7. What number will it point to in 10 minutes?

Answer :  9
Each number the minutes hand points, indicates 5 minutes.
So,10 minutes indicates 2 more numbers, the minutes hand moves on to.
That means, 7 + 2 = 9.
So, after 10 minutes, the minutes hand points on to 9.

Question 8.
Precision
The hour hand points between the 11 and the 12. In 25 minutes it will be the next hour. What time is it now?

Answer: 11 hours 35 minutes.

Explanation:
Each number the minutes hand point, indicates 5 minutes.
So, 25  minutes indicates 5  more numbers, the minutes hand moves on to.
that means ,5 x 5 = 25
So, after 25 minutes, the time is 12: 00
now, the time is 11 hours 35 minutes .

Question 9.
Modeling Real Life
Your walk to school lasts 15 minutes. Show and write the time your walk ends.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 157

Answer: 8 hours 25 minutes

Question 10.
DIG DEEPER!
Your swimming lesson starts at 5:30. It lasts 35 minutes. What time does the lesson end?
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 158

Answer:
6 hours 5 minutes.

Explanation:
Each number the minutes hand points, indicates 5 minutes.
So,35 minutes indicates 7 more numbers, the minutes hand moves on to.
That means, 7 x 5= 35
So, after 35 minutes, the lesson ends on 6 hours 5 minutes.

Review & Refresh

Question 11.
The crayon is about 7 centimeters long. What is the best estimate of the length of the toothpick?
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 159

Answer: 4 Centimeters.

Explanation:
Given,
The length of crayon is 7 meters long.
4 centimeters is the best estimate of length of the toothpick .

Lesson 14.9 Tell Time Before and After the Hour

Explore and Grow

Write each time on the digital clocks. How much time has passed?
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 160

Answer:
1) the time is  11hours 45 minutes
2)  the time is 12: 00
on comparing both the clocks 15 minutes time has been passed.

Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 161

Answer:
1) In the first clock the time is 1:00
2) In the second clock the time is 1:15 on comparing both the clocks 15 minutes has been passed

Show and Grow

Write the time. Circle another way to say the time.

Question 1.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 162

Answer:
4 hours 15 minutes.
As the hours hand is in between 4 and 5 and the minutes hand on 3 . So, the time is near to 15 minutes.
That means, the time is 4 hours and 15 minutes.
Also known as quarter past 4.

Question 2.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 163

Answer:
11 hours 30 minutes.
As the hours hand is in between 11 and 12 and the minutes hand is on 6. So, the time is near to 30 minutes.
That means, the time is 11 hours and 30 minutes.
Also known as half past 11.

Question 3.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 164

 

Answer:
2 hours 45 minutes.
As the hours hand is in between 2 and 3 and the minutes hand is near to 9. So, the time is near to 45 minutes.
That means, the time is 2 hours and 45 minutes.
Also known as quarter to 3.

Question 4.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 165

Answer:
9 hours 20 minutes.
As the hours hand is in between 9 and 10 and the minutes hand is near to 4. So, the time is near to 20 minutes.
That means, the time is 9 hours and 20 minutes.
also known as 20 minutes after 9.

Apply and Grow: Practice

Write the time. Circle another way to say the time.

Question 5.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 166

Answer: 1 hours 30 minutes.
As the hours hand is in between 1 and 2 and the minutes hand is near to 6. So, the time is near to 30 minutes.
That means, the time is 1 hours and 30 minutes.
Also known as half past 1.

Question 6.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 167

Answer:
7 hours 10 minutes.
As the hours hand is in between 7 and 8 and the minutes hand is near to 2. So, the time is near to 10 minutes.
That means, the time is 7 hours and 10 minutes.
also known as ,10 minutes after 7

Show and write the time.

Question 7.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 168

Answer: 10 : 45

Question 8.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 169

Answer: 5 hours 15 minutes

Question 9.
Which One Doesn’t Belong? Which time does not belong with the other three?
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 170

Answer: Quarter past 7

Explanation:
6 : 45
45 minutes after 6 means 6 :45
Quarter to 7 means 6 : 45
These three are same .
But , quarter past 7 means 7 : 30
So, quarter past 7 is different.

Question 10.
Precision
Is it time for homework or dinner?
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 171

Answer: Dinner
Quarter after 6 is also known as 6 hours 15 minutes.
6: 15 p.m.- dinner

Think and Grow: Modeling Real Life

School starts at quarter past 8. Are you early or late to school? Explain.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 172
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 173

Answer:
Early to school
now, the time is 7 hours 55 minutes
the school starts at 8 hours 15 minutes ;it means quarter past 8 .
so, I am early to school

Show and Grow

Question 11.
A movie starts at quarter to 6. Are you early or late to the movie? Explain.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 174
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 175

Answer:
I am late to the movie .
Now, the time is 6 hours 5 minutes .
but, the movie starts at quarter to 6
which means , 5 hours 45 minutes
so I am late to the movie .

Question 12.
DIG DEEPER!
You arrive at the bus station 20 minutes before 12. Which is the first bus you can board? How many minutes are there until it leaves?
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 176Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 177

Answer:
The time when i arrived bus station is 11 hours 40 minutes
So, I can board blue bus.
Blue bus leaves at 11 hours 55 minutes .
So , there are 15 more minutes until the blue bus leaves the bus station.

Tell Time Before and After the Hour Homework & Practice 14.9

Write the time. Circle another way to say the time.

Question 1.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 178

Answer: quarter to 12 means 11 hours 45 minutes

 

As the hours hand is between 11 and 12, the minutes hand is on 9. So, the time is near to  45 minutes.
That means, the time is 11 hours and 45 minutes.

Question 2.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 179

Answer: 40 minutes after 5 which means 5 hours 40 minutes

As the hours hand is between 5 and 6, the minutes hand is on 8. So, the time is near to  40 minutes.
That means, the time is 5 hours and 40 minutes.
Show and write the time.

Question 3.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 180

Answer: 3 hours 30 minutes

As the hours hand is between 3 and 4, the minutes hand is on 6. So, the time is near to 30 minutes.
That means, the time is 3 hours and 30 minutes.

Question 4.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 181

Answer: 12 hours 15 minutes

As the hours hand is between 12 and 1, the minutes hand is on 3. So, the time is near to  15 minutes.
That means, the time is 12 hours and 15 minutes.

Question 5.
YOU BE THE TEACHER
Newton says it is 2:45, or quarter to 3. Is he correct? Explain.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 182

Answer:
Yes, he is correct.

Explanation:
quarter means 15 minutes .
Each number the minutes hand points, indicates 5 minutes.
That means, 3 x 5= 15
so, 2:45 and quarter to 3 are both same
So, newton is correct

Question 6.
Modeling Real Life
A show starts at quarter to 7. Are you early or late to the show? Explain.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 183

Answer:
I am  early to the show
Now ,the time is 6 hours 30 minutes
but the show starts at quarter to 7 which means 6 hours 45 minutes
So ,I am 15 minutes early to the show

Question 7.
DIG DEEPER!
You arrive at the metro station 10minutes after 2. Which is the first train you can board? How many minutes are there until it leaves?
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 184

Answer:
The time 10 minutes after 2 is 2 hours 10 minutes
So ,I can board the yellow train which will leaves the railway station at 2: 30
There are 20 minutes more until the yellow train leaves the railway station.

Review & Refresh

Question 8.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 185

Answers: 162.

 

Question 9.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 186

Answer: 177

Explanation:

Question 10.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 187

Answer: 446.

Explanation:

Lesson 14.10 Relate A.M. and P.M.

Explore and Grow

Describe what you do in the morning. Show and write the time. Describe what you do in the evening. Show and write the time.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 188

Show and Grow

Write the time. Circle a.m. or p.m.

Question 1.
Eat breakfast
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 189

Answer: 7 hours 30 minutes a.m.
As the hours hand is on 7, the minutes hand is on 6. So, the time is near to 30  minutes.
That means, the time is 7 hours and 30 minutes.

Question 2.
Eat dinner
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 190

Answer:
6 hours 15 minutes p.m.
As the hours hand is between 6 and 7, the minutes hand is on 3. So, the time is near to  15 minutes.
That means, the time is 6 hours and 15 minutes.

Question 3.
Go to art class
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 191

Answer :
10 hours 25 minutes a.m.
As the hours hand is between 10 and11, the minutes hand is on 5. So, the time is near to 25  minutes.
That means, the time is 10 hours and 25 minutes.

Question 4.
Do homework
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 192

Answer:
4 hours 40 minutes p.m.
As the hours hand is between 4 and 5, the minutes hand is on 8. So, the time is near to  40  minutes.
That means, the time is 4 hours and 40 minutes.

Apply and Grow: Practice

Write the time. Circle a.m. or p.m.

Question 5.
Ride the bus to school
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 193

Answer:
8:00 a.m.
As the hours hand is on 8, the minutes hand is on 12.
That means, the time is 8 hours

Question 6.
Go to a party
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 194

Answer:
6hours 30 minutes p.m.
As the hours hand is between 6 and 7, the minutes hand is on 6. So, the time is 30  minutes.
That means, the time is 6 hours and 30 minutes.

Draw to show the time. Circle a.m. or p.m.

Question 7.
Read before bed
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 195

Answer:
7 hours and 50 minutes p.m.

Question 8.
Sunrise
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 196

Answer:
5 hours 20 minutes a.m.

Question 9.
Reasoning
Use the times in the list to complete the story.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 197

Answer:
You arrive at school at 8: 30 a.m. Your class goes to music at 10:15 a.m. After school,
you read a book at 5:20 p.m.

Think and Grow: Modeling Real Life

Use the times to complete the timeline. Write something you might do at those times.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 198

Answer:
1) 8: 45 a.m. –  School time
2)2:50 p.m. –  Music class time
3) 4:10 p.m. –  Home work

Show and Grow

Question 10.
Use the times to complete the timeline. Write something you might do at those times.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 199
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 200

Answer:
1) 7: 30 a.m. – Yoga class
2) 9:55 a.m.-  Art class
3) 7:35 p.m. – Dinner

Question 11.
DIG DEEPER!
Use the times to complete the timeline. Then rewrite each time digitally below, including a.m. or p.m
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 201
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 202

Answer:
1) 20 minutes before 7 –      6:40 a.m
2) 10 minutes after 11   –     11:10 a.m
3) noon                         –      12:00 p.m
4) quarter past 3           –        3:15 p.m
5)  half past 6                 –        6:30 p.m
6)2  0 minutes after 8    –        8:20 p.m

Relate A.M. and P.M. Homework & Practice 14.10

Write the time. Circle a.m. or p.m.

Question 1.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 203

Answer:
9 hours 25 minutes p.m
As the hours hand is between 9 and 10, the minutes hand is on 5. So, the time is near to 25 minutes.
That means, the time is 9 hours and 25 minutes.

Question 2.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 204

 

Answer:
7hours 15 minutes A.m.
As the hours hand is between 7 and 8, the minutes hand is on 3. So, the time is 15  minutes.
That means, the time is 7 hours and 15 minutes.

Draw to show the time. Circle a.m. or p.m.

Question 3.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 205

Answer: 12:00 P.m.

Question 4.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 206

Answer: 4: 40 P.m.

Question 5.
Reasoning
Right now, it is p.m. In 10 minutes it will be a.m. What time is it now? Explain.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 207

 

Answer:
Time is 11 hours 50 minutes.
Each minute hand point indicates 5 minutes
That means 2x 5= 10 minutes
a.m. starts imediatly after 12:00 pm
So, now the time is 11: 50 p.m.

Question 6.
Modeling Real Life
Use the times to complete the timeline. Write something you might do at those times.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 208

Answer:
1 ) 8:15 a.m. – break fast
2) 11: 55 a.m. – art class
3)  8: 15 p.m.- drawing time

Question 7.
DIG DEEPER!
Use the times to complete the timeline. Then rewrite each time digitally below, including a.m. or p.m.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 209

Answer:
1 ) Quarter to 8 –            7:45 a.m.
2) 20 minutes after 8 –  8: 20 a.m.
3) Quarter past 11 –      11: 15 a.m.
4) Noon –                       12:00 p.m.
5) 10 minutes after 2 –   2 : 10 p.m.
6) Half past 9-                 9:30 p.m.

Review & Refresh

Question 8.
65 + 36 = ____

Answer:
65 +36 = 101

Question 9.
56 + 18 = ____
Answer: 74

Money and Time Performance Task 14

Question 1.
a. You have two $1 bills, 1 quarter, 5 dimes, 3 nickels, and 2 pennies. How much more money do you need to buy a subway pass?
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 210

Answer:
The cost of  subway pass is $3
$3 is equals to 300 cents
$1 is equals to 100 cents
Two $1 = 100 + 100 = 200cents
1 quarter = 25 cents
1 dime = 10 cents
Which means , 5 x 10 = 50
1 nickel = 5 cents
Which means, 3 x 5 = 15
and 2 pennies = 2  cents
Total :  200 +25+50+15+2 = 292
300 – 292 = 8 cents
I need 8 more cents to buy a pass.
So, I do not have enough money to buy a subway pass.

b. You find a dime. Do you have enough money to buy the pass now?
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 211

Answer: Yes

Explanation:
1 dime = 10 cents
292 +10 = 302 cents
The cost of subway pass is $3 which means 300 cents
If I find a dime, I have enough money to buy a pass

Question 2.
A weekly subway pass is $32. A customer pays with a $50 bill. Use tally marks to show three different ways that the customer can receive change. What is the total change?
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 212Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 212.2

Answer:
The cost of weekly subway pass is $32 .
Given that ,
A customer paid a $50 bill
Total change :
$50 -$32 = $18
$18 can be shown as
1) 1 $10 bill , 1 $5 bill and 3 $1 bill
2) 1 $10 bill, 8 $1 bills
3) 3 $ 5 bills and 3 $1 bill .

Question 3.
You arrive at the subway station at quarter to 3. What times will the subways arrive?
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 213

Answer: car A arrives subway

Money and Time Activity

To Play: Place the Flip and Find Cards face down in the boxes. Take turns flipping 2 cards. If your cards show the same time or value, keep the cards. If your cards show different times or values, flip the cards back over. Play until all matches are made.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 214

Money and Time Chapter Practice

4.1 Find Total Values of Coins

Question 1.
Count on to find the total value.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 215

Answer: 46 cents

Explanation
1 quarter = 25 cents
1 dime = 10 cents
1 nickel = 5 cents
Which means , 5 x 2 =10 cents
1 penny = 1 cent
Total : 25 +10+5+5+1 = 46 cents
Therefore the total value of coins 46 cents.

Question 2.
Modeling Real Life
You have 2 quarters, 2 dimes, and 1 penny. How many cents do you have? Do you have enough money to buy the frozen fruit bar?
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 216
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 217

Answer:
The cost of frozen fruit bar is 75 cents
1 quarter = 25 cents
which means , 2 quarter = 2 x 25 = 50 cents
1 dime = 10 cents
which means , 2 dimes = 2 x 10 = 20 dimes
1 penny = 1 cent
total amount I have is 50+20+1 = 71 cents
therefore I have 71 cents
So ,I do not have enough money to buy the frozen fruit bar.

14.2 Order to Find Total Values of Coins

Question 3.
Draw and label the coins from the greatest value to the least value. Then find the total value.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 218

Answer:
43 cents

Explanation
1 quarter = 25 cents
1 dime = 10 cents
1 nickel = 5 cents
3 pennies = 3cents
Total value = 25 +10 +5 + 3 = 43 cents

>>>

14.3 Show Money Amounts in Different Ways

Question 4.
Draw and label coins to show the amount in two different ways.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 219

Answer:
56 cents can be shown as
2 quarter, 1 nickel and 1 penny
1 quarter = 25 cents
Which means, 2 x 25 = 50 cents
1 nickel = 5 cents
and 1 penny = 1 cents
Total : 50+5+1 = 56 cents

Another way :
56  cents can be shown as 5 dime, 1 nickel and 1 penny
1 dime = 10 cents
5 dimes = 5 x 10 = 50 cents
1 nickel = 5 cents
and 1 penny = 1 cent
Total value :
50 + 5+1 = 56 cents

14.4 Make One Dollar

Draw coins to make $1. How many cents do you need?

Question 5.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 220

Answer: $ 1 is equals to 100 cents
100 cents – 79 cents = 21 cents.
So , I need 21 cents to make $ 1
21 cents can be shown in coins as 2 dime , 1 penny
1 dime = 10 cents
Which means , 2 dime = 2 x 10 = 20 cents
1 penny = 1 cent
Total : 20 + 1 = 21 cents

Question 6.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 221

Answer:
$1 is equals to 100 cents
1 dime = 10 cents
Which means , 2 cents= 2 x 10 = 20 cents
3 pennies = 3 cents
Total value = 20+3 = 23 cents
Now, 100 cents- 23 cents = 77 cents
77 cents can be shown as 3 quarters , 2 pennies
1 quarter = 25 cents
Which means , 3 quarters = 3 x 25 = 75 cents
2 pennies = 2 cents
Total value : 75 + 2 = 77 cents

14.5 Make Change from One Dollar

Question 7.
Reasoning
Newton buys a toy for 21¢. Descartes buys one for 94¢. You buy one for 57¢. You each pay with $1. Who gets back the least amount of money? How do you know?
_____________________________________
_____________________________________

Answer:
$1 is equals to 100 cents
The cost of newtons toy is 21 cents
100 – 21 =79 cents
So, 79 cents is change newtons gets back.
Now, the cost of descartes toy is 94 cents
100 – 97 = 3 cents
So Descartes change is 3 cents
On comparing both newtons and Descartes money
79 cents is greater than 3cents;
79 > 3
Therefore, Descartes gets back the least amount of money.

14.6 Find Total Values of Bills

Question 8.
Count on to find the total value.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 222

Answer: $ 38

Explanation :
We have , Three $ 1 bills
which means , $1 +$1 $1 = $3
One $ 5 bill
One $ 10 bill
One $ 20 bill
Total value :
$ 3+ $5 +$10 + $20 =$38
Therefore the total value of the dollar bills is $ 38.

Question 9.
Newton has five $10 bills. He has $32 more than Descartes. How much money does Descartes have?
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 223

Answer:  $18

Explanation :
five $ 10 bills is equals to
5 x 10 = $ 50
Given that , Newton has $32 more than Descartes
We know that, Newton have $50
Now, $50 – $32 = $18
Therefore , Descartes have $18

14.8 Tell Time to the Nearest Five Minutes

Write the time.

Question 10.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 224

Answer: 7 hours 20 minutes

Explanation :
As the hours hand is between 7 and 8, the minutes hand is on 4. So, the time is 20 minutes.
That means, the time is 7 hours and 20 minutes.

Question 11.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 225

Answer: 3 hours 45 minutes

Explanation :
As the hours hand is between 3 and 4, the minutes hand is on 9. So, the time is 45 minutes.
That means, the time is 3 hours and 45 minutes.

Question 12.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 226

Answer: 10 hours 15 minutes

Explanation :
As the hours hand is between 10 and 11, the minutes hand is on 3. So, the time is 15 minutes.
That means, the time is 10 hours and 15 minutes.

14.9 Tell Time Before and After the Hour

Question 13.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 227

Answer: quarter past 5
which means 5 hours 15 minutes
quarter to 5 means 4 hours 45 minutes

Question 14.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 228

Answer: quarter to 10
which means ,9 hours 45 minutes
half past 9 means 9 hours 30 minutes

Question 15.
Modeling Real Life
Soccer practice starts at half past 1. Are you early or late to soccer practice? Explain.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 229

Answer:  I am late to soccer practice.

Explanation:
now the time is 1 hour 45 minutes .
The class starts at half past one
Which means ,1 hour 30 minutes
So ,I am late by 15 minutes to the class.

14.10 Relate A.M. and P.M.

Draw to show the time. Circle a.m. or p.m.

Question 16.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 230

Answer : 9hours 30 minutes p.m

Question 17.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 231

 

Answer:7 hours 25 minutes a.m.

Conclusion:

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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 2 Chapter 7 Understand Place Value to 1,000

Big Ideas Math Answers Grade 2 Chapter 7

Big Ideas Math Grade 2 Chapter 7 Understand Place Value to 1,000 Answer Key is provided in a comprehensive manner for better understanding. Students must solve as many questions as possible in BIM 2nd Grade Chapter 7 Understand Place Value to 1,000 Book and cross-check method and solutions here. If you wanted to become a pro in the place value concept, then refer to Big Ideas Math Answers Grade 2 Chapter 7 Understand Place Value to 1,000.

Big Ideas Math Book 2nd Grade Answer Key Chapter 7 Understand Place Value to 1,000

The list of concepts covered in Big Ideas Math Book 2nd Grade 7th Chapter Understand Place Value to 1,000 is given in the below section. It is one of the important concepts that must be learned at the elementary. So, no student is supposed to skip this chapter. Download BIM Grade 2 Chapter 7 Understand Place Value to 1,000 PDF for free of cost and start preparation.

Parents and teachers can have a look at this free Big Ideas Math 2nd Grade Chapter 7 Understand Place Value to 1,000 Answers for a better understanding. After solving the questions available here, you will be able to solve any kind of question framed on the concept and clear your tests with a good score. If you have a doubt just tap on the respective link you wanted to prepare and learn the fundamental topics included within it easily.

Vocabulary

Lesson: 1 Hundreds

Lesson: 2 Model Numbers to 1,000

Lesson: 3 Understand Place Value

Lesson: 4 Write Three-Digit Numbers

Lesson: 5 Represent Numbers in Different Ways

Understand Place Value to 1,000

Understand Place Value to 1,000 Vocabulary

Big Ideas Math Answer Key Grade 2 Chapter 7 Understand Place Value to 1,000 v 1

Organize It
Use the review words to complete the graphic organizer.
Big Ideas Math Answer Key Grade 2 Chapter 7 Understand Place Value to 1,000 v 2

Answer:
Big-Ideas-Math-Answer-Key-Grade-2-Chapter-7-Understand-Place-Value-to-1000-v-2

Explanation:
Tens Place: The first digit on the right of the decimal point means tenths.
Ones Place The ones place is just to the left of the decimal point.

Define It
Use your vocabulary cards to identify the word.
Big Ideas Math Answer Key Grade 2 Chapter 7 Understand Place Value to 1,000 v 3

Answer:
Big-Ideas-Math-Answer-Key-Grade-2-Chapter-7-Understand-Place-Value-to-1000-v-3

Big Ideas Math Answer Key Grade 2 Chapter 7 Understand Place Value to 1,000 v 4
Big Ideas Math Answer Key Grade 2 Chapter 7 Understand Place Value to 1,000 v 5

Big Ideas Math Answer Key Grade 2 Chapter 7 Understand Place Value to 1,000 v 6
Big Ideas Math Answer Key Grade 2 Chapter 7 Understand Place Value to 1,000 v 7

Lesson 7.1 Hundreds

Explore and Grow
How many unit cubes and rods are in a flat?
Big Ideas Math Answer Key Grade 2 Chapter 7 Understand Place Value to 1,000 7.1 1
Answer: 100 cubes and 10 rods
Explanation
Given that, For counting cubes, should count all the rows and columns. By counting all the boxes there are
10 rows and 10 columns= 100 cubes To find out the rods count only the columns, by counting columns there are 10 rods.
Show and Grow
Write how many tens. Circle groups of 10 tens. Write how many hundreds. Then write the number.
Question 1.
Big Ideas Math Answer Key Grade 2 Chapter 7 Understand Place Value to 1,000 7.1 2
______ tens
______ hundred
______
Answer: 20 tens & 2 hundred (200)
Explanation
Given that 20 tens are there by circling 10 tens
For counting tens, count columns and for counting hundreds count all the boxes.
Question 2.
Big Ideas Math Answer Key Grade 2 Chapter 7 Understand Place Value to 1,000 7.1 3
______ tens
______ hundred
______
Answer: 40 tens & 4 hundred (400)
Explanation
Given that 40 tens are there by circling 10 tens
For counting tens, count columns and for counting hundreds count all the boxes.

Apply and Grow: Practice

Write how many tens. Circle groups of 10 tens. Write how many hundreds. Then write the number.
Question 3.
Big Ideas Math Answer Key Grade 2 Chapter 7 Understand Place Value to 1,000 7.1 4
______ tens
______ hundred
______
Answer: 30 tens &  3 hundreds (300)
Explanation
Given that 30 tens are there by circling 10 tens
for counting tens, count columns and for counting hundreds count all the boxes.

Question 4.
Big Ideas Math Answer Key Grade 2 Chapter 7 Understand Place Value to 1,000 7.1 5
______ tens
______ hundred
______
Answer: 60 tens & 6 hundreds
Explanation
Given that 60 tens are there by circling 10 tens.
for counting tens, count columns and for counting hundreds count all the boxes.

Question 5.
DIG DEEPER!
How many hundreds are in 700? How many tens?
_____ hundreds
______ tens
Answer: 7 hundreds are in 700 and 70 tens are in 700.
Explanation
there are 7 hundreds in 700 and 70 tens in 700. 10 tens gives  1 hundreds totally there are 70 tens is 700.

Question 6.
You have 80 bags of crayons. There are 10 crayons in each bag. How many crayons do you have in all?
_____ crayons
Answer: 800 crayons
Explanation
Given that, There are 80 bags of crayons each bags contains 10 crayons.10 x 80=800 ( 10 crayons in each bags x 80 totally bags = 800 )
Thus, There are 800 crayons do we have in all.

Think and Grow: Modeling Real Life

A store sells oranges in bags of 10. The store sells 500 oranges. How many bags do they sell?
Big Ideas Math Answer Key Grade 2 Chapter 7 Understand Place Value to 1,000 7.1 6
Make a quick sketch:
_____ bags
Answer: 50 bags
Explanation
Given that, The store sells 500 oranges each bags contains 10 oranges.10 X 50 = 500  (10 oranges in each bags x 50 bags = 500 oranges)
Thus, The store sell the 50 bags of oranges.

Question 7.
A store sells bottles of glitter glue in boxes of 10. The store sells 600 bottles. How many boxes do they sell?
Big Ideas Math Answer Key Grade 2 Chapter 7 Understand Place Value to 1,000 7.1 7
_____ boxes
Answer:  60 boxes
Explanation
Given that, The store sells 600 bottles of glitter glue each boxes contains 10 glitter glue bottles.
10 X 60 = 600  (10 bottle of glitter glue in each boxes x 60 boxes = 600 bottles of glitter glue)
Thus, The store sell the 60 boxes of glitter glue.

Question 8.
DIG DEEPER!
You have 10 packages of invitations. Each package has 10 invitations. You need 300 invitations. How many more packages do you need?
Big Ideas Math Answer Key Grade 2 Chapter 7 Understand Place Value to 1,000 7.1 8
_____ more packages
Answer: 30 packages
Explanation
Given that, There are 10 packages of invitations each packages has 10 invitations you needed 300 invitations
10 x 30 = 300 (10 invitations of  packages x 30 packages = 300 invitations)
Thus, 30 packages of invitations are needed.

Hundreds Homework & Practice 7.1

Write how many tens. Circle groups of 10 tens. Write how many hundreds. Then write the number.
Question 1.
Big Ideas Math Answer Key Grade 2 Chapter 7 Understand Place Value to 1,000 7.1 9
_____ tens
______ hundreds
______
Answer: 50 tens & 5 hundreds.
Explanation
Given that 50 tens are there by circling 10 tens
For counting tens, count columns and for counting hundreds count all the boxes.

Question 2.
Big Ideas Math Answer Key Grade 2 Chapter 7 Understand Place Value to 1,000 7.1 10
_____ tens
______ hundreds
______
Answer: 80 tens & 8 hundreds
Explanation
Given that 80 tens are there by circling 10 tens
For counting tens, count columns and for counting hundreds count all the boxes.

Question 3.
Which One Doesn’t Belong?
Which does not belong with the other three?
Big Ideas Math Answer Key Grade 2 Chapter 7 Understand Place Value to 1,000 7.1 11
Answer: 70

Question 4.
Modeling Real Life
A class has 30 boxes of pencils. Each box has 10 pencils. The class needs 600 pencils. How many more boxes does the class need?
Big Ideas Math Answer Key Grade 2 Chapter 7 Understand Place Value to 1,000 7.1 12
______ boxes
Answer: 30 boxes
Explanation
Given that, the class has 30 boxes of pencils each boxes contains 10 pencils. The class needs 600 pencils.
10 x 60 (30 +30)=600 (10 Pencils in each boxes X 60 boxes (30 boxes which you have + 30 boxes of pencils you need)= 600 pencils)
Thus, the class needs another 30 boxes.

Question 5.
Modeling Real Life
You have 10 packages of cards. Each package has 10 cards. You need 200 cards. How many more packages do you need?
Big Ideas Math Answer Key Grade 2 Chapter 7 Understand Place Value to 1,000 7.1 13
______ more packages
Answer: 10 package
Explanation
Given that, 10 packages of cards each packages contains 10 cards, you need 200 cards
10 x 20= 200 (10 cards in each packages X 20 packages of cards (10 packages which you have+10 packages you needed) =200 cards)
Thus, you need another 10 packages of cards.

Review & Refresh

Question 6.
Big Ideas Math Answer Key Grade 2 Chapter 7 Understand Place Value to 1,000 7.1 14
Answer: 99

Question 7.
Big Ideas Math Answer Key Grade 2 Chapter 7 Understand Place Value to 1,000 7.1 15
Answer: 75

Lesson 7.2 Model Numbers to 1,000

Explore and Grow

Model the number. Make a quick sketch to match.
Big Ideas Math Answers 2nd Grade Chapter 7 Understand Place Value to 1,000 7.2 1
Answer:
Big-Ideas-Math-Answers-2nd-Grade-Chapter-7-Understand-Place-Value-to-1000-7.2-2

Show and Grow

Question 1.
Big Ideas Math Answers 2nd Grade Chapter 7 Understand Place Value to 1,000 7.2 2
______ hundreds, _______ tens, and ______ ones is ______.
Answer: 7 hundreds, 4 tens and 5 ones is 745
Explanation
There are 7 unit of cubes each unit of cubes contains 10 tens which gives 700(10 tens in one unit of cubes x 70 tens in the whole unit of cubes = 700)
and there are 4 rod each contains 10 cubes that gives 4 tens is 40 and plus the ones 5 is 745.

Apply and Grow: Practice

Question 2.
Big Ideas Math Answers 2nd Grade Chapter 7 Understand Place Value to 1,000 7.2 3
______ hundreds, _______ tens, and ______ ones is ______.
Answer: 4 hundreds, 6 tens and 3 ones is 463.
Explanation
There are 4 unit of cubes each unit of cubes contains 10 tens which gives 400(10 tens in one unit of cubes x 40 tens in the whole unit of cubes = 400)
and there are 6 rod each contains 10 cubes that gives 6 tens is 60 and plus the ones 3 is 463.

Question 3.
Big Ideas Math Answers 2nd Grade Chapter 7 Understand Place Value to 1,000 7.2 4
______ hundreds, _______ tens, and ______ ones is ______.
Answer: 6 hundreds, 0 tens and 9 ones is 609.
Explanation
There are 6 unit of cubes each unit of cubes contains 10 tens which gives 600(10 tens in one unit of cubes x 70 tens in the whole unit of cubes = 700)
The place value of 0 (0 x 10 tens = 0) and plus the ones 9 is 609.

Question 4.
Big Ideas Math Answers 2nd Grade Chapter 7 Understand Place Value to 1,000 7.2 5
______ hundreds, _______ tens, and ______ ones is ______.
Answer: 3 hundreds, 4 tens and 6 ones is 346.
Explanation
There are 3 unit of cubes each unit of cubes contains 10 tens which gives 300(10 tens in one unit of cubes x 30 tens in the whole unit of cubes = 300)
and there are 4 rod each contains 10 cubes that gives 4 tens is 40 and plus the ones 6 is 346.

Question 5.
DIG DEEPER!
What number is Newton thinking about?
Big Ideas Math Answers 2nd Grade Chapter 7 Understand Place Value to 1,000 7.2 6
Answer:58

Think and Grow: Modeling Real Life

You buy the markers shown. How many markers do you buy?
Big Ideas Math Answers 2nd Grade Chapter 7 Understand Place Value to 1,000 7.2 7
Write the missing numbers:
_______ hundreds, _______ tens, and ______ ones
_______ markers
Answer: 3 hundreds, 2 tens and 4 ones is 324 markers.
Explanation
The above images shows that 3 packs of 100 markers(100+100+100=300) 2 packs of 10 markers(10+10=20) with additional of 4 markers, that gives
(300+20+4=324)Totally 324 markers you buy.

Show and Grow

Question 6.
You buy the balloons shown. How many balloons do you buy?
Big Ideas Math Answers 2nd Grade Chapter 7 Understand Place Value to 1,000 7.2 8
______ balloons
Answer:  570 balloons
Explanation
The above images shows that 5 packs of 100 balloons(100+100+100+100+100= 500) 7 packs of 10 balloons(10+10+10+10+10+10+10=70),
that gives ( 500+70=570)Totally 570 balloons you buy.

Model Numbers to 1,000 Homework & Practice 7.2

Question 1.
Big Ideas Math Answers 2nd Grade Chapter 7 Understand Place Value to 1,000 7.2 9
______ hundreds, _______ tens, and ______ ones is ______.
Answer: 4 hundreds,7 tens 2 ones is 472
Explanation
There are 4 unit of cubes each unit of cubes contains 10 tens which gives 400(10 tens x 40 tens of whole cubes =400)
There are 7 rod gives you a 7 tens is 70 and plus 2 ones is 472.

Question 2.
Big Ideas Math Answers 2nd Grade Chapter 7 Understand Place Value to 1,000 7.2 10
______ hundreds, _______ tens, and ______ ones is ______.
Answer: 7 hundreds 4 tens 0 ones is 740.
Explanation
There are 7 unit of cubes each unit of cubes contains 10 tens which gives 700(10 tens in one unit of cubes x 70 tens in the whole unit of cubes = 700)
and there are 4 rod each contains 10 cubes that gives 4 tens is 40 and plus the ones place above picture there were no ones 0 is 740

Question 3.
Big Ideas Math Answers 2nd Grade Chapter 7 Understand Place Value to 1,000 7.2 11
______ hundreds, _______ tens, and ______ ones is ______.
Answer: 2 hundreds 8 tens and 6 ones is 286.
Explanation
There are 2 unit of cubes each unit of cubes contains 10 tens which gives you 200 (10 tens in one unit of cubes x 20 tens in the whole unit of cubes = 200) and there are 8 rod each contains 10 cubes that gives 8 tens is 80 and plus the ones 6 is 286.

Question 4.
YOU BE THE TEACHER
Your friend writes 8 hundreds and 3 ones as 83. Is your friend correct? Explain.
_________________________
_________________________
Answer:  No, 8 is tens and 3 is ones as 83.
Explanation
The first digit is the tens place it tells you that there are 8 tens in the number 83. The next digit is the ones place which is 3.
Therefore there are 8 sets of 10 tens and plus 3 ones in the number 83.

Question 5.
Modeling Real Life
You buy the stickers shown. How many stickers do you buy?
Big Ideas Math Answers 2nd Grade Chapter 7 Understand Place Value to 1,000 7.2 12
______ stickers
Answer: 620 stickers
Explanation
Given that, There are 6 packs of stickers each contains 100 stickers that is (100+100+100+100+100+100=600
and 2 packs of stickers each contains 10 stickers that is (10+10= 20) Therefore 600 +20= 620
Thus, 620 sticker you buy.

Review & Refresh

Question 6.
Big Ideas Math Answers 2nd Grade Chapter 7 Understand Place Value to 1,000 7.2 13
Answer: 82

Question 7.
Big Ideas Math Answers 2nd Grade Chapter 7 Understand Place Value to 1,000 7.2 14
Answer: 83

Question 8.
Big Ideas Math Answers 2nd Grade Chapter 7 Understand Place Value to 1,000 7.2 15
Answer: 71

Lesson 7.3 Understand Place Value

Explore and Grow

Make quick sketches to model each number.
Big Ideas Math Answers Grade 2 Chapter 7 Understand Place Value to 1,000 7.3 1
What do you notice about the numbers?
_________________________
_________________________
Answer:
Big-Ideas-Math-Answers-Grade-2-Chapter-7-Understand-Place-Value-to-1000-7.3-1

Show and Grow

Circle the value of the underlined digit.
Question 1.
483
300
30
3
Answer: 3
Explanation
Given that there are 3 digit number which is 483. A number can have many digits and each
digit has a special place and value. the 8 is in ones place and its place value is 8.

Question 2.
791
9 hundreds
9 tens
9 ones
Answer: 9 tens
Explanation
Given that there are 3 digit number which is 791. A number can have many digits and each
digit has a special place and value. the second digit at tens place. The 9 is in tens place value is 90.
Question 3.
612
6
60
600
Answer: 600
Explanation
Given that there are 3 digit number which is 612. A number can have many digits and each
digit has a special place and value. the first digit at hundreds place. The 6 is in hundreds place value is 600.

Question 4.
578
7 hundreds
7 tens
7 ones
Answer: 7 tens
Explanation
Given that there are 3 digit number which is 578. A number can have many digits and each
digit has a special place and value. the second digit at tens place. The 7 is in tens place value is 70.

Apply and Grow: Practice

Circle the value of the underlined digit.
Question 5.
354
300
30
3
Answer: 300
Explanation
Given that there are 3 digit number which is 354. A number can have many digits and each
digit has a special place and value. the first digit at hundreds place. The 3 is in hundreds place value is 300.

Question 6.
726
2 hundreds
2 tens
2 ones
Answer: 2 tens
Explanation
Given that there are 3 digit number which is 726. A number can have many digits and each
digit has a special place and value. the second digit at tens place. The 2 is in tens place value is 20.

Question 7.
594
4
40
400
Answer: 4
Explanation
Given that there are 3 digit number which is 594. A number can have many digits and each
digit has a special place and value. the 4 is in ones place and its place value is 4.

Question 8.
475
5 hundreds
50
5 ones
Answer: 5 ones
Explanation
Given that there are 3 digit number which is 475. A number can have many digits and each
digit has a special place and value. the 5 is in ones place and its place value is 5.

Circle the values of the underlined digit.
Question 9.
639
3 tens
300
30
Answer: 3 tens
Explanation
Given that there are 3 digit number which is 639. A number can have many digits and each
digit has a special place and value. the second digit at tens place. The 3 is in tens place value is 30.

Question 10.
872
8 hundreds
80
800
Answer: 8 hundreds
Explanation
Given that there are 3 digit number which is 872. A number can have many digits and each
digit has a special place and value. the first digit at hundreds place. The 8 is in hundreds place value is 800.

Question 11.
Number Sense
Write the number that has the following values.
The tens digit has a value of 40.
The ones digit has a value of 2.
The hundreds digit has a value of 600. ______
Answer: 642
Explanation
The combination of hundreds, tens, ones of 600, 40, 2 respectively.

Think and Grow: Modeling Real Life

How many points is one ball worth in each bucket?
Big Ideas Math Answers Grade 2 Chapter 7 Understand Place Value to 1,000 7.3 2
Write the score:
______ hundreds, ______ tens, and _______ ones
Blue bucket: ______ points
Yellow bucket: _____ point
Red bucket: ______ points
Answer: Blue bucket 3 points,yellow bucket 4 points and red bucket 5 points

Show and Grow

Question 12.
How many points is one ring worth on each peg?
Big Ideas Math Answers Grade 2 Chapter 7 Understand Place Value to 1,000 7.3 3
Green peg: ______ points
Blue peg: _______ point
Purple peg: ______ points
Answer:
Green peg: 1 points
Blue peg: 3 point
Purple peg: 2 points

Understand Place Value Homework & Practice 7.3

Circle the value of the underlined digit.
Question 1.
523
500
50
5
Answer: 500
Explanation
Given that there are 3 digit number which is 523. A number can have many digits and each
digit has a special place and value. the first digit at hundreds place. The 5 is in hundreds place value is 500.

Question 2.
738
8 hundreds
8 ones
8 tens
Answer: 8 ones
Explanation
Given that there are 3 digit number which is 738. A number can have many digits and each
digit has a special place and value. the 8 is in ones place and its place value is 8.

Question 3.
364
60
6 ones
6 hundreds
Answer: 60
Explanation
Given that there are 3 digit number which is 364. A number can have many digits and each
digit has a special place and value. the second digit at tens place. The 6 is in tens place value is 60.

Circle the values of the underlined digit.
Question 4.
434
4
4 ones
4 hundreds
Answer: 4 ones
Explanation
Given that there are 3 digit number which is 434. A number can have many digits and each
digit has a special place and value. the 4 is in ones place and its place value is 4.

Question 5.
920
2
2 tens
20
Answer: 2 tens
Explanation
Given that there are 3 digit number which is 920. A number can have many digits and each
digit has a special place and value. the second digit at tens place. The 2 is in tens place value is 20.

Question 6.
DIG DEEPER!
Write the number that matches the clues.
The value of the hundreds digit is 800.
The value of the tens digit is 10 less than 70.
The value of the ones digit is an even number greater than 7.
_________
Answer:
i. 8
80
8

Question 7.
Structure
Write each number in the correct circle.
Big Ideas Math Answers Grade 2 Chapter 7 Understand Place Value to 1,000 7.3 4
Answer:
Big-Ideas-Math-Answers-Grade-2-Chapter-7-Understand-Place-Value-to-1000-7.3-4

Question 8.
Modeling Real Life
How many points is one ball worth in each hoop?
Big Ideas Math Answers Grade 2 Chapter 7 Understand Place Value to 1,000 7.3 5
Blue hoop: ______ points
Orange hoop: _______ points
Red hoop: _______ point
Answer:
Blue hoop: 4 points
Orange hoop: 2 points
Red hoop: 5 points

Review & Refresh

Question 9.
Big Ideas Math Answers Grade 2 Chapter 7 Understand Place Value to 1,000 7.3 6
Answer: 19

Question 10.
Big Ideas Math Answers Grade 2 Chapter 7 Understand Place Value to 1,000 7.3 7
Answer: 42

Question 11.
Big Ideas Math Answers Grade 2 Chapter 7 Understand Place Value to 1,000 7.3 8
Answer: 38

Lesson 7.4 Write Three-Digit Numbers

Explore and Grow

Identify the value of the base ten blocks.
Big Ideas Math Solutions Grade 2 Chapter 7 Understand Place Value to 1,000 7.4 1
What is the total value of the base ten blocks? ________

How can you write the value of the base ten blocks as an equation?
Big Ideas Math Solutions Grade 2 Chapter 7 Understand Place Value to 1,000 7.4 2
Answer:
Standard Form is 246
Expanded Form is 200 + 40 + 6 = 246
Word Form is Two Hundred and Forty Six

Show and Grow

Write the number in standard form, expanded form, and word form.
Question 1.
Big Ideas Math Solutions Grade 2 Chapter 7 Understand Place Value to 1,000 7.4 3
Answer: Standard form 528,
Expanded form 500+20+8,
Word form Five Hundred Twenty Eight.

Question 2.
Big Ideas Math Solutions Grade 2 Chapter 7 Understand Place Value to 1,000 7.4 4
Answer: Standard form 709,
Expanded form 700+9,
Word form Seven Hundred Nine.

Apply and Grow: Practice 

Write the number in expanded form and word form.
Question 3.
837
_____ + _____ + _____
______________
Answer: 800+30+7 and Eight hundred thirty seven
Explanation
In expanded form, write the number by showing the value of each digit (800+30+7)
In Word form write the number by using words but no numbers (Eight hundred thirty seven)
Question 4.
954
_____ + _____ + ______
_____________
Answer:900+50+4 and Nine hundred fifty four
Explanation
In expanded form, write the number by showing the value of each digit (900+50+4)
In Word form write the number by using words but no numbers ( Nine hundred fifty four)
Write the number in standard form and word form.
Question 5.
500 + 60
______
__________
Answer: 560 and Five hundred sixty
Explanation
In standard form, numbers are written using only numbers. There are no words (560)
In Word form write the number by using words but no numbers ( Nine hundred fifty four)
Question 6.
700 + 20 + 1
_____
____________
Answer:721 and Seven hundred twenty one
Explanation
In standard form, numbers are written using only numbers. There are no words (721)
In Word form write the number by using words but no numbers ( Seven hundred twenty one)

Write the number in expanded form and standard form.
Question 7.
six hundred seventy-four
_____ + ______ + ______
__________
Answer: 600+70+4 and 674
Explanation
In expanded form, write the number by showing the value of each digit (600+70+4)
In standard form, numbers are written using only numbers. There are no words (674)
Question 8.
four hundred seven
_____ + ______ + ______
__________
Answer: 400+7 and 407
Explanation
In expanded form, write the number by showing the value of each digit (400+7)
In standard form, numbers are written using only numbers. There are no words (407)
Question 9.
Structure
Which number did Newton model?
Big Ideas Math Solutions Grade 2 Chapter 7 Understand Place Value to 1,000 7.4 5

Think and Grow: Modeling Real Life

There are 819 pets in a pet store. 800 are fish. 9 are cats. The rest are birds. How many birds are there?
Big Ideas Math Solutions Grade 2 Chapter 7 Understand Place Value to 1,000 7.4 6
Expanded form:
_____ + _____ + ______
_____ birds
Answer: 10 birds are there and 800+10+9
Explanation
Given that, There are 819 pets in pets store
800 are fish, 9 are cats and rest are birds.
Thus,819 the right the first digit will be at ones place 9 is in ones place and its place value is 9 ,the second digit at tens place 1 is in tens place its place value is 10 and the third digit at hundreds place.8 is in hundreds place its place value is 800.

Show and Grow

Question 10.
There are 21
7 flowers at a flower stand. 200 are roses. 7 are sunflowers. The rest are tulips. How many tulips are there?
Big Ideas Math Solutions Grade 2 Chapter 7 Understand Place Value to 1,000 7.4 7
______ tulips
Answer: 10 are tulips in flower stand.
Explanation
Given that, There are 217 flowers in flowers stand
200 are roses, 7 are sunflowers and rest are tulips (200 – 7 = 10)
Thus,10 are the tulips.
Question 11.
There are 185 books at a book fair. 80 are chapter books. 5 are comic books. The rest are picture books. How many picture books are there?
Big Ideas Math Solutions Grade 2 Chapter 7 Understand Place Value to 1,000 7.4 8
______ picture books
Answer: 100 books are picture books.
Explanation
Given that, 185 books are at book fair.
80 books are chapter books and 5 are comic books and rest books are picture books
(80 – 5 = 100)
Thus, 100 books are picture books in book fair.
Question 12.
You sell 326 candles. You sell 6 small candles. You sell 20 medium candles. The rest are large candles. How many large candles did you sell?
Big Ideas Math Solutions Grade 2 Chapter 7 Understand Place Value to 1,000 7.4 9
______ large candles
Answer: 300 are large candles
Explanation:
Given that, 326 candles are sell by you,
6 are small candles and 20 are medium candles and rest are large candles
326-6(small candles)=320
320-20(medium candles)=300
Thus,300 large candles are you sell.
Three-Digit Numbers Homework & Practice 7.4

Write the number in expanded form and word form.
Question 1.
137
______ + _____ + _____
____________
Answer: 100+30+7 and One hundred thirty seven.
Explanation:
In expanded form, write the number by showing the value of each digit (100+30+7)
In Word form write the number by using words but no numbers ( One hundred twenty four)
Question 2.
280
_____ + _____ + ______
___________
Answer: 200+80 and Two hundred eighty
Explanation
In expanded form, write the number by showing the value of each digit (200+80)
In Word form write the number by using words but no numbers ( Two hundred eighty )
Write the number in standard form and word form.
Question 3.
600 + 10 + 5 ______
___________
Answer:  615 and Six hundred fifteen
Explanation
In standard form, numbers are written using only numbers. There are no words (615)
In Word form write the number by using words but no numbers ( Six hundred fifteen )
Question 4.
900 + 70 + 6 _____
___________
Answer: 976 and Nine hundred seventy six
In standard form, numbers are written using only numbers. There are no words (976)
In Word form write the number by using words but no numbers ( Nine hundred seventy six )
Write the number in expanded form and standard form.
Question 5.
three hundred nine
_____ + ______ + _____
______
Answer: 300+9 and 309
Explanation
In expanded form, write the number by showing the value of each digit (300+9)
In standard form, numbers are written using only numbers. There are no words (309)
Question 6.
eight hundred sixty-two
_____ + ______ + ______
_______
Answer: 800+60+2 and 862
Explanation
In expanded form, write the number by showing the value of each digit (800+60+2)
In standard form, numbers are written using only numbers. There are no words (862))
Question 7.
Which One Doesn’t Belong?
Which does not belong with the other three?
Big Ideas Math Solutions Grade 2 Chapter 7 Understand Place Value to 1,000 7.4 10
Answer: 1 + 9 + 6 does not belong to the other three.

Question 8.
DIG DEEPER!
A number has 7 hundreds. The tens digit is 5 less than the hundreds digit. The ones digit is 2 more than the hundreds digit. What is the number?
______
Answer: 729
Explanation
given that, a number has 7 hundred
The tens digit is 5 less than hundreds digit

Question 9.
Modeling Real Life
There are 438 vegetables planted. 400 are carrots. 8 are beets. The rest are onions. How many onions are there?
Big Ideas Math Solutions Grade 2 Chapter 7 Understand Place Value to 1,000 7.4 11
______ onions
Answer: 30 onions are there
Explanation
As per the statement, There are 438 vegetables are planted
400 are carrot (438-400=38)
8 are beets (38 – 8=30)
Thus, the Remaining 30 are onions.

Question 10.
Modeling Real Life
There are 593 students in after-school programs. 3 students take dance class. 90 students take art class. The rest take karate class. How many students take karate class?
Big Ideas Math Solutions Grade 2 Chapter 7 Understand Place Value to 1,000 7.4 12
______ students
Answer: 500 students are in karate class.
Explanation:
As per statements, Totally there are 593 students
3 students are in dance class (593-3=590)
90 students are in art class (590-90=500)
Thus, the Remaining 500 students are in karate class.

Review & Refresh

Question 11.
Find the sum. Then change the order of the addends. Write the new equation.
4 + 7 = ______
_____ + _____ = ______
Answer: 11

Lesson 7.5 Represent Numbers in Different Ways

Explore and Grow

Circle the models that show each number.
Big Ideas Math Answer Key Grade 2 Chapter 7 Understand Place Value to 1,000 7.5 1
Answer:

Show and Grow

Question 1.
Show 261 two ways.
Big Ideas Math Answer Key Grade 2 Chapter 7 Understand Place Value to 1,000 7.5 2
Answer:

Question 2.
Show 345 two ways.
Big Ideas Math Answer Key Grade 2 Chapter 7 Understand Place Value to 1,000 7.5 3
Answer:

Apply and Grow: Practice

Question 3.
Show 432 two ways.
Big Ideas Math Answer Key Grade 2 Chapter 7 Understand Place Value to 1,000 7.5 4
Answer:

Question 4.
Show 527 two ways.
Big Ideas Math Answer Key Grade 2 Chapter 7 Understand Place Value to 1,000 7.5 5
Answer:

Question 5.
YOU BE THE TEACHER
Your friend says that 800 + 40 + 11 is the same as 800 + 50 +1. Is your friend correct? Explain.
_________________________
__________________________
Answer:

Think and Grow: Modeling Real Life

The models show how many dinosaur toys you and your friend have. Does your friend have the same number of dinosaur toys as you? Explain.
Big Ideas Math Answer Key Grade 2 Chapter 7 Understand Place Value to 1,000 7.5 6
Big Ideas Math Answer Key Grade 2 Chapter 7 Understand Place Value to 1,000 7.5 7
_________________________
__________________________
Answer:

Show and Grow

Question 6.
The models show how many trading cards you and your friend have. Does your friend have the same number of trading cards as you? Explain.
Big Ideas Math Answer Key Grade 2 Chapter 7 Understand Place Value to 1,000 7.5 8
______________________
______________________
Answer:

Represent Numbers in Different Ways Homework & Practice 7.5

Question 1.
Show 134 two ways.
Big Ideas Math Answer Key Grade 2 Chapter 7 Understand Place Value to 1,000 7.5 9
Answer:

Question 2.
Show 319 two ways.
Big Ideas Math Answer Key Grade 2 Chapter 7 Understand Place Value to 1,000 7.5 10
Answer:

Question 3.
Number Sense
Which ways show 948?
8 hundreds, 4 tens, and 8 ones
9 hundreds, 4 tens, and 8 ones
8 hundreds, 4 tens, and 9 ones
8 hundreds, 14 tens, and 9 ones
94 tens and 8 ones
948 ones
Answer:

Question 4.
Modeling Real Life
The models show how many bouncy balls you and your friend have. Does your friend have the same number of bouncy balls as you? Explain.
Big Ideas Math Answer Key Grade 2 Chapter 7 Understand Place Value to 1,000 7.5 11
________________________
__________________________
Answer:

Review & Refresh

Question 5.
70 + 30 = ______
Answer: 100

Question 6.
53 + 19 = _____
Answer: 72

Question 7.
90 − 50 = _____
Answer: 40

Question 8.
64 − 40 = _______
Answer: 24

Understand Place Value to 1,000 Performance Task

You make trail mix using the sunflower seeds, almonds, and raisins shown.
Big Ideas Math Answers 2nd Grade Chapter 7 Understand Place Value to 1,000 1
Question 1.
How many sunflower seeds, almonds, and raisins do you use?
______ + _____ + _____
_____
Answer: 400+ 30+ 5 is equal to 435.
Explanation
Given that, 400 sunflower seeds, 30 almonds and 5 raisins  for trail mix
thus, 400+30+5= 435

Question 2.
You add another bag of almonds and another bag of raisins to your trail mix. How many sunflower seeds, almonds, and raisins are there now?
_____ + _____ + _____
______
Answer: 400+40+10= 450
Explanation
Given that, 400 sunflower seeds, 30 of almonds add another bags of almonds is contains 10 totally 40 and 5 of raisins add another bags of raisins contains 5 totally 10 raisins.
400+40+10=450

Question 3.
Your friend makes trail mix with 3 bags of sunflower seeds,13 bags of almonds, and 4 bags of raisins. Does your friend use the same number of ingredients as you? Explain.
Yes No
__________________
___________________
Answer: No, friend makes a trail mix with only 3 bags of sunflower 13 bags of almonds and 4 bag of raisins. The ingredients you used is 400 sunflower seeds,40 almonds bags and 10 raisins bags. Friend and your’s are not the same ingredients.

Understand Place Value to 1,000 Activity

Naming Numbers Flip and Find
To Play: Place the Naming Numbers Flip and Find Cards face down in the boxes. Take turns flipping 2 cards. If your cards show the same number, keep the cards. If your cards show different numbers, flip the cards back over. Play until all matches are made.
Big Ideas Math Answers 2nd Grade Chapter 7 Understand Place Value to 1,000 2

7.1 Hundreds

Write how many tens. Circle groups of 10 tens. Write how many hundreds. Then write the number.
Question 1.
Big Ideas Math Answers 2nd Grade Chapter 7 Understand Place Value to 1,000 chp 1
_____ tens
_______ hundreds
________
Answer: 40 tens and 4 hundreds
Explanation
Given that 40 tens are there by circling 10 tens
For counting tens, count columns and for counting hundreds count all the boxes.

Question 2.

Big Ideas Math Answers 2nd Grade Chapter 7 Understand Place Value to 1,000 chp 2
_____ tens
_______ hundreds
________
Answer:  50 tens and 5 hundred
Explanation
Given that 50 tens are there by circling 10 tens
For counting tens, count columns and for counting hundreds count all the boxes.

Question 3.

Modeling Real Life
A cafeteria has 80 bags of pretzels. Each bag has 10 pretzels. The cafeteria needs 900 pretzels. How many more bags does the cafeteria need?
______ more bags
Answer: 10 bags

Explanation:
Given that, the cafeteria has 80 bags of pretzels  each bags contains 10 pretzels, cafeteria needs 900 pretzels,
10 bags x 10 pretzels =100
Thus 10 bags of pretzels needs for the cafeteria.

7.2 Model Numbers to 1,000

Question 4.
Big Ideas Math Answers 2nd Grade Chapter 7 Understand Place Value to 1,000 chp 4
______ hundreds, _____ tens, and _____ ones is ______.
Answer:5 hundreds 2 tens, 5 ones is 525.
Explanation
There are 5 unit of cubes each unit of cubes contains 10 tens which gives 500(10 tens in one unit of cubes x 50 tens in the whole unit of cubes = 500)
and there are 2 rod each contains 10 cubes that gives 2 tens is 20 and plus the ones 5 is 525.

Question 5.
Big Ideas Math Answers 2nd Grade Chapter 7 Understand Place Value to 1,000 chp 5
______ hundreds, _____ tens, and _____ ones is ______.
Answer: 8 hundred,0 tens, and 3 ones is 803.
Explanation
There are 8 unit of cubes each unit of cubes contains 10 tens which gives 800(10 tens in one unit of cubes x 80 tens in the whole unit of cubes = 800)
the second digit at tens place 0 in the tens place and the right first digit is ones plus the ones 3 is 803.

Question 6.
Number Sense
What number is Descartes thinking about?
Big Ideas Math Answers 2nd Grade Chapter 7 Understand Place Value to 1,000 chp 6
______
Answer:425
Explanation
The above answer represents that 4 hundreds is 400 and the second digit at tens place. The 2 is in tens place value is
20 and the right first digit is in ones place the 5 is in ones place value 5 is 425.

7.3 Understand Place Value

Circle the value of the underlined digit
Question 7.
429
200
2
2 tens
Answer: 2 tens
Explanation
Given that there is 3 digit number which is 429. A number can have many digits and each
digit has a special place and value. the second digit at tens place. The 2 is in tens place value is 20.
Question 8.
751
70
700
7 tens
Answer: 700
Explanation
Given that there is 3 digit number which is 751. A number can have many digits and each
digit has a special place and value. the first digit at hundreds place. The 7 is in hundreds place value is 700

Question 9.
Number Sense
Write the number that has the following values.
The tens digit has a value of 60.
The ones digit has a value of 3.
The hundreds digit has a value of 900. _______
Answer:963
Explanation:
A number can have many digits and each digit has a special place and value. the first digit at hundreds place.
The 9 is in hundreds place value is 900. The second digit at tens place the 6 is in tens place value is 60 and
the right first digit is in ones place the 3 is in ones place value 3 is 963.

7.4 Write Three-Digit Numbers

Write the number in expanded form and word form.
Question 10.
605
_____ + _____ + ______
________
Answer: 600+5 and Six hundred five
Explanation
In expanded form, write the number by showing the value of each digit (600+05)
In Word form write the number by using words but no numbers ( Six hundred five)

Question 11.
541
______ + ______ + ______

___________
Answer: 500+40+1 and Five hundred forty-one
Explanation
In expanded form, write the number by showing the value of each digit (500+40+1)
In Word form write the number by using words but no numbers (Five hundred forty-one)

Write the number in standard form and word form.
Question 12.
100 + 20 + 4
______
___________
Answer: 124 and One hundred twenty-four
Explanation
In standard form, numbers are written using only numbers. There are no words (124)
In Word form write the number by using words but no numbers (One hundred twenty-four)

Question 13.
700 + 8
_______
_____________
Answer: 708 and Seven hundred eight
Explanation
In standard form, numbers are written using only numbers. There are no words (708)
In Word form write the number by using words but no numbers (Seven hundred eight)

Write the number in expanded form and standard form.
Question 14.
three hundred thirty
_____ + _____ + _____
_____
Answer: 300+30 and 330
Explanation
In expanded form, write the number by showing the value of each digit (300+30)
In standard form, numbers are written using only numbers. There are no words. (330)
Question 15.
two hundred fifty-six
_____ + ______ + _____
______
Answer: 200+50+6 and 256
Explanation
In expanded form, write the number by showing the value of each digit (200+50+6)
In standard form, numbers are written using only numbers. There are no words. (256)
7.5 Represent Numbers in Different Ways

Question 16.
Show 345 two ways.
Big Ideas Math Answers 2nd Grade Chapter 7 Understand Place Value to 1,000 chp 16
Answer: Three hundred forty-five and 300+40+5
Explanation
There are 3 digit number(345)  1 digit is for hundreds which is 3, 2nd digit is for tens which is 2 and the last digit for once 5 is 345
In word form for 345 is Three hundreds forty-five,
The expanded form for 345 is 300+40+5 which gives 345.

Question 17.
Show 562 two ways.
Big Ideas Math Answers 2nd Grade Chapter 7 Understand Place Value to 1,000 chp 17
Answer: Five hundred sixty-two and 500+60+2
Explanation
There are 3 digit numbers 1 digit number is hundreds which are 5, 2nd digit number is tens which is 6 tens and the last number 2 ones is 562
In word form for 562 is Five hundred sixty-two,
The expanded form for 562 is 500+60+2 is 562.

Question 18.
Your friend says that 600 + 30 + 1 is the same as 500 + 130 + 1. Is your friend correct? Explain.
___________________
___________________
Answer: Yes correct, given that 600+30+1= 631 ( which gives) and 500+130+1 = 631( which also gives the same answer)
both equation shows the same answers.

Conclusion:

In this chapter, a brief explanation of the lessons are discussed in Answer Key of Big Ideas Math book Grade 2 Chapter 7 Understand Place Value to 1,000. Here we have provided the exercise problems and the solutions to help in practising the lessons. you can get the different and simple methods of solving problems in Big Ideas Math Grade 2 Answer Key Chapter 7 Understand Place Value to 1,000. Stay in touch with our site to get the latest edition solutions of all other Big Ideas Math Grade 2 Answers.

Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area

Big Ideas Math Answers Grade 3 Chapter 15

Find Big Ideas Math Grade 3 Chapter 15 Find Perimeter and Area Questions with Solutions here. The students can get the answer along with the detailed explanation in the following sections. This BIG Math Answers Grade 3 Chapter 15 Find Perimeter and Area is helpful to complete the homework in time. Hence, download Big Ideas Math Book Grade 3 Chapter 15 Find Perimeter and Area Solutions pdf for free of cost.

Big Ideas Math Book 3rd Grade 15th Chapter Find Perimeter and Area Answer Key

Students have to prepare all the topics of Big Ideas Math Answers 3rd Grade 15th Chapter Find Perimeter and Area to score good marks in the exam. The different topics included in BIM Book Grade 3 Chapter 15 Find Perimeter and Area Answers are Understand Perimeter, Find Perimeters of Polygons, Find Unknown Side Lengths, Same Perimeter, Different Areas, and Same Area, Different Perimeters. After preparing these topics, check your skills at the performance task provided at the end.

The success criteria of Find Perimeter and Area is to understand the area and perimeter of polygons. The topic wise quick links to access Big Ideas Math Book Grade 3 Chapter 15 Find Perimeter and Area Answer Key is provided below. So, just tap on them and get the solutions easily.

Lesson 1 Understand Perimeter

Lesson 2 Find Perimeters of Polygons

Lesson 3 Find Unknown Side Lengths

Lesson 4 Same Perimeter, Different Areas

Lesson 5 Same Area, Different Perimeters

Performance Task

Lesson 15.1 Understand Perimeter

Explore and Grow

Question 1.
Model a rectangle on your geoboard. Draw the rectangle and label its side lengths. What is the distance around the rectangle?
Big Ideas Math Answer Key Grade 3 Chapter 15 Find Perimeter and Area 1
Answer:
The units around the rectangle are 15 units.

Explanation:
Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area
In the above figure, the length of the rectangle is 4 units, and
the breadth of the rectangle is 3 units.
The units around the rectangle are 15 units.

Structure
Change the side lengths of the rectangle on your geoboard. What do you notice about the distance around your rectangle compared to the distance around the rectangle above? Explain

Answer:
The distance around the rectangle is 15 units

Explanation:
Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area
In the above figure, we can see the length of the rectangle is 3 units, and
the breadth of the rectangle is 4 units.
There is no change in the distance around the rectangle which is 15 units.

Think and Grow: Understand Perimeter
Perimeter is the distance around a figure. You can measure perimeter using standard units, such as inches, feet, centimeters, and meters.
Example
Find the perimeter of the rectangle.
Choose a unit to begin counting. Count each unit around the rectangle.
Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area 2
Answer:
The perimeter of the rectangle is 20 in.

Explanation:
To find the perimeter of the triangle, we need to know the length and the breadth of the rectangle.
Here, the length of the rectangle is 7 in,
and the breadth is 3 in,
So the perimeter of the rectangle is 2(Length + Breadth)
= 2(7+3)
= 2(10)
= 20 in.

Show and Grow

Question 1.
Find the perimeter of the figure.
Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area 4
There are ___ units around the figure.
Answer:
There are 22 units around the figure.

Explanation:
To find the perimeter of the rectangle, we need to know the length and the breadth of the rectangle.
Here, the length of the rectangle is 6 m,
and the breadth is 5 m,
So the perimeter of the rectangle is 2(Length + Breadth)
= 2(6+5)
= 2(11)
= 22 m.

Question 2.
Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area 5
There are ___ units around the figure.
So, the perimeter is ____ feet.
Answer:
The perimeter of the figure is 24 ft.

Explanation:
To find the perimeter of the figure, we will add all the sides of the figure
So the sides of the figure are 5,3,2,4,3,7.
and the perimeter of the figure is 5+3+2+4+3+7
= 24 ft.

Question 3.
Draw the figure that has a perimeter of 16 centimeters.
Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area 6
Answer:
The sides of the figure are 4 cm.

Explanation:
As the perimeter of the figure is 16 cm,
So let the sides of the figure be 4 cm,
by that, we can get the perimeter as 16 cm, i.e
4+4+4+4= 16 cm.
Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area

Apply and Grow: Practice

Question 4.
Find the perimeter of the figure.
Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area 7
There are ___ units around the figure.
So, the perimeter is ___
Answer:
There are 20 units around the figure.
So, the perimeter is 20 cm.

Explanation:
To find the perimeter of the figure, we will add all the sides of the figure
So the sides of the figure are 3,3,2,2,5,5.
and the perimeter of the figure is 3+3+2+2+5+5
= 20 units,
So the perimeter is 20 cm.

Question 5.
Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area 8
There are ___ units around the figure.
So, the perimeter is ___
Answer:
There are 26 units around the figure.
So, the perimeter is 26 ft.

Explanation:
To find the perimeter of the figure, we will add all the sides of the figure
So the sides of the figure is 5,5,2,3,1,3,2,5
and the perimeter of the figure is 5+5+2+3+1+3+2+5
= 26 units,
So the perimeter is 26 ft.

Question 6.
Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area 9
Perimeter = ___
Answer:
The perimeter of the figure is 22 in.

Explanation:
To find the perimeter of the figure, we will add all the sides of the figure
So the sides of the figure is 4,3,2,3,3,3,1,3
and the perimeter of the figure is 4+3+2+3+3+3+1+3
= 22 units,
So the perimeter is 22 in.

Question 7.
Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area 10
Perimeter = ___
Answer:
The perimeter is 24 m.

Explanation:
To find the perimeter of the figure, we will add all the sides of the figure
So the sides of the figure is 5,7,2,2,1,2,1,1,1,2
and the perimeter of the figure is 5+7+2+2+1+2+1+1+1+2
= 24 units,
So the perimeter is 24 m.

Question 8.
Draw a figure that has a perimeter likely measurement for the of 14 centimeters.
Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area 11
Answer:
The length of the rectangle is 2 cm, and
the breadth is 5 cm.

Explanation:
Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area
In the above figure, we can see the length of the rectangle is 2 cm
and the breadth of the rectangle is 5 cm
perimeter = 2(length + breadth)
= 2(5+2)
= 2(7)
= 14 cm.
so the perimeter of the rectangle is 14 cm.

Question 9.
Precision Which is the most likely measurement for the perimeter of a photo?
20 inches
100 meters
5 centimeters
2 inches
Answer:
2 inches.

Explanation:
The most likely measurement for the perimeter of a photo is 2 inches.

Question 10.
You be the teacher Your friend counts the units around the figure and says the perimeter is 12 units. Is your friend correct? Explain.
Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area 12
Answer:
No, he is not correct.

Explanation:
No, he is not correct. As there are 2+2+1+1+4+1+1+2= 14 units, but not 12 units. So he is not correct.

Think and Grow: Modeling Real Life
Use a centimeter ruler to find the perimeter of the bookmark.
Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area 13
The perimeter is ___
Answer:
The perimeter of the bookmark is 26 cm.

Explanation:
On measuring, the length of the sides of the bookmark are 4 cm, 9 cm, 2 cm, 2 cm, 9 cm
to find the perimeter of the bookmark, we will add all the length of the sides
so the perimeter of the bookmark is
p = 4 cm+ 9 cm+ 2 cm+ 2 cm+ 9 cm
= 26 cm.
The perimeter of the bookmark is 26 cm.

Show and Grow

Question 11.
Use an inch ruler to find the perimeter of the decal.
Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area 14
Answer:
The perimeter of the decal is 17 in.

Explanation:
On measuring, the length of the sides of the decal is 5in, 2in, 5in, 5in
to find the perimeter of the decal, we will add all the length of the sides
so the perimeter of the decal is
p= 5in+2in+5in+5in
= 17 in
The perimeter of the decal is 17 in.

Question 12.
How much greater is the perimeter of your friend’s desk than the perimeter of your desk?
Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area 15
Answer:
Friend’s perimeter is 22-16= 6 times greater.

Explanation:
As we can see figure 1 is a square and the sides of the square are 4 ft.
So the perimeter of the square is a+a+a+a,
= 4+4+4+4
= 16 ft.
And now let’s find the perimeter of the friend’s figure,
So the sides of the figure is 2,6,5,2,3,4
and the perimeter is 2+6+5+2+3+4= 22 ft.
By this, we can see that the friend’s figure has a greater perimeter,
and friend’s perimeter is 22-16= 6 times greater.

Question 13.
DIG DEEPER!
Explain how you might use a centimeter ruler and string to estimate the perimeter of the photo of the window.
Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area 16
Answer:
Here, we will use the ruler to find the length of the bottom part of the window and the sides which are straight we can find the length using the ruler. And the curve sides we will measure using the string.

Understand Perimeter Homework & Practice 15.1

Question 1.
Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area 17
There are ___ units around the figure.
So, the perimeter is ___ inches.
Answer:
There are 20 units around the figure,
So the perimeter is 20 in.

Explanation:
To find the perimeter of the figure, we will add all the sides of the figure,
As we can see the above image is a square,
and the perimeter of the square is s+s+s+s
= 5+5+5+5
= 20 in.
As there are 20 units around the figure,
So the perimeter is 20 in.

Question 2.
Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area 18
There are ___ units around the figure.
So, the perimeter is ___ feet.

Answer:
There are 32 units around the figure,
So the perimeter is 32 ft.

Explanation:
To find the perimeter of the figure, we will add all the sides of the figure,
The sides of the above figure is 2,3,4,3,2,5,8,5
and the perimeter of the figure is 2+3+4+3+2+5+8+5
= 32 units.
As there are 32 units around the figure,
So the perimeter is 32 ft.

Question 3.
Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area 19
Perimeter = ___
Answer:
There are 26 units around the figure,
So the perimeter is 26 cm.

Explanation:
To find the perimeter of the figure, we will add all the sides of the figure,
The sides of the above figure is 2,2,2,3,1,3,5,8
and the perimeter of the figure is 2+2+2+3+1+3+5+8
= 26 units.
As there are 26 units around the figure,
So the perimeter is 26 cm.

Question 4.
Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area 20
Perimeter = ___
Answer:
There are 38 units around the figure,
So the perimeter is 38 m.

Explanation:
To find the perimeter of the figure, we will add all the sides of the figure,
The sides of the above figure is 2,8,2,3,1,3,2,8,2,3,1,3
and the perimeter of the figure is 2+8+2+3+1+3+2+8+2+3+1+3
= 38 units.
As there are 38 units around the figure,
So the perimeter is 38 m.

Question 5.
Draw a figure that has a perimeter of 18 inches.
Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area 21
Answer:
The length of the rectangle is 5 in
and the breadth of the rectangle is 4 in.

Explanation:
Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area
Let the length of the rectangle be 5 in
and the breadth of the rectangle be 4 in
so the perimeter of the rectangle is
p = 2(length+breadth)
= 2(5+4)
= 2(9)
= 18 in.

Question 6.
Reasoning
Which color represents the perimeter of the rectangle? What does the other color represent?
Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area 22
Answer:
Blue color represents the perimeter of the rectangle and
the other color yellow represents the area of the figure.

Explanation:
In the above figure, the blue color represents the perimeter of the rectangle.
Because the perimeter represents the distance around the edge of the shape.
And the other color yellow represents the area of the figure,
as area represents the amount of space inside a shape.

Question 7.
Modeling Real Life
Use a centimeter ruler tofind the perimeter of the library card.
Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area 24
Answer:
The perimeter of the library card is 24 cm.

Explanation:
On measuring, the length of the sides of the library card is 5cm, 7cm, 5cm, 7cm
to find the perimeter of the library card, we will add all the length of the sides
so the perimeter of the library card is
p= 5cm+7cm+5cm+7cm
= 24 cm
The perimeter of the library card is 24 cm.

Question 8.
Modeling Real Life
How much greater is the perimeter of your piece of fabric than the perimeter of your friend’ spiece of fabric?
Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area 25
Answer:
My fabric is 4 inches greater than my friend’s fabric.

Explanation:
To find the perimeter of the figure 1, we will add all the sides of the figure,
The sides of the above figure is 2,1,1,2,2,2,8,2,2,2,1,1
and the perimeter of the figure is 2+1+1+2+2+2+8+2+2+2+1+1
= 26 in.
So the perimeter of the figure is 26 in.
And now let’s find the perimeter of the friend’s figure,
So the sides of the figure is 1,2,2,4,1,5,4,3
and the perimeter is 1+2+2+4+1+5+4+3
= 22 in.
By this, we can see that my fabric has the highest perimeter than the friend’s fabric
which is 26-22= 4 in greater.

Review & Refresh

Write two equivalent fractions for the whole number.

Question 9.
Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area 26
Answer:
1= 4/4 = 6/6

Explanation:
Here, the equivalent fraction for the whole number means if the numerator was divided by the denominator without any reminder then the fraction is equivalent to a whole number. So 1= 4/4 = 6/6.

Question 10.
Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area 27
Answer:
4= 4/1 = 8/2.

Explanation:
Here, the equivalent fraction for the whole number means if the numerator was divided by the denominator without any reminder then the fraction is equivalent to a whole number. So 4= 4/1 = 8/2.

Question 11.
Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area 28
Answer:
6= 24/4 = 36/6.

Explanation:
Here, the equivalent fraction for the whole number means if the numerator was divided by the denominator without any reminder then the fraction is equivalent to a whole number. So 6= 24/4 = 36/6.

Find Perimeters of Polygons 15.2

Explore and Grow

Model a rectangle on your geoboard. Draw the rectangle and label its side lengths. Then find the perimeter in more than one way.
Big Ideas Math Answers 3rd Grade Chapter 15 Find Perimeter and Area 29

Answer:
The perimeter of the rectangle is 14 units.

Explanation:
Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area
In  the above figure, we can see the length of the rectangle is 4 units
and the breadth of the rectangle is 3 units,
so the perimeter of the rectangle is
p =2(length + breadth)
= 2(4+3)
= 2(7)
= 14 units.
The perimeter of the rectangle is 14 units.
Critique the Reasoning of Others
Compare your methods of finding the perimeter to your partner’s methods. Explain how they are alike or different.
Answer:
There are two methods to find the perimeter explained below.

Explanation:
There are two ways to find the perimeter.
The first method is
Perimeter = 2(length + breadth)
here, we will add length and breadth, and then we will multiply the result by 2.
and the second method is
Perimeter = a+b+c+d
here, we will add all the sides of the figure.

Think and Grow: Find Perimeter

Example
Find the perimeter of the trapezoid.
Big Ideas Math Answers 3rd Grade Chapter 15 Find Perimeter and Area 30
You can find the perimeter of a figure by adding all of the side lengths.
Big Ideas Math Answers 3rd Grade Chapter 15 Find Perimeter and Area 31
Write an equation. The letter P represents the unknown perimeter. Add the side lengths.
So, the perimeter is ___.
Answer:
So, the perimeter is 36 in.

Explanation:
To find the perimeter of the trapezoid, we will add all the sides of the trapezoid,
so the sides of the trapezoid is 6 in,12 in,8 in,10 in
The perimeter is 6+12+8+10
= 36 in.

Example
Find the perimeter of the rectangle
Big Ideas Math Answers 3rd Grade Chapter 15 Find Perimeter and Area 33
Because a rectangle has two pairs of equal sides, you can also use multiplication to solve.
Big Ideas Math Answers 3rd Grade Chapter 15 Find Perimeter and Area 34
One Way:
Big Ideas Math Answers 3rd Grade Chapter 15 Find Perimeter and Area 35
Another Way:
Big Ideas Math Answers 3rd Grade Chapter 15 Find Perimeter and Area 36
So, the perimeter is ___.

Answer:
Perimeter = 7+9+7+9
= 32 cm.
Perimeter= 2×9 + 2× 7
= 32 cm.
So, the perimeter is 32 cm.

Explanation:
To find the perimeter of a rectangle, we have two ways,
One way is to add all the sides of the rectangle, which is
7+9+7+9= 32 cm.
And the other way is, as the two sides of the rectangle are equal, we wil use formula
Perimeter = 2( Length + Breadth)
= 2× Length + 2 × Breadth
= 2×9 + 2× 7
= 18+ 14
= 32 cm.

Show and Grow

Find the perimeter of the polygon.
Answer:

Explanation:
To find the perimeter of the polygon, we will add the length of the all sides.

Question 1.
Big Ideas Math Answers 3rd Grade Chapter 15 Find Perimeter and Area 37
The perimeter is ___.
Answer:
The perimeter is 12 m.

Explanation:
To find the perimeter of the polygon, we will add the length of all sides of the polygon.
So the length of the sides is 5 m, 3 m, 3 m, 2 m
The perimeter is 5+3+3+2
= 12 m.

Question 2.
Big Ideas Math Answers 3rd Grade Chapter 15 Find Perimeter and Area 38
The perimeter is ___.

Answer:
The perimeter is 24 ft.

Explanation:
As we can see in the above figure which has all sides are equal,
so the perimeter of the square is 4s
= 4×6
= 24 ft.

Apply and Grow: Practice

Find the perimeter of the polygon.

Question 3.
Big Ideas Math Answers 3rd Grade Chapter 15 Find Perimeter and Area 39
Perimeter = ___
Answer:
The perimeter of the polygon is 28 m.

Explanation:
To find the perimeter of the polygon, we will add the length of the all sides of the polygon.
So the length of the sides is 4 m, 9 m, 10 m, 5 m
The perimeter is 4+9+10+5
= 28 m.

Question 4.
Big Ideas Math Answers 3rd Grade Chapter 15 Find Perimeter and Area 40
Perimeter = ___
Answer:
The perimeter of the figure is 40 in.

Explanation:
To find the perimeter of the figure, we will add the length of all sides of the figure.
So the length of the sides is 12 in, 4 in, 8 in, 7 in,9 in.
The perimeter is 12+4+8+7+9
= 40 in.

Question 5.
Rectangle
Big Ideas Math Answers 3rd Grade Chapter 15 Find Perimeter and Area 41
Perimeter = ___
Answer:
The perimeter of the rectangle is 36 cm.

Explanation:
To find the perimeter of the rectangle, we need to know the length and the breadth of the rectangle.
Here, the length of the rectangle is 9 cm,
and the breadth is 10 cm,
So the perimeter of the rectangle is 2(Length + Breadth)
= 2(10+9)
= 2(19)
= 38 cm.

Question 6.
Rhombus
Big Ideas Math Answers 3rd Grade Chapter 15 Find Perimeter and Area 42
Perimeter = ___
Answer:
The perimeter of the Rhombus is 4 in.

Explanation:
As all sides of the Rhombus are equal, so the formula of the Rhombus is 4a
and the side of the Rhombus is 1 in,
so the perimeter is 4a
= 4×1
= 4 in.

Question 7.
Parallelogram
Big Ideas Math Answers 3rd Grade Chapter 15 Find Perimeter and Area 43
Perimeter = ___
Answer:
The perimeter of the parallelogram is 22 cm.

Explanation:
As the opposite sides of the parallelogram are equal,
so the perimeter of the parallelogram is 2(a+b)
the length of the parallelogram is 3 cm,
and the breadth of the parallelogram is 8 cm
So the perimeter of the parallelogram is
= 2(3+8)
= 2(11)
= 22 cm.

Question 8.
Square
Big Ideas Math Answers 3rd Grade Chapter 15 Find Perimeter and Area 44
Perimeter = ___
Answer:
The perimeter of the square is 16 ft.

Explanation:
As we can see in the above figure which has all sides are equal,
so the perimeter of the square is 4s
= 4×4
= 16 ft.

Question 9.
You build a pentagon out of wire for a social studies project. Each side is 8 centimeters long. What is the perimeter of the pentagon?
Answer:
The perimeter of the pentagon is 40 cm.

Explanation:
The length of the pentagon is 8 cm,
so the perimeter of the pentagon is 5a,
which is 5×8
= 40 cm

Question 10.
Number Sense
The top length of the trapezoid is 4 feet. The bottom length is double the top. The left and right lengths are each 2 feet less than the bottom. Label the side lengths and find the perimeter of the trapezoid.
Big Ideas Math Answers 3rd Grade Chapter 15 Find Perimeter and Area 45
Answer:
The sides of the trapezoid are 4ft, 6ft, 8ft, 6ft, and the perimeter is 24 ft.

Explanation:
As the top length of the trapezoid is 4 feet and the bottom length is double the top,
which is 4×2= 8 feet. And the left and right lengths are each 2 feet less than the bottom,
which means 8 – 2 = 6 ft each.

Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area

So to find the perimeter of the trapezoid is we will add the length of all sides.
The perimeter of the trapezoid is 4+6+8+6= 24 ft.

Question 11.
Writing
Explain how finding the perimeter of a rectangle is different from finding its area.
Answer:
Perimeter= 2(length + breadth)
Area = length × breadth.

Explanation:
To find the perimeter of the rectangle,
we will add the length and breadth and will multiply the result with 2
Perimeter= 2(length + breadth)
and to find the area of the rectangle,
we will multiply the length and breadth of the rectangle.
Area = length × breadth.

Question 12.
Dig Deeper!
A rectangle has a perimeter of 12 feet. What could its side lengths be ?
Answer:
The length of the rectangle is 4 feet
and the breadth of the rectangle is 2 feet

Explanation:
Given the perimeter is 12 feet,
so the let the length be 4 feet
and the breadth be 2 feet
Let’s check on the length and breadth is correct are not
perimeter = 2( length + breadth)
= 2( 4+2)
= 2(6)
= 12 feet.

Think and Grow: Modeling Real Life

The rectangular sign is 34 feet longer than it is wide. What is the perimeter of the sign?
Big Ideas Math Answers 3rd Grade Chapter 15 Find Perimeter and Area 46
Understand the problem:
Make a plan:
Solve:
The perimeter is ___.
Answer:
The perimeter of the rectangular sign is 124 ft.

Explanation:
Given the rectangular sign is 34 feet longer than it’s wide
and the wide is 14 ft,
so the length is 34ft + 14ft= 48 ft.
the perimeter of the rectangular sign is
P= 2(48ft + 14ft)
= 2(62ft)
= 124 ft.
The perimeter of the rectangular sign is 124 ft.

Show and Grow

Question 13.
A city has a rectangular sidewalk in a park. The sidewalk is 4 feet wide and is 96 feet longer than it is wide. What is the perimeter of the sidewalk?
Answer:
The perimeter of the sidewalk is 208 feet.

Explanation:
As given the rectangular sidewalk’s wide is 4 feet and the length is 96 feet longer than it’s wide,
which means 96+4= 100 feet is the length of the rectangular sidewalk,
so the perimeter of the rectangular sidewalk is 2( length + breadth)
= 2( 4+100)
= 2(104)
= 208 feet.

Question 14.
A team jogs around a rectangular field three times. The field is 80 yards long and 60 yards wide. How many yards does the team jog?
Answer:
The number of yards does the team jog is 3×280= 840 yards.

Explanation:
The length of the rectangular field is 80 yards,
The breadth of the rectangular field is 60 yards,
So, the perimeter of the rectangular field is 2(length + breadth)
= 2(80+60)
= 2(140)
= 280 yards.
As the team jogs around a rectangular field three times,
so the number of yards does the team jog is 3×280= 840 yards.

Question 15.
Each side of the tiles is 8 centimeters long. What is the sum of the perimeters?
Big Ideas Math Answers 3rd Grade Chapter 15 Find Perimeter and Area 47
You put the tiles together as shown. Is the perimeter of this new shape the same as the sum of the perimeters above? Explain.
Big Ideas Math Answers 3rd Grade Chapter 15 Find Perimeter and Area 48
Answer:
The sum of the perimeters is 96 cm.
Yes, there will be a change in the perimeter of the new figure. The perimeter is 80 cm.

Explanation:
In the above figure, we can see a hexagon that has six sides.
So the formula for the perimeter of a hexagon is 6a,
the perimeter of the tiles is 6×8
= 48 cm.
So the sum of the tiles 48+48= 96 cm.
Yes, there will be a change in the perimeter of the new figure. As there are ten sides in the new figure, so the perimeter of the new figure is 10×8= 80 cm.

Find Perimeters of Polygons Homework & Practice 15.2

Find the perimeter of the polygon

Question 1.
Big Ideas Math Answers 3rd Grade Chapter 15 Find Perimeter and Area 49
Perimeter = ___
Answer:
The perimeter of the polygon is 29 in.

Explanation:
The perimeter of the polygon is the sum of the length of its sides,
So the sides of the polygon are 9 in, 4 in, 6 in, 10 in,
and the perimeter of the polygon is 9+4+6+10= 29 in.

Question 2.
Big Ideas Math Answers 3rd Grade Chapter 15 Find Perimeter and Area 50
Perimeter = ___
Answer:
The perimeter of the figure is 38 cm.

Explanation:
The perimeter of the figure is the sum of the length of its sides,
So the sides of the figure are 6 cm, 5 cm, 7 cm, 8 cm, 7 cm, 5 cm,
and the perimeter of the figure is 6+5+7+8+7+5= 38 cm.

Question 3.
Square
Big Ideas Math Answers 3rd Grade Chapter 15 Find Perimeter and Area 51
Perimeter = ___
Answer:
The perimeter of the square is 8 ft.

Explanation:
The length of the square is 2 ft,
and the perimeter of the square is 4a
= 4×2
= 8 ft.
so the perimeter of the square is 8 ft.

Question 4.
Parallelogram
Big Ideas Math Answers 3rd Grade Chapter 15 Find Perimeter and Area 52
Perimeter = ___
Answer:
The perimeter of the parallelogram is 8m.

Explanation:
The length of the parallelogram is 1 m,
and the breadth of the parallelogram is 3 m,
so the perimeter of the parallelogram is 2(length + breadth)
= 2(1+3)
= 2(4)
= 8 m.

Question 5.
Rhombus
Big Ideas Math Answers 3rd Grade Chapter 15 Find Perimeter and Area 53
Perimeter = ___
Answer:
The perimeter of the rhombus is 40 ft.

Explanation:
The length of the side of the rhombus is 10 ft,
and the perimeter of the rhombus is 4a
= 4×10
= 40 ft.

Question 6.
Big Ideas Math Answers 3rd Grade Chapter 15 Find Perimeter and Area 54
Perimeter = ___
Answer:
The perimeter of the rectangle is 22 in.

Explanation:
To find the perimeter of the rectangle, we need to know the length and the breadth of the rectangle.
Here, the length of the rectangle is 7 in,
and the breadth is 4 in,
So the perimeter of the rectangle is 2(Length + Breadth)
= 2(7+4)
= 2(11)
= 22 in.

Question 7.
Each side of a triangle is 5 centimeters long. What is the perimeter of the triangle?
Answer:
The perimeter of the triangle is 15 cm.

Explanation:
Given the length of the side of the triangle is 5 cm,
and the perimeter of the triangle is 3a
= 3×5
= 15 cm.

Question 8.
You Be The Teacher
Descartes says that a square will always have a greater perimeter than a triangle because it has more sides. Is he correct? Explain.
Answer:
Yes, Descartes is correct.

Explanation:
Yes, Descartes is correct. As the square has four sides and the perimeter of the square is 4a.
Whereas the triangle has 3 sides and the perimeter of the triangle is 3a.
So the square will always have a greater perimeter than a triangle.

Question 9.
Structure
Draw a pentagon and label its sides so that it has the same perimeter as the rectangle.
Big Ideas Math Answers 3rd Grade Chapter 15 Find Perimeter and Area 55
Answer:
The length of the side of the pentagon is 4m.

Explanation:
Given the length of the rectangle is 7 m,
and the breadth of the rectangle is 3 m,
So the perimeter of the rectangle is 2( length + breadth)
= 2(7+3)
= 2(10)
= 20 m.
Here we have the perimeter of the pentagon which is 20 m,
so we should find the sides of the pentagon,
As we know the perimeter of the pentagon is 5a
5a= 20
a= 20/5
= 4 m.
So the length of the side of the pentagon is 4m.
Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area

Question 10.
Modeling Real Life
An Olympic swimming pool is 82 feet longer than it is wide. What is the perimeter of the swimming pool?
Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area 150
Answer:
The perimeter of the swimming pool is 492 ft.

Explanation:
Given the Olympic swimming pool is 82 feet longer than it’s wide
and the wide is 82 ft,
so the length of the swimming pool is 82+82= 164 ft
the perimeter of the swimming pool is
p= 2(length+breadth)
= 2(164+82)
= 2(246)
= 492 ft.
The perimeter of the swimming pool is 492 ft.

Question 11.
Modeling Real Life
You put painter’s tape around two rectangular windows. The windows are each 52 inches long and 28 inches wide. How much painter’s tape do you need?
Answer:
The painter’s tape 160 inches.

Explanation:
Given the length of the window is 52 inches
and the width of the window is 28 inches
so the perimeter of the window is
p= 2(length + breadth)
= 2(52+28)
= 2(80)
= 160 inches.
So the painter’s tape 160 inches.

Review & Refresh

Question 12.
Big Ideas Math Answers 3rd Grade Chapter 15 Find Perimeter and Area 56
Answer:
737.

Explanation:
On adding 590+147 we will get 737.

Question 13.
Big Ideas Math Answers 3rd Grade Chapter 15 Find Perimeter and Area 57
Answer:
894.

Explanation:
On adding 636+258 we will get 894

Question 14.
Big Ideas Math Answers 3rd Grade Chapter 15 Find Perimeter and Area 58
Answer:
805.

Explanation:
On adding 476+329 we will get 805

Lesson 15.3 Find Unknown Side Lengths

Explore and Grow

You have a map with the three side lengths shown. The perimeter of the map is 20 feet. Describe how you can find the fourth side length of your map without measuring.
Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area 59

Answer:
The fourth side of the map is 4ft.

Explanation:
We can find the value of the fourth side in two methods
First method:
Given the perimeter of the map is 20 feet,
and the sides of the map is 6 ft, 4 ft, 6ft, X ft.
As we know the perimeter of the rectangle is
6+4+6+X= 20 feet
16+X= 20
X= 20- 16
= 4 ft.
Second method:
As we know that the opposite sides of the rectangle are equal, as we know that the length of the side is 4 ft so the other side will also be 4 ft.

Repeated Reasoning
How is finding the unknown side length of a square different from finding the unknown side length of a rectangle?
Answer:
Refer below for a detailed explanation.

Explanation:
To find the unknown side length of the square
if we know the perimeter of the square then
the perimeter of the square is
p= 4a
we will substitute the value of p, on solving we will get the length of the square.
and to find the unknown side length of the rectangle,
we need to know the area or perimeter of the rectangle
and the other side of the rectangle.
so the formula of the perimeter of the rectangle is
p = 2(length + breadth)
we will substitute the perimeter value and the other side value
then we can find the length of the rectangle.

Think and Grow: Find Unknown Side Lengths
Example
The perimeter of the trapezoid is 26 feet. Find the unknown side length.
Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area 60
Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area 61
Write an equation for the perimeter.
Add the known side lengths.
What number plus 16 equals 26?
The unknown side length is ___.
Answer:
K= 10,
The number 16+10 equals 26.
The unknown side length is 10.

Explanation:
Given the perimeter of the trapezoid is 26 ft,
So the perimeter of the trapezoid is
K+5+6+5= 26
K+16= 26
K= 10.
The number 16+10 equals 26.
The unknown side length is 10.

Example
Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area 62
The perimeter of the square is 32 centimeters. Find the length of each side of the square.
Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area 63
Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area 64
Write an equation for the perimeter 4 times what number equals 32?
So, the length of each side is ___.
Answer:
n= 8.
The length of each side is 8 cm.

Explanation:
The perimeter of the square is 32 cm
So to find the sides of the square
4a= 32
a= 32/4
= 8 cm.
So, the length of each side is 8 cm.

Show and Grow

Find the unknown side length.

Question 1.
Perimeter = 34 inches
Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area 65
y = ___
Answer:
y = 13 in.

Explanation:
Given the perimeter is 34 inches,
and the sides of the figure are 10 in, 7 in, 4 in, y in.
so the perimeter of the figure is
34 in = 10+7+4+y
34 = 21+ y
y = 34 – 21
y = 13.

Question 2.
Perimeter = 20 meters
Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area 66
j = ___
Answer:
The length of the sides of the square is 5 m.

Explanation:
As we can see the above figure is a square and the perimeter of the square 20 meters,
so the sides of the square are
perimeter = 4a
20 = 4 j
j= 5 m.

Apply and Grow: Practice

Find the unknown side length.

Question 3.
Perimeter = 19 feet
Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area 67
y = ___
Answer:
The perimeter of the figure is 8 ft.

Explanation:
The perimeter of the figure is 19 feet,
and the length of the sides of the figure is 8 ft, 3 ft, y ft.
so perimeter = 8 ft + 3 ft + y ft
19= 11 ft + y ft
y= 19 ft – 11 ft
= 8 ft.

Question 4.
Perimeter = 26 centimeters
Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area 68
d = ___
Answer:
d= 4 cm.

Explanation:
The perimeter of the figure is 26 cm,
and the length of the sides of the figure is 10 cm, 5 cm, 7 cm, d cm.
so the perimeter of the figure is
p = 10+5+7+d
26 = 22 + d
d= 26-22
= 4

Question 5.
Perimeter = 30 feet
Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area 69
k = ___
Answer:
k = 11 ft.

Explanation:
Given the perimeter of the figure is 30 feet,
and the length of the sides is 5ft, 12 ft, 2 ft, k ft
So the perimeter of the figure is
p = 5 + 12 + 2 + k
30 ft = 19 ft + k
k = 11 ft.

Question 6.
Perimeter = 32 inches
Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area 70
k = ___
Answer:
k= 4 in.

Explanation:
Given the perimeter of the figure is 32 inches,
and the lengths of all sides is 10 in, 4 in, 5 in, 4 in, 5 in, k in.
So the perimeter of the figure is
p= 10+4+5+4+5+k
32 in =  28 in + k
k= 4 in.

Question 7.
Perimeter = 8 meters
Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area 71
y = ___
Answer:
y = 2 m.

Explanation:
Given the perimeter of the rhombus is 8 feet,
and the length of the side is y m,
So the perimeter of the rhombus is
p = 4a
8 m = 4×y
y = 2 m.

Question 8.
Perimeter = 48 inches
Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area 72
d = ___
Answer:
d = 8 in.

Explanation:
Given the perimeter of the Hexagon is 48 inches,
and the length of the sides is d in
So the perimeter of the hexagon is
p = 6 a
48 in = 6 × d in
d= 8 in.

Question 9.
Number Sense
A rectangle has a perimeter of 30 centimeters. The left side is 7 centimeters long. What is the length of the top side?
Answer:
The length of the top side is 8 cm.

Explanation:
Given the perimeter of the rectangle is 30 cm,
and the length of the left side of the rectangle is 7 cm,
So let the length of the top side be X,
Perimeter of the rectangle is
P = 2 (Length + breadth)
30 = 2 ( 7 cm + X cm)
30 / 2 = 7 cm + X cm
15 = 7 cm + X cm
X = 15 cm – 7 cm
X = 8 cm.
so, the length of the top side is 8 cm.

Question 10.
Writing
A triangle has three equal sides and a perimeter of 21 meters. Explain how to use division to find the side lengths.
Answer:
The length of the side is 7 m.

Explanation:
Given the perimeter of the triangle is 21 m,
and we need to find the side of the lengths,
so the perimeter of the triangle is
p = 3a
21 m = 3×a
a = 21/3
= 7 m
So the length of the side is 7 m.

Question 11.
DIG DEEPER!
Newton draws and labels the square and rectangle below. The perimeter of the combined shape is 36 feet. Find the unknown side length.
Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area 73
Answer:
The unknown side of the length is 14 ft.

Explanation:
As the perimeter of the combined shape is 36 feet,
and the length of the side of the rectangle is 4 ft, and the other side be X ft
and the perimeter of the rectangle is 36 ft,
so perimeter = 2 (length + breadth)
36 ft  = 2( 4 ft + X ft)
36/2 = 4 ft + X ft
18 = 4 ft + X ft
X= 14 ft.
The unknown side of the length is 14 ft.

Think and Grow: Modeling Real Life

The perimeter of the rectangular vegetable garden is 30 meters. What are the lengths of the other three sides?
Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area 74
Understand the problem:
Make a plan:
Solve:
The lengths of the other three sides are ___, ___, and ___.
Answer:
The lengths of other three sides is 6 m, 9 m, 9 m.

Explanation:
The perimeter of the rectangular vegetable garden is 30 m
as it is in a rectangular shape, so the opposite sides are equal,
and the length of the side of the rectangular vegetable garden is 6m,
let the other side be X m
so the perimeter is
p = 2( length + breadth)
30 m = 2( 6 m+ X m)
30/2 = 6 + X
15 = 6 + X
X= 15 – 6
= 9 m.
So the lengths of other three sides is 6 m, 9 m, 9 m.

Show and Grow

Question 12.
The perimeter of the rectangular zoo enclosure is 34 meters. What are the lengths of the other three sides?
Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area 75
Answer:
The lengths of the other three sides are 12 m, 11 m, 11 m.

Explanation:
The perimeter of the rectangular zoo is 34 m
as it is in rectangular shape, so the opposite sides are equal,
and the length of the side of the rectangular zoo is 12 m,
let the other side be X m
so the perimeter is
p = 2( length + breadth)
34 m = 2( 12 m+ X m)
34/2 = 12 + X
17 = 6 + X
X= 17 – 6
= 11 m.
So the lengths of the other three sides is 12 m, 11 m, 11 m.

Question 13.
The floor of an apartment is made of two rectangles. The Perimeter is 154 feet. What are the lengths of the other three sides?
Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area 76
Answer:
The other length of the side of the small rectangle is 7 ft.
The other length of the side of the big rectangle is 30 ft.

Explanation:
Given the perimeter of the apartment is 154 feet,
first, we will take the big rectangle,
as the opposite sides of the rectangle are equal and the length of the big rectangle is 30 ft
so the other length is also 30 ft.
as the perimeter of the small rectangle is 38 ft
and the length of the one side of the rectangle is 12 ft
so the other length of the small rectangle is
p = 2(length+breadth)
38 = 2(12 + breadth)
38/2 = 12+ breadth
19= 12 + breadth
breadth= 19 – 12
= 7 ft.
The other length of the side is 7 ft.

Question 14.
DIG DEEPER!
You want to make a flower bed in the shape of a pentagon. Two sides of the flower bed are each 7 inches long, and two sides are each 16 inches long. The perimeter is 57 inches. Sketch the flower bed and label all of the side lengths.
Answer:
The length of the other side is 11 ft.

Explanation:
Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area
Given the perimeter of the flower bed shaped pentagon is 57 inches
and the two sides of the flower bed each is 7 inches long
and the other two sides of the flower bed each is 16 inches long
the other side of the flower bed be X
the perimeter of the flower bed is
p = 7+7+16+16+X
57= 46+X
X= 11 in.
The length of the other side is 11 ft.

Find Unknown Side Lengths Homework & Practice 15.3

Find the unknown side length.

Question 1.
Perimeter = 24 feet
Big Ideas Math Solutions Grade 3 Chapter 15 Find Perimeter and Area 77
d = ___
Answer:
The length of the unknown side is 8 ft.

Explanation:
The perimeter of the triangle is 24 ft,
and the lengths of the sides is 10 ft, 6 ft, d ft
so the perimeter of the triangle is
p = 10+6+d
24 = 10+6+d
24 = 16+d
d = 24 – 16
=  8 ft.
The length of the unknown side is 8 ft.

Question 2.
Perimeter = 46 inches
Big Ideas Math Solutions Grade 3 Chapter 15 Find Perimeter and Area 78
k = ___
Answer:
The length of the unknown side is 15 in.

Explanation:
The perimeter of the figure is 46 inches,
and the lengths of the sides is 13 in, 5 in, 13 in, k in
so the perimeter of the figure is
p = 13+5+13+k
46 = 31+k
46-31 = k
k = 15
=  15 in.
The length of the unknown side is 15 in.

Question 3.
Perimeter = 21 centimeters
Big Ideas Math Solutions Grade 3 Chapter 15 Find Perimeter and Area 79
y = ___
Answer:
The length of the unknown side is 7 cm.

Explanation:
The perimeter of the figure is 21 cm,
and the lengths of the sides is 4 cm, 1 cm, 9 cm, y cm
so the perimeter of the figure is
p = 4 cm+ 1 cm+ 9 cm+ y cm
21 = 14+ y
y = 21 – 14
y = 7
=  7 cm.
The length of the unknown side is 7 cm.

Question 4.
Perimeter = 41 meters
Big Ideas Math Solutions Grade 3 Chapter 15 Find Perimeter and Area 80
y = ___
Answer:
The length of the unknown side is 8 m.

Explanation:
The perimeter of the figure is 41 m,
and the lengths of the sides is 3 m, 12 m, 10 m, 8 m, y m
so the perimeter of the figure is
p = 3+12+10+8+y
41 = 33 + y
y = 41-33
y = 8
=  8 m.
The length of the unknown side is 8 m.

Question 5.
Perimeter = 12 feet
Big Ideas Math Solutions Grade 3 Chapter 15 Find Perimeter and Area 81
d = ___
Answer:
The length of the sides of the triangle is 4 ft.

Explanation:
The perimeter of the triangle is 12 feet,
and the length of the side of the triangle is d ft,
so the perimeter of the triangle is
p = 3 a
12 = 3 × d
d = 12/3
= 4 ft.
The length of the sides of the triangle is 4 ft.

Question 6.
Perimeter = 50 inches
Big Ideas Math Solutions Grade 3 Chapter 15 Find Perimeter and Area 82
k = ___
Answer:
The length of the sides of the Hexagon is 4 in.

Explanation:
The perimeter of the Hexagon is 50 inches,
and the length of the side of the Hexagon is k in,
so the perimeter of the hexagon is
p = 5 a
50 = 5 × k
k = 50/5
= 10 in.
The length of the sides of the triangle is 10 in.

Question 7.
DIG DEEPER!
Each polygon has equal side lengths that are whole numbers. Which polygon could have a perimeter of 16 centimeters? Explain.
Big Ideas Math Solutions Grade 3 Chapter 15 Find Perimeter and Area 83
Answer:
The length of the sides of the octagon is 2 cm.

Explanation:
In the above three polygons, the second figure is an octagon, which has eight sides.
and the perimeter of the octagon is
p = 8 a
16 cm = 8 a
a = 16 /8
= 2 cm.

Question 8.
Number Sense
The area of a square is 25 square inches. What is its perimeter?
Answer:
The perimeter of the square is 20 inches.

Explanation:
The area of the square is 25 square inches, so
area = s^2
25 = s^2
s= 5 inches
so the perimeter of the square is
p = 4s
= 4×5
= 20 inches.

Question 9.
Modeling Real Life
The perimeter of the rectangular side walk is 260 meters. What are the lengths of the other three sides?
Big Ideas Math Solutions Grade 3 Chapter 15 Find Perimeter and Area 84
Answer:
The length of the other three sides is 10 m,10m,120 m.

Explanation:
The perimeter of the rectangular side walk is 260 meters,
and the length of the one side of the side walk is 120 m,
so the perimeter of the rectangular side walk is
p = 2( length + breadth)
260 = 2 ( 120 + breadth)
260/2 = 120 + breadth
130 = 120 + breadth
breadth = 130 – 120
= 10 m.
The length of the other three sides is 10 m,10m,120 m.

Question 10
Modeling Real Life
Two rectangular tables are pushed together. The perimeter is 40 feet. What are the lengths of the other three sides?
Big Ideas Math Solutions Grade 3 Chapter 15 Find Perimeter and Area 85
Answer:
The other length of the side of the small rectangle is 3 ft.
The other length of the side of the big rectangle is 5 ft.

Explanation:
Given the perimeter of the apartment is 40 feet,
first, we will take the big rectangle,
as the opposite sides of the rectangle are equal
and the length of the big rectangle is 5 ft
so the other length is also 5 ft.
as the perimeter of the small rectangle is 10 ft
and the length of the one side of the rectangle is 2 ft
so the other length of the small rectangle is
p = 2(length+breadth)
10 = 2(2 + breadth)
10/2 = 2+ breadth
5= 2 + breadth
breadth= 5 – 2
= 3 ft.
The other length of the side is 3 ft.

Review & Refresh

Write the time. Write another way to say the time.

Question 11.
Big Ideas Math Solutions Grade 3 Chapter 15 Find Perimeter and Area 86
Answer:
06: 48

Explanation:
Another way to say time is 06: 48

Question 12.
Big Ideas Math Solutions Grade 3 Chapter 15 Find Perimeter and Area 87
Answer:
03: 24

Explanation:
Another way to say time is 03: 24

Question 13.
Big Ideas Math Solutions Grade 3 Chapter 15 Find Perimeter and Area 88
Answer:

Explanation:
Another way to say the time is 11: 48

Lesson 15.4 Same Perimeter, Different Areas

Use color tiles to create two different rectangles that each have a perimeter of 16 units. Then draw your rectangles and label their dimensions. Do the rectangles have the same area? Explain how you know.
Big Ideas Math Solutions Grade 3 Chapter 15 Find Perimeter and Area 89

Answer:
No, the area of rectangle 1 and rectangle 2 is not the same.

Explanation:

Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area
Given the perimeter of the rectangle is 16 units
The length of rectangle 1 is 5 units
and the breadth of rectangle 1 is 3 units
so the area of rectangle 1 is
area = length × breadth
= 5×3
= 15 square units.
The length of rectangle 2 is 6 units
and the breadth of rectangle 2 is 2 units
so the area of rectangle 2 is
area = length × breadth
= 6×2
= 12 square units.
No, the area of rectangle 1 and rectangle 2 is not the same.

Repeated Reasoning
Draw another rectangle that has the same perimeter but different dimensions. Compare the area of the new rectangle to the rectangles above. What do you notice?
Answer:
No, the area of rectangle 1 and rectangle 2 is not the same.

Explanation:
Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area
The length of rectangle 1 is 4 units
and the breadth of rectangle 1 is 3 units
the perimeter of rectangle 1 is
p= 2(length+breadth)
= 2(4+3)
= 2(7)
= 14 units.
so the area of rectangle 1 is
area = length × breadth
= 4×3
= 12 square units.
The length of rectangle 2 is 5 units
and the breadth of rectangle 2 is 2 units
the perimeter of rectangle 1 is
p= 2(length+breadth)
= 2(5+2)
= 2(7)
= 14 units.
so the area of rectangle 2 is
area = length × breadth
= 5×2
= 10 square units.
No, the area of rectangle 1 and rectangle 2 is not the same.

Think and Grow : Same Perimeter, Different Areas

Example :
Find the perimeter and the area of Rectangle A. Draw a different rectangle that has the same perimeter. Which rectangle has the greater area?
Big Ideas Math Solutions Grade 3 Chapter 15 Find Perimeter and Area 90
Big Ideas Math Solutions Grade 3 Chapter 15 Find Perimeter and Area 91
Rectangle ___ has a greater area.
Answer:
The perimeter of the rectangle A is 20 m
and the area of the rectangle A is 24 m2
The perimeter of rectangle B is 20 m
and the area of the rectangle B is 16 m2
The rectangle A has a greater area.

Explanation:
Given the length of the rectangle is 6m
and the breadth of the rectangle is 4m,
so the perimeter of the rectangle is
p = 2( length + breadth)
= 2(6 + 4)
= 2(10)
= 20 m.
And the area of the rectangle is
a = length × breadth
= 6 × 4
= 24 m2

Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area

we can see in the above figure another rectangle with a length of 8m,
and the breadth is 2m,
so the perimeter of the rectangle is
p = 2(length + breadth)
= 2(8 + 2)
= 2(10)
= 20 m.
And the area of the rectangle is
area = length × breadth
= 8×2
= 16 m2
So the rectangle A has a greater area.

Show and Grow

Question 1.
Find the perimeter and area of Rectangle A. Draw a different rectangle that has the same perimeter. Which rectangle has the greater area?
Big Ideas Math Solutions Grade 3 Chapter 15 Find Perimeter and Area 92
Perimeter = ___
Area = ___
Big Ideas Math Solutions Grade 3 Chapter 15 Find Perimeter and Area 93
Perimeter = ___
Area = ___
Rectangle ___ has the greater area.
Answer:The perimeter of the rectangle A is 14 in
and the area of the rectangle A is 10 in2
The perimeter of rectangle B is 14 in
and the area of the rectangle B is 12 in2
The rectangle B has a greater area.

Explanation:
Given the length of the rectangle is 5 in
and the breadth of the rectangle is 2 in,
so the perimeter of the rectangle is
p = 2( length + breadth)
= 2(5 + 2)
= 2(7)
= 14 in.
And the area of the rectangle is
a = length × breadth
= 5×2
= 10 in2

Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area

we can see in the above figure another rectangle with a length of 4 in,
and the breadth is 3 in,
so the perimeter of the rectangle is
p = 2(length + breadth)
= 2(4 + 3)
= 2(7)
= 14 in.
And the area of the rectangle is
area = length × breadth
= 4×3
= 12 in2
So the rectangle B has a greater area.

Apply and Grow: Practice

Find the perimeter and area of Rectangle A. Draw a different rectangle that has the same perimeter. Which rectangle has the greater area?

Question 2.
Rectangle A
Big Ideas Math Solutions Grade 3 Chapter 15 Find Perimeter and Area 94
Perimeter = ___
Area = ___
Big Ideas Math Solutions Grade 3 Chapter 15 Find Perimeter and Area 95
Perimeter = ___
Area = ___
Rectangle ___ has the greater area.
Answer:
The perimeter of the rectangle A is 22 cm
and the area of the rectangle A is 11 cm2
The perimeter of rectangle B is 22 cm
and the area of the rectangle B is 30 cm2
The rectangle B has a greater area.

Explanation:
Given the length of the rectangle is 10 cm,
and the breadth of the rectangle is 1 cm,
so the perimeter of the rectangle is
p = 2( length + breadth)
= 2(10 + 1)
= 2(11)
= 22 cm.
And the area of the rectangle is
a = length × breadth
= 11 × 1
= 11 cm2

Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area

we can see in the above figure another rectangle with a length of 6 cm,
and the breadth is 5 cm,
so the perimeter of the rectangle is
p = 2(length + breadth)
= 2(6 + 5)
= 2(11)
= 22 cm.
And the area of the rectangle is
area = length × breadth
= 6×5
= 30 cm2
So the rectangle B has a greater area.

Question 3.
Rectangle A
Big Ideas Math Solutions Grade 3 Chapter 15 Find Perimeter and Area 96
Perimeter = ___
Area = ___

Rectangle B
Big Ideas Math Solutions Grade 3 Chapter 15 Find Perimeter and Area 97
Perimeter = ___
Area = ___
Rectangle ___ has the greater area
Answer:
The perimeter of the rectangle A is 20 m
and the area of the rectangle A is 21 m2
The perimeter of rectangle B is 20 m
and the area of the rectangle B is 24 m2
The rectangle B has a greater area.

Explanation:
Given the length of the rectangle is 7 m,
and the breadth of the rectangle is 3 m,
so the perimeter of the rectangle is
p = 2( length + breadth)
= 2(7 + 3)
= 2(10)
= 20 m.
And the area of the rectangle is
a = length × breadth
= 7 × 3
= 21 m2

Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area

we can see in the above figure another rectangle with a length of 6 m,
and the breadth is 4 m,
so the perimeter of the rectangle is
p = 2(length + breadth)
= 2(6 + 4)
= 2(10)
= 20 m.
And the area of the rectangle is
area = length × breadth
= 6×4
= 24 m2
So the rectangle B has a greater area.

Question 4.
MP Structure
Draw a rectangle that has the same perimeter as the one shown, but with a lesser area. What is the area ?
Big Ideas Math Solutions Grade 3 Chapter 15 Find Perimeter and Area 98

Answer:
The perimeter of the rectangle A is 26 ft
and the area of the rectangle A is 40 ft2
The perimeter of rectangle B is 26 ft
and the area of the rectangle B is 30 ft2
The rectangle B has a greater area.

Explanation:
Given the length of the rectangle is 5 ft,
and the breadth of the rectangle is 8 ft,
so the perimeter of the rectangle is
p = 2( length + breadth)
= 2(5 + 8)
= 2(13)
= 26 ft.
And the area of the rectangle is
a = length × breadth
= 5 × 8
= 40 ft

Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area

we can see in the above figure another rectangle with a length of 10 ft,
and the breadth is 3 ft,
so the perimeter of the rectangle is
p = 2(length + breadth)
= 2(10 + 3)
= 2(13)
= 26 ft.
And the area of the rectangle is
area = length × breadth
= 10×3
= 30 ft2
So the rectangle A has a greater area.

Think and Grow: Modeling Real Life

A paleontologist has 12 meters of twine to rope off a rectangular section of the ground. How long and wide should she make the roped-off section so it has the greatest possible area?
Draw to show:
She should make the roped-off section ___ meters long and ___ meters wide.
Answer:
She should make the roped-off section 4 meters long and 2 meters wide.

Explanation:
Given that a paleontologist has 12 meters of twine to rope off a rectangular section,
so if we take the length as 4 m and width as 2 m then we can get the greatest possible area,
so the area of the rectangular section is
area = length × breadth
= 4 m ×2 m
= 8 m2

Show and Grow

Question 5.
Newton has 16 feet of wood to make a rectangular sandbox. How long and wide should he make the sandbox so it has the greatest possible area?
Big Ideas Math Solutions Grade 3 Chapter 15 Find Perimeter and Area 99
Answer:
The greatest possible area of the rectangular sandbox is 15 ft2

Explanation:
As Newton has 16 feet of wood to make a rectangular sandbox,
so let the length be 5 ft and the wide be 3 ft to get the greatest possible area,
so the area of the rectangular sandbox is
area = length × breadth
= 5 ft × 3 ft
= 15 ft2
The greatest possible area of the rectangular sandbox is 15 ft2

Question 6.
DIG DEEPER!
You and Newton are building forts. You each have the same length of rope to make a rectangular perimeter for the forton the ground. Your roped-off section is shown. Newton’s section has a greater area than yours. Draw one way Newton could rope off his fort.
Big Ideas Math Solutions Grade 3 Chapter 15 Find Perimeter and Area 100

Descartes also builds a fort. He has the same length of rope as you to make a perimeter around his fort. Descartes’s roped-off section has a lesser area than yours. Draw one way Descartes could rope off his fort.

Answer:
Refer the below for detailed explanation.

Explanation:
The length of the rope is 7 ft
and the breadth of the rope is 3 ft
the perimeter is
p = 2(length+breadth)
= 2(7+3)
= 2(10)
= 20 ft.
The area of the rectangle is
area= length×breadth
= 7×3
= 21 square feet.

Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area
Let the length of the Newton’s rope is 6 ft
and the breadth of the Newton’s rope is 4 ft
the perimeter is
p = 2(length+breadth)
= 2(6+4)
= 2(10)
= 20 ft.
The area of the rectangle is
area= length×breadth
= 6×4
= 24 square feet.
And here Newton’s area is greater.

Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area

Let the length of the Descarte’s rope is 8 ft
and the breadth of the Descarte’s rope is 2 ft
the perimeter is
p = 2(length+breadth)
= 2(8+2)
= 2(10)
= 20 ft.
The area of the rectangle is
area= length×breadth
= 8×2
= 16 square feet.
And here Descartes area is lesser.

Same Perimeter, Different Areas Homework & Practice 15.4

Question 1.
Find the perimeter and the area of Rectangle A. Draw a different rectangle that has the same perimeter? Which rectangle has the greater area?
Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area 101
Perimeter = ___
Area = ___
Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area 102
Perimeter = ___
Area = ___
Rectangle __ has the greater area.
Answer:
The perimeter of the rectangle A is 7 cm
and the area of the rectangle A is 24 cm2
The perimeter of rectangle B is 26 ft
and the area of the rectangle B is 30 ft2
The rectangle B has a greater area.

Explanation:
Given the length of the rectangle is 7 cm,
and the breadth of the rectangle is 5 cm,
so the perimeter of the rectangle is
p = 2( length + breadth)
= 2(7 + 5)
= 2(12)
= 24 cm.
And the area of the rectangle is
a = length × breadth
= 7 × 5
= 35 cm2.

Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area

we can see in the above figure rectangle with a length of 6.5 cm,
and the breadth is 5.5 cm,
so the perimeter of the rectangle is
p = 2(length + breadth)
= 2(6.5 + 5.5)
= 2(12)
= 24 ft.
And the area of the rectangle is
area = length × breadth
= 6.5×5.5
= 35.75 square feet
So the rectangle B has a greater area.

Question 2.
Patterns
Complete the pattern. Find the area of each rectangle.
Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area 103
Each rectangle has the same perimeter. As the area increases, what do you notice about the shape of the rectangle?
Answer:
As the area increases the shape of the figure was changed, we can see the figure was changed from rectangle to square.

Explanation:
As we know that the perimeter of the above figure is the same,
so the perimeter of the above figures is
p = 2 (length + breadth)
= 2 (1 m+9 m)
= 2(10 m)
= 20 m.
So, the perimeter of the above figures is 20 m.
The area of figure 1 is
area = length × breadth
= 1 m × 9 m
= 9 m2.
The area of figure 2 is
= 8m × 2m
= 16 m2.
The area of figure 3 is
= 7m × 3m
= 21 m2.
Let the length of figure 4 be 6m and the breadth be 4m,
The area of figure 4 is
= 6m × 4m
= 24 m2.
Let the length of figure 5 be 5m and the breadth be 5m,
The area of figure 5 is
= 5m × 5m
= 25 m2.

Question 3.
Modeling Real Life
You are making a card with a 36-centimeter ribbon border. How long and wide should you make the card so you have the greatest possible area to write?
Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area 104
Answer:
The length and the breadth of the card is 9 cm.

Explanation:
Given the perimeter of the card with a ribbon border is 36 cm
so the length of the card is
p =4a
36 = 4a
a= 36/4
= 9 cm.
The length and the breadth of the card is 9 cm.
The area of the card is
a = length×breadth
= 9×9
= 81 square cm.

Question 4.
DIG DEEPER!
A school has two rectangular playgrounds that each have the same perimeter. The first playground is shown. The second has a lesser area than the first. Draw one way the second playground could look.
Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area 105

The school builds another playground. It has the same perimeter as the first. The third playground has a greater area than the first. Draw one way the third playground could look
Answer:

Explanation:
Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area

Review & Refresh

Question 5.
2 × 30 = ___
Answer:
60

Explanation:
On multiplying 2 × 30 we will get 60.

Question 6.
6 × 20 = ___
Answer:
120

Explanation:
On multiplying 6 × 20 we will get 120.

Question 7.
3 × 90 = ___
Answer:
270

Explanation:
On multiplying 3 × 90 we will get 270.

Lesson 15.5 Same Area, Different Perimeters

Explore and Grow

Use color tiles to create two different rectangles that each have an area of 18 square units. Then draw your rectangles and label their dimensions. Do the rectangles have the same perimeter? Explain how you know
Big Ideas Math Answers 3rd Grade Chapter 15 Find Perimeter and Area 106

Answer:
By comparing the perimeters of both rectangles, we can see that the perimeters are not the same.

Explanation:
Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area
In the above figure, we can see two colored rectangles
the length of rectangle 1 is 6 units
and the breadth of rectangle 1 is 3 units
so the perimeter of rectangle 1 is
p = 2(length+breadth)
= 2(6+3)
= 2(9)
= 18 units
and the area of rectangle 1 is
area= length×breadth
= 6×3
= 18 square units
The length of rectangle 2 is 9 units
and the breadth of rectangle 2 is 2 units
so the perimeter of rectangle 2 is
p = 2(length+breadth)
= 2(9+2)
= 2(11)
= 22 units.
and the area of rectangle 2 is
area= length×breadth
= 9×2
= 18 square units.
By comparing the perimeters of both rectangles, we can see that the perimeters are not the same.

Repeated Reasoning
As the perimeter increases and the area stays the same, what do you notice about the shape of the rectangle?

Think and Grow : Same Area, Different Perimeters

Example
Find the area and the perimeter of Rectangle A. Draw a different rectangle that has the same area. Which rectangle has the lesser perimeter?
Big Ideas Math Answers 3rd Grade Chapter 15 Find Perimeter and Area 107
Area = 2 × 6
= _____
Perimeter = 6 + 2 + 6 + 2
= ______
Big Ideas Math Answers 3rd Grade Chapter 15 Find Perimeter and Area 108
Area = ___ × ___
= ____
Perimeter = ___ + ___ + ___ + ___
= ____
Rectangle ___ has the lesser perimeter.
Answer:
The perimeter of the rectangle A is 16 ft
and the area of the rectangle A is 12 ft2
The perimeter of rectangle B is 14 ft
and the area of the rectangle B is 12 ft2
The rectangle A has a lesser area.

Explanation:
Given the length of the rectangle is 6 ft,
and the breadth of the rectangle is 2 ft,
so the perimeter of the rectangle is
p = 2( length + breadth)
= 2(6 + 2)
= 2(8)
= 16 ft.
And the area of the rectangle is
a = length × breadth
= 6 × 2
= 12 ft2

Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area
we can see in the above figure rectangle B and the length be 4 ft,
and the breadth be 3ft,
so the perimeter of the rectangle is
p = 2(length + breadth)
= 2(4 + 3)
= 2(7)
= 14 ft.
And the area of the rectangle is
area = length × breadth
= 4×3
= 12 ft2
So the rectangle A has a lesser area.

Show and Grow

Question 1.
Find the area and the perimeter of Rectangle A. Draw a different rectangle that has the same area. Which rectangle has the lesser perimeter?
Big Ideas Math Answers 3rd Grade Chapter 15 Find Perimeter and Area 109
Area = ___
Perimeter = ___
Big Ideas Math Answers 3rd Grade Chapter 15 Find Perimeter and Area 110
Area = ___
Perimeter = ___
Rectangle __ has the lesser perimeter.
Answer:
The perimeter of the rectangle A is 16 ft
and the area of the rectangle A is 12 ft2
The perimeter of rectangle B is 14 ft
and the area of the rectangle B is 12 ft2
The rectangle A has a lesser area.

Explanation:
Given the length of the rectangle is 6 cm,
and the breadth of the rectangle is 6 cm,
so the perimeter of the rectangle is
p = 2( length + breadth)
= 2(6 + 6)
= 2(12)
= 24 cm.
And the area of the rectangle is
a = length × breadth
= 6 × 6
= 36 cm2

Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area

we can see in the above figure rectangle B and the length be 9 cm,
and the breadth be 4 cm,
so the perimeter of the rectangle is
p = 2(length + breadth)
= 2(9 + 4)
= 2(13)
= 26 cm.
And the area of the rectangle is
area = length × breadth
= 9×4
= 36 cm2
So the rectangle A has a lesser perimeter.

Apply and Grow: Practice

Find the area and the perimeter of Rectangle A. Drawa different rectangle that has the same area. Which rectangle has the lesser perimeter?
Answer:

Question 2.
Rectangle A
Big Ideas Math Answers 3rd Grade Chapter 15 Find Perimeter and Area 111
Area = ___
Perimeter = ___
Rectangle B
Big Ideas Math Answers 3rd Grade Chapter 15 Find Perimeter and Area 112
Area = ___
Perimeter = ___
Rectangle ___ has the lesser perimeter.
Answer:
The perimeter of the rectangle A is 24 in
and the area of the rectangle A is 20 in2
The perimeter of rectangle B is 18 in
and the area of the rectangle B is 20 in2
The rectangle B has a lesser perimeter.

Explanation:
Given the length of the rectangle is 10 in,
and the breadth of the rectangle is 2 in,
so the perimeter of the rectangle is
p = 2( length + breadth)
= 2(10 + 2)
= 2(12)
= 24 in.
And the area of the rectangle is
a = length × breadth
= 10 × 2
= 20 in2

Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area

we can see in the above figure rectangle B and the length be 5 in,
and the breadth be 4 in,
so the perimeter of the rectangle is
p = 2(length + breadth)
= 2(5 + 4)
= 2(9)
= 18 in.
And the area of the rectangle is
area = length × breadth
= 5×4
= 20 in2
So the rectangle B has a lesser perimeter.

Question 3.
Rectangle A
Big Ideas Math Answers 3rd Grade Chapter 15 Find Perimeter and Area 113
Area = ___
Perimeter = ___
Rectangle B
Big Ideas Math Answers 3rd Grade Chapter 15 Find Perimeter and Area 114
Area = ___
Perimeter = ___
Rectangle ___ has the lesser perimeter.
Answer:
The perimeter of the rectangle A is 12 m
and the area of the rectangle A is 8 m2
The perimeter of rectangle B is 18 m
and the area of the rectangle B is 8 m2
The rectangle A has a lesser perimeter.

Explanation:
Given the length of the rectangle is 4 m,
and the breadth of the rectangle is 2 m,
so the perimeter of the rectangle is
p = 2( length + breadth)
= 2(4 + 2)
= 2(6)
= 12 m.
And the area of the rectangle is
a = length × breadth
= 4 × 2
= 8 m2

Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area

we can see in the above figure rectangle B and the length be 8 m,
and the breadth be 1 m,
so the perimeter of the rectangle is
p = 2(length + breadth)
= 2(8+ 1)
= 2(9)
= 18 m.
And the area of the rectangle is
area = length × breadth
= 8×1
= 8 m2
So the rectangle A has a lesser perimeter.

Question 4.
DIG DEEPER!
The perimeter of a blue rectangle is 10 feet. The perimeter of a green rectangle is 14 feet. Both rectangles have the same area. Find the area and the dimensions of each rectangle.
Answer:

Explanation:
Given the perimeter of the blue rectangle is 10 ft and
the perimeter of the green rectangle is 14 ft

Think and Grow: Modeling Real Life

You have 40 square patio bricks that are each 1 foot long and 1 foot wide. You want to make a rectangular patio with all of the bricks. How long and wide should you make the patio so it has the least possible perimeter?
Big Ideas Math Answers 3rd Grade Chapter 15 Find Perimeter and Area 115
Draw to show:
You should make the patio ___ feet long and ___ feet wide.
Answer:
You should make the patio 8 feet long and 5 feet wide.
The least possible perimeter is 26 feet.

Explanation:
Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area
As there are 40 square patio bricks and each brick is 1 foot long and 1 foot wide
so to make a rectangular patio we need
the length of the rectangular patio be 8 feet
and the breadth of the rectangular patio be 5 feet
so the perimeter of the rectangular patio is
p= 2(length+breadth)
= 2(8+5)
= 2(13)
= 26 feet.
The least possible perimeter is 26 feet.

Show and Grow

Question 5.
Your friend has 16 square foam tiles that are each 1 foot long and 1 foot wide. He wants to make a rectangular exercise space with all of the tiles. How long and wide should he make the exercise space so it has the least possible perimeter?
Big Ideas Math Answers 3rd Grade Chapter 15 Find Perimeter and Area 116
Answer:
The least possible perimeter is 16 feet.

Explanation:
As there are 16 square foam tiles and each foam tile is 1 foot long and 1 foot wide
so to make an exercise space we need
the length of the exercise space be 4 feet
and the breadth of the exercise space be 4 feet
so the perimeter of the exercise space is
p= 2(length+breadth)
= 2(4+4)
= 2(8)
= 16 feet.
The least possible perimeter is 16 feet.

Question 6.
DIG DEEPER!
You and your friend each use fencing to make a rectangular playpen for a puppy. Each pen has the same area. Your pen is shown. Your friend’s pen uses less fencing than yours. Draw one way your friend could make her pen.
Big Ideas Math Answers 3rd Grade Chapter 15 Find Perimeter and Area 117

Your cousin makes a playpen for a puppy. His pen has the same area as your pen. Your cousin’s pen uses more fencing than yours. Draw one way your cousin could make his pen.
Answer:

 

Same Area, Different Perimeters Homework & Practice 15.5

Question 1.
Find the area and the perimeter of Rectangle A. Drawa different rectangle that has the same area. Which rectangle has the lesser perimeter?
Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area 118
Area = ___
Perimeter = ___
Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area 119
Area = ___
Perimeter = ___
Rectangle __ has the lesser perimeter.
Answer:
The perimeter of rectangle A is 4 in
and the area of rectangle A is 20 square inches.
It is not possible to draw a rectangle that has the same area and different perimeter.

Explanation:
Given the length of the rectangle is 4 in,
and the breadth of the rectangle is 4 in,
so the perimeter of the rectangle is
p = 2( length + breadth)
= 2(4+4)
= 2(8)
= 16 in.
And the area of the rectangle is
a = length × breadth
= 4 × 4
= 16 square inches.

Question 2.
Structure
The dimensions of a rectangle are 4 feet by 10 feet. Which shape has the same area, but a different perimeter?
Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area 120
Answer:
The yellow shape rectangle has the same area and different perimeter.

Explanation:
Given the dimensions of the rectangle are 4 feet by 10 feet
so the perimeter of the rectangle is
p= 2(length+breadth)
= 2(4+10)
= 2(14)
= 28 feet.
The area of the rectangle is
area = length×breadth
= 10×4
= 40 square feet.
Here, we can see the yellow rectangle has a length of 8 feet
and the breadth of the rectangle is 5 feet
so the perimeter of the rectangle is
p = 2(length+beadth)
= 2(8+5)
= 2(13)
= 26 feet.
and the area of the rectangle is
area = length×breadth
= 8×5
= 40 square feet.

Question 3.
MP Reasoning
The two fields have the same area. Players run one lap around each field. At which field do the players run farther?
Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area 121
Answer:
As the perimeter of field A is greater than field B so in field A the players run father.

Explanation:
Let the length of field A is 10m
and the breadth of field A is 2m
so the perimeter of field A is
p = 2(length+breadth)
= 2(10+2)
= 2(12)
= 24 m.
and the area of field A is
area= length×breadth
= 10×2
= 20 square meters.
Let the length of field B is 5m
and the breadth of field B is 4m
so the perimeter of field B is
p = 2(length+breadth)
= 2(5+4)
= 2(9)
= 18 m.
and the area of field A is
area= length×breadth
= 5×4
= 20 square meters.
As the perimeter of field A is greater than field B so in field A the players run father.

Question 4.
Modeling Real Life
You have 24 square pieces of T-shirt that are each 1 foot long and 1 foot wide. You want to make a rectangular T-shirt quilt with all of the pieces. How long and wide should you make the quilt so it has the least possible perimeter?
Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area 122
Answer:
The least possible perimeter is 20 feet.

Explanation:
As there are 24 square pieces of T-shirt and each was 1 foot long and 1 foot wide
so to make a rectangular T-shirt quilt with all of the pieces we need
the length of the rectangular T-shirt quilt be 6 feet
and the breadth of the rectangular T-shirt quilt be 4 feet
so the perimeter of the rectangular T-shirt quilt is
p= 2(length+breadth)
= 2(6+4)
= 2(10)
= 20 feet.
The least possible perimeter is 20 feet.

Question 5.
DIG DEEPER!
You and Descartes each have40 cobblestone tiles to arrange in to a rectangular pathway. Your pathway is shown. Descartes’s pathway has a lesser perimeter than yours. Draw one way Descartes could make his pathway.
Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area 123
Newton also makes a rectangular pathway with 40 cobblestone tiles. His pathway has a greater perimeter than yours. Draw one way Newton could make his pathway.
Answer:

Review & Refresh

Identify the number of right angles and pairs of parallel sides.

Question 6.
Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area 124
Right angles: ___
Pairs of Parallel sides: ___
Answer:
Right angles: 1
Pairs of Parallel sides: 2.

Explanation:
In the above figure, we can see there are the right angle is 1, and the pairs of parallel sides are 2.

Question 7.
Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area 125
Right angles: ___
Pairs of parallel sides: ____
Answer:
Right angles: 4.
Pairs of Parallel sides: 2.

Explanation:
In the above figure, we can see there are the right angle is 4, and the pairs of parallel sides are 2.

Find Perimeter and Area Performance Task

You and your cousin build a tree house.

Question 1.
The floor of the tree house is in the shape of a quadrilateral with parallel sides that are 4 feet long and 10 feet long. The other 2 sides are equal in length. The perimeter is 24 feet. Sketch the floor and label all of the side lengths.
Big Ideas Math Answer Key Grade 3 Chapter 15 Find Perimeter and Area 151
Answer:
The length of the sides is 6 feet.

Explanation:
The perimeter of the floor is 24 feet
so the length of the floor be 6 feet
as the two other sides are also equal
so the other side length also be 6 feet
and let’s check the perimeter
p = 2(length+breadth)
= 2(6+6)
= 2(12)
= 24 feet.

Question 2.
Each rectangular wall of the tree house is 5 feet tall. How many square feet of wood is needed for all of the walls?
Answer:

Explanation:

Question 3.
You cut out a door in the shape of a rectangle with sides that are whole numbers. Its area is 8 square feet. What is the height of the door?
Answer:
The height of the rectangular door is 4 feet.

Explanation:
The area of the rectangular shape door is 8 square feet
as the sides of the rectangular door are whole numbers
so the length rectangular door be 4 feet
and the breadth be 2 feet
then we can get the area 8 square feet
let’s check the area
area = length×breadth
= 4×2
= 8 square feet.
So the height of the rectangular door is 4 feet.

Question 4.
You want to paint the floor and walls on the inside of your tree house. The area of the floor is 28 square feet. Each quart of paint covers 100 square feet.
a. How many quarts of paint do you need to buy?
b. Do you have enough paint to paint the outside walls of the tree house? Explain.
Big Ideas Math Answer Key Grade 3 Chapter 15 Find Perimeter and Area 152
Answer:
a. 2,800 square feet quarts of paint we need to buy.

Explanation:
a. The area of the floor is 28 square feet and each quart of paint covers 100 square feet, so we need to buy
28×100= 2,800 square feet quarts of paint.

b.

Find Perimeter and Area Activity

Perimeter Roll and Conquer
Big Ideas Math Answer Key Grade 3 Chapter 15 Find Perimeter and Area 153
Directions:
1. Players take turns rolling two dice.
2. On your turn, draw a rectangle on the board using the numbers on the dice as the side lengths. Your rectangle cannot cover another rectangle.
3. Write an equation tofind the perimeter of the rectangle.
4. If you cannot fit a rectangle on the board, then you lose your turn. Play 10 rounds, if possible.
5. Add all of your rectangles’ perimeters together. The player with the greatest sum wins!
Big Ideas Math Answer Key Grade 3 Chapter 15 Find Perimeter and Area 154
Answer:

Explanation:

Find Perimeter and Area Chapter Practice

15.1 Understand Perimeter

Find the perimeter of the figure

Question 1.
Big Ideas Math Answer Key Grade 3 Chapter 15 Find Perimeter and Area 156
Perimeter = ___
Answer:
The perimeter of the rectangle is 18 cm.

Explanation:
In the above figure, we can see the rectangle
with a length of 5 cm,
and the breadth of 4 cm
the perimeter of the rectangle is
p = 2 (length + breadth)
= 2 (5+4)
= 2(9)
= 18 cm.

Question 2.
Big Ideas Math Answer Key Grade 3 Chapter 15 Find Perimeter and Area 157
Perimeter = ___
Answer:
The perimeter of the figure is 26 ft.

Explanation:
To find the perimeter of the above figure,
we will add the lengths of all sides of the figure
the sides of the above figure is 2 ft, 8 ft, 3 ft, 2 ft, 2 ft, 2 ft, 1 ft, 1 ft, 2 ft, 3 ft
the perimeter of the above figure is
p = 2+8+3+2+2+2+1+1+2+3
= 26 ft.

Question 3.
Draw a figure that has a perimeter of 10 inches.
Big Ideas Math Answer Key Grade 3 Chapter 15 Find Perimeter and Area 158
Answer:

15.2 Find Perimeter of Polygons

Find the perimeter of the polygon

Question 4.
Big Ideas Math Answer Key Grade 3 Chapter 15 Find Perimeter and Area 159
Perimeter = ___
Answer:
The perimeter of the polygon is 33 cm.

Explanation:
To find the perimeter of the polygon, we will add all the sides of the polygon
so the sides of the polygon are 9 cm, 6 cm, 8 cm, 10 cm
the perimeter of the polygon is
p = 9 cm +6 cm +8 cm +10 cm
= 33 cm.

Question 5.
Big Ideas Math Answer Key Grade 3 Chapter 15 Find Perimeter and Area 160
Perimeter = ___
Answer:
The perimeter of the figure is 27 ft.

Explanation:
To find the perimeter of the figure, we will add all the sides of the figure
so the sides of the perimeter is 5 ft, 11 ft, 7 ft, 3 ft, 1 ft
the perimeter of the figure is
p = 5 ft+11 ft+ 7 ft+3 ft+1 ft
= 27 ft.

Question 6.
Parallelogram
Big Ideas Math Answer Key Grade 3 Chapter 15 Find Perimeter and Area 161
Perimeter = ___
Answer:
The perimeter of the parallelogram is 12 m.

Explanation:
Given the length of the parallelogram is 4 m
and the breadth of the parallelogram is 2 m
the perimeter of the parallelogram is
p = 2 (length + breadth)
= 2( 4 m+ 2 m)
= 2(6 m)
= 12 m

Find the perimeter of the polygon

Question 7.
Rhombus
Big Ideas Math Answer Key Grade 3 Chapter 15 Find Perimeter and Area 162
Perimeter = ___
Answer:
The perimeter of the rhombus is 36 cm.

Explanation:
Given the length of the side of the rhombus is 9 cm
and the perimeter of the rhombus is
p = 4a
= 4× 9 cm
= 36 cm.

Question 8.
Rectangle
Big Ideas Math Answer Key Grade 3 Chapter 15 Find Perimeter and Area 163
Perimeter = ___
Answer:
The perimeter of the rectangle is 26 in.

Explanation:
The length of the rectangle is 5 inch
and the breadth of the rectangle is 8  inch
the perimeter of the rectangle is
p = 2( length + breadth)
= 2( 5 in+ 8 in)
= 2(13 in)
= 26 in.
So the perimeter of the rectangle is 26 in.

Question 9.
Square
Big Ideas Math Answer Key Grade 3 Chapter 15 Find Perimeter and Area 164
Perimeter = ___
Answer:
The perimeter of the square is 28 ft.

Explanation:
The length of the square is 7 ft
so the perimeter of the square is
perimeter= 4a
= 4×7
= 28 ft.

Question 10.
Modeling Real Life 
You want to put lace around the tops of the two rectangular lampshades. How many centimeters of lace do you need?
Big Ideas Math Answer Key Grade 3 Chapter 15 Find Perimeter and Area 165
Answer:
We need 1,120 square centimeters.

Explanation:
The length of the rectangular lampshades is 35 cm
The breadth of the rectangular lampshades is 32 cm
and the area of the rectangular lampshades is
area= length×breadth
= 32×35
= 1,120 square cm.
So 1,120 square centimeters of lace you need.

15.3 Find Unknown Side Lengths

Find the unknown side length.

Question 11.
Perimeter = 22 feet
Big Ideas Math Answers 3rd Grade Chapter 15 Find Perimeter and Area 166
d = ____
Answer:
The length of the other side is 9 ft.

Explanation:
Given the perimeter of the above figure is 22 feet
and the length of the sides of the figure is 6 ft, 7 ft, d ft
so the perimeter of the figure is
p = 6 ft+ 7 ft+ d ft
22  ft = 13 ft + d ft
d = 22 ft – 13 ft
= 9 ft.
So, the length of the other side is 9 ft.

Question 12.
Perimeter = 31 inches
Big Ideas Math Answers 3rd Grade Chapter 15 Find Perimeter and Area 167
k = ___
Answer:
The length of the other side is 5 in.

Explanation:
Given the perimeter of the above figure is 31 inches
and the length of the sides of the figure is 10 in, 4 in, 12 in and k in.
so the perimeter of the figure is
p = 10 in+ 4 in+ 12 in+k in
31 in = 26 in + k in
k = 31 in – 26 in
= 5 in.
So, the length of the other side is 5 in.

Question 13.
Perimeter = 34 meters
Big Ideas Math Answers 3rd Grade Chapter 15 Find Perimeter and Area 168
y = ___
Answer:
The length of the other side is 7 m.

Explanation:
Given the perimeter of the above figure is 34 meters
and the length of the sides of the figure is 11 m, 8 m, 2 m, 1 m, 5 m, y m.
so the perimeter of the figure is
p = 11 m+ 8 m+ 2 m+ 1 m+ 5 m+ y m
34 m = 27 m + y m
y = 34 m – 27 m
= 7 m.
So, the length of the other side is 7 m.

Find the unknown side length.

Question 14.
Perimeter = 24 feet
Big Ideas Math Answers 3rd Grade Chapter 15 Find Perimeter and Area 169
k = ___
Answer:
The length of the sides of the triangle is 8 feet.

Explanation:
The perimeter of the triangle is 24 feet
and the perimeter of the triangle is
p = 3a
24 feet = 3a
a= 24/3
= 8 feet.
So, the length of the sides of the triangle is 8 feet.

Question 15.
Perimeter = 16 meters
Big Ideas Math Answers 3rd Grade Chapter 15 Find Perimeter and Area 170
y = ___
Answer:
The length of the sides of the square is 4 meters.

Explanation:
The perimeter of the triangle is 16 meters
and the perimeter of the triangle is
p = 4a
16 meters = 4a
a= 16/4
= 4 meters.
So, the length of the sides of the triangle is 4 meters.

Question 16.
Perimeter = 30 inches
Big Ideas Math Answers 3rd Grade Chapter 15 Find Perimeter and Area 171
d = ___
Answer:
The length of the sides of the pentagon is 6 inches.

Explanation:
The perimeter of the pentagon is 30 inches
and the perimeter of the pentagon is
p = 5a
30 inches = 5a
d= 30/5
= 6 inches.
So, the length of the sides of the pentagon is 6 inches.

Question 17.

Number Sense
A rectangle has a perimeter of 38 centimeters. The left side length is 10 centimeters. What is the length of the top side?
Answer:
The length of the top side is 9 cm.

Explanation:
Given the perimeter of the rectangle is 38 cm and
the left side length is 10 cm
Let the length of the top side be X, so
perimeter of the rectangle is
p = 2( length +breadth)
38 = 2(10+X)
38/2 = 10 + X
19 = 10 + X
X = 19 – 10
= 9 cm.
So the length of the top side is 9 cm.

15.4 Same Perimeter, Different Area

Question 18.
Find the perimeter and area of Rectangle A. Drawa different rectangle that has the same perimeter. Which rectangle has the greater area?
Big Ideas Math Answers 3rd Grade Chapter 15 Find Perimeter and Area 172
Perimeter = ____
Area = ___
Big Ideas Math Answers 3rd Grade Chapter 15 Find Perimeter and Area 173
Perimeter = ____
Area = ___
Rectangle ___ has the greater area.
Answer:
The perimeter of the rectangle A is 14 m
The area of the rectangle A is 10 square meters
The perimeter of the rectangle B is 14 m
The area of the rectangle B is 12 square meters
The rectangle B has greater area.

Explanation:
The length of the rectangle is 5m
and the breadth of the rectangle is 2m
the perimeter of the rectangle is
p= 2(length+breadth)
= 2(5+2)
= 2(7)
= 14 m
and the area of the rectangle is
area = length×breadth
= 2×5
= 10 square meters.

Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area img 27

In the above image, we can see the length of the rectangle is 4 m
and the breadth of the rectangle is 3 m
so the perimeter of the rectangle is
p= 2(length+breadth)
= 2(4+3)
= 2(7)
= 14 m.
and the area of the rectangle is
area = length×breadth
= 4×3
= 12 square meters.
The rectangle B has greater area.

Question 19.
Patterns
Each Rectangle has the same perimeter. Are the areas increasing or decreasing ? Explain.
Big Ideas Math Answers 3rd Grade Chapter 15 Find Perimeter and Area 174
Answer:
As we can see in the above images the length of the images was increasing one by one and the breadth is decreasing, so the areas increasing or decreasing will depend upon the breadth of the rectangle. So, if the breadth is also increasing then the area will also be increasing. And if the breadth was decreasing then the area will also be decreasing.

15.5 Same Area, Different Perimeters

Question 20.
Find the area and the perimeter of Rectangle A. Drawa different rectangle that has the same area. Which rectangle has the lesser perimeter?
Big Ideas Math Answers 3rd Grade Chapter 15 Find Perimeter and Area 175
Area = ___
Perimeter = ___
Big Ideas Math Answers 3rd Grade Chapter 15 Find Perimeter and Area 176
Area = ___
Perimeter = ___
Rectangle ___ has the lesser perimeter.
Answer:
The perimeter of the rectangle A is 14 m
The area of the rectangle A is 10 square meters
The perimeter of the rectangle B is 14 m
The area of the rectangle B is 12 square meters
The rectangle B has greater area.

Explanation:
The length of the rectangle is 10 in
and the breadth of the rectangle is 5 in
the perimeter of the rectangle is
p= 2(length+breadth)
= 2(10+5)
= 2(15)
= 30 in
and the area of the rectangle is
area = length×breadth
= 10×5
= 50 square inches.

Question 21.
Reasoning
The two dirt-bike parks have the same area. Kids ride dirt bikes around the outside of each park. At which park do the kids ride farther ? Explain.
Big Ideas Math Answers 3rd Grade Chapter 15 Find Perimeter and Area 177
Answer:
As the length of the park B is longer, so at the park B kids rid farther than the park A.

Find Perimeter and Area Cumulative practice 1 – 15

Question 1.
A mango has a mass that is 369 grams greater than the apple. What is the mass of the mango?
A. 471 grams
B. 369 grams
C. 267 grams
D. 461 grams
Big Ideas Math Answers 3rd Grade Chapter 15 Find Perimeter and Area 178
Answer:
B.

Explanation:
The mass of the mango is 369 grams greater than apple

Question 2.
Which term describes two of the shapes shown, but all three of the shapes?
Big Ideas Math Answers 3rd Grade Chapter 15 Find Perimeter and Area 179
A. polygon
B. rectangle
C. square
D. parallelogram
Answer:
B, C, D.

Explanation:
In the above figures, we can see the parallelogram and the square. As the square is also known as a rectangle so we will choose option B also.

Question 3.
A rectangular note card has an area of 35 square inches. The length of one of its sides is 7 inches. What is the perimeter of the note card?
A. 5 inches
B. 24 inches
C. 84 inches
D. 12 inches
Answer:
The breadth of the rectangular note card is 24 inches.

Explanation:
The area of the rectangular note card is 35 square inches and the length of one of its sides is 7 inches
so the breadth of the rectangular note card is
area = length × breadth
35 = 7 × breadth
breadth = 35/7
= 5 inches.
The perimeter of the rectangular note card is
p = 2(length+breadth)
= 2(7+5)
= 2(12)
= 24 inches.
The breadth of the rectangular note card is 24 inches.

Question 4.
How many minutes are equivalent to4 hours?
A. 400 minutes
B. 240 minutes
C. 24 minutes
D. 40 minutes
Answer:
B

Explanation:
The number of minutes is equivalent to 4 hours is
4× 60= 240 minutes.

Question 5.
A balloon artist has 108 balloons. He has 72 white balloons, and an equal number of red, blue, green, and purple balloons. How many purple balloons does he have?
A. 36
B. 180
C. 9
D. 32
Big Ideas Math Answers 3rd Grade Chapter 15 Find Perimeter and Area 180
Answer:
9 balloons.

Explanation:
As a balloon artist has 108 balloons and he has 72 white balloons
and the remaining balloons are 108 – 72= 36 balloons
and an equal number of red, blue, green, and purple balloons
which means 36 balloons are equally divided by 4 colors of balloons, so
36÷4 = 9 balloons.
So the purple balloons are 9.

Question 6.
Which statements about the figures are true?
Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area 181
Answer:
The shapes have different perimeters.
The shapes have the same area.

Explanation:
The length of the side of the square is 6 in,
and the perimeter of the square is
p = 4a
= 4×6 in
= 24 in
The area of the square is a^2
= 6 in×6 in
36 in^2.

Question 7.
The graph show many students ordered each lunch option.
Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area 182
Part A How many students ordered lunch?
Part B Choose a lesser value for the key. How will the graph change?
Answer:
Part A: 60 students ordered lunch.
Part B: Turkey hot dog has a lesser value.

Explanation:
Part A:
The number of students who ordered lunch is
the grilled chicken was ordered by 21 students
Turkey hot dog was ordered by 9 students
A peanut butter and jelly sandwich was ordered by 12 students
the salad bar was ordered by 18 students
so the number of students who ordered lunch is
21+9+12+18= 60 students.

Part B:
The turkey hot dog was ordered by 9 students which is a lesser value.

Question 8.
Find the sum
Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area 183
Answer:
935

Explanation:
The sum of the above given numbers is 935

Question 9.
What is the perimeter of the figure?
Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area 184
A. 26 units
B. 22 units
C. 20 units
D. 16 unit
Answer:
B

Explanation:
The sides of the figure is 2,4,1,2,1,1,2,2,1,1,2,1,1,1
and the perimeter of the figure is
p = 2+4+1+2+1+1+2+2+1+1+2+1+1+1
= 22 units.

Question 10.
Which bar graph correctly shows the data?
Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area 185
Answer:
Graph B.

Explanation:
Graph B shows the correct graph data.

Question 11.
Which polygons have at least one pair of parallel sides?
Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area 186
Answer:
Red color polygon.

Explanation:
The red color polygon has one pair of parallel sides, as it is a trapezoid.

Question 12.
The perimeter of the polygon is 50 yards. What is the missing side length?
A. 41 yards
B. 10 yards
C. 91 yards
D. 9 yards
Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area 187
Answer:
The missing side length is 9 yards.

Explanation:
Given the perimeter of the polygon is 50 yards
Let the missing side length be X yd
and the lengths of the sides of the polygon is 15 yds, 6 yds, 13 yds, 7 yds, and X yd,
So the perimeter of the polygon is
p = 15 yd+6 yd+13 yd+ 7 yd+ X yd
50 yards = 41 yards + X Yards
X = 9 yards.
So the missing side length is 9 yards.

Question 13.
Which line plot correctly shows the data?
Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area 188
Answer:
A

Explanation:
Option A line plot shows the correct data.

Question 14.
Your friend is asked to draw a quadrilateral with four right angles. She says it can only be a square. Is she correct?
A. Yes, there is no other shape it can be.
B. No, it could also be a rectangle.
C. No, it could also be a hexagon.
D. No, it could also be a trapezoid.
Answer:
Yes, there is no other shape it can be.

Explanation:
Yes, she is correct. There is no other shape than the square with four right angles.

Question 15.
Which numbers round to480 when rounded to the nearest ten?
Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area 189
Answer:
484, 480, 478

Explanation:
The numbers that are rounded to 480 to the nearest ten is 484, 480, 478.

Find Perimeter and Area Cumulative Steam Performance Task 1 – 15

Question 1.
Use the Internet or some other resource to learn more about crested geckos.
a. Write three interesting facts about geckos.
b. Geckos need to drink water every day. Is this amount of water milliliters or liters? Explain.
c. Geckos can live in a terrarium. Is the capacity of this terrarium milliliters liters measured in or?
Big Ideas Math Solutions Grade 3 Chapter 15 Find Perimeter and Area 190
Answer:
a) The three interesting facts about geckos are:
i) Geckos are a type of lizards and their toes help them to stick to any surface except Teflon.
ii) Gecko’s eyes are 350 times more sensitive than human eyes to light.
iii) Some of the pieces of Geckos have no legs and look more like snakes.

b) The number of water Geckos will have is in milliliters only as Geckos will not often drink water.

c)Yes, geckos can live in a terrarium and the capacity of this terrarium is between 120 liters to 200 liters.

Question 2.
Your class designs a terrarium for a gecko.
a. The base of the terrarium is a hexagon. Each side of the hexagon is 6 inches long. What is the perimeter of the base?
b. The terrarium is 20 inches tall. All of the side walls are made of glass. How many square inches of glass is needed for the terrarium?
c. Another class designs a terrarium with a rectangular base. All of its sides are equal in length. The base has the same perimeter as the base your class designs. What is the perimeter of the base? What is the area?
Big Ideas Math Solutions Grade 3 Chapter 15 Find Perimeter and Area 191
Answer:
a. 36 inches.
b.

Explanation:

a.
Given the length of the sides of the hexagon is 6 inches, and
the perimeter of the hexagon is
p = 6a
= 6 × 6
= 36 inches.

Question 3.
An online store sells crested geckos. The store owner measures the length of each gecko in the store. The results are shown in the table.
Big Ideas Math Solutions Grade 3 Chapter 15 Find Perimeter and Area 192
Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area 193
a. Use the table to complete the line plot.
Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area 194
b. How many geckos did the store owner measure?
c. What is the difference in the lengths of the longest gecko and the shortest gecko?
d. How many geckos are shorter than 6\(\frac{1}{4}\) inches?
e.The length of a gecko’s tail is about 3 inches. How would the line plot change if the store owner measured the length of each gecko without its tail?
Answer:
b. The number of geckos the store owner measures is 24.

d. 12

Explanation:

a.
Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area

b. The number of geckos the store owner measures is 24.

d. The number of geckos shorter than 6\(\frac{1}{4}\) inches are 12.

Conclusion:

We hope that the details provided here regarding Big Ideas Math Answers Grade 3 Chapter 15 Find Perimeter and Area is helpful to the students for better preparation. One who wants to become a pro in maths can download BIM Book Grade 3 Chapter 15 Find Perimeter and Area Answers in pdf format. Bookmark our site to get the answers for other chapters of grade 3.

Big Ideas Math Answers Grade 4 Chapter 1 Place Value – Detailed Concepts

Big Ideas Math Answers Grade 4 Chapter 1

Big Ideas Math Answers Grade 4 Chapter 1 Place Value Concepts helps thousands of students to get excellent grades. Help your students keep up to do the math with the reference of Big Ideas 4th Grade Math Book Answer Key Chapter 1 Place Value. Students can effectively solve all problems and grab knowledge and skills by using the 4th standard Big Ideas Math Answer Key. Big Ideas Math Answers Grade 4 Chapter 1 Place Value Concepts will give all that flexibility to build the confidence to do math in an easy way. It’s important to aware of the procedure of solving a problem more than finding the correct answer to improve the skills of your students.

Big Ideas Math Book 4th Grade Chapter 1 Place Value Concepts Answer Key

Help your students strengthen their knowledge by practicing all the Chapter 1 Place Value Concepts. All the questions and answers are explained with images, graphs for easy understanding of students. Do all the given activities and check your answers to test your knowledge. Students can easily promote to the next grade by practicing with the Big Ideas 4th Grade Answer Key.

Lesson 1: Understand Place Value

Lesson 2: Read and Write Multi-Digit Numbers

Lesson 3: Compare Multi-Digit Numbers

Lesson 4: Round Multi-Digit Numbers

Performance Task

Lesson 1.1 Understand Place Value

Explore and Grow

Model the number. Draw to show your model. Then write the value of each digit.
Big Ideas Math Answer Key Grade 4 Chapter 1 Place Value Concepts 1.1 1
Explanation: Every Digit in a number has a place value according to the position of a Digit. We define the place value of a digit by considering the Rightmost digit’s position as “1’s place.” After 1’s place, 10’s place, 100’s place, and so on.

Answer (i):
As per the Explanation,
The Position Value of ‘1’ is: 1000
The position value of ‘2’ is: 200
The position value of ‘7’ is: 70
The position value of ‘5’ is: 5

Answer (ii): As per Explanation,
The position value of ‘3’ is: 3000
The position value of ‘3’ is: 300
The position value of ‘3’ is: 30
The position value of ‘3’ is: 3

Repeated Reasoning
Compare the value of the tens digit to the value of the ones digit. Then do the same with the hundreds and tens digits, and the thousands and hundred’s digits. What do you notice?

Answer:  When we compare the ten’s digit to the value of one’s digit, we can notice that we got the 10’s digit by multiplying ’10’ to the 1’s digit. For comparison of 100’s and 1000’s digit also, we will get 100’s digit is “10 Times” of the 1000’s digit.

Think and Grow: Understand Place Value

A place value chart shows the value of each digit in a number.

The value of each place is 10 times the value of the place to the right.

The place value chart shows how the place values are grouped.

Each group is called a period. In a number, periods are separated by commas.

Big Ideas Math Answer Key Grade 4 Chapter 1 Place Value Concepts 1.1 2

Example
Big Ideas Math Answer Key Grade 4 Chapter 1 Place Value Concepts 1.1 3

• The number, in standard form, is _______.

Answer: 2,75,449

Explanation: The Position of each digit is defined in a number according to place value and we can count “place-value” up to “lakhs ” starting from “1’s place” from the Rightmost position and count from thereon.

• The value of the digit 7 is 7 ten thousand or ______.

Answer: 70,000

Explanation: The Position of each digit is defined in a number according to place value and we can count “place-value” up to “lakhs ” starting from “1’s place” from the Rightmost position and count from thereon.

• The value of the digit 4 in the hundreds place is ______.

Answer: 400

Explanation: The Position of each digit is defined in a number according to place value and we can count “place-value” up to “lakhs ” starting from “1’s place” from the Rightmost position and count from thereon.

• The value of the digit 4 in the tens place is ______.

Answer: 40

Explanation: The Position of each digit is defined in a number according to place value and we can count “place-value” up to “lakhs ” starting from “1’s place” from the Rightmost position and count from thereon.

• The value of the digit 4 in the hundreds place is ______ times the value of the digit 4 in the tens place.

Answer: 10 Times

Explanation: We know the value of ‘4’ in 100’s place: 400

We know the value of ‘4’ in 10’s place: 40

So, we can get ‘400’ when we multiply ’40’ with its place value ’10’.

Show and Grow

Write the value of the underlined digit.

Question 1.

93,517

Answer: The place value of ‘5’ in ‘93,517’ is:  500

Explanation: The position of each digit in a number is defined and it is the value based on the position of the digit. So, the place value of ‘5’ in ‘93,517’ is:  500

Question 2.

685,726

Answer: The place value of ‘8’ in ‘6,85,26’ is :  80,000

Explanation: The position of each digit in a number is defined and it is the value based on the position of the digit. So, the  place value of ‘8’ in ‘6,85,26’ is :  80,000

Question 3.

359,842

Answer: The place value of ‘9’ in ‘359,842’ is:  9000

Explanation: The position of each digit in a number is defined and it is the value based on the position of the digit. So, the place value of ‘9’ in ‘359,842’ is:  9000

Question 4.

483,701

Answer: The place value of ‘4’ in ‘483,701’ is :  400,000

Explanation: The position of each digit in a number is defined and it is the value based on the position of the digit. So, the place value of ‘4’ in ‘483,701’ is :  400,000

Compare the values of the underlined digits.

Question 5.

20 and 200

Answer: The value of ‘2’ in ‘200’ is 10 times the value of ‘2’ in ’20’.

Explanation: The position value of 2 in 200: 200

The position value of 2 in 20: 20

As we notice, the value of 2 in 20 is “10 Times” of the value of 2 in 200.

Question 6.

1,000 and 100

Answer: The value of ‘1’ in ‘100’ is 10 times the value of ‘1’ in ‘1000’.

Explanation: The position value of 1 in 1000: 1000

The position value of 1 in 100: 100

As we notice, the value of 1 in 100 is “10 Times” of the value of 1 in 1000.

Apply and Grow: Practice

Write the value of the underlined digit.

Question 7.

45,802

Answer: The place value of ‘2’ in ‘45,802’ is :  2

Explanation: The position of each digit in a number is defined and it is the value based on the position of the digit. So, the place value of ‘2’ in ‘45,802’ is :  2

Question 8.

97,361

Answer: The place value of ‘7’ in ‘97,361’ is :  7000

Explanation: The position of each digit in a number is defined and it is the value based on the position of the digit. So, the place value of ‘7’ in ‘97,361’ is:  7000

Question 9.

168,392

Answer: The place value of ‘9’ in ‘168,392’ is:  90

Explanation: The position of each digit in a number is defined and it is the value based on the position of the digit. So, the place value of ‘9’ in ‘168,392’ is:  90

Question 10.

807,516

Answer: The place value of ‘8’ in ‘807,516’ is :  800,000

Explanation: The position of each digit in a number is defined and it is the value based on the position of the digit. So, the place value of ‘8’ in ‘807,516’ is :  800,000

Question 11.

400,532

Answer: The place value of ‘5’ in ‘400,532’ is :  500

Explanation: The position of each digit in a number is defined and it is the value based on the position of the digit. So, the place value of ‘5’ in ‘400,532’ is:  500

Question 12.

749,263

Answer: The place value of ‘4’ in ‘749,263’ is :  40,000

Explanation: The position of each digit in a number is defined and it is the value based on the position of the digit. So, the place value of ‘4’ in ‘749,263’ is :  40,000

Question 13.

619,457

Answer: The place value of ‘6’ in ‘619,457’ is :  600,000

Explanation: The position of each digit in a number is defined and it is the value based on the position of the digit. So, the place value of ‘6’ in ‘619,457’ is :  600,000

Question 14.

301,882

Answer: The place value of ‘1’ in ‘301,882’ is :  1,000

Explanation: The position of each digit in a number is defined and it is the value based on the position of the digit. So, the place value of ‘1’ in ‘301,882’ is :  1,000

Compare the values of the underlined digits.

Question 15.

4 and 47

Answer: The value of 4 in 47 is 10 times the value of 4 in 4.

Explanation: The position value of 4 in 47 is: 40

The position value of 4 in 4: 4

As we notice, the value of 4 in 47 is “10 Times” of the value of 4 in 4.

Question 16.

35,649 and 23,799

Answer: The value of 3 in 35,649 is 10 times the value of 3 in 23,799.

Explanation: The position value of 3 in 35,649 is: 30,000

The position value of 3 in 23,799: 3,000

As we notice, the value of 3 in 35,649 is 10 times the value of 3 in 23,799.

Question 17.

Your friend is 9 years old. Your neighbor is 90 years old. Your neighbor is how many times as old as your friend?

Answer: Given that your friend is 9 years old and your neighbor is 90 years old. From these two, we can conclude that the neighbor is 10 times the age of your friend.

Question 18.

YOU BE THE TEACHER

Is Newton correct? Explain.
Big Ideas Math Answer Key Grade 4 Chapter 1 Place Value Concepts 1.1 4

Answer:

Question 19.

Writing

Explain the relationship between the place values when the same two digits are next to each other in a multi-digit number.

Answer: We know that the place value of a Digit is dependent on the Position of the Digit in a given number. Hence, even those are the same numbers or different numbers placed next to each other, the preceding digit is 10 times the value of the former number.

Think and Grow: Modeling Real Life

Example

What is the value of the digit 3 in the distance around Saturn? in the distance around Jupiter? How do these values relate to each other?
Big Ideas Math Answer Key Grade 4 Chapter 1 Place Value Concepts 1.1 5

Write each number in a place value chart.

Big Ideas Math Answer Key Grade 4 Chapter 1 Place Value Concepts 1.1 6

Compare the values of the digit 3 in each number.

Answer: The value of ‘3’ in Jupiter: 30,000

The value of ‘3’ in Saturn: 300,000

So, we can say that the value of ‘3’ in Saturn is 10 times the value of 3 in Saturn.

Saturn: The value of the digit 3 is ______.

Answer: 300,000

Explanation: As we can see, the value of 3 in Saturn is: 300,000

Jupiter: The value of the digit 3 is ______.

Answer: 30,000

Explanation: As we can see, the value of 3 in Jupiter is: 30,000

The value of the digit 3 in the distance around Saturn is ______ times the value of the digit 3 in the distance around Jupiter.

Answer: 10 times

Explanation: The value of ‘3’ in Jupiter: 30,000

The value of ‘3’ in Saturn: 300,000

So, we can say that the value of ‘3’ in Saturn is 10 times the value of 3 in Saturn.

Show and Grow

Use the table above.

Question 20.

The distance around which planet has an 8 in the thousands place?

Answer: Venus

Explanation: From the Table above, we see that the distance around Venus has ‘8’ in 1000’s place.

Question 21.

Compare the value of the 8s in the distance around Saturn.

Answer: The value of 1st 8 in Saturn: 800

The value of 2nd 8 in Saturn: 80

Explanation: From the above values, we can say that the value of 1st 8 is 10 times the value of 2nd 8 in Saturn.

Question 22.

What is the value of the digit 4 in the distance around Earth? in the distance around Neptune? How do these values relate to each other?

Answer: The value of 4 on Earth is 10 times the value of 4 in Neptune.

Explanation: The value of 4 on Earth is: 40,000

The value of 4 in Neptune is: 4,000

By comparing these two values, we can conclude that the value of 4 on Earth is 10 times the value of  4 in Neptune.

Question 23.
Compare the values of the first digits in the distances around Earth and Jupiter. Explain how you can use the values to compare the sizes of these two planets.

Answer:
The value of the 1st digit in Jupiter is 10 times the value of the 1st digit on Earth.

Explanation:
The value of 4 on Earth is: 40,000
The value of 4 in Jupiter is: 400,000
From these 2 values, we can conclude that the value of the 1st digit in Jupiter is 10 times the value of the 1st digit on Earth.

Understand Place Value Homework & Practice 1.1

Write the value of the underlined digit.

Question 1.

79,043

Answer: The place value of 4 in 79,043 is: 40

Explanation: The position of each digit in a number is defined and it is the value based on the position of the digit. So, the place value of 4 in 79,043 is: 40

Question 2.

52,618

Answer: The place value of 8 in 52,618 is: 8

Explanation: The position of each digit in a number is defined and it is the value based on the position of the digit. So, the place value of 8 in 52,618 is: 8

Question 3.

379,021

Answer: The place value of 9 in 379,021 is: 9000

Explanation: The position of each digit in a number is defined and it is the value based on the position of the digit. So, the place value of 9 in 379,021 is: 9000

Question 4.

958,641

Answer: The place value of 6 in 958,641 is: 600

Explanation: The position of each digit in a number is defined and it is the value based on the position of the digit. So, the place value of 6 in 958,641 is: 600

Question 5.

203,557

Answer: The place value of 2 in 203,557 is: 200,000

Explanation: The position of each digit in a number is defined and it is the value based on the position of the digit. So, the place value of 2 in 203,557 is: 200,000

Question 6.

145,860

Answer: The place value of 5 in 145,860 is: 5,000

Explanation: The position of each digit in a number is defined and it is the value based on the position of the digit. So, the place value of 5 in 145,860 is: 5,000

Question 7.

497,384

Answer: The place value of 3 in 497,384 is: 300

Explanation: The position of each digit in a number is defined and it is the value based on the position of the digit. So, the place value of 3 in 497,384 is: 300

Question 8.

612,739

Answer: The place value of 1 in 612,739 is: 10,000

Explanation: The position of each digit in a number is defined and it is the value based on the position of the digit. So, the place value of 1 in 612,739 is: 10,000

Compare the values of the underlined digits.

Question 9.

603 and 6,425

Answer: The value of 6 in 6,425 is 10 times the value of 6 in 603

Explanation: The value of 6 in 603 is: 600

The value of 6 in 6,425 is: 6,000

From these values, we can conclude that the value of 6 in 6,425 is 10 times the value of 6 in 603

Question 10.

930,157 and 89,216

Answer: The value of 9 in 930,157  is 100 times the value of 9 in 89,216

Explanation: The value of 9 in 930,157 is: 900,000

The value of 9 in 89,216 is: 9,000

From these values, we can conclude that the value of 9 in 930,157  is 100 times the value of 9 in 89,216.

Question 11.

A car can travel 50 miles per hour. A tsunami can travel 500 miles per hour. The tsunami is how many times faster than the car?

Answer: The Tsunami is 10 times faster than the car.

Explanation: Given that,

A car can travel 50 miles per hour

A tsunami can travel 500 miles per hour.

From these 2 values, we can conclude that the tsunami is 10 times faster than the car.

Question 12.

Number Sense

In the number 93,825, is the value in the ten thousand’s place 10 times the value in the thousands place? Explain.

Answer: Yes, the value in the ten thousand’s place is 10 times the value in the thousands place.

Explanation: The given number is 93,825.

But, we know how big is the number, the value in the ten thousand’s place is always 10 times the value in the thousands place.

Question 13.

Reasoning

Write the greatest number possible using each number card once. Then write the least six-digit number possible.

Big Ideas Math Answer Key Grade 4 Chapter 1 Place Value Concepts 1.1 7

Answer: The Greatest 6-digit  number is: 986531

The least 6-digit number is: 135689

Explanation: Given digits are : 6 , 1 , 3 , 8 , 9 , 5

By using each digit only once,

The Greatest 6-digit  number is: 986531

The least 6-digit number is: 135689

Question 14.

Modeling Real Life

Use the table.

Big Ideas Math Answer Key Grade 4 Chapter 1 Place Value Concepts 1.1 8

Question 14.

The height of which mountain has a 3 in the thousands place?

Big Ideas Math Answer Key Grade 4 Chapter 1 Place Value Concepts 1.1 9

Answer: Mount Kinabalu

Explanation: From the given table, we can conclude that Mount Kinabalu has ‘3’ in the thousands place.

Question 15.

What is the value of the digit 5 in the height ht of K2? in the height of Mount Everest? How do these values relate to each other?

Answer: The value of the 5 in the height of K2 is: 50

The value of the 5 in the height of Mount. Everest is: 5

Hence,

The value of 5 in K2 is 10 times the value of 5 in Mount. Everest

Explanation: From the given table, we can conclude that

The value of the 5 in the height of K2 is: 50

The value of the 5 in the height of Mount. Everest is: 5

Hence,

The value of 5 in K2 is 10 times the value of 5 in Mount. Everest

Question 16.

The tallest mountain in the world is shown in the table. Which mountain is it?

Answer: Mount. Everest

Explanation: From the given table, the Tallest Mountain in the World is: Mount. Everest

Review & Refresh

Question 17.

Use the graph to answer the questions.

How many seconds did the Peach Street traffic light stay red?

How many more seconds did the Valley Road traffic light stays red than the Elm Street traffic light?

 

Big Ideas Math Answer Key Grade 4 Chapter 1 Place Value Concepts 1.1 10

Answer: The time which the Peach Street traffic light stays red is: 55 seconds

The time which the Valley road traffic light stays red than the Elm Street traffic light is: 15 seconds

Explanation: Let us  consider, Half circle= 5 seconds

From the given graph,

The time which the Peach Street traffic light stays red is: 55 seconds

The time which the valley road traffic light stays red is: 35 seconds

The time which the Elm Street traffic light stays red is: 20 seconds

So,

The time which the Valley road traffic light stays red than the Elm Street traffic light is 15 seconds.

Lesson 1.2 Read and Write Multi-Digit Numbers

Explore and Grow

Model each number. Draw to show your models. Then write each number a different way.

2,186

Answer: Word Form: Two Thousand, One hundred eighty-six

Expanded Form: 2,000 + 100 + 80 + 6

Explanation: We can write the given number in 3 forms. They are:

A) Standard Form  B) Word Form  C) Expanded Form

Now, the Given number is 2,186 which is in Standard Form. So, we have to write the given number in the remaining 2 forms.

Five thousand, two hundred thirteen

Answer:

Standard Form: 5,213

Explanation:

We can write the given number in 3 forms. They are:

A) Standard Form  B) Word Form  C) Expanded Form

Now, the Given number is 5,213 which is in Standard Form. So, we have to write the given number in the remaining 2 forms.

Expanded Form:

Answer: 5,000 + 200 + 10 + 3

Explanation: We can write the given number in 3 forms. They are:

A) Standard Form  B) Word Form  C) Expanded Form

Now, the Given number is Five thousand, two hundred thirteen which is in word Form. So, we have to write the given number in the remaining 2 forms.

3,000 + 600 + 90 + 4

Answer: Standard Form: 3,694

Explanation: Now, the Given number is 3,000 + 600 + 90 + 4 which is in Expanded Form. So, we have to write the given number in the remaining 2 forms.

Word Form:

Answer: Three thousand, Six hundred ninety- four

Explanation: We can write the given number in 3 forms. They are:

A) Standard Form  B) Word Form  C) Expanded Form

Now, the Given number is 3,000 + 600 + 90 + 4  which is in Expanded Form. So, we have to write the given number in the remaining 2 forms.

Structure

Model a different four-digit number. Write the number in many different ways as possible. Compare your work to your partner’s.

Answer: Consider a 4- digit number ” 9,836″

So, we can write “9,836” in:

Standard Form: 9,836

Word Form: Nine thousand, Eight hundred thirty-six

Expanded Form: 9,000 + 800 + 30 + 6

Explanation: We can write the given number in 3 forms. They are:

A) Standard Form  B) Word Form  C) Expanded Form

Now, the given number which is 9,836 is in Standard Form. So, we have to write the given number in the remaining 2 forms.

Think and Grow: Read and Write Multi-Digit Numbers

Example

Write the number in standard form, word form, and expanded form.

Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 8

Standard form:

Answer: 427, 561

Explanation: We can write the given number in 3 forms. They are:

A) Standard Form  B) Word Form  C) Expanded Form

Now, the given number which is 427,561 is in Standard Form. So, we have to write the given number in the remaining 2 forms.

Word form:

Answer: Four hundred Twenty-seven thousand, five hundred sixty-one

Explanation:

We can write the given number in 3 forms. They are:

A) Standard Form  B) Word Form  C) Expanded Form

Now, the given number which is 427,561 is in Standard Form. So, we have to write the given number in the remaining 2 forms.

Expanded form: _____ + _______ + ______ + ______ + ______ + ______

Answer: 400,000 + 20,000 + 7,000 + 500 + 60 + 1

Explanation:

We can write the given number in 3 forms. They are:

A) Standard Form  B) Word Form  C) Expanded Form

Now, the given number which is 427,561 is in Standard Form. So, we have to write the given number in the remaining 2 forms.

Example

Use the place value chart to write the number below in standard form and expanded form.

Big Ideas Math Answers 4th Grade Chapter 1 Place Value Concepts 1.2 2
Standard form:

Word form: fifty-four thousand, two

Expanded form: ____ + ______ + ______

Answer: Standard Form: 54,002

Expanded Form: 50,000 + 4,000 + 0 + 0 +2

Explanation:

We can write the given number in 3 forms. They are:

A) Standard Form  B) Word Form  C) Expanded Form

Now, the given number which is fifty-four thousand, two is in Word Form. So, we have to write the given number in the remaining 2 forms.

Show and Grow

Write the number in two other forms.

Question 1. 1

Standard form:

Answer: 38,650

Explanation:

We can write the given number in 3 forms. They are:

A) Standard Form  B) Word Form  C) Expanded Form

Now, the given number which is 38,650 is in Standard Form. So, we have to write the given number in the remaining 2 forms.

Word form:

Answer: Thirty-eight thousand, six hundred and fifty

Explanation:

We can write the given number in 3 forms. They are:

A) Standard Form  B) Word Form  C) Expanded Form

Now, the given number which is 38,650 is in Standard Form. So, we have to write the given number in the remaining 2 forms.

Expanded form:
Answer: 30,000 + 8,000 + 600 + 50 + 0

Explanation:

We can write the given number in 3 forms. They are:

A) Standard Form  B) Word Form  C) Expanded Form

Now, the given number which is 38,650 is in Standard Form. So, we have to write the given number in the remaining 2 forms.

Question 2.

Standard form:

Answer: 105,098

Explanation:

We can write the given number in 3 forms. They are:

A) Standard Form  B) Word Form  C) Expanded Form

Now, the given number which is 105,098 is in Standard Form. So, we have to write the given number in the remaining 2 forms.

Word form: one hundred five thousand, ninety-eight

Explanation:

We can write the given number in 3 forms. They are:

A) Standard Form  B) Word Form  C) Expanded Form

Now, the given number which is 105,098 is in Standard Form. So, we have to write the given number in the remaining 2 forms.

Expanded form:

Answer: 100,000 + 0 + 5,000 + 0 + 90 + 8

Explanation:

We can write the given number in 3 forms. They are:

A) Standard Form  B) Word Form  C) Expanded Form

Now, the given number which is 105,098 is in Standard Form. So, we have to write the given number in the remaining 2 forms.

Apply and Grow: Practice

Write the number in two other forms.

Question 3.

Standard form:

Answer: 642,050

Explanation:

We can write the given number in 3 forms. They are:

A) Standard Form  B) Word Form  C) Expanded Form

Now, the given number which is 642,050 is in Standard Form. So, we have to write the given number in the remaining 2 forms.

Word form:

Answer: Six hundred forty- two thousand and fifty

Explanation:

We can write the given number in 3 forms. They are:

A) Standard Form  B) Word Form  C) Expanded Form

Now, the given number which is 642,050 is in Standard Form. So, we have to write the given number in the remaining 2 forms.

Expanded form:

Answer: 600,000 + 40,000 + 2,000 + 50

Explanation:

We can write the given number in 3 forms. They are:

A) Standard Form  B) Word Form  C) Expanded Form

Now, the given number which is 642,050 is in Standard Form. So, we have to write the given number in the remaining 2 forms.

Question 4.

Standard form:  134,078

Word form:

Answer: One hundred thirty-four thousand and seventy-eight

Explanation:

We can write the given number in 3 forms. They are:

A) Standard Form  B) Word Form  C) Expanded Form

Now, the given number which is 134,078 is in Standard Form. So, we have to write the given number in the remaining 2 forms.

Expanded form:

Answer:100,000 + 30,000 + 4,000 + 0 + 70 + 8

Explanation:

We can write the given number in 3 forms. They are:

A) Standard Form  B) Word Form  C) Expanded Form

Now, the given number which is 134,078 is in Standard Form. So, we have to write the given number in the remaining 2 forms.

Question 5.

Complete the table.

Big Ideas Math Answers 4th Grade Chapter 1 Place Value Concepts 1.2 3

Answer:

A) Given Word Form is: three thousand, four hundred ninety-seven

Standard Form: 3,497

Expanded Form: 3,000 + 400 + 90 + 7

Explanation:

We can write the given number in 3 forms. They are:

A) Standard Form  B) Word Form  C) Expanded Form

Now, the given number which is three thousand, four hundred ninety-seven is in Word Form. So, we have to write the given number in the remaining 2 forms.

B) Given Expanded Form is: 50,000 + 2,000 + 400 + 80

Standard Form: 52,480

Word Form: Fifty-two thousand four hundred eighty

Explanation:

We can write the given number in 3 forms. They are:

A) Standard Form  B) Word Form  C) Expanded Form

Now, the given number which is 50,000 + 2,000 + 400 + 80 is in Expanded Form. So, we have to write the given number in the remaining 2 forms.

C) Given Standard Form is:  610,010

Word Form: Six hundred ten thousand, ten

Expanded Form: 600,000 + 10,000 + 0 + 10 + 0

Explanation:

We can write the given number in 3 forms. They are:

A) Standard Form  B) Word Form  C) Expanded Form

Now, the given number which is 610,010 is in Standard Form. So, we have to write the given number in the remaining 2 forms.

Question 6.

YOU BE THE TEACHER

The fangtooth fish lives about 6,500 feet underwater. Newton reads this number as “six thousand five hundred.”Descartes reads the number as “sixty-five hundred.”Did they both read the number correctly? Explain.

Big Ideas Math Answers 4th Grade Chapter 1 Place Value Concepts 1.2 4

Answer: Yes

Explanation:

The given number is “6,500”. Newton read it as “Six thousand five hundred” and Descartes read it as “Sixty-five hundred” because “One thousand is equal to 100 hundred”.

Question 7.

Which One Doesn’tBelong?

Which one not does belong with the other three?

eight hundred sixteen thousand, nine

800,000 + 10,000 + 6,000 + 900

816,009

Big Ideas Math Answers 4th Grade Chapter 1 Place Value Concepts 1.2 5

Answer: The “Expanded Form” does not belong among the given three.

Explanation:

We can write any number in 3 Forms. They are;

A) Standard Form  B) Word Form  C) Expanded Form

Now, the given “period Table” can read as “816,009” and ‘9’ has the place-value of 1 but in the Expanded Form, ‘9’ has a place-value of 100.

Think and Grow: Modeling Real Life

Example

Morse code is a code in which numbers and letters are represented by a series of dots and dashes. Use the table to write the number in standard form, word form, and expanded form.

Big Ideas Math Answers 4th Grade Chapter 1 Place Value Concepts 1.2 6

Big Ideas Math Answers 4th Grade Chapter 1 Place Value Concepts 1.2 7
Standard form:

Word form:

Expanded form:

Show and Grow

Question 8.

Use the table above to write the number in standard form, word form, and expanded form.

Big Ideas Math Answers 4th Grade Chapter 1 Place Value Concepts 1.2 8
Answer:

Question 9.

Use the number 20,000 + 9,000 + 400 + 50 to complete the check.

Big Ideas Math Answers 4th Grade Chapter 1 Place Value Concepts 1.2 9

Answer:

Standard Form: 29,450

Word Form: Twenty-nine thousand four hundred fifty

Explanation:

We can write any number in 3 Forms. They are;

A) Standard Form  B) Word Form  C) Expanded Form

Now, the given number is 20,000 + 9,000 + 400 + 50 which is in Expanded Form. So, we have to write the given number in the remaining 2 forms.

Read and Write Multi-Digit Numbers Homework & Practice 1.2

Write the number in two other forms.

Question 1.

Standard form:

Word form:

Expanded form: 500,000 + 40,000 + 3,000 + 200 + 90 + 8

Answer: Standard Form: 543,298

Word Form: Five hundred forty-three thousand, two hundred Ninety-eight

Explanation:

We can write any number in 3 Forms. They are;

A) Standard Form  B) Word Form  C) Expanded Form

Now, the given number is 500,000 + 40,000 + 3,000 + 200 + 90 + 8  which is in Expanded Form. So, we have to write the given number in the remaining 2 forms.

Question 2.

Standard form:

Word form: forty-eight thousand, six hundred three

Expanded form:

Answer: Standard Form: 48,603

Expanded Form: 40,000 + 8,000 + 600 + 0 + 3

Explanation:

We can write any number in 3 Forms. They are;

A) Standard Form  B) Word Form  C) Expanded Form

Now, the given number is forty-eight thousand, six hundred three which is in Word Form. So, we have to write the given number in the remaining 2 forms.

Question 3.

Complete the table.

Big Ideas Math Answers 4th Grade Chapter 1 Place Value Concepts 1.2 10

Answer:

A) 9,629

Word Form: Nine thousand, six hundred and twenty-nine

Expanded Form: 9,000 + 600 + 20 + 9

Explanation:

We can write any number in 3 Forms. They are;

A) Standard Form  B) Word Form  C) Expanded Form

Now, the given number is 9,629 which is in Standard Form. So, we have to write the given number in the remaining 2 forms.

B) 30,000 + 7,000 + 800 + 2

Standard Form: 37,802

Word Form: Thirty-seven thousand, Eight hundred two

Explanation:

We can write any number in 3 Forms. They are;

A) Standard Form  B) Word Form  C) Expanded Form

Now, the given number is 30,000 + 7,000 + 800 + 2 which is in Expanded Form. So, we have to write the given number in the remaining 2 forms.

C) four hundred sixteen thousand, eighty-seven

Standard Form: 416,087

Expanded Form: 400,000 + 16,000 + 0 + 80 + 7

Explanation:

We can write any number in 3 Forms. They are;

A) Standard Form  B) Word Form  C) Expanded Form

Now, the given number is four hundred sixteen thousand, eighty-seven which is in Word Form. So, we have to write the given number in the remaining 2 forms.

Question 4.

Reasoning

Your teacher asks the class to write forty-two thousand, ninety-three in standard form. Which student wrote the correct number? What mistake did the other student make?

Big Ideas Math Answers 4th Grade Chapter 1 Place Value Concepts 1.2 11

Answer: Student B

Explanation: The teacher asked Student A and B to write “forty-two thousand, ninety-three” in Standard form. That means ‘9’ must be in the tens place. But, Student A put the ‘9’ in the hundreds place whereas Student B put the ‘9’ in the tens place.

Question 5.

Logic

The number has two periods. The thousands period is written as six hundred eight thousand in word form. The one period is written as 600 + 80 in expanded form. What is the number?

Answer: 608,680

Explanation: Given that the number has 2 periods. They are 1) The thousands period   2) The hundreds period

The given Thousands period is: Six hundred eight thousand ( Word Form)

The given Hundreds period is: 600 + 80 ( Expanded Form)

On combining the 2 periods, we get the number “608,680” in the Standard Form.

Question 6.

Modeling Real Life

Use the table to write the number in standard form, word form, and expanded form.

Big Ideas Math Answers 4th Grade Chapter 1 Place Value Concepts 1.2 12

Big Ideas Math Answers 4th Grade Chapter 1 Place Value Concepts 1.2 13

Answer:

Question 7.

Modeling Real Life

Use the number 3,000 + 70 + 1 to complete the check.

Big Ideas Math Answers 4th Grade Chapter 1 Place Value Concepts 1.2 14

Answer: Standard Form: 3,071

Word Form: Three thousand, Seventy-one

Explanation:

We can write any number in 3 Forms. They are;

A) Standard Form  B) Word Form  C) Expanded Form

Now, the given number is 3,000 + 70 + 1 which is in Expanded Form. So, we have to write the given number in the remaining 2 forms.

Review & Refresh

Compare

Question 8.

Big Ideas Math Answers 4th Grade Chapter 1 Place Value Concepts 1.2 15

Answer: \(\frac{1}{6} \) is less than \(\frac{2}{6} \)

Explanation:

\(\frac{1}{6} \) = 0.1666

\(\frac{2}{6} \) = 0.333

Hence, \(\frac{1}{6} \) is less than \(\frac{2}{6} \)

Question 9.

Big Ideas Math Answers 4th Grade Chapter 1 Place Value Concepts 1.2 16

Answer: \(\frac{2}{2} \) is greater than \(\frac{2}{3} \)

Explanation:

\(\frac{2}{2} \) = 1

\(\frac{2}{3} \) = 0.666

Hence, \(\frac{2}{2} \) is greater than \(\frac{2}{3} \)

Question 10.

Big Ideas Math Answers 4th Grade Chapter 1 Place Value Concepts 1.2 17

Answer: \(\frac{1}{2} \) is less than \(\frac{3}{4} \)

Explanation:

\(\frac{1}{2} \) = 0.5

\(\frac{3}{4} \) = 0.75

Hence, \(\frac{1}{2} \) is less than \(\frac{3}{4} \)

Question 11.

Big Ideas Math Answers 4th Grade Chapter 1 Place Value Concepts 1.2 18

Answer: \(\frac{1}{4} \) is equal to \(\frac{2}{8} \)

Explanation:

\(\frac{1}{4} \) = 0.25

\(\frac{2}{8} \) = 0.25

Hence, \(\frac{1}{4} \) is equal to \(\frac{2}{8} \)

Lesson 1.3 Compare Multi-Digit Numbers

Explore and Grow

Goal: Make the greatest number possible.

Draw a Number Card. Choose a place value for the digit. Write the digit in the place value chart. Continue until the place value chart is complete.

Big Ideas Math Answers Grade 4 Chapter 1 Place Value Concepts 1.3 1

Compare your number with your partner’s. Whose number is greater?

Answer: Let me consider my number as “A” and my partner number as “B”

Let,

A = 999999 and B = 999998

From  the 2 numbers, we can conclude that

A > B

Construct Arguments

Explain your strategy to your partner. Compare your strategies.

Answer:

Think and Grow: Compare Multi-Digit Numbers

Example

Compare 8,465 and 8,439.

Use a place value chart.

Big Ideas Math Answers Grade 4 Chapter 1 Place Value Concepts 1.3 2

Start at the left. Compare the digits in each place until the digits differ.

Big Ideas Math Answers Grade 4 Chapter 1 Place Value Concepts 1.3 3

Answer:

Step 1: 8 thousand = 8 thousand

Step 2: 4 hundred = 4 hundred

Step 3: 6 tens > 3 tens

So,

8,465 > 8,439

Explanation:

The given 2 numbers are 8,465 and 8,439

By comparing the 2 numbers, we can conclude that

8,465 > 8,439

Show and Grow

Write which place to use when comparing the numbers.

Question 1.

2,423

2,324

Answer: 2,423 is greater than 2,324

Explanation:

The given 2 numbers are 2,423 and 2,324

From these 2 numbers, we can conclude that 2,423 is greater than 2,324.

Question 2.

9,631

9,637

Answer: 9,631 is less than 9,637

Explanation:

The given 2 numbers are 9,631 and 9,637.

From these 2 numbers, we can conclude that 9,631 is less than 9,637

Question 3.

15,728

16,728

Answer: 15,728 is less than 16,728

Explanation:

The given 2 numbers are 15,728 and16,728.

From these 2 numbers, we can conclude that 15,728 is less than 16,728

Compare

Question 4.

Big Ideas Math Answers Grade 4 Chapter 1 Place Value Concepts 1.3 4

Answer: 5,049 is less than 5,082

Explanation:

The given 2 numbers are 5,049 and 5,082.

From these 2 numbers, we can conclude that5,049 is less than 5,082

Question 5.

Big Ideas Math Answers Grade 4 Chapter 1 Place Value Concepts 1.3 5

Answer: 735,283 is equal to 735,283

Explanation:

The given 2 numbers are 735,283 and 735,283.

From these 2 numbers, we can conclude that 735,283 is equal to 735,283

Question 6.

Big Ideas Math Answers Grade 4 Chapter 1 Place Value Concepts 1.3 6

Answer: 43,694 is greater than 3,694

Explanation:

The given 2 numbers are 43,694 and 3,694.

From these 2 numbers, we can conclude that 43,694 is greater than 3,694

Question 7.

Big Ideas Math Answers Grade 4 Chapter 1 Place Value Concepts 1.3 7

Answer: 88,195 is greater than 78,195

Explanation:

The given 2 numbers are88,195 and 78,195.

From these 2 numbers, we can conclude that 88,195 is greater than 78,195

Question 8.

Big Ideas Math Answers Grade 4 Chapter 1 Place Value Concepts 1.3 8

Answer: 6,480 is less than 6,508

Explanation:

The given 2 numbers are 6,480 and 6,508.

From these 2 numbers, we can conclude that 6,480 is less than 6,508

Question 9.

Big Ideas Math Answers Grade 4 Chapter 1 Place Value Concepts 1.3 9

Answer: 321,817 is greater than 312,827

Explanation:

The given 2 numbers are 321,817 and 312,827.

From these 2 numbers, we can conclude that 321,817 is greater than 312,827

Apply and Grow: Practice

Compare.

Question 10.

Big Ideas Math Answers Grade 4 Chapter 1 Place Value Concepts 1.3 10
Answer: 6,052 is less than 6,520

Explanation:

The given 2 numbers are 6,052 and 6,520.

From these 2 numbers, we can conclude that 6,052 is less than 6,520

Question 11.

Big Ideas Math Answers Grade 4 Chapter 1 Place Value Concepts 1.3 11
Answer: 891,634 is greater than 871,634

Explanation:

The given 2 numbers are 891,634 and 871,634

From these 2 numbers, we can conclude that 891,634 is greater than 871,634

Question 12.

Big Ideas Math Answers Grade 4 Chapter 1 Place Value Concepts 1.3 12

Answer: 28,251 is greater than 26,660

Explanation:

The given 2 numbers are28,251 and 26,660

From these 2 numbers, we can conclude that 28,251 is greater than 26,660

Question 13.

Big Ideas Math Answers Grade 4 Chapter 1 Place Value Concepts 1.3 13
Answer: 324,581 is greater than 32,458

Explanation:

The given 2 numbers are 324,581 and 32,458

From these 2 numbers, we can conclude that 324,581 is greater than 32,458

Question 14.

Big Ideas Math Answers Grade 4 Chapter 1 Place Value Concepts 1.3 14
Answer: 230,611 is greater than 230,610

Explanation:

The given 2 numbers are 230,611 and 230,610

From these 2 numbers, we can conclude that 230,611 is greater than 230,610

Question 15.

Big Ideas Math Answers Grade 4 Chapter 1 Place Value Concepts 1.3 15
Answer: 909,900 is greater than 909,009

Explanation:

The given 2 numbers are 909,900 and 909,009

From these 2 numbers, we can conclude that 909,900 is greater than 909,009

Question 16.

Big Ideas Math Answers Grade 4 Chapter 1 Place Value Concepts 1.3 16
Answer: 7,134 is less than 7,634

Explanation:

The given Expanded Form is: 7,000 + 100 + 30 + 4 = 7,134

The given 2 numbers are 7,134 and 7,634

From these 2 numbers, we can conclude that 7,134 is less than 7,634

Question 17.

Big Ideas Math Answers Grade 4 Chapter 1 Place Value Concepts 1.3 17
Answer: 100,003 is greater than 10,003

Explanation:

The given Word Form is: ten thousand, three = 10,003

The given 2 numbers are 100,003 and 10,003

From these 2 numbers, we can conclude that 100,003 is greater than 10,003

Question 18.

Big Ideas Math Answers Grade 4 Chapter 1 Place Value Concepts 1.3 18
Answer: 16,409 is less than 16,490

Explanation:

The given word Form is: Sixteen thousand, four hundred nine = 16,409

The given 2 numbers are 16,409 and16,490

From these 2 numbers, we can conclude that 16,409 is less than 16,490

Question 19.

Big Ideas Math Answers Grade 4 Chapter 1 Place Value Concepts 1.3 19
Answer: 461,329 is equal to 461,329

Explanation:

The given Expanded Form is: 400,000 + 60,000 + 1,000 + 300 + 20 + 9 =461,329

The given 2 numbers are 461,329 and 461,329

From these 2 numbers, we can conclude that 461,329 is equal to 461,329

Question 20.

Two brands of televisions cost $1,598 and $1,998. Which is the lesser price?

Big Ideas Math Answers Grade 4 Chapter 1 Place Value Concepts 1.3 20

Answer: The lesser price is: $1,598

Explanation:

The given costs of two brands of Television are $1,598 and $1,998

From the two costs of Television, we can conclude that the lesser price is: $1,598

Question 21.

DIG DEEPER!

Your friend says she can tell which sum is greater without adding the numbers. How can she tell?

34,593 + 6,781

34,593 + 6,609

Answer: 6,781 is greater tha 6,609

Explanation:

The given sums are:

34,593 + 6,781

34,593 + 6,609

From these two sums, we can see that 6,781 is greater tha 6,609

Question 22.

Number Sense

Write all of the digits that make the number greater than 23,489 and less than 26,472.

2?,650

Answer: 3 and 4

Explanation:

The given numbers are: 23,489 and 26,472

The digits that make the number greater than 23,489 and less than 26,472 are: 3 and 4

Question 23.

YOU BE THE TEACHER

Your friend says 38,675 is less than 9,100 because 3 is less than 9. Is your friend correct? Explain.

Answer: No

Explanation:

The given numbers are 38,675 and 9,100

In 38,675, the highest place-value is: 30,000

In 9,100, the highest place-value is: 9,000

From these, we can conclude that 38,675 is greater than 9,100

Think and Grow: Modeling Real Life

Example

Which playground costs the least?

Write each number in a place value chart.

Big Ideas Math Answers Grade 4 Chapter 1 Place Value Concepts 1.3 21

Big Ideas Math Answers Grade 4 Chapter 1 Place Value Concepts 1.3 22

Answer:

The playground which costs the least is: Green

Explanation: From the given table, we can conclude that “Green” has the least cost which is $35,872

Order the numbers from least to greatest.
_______, ______, _______

Answer: Green, Yellow, Blue

Explanation:

Compare the costs of 3 playgrounds

Green: $35,872    Yellow: $36,827  Blue: $36,927

From these costs, we can conclude that the order is: Green, Yellow, Blue

The ______ playground costs the least.

Answer: Green

Explanation:

Compare the costs of 3 playgrounds

Green: $35,872    Yellow: $36,827  Blue: $36,927

From these costs, we can conclude that the “Green Playground” has the least cost.

Show and Grow

Question 24.

Who received the highest score? Did anyone beat the high score?

Big Ideas Math Answers Grade 4 Chapter 1 Place Value Concepts 1.3 23

Big Ideas Math Answers Grade 4 Chapter 1 Place Value Concepts 1.3 24

Answer: You

Explanation:

The given high score is: 252,980

The score of You is: 254,020

From the given scores, we can conclude that “You” scored the highest score than the given high score.

Question 25.

Name two cities that have a greater population than St. Louis. Name two cities that do not have a greater population than Oakland.

Big Ideas Math Answers Grade 4 Chapter 1 Place Value Concepts 1.3 25

Answer:

The two cities that have a greater population than St. Louis: Tulsa, OK; Cleveland, OH; Oakland, CA

The two cities that do not have a greater population than Oakland: St. Louis, MO; Anchorage, AK

Compare Multi-Digit Numbers Homework & Practice 1.3

Write which place to use when comparing the numbers

Question 1.

396,241

386,201

Answer: Ten thousand’s  place

Explanation: The given numbers are: 396,241 and 386,201

When we want to compare the numbers directly, compare Ten thousand’s place to get the answer.

Question 2.

50,718

50,798

Answer: Tens place

Explanation: The given numbers are: 50,718 and 50,798

When we want to compare the numbers directly, compare Tens place to get the answer.

Question 3.

159,624

459,623

Answer: Hundred thousand place

Explanation: The given numbers are: 159,624 and 459,623

When we want to compare the numbers directly, compare a Hundred thousand place to get the answer.

Compare

Question 4.

Big Ideas Math Answers Grade 4 Chapter 1 Place Value Concepts 1.3 26

Answer: 7,902 is less than 7,912

Explanation:

The given 2 numbers are 7,902 and 7,912

From these 2 numbers, we can conclude that 7,902 is less than 7,912

Question 5.

Big Ideas Math Answers Grade 4 Chapter 1 Place Value Concepts 1.3 27

Answer: 63,871 is less than 63,902

Explanation:

The given 2 numbers are63,871 and 63,902

From these 2 numbers, we can conclude that 63,871 is less than 63,902

Question 6.

Big Ideas Math Answers Grade 4 Chapter 1 Place Value Concepts 1.3 28

Answer: 839,304 is greater than 829,001

Explanation:

The given 2 numbers are 839,304 and 829,001

From these 2 numbers, we can conclude that 839,304 is greater than 829,001

Question 7.

Big Ideas Math Answers Grade 4 Chapter 1 Place Value Concepts 1.3 29

Answer: 48,931 is greater than 48,930

Explanation:

The given 2 numbers are 48,931 and 48,930

From these 2 numbers, we can conclude that 48,931 is greater than 48,930

Question 8.

Big Ideas Math Answers Grade 4 Chapter 1 Place Value Concepts 1.3 30

Answer: 5,739 is equal to 5,739

Explanation:

The given 2 numbers are 5,739 and 5,739

From these 2 numbers, we can conclude that5,739 is equal to 5,739

Question 9.

Big Ideas Math Answers Grade 4 Chapter 1 Place Value Concepts 1.3 31

Answer: 10,000 is less than 100,000

Explanation:

The given 2 numbers are 10,000 and 100,000

From these 2 numbers, we can conclude that 10,000 is less than 100,000

Compare

Question 10.

Big Ideas Math Answers Grade 4 Chapter 1 Place Value Concepts 1.3 32

Answer: 16,420 is greater than 16,402

Explanation:

The given Expanded Form is: 10,000 + 6,000 + 400 + 2 = 16,402

The given 2 numbers are 16,420 and 16,402

From these 2 numbers, we can conclude that 16,420 is greater than 16,402

Question 11.

Big Ideas Math Answers Grade 4 Chapter 1 Place Value Concepts 1.3 33

Answer: 14.000 is less than 114,000

Explanation:

The given Word Form is fourteen thousand = 14,000

The given 2 numbers are 14.000 and 114,000

From these 2 numbers, we can conclude that 14.000 is less than 114,000

Question 12.

Big Ideas Math Answers Grade 4 Chapter 1 Place Value Concepts 1.3 34

Answer: 800,095 is less than 800,695

Explanation:

The given Word Form is eight hundred thousand, ninety-five = 800,095

The given 2 numbers are 800,095 and 800,695

From these 2 numbers, we can conclude that 800,095 is less than 800,695

Question 13.

Big Ideas Math Answers Grade 4 Chapter 1 Place Value Concepts 1.3 35

Answer: 509,733 is equal to 509,733

Explanation:

The given Expanded Form is 500,000 + 9,000 + 700 + 30 + 3 = 509,733

The given 2 numbers are 509,733 and509,733

From these 2 numbers, we can conclude that 509,733 is equal to 509,733

Question 14.

Two different laptops cost $1,050 and$1,150. Which is the lesser price?

Big Ideas Math Answers Grade 4 Chapter 1 Place Value Concepts 1.3 36

Answer: $1,050 is the lesser price.

Explanation:

The cost of two different Laptops are: $1,050 and $1,150

When compare $1,050 and $1,150, we can conclude that $1,050 is the lesser price.

Question 15.

DIG DEEPER!

If the leftmost digits of two multi-digit numbers are 6and 9, can you tell which number is greater? Explain.

Answer: 9 is a greater number

Explanation:

The given leftmost digits of two numbers are 6 and 9.

When we compare 6 and 9, we can conclude that 9 is the greater number.

Question 16.

Writing

Explain why you start comparing digits on the left when comparing multi-digit numbers.

Answer: When we compare multi-digit numbers, we start comparing digits on the leftmost side because those numbers’ positions have the highest place-value.

Question 17.

Modeling Real Life

Use the table to answer the questions.

Name two museums that had a yearly attendance greater than Children’sMuseum of Denver. Name two museums

that had a yearly attendance less than Discovery Place.

Big Ideas Math Answers Grade 4 Chapter 1 Place Value Concepts 1.3 37

Answer:

The museums that had a yearly attendance greater than the Children’s Museum of Denver are:

A) Boston Children’s Museum  B) Children’s Museum of Houston  C) Discovery palace  D) Exploratorium

The museums that had a yearly attendance less than Discovery Place are:

A) Boston Children’s Museum  B) Children’s Museum of Denver  C) Please Touch Museum

Review & Refresh

Round.

Question 18.

654

Nearest ten: _____

Nearest hundred: _____

Answer: 50, 600

Explanation:

The given number is: 654

When we round off the tens place, the nearest ten is: 50

When we round off the hundreds place, the nearest hundred is: 600

Question 19.

709

Nearest ten: _____

Nearest hundred: ______

Answer: 0, 700

Explanation:

The given number is: 709

When we round off the tens place, the nearest ten is: 0 ( Since the tens place is 0)

When we round off the hundreds place, the nearest hundred is: 700

Question 20.

25

Nearest ten: _____

Nearest hundred: _______

Answer: 20, 0

Explanation:

The given number is: 25

When we round off the tens place, the nearest ten is: 20

When we round off the hundreds place, the nearest hundred is: 0 ( Since the hundreds place is 0)

Lesson 1.4 Round Multi-Digit Numbers

Explore and Grow

Write five numbers that round to 250 when rounded to the nearest ten.

Big Ideas Math Solutions Grade 4 Chapter 1 Place Value Concepts 1.4 1

Answer: 210, 220, 230, 240, 250

Explanation:

The five numbers that round to 250 when rounded to the nearest ten are: 210, 220, 230, 240, 250

 

Write five numbers that round to 500 when rounded to the nearest hundred.

Big Ideas Math Solutions Grade 4 Chapter 1 Place Value Concepts 1.4 2

Answer: 100, 200, 300, 400, 500

Explanation:

The five numbers that round to 500 when rounded to the nearest hundred are: 100, 200, 300, 400, 500

Repeated Reasoning

Explain your strategy. Then explain how you could use the strategy to round a four-digit number to the nearest thousand.

Answer:

Think and Grow: Round Multi-Digit Numbers

To round a number, find the multiple of 10, 100, 1,000, and so on, that is closest to the number. You can use a number line or place value to round numbers.

Example

Use a number line to round 4,276 to the nearest thousand.
Big Ideas Math Solutions Grade 4 Chapter 1 Place Value Concepts 1.4 3

4,276 is closer to 4,000 than it is to 5,000.

So, 4,276 rounded to the nearest thousand is _______.

Answer: 4,000

Explanation: The number 4276 can be round off to 4,000 when rounded to the nearest thousand.

Example

Use place value to round 385,617 to the nearest ten thousand.

Big Ideas Math Solutions Grade 4 Chapter 1 Place Value Concepts 1.4 4

So, 385,617 rounded to the nearest thousand is _______.

Answer: 390,000

Explanation:

The number 385,617 can be round off to 390,000 when rounded to the nearest thousand. ( Since the thousands position is greater than 5)

Show and Grow

Round the number to the place of the underlined digit.

Question 1.

6,912

Answer: 7000

Explanation:

The number 6,912 can be round off to 7,000 when rounded to the nearest thousand. ( Since the thousands position is greater than 5)

Question 2.

43,215

Answer: 43,000

Explanation:

The number 43,215 can be round off to 43,000 when rounded to the nearest thousand. ( Since the thousands position is less than 5)

Question 3.

25,883

Answer: 25,000

Explanation:

The number 25,883 can be round off to 25000 when rounded to the nearest thousand. ( Since the thousands position is less than 5)

Question 4.

148,796

Answer: 140,000

Explanation:

The number 148,796 can be round off to 140000 when rounded to the nearest thousand. ( Since the thousands position is less than 5)

Question 5.

Round 5,379 to the nearest thousand.

Answer: 5,000

Explanation:

The number 5,379 can be round off to 5,000 when rounded to the nearest thousand. ( Since the thousands position is equal to 5)

Question 6.

Round 70,628 to the nearest ten thousand.

Answer: 70,000

Explanation:

The number 70,628 can be round off to 70,000 when rounded to the nearest ten thousand.

Question 7.

Round 362,113 to the nearest hundred thousand.

Answer:360,000

Explanation:

The number 362,113 can be round off to 360000 when rounded to the nearest hundred thousand.

Question 8.

Round 982,638 to the nearest thousand.

Answer:980,000

Explanation:

The number 982,638 can be round off to 980,000 when rounded to the nearest thousand. ( Since the thousands position is less than 5)

Apply and Grow: Practice

Round the number to the place of the underlined digit.

Question 9.

3,641

Answer: 3,700

Explanation:

The number 3,641 can be round off to 3700 when rounded to the nearest hundred. ( Since the hundreds position is greater than 5)

Question 10.

17,139

Answer: 18,000

Explanation:

The number 17,139can be round off to 18000 when rounded to the nearest thousand. ( Since the thousands position is greater than 5)

Question 11.

426,384

Answer: 426,300

Explanation:

The number 426,384 can be round off to 426300 when rounded to the nearest hundred ( Since the hundreds position is greater than 5)

Question 12.

542,930

Answer: 542,000

Explanation:

The number 542,930 can be round off to 542,000 when rounded to the nearest thousand. ( Since the thousands position is greater than 5)

Round the number to the nearest thousand.

Question 13.

9,426

Answer: 9,000

Explanation:

The number 9,426 can be round off to 9000 when rounded to the nearest thousand.

Question 14.

57,496

Answer: 58,000

Explanation:

The number 57,496 can be round off to 58000 when rounded to the nearest thousand. ( Since the thousands position is greater than 5)

Question 15.

360,491

Answer: 360,000

Explanation:

The number 360,491 can be round off to 360000 when rounded to the nearest thousand.

Question 16.

824,137

Answer: 824,000

Explanation:

The number 824,137 can be round off to 824,000 when rounded to the nearest thousand. ( Since the thousands position is less than 5)

Round the number to the nearest hundred thousand.

Question 17.

226,568

Answer: 200,000

Explanation:

The number 226,568 can be round off to 200,000 when rounded to the nearest hundred thousand. ( Since the hundred thousand position is less than 5)

Question 18.

457,724

Answer: 400,000

Explanation:

The number 457,724 can be round off to 400,000 when rounded to the nearest hundred thousand. ( Since the hundred thousand  position is less than 5)

Question 19.

108,665

Answer: 100,000

Explanation:

The number 108,665 can be round off to 100,000 when rounded to the nearest hundred thousand. ( Since the hundred thousand  position is less than 5)

Question 20.

75,291

Answer:

Question 21.

Number Sense

Write a five-digit number that has the digits 6, 0, 4, 2, and 8 and rounds to 70,000 when rounded to the nearest ten thousand.

Answer: 68,420

Explanation:

The given 5-digits are: 6, 0, 4, 2 and 8

The 5-git number which can be formed by using the given 5 digits are 68,420 and 68,420 can be round off to 70,000 when rounded to the nearest ten thousand.

Question 22.

Number Sense

When finding the United States census, should you round or find an exact answer? Explain.

Answer:

Question 23.

Open-Ended

A lightning strike can reach a temperature of about 54,000 degrees Fahrenheit. Write four possible temperatures for a lightning strike.

Answer:

Think and Grow: Modeling Real Life

Example

When the results are rounded to the nearest ten thousand, which token received about 160,000 votes?

Big Ideas Math Solutions Grade 4 Chapter 1 Place Value Concepts 1.4 5

Answer:

When the results are rounded to the nearest ten thousand, the tokens which received 1,60,000 votes are:

A) Race Car  B) Rubber Duck

Round each number to the nearest ten thousand.

Big Ideas Math Solutions Grade 4 Chapter 1 Place Value Concepts 1.4 6

Answer:

A) 134,704 : 130,000  B) 154,165 : 150,000  C) 207,954 : 208,000  D) 212,476 : 212,000

E) 167,582 : 170,000  F) 146,661 : 150,000  G) 165,083 : 170,000 H) 160,485 : 160,000

So, the _______ received about 160,000 votes.

Answer:

A) Race Car  B) Rubber Duck

Show and Grow

Use the table above.

Question 24.

When the results are rounded to the nearest ten thousand, which tokens received about 150,000 votes?

Answer:

A) Cat  B)  Penguin

Explanation:

From the above table, we can conclude that Cat and the Penguin have 160,000 votes when rounded to the nearest ten thousand.

Question 25.

When the results are rounded to the nearest hundred thousand, which tokens received about 100,000 votes?

Answer:

A) Cat  B) Penguin

Explanation:

From the above table, we can conclude that Cat and the Penguin have 160,000 votes when rounded to the nearest hundred thousand

Question 26.

DIG DEEPER!

To which place should you round each number of votes so that you can order the numbers from greatest to least? Explain.

Answer:

Question 27.
A car battery should be replaced when the odometer shows about 50,000 miles. None of the cars below have had a battery replacement. Which cars might need a new battery?
Big Ideas Math Solutions Grade 4 Chapter 1 Place Value Concepts 1.4 7
Answer: ODO

Explanation:

From the given Fig.,

we can conclude that ODO which has above 50,000 miles needs a battery replacement.

Round Multi-Digit Numbers Homework & Practice 1.4

 

Round the number to the place of the underlined digit.

Question 1.

6,251

Answer: 6,250

Explanation:

The given number is 6,251 and the underlined digit is 5.

Since the underlined digit is equal to 5, we can conclude that the given number can round off to 6,250

Question 2.

8,962

Answer: 9,000

Explanation:

The given number is 8,962 and the underlined digit is 8.

Since the underlined digit is greater than 5, we can conclude that the given number can round off to 9,000

Question 3.

3,951

Answer: 3,950

Explanation:

The given number is 3,951 and the underlined digit is 9.

we can conclude that the given number can round off to 3,950

Question 4.

79,064

Answer: 80,000

Explanation:

The given number is 79,064 and the underlined digit is 9.

Since the underlined digit is greater than 5, we can conclude that the given number can round off to 80,000

Question 5.

43,976

Answer: 43,000

Explanation:

The given number is 43,976 and the underlined digit is 3.

Since the underlined digit is greater than 5, we can conclude that the given number can round off to 43,000

Question 6.

24,680

Answer: 24,000

Explanation:

The given number is 24,680 and the underlined digit is 2.

Since the underlined digit is less than 5, we can conclude that the given number can round off to 24,000

Question 7.

726,174

Answer: 800,000

Explanation:

The given number is 726,174 and the underlined digit is 7.

Since the underlined digit is greater than 5, we can conclude that the given number can round off to 800,000

Question 8.

138,691

Answer: 140,000

Explanation:

The given number is 138,691 and the underlined digit is 8.

Since the underlined digit is greater than  5, we can conclude that the given number can round off to 140,000

Round the number to the nearest thousand.

Question 9.

5,517

Answer: 5,000

Explanation:

The given number is 5,517. Since the thousands position is equal to 5, we can round it off to 5,000

Question 10.

70,628

Answer: 70,000

Explanation:

The given number is 70,628. Since the thousands position is less than 5, we can round it off to 70,000

Round the number to the nearest ten thousand.

Question 11.

114,782

Answer: 110,000

Explanation:

The given number is 114,782. Since the ten thousand’s  position is less than 5, we can round it off to 110,000

Question 12.

253,490

Answer: 250,000

Explanation:

The given number is 253,490. Since the ten  thousand’s position is equal to 5, we can round it off to 250,000

Round the number to the nearest hundred thousand.

Question 13.

628,496

Answer: 700,000

Explanation:

The given number is 628,496. Since the hundred  thousand’s position is greater than 5, we can round it off to 700,000

Question 14.

90,312

Answer: 90,312

Explanation:

The given number is 90,312. There is no hundred thousand’s position in the given number. So, there is no effect on the given number.

Question 15.

Structure

Round * to the nearest thousand and to the nearest ten thousand.

Big Ideas Math Solutions Grade 4 Chapter 1 Place Value Concepts 1.4 8

Nearest thousand: ______

Nearest ten thousand: ______

Answer: Nearest Thousand: 8000

Nearest ten thousand: 10,000

Explanation:

In the given fig., the * sign is present after 7,500.

From this, we can conclude that

Nearest Thousand: 8000

Nearest ten thousand: 10,000

Question 16.

Number Sense

Which numbers round to 300,000 when rounded to the nearest hundred thousand?

Big Ideas Math Solutions Grade 4 Chapter 1 Place Value Concepts 1.4 9

Answer: 302,586

Explanation:

From the given fig., when we observe the positions of hundred’s  and ten thousand’s positions, it is clear that “302,586” can round off to 300,000

Question 17.

Number Sense

When discussing the price of a laptop, should you round to the nearest thousand or the nearest ten? Explain.

Answer: When you are discussing the price of a laptop, you should round off to the nearest thousand instead of the nearest ten.

Example: Suppose you want to buy a laptop that costs $152,236. When the dealer told you the price of a laptop, you will ask to round off the price of the laptop so that it will be easier for you to buy that laptop.

Question 18.

YOU BE THE TEACHER

Your friend says 5,953 rounds to 5,053 when rounded to the nearest hundred. Is your friend correct? Explain.

Answer: Wrong

Explanation:

The given number is 5,953. When rounded off to the nearest hundred, the given number becomes 6,000 since the hundreds position is greater than 5.

Question 19.

Modeling Real Life

A glassblower is using a furnace to melt glass. When the furnace reaches about 2,000 degrees Fahrenheit, when

rounded to the nearest hundred, she can put the glass in. At which temperatures could she put the glass into the

furnace?

Big Ideas Math Solutions Grade 4 Chapter 1 Place Value Concepts 1.4 10

Big Ideas Math Solutions Grade 4 Chapter 1 Place Value Concepts 1.4 11
Answer: 1925

Explanation:

Given that the furnace reached 2000 degrees Fahrenheit when rounded off to the nearest hundred.

So, the given options are: A) 2005   B) 1899  C) 1925  D) 275  E) 1002

From the given options, it is clear that 1925 degrees can round off to 2000 degrees.

Note: The option 2005 degrees can’t become the answer because when rounding off to the nearest number, the number should be less than the desired number.

Review & Refresh

Find the perimeter of the polygon.

Question 20.

Big Ideas Math Solutions Grade 4 Chapter 1 Place Value Concepts 1.4 12

Perimeter: _____

Answer: 23 in.

Explanation:

The “Perimeter” is defined as the total length of the geometrical figure covered on all sides.

So, the perimeter of the Polygon can be calculated as :

10 in. + 4 in. + 6 in. + 3 in. = 23 in.

Question 21.

Big Ideas Math Solutions Grade 4 Chapter 1 Place Value Concepts 1.4 13

Perimeter: _____

Answer: 26 m

Explanation:

The “Perimeter” is defined as the total length of the geometrical figure covered on all sides.

In a Parallelogram, the parallel sides are equal.

So, the length of all the sides of a parallelogram is: 8 m, 8 m, 5 m, 5 m

So,

The perimeter of the parallelogram can be calculated as:

8 m + 8 m + 5 m + 5 m = 26 m

Question 22

Big Ideas Math Solutions Grade 4 Chapter 1 Place Value Concepts 1.4 14

Perimeter: ______

Answer: 28 ft

Explanation:

The “Perimeter” is defined as the total length of the geometrical figure covered on all sides.

In a Square, the length of all the sides is equal. So, even in the fig., the length of 1 side is given, we can take it as the length of each of the remaining sides.

So, the perimeter of the square can be calculated as:

7 ft + 7 ft + 7ft + 7 ft = 28 ft.

Place Value Concepts Performance Task

You hike from Point A through Point F along the orange path shown on the map.

Big Ideas Math Answer Key Grade 4 Chapter 1 Place Value Concepts 1

Question 1.

What is the distance in elevation between each contour line?

Answer: 100 ft.

Explanation:

When we observe the Contour lines from 2000 ft. to 3000 ft., we can conclude that each Contour line can represent 100 ft.

Question 2.

As you walk from A to C, are you walking uphill or downhill? Explain.

Answer: Downhill

Explanation: When we observe the orange path in the given Contour lines, we can conclude that from A to C, you are going downhill.

Question 3.

Which letter represents the highest point? Estimate the height.

Answer: C

Explanation:

If we observe the Orange path, then we can see that Point C represents the highest point.

Question 4.

A water station has an elevation of 2,763 feet. Which letter represents the location of the water station?

Answer: Point B

Explanation:

From the Orange path, we can see that between 2,500 ft and 3,000 ft., Point is present which can represent the location of the water Station that has an elevation of 2,763 feet.

Question 5.

You take a break when you are at two thousand eighty feet. At which letter do you take a break?

Answer: Point A

Explanation:

Point A is present after 2,000 ft which is the nearest to 2,080 ft. So, Point A takes a break here.

Question 6.

About how much higher do you think Point C is than Point D? Explain.

Answer:

Place Value Concepts Activity

Place Value Plug-In

Directions:

1.Players take turns.

2.On your turn, roll six dice.Arrange the dice into a six-digit number that matches one of the descriptions.

3.Write your number on the lines.

4. The first player to complete all of the numbers wins!

Big Ideas Math Answer Key Grade 4 Chapter 1 Place Value Concepts 2

Place Value Concepts Chapter Practice

1.1 Understand Place Value

Write the value of the underlined digit.

Question 1.

26,490

Answer: 90

Explanation;

The position of each digit in a number is defined and it is the value based on the position of the digit. So, the place value of ‘9’ in 26,490’ is:  90

Question 2.

57,811

Answer: 50,000

Explanation:

The position of each digit in a number is defined and it is the value based on the position of the digit. So, the place value of ‘5’ in ‘57,811’ is:  50,000

Question 3.

308,974

Answer: 8,000

Explanation:

The position of each digit in a number is defined and it is the value based on the position of the digit. So, the place value of ‘8’ in ‘308,974’ is: 8,000

Question 4.

748,612

Answer: 700,000

Explanation:

The position of each digit in a number is defined and it is the value based on the position of the digit. So, the place value of ‘7’ in ‘748,612’ is: 700,000

Compare the values of the underlined digits

Question 5.

94 and 982

Answer: The value of  9 in 982 is 10 times the value of 9 in 94

Explanation:

The value of 9 in 982 is: 900

The value of 9 in 94 is: 90

From these values, we can conclude that the value of  9 in 982 is 10 times the value of 9 in 94

Question 6.

817,953 and 84, 006

Answer: The value of 8 in 817,953 is 10 times the value of 8 in 84,006

Explanation:

The value of 8 in 817,953 is:800,000

The value of 8 in 84,006 is:80,000

From these values, we can conclude that the value of 8 in 817,953 is 10 times the value of 8 in 84,006

1.2 Read and Write Multi-Digit Numbers

Question 7.

Complete the table.

Big Ideas Math Answer Key Grade 4 Chapter 1 Place Value Concepts chp 7

Answer:

A)50,000 + 600 + 90 + 1

Word Form: Fifty thousand, Six hundred Ninety-one

Standard Form: 50,691

Explanation:

We can write any number in 3 Forms. They are;

A) Standard Form  B) Word Form  C) Expanded Form

Now, the given number is 50,000 + 600 + 90 + 1 which is in Expanded Form. So, we have to write the given number in the remaining 2 forms.

B) Seven hundred two thousand, five hundred

Standard Form:702,500

Expanded Form: 700,000 + 0 + 2,000 + 500 + 0 + 0

Explanation:

We can write any number in 3 Forms. They are;

A) Standard Form  B) Word Form  C) Expanded Form

Now, the given number is Seven hundred two thousand, five hundred which is in Word Form. So, we have to write the given number in the remaining 2 forms.

C) 993,004

Word Form: Nine hundred Ninety-three thousand, four

Expanded Form: 900,000 + 90,000 + 3,000 + 0 + 0 +4

Explanation:

We can write any number in 3 Forms. They are;

A) Standard Form  B) Word Form  C) Expanded Form

Now, the given number is 993,004 which is in Standard Form. So, we have to write the given number in the remaining 2 forms.

Question 8.

Modeling Real Life

Use the number 10,000 + 4,000 + 300 + 90 + 9 to complete the check.

Big Ideas Math Answer Key Grade 4 Chapter 1 Place Value Concepts chp 8

Answer:

Word Form: Fourteen thousand, Three hundred Ninety-nine

Standard Form: 14,399

Explanation:

We can write any number in 3 Forms. They are;

A) Standard Form  B) Word Form  C) Expanded Form

Now, the given number is10,000 + 4,000 + 300 + 90 + 9 which is in Expanded Form. So, we have to write the given number in the remaining 2 forms.

1.3 Compare Multi-Digit Numbers

Write which place to use when comparing the numbers.

Question 9.

46,027

46,029

Answer: 46,027 is less than 46,029

Explanation:

The given 2 numbers are 46,027 and 46,029.

From these 2 numbers, we can conclude that 46,027 is less than 46,029

Question 10.

548,003

545,003

Answer: 548,003 is greater than 545,003

Explanation:

The given 2 numbers are 548,003 and 545,003.

From these 2 numbers, we can conclude that 548,003 is greater than 545,003

Question 11.

619,925

630,982

Answer: 619,925 is less than 630,982

Explanation:

The given 2 numbers are 619,925 and 630,982.

From these 2 numbers, we can conclude that 619,925 is less than 630,982

Compare.

Question 12.

Big Ideas Math Answer Key Grade 4 Chapter 1 Place Value Concepts chp 12

Answer: 4,021 is less than 4,210

Explanation:

The given 2 numbers are 4,021 and 4,210.

From these 2 numbers, we can conclude that 4,021 is less than 4,210

Question 13.

Big Ideas Math Answer Key Grade 4 Chapter 1 Place Value Concepts chp 13

Answer: 78,614 is equal to 78,614

Explanation:

The given 2 numbers are 78,614 and 78,614.

From these 2 numbers, we can conclude that 78,614 is equal to 78,614

Question 14.

Big Ideas Math Answer Key Grade 4 Chapter 1 Place Value Concepts chp 14

Answer: 816,532 is greater than 816,332

Explanation:

The given 2 numbers are 816,532 and 816,332.

From these 2 numbers, we can conclude that 816,532 is greater than 816,332

Question 15.

Big Ideas Math Answer Key Grade 4 Chapter 1 Place Value Concepts chp 15

Answer: 55,002 is less than 65,002

Explanation:

The given 2 numbers are 55,002 and 65,002.

From these 2 numbers, we can conclude that 55,002 is less than 65,002

Question 16.

Big Ideas Math Answer Key Grade 4 Chapter 1 Place Value Concepts chp 16

Answer: 3,276 is greater than 3,275

Explanation:

The given 2 numbers are 3,276 and 3,275.

From these 2 numbers, we can conclude that 3,276 is greater than 3,275

Question 17.

Big Ideas Math Answer Key Grade 4 Chapter 1 Place Value Concepts chp 17

Answer: 45,713 is less than 457,130

Explanation:

The given 2 numbers are45,713 and 457,130.

From these 2 numbers, we can conclude that 45,713 is less than 457,130

Question 18.

Big Ideas Math Answer Key Grade 4 Chapter 1 Place Value Concepts chp 18

Answer: 569,021 is greater than 56,902

Explanation:

The given Expanded Form is: 50,000 + 6,000 + 900 + 2 =56,902

The given 2 numbers are569,021 and 56,902.

From these 2 numbers, we can conclude that 569,021 is greater than 56,902

Question 19.

Big Ideas Math Answer Key Grade 4 Chapter 1 Place Value Concepts chp 19

Answer: 37,000 is less than 307,000

Explanation:

The given Word Form is: Thirty – seven thousand = 37,000

The given 2 numbers are 37,000 and 307,000.

From these 2 numbers, we can conclude that37,000 is less than 307,000

Question 20.

Two different hot tubs cost $4,179 and $4,139. Which is the lesser price?

Answer: The lesser price is $4,139

Explanation:

The costs of two different hot tubs are: $4,179 and $4,139

On comparing the costs of two different hot tubs, we can conclude that the lesser price is $4,139

Question 21.

Number Sense

Write all of the digits that make the number greater than 47,068 and less than 47,468.?

47,?68

Answer: 1, 2, 3

Explanation:

The digits that make the number greater than 47,068 and less than 47,468 are: 1, 2, 3

1.4 Round Multi-Digit Numbers

Round the number to the place of the underlined digit.

Question 22.

8,614

Answer: 8,610

Explanation:

The number 8,614 can be round off to 8,610 when rounded to the nearest ten. ( Since the tens position is less than 5)

Question 23.

2,725

Answer: 2,000

Explanation:

The number 2,725 can be round off to 2,000 when rounded to the nearest thousand. ( Since the thousands position is less than 5)

Question 24.

27,602

Answer: 27,700

Explanation:

The number 27,602 can be round off to 27,700 when rounded to the nearest hundred. ( Since the hundreds position is greater than 5)

Question 25.

906,154

Answer: 906,100

Explanation:

The number 906,154 can be round off to 906,100 when rounded to the nearest hundred. ( Since the hundreds position is less than 5)

Round the number to the nearest thousand.

Question 26.

1,358

Answer: 1,000

Explanation:

The number 1,358 can be round off to 1,000  when rounded to the nearest thousand. ( Since the thousands position is less than 5)

Question 27.

57,094

Answer: 58,000

Explanation:

The number 57,094 can be round off to 58,000 when rounded to the nearest thousand. ( Since the thousands position is greater than 5)

Round the number to the nearest ten thousand.

Question 28.

431,849

Answer: 431,000

Explanation:

The number 431,849 can be round off to 431,000  when rounded to the nearest ten thousand. ( Since the ten thousand’s  position is less than 5)

Question 29.

60,995

Answer: 70,000

Explanation:

The number 60,995 can be round off to 70,000 when rounded to the nearest ten thousand. ( Since the ten thousand’s position is greater than 5)

Final Words:

We wish the information provided in the Big Ideas Math Answers Grade 4 Chapter 1 Place Value Concepts is helpful for you. Make use of the above links and start your preparation. Keep in touch with us to get the updates regarding all the chapters of Big Ideas Math Answers Grade 4.

Big Ideas Math Answers Grade 6 Chapter 10 Data Displays

Big Ideas Math Answers Grade 6 Chapter 10 Data Displays

Big Ideas Math Answers Grade 6 Chapter 10 Data Displays: Free Download Big Ideas Math Answers Grade 6 Chapter 10 Data Displays Pdf here. Get step by step solution to all the concepts of Stem-and-Leaf Plots, Histograms, Shapes of Distributions, Choosing Appropriate Measures, and so on. Download Answer Key of Big Ideas Math 6th Grade Chapter 10 Data Displays Pdf for free of cost.

Start your preparation with the given material and know-how to manage your time and your strengths and weaknesses. Big Ideas Math Answers Grade 6 Data Displays problems will help you to score more marks in the exam. Follow the below sections to get the detailed material and concepts of data displays.

Big Ideas Math Book 6th Grade Answer Key Chapter 10 Data Displays

Do you wanna master in data displays concept? It requires a lot of preparation and dedication to get perfect in this concept. To master this concept, you have to check BIM Math 6th Grade Answer Key Chapter 10 Data Displays pdf. With the help of this article, you can easily prepare the timetable.

If you prepare the timetable, you can easily know the time available to cover each chapter of the concept. There are several chapters involved in this concept. Make use of the Big Ideas Math Grade 6 Solution Key which acts as a teacher for you in your complete preparation. Follow the Big Ideas Math Book 6th Grade Answer Key Chapter 10 Data Displays and overcome your weaknesses.

Performance Task

Lesson 1 – Stem-and-Leaf Plots

Lesson 2 – Histograms

Lesson 3 – Shapes of Distributions

Lesson 4 – Choosing Appropriate Measures

Lesson 5 – Box-and-Whisker Plots

Data Displays

Data Displays STEAM Video/Performance Task

STEAM Video
Choosing a Dog
Different animals grow at different rates. Given a group of puppies, describe an experiment that you can perform to compare their growth rates. Describe a real-life situation where knowing an animal’s growth rate can be useful.

Watch the STEAM Video “Choosing a Dog.” Then answer the following questions.
1. Using Alex and Tony’s stem-and-leaf plots below, describe the weights of most dogs at 3 months of age and 6 months of age.
Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 1

Answer:
Weight of dogs at 3 months:
29, 34, 40, 40, 41, 42, 44, 46, 47, 48, 48, 53
Weight of dogs at 6 months:
57, 58, 61, 61, 63, 64, 65, 65, 65, 66, 67, 73
2. Make predictions about how the stem-and-leaf plot will look after 9 months and after 1 year.
Weight of dogs at 9 months
77, 78, 81, 81, 83, 84, 85, 85, 85, 86, 87, 91
Weight of dogs at 1 year
87, 88, 89, 93, 94, 95, 95, 95, 95, 96, 97, 99

Performance Task
Classifying Dog Breeds by Size
Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 2
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 names, breeds, and weights of full-grown dogs at a shelter.
Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 3
You will use a data display to make conclusions about the sizes of dogs at the shelter. Why might someone be interested in knowing the sizes of dogs at a shelter?

Answer:
Because they need time to adjust.
You can buy the dog shelter based on the height and weight of the dogs.

Data Displays Getting Ready for Chapter 10

Chapter Exploration
Work with a partner. A famous data set was collected in Scotland in the mid-1800s. It contains the chest sizes(in inches) of 5738 men in the Scottish Militia.
Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 4
1. Describe the shape of the bar graph shown above.

Answer: The shape of the above graph is Histogram.

2. Which of the following data sets have a bar graph that is similar in shape to the bar graph shown above? Assume the sample is selected randomly from the population. Explain your reasoning.
a. the heights of 500 women
b. the ages of 500 dogs
c. the last digit of 500 phone numbers
d. the weights of 500 newborn babies

Answer: The last digit of 500 phone numbers is similar in shape to the bar graph shown above.

3. Describe two other real-life data sets, one that is similar in shape to the bar graph shown above and one that is not.

Answer: The height of 500 students in the school and age of students in the classroom.

Vocabulary
The following vocabulary terms are defined in this chapter. Think about what each term might mean and record your thoughts.
stem-and-leaf plot
box-and-whisker plot
frequency table
five-number summary

Answer:
i. stem-and-leaf plot: A stem-and-leaf display or stem-and-leaf plot is a device for presenting quantitative data in a graphical format, similar to a histogram, to assist in visualizing the shape of a distribution.
ii. A box and whisker plot—also called a box plot—displays the five-number summary of a set of data. The five-number summary is the minimum, first quartile, median, third quartile, and maximum. In a box plot, we draw a box from the first quartile to the third quartile. A vertical line goes through the box at the median.
iii. In statistics, a frequency distribution is a list, table, or graph that displays the frequency of various outcomes in a sample. Each entry in the table contains the frequency or count of the occurrences of values within a particular group or interval.
iv. The five-number summary is a set of descriptive statistics that provides information about a dataset.

Lesson 10.1 Stem-and-Leaf Plots

EXPLORATION 1

Making a Data Display
Workwith a partner. The list below gives the ages of women when they became first ladies of the United States.
Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.1 1
a. The incomplete data display shows the ages of the first ladies in the left column of the list above. What do the numbers on the left represent? What do the numbers on the right represent?
b. This data display is called a stem-and-leaf plot. What numbers do you think represent the stems? leaves? Explain your reasoning.
c. Complete the stem-and-leaf plot using the remaining ages.

Answer:
Big-Ideas-Math-Answer-Key-Grade-6-Chapter-10-Data-Displays-10.1-1
The tens place represents the stem and the ones place represents the leaf.
d. REASONING
Write a question about the ages of first ladies that is easier to answer using a stem-and-leaf plot than a dot plot.
Answer: Make the stem and leaf plot to find the ages of the first ladies.
By using the above data you can make the stem and leaf plot easily.

Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.1 2

Key Idea
Stem-and-Leaf Plots
A stem-and-leaf plot uses the digits of data values to organize a data set. Each data value is broken into a stem(digit or digits on the left) and a leaf(digit or digits on the right).
A stem-and-leaf plot shows how data are distributed.
Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.1 4

EXAMPLE 1

Making a Stem-and-Leaf Plot
Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.1 5
Make a stem-and-leaf plot of the lengths of the 12 phone calls.
Step 1: Order the data.
2, 3, 5, 6, 10, 14, 18, 23, 23, 30, 36, 55
Step 2: Choose the stems and the leaves. Because the data values range from 2 to 55, use the tens digits for the stems and the ones digits for the leaves. Be sure to include the key.
Step 3: Write the stems to the left of the vertical line.
Step 4: Write the leaves for each stem to the right of the vertical line.
Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.1 6

Try It
Question 1.
Make a stem-and-leaf plot of the hair lengths.
Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.1 7
Answer:
Step 1: Order the data.
1, 1, 1, 2, 2, 4, 5, 5, 7, 12, 20, 23, 27, 30, 32, 33, 38, 40, 44, 47
Step 2: Choose the stems and the leaves. Because the data values range from 1 to 47, use the tens digits for the stems and the ones digits for the leaves. Be sure to include the key.
Step 3: Write the stems to the left of the vertical line.
Step 4: Write the leaves for each stem to the right of the vertical line.
Big Ideas Math Answers Grade 6 Chapter 12 Data Displays img_5

EXAMPLE 2
Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.1 8
The stem-and-leaf plot shows student quiz scores. (a) How many students scored less than 8 points? (b) How many students scored at least 9 points? (c) How are the data distributed?
a. There are five scores less than 8 points:
6.6, 7.0, 7.5, 7.7, and 7.8.
Five students scored less than 8 points.10
b. There are four scores of at least 9 points:
9.0, 9.2, 9.9, and 10.0.
Four students scored at least 9 points.
c. There are few low quiz scores and few high quiz scores. So, most of the scores are in the middle, from 8.1 to 8.9 points.

Try It
Question 2.
Use the grading scale at the right.
Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.1 9
a. How many students received a B on the quiz?
Answer: There are 9 students who received a B on the quiz.
b. How many students received a C on the quiz?
Answer: There are 4 students who received a C on the quiz.

Self – Assessment for Concepts & Skills

Solve each exercise. Then rate your understanding of the success criteria in your journal.
Question 3.
MAKING A STEM-AND-LEAF PLOT
Make a stem-and-leaf plot of the data values 14, 22, 9, 13, 30, 8, 25, and 29.
Answer:
The ones represent the leaf and the tens place represent the stem.
Big Ideas Math Grade 6 Chapter 10 Data Displays img_6

Question 4.
WRITING
How does a stem-and-leaf plot show the distribution of a data set?
Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.1 10
Answer:
02, 03, 1, 21, 26, 30, 34, 36, 44, 45, 48, 48, 49

Explanation:
A stem-and-leaf display or stem-and-leaf plot is a device for presenting quantitative data in a graphical format, similar to a histogram, to assist in visualizing the shape of a distribution.

Question 5.
REASONING
Consider the stem-and-leaf plot shown.
a. How many data values are at most 10?
Answer: By seeing the above stem and leaf plot we can find the data values of at most 10.
The data values less than or equal to 10 are 3.
b. How many data values are at least 30?
Answer: By seeing the above stem and leaf plot we can find the data values of at least 30.
The data values of less than 30 are 5.
c. How are the data distributed?
Answer: The data is distributed according to the stem and leaf plot. The tens place is given to the stem and the ones place is given to the leaf.

Question 6.
CRITICAL THINKING
How can you display data whose values range from 82 through 129 in a stem-and-leaf plot?
Answer:
Given data range from 82 to 129
Considering 9 random values between 82 and 129.
From the data 86, 91, 93, 100, 107, 109, 113, 122, 124, stem and leaf are calculated for each number.
86 is split into 8 (stem) and 6 (leaf)
91 is split into 9 (stem) and 1 (leaf)
93 is split into 9 (stem) and 3 (leaf)
100 is split into 10 (stem) and 0 (leaf)
107 is split into 10 (stem) and 7 (leaf)
109 is split into 10 (stem) and 9 (leaf)
113 is split into 11 (stem) and 3 (leaf)
122 is split into 12 (stem) and 2 (leaf)
124 is split into 12 (stem) and 4 (leaf)

Big Ideas Math Grade 6 Chapter 10 Data Displays img_7

EXAMPLE 3
Modeling Real Life
The stem-and-leaf plot shows the heights of several houseplants. Use the data to answer the question, “What is a typical height of a houseplant?
Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.1 11
Find the mean, median, and mode of the data. Use the measure that best represents the data to answer the statistical question.
Mean: \(\frac{162}{15}\) = 10.8
Median: 11
Mode: 11
The mean is slightly less than the median and mode, but all three measures can be used to represent the data.
So, the typical height of a houseplant is about 11 inches.

Self-Assessment for Problem Solving

Solve each exercise. Then rate your understanding of the success criteria in your journal.
Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.1 12
Question 7.
DIG DEEPER!
Work with a partner. Use two number cubes to conduct the following experiment. Then use a stem-and-leaf plot to organize your results and describe the distribution of the data.
• Toss the cubes and find the product of the resulting numbers. Record your results.
• Repeat this process 30 times.
Answer:
From the data 15,4,6,30,6,8,36,12,12,6,6,4,15,10,4,2,20,3,15,6,4,6,3,10,3,20,4,12,4,20.
Stem and leaf are calculated for each number.
15 is split into 1 (stem) and 5 (leaf)
04 is split into 0 (stem) and 4 (leaf)
06 is split into 0 (stem) and 6 (leaf)
30 is split into 3 (stem) and 0 (leaf)
06 is split into 0 (stem) and 6 (leaf)
08 is split into 0 (stem) and 8 (leaf)
36 is split into 3 (stem) and 6 (leaf)
12 is split into 1 (stem) and 2 (leaf)
12 is split into 1 (stem) and 2 (leaf)
06 is split into 0 (stem) and 6 (leaf)
06 is split into 0 (stem) and 6 (leaf)
04 is split into 0 (stem) and 4 (leaf)
15 is split into 1 (stem) and 5 (leaf)
10 is split into 1 (stem) and 0 (leaf)
04 is split into 0 (stem) and 4 (leaf)
02 is split into 0 (stem) and 2 (leaf)
20 is split into 2 (stem) and 0 (leaf)
03 is split into 0 (stem) and 3 (leaf)
15 is split into 1 (stem) and 5 (leaf)
06 is split into 0 (stem) and 6 (leaf)
04 is split into 0 (stem) and 4 (leaf)
06 is split into 0 (stem) and 6 (leaf)
03 is split into 0 (stem) and 3 (leaf)
20 is split into 2 (stem) and 0 (leaf)
04 is split into 0 (stem) and 4 (leaf)
12 is split into 1 (stem) and 2 (leaf)
04 is split into 0 (stem) and 4 (leaf)
20 is split into 2 (stem) and 0 (leaf)
Big ideas Math Grade 6 Chapter 10 Data Displays img_8

Question 8.
The stem-and-leaf plot shows the weights (in pounds) of several puppies at a pet store. Use the data to answer the question, “How much does a puppy at the pet store weigh?
Answer:
We can use the mean of the data. To find the mean, add the data then divide the sum of the number of data
(8+12+15+17+18+24+24+31)/8 = 149/8 = 18.625
To the nearest pound, a puppy weighs about 19 pounds

Stem-and-Leaf Plots Homework & Practice 10.1

Review & Refresh

Find and interpret the mean absolute deviation of the data.
Question 1.
8, 6, 8, 5, 3, 10, 11, 5, 7
Answer:
First, arrange the given values in the ascending order.
3, 5, 5, 6, 7, 8, 8, 10, 11
We find the mean of the data
mean = (3 + 5 + 5 + 6 + 7 + 8 + 8 + 10 + 11)/9
mean = 7

Question 2.
55, 46, 39, 62, 55, 51, 48, 60, 39, 45
Answer:
First, arrange the given values in the ascending order.
39, 39, 45, 46, 48, 51, 55, 55, 60, 62
We find the mean of the data
mean = (39 + 39 + 45 + 46 + 48 + 51 + 55 + 55 + 60 + 62)/10
mean = 50

Question 3.
37, 54, 41, 18, 28, 32
Answer:
First, arrange the given values in the ascending order.
18, 28, 32, 37, 41, 54
We find the mean of the data
mean = (18+28+32+37+41+54)/6
mean = 35

Question 4.
12, 25, 8, 22, 6, 1, 10, 4
Answer:
First, arrange the given values in ascending order.
1, 4, 6, 8, 10,12, 22, 25
mean = (1+ 4 + 6 + 8 + 10 + 12 + 22 + 25)/8
mean = 11

Use the Distributive Property to simplify the expression.
Question 5.
5(n + 8)
Answer: 5n + 40

Explanation:
5(n + 8) = 5 × n + 5 × 8
5n + 40

Question 6.
7(y – 6)
Answer: 7y – 42

Explanation:
7(y – 6) = 7 × y – 7 × 6
7y – 42

Question 8.
14(2b + 3)
Answer: 28b + 42

Explanation:
14(2b + 3) = 14 × 2b + 14 × 3
28b + 42

Question 9.
11(9 + s)
Answer: 99 + 11s

Explanation:
11(9 + s) = 11 × 9 + 11 × s
99 + 11s

Solve the equation.
Question 9.
\(\frac{p}{2}\) = 8
Answer: 16

Explanation:
\(\frac{p}{2}\) = 8
p = 8 × 2
p = 16

Question 10.
28 = 6g
Answer: 4.66

Explanation:
28 = 6g
g = 28/6 = 4.66
Thus g = 4.66

Question 11.
3d ÷ 4 = 9
Answer: 12

Explanation:
3d ÷ 4 = 9
3d = 9 × 4
3d = 36
d = 36/3
d = 12
Thus d = 12

Question 12.
10 = \(\frac{2z}{3}\)
Answer:

Explanation:
10 = \(\frac{2z}{3}\)
10 × 3 = 2z
2z = 30
z = 30/2
z = 15
So, z = 15

Concepts, Skills, & Problem Solving

REASONING
Write a question that is easier to answer using the stem-and-leaf plot than a dot plot. (See Exploration 1, p. 457.)
Question 13.
Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.1 13
Answer:
Make a stem leaf plot to find the number of customers who visit your shop.
12, 13, 16, 17, 20, 21, 21, 23, 23, 28, 28, 32, 33, 34, 34, 35, 35, 36, 39, 39, 40, 41, 41, 42, 44, 46, 47, 48, 49, 49

Question 14.
Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.1 14
Answer:
Make the stem leaf plot to find the number of text messages you received per hour.
40, 40, 42, 46, 46, 49, 51, 51, 53, 53, 57, 57, 57, 59, 59, 59, 61, 62, 62, 65, 65, 66, 67, 68, 68, 70, 72, 72, 73, 74.

MAKING A STEM-AND-LEAF PLOT Make a stem-and-leaf plot of the data.
Question 15.
Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.1 15
Answer:
We have to write the stem and leaf plot for the above table.
Big Ideas Math Grade 6 Chapter 10 Data Displays img_9

Question 16.
Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.1 16
Answer:
We have to write the stem and leaf plot for the above table.
Big Ideas Math Grade 6 Chapter 10 Data Displays img_10

Question 17.
Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.1 17
Answer:
We have to write the stem and leaf plot for the above table.
Big Ideas Math Grade 6 Chapter 10 Data Displays img_10

Question 18.
Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.1 18
Answer:
We have to write the stem and leaf plot for the above table.
Big Ideas Math Grade 6 Chapter 10 Data Displays img_12

Question 19.
Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.1 19
Answer:
We have to write the stem and leaf plot for the above table.
Big Ideas Math Grade 6 Chapter 10 Data Displays img_13

Question 20.
Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.1 20
Answer:
We have to write the stem and leaf plot for the above table.
Big Ideas Math Grade 6 Chapter 10 Data Displays img_14

Question 21.
YOU BE THE TEACHER
Your friend makes a stem-and-leaf plot of the data. Is your friend correct? Explain your reasoning.
Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.1 21
51, 25, 47, 42, 55, 26, 50, 44, 55
Answer: your friend is correct

Explanation:
In stem and leaf plot the tens place represent stem and the ones place represent the leaf.

MODELING REAL LIFE
The stem-and-leaf plot shows the numbers of confirmed cases of a virus in 15 countries.
Question 22.
How many of the countries have more than 60 confirmed cases?
Answer: 6 countries

Explanation:
By seeing the above stem and leaf plot we can find the number of cases more than 60.
The number of leaf represents the number of countries.
62, 63, 63, 67, 75, 97.
Thus there are 6 countries that have more than 60 confirmed cases.

Question 23.
Find the mean, median, mode, range, and interquartile range of the data.
Answer:
41, 41, 43, 43, 45, 50, 52, 53, 54, 62, 63, 63, 67, 75, 97
In its simplest mathematical definition regarding data sets, the mean used is the arithmetic mean, also referred to as mathematical expectation, or average.
Mean:
mean = (41+41+43+43+45+50+52+53+54+62+63+63+67+75+97)/15
mean = 56.6
Median:
In the odd cases where there are only two data samples or there is an even number of samples where all the values are the same, the mean and median will be the same.
41, 41, 43, 43, 45, 50, 52, 53, 54, 62, 63, 63, 67, 75, 97
So, the median of the given data is 53.
Mode:
The mode is the value in a data set that has the highest number of recurrences.
41, 41, 43, 43, 45, 50, 52, 53, 54, 62, 63, 63, 67, 75, 97
mode = 41, 43, 63 (Repeated 2 times)

Question 24.
How are the data distributed?
Answer:
The distribution of a data set is the shape of the graph when all possible values are plotted on a frequency graph. Usually, we are not able to collect all the data for our variable of interest.

Question 25.
Which data value is an outlier? Describe how the outlier affects the mean.
Answer:
Outliers affect the mean value of the data but have little effect on the median or mode of a given set of data.
Example: A student receives a zero on a quiz and subsequently. has the following scores: 0, 70, 70, 80, 85, 90, 90, 90, 95, 100. Outlier: 0.

Question 26.
REASONING
Each stem-and-leaf plot below has a mean of 39. Without calculating, determine which stem-and-leaf plot has the lesser mean absolute deviation. Explain your reasoning.
Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.1 23
Answer:
i. 23, 27, 30, 32, 36, 39, 41, 42, 45, 48, 51, 54
Mean = (23+27+30+32+36+39+41+42+45+48+51+54)/12
Mean = 39
The mean absolute deviation is 7.833
ii. 22, 24, 25, 28, 29, 33, 38, 45, 53, 56, 57, 58
Mean = (22+24+25+28+29+33+38+45+53+56+57+58)/12
Mean = 39
The mean absolute deviation is 12.333
Thus the first stem and leaf plot has the lesser mean absolute deviation.

Question 27.
DIG DEEPER!
The stem-and-leaf plot shows the daily high temperatures (in degrees Fahrenheit) for the first 15 days of June.
Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.1 24
a. When you include the daily high temperatures for the rest of the month, the mean absolute deviation increases. Draw a stem-and-leaf plot that could represent all of the daily high temperatures for the month.

Answer:
Big Ideas Math Answers Grade 6 Chapter 12 Data Displays img_6
b. Use your stem-and-leaf plot from part(a) to answer the question, “What is a typical daily high temperature in June?”
Answer: 89°F is the high temperature in the month of June.

Question 28.
CRITICAL THINKING
The back-to-back stem-and-leaf plot shows the 9-hole golf scores for two golfers. Only one of the golfers can compete in a tournament as your teammate. Use measures of center and measures of variation to support choosing either golfer.
Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.1 25
Answer:
The scores of Rich are
35, 37, 41, 42, 43, 44, 45, 48
The scores of Will are
42, 43, 44, 44, 46, 47, 47, 48, 49
Will can compete in the tournament.

Lesson 10.2 Histograms

EXPLORATION 1

Performing an Experiment
Work with a partner.
a. Make the airplane shown from a single sheet of 8\(\frac{1}{2}\) by-11-inch paper. Then design and make your own paper airplane.
Big Ideas Math Answers 6th Grade Chapter 10 Data Displays 10.2 1
b. PRECISION
Fly each airplane 20 times. Keep track of the distance flown each time. Specify Units. What units will you use to measure the distance flown? Will the units you use affect the results in your frequency table? Explain.
c. A frequency table groups data values into intervals. The frequency is the number of values in an interval. Use a frequency table to organize the results for each airplane.
d. MODELING Represent the data in the frequency tables graphically. Which airplane flies farther? Explain your reasoning.
Answer:

Big Ideas Math Answers 6th Grade Chapter 10 Data Displays 10.2 3

Key Idea
Histograms
p. 463 frequency, A histogram is a bar graph that shows the frequencies of data values in intervals of the same size.
Big Ideas Math Answers 6th Grade Chapter 10 Data Displays 10.2 4
The height of a bar represents the frequency of the values in the interval.

EXAMPLE 1
Making a Histogram
The frequency table shows the numbers of laps that people in a swimming class completed today. Display the data in a histogram.
Big Ideas Math Answers 6th Grade Chapter 10 Data Displays 10.2 5
Step 1: Draw and label the axes.
Step 2: Draw a bar to represent the frequency of each interval.
Big Ideas Math Answers 6th Grade Chapter 10 Data Displays 10.2 6

Try It
Question 1.
The frequency table shows the ages of people riding a roller coaster. Display the data in a histogram.
Big Ideas Math Answers 6th Grade Chapter 10 Data Displays 10.2 7
Answer:
BIM Grade 6 Answers Chapter 10 Data Displays img_21

EXAMPLE 2
Using a Histogram
The histogram shows winning speeds at the Daytona 500.
(a) Which interval contains the most data values?
(b) How many of the winning speeds are less than 140 miles per hour?
(c) How many of the winning speeds are at least 160 miles per hour?
Big Ideas Math Answers 6th Grade Chapter 10 Data Displays 10.2 8.1
a. The interval with the tallest bar contains the most data values.
So, the 150−159 miles per hour interval contains the most data values.
b. One winning speed is in the 120−129 miles per hour interval, and eight winning speeds are in the 130−139 miles per hour interval.
So, 1 + 8 = 9 winning speeds are less than 140 miles per hour.
c. Eight winning speeds are in the 160−169 miles per hour interval, and five winning speeds are in the 170−179 miles per hour interval.
So, 8 + 5 =13 winning speeds are at least 160 miles per hour.

Try It
Question 2.
The histogram shows the numbers of hours that students in a class slept last night.
a. How many students slept at least 8 hours?
b. How many students slept less than 12 hours?
Big Ideas Math Answers 6th Grade Chapter 10 Data Displays 10.2 8

Answer:

A. Number of students who slept for 8 to 11 hours is 8.
Number of students who slept for 12 to 15 hours is 3.
The total number of students who slept for atleast 8 hours is 8.

B. Number of students who slept for 8 to 11 hours is 8.
Number of students who slept for 4 to 7 hours is 8.
Number of students who slept for 0 to 3 hours is 2.
Thus the number of students who slept for less than 12 hours is 8 + 8 + 2 = 18 students

EXAMPLE 3
Comparing Data Displays
The data displays show how many push-ups students in a class completed for a physical fitness test. Which data display can you use to find how many students are in the class? Explain.
Big Ideas Math Answers 6th Grade Chapter 10 Data Displays 10.2 9
You can use the histogram because it shows the number of students in each interval. The sum of these values represents the number of students in the class. You cannot use the circle graph because it does not show the number of students in each interval.

Try It
Question 3.
Which data display should you use to describe the portion of the entire class that completed 30−39 push-ups? Explain.
Answer: You should use the percentage of the number of students in the interval of 30-39 to find the completed push-ups.
The portion of the entire class that completed 30−39 push-ups is 24%

Self-Assessment for Concepts & Skills

Solve each exercise. Then rate your understanding of the success criteria in your journal
Question 4.
MAKING A HISTOGRAM
The table shows the numbers of siblings of students in a class.
Big Ideas Math Answers 6th Grade Chapter 10 Data Displays 10.2 10
a. Display the data in a histogram.b. Explain how you chose reasonable intervals for your histogram in part
Answer:

Question 5.
NUMBER SENSE
Can you find the range and the interquartile range of the data in the histogram? If so, find them. If you cannot find them, explain why not.
Big Ideas Math Answers 6th Grade Chapter 10 Data Displays 10.2 11
Answer:

EXAMPLE 4
Modeling Real Life made using the data displays in Example 3?
A. Twelve percent of the class completed 9 push-ups.
B. Five students completed at least 10 and at most 19 push-ups.
C. At least one student completed more than 39 push-ups.
D. Less than \(\frac{1}{4}\) of the class completed 30 or more push-ups.
The circle graph shows that12% completed 0−9 push-ups, but you cannot determine how many completed exactly 9. So, Statement A cannot be made.
Big Ideas Math Answers 6th Grade Chapter 10 Data Displays 10.2 12
In the histogram, the bar height for the 10−19 interval is 5, and the bar height for the 40−49 interval is 1. So, Statements B and C can be made.
The circle graph shows that24% completed 30−39 push-ups, and 4% completed 40−49 push-ups. So, 24% + 4% =28% completed 30 or more push-ups. Because \(\frac{1}{4}\) = 25% and 28% > 25%, Statement D cannot be made.
The correct answers are A and D.

Self-Assessment for Problem Solving
Solve each exercise. Then rate your understanding of the success criteria in your journal.
Big Ideas Math Answers 6th Grade Chapter 10 Data Displays 10.2 13
Question 6.
The histogram shows the numbers of rebounds per game for a middle school basketball player in a season.
a. Which interval contains the most data values?
b. 54 How many games did the player play during the season?
c. In what percent of the games did the player have 4 or more rebounds?
Answer:

Question 7.
Determine whether you can make each statement by using the histogram in the previous exercise.Explain.Rebounds
a. The basketball player had 2 rebounds in 6 different games.
b. The basketball player had more than 1 rebound in 9 different games
Answer:

Histograms Homework & Practice 10.2

Review & Refresh

Make a stem-and-leaf plot of the data.
Question 1.
Big Ideas Math Answers 6th Grade Chapter 10 Data Displays 10.2 14
Answer:
Big Ideas Math Answers Grade 6 Chapter 10 Data Displays img_11

Question 2.
Big Ideas Math Answers 6th Grade Chapter 10 Data Displays 10.2 15
Answer:
Big Ideas Math Answers Grade 6 Chapter 10 Data Displays img_12

Find the percent of the number.
Question 3.
25% of 180
Answer: 45

Explanation:
25% = 25/100
25/100 × 180
We get 45
So, 25% of 180 is 45.

Question 4.
30% of 90
Answer: 27

Explanation:
30% = 30/100
30/100 × 90 = 27
So, 30% of 90 is 27

Question 5.
16% of 140
Answer: 22.4

Explanation:
16% = 16/100
16/100 × 140 = 22.4
So, 16% of 140 is 22.4

Question 6.
64% of 807.
Answer: 516.48

Explanation:
64% = 64/100
64/100 × 807 = 516.48
So, 64% of 807 is 516.48

Question 7.
What is the least common multiple of 7 and 12?
A. 28
B. 42
C. 84
D. 168
Answer: 84

Explanation:
Find and list multiples of each number until the first common multiple is found. This is the lowest common multiple.
Multiples of 7:
7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, 98
Multiples of 12:
12, 24, 36, 48, 60, 72, 84, 96, 108
Therefore,
LCM(7, 12) = 84
Thus the correct answer is option c.

Concepts, Skills, & Problem Solving
MAKING A FREQUENCY TABLE Organize the data using a frequency table. (See Exploration 1, p. 463.)
Question 8.
Big Ideas Math Answers 6th Grade Chapter 10 Data Displays 10.2 16
Answer:

BIM Grade 6 Answer Key Chapter 10 Data Displays img_13

Question 9.
Big Ideas Math Answers 6th Grade Chapter 10 Data Displays 10.2 17
Answer:
BIM Grade 6 Answer Key Chapter 10 Data Displays img_14

MAKING A HISTOGRAM Display the data in a histogram.
Question 10.
Big Ideas Math Answers 6th Grade Chapter 10 Data Displays 10.2 18
Answer:
BIM Grade 6 Answer Key Chapter 10 Data Displays img_15

Question 11.
Big Ideas Math Answers 6th Grade Chapter 10 Data Displays 10.2 19
Answer:
BIM Grade 6 Answer Key Chapter 10 Data Displays img_16

Question 12.
Big Ideas Math Answers 6th Grade Chapter 10 Data Displays 10.2 20
Answer:
BIM Grade 6 Answer Key Chapter 10 Data Displays img_17

Question 13.
Big Ideas Math Answers 6th Grade Chapter 10 Data Displays 10.2 21
Answer:
BIM Grade 6 Answer Key Chapter 10 Data Displays img_18

Question 14.
Big Ideas Math Answers 6th Grade Chapter 10 Data Displays 10.2 22
Answer:
BIM Grade 6 Answer Key Chapter 10 Data Displays img_19

Question 15.
Big Ideas Math Answers 6th Grade Chapter 10 Data Displays 10.2 23
Answer:
BIM Grade 6 Answer Key Chapter 10 Data Displays img_20

Question 16.
YOU BE THE TEACHER
Your friend displays the data in a histogram. Is your friend correct? Explain your reasoning.
Big Ideas Math Answers 6th Grade Chapter 10 Data Displays 10.2 24
Answer: yes your friend is correct.
The frequency table matches the histogram.

Question 17.
MODELING REAL LIFE
The histogram shows the numbers of magazines read last month by the students in a class.
Big Ideas Math Answers 6th Grade Chapter 10 Data Displays 10.2 25
a. Which interval contains the fewest data values?
Answer: The interval 4-5 has the fewest data values.
b. How many students are in the class?
Answer:
0-1 = 2
2-3 = 15
4-5 = 0
6-7 = 3
2 + 15 + 3 = 20
c. What percent of the students read fewer than six magazines?
Answer: By seeing the above histogram we can say that 25% of the students read fewer than six magazines.

Question 18.
YOU BE THE TEACHER
Your friend interprets the histogram. Is your friend correct? Explain your reasoning.
Big Ideas Math Answers 6th Grade Chapter 10 Data Displays 10.2 26
Answer:
Compare your friend with the above histogram.
By seeing the above histogram we can say that it took 12 seconds to download songs.
So, your friend is correct.

Question 19.
REASONING
The histogram shows the percent of the voting-age population in each state who voted in a presidential election. Explain whether the graph supports each statement.
Big Ideas Math Answers 6th Grade Chapter 10 Data Displays 10.2 27
a. Only 40% of one state voted.
b. In most states, between 50% and 64.9% voted.
c. The mode of the data is between 55% and 59.9%
Answer:

Question 20.
PROBLEM SOLVING
The histograms show the areas of counties in Pennsylvania and Indiana. Which state do you think has the greater area? Explain.
Big Ideas Math Answers 6th Grade Chapter 10 Data Displays 10.2 28
Answer:

Question 21.
MODELING REAL LIFE
The data displays show how many pounds of garbage apartment residents produced in 1 week. Which data display can you use to find how many residents produced more than 25 pounds of garbage? Explain.
Big Ideas Math Answers 6th Grade Chapter 10 Data Displays 10.2 29
Answer:

Question 22.
REASONING
Determine whether you can make each statement by using the data displays in Exercise 21. Explain your reasoning.
a. One resident produced 10 pounds of garbage.
b. Twelve residents produced between 20 and 29 pounds of garbage.
Answer:

Question 23.
DIG DEEPER!
The table shows the lengths of some whales in a marine sanctuary.
Big Ideas Math Answers 6th Grade Chapter 10 Data Displays 10.2 30
a. Make a histogram of the data starting with the interval 51−55.
b. Make another histogram of the data using a different-sized interval.
c. Compare and contrast the two histograms.
Answer:

Question 24.
LOGIC
Can you find the mean or the median of the data in Exercise 17? Explain.
Answer:

Lesson 10.3 Shapes of Distributions

Big Ideas Math Answers Grade 6 Chapter 10 Data Displays 10.3 1

EXPLORATION
Describing Shapes of Distributions
Work with a partner. The lists show the first three digits and last four digits of several phone numbers in the contact list of a cell phone.
Big Ideas Math Answers Grade 6 Chapter 10 Data Displays 10.3 2
a. Compare and contrast the distribution of the last digit of each phone number to the distribution of the first digit of each phone number. Describe the shapes of the distributions.
b. Describe the shape of the distribution of the data in the table below. Compare it to the distributions in part(a).
Answer:

You can use dot plots and histograms to identify shapes of distributions.

Key Ideas
Symmetric and Skewed Distributions
Big Ideas Math Answers Grade 6 Chapter 10 Data Displays 10.3 3

EXAMPLE 1
Describing Shapes of Distributions

Describe the shape of each distribution.
Big Ideas Math Answers Grade 6 Chapter 10 Data Displays 10.3 4

Try It
Question 1.
Describe the shape of the distribution.
Big Ideas Math Answers Grade 6 Chapter 10 Data Displays 10.3 5
Answer:
BIM 6th Grade Chapter 10 Data Displays Answer Key img_4
A symmetric distribution has a graph in which the left side is a mirror image of the right side.
A skewed distribution has a graph in which a “tail” extends to the left and most data are on the right OR a “tail” extends to the right and most data are on the left.

EXAMPLE 2
Describing the Shape of a Distribution
The frequency table shows the ages of people watching a comedy in a theater. Display the data in a histogram. Then describe the shape of the distribution.
Big Ideas Math Answers Grade 6 Chapter 10 Data Displays 10.3 6
Draw and label the axes. Then draw a bar to represent the frequency of each interval.
Most of the data are on the right, and the tail extends to the left.
So, the distribution is skewed left.
Answer:
For a distribution that is skewed right, the tail extends to the right and most of the data are on the left side of the graph.

Try It
Question 2.
The frequency table shows the ages of people watching a historical movie in a theater. Display the data in a histogram. Describe the shape of the distribution.
Big Ideas Math Answers Grade 6 Chapter 10 Data Displays 10.3 7
Answer:

Self-Assessment for Concepts & Skills

Solve each exercise. Then rate your understanding of the success criteria in your journal.
Question 3.
WRITING
Explain in your own words what it means for a distribution to be (a) skewed left, (b) symmetric, and (c) skewed right.
Answer:

Question 4.
DESCRIBING A DISTRIBUTION
Display the data shown in a histogram. Describe the shape of the distribution.
Big Ideas Math Answers Grade 6 Chapter 10 Data Displays 10.3 8
Answer:

Question 5.
WHICH ONE DOESN’T BELONG?
Which histogram does not belong with the other three? Explain your reasoning.
Big Ideas Math Answers Grade 6 Chapter 10 Data Displays 10.3 9
Answer:

EXAMPLE 3
Modeling Real Life
The histogram shows the ages of people watching an animated movie in the same theater as in Example 2. Which movie has an older audience?
Big Ideas Math Answers Grade 6 Chapter 10 Data Displays 10.3 10
Understand the problem
You are given histograms that display the ages of people watching two movies. You are asked to determine which movie has an older audience.

Make a plan
Use the intervals and distributions of the data to determine which movie has an older audience.

Solve and check
The intervals in the histograms are the same. Most of the data for the animated movie are on the left, while most of the data for the comedy are on the right. This means that the people watching the comedy are generally older than the people watching the animated movie.

So, the comedy has an older audience.
Big Ideas Math Answers Grade 6 Chapter 10 Data Displays 10.3 11
Check Reasonableness
The movies have similar attendance. However,only4 people watching the comedy are 17 or under. A total of 35 people watching the animated movie are 17 or under. So, it is reasonable to conclude that the comedy has an older audience.

Self-Assessment for Problem Solving
Solve each exercise. Then rate your understanding of the success criteria in your journal.
Question 6.
The frequency table shows the numbers of visitors each day to parks in Aurora and Grover in one month. Which park generally has more daily visitors? Justify your answer.
Big Ideas Math Answers Grade 6 Chapter 10 Data Displays 10.3 12
Answer:

Question 7.
DIG DEEPER!
The frequency tables below show the ages of guests on two cruises. Can you make accurate comparisons of the ages of the guests? Explain your reasoning.
Big Ideas Math Answers Grade 6 Chapter 10 Data Displays 10.3 13
Answer:

Shapes of Distributions Homework & Practice 10.1

Review & Refresh

Display the data in a histogram.
Question 1.
Big Ideas Math Answers Grade 6 Chapter 10 Data Displays 10.3 14
Answer:
Big Ideas Math Answers Grade 6 Chapter 12 Data Displays img_1
On the vertical axis, place frequencies. Label this axis “Frequency”.
On the horizontal axis, place the lower value of each interval.
Draw a bar extending from the lower value of each interval to the lower value of the next interval.

Question 2.
Big Ideas Math Answers Grade 6 Chapter 10 Data Displays 10.3 15
Answer:
BIM Grade 6 Answer Key Chapter 10 Data Displays img_2

On the vertical axis, place frequencies. Label this axis “Frequency”.
On the horizontal axis, place the lower value of each interval.
Draw a bar extending from the lower value of each interval to the lower value of the next interval.

Question 3.
Big Ideas Math Answers Grade 6 Chapter 10 Data Displays 10.3 16
Answer:
Big Ideas Math Answers Grade 6 Chapter 12 Data Displays img_3

On the vertical axis, place frequencies. Label this axis “Frequency”.
On the horizontal axis, place the lower value of each interval.
Draw a bar extending from the lower value of each interval to the lower value of the next interval.

Write a unit rate for the situation.
Question 4.
$200 per 8 days
Answer:
200/8 = 25
Thus $25 per day.

Question 5.
60 kilometers for every 1.5 hours
Answer:
Your average speed is 60 km per 1.5 hours.
60/1.5 = 40 km/hr

Concepts, Skills, &Problem Solving

DESCRIBING SHAPES OF DISTRIBUTIONS Describe the shape of the distribution of the data in the table. (See Exploration 1, p. 471.)
Question 6.
Big Ideas Math Answers Grade 6 Chapter 10 Data Displays 10.3 17
Answer:
Step 1:
Order the data
0, 0, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 5, 5, 6

Question 7.
Big Ideas Math Answers Grade 6 Chapter 10 Data Displays 10.3 18
Answer:
Step 1:
Order the data
12, 12, 12, 13, 13, 13, 14, 14, 14, 15, 15, 15, 16, 16, 16.

DESCRIBING SHAPES OF DISTRIBUTIONS

Describe the shape of the distribution.
Question 8.
Big Ideas Math Answers Grade 6 Chapter 10 Data Displays 10.3 19
Answer: 25, 26, 27, 27, 28, 28, 28, 28, 29, 29, 29, 29, 29, 30, 30, 30

Question 9.
Big Ideas Math Answers Grade 6 Chapter 10 Data Displays 10.3 20
Answer:

15, 16, 16, 17, 17, 17, 18, 18, 18, 18, 18, 19, 19, 19, 20, 20, 21

Question 10.
Big Ideas Math Answers Grade 6 Chapter 10 Data Displays 10.3 21
Answer:

Question 11.
Big Ideas Math Answers Grade 6 Chapter 10 Data Displays 10.3 22
Answer:

Question 12.
MODELING REAL LIFE
The frequency table shows the years of experience for the medical states in Jones County and Pine County. Display the data for each county in a histogram. Which county’s medical state has less experience? Explain.
Big Ideas Math Answers Grade 6 Chapter 10 Data Displays 10.3 23
Answer:

Question 13.
REASONING
What is the shape of the distribution of the restaurant waiting times? Explain your reasoning.
Big Ideas Math Answers Grade 6 Chapter 10 Data Displays 10.3 24
Answer:

Question 14.
LOGIC
Are all distributions either approximately symmetric or skewed? Explain. If not, give an example.
Answer:

Question 15.
REASONING
Can you use a stem-and-leaf plot to describe the shape of a distribution? Explain your reasoning.
Answer:

Question 16.
DIG DEEPER!
The table shows the donation amounts received by a charity in one day.
Big Ideas Math Answers Grade 6 Chapter 10 Data Displays 10.3 25
a. Make a histogram of the data starting with the interval 0–14. Describe the shape of the distribution.
b. A company adds $5 to each donation. Make another histogram starting with the same interval as in part(a). Compare the shape of this distribution with the distribution in part(a). Explain any differences in the distributions.
Answer:

Question 17.
CRITICAL THINKING
Describe the shape of the distribution of each bar graph. Match the letters A, B, and C with the mean, the median, and the mode of each data set. Explain your reasoning.
Big Ideas Math Answers Grade 6 Chapter 10 Data Displays 10.3 26
Answer:

Lesson 10.4 Choosing Appropriate Measures

EXPLORATION 1
Using Shapes of Distributions
Work with a partner.
In Section 10.3 Exploration 1(a), you described the distribution of the first digits of the numbers at the right. In Exploration 1(b), you described the distribution of the data set below.
Big Ideas Math Solutions Grade 6 Chapter 10 Data Displays 10.4 1
What do you notice about the measures of center, measures of variation, and the shapes of the distributions? Explain.
b. Which measure of center best describes each data set? Explain your reasoning.
Big Ideas Math Solutions Grade 6 Chapter 10 Data Displays 10.4 2
c. which measures best describe the data. Which measure of variation best describes each data set? Explain your reasoning.
Answer:

You can use a measure of center and a measure of variation to describe the distribution of a data set.e shape of the distribution can help you choose which measures are the most appropriate to use.

Key Idea

Choosing Appropriate Measures
The mean absolute deviation (MAD) uses the mean in its calculation. So, when a data distribution is symmetric,
• use the mean to describe the center and
• use the MAD to describe the variation.

The interquartile range (IQR) uses quartiles in its calculation. So, when a data distribution is skewed,
• use the median to describe the center and
• use the IQR to describe the variation.

EXAMPLE 1
Choosing Appropriate Measures
The frequency table shows the number of states that border each state in the United States. What are the most appropriate measures to describe the center and the variation?
Big Ideas Math Solutions Grade 6 Chapter 10 Data Displays 10.4 3
To see the distribution of the data, display the data in a histogram.
The left side of the graph is approximately a mirror image of the right side of the graph. The distribution is symmetric.
So, the mean and the mean absolute deviation are the most appropriate measures to describe the center and the variation.

Try It
Question 1.
The frequency table shows the gas mileages of several motorcycles made by a company. What are the most appropriate measures to describe the center and the variation?
Big Ideas Math Solutions Grade 6 Chapter 10 Data Displays 10.4 4
Answer:
To see the distribution of the data, display the data in a histogram.
Big Ideas Math Grade 6 Chapter 10 Data Displays img_15

EXAMPLE 2
Describing a Data Set
The dot plot shows the average numbers of hours students in a class sleep each night. Describe the center and the variation of the data set.
Big Ideas Math Solutions Grade 6 Chapter 10 Data Displays 10.4 5
Most of the data values are on the right, clustered around 9, and the tail extends to the left. The distribution is skewed left, so the median and the interquartile range are the most appropriate measures to describe the center and the variation.
The median is 8.5 hours. The first quartile is 7.5, and the third quartile is 9. So, the interquartile range is 9 − 7.5 = 1.5 hours.
The data are centered around 8.5 hours. The middle half of the data varies by no more than 1.5 hours.

Try It
Question 2.
The dot plot shows the numbers of hours people spent at the gym last week. Describe the center and the variation of the data set.
Big Ideas Math Solutions Grade 6 Chapter 10 Data Displays 10.4 6
Answer:
Most of the data values are on the right, clustered around 6, and the tail extends to the left. The distribution is skewed left, so the median and the interquartile range are the most appropriate measures to describe the center and the variation.
The median is 5 hours. The first quartile is 2, and the third quartile is 4.
So, the interquartile range is 4 – 2 = 2 hours
The data are centered around 5 hours. The middle half of the data varies by no more than 2 hours.

Self-Assessment for Concepts & Skills

Solve each exercise. Then rate your understanding of the success criteria in your journal.
Question 3.
OPEN-ENDED
Construct a dot plot for which the mean is the most appropriate measure to describe the center of the distribution.
Answer:

CHOOSING APPROPRIATE MEASURES
Choose the most appropriate measures to describe the center and the variation. Explain your reasoning. Then find the measures you chose.
Question 4.
Big Ideas Math Solutions Grade 6 Chapter 10 Data Displays 10.4 7
Answer:
20, 28, 32, 32, 36, 36, 40, 40, 40, 40, 44, 44, 44, 48
Order your data set from lowest to highest values
Find the median. This is the second quartile Q2.
At Q2 split the ordered data set into two halves.
The lower quartile Q1 is the median of the lower half of the data.
Q1 = 32
The upper quartile Q3 is the median of the upper half of the data.
Q3 = 44
Median is the average of the data values.
So, the median, Q2 is 40.
Interquartile Range = Q3 – Q1
IQR = 44 – 32
IQR = 12

Question 5.
Big Ideas Math Solutions Grade 6 Chapter 10 Data Displays 10.4 8
Answer:
8, 10, 10, 12, 12, 12, 14, 14, 14, 14, 16, 16, 16, 18, 18, 20
8, 10, 12, 14, 16, 18, 20
Order your data set from lowest to highest values
Find the median. This is the second quartile Q2.
At Q2 split the ordered data set into two halves.
The lower quartile Q1 is the median of the lower half of the data.
Q1 = 12
The upper quartile Q3 is the median of the upper half of the data.
Q3 = 16
Median is the average of the data values.
So, the median, Q2 is 14
Interquartile Range = Q3 – Q1
IQR = 16 – 12
IQR = 4

Question 6.
WRITING
Explain why the most appropriate measures to describe the center and the variation of a data set are determined by the shape of the distribution.
Answer:
You can use a measure of center and a measure of variation to describe the distribution of a data set. The shape of the distribution can help you choose which measures are the most appropriate to use. The dot plot shows the average number of hours students in a class sleep each night.

EXAMPLE 3
Modeling Real Life
Two baskets each have16 envelopes with money inside, as shown in the tables. How much does a typical envelope in each basket contain? Why might a person want to pick from Basket B instead of Basket A?
Big Ideas Math Solutions Grade 6 Chapter 10 Data Displays 10.4 9
In each graph, the left side is a mirror image of the right side. Because both distributions are symmetric, the mean and the mean absolute deviation are the most appropriate measures to describe the center and the variation.
The mean of each data set is \(\frac{800}{16}\) = $50. The MAD of Basket A is \(\frac{320}{16}\) = $20, and the MAD of Basket B is \(\frac{120}{16}\) = $7.50. So, Basket A has more variability.

A typical envelope in each basket contains about $50. A person may choose from Basket B instead of Basket A because there is less variability. This means it is more likely to get an amount near $50 by choosing an envelope from Basket B than by choosing an envelope from Basket A.
Answer:

Self-Assessment for Problem Solving

Solve each exercise. Then rate your understanding of the success criteria in your journal.
Question 7.
Why might a person want to pick from Basket A instead of Basket B in Example 3? Explain your reasoning.
Answer:

Question 8.
In a video game, two rooms each have 12 treasure chests containing gold coins. The tables show the numbers of coins in each chest. You pick one chest and are rewarded with the coins inside. From which room would you choose? Explain your reasoning.
Big Ideas Math Solutions Grade 6 Chapter 10 Data Displays 10.4 10
Answer:

Question 9.
Create a dot plot of the numbers of pets that students in your class own. Describe the center and the variation of the data set.
Answer:

Choosing Appropriate Measures Homework & Practice 10.4

Review & Refresh

Describe the shape of the distribution.
Question 1.
Big Ideas Math Solutions Grade 6 Chapter 10 Data Displays 10.4 11
Answer:
Order the data
5, 6, 7, 7, 8, 8, 8, 9, 9, 9, 9, 9, 10, 10, 10, 10
The shape of the distribution for the above dot plot is
Big-Ideas-Math-Solutions-Grade-6-Chapter-10-Data-Displays-10.4-11

Question 2.
Big Ideas Math Solutions Grade 6 Chapter 10 Data Displays 10.4 12
Answer:

Find the median, first quartile, third quartile, and interquartile range of the data.
Question 3.
68, 74, 67, 72, 63, 70, 78, 64, 76
Answer:
Order the data
63, 64, 67, 68, 70, 72, 74, 76, 78
The median is nothing but the average value of the data.
70 is the average of the data values.
Order your data set from lowest to highest values
Find the median. This is the second quartile Q2.
Thus the second quartile Q2 is 70.
At Q2 split the ordered data set into two halves.
The lower quartile Q1 is the median of the lower half of the data.
Thus Q1 is 65.5
The upper quartile Q3 is the median of the upper half of the data.
Q3 is 75
Interquartile Range IQR = 9.5
If the size of the data set is odd, do not include the median when finding the first and third quartiles.
If the size of the data set is even, the median is the average of the middle 2 values in the data set. Add those 2 values, and then divide by 2. The median splits the data set into lower and upper halves and is the value of the second quartile Q2.

Question 4.
39, 48, 33, 24, 30, 44, 36, 41, 28, 53
Answer:
Order the data
24, 28, 30, 33, 36, 39, 41, 44, 48, 53
If the size of the data set is even, the median is the average of the middle 2 values in the data set. Add those 2 values, and then divide by 2. The median splits the data set into lower and upper halves and is the value of the second quartile Q2.
Median is (36+39)/2 = 37.5
Order your data set from lowest to highest values
Find the median. This is the second quartile Q2.
Thus the second quartile Q2 is 37.5
At Q2 split the ordered data set into two halves.
The lower quartile Q1 is the median of the lower half of the data.
Thus Q1 is 30
The upper quartile Q3 is the median of the upper half of the data.
Q3 is 44
Interquartile Range IQR = 14
If the size of the data set is odd, do not include the median when finding the first and third quartiles.

Divide. Write the answer in simplest form.
Question 5.
4\(\frac{2}{5}\) ÷ 2
Answer: 2 \(\frac{1}{5}\)

Explanation:
Convert any mixed numbers to fractions.
4\(\frac{2}{5}\) = \(\frac{22}{5}\)
\(\frac{22}{5}\) × \(\frac{1}{2}\) = \(\frac{22}{10}\)
Now convert from improper fraction to the mixed fraction.
\(\frac{22}{10}\) = 2 \(\frac{1}{5}\)

Question 6.
5\(\frac{1}{8}\) ÷ \(\frac{7}{8}\)
Answer: 5 \(\frac{6}{7}\)

Explanation:
Convert any mixed numbers to fractions.
5\(\frac{1}{8}\) = \(\frac{41}{8}\)
\(\frac{41}{8}\) ÷ \(\frac{7}{8}\)
\(\frac{41}{8}\) × \(\frac{8}{7}\) = \(\frac{328}{56}\)
Now convert from improper fraction to the mixed fraction.
\(\frac{328}{56}\) = 5 \(\frac{6}{7}\)

Question 7.
2\(\frac{3}{7}\) ÷ 1\(\frac{1}{7}\)
Answer: 2 \(\frac{1}{8}\)

Explanation:
Convert any mixed numbers to fractions.
2\(\frac{3}{7}\) = \(\frac{17}{7}\)
1\(\frac{1}{7}\) = \(\frac{8}{7}\)
\(\frac{17}{7}\) ÷ \(\frac{8}{7}\) = \(\frac{119}{56}\)
Simplify the fraction
\(\frac{119}{56}\) = 2 \(\frac{1}{8}\)

Question 8.
\(\frac{4}{5}\) ÷ 7\(\frac{1}{2}\)
Answer: \(\frac{8}{75}\)

Explanation:
Convert any mixed numbers to fractions.
7\(\frac{1}{2}\) = \(\frac{15}{2}\)
\(\frac{4}{5}\) ÷ \(\frac{15}{2}\) = \(\frac{8}{75}\)

Concepts, Skills, & Problem Solving

USING SHAPES OF DISTRIBUTIONS
Find the mean and the median of the data set. Which measure of center best describes the data set? Explain your reasoning. (See Exploration 1, p. 477.)
Question 9.
9, 3, 7, 7, 9, 2, 8, 9, 6, 7, 8, 9
Answer:

Question 10.
24, 25, 27, 27, 23, 29, 26, 26, 26, 25, 28
Answer:

CHOOSING APPROPRIATE MEASURES
Choose the most appropriate measures to describe the center and the variation.
Question 11.
Big Ideas Math Solutions Grade 6 Chapter 10 Data Displays 10.4 13
Answer:

Question 12.
Big Ideas Math Solutions Grade 6 Chapter 10 Data Displays 10.4 14
Answer:

Question 13.
Big Ideas Math Solutions Grade 6 Chapter 10 Data Displays 10.4 15
Answer:

Question 14.
Big Ideas Math Solutions Grade 6 Chapter 10 Data Displays 10.4 16
Answer:

Question 15.
DESCRIBING DATA SETS
Describe the centers and the variations of the data sets in Exercises 11 and 12.
Answer:

Question 16.
MODELING REAL LIFE
The frequency table shows the numbers of eggs laid by several octopi. What are the most appropriate measures to describe the center and the variation? Explain your reasoning.
Big Ideas Math Solutions Grade 6 Chapter 10 Data Displays 10.4 17
Answer:

Question 17.
MODELING REAL LIFE
The dot plot shows the vertical jump heights (in inches) of several professional athletes. Describe the center and the variation of the data set.
Big Ideas Math Solutions Grade 6 Chapter 10 Data Displays 10.4 18
Answer:

Question 18.
OPEN-ENDED
Describe a real-life situation where the median and the interquartile range are likely the best measures of center and variation to describe the data. Explain your reasoning.
Answer:

Question 19.
PROBLEM SOLVING
You play a board game in which you draw from one of two piles of cards. Each card has a number that says how many spaces you will move your piece forward on the game board. The tables show the numbers on the cards in each pile. From which pile would you choose? Explain your reasoning.
Big Ideas Math Solutions Grade 6 Chapter 10 Data Displays 10.4 19
Answer:

Question 20.
DIG DEEPER!
The frequency table shows the numbers of words that several students can form in 1 minute using the letters P, S, E, D, A. What are the most appropriate measures to describe the center and variation? Can you find the exact values of the measures of center and variation for the data? Explain.
Big Ideas Math Solutions Grade 6 Chapter 10 Data Displays 10.4 20
Answer:

Question 21.
REASONING
A bag contains 20 vouchers that can be redeemed for different numbers of tokens at an arcade, as shown in the table.
Big Ideas Math Solutions Grade 6 Chapter 10 Data Displays 10.4 21
a. Find the most appropriate measure to describe the center of the data set.
b. You randomly select a voucher from the bag. How many tokens are you most likely to receive? Explain.
c. Are your answers in parts (a) and (b) the same? Explain why or why not.
Answer:

Lesson 10.5 Box-and-Whisker Plots

EXPLORATION 1
Drawing a Box-and-Whisker Plot
Work with a partner. Each student in a sixth-grade class is asked to choose a number from 1 to 20. The results are shown below.
Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.5 1
a. The box-and-whisker plot below represents the data set. Which part represents the box? the whiskers? Explain.
Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.5 2
b. What does each of the five plotted points represent?
c. In your own words, describe what a box-and-whisker plot is and what it tells you about a data set.
d. Conduct a survey in your class. Have each student write a number from 1 to 20 on a piece of paper. Collect all of the data and draw a box-and-whisker plot that represents the data. Compare the data with the box-and-whisker plot in part(a).
Answer:

Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.5 3.1

Key Idea
Box-and-Whisker Plot
A box-and-whisker plot represents a data set along a number line by using the least value, the greatest value, and the quartiles of the data. A box-and-whisker plot shows the variability of a data set.
Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.5 3
The five numbers that make up the box-and-whisker plot are called the five-number summary of the data set.

EXAMPLE 1
Making a Box-and-Whisker Plot
Make a box-and-whisker plot for the ages(in years) of the spider monkeys at a zoo.
15, 20, 14, 38, 30, 36, 30, 30, 27, 26, 33, 35
Step 1: Order the data. Find the quartiles.
Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.5 4
Step 2: Draw a number line that includes the least and greatest values. Graph points above the number line that represent the five-number summary.
Step 3: Draw a box using the quartiles. Draw a line through the median. Draw whiskers from the box to the least and the greatest values.
Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.5 5
Answer:

Try It
Question 1.
A group of friends spent 1, 0, 2, 3, 4, 3, 6, 1, 0, 1, 2, and 2 hours online last night.Make a box-and-whisker plot for the data.
Answer:

The figure shows how data are distributed in a box-and-whisker plot.
Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.5 6

EXAMPLE 2
Analyzing a Box-and-Whisker Plot
The box-and-whisker plot shows the body mass index (BMI) of a sixth-grade class.
Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.5 7
a. What fraction of the students have a BMI of at least 22?
The right whisker represents students who have a BMI of at least 22.
So, about \(\frac{1}{4}\) of the students have a BMI of at least 22.
b. Are the data more spread out below the first quartile or above the third quartile? Explain.
The right whisker is longer than the left whisker.
So, the data are more spread out above the third quartile than below the first quartile.
c. Find and interpret the interquartile range of the data.
interquartile range = third quartile − first quartile
= 22 – 19 = 3
So, the middle half of the students’ BMIsvaries by no more than 3.

Try It
Question 2.
The box-and-whisker plot shows the heights of the roller coasters at an amusement park.
(a) What fraction of the roller coasters are between 120 feet tall and 220 feet tall?
(b) Are the data more spread out below or above the median? Explain.
(c) Find and interpret the interquartile range of the data.
Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.5 8
Answer:

A box-and-whisker plot also shows the shape of a distribution.
Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.5 9

EXAMPLE 3
Identifying Shapes of Distributions
The double box-and-whisker plot represents the life spans of crocodiles and alligators at a zoo. Identify the shape of the distribution of the lifespans of alligators.
Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.5 10
For alligator life spans, the whisker lengths are equal. The median is in the middle of the box. The left side of the box-and-whisker plot is a mirror image of the right side of the box-and-whisker plot.
So, the distribution is symmetric.
Answer:

Try It
Question 3.
Identify the shape of the distribution of the life spans of crocodiles.
Answer:

Self-Assessment for Concepts & Skills

Solve each exercise. Then rate your understanding of the success criteria in your journal.
Question 4.
VOCABULARY
Explain how to find the five-number summary of a data set.
Answer:

MAKING A BOX-AND-WHISKER PLOT
Make a box-and-whisker plot for the data. Identify the shape of the distribution.
Question 5.
Ticket prices (dollars): 39, 42, 40, 47, 38, 39, 44, 55, 44, 58, 45
Answer:

Question 6.
Number of sit-ups: 20, 20, 23, 25, 25, 26, 27, 29, 30, 30, 32, 34, 37, 38
Answer:

Question 7.
NUMBER SENSE
In a box-and-whisker plot, what fraction of the data is greater than the first quartile?
Answer:

EXAMPLE 4
Modeling Real Life
The double box-and-whisker plot represents the prices of snowboards at two stores.
Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.5 11
a. Which store’s prices are more spread out? Explain. Both boxes appear to be the same length. So, the interquartile range of each data set is equal. The range of the prices in Store B, however, is greater than the range of the prices in Store A.
So, the prices in Store B are more spread out.
b. Which store’s prices are generally higher? Explain.
For Store A,the distribution is symmetric with about one-half of the prices above $300.
For Store B, the distribution is skewed right with about three-fourths of the prices above $300.
So, the prices in Store B are generally higher.
Answer:

Self-Assessment for Problem Solving

Solve each exercise. Then rate your understanding of the success criteria in your journal.
Question 8.
The tables at the left show the test scores of two sixth-grade achievement tests. The same group of students took both tests. The students took one test in the fall and the other in the spring.
a. Analyze each distribution. Then compare and contrast the test results.
b. Which table likely represents the results of which test? Explain your reasoning.
Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.5 12
Answer:

Question 9.
Make a box-and-whisker plot that represents the heights of the boys in your class. Then make a box-and-whisker plot that represents the heights of the girls in your class. Compare and contrast the distributions.
Answer:

Box-and-Whisker Plots Homework & Practice 10.5

Review & Refresh

Choose the most appropriate measures to describe the center and the variation.
Question 1.
Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.5 13
Answer:

Question 2.
Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.5 14
Answer:

Copy and complete the statement using < or >.
Question 3.
Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.5 15
Answer: –\(\frac{2}{3}\) > –\(\frac{3}{4}\)

Explanation:
Compare fractions to find which fraction is larger and which is smaller.
The least common denominator (LCD) is 12
Rewriting as equivalent fractions with the LCD:
\(\frac{2}{3}\) = \(\frac{8}{12}\)
\(\frac{3}{4}\) = \(\frac{9}{12}\)
Now compare the fractions
–\(\frac{8}{12}\) >-\(\frac{9}{12}\)
Thus we can say that –\(\frac{2}{3}\) > –\(\frac{3}{4}\)

Question 4.
Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.5 16
Answer: -2 \(\frac{1}{5}\) < -2 \(\frac{1}{6}\)

Explanation:
Compare fractions to find which fraction is larger and which is smaller.
Rewriting these inputs as fractions:
2 \(\frac{1}{5}\) = \(\frac{11}{5}\)
2 \(\frac{1}{6}\) = \(\frac{13}{6}\)
The LCM is 30
Rewriting as equivalent fractions with the LCD
\(\frac{11}{5}\) = \(\frac{66}{30}\)
\(\frac{13}{6}\) = \(\frac{65}{30}\)
– \(\frac{66}{30}\) < – \(\frac{65}{30}\)
-2 \(\frac{1}{5}\) < -2 \(\frac{1}{6}\)

Question 5.
Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.5 17
Answer: -5.3 > -5.5

Explanation:
Compare fractions to find which fraction is larger and which is smaller.
The smallest number with the negative sign will be the greater number
Thus -5.3 > -5.5

Factor the expression using the GCF.
Question 6.
42 + 14
Answer

Question 7.
12x – 18
Answer:

Question 8.
28n + 20
Answer:

Question 9.
60g – 25h
Answer:

Concepts, Skills, & Problem Solving

COMPARING DATA Compare the data in the box-and-whisker plots. (See Exploration 1, p. 483.)
Question 10.
Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.5 18
Answer:

Question 11.
Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.5 19
Answer:

MAKING A BOX-AND-WHISKER PLOT
Make a box-and-whisker plot for the data.
Question 12.
Ages of teachers (in years): 30, 62, 26, 35, 45, 22, 49, 32, 28, 50, 42, 35
Answer:

Question 13.
Quiz scores: 8, 12, 9, 10, 12, 8, 5, 9, 7, 10, 8, 9, 11
Answer:

Question 14.
Donations (in dollars): 10, 30, 5, 15, 50, 25, 5, 20, 15, 35, 10, 30, 20
Answer:

Question 15.
Science test scores: 85, 76, 99, 84, 92, 95, 68, 100, 93, 88, 87, 85
Answer:

Question 16.
Shoe sizes: 12, 8.5, 9, 10, 9, 11, 11.5, 9, 9, 10, 10, 10.5, 8
Answer:

Question 17.
Ski lengths (in centimeters): 180, 175, 205, 160, 210, 175, 190, 205, 190, 160, 165, 195
Answer:

Question 18.
YOU BE THE TEACHER
Your friend makes a box-and-whisker plot for the data shown. Is your friend correct? Explain your reasoning.
Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.5 20
2, 6, 4, 3, 7, 4, 6, 9, 6, 8, 5, 7
Answer:

Question 19.
MODELING REAL LIFE
The numbers of days 12 friends went camping during the summer are 6, 2, 0, 10, 3, 6, 6, 4, 12, 0, 6, and 2. Make a box-and-whisker plot for the data. What is the range of the data?
Answer:

Question 20.
ANALYZING A BOX-AND-WHISKER PLOT
The box-and-whisker plot represents the numbers of gallons of water needed to fill different types of dunk tanks offered by a company.
Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.5 21
a. What fraction of the dunk tanks requires at least 500 gallons of water?
b. Are the data more spread out below the first quartile or above the third quartile? Explain.
c. Find and interpret the interquartile range of the data.
Answer:

Question 21.
MODELING REAL LIFE
The box-and-whisker plot represents the heights (in meters) of the tallest buildings in Chicago.
Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.5 22
a. What percent of the buildings are no taller than 345 meters?
b. Is there more variability in the heights above 345 meters or below 260.5 meters? Explain.
c. Find and interpret the interquartile range of the data.
Answer:

Question 22.
CRITICAL THINKING
The numbers of spots on several frogs in a jungle are shown in the dot plot.
Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.5 23
a. Make a box-and-whisker plot for the data.
b. Compare the dot plot and the box-and-whisker plot. Describe the advantages and disadvantages of each data display.
Answer:

SHAPES OF BOX-AND-WHISKER PLOTS
Identify the shape of the distribution. Explain.
Question 23.
Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.5 24
Answer:

Question 24.
Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.5 25
Answer:

Question 25.
Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.5 26
Answer:

Question 26.
Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.5 27
Answer:

Question 27.
MODELING REAL LIFE
The double box-and-whisker plot represents the start times of recess for classes at two schools.
Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.5 28
a. Identify the shape of each distribution.
b. Which school’s start times for recess are more spread out? Explain.
c. You randomly pick one class from each school. Which class is more likely to have recess before lunch? Explain.
Answer:

MAKING A BOX-AND-WHISKER PLOT
Make a box-and-whisker plot for the data.
Question 28.
Temperatures (in °C): 15, 11, 14, 10, 19, 10, 2, 15, 12, 14, 9, 20, 17, 5
Answer:

Question 29.
Checking account balances (in dollars): 30, 0, 50, 20, 90, −15, 40, 100, 45, −20, 70, 0
Answer:

Question 30.
REASONING
The data set in Exercise 28 has an outlier. Describe how removing the outlier affects the box-and-whisker plot.
Answer:

Question 31.
OPEN-ENDED
Write a data set with 12 values that has a symmetric box-and-whisker plot.
Answer:

Question 32.
CRITICAL THINKING
When does a box-and-whisker plot not have one or both whiskers?
Answer: A simpler formulation is this: no whisker will be visible if the lower quartile is equal to the minimum, or if the upper quartile is equal to the maximum.

Question 33.
STRUCTURE Draw a histogram that could represent the distribution shown in Exercise 25.
Answer:

Question 34
DIG DEEPER!
The double box-and-whisker plot represents the goals scored per game by two lacrosse teams during a 16-game season.
Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays 10.5 29
a. Which team is more consistent? Explain.
b. Team 1 played Team 2 once during the season. Which team do you think won? Explain.
c. Can you determine the number of games in which Team 2 scored 10 goals or less? Explain your reasoning.
Answer:

Question 35.
CHOOSE TOOLS
A market research company wants to summarize the variability of the SAT scores of graduating seniors in the United States. Should the company use a stem-and-leaf plot, a histogram, or a box-and-whisker plot? Explain.
Answer:

Data Displays Connecting Concepts

Using the Problem-Solving Plan
1. The locations of pitches in an at-bat are shown in the coordinate plane, where the coordinates are measured in inches. Describe the location of a typical pitch in the at-bat.
Understand the problem
You know the locations of the pitches. You are asked to find the location of a typical pitch in the at-bat.

Make a plan
First, use the coordinates of the pitches to create two data sets, one for the x-coordinates of the pitches and one for the y-coordinates of the pitches. Next, make a box-and-whisker plot for each data set. Then use the most appropriate measure of center for each data set to find the location of a typical pitch.

Solve and check
Use the plan to solve the problem. Then check your solution.
Big Ideas Math Answers 6th Grade Chapter 10 Data Displays cc 1

2. A set of 20 data values is described below. Sketch a histogram that could represent the data set. Explain.
Big Ideas Math Answers 6th Grade Chapter 10 Data Displays cc 2
• least value: 10
• first quartile: 25
• mean: 29
• third quartile: 34
• greatest value: 48
• MAD: 7

3. The chart shows the dimensions (in inches) of several flat-rate shipping boxes. Each box is in the shape of a rectangular prism. Describe the distribution of the volumes of the boxes. Then find the most appropriate measures to describe the center and the variation of the volumes.

Performance Task
Classifying Dog Breeds by Size
At the beginning of this chapter, you watched a STEAM Video called “Choosing a Dog.” 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 6th Grade Chapter 10 Data Displays cc 3

Data Displays Chapter Review

Review Vocabulary
Write the definition and give an example of each vocabulary term.
Big Ideas Math Answers 6th Grade Chapter 10 Data Displays cr 1

Graphic Organizers
You can use an Information Frame to help you organize and remember concepts. Here is an example of an Information Frame for the vocabulary term histogram.
Big Ideas Math Answers 6th Grade Chapter 10 Data Displays cr 2
Choose and complete a graphic organizer to help you study the concept.
1. stem-and-leaf plot
2. frequency table
3. shapes of distributions
4. box-and-whisker plot
Answer:

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 6th Grade Chapter 10 Data Displays cr 3

10.1 Stem-and-Leaf Plots
Learning Target: Display and interpret data in stem-and-leaf plots.

Make a stem-and-leaf plot of the data.
Question 1.
Big Ideas Math Answers 6th Grade Chapter 10 Data Displays crr 1
Answer:
Big Ideas Math Answers Grade 6 Chapter 10 Data Displays img_16

Question 2.
Big Ideas Math Answers 6th Grade Chapter 10 Data Displays crr 2
Answer:
Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays img_17

Question 3.
The stem-and-leaf plot shows the weights (in pounds) of yellowfin tuna caught during a fishing contest.
Big Ideas Math Answers 6th Grade Chapter 10 Data Displays crr 3
a. How many tuna weigh less than 90 pounds?
b. Find the mean, median, mode, range, and interquartile range of the data.
c. How are the data distributed?
Answer:

Question 4.
The stem-and-leaf plot shows the body mass index (BMI) for adults at a recreation center. Use the data to answer the question, “What is the typical BMI for an adult at the recreation center?” Explain.
Big Ideas Math Answers 6th Grade Chapter 10 Data Displays crr 4
Answer:

Question 5.
Write a statistical question that can be answered using the stem-and-leaf plot.
Big Ideas Math Answers 6th Grade Chapter 10 Data Displays crr 5
Answer:

10.2 Histograms (pp. 463-470)
Learning Target: Display and interpret data in histograms.

Display the data in a histogram.
Question 6.
Big Ideas Math Answers 6th Grade Chapter 10 Data Displays crr 6
Answer:

Question 7.
Big Ideas Math Answers 6th Grade Chapter 10 Data Displays crr 7
Answer:
Big Ideas Math Answer Key Grade 6 Chapter 10 Data Displays img_19

Question 8.
The histogram shows the number of crafts each member of a craft club made for a fundraiser.
Big Ideas Math Answers 6th Grade Chapter 10 Data Displays crr 8
a. Which interval contains the most data values?
b. Frequency How many members made at least 6 crafts?
c. Can you use the histogram to determine the total number of crafts made? Explain.
Answer:

10.3 Shapes of Distributions (pp. 471–476)
Learning Target: Describe and compare shapes of distributions.

Question 9.
Describe the shape of the distribution.
Answer: The shape of a distribution is described by its number of peaks and by its possession of symmetry, its tendency to skew, or its uniformity.

Question 10.
The frequency table shows the math test scores for the same class of students as Exercise 9. Display the data in a histogram. Which test has higher scores?
Big Ideas Math Answers 6th Grade Chapter 10 Data Displays crr 10
Answer:

Question 11.
The table shows the numbers of neutrons for several elements in the nonmetal group of the periodic table. Make a histogram of the data starting with the interval 0–9. Describe the shape of the distribution.
Big Ideas Math Answers 6th Grade Chapter 10 Data Displays crr 11
Answer:

10.4 Choosing Appropriate Measures (pp. 477–482)
Learning Target: Use the shape of the distribution of a data set to determine which measures of center and variation best describe the data.

Choose the most appropriate measures to describe the center and the variation. Students’ Heights
Question 12.
Big Ideas Math Answers 6th Grade Chapter 10 Data Displays crr 12
Answer:

Question 13.
Big Ideas Math Answers 6th Grade Chapter 10 Data Displays crr 13
Answer:

Question 14.
Describe the center and the variation of the data set in Exercise 13.
Answer:

10.5 Box-and-Whisker Plots (pp. 483–490)
Learning Target: Display and interpret data in box-and-whisker plots.

Make a box-and-whisker plot for the data.
Question 15.
Ages of volunteers at a hospital:
14, 17, 20, 16, 17, 14, 21, 18
Answer:

Question 16.
Masses (in kilograms) of lions:
120, 200, 180, 150, 200, 200, 230, 160
Answer:

Question 17.
The box-and-whisker plot represents the lengths (in minutes) of movies being shown at a theater.
Big Ideas Math Answers 6th Grade Chapter 10 Data Displays crr 17
a. What percent of the movies are no longer than 120 minutes?
b. Is there more variability in the movie lengths longer than 130 minutes or shorter than 110 minutes? Explain.
c. Find and interpret the interquartile range of the data.
Answer:

Question 18.
The double box-and-whisker plot represents the heights of students in two math classes.
Big Ideas Math Answers 6th Grade Chapter 10 Data Displays crr 18
a. Identify the shape of each distribution.Height(cm)
b.Which class has heights that are more spread out? Explain.
c.You randomly pick one student from each class. Which student is more likely to be taller than 170 centimeters? Explain.
Answer:

Data Displays Practice Test

Make a stem-and-leaf plot of the data.
Question 1.
Big Ideas Math Answers Grade 6 Chapter 10 Data Displays pt 1
Answer:
Big Ideas Math Answers Grade 6 Chapter 10 Data Displays img_8

Question 2.
Big Ideas Math Answers Grade 6 Chapter 10 Data Displays pt 2
Answer:

Big Ideas Math Answers Grade 6 Chapter 10 Data Displays img_9

Question 3.
Find the mean, median, mode, range, and interquartile range of the data.
Big Ideas Math Answers Grade 6 Chapter 10 Data Displays pt 3
Answer:
Given data 35, 38, 40, 41, 48, 50, 54, 54, 54, 55, 59, 60
Mean:
The mean refers to an intermediate value between a discrete set of numbers, namely, the sum of all values in the data set, divided by the total number of values.
x̄ = (35+38+40+41+48+50+54+54+54+55+59+60)/12
x̄ = 49
Thus mean of the given data is 49.
Median:
Given data 35, 38, 40, 41, 48, 50, 54, 54, 54, 55, 59, 60
In the case where the total number of values in a data sample is odd, the median is simply the number in the middle of the list of all values. When the data sample contains an even number of values, the median is the mean of the two middle values.
Median = (50+54)/2 = 104/2 = 52
Thus the median of the given data is 52.
Mode:
The mode is the value in a data set that has the highest number of recurrences.
35, 38, 40, 41, 48, 50, 54, 54, 54, 55, 59, 60
Mode = 54 (repeated 3 times)

Question 4.
Display the data in a histogram. How many people watched less than 20 hours of television per week?
Big Ideas Math Answers Grade 6 Chapter 10 Data Displays pt 4
Answer:
BIM Grade 6 Chapter 10 Data Displays Answer Key img_10
By seeing the above histogram we can find the number of people who watched less than 20 hours of television per week.
14 + 16 = 30
Therefore 30 people watched less than 20 hours per week.

Question 5.
The dot plot shows the numbers of glasses of water Water Consumed that the students in a class drink in one day.
Big Ideas Math Answers Grade 6 Chapter 10 Data Displays pt 5
a. Describe the shape of the distribution.
b. Choose the most appropriate measures to describe the center and the variation. Find the measures you chose.
Answer:

Question 6.
Make a box-and-whisker plot for the lengths (in inches) of fish in a pond: 12, 13, 7, 8, 14, 6, 13, 10.
Answer:

Question 7.
The double box-and-whisker plot compares the battery lives (in hours) of two brands of cell phones.
Big Ideas Math Answers Grade 6 Chapter 10 Data Displays pt 7
a. What is the range of the upper 75% of battery life for each brand of cell phone?
b. Which brand of cell phone typically has a longer battery life? Explain.
c. In the box-and-whisker plot, there are 190 cell phones of Brand A that have at most 10.5 hours of battery life. About how many cell phones are represented in the box-and-whisker plot for Brand A?
Answer:

Data Displays Cumulative Practice

Question 1.
Research scientists are measuring the numbers of days lettuce seeds take to germinate. In a study, 500 seeds were planted. Of these,473 seeds germinated. The box-and-whisker plot summarizes the numbers of days it took the seeds to germinate. What can you conclude from the box-and-whisker plot?
Big Ideas Math Solutions Grade 6 Chapter 10 Data Displays cp 1
A. The median number of days for the seeds to germinate is 12.
B. 50% of the seeds took more than 8 days to germinate.
C. 50% of the seeds took less than 5 days to germinate.
D. The median number of days for the seeds to germinate was 6.
Answer:

Question 2.
Find the interquartile range of the data.
15 7 5 8 9 20 12 7 11 7 15
F. 8
G. 11
H. 12
I. 20
Answer: 8

Question 3.
There are seven different integers in a set. When they are listed from least to greatest, the middle integer is −1. Which statement below must be true?
A. There are three negative integers in the set.
B. There are three positive integers in the set.
C. There are four negative integers in the set.
D. The integer in the set after −1 is positive.
Answer:

Question 4.
What is the mean number of seats?
Big Ideas Math Solutions Grade 6 Chapter 10 Data Displays cp 4
F. 2.4 seats
G. 5 seats
H. 6.5 seats
I. 7 seats5.
Answer:

Question 5.
On Wednesday, a town received 17 millimeters of rain. This was x millimeters more rain than the town received on Tuesday. Which expression represents the amount of rain, in millimeters, the town received on Tuesday?
A. 17x
B. 17 – x-c
C. x + 17
D. x – 17
Answer:

Question 6.
One of the leaves is missing in the stem-and-leaf plot.
Big Ideas Math Solutions Grade 6 Chapter 10 Data Displays cp 6
The median of the data set represented by the stem-and-leaf plot is 38. What is the value of the missing leaf?
Answer:

Question 7.
Which property is demonstrated by the equation?
723 + (y + 277) = 723 + (277 + y)
F. Associative Property of Addition
G. Commutative Property of Addition
H. Distributive Property
I. Addition Property of Zero
Answer: Associative Property of Addition

Explanation:
Associative property of addition: Changing the grouping of addends does not change the sum
Thus the correct answer is option F.

Question 8.
A student took five tests and had a mean score of 92. Her scores on the first 4 tests were 90, 96, 86, and 92. What was her score on the fifth test?
A. 92
B. 93
C. 96
D. 98
Answer: 86

Explanation:
Given that,
A student took five tests and had a mean score of 92.
Her scores on the first 4 tests were 90, 96, 86, and 92.
(90+96+86+92+s)/5=90
(364+s)/5=90
364+s=450
s=86
So she scored an 86 on the fifth test.

Question 9.
At the end of the school year, your teacher counted the number of absences for each student. The results are shown in the histogram. How many students had fewer than 10 absences?
Big Ideas Math Solutions Grade 6 Chapter 10 Data Displays cp 9
Answer:

Question 10.
The ages of the 16 members of a camera club are listed below.
40, 22, 24, 58, 30, 31, 37, 25, 62, 40, 39, 37, 28, 28, 51, 44
Big Ideas Math Solutions Grade 6 Chapter 10 Data Displays cp 10.1
Big Ideas Math Solutions Grade 6 Chapter 10 Data Displays cp 10
Part A Order the ages from youngest to oldest.
Part B Find the median of the ages.
Part C Make a box-and-whisker plot for the ages.
Answer:

Final Words:

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Big Ideas Math Answers Grade 5 | Big Ideas Math Book 5th Grade Answer Key

Big Ideas Math Answers Grade 5 | Big Ideas Math Book 5th Grade Answer Key

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Big Ideas 5th Grade Math Book Answer Key | Big Ideas 5th Grade Math Answers Solutions Pdf

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Big Ideas Math Answers Grade 6 Advanced Chapter 1 Numerical Expressions and Factors

Big Ideas Math Answers Grade 6 Advanced Chapter 1

Do you want to become proficient in Math and get a good hold of the Big Ideas Math Book Grade 6 Advanced? If so, this is the one-stop destination for all your needs. You can access the Topicwise BIM Grade 6 Advanced Ch 1 Numerical Expressions and Factors Solutions PDF available as a part of your preparation. Test your knowledge of the concepts by solving from the Big Ideas Math Answers Grade 6 Advanced Chapter 1 and identify the knowledge gap accordingly.

Big Ideas Math Book 6th Grade Advanced Answer Key Chapter 1 Numerical Expressions and Factors

Use the ultimate guide over here for all your doubts on Big Ideas Math 6th Grade Advanced Ch 1 Numerical Expressions and Factors Concepts. Big Ideas Math Book 6th Grade Advanced Solutions provided here covers questions from Chapter 1 aligned as per the BIm Textbooks.  All you have to do is simply tap on the quick links available to avail the respective topic in no time. Begin your preparation right away taking the help of the Numerical Expressions and Factors Big Ideas Math Grade 6 Solutions and clear the exam with flying colors.

Numerical Expressions and Factors STEAM Video/ Performance Task

STEAM Video

Big Ideas Math Answer Key Grade 6 Advanced Chapter 1 Numerical Expressions and Factors 1

Filling Piñatas

Common factors can be used to make identical groups of objects. Can you think of any situations in which you would want to separate objects into equal groups? Are there any common factors that may be more useful than others? Can you think of any other ways to use common factors?

Watch the STEAM Video “Filling Piñatas.” Then answer the following questions. The table below shows the numbers of party favors that Alex and Enid use to make piñatas
Big Ideas Math Answer Key Grade 6 Advanced Chapter 1 Numerical Expressions and Factors 2

Question 1.
When finding the number of identical piñatas that can be made, why is it helpful for Alex and Enid to list the factors of each number given in the table?
Answer:

Question 2.
You raIt to create 6 identical piñatas. How can you change the numbers of party favors in the table to make this happen? Can you do this without changing the total number of party favors?
Answer:

Performance Task

Big Ideas Math Answer Key Grade 6 Advanced Chapter 1 Numerical Expressions and Factors 5

Setting the Table

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 asked to plan a fundraising event with the items below.

72 chairs
48 balloons
24 flowers
32 candles

You will find the greatest number of identical tables that can be prepared, and what will be in each centerpiece.When making arrangements for a party, should a part planner always use the greatest number of identical tables possible? Explain why or why not.
Answer:

Getting Ready for Numerical Expressions and Factors

Chapter Exploration

Work with a partner. In Exercises 1 and 2, use the table.
Big Ideas Math Answer Key Grade 6 Advanced Chapter 1 Numerical Expressions and Factors 3

Big Ideas Math Answer Key Grade 6 Advanced Chapter 1 Numerical Expressions and Factors 4
Eratosthenes
(c. 276 – c. 194 B.c.)
This table is called the Sieve of Eratosthenes. Eratosthenes was a Greek mathematician who was the chief librarian at theLibrary of Alexandria in Egypt. He was the first person to calculate the circumference of Earth.

Question 1.
Cross out the multiples of 2 that are greater than 2. Do the same for 3, 5, and 7.
Answer:

Question 2.
The numbers that are not crossed out are called prime numbers. ‘The numbers that are crossed out are called composite numbers. In your own words, describe the characteristics of prime numbers and composite numbers.
Answer:

Question 3.
MODELING REAL LIFE
Work with a partner. Cicadas are insects that live underground and emerge from the ground after x or x + 4 years. Is it possible that both x and x + 4 are prime? Give some examples.
Answer:

Numerical Expressions and Factors Vocabulary

The following vocabulary terms are defined in this chapter. think about what each term might rucan and record your thoughts.
exponent                                                common factors                               common multiples
numerical expression                             greatest common factor                   least common multiple
order of operations
Answer:

Section 1.1 Powers and Exponents

Exploration 1

Writing Expressions Using Exponents

Work with a partner. Copy and complete the table.

Repeated FactorsUsing an ExponentValue
a. 10 × 10
b. 4 × 4
c. 6 × 6
d. 10 × 10 × 10
e. 100 × 100 × 100
f. 3 × 3 × 3 × 3
g. 4 × 4 × 4 × 4 × 4
h. 2 × 2 × 2 × 2 × 2 × 2

Answer:

Math Practice
Repeat Calculations
What patterns do you notice in the expressions? How does this help you write exponents?
Answer:

i. In your own words, describe what the two numbers in the expression 3<sup>5</sup> mean.
Answer:

Exploration 2

Using a Calculator to Find a Pattern

Work with a partner. Copy the diagram. Use a calculator to find each value. Write one digit of the value in each box. Describe the pattern in the digits of the values.
Big Ideas Math Answers 6th Grade 6 Advanced Chapter 1 Numerical Expressions and Factors 5
Answer:

Lesson 1.1 Powers and Exponents

Try It Write the product as a power.

Question 1.
2 × 2 × 2
Answer:

Question 2.
6 • 6 • 6 • 6 • 6 • 6
Answer:

Question 3.
15 × 15 × 15 × 15
Answer:

Question 4.
20 • 20 • 20 • 20 • 20 • 20 • 20
Answer:

Try It Find the value of the power.

Question 5.
63
Answer:

Question 6.
92
Answer:

Question 7.
34
Answer:

Question 8.
182
Answer:

Try It Determine whether the number is a perfect square.

Question 9.
25
Answer:

Question 10.
2
Answer:

Question 11.
99
Answer:

Question 12.
36
Answer:

FINDING VALUES OF POWERS
Find the value of the power.
Question 13.
82
Answer:

Question 14.
35
Answer:

Question 15.
113
Answer:

Question 16.
VOCABULARY
How are exponents and powers different
Answer:

Question 17.
VOCABULARY
Is 10 a perfect square? Is 100 a perfect square? Explain.
Answer:

Question 18.
WHICH ONE DOESN’T BELONG?
Which one does not belong with the other three? Explain your reasoning.
24 = 2 × 2 × 2 × 2
32 = 3 × 3
3 + 3 + 3 + 3 = 3 × 4
5 • 5 • 5 = 53
Answer:

Self-Assessment for Problem Solving

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

Question 19.
A square solar panel has an area of 16 square feet. Write the area as a power. Then find the side lengths of the Panel.
Answer:

Question 20.
The four-square court shown is a square made up of four identical smaller squares. What is the area of the court?
Big Ideas Math Answers 6th Grade 6 Advanced Chapter 1 Numerical Expressions and Factors 6
Answer:

Question 21.
DIG DEEPER!
Each face of a number cube is a square with a side length of 16 millimeters. What is the total area of all of the faces of the number cube?
Answer:

Powers and Exponents Practice 1.1

Review & Refresh

Multiply

Question 1.
150 × 2
Answer:

Question 2.
175 × 8
Answer:

Question 3.
123 × 3
Answer:

Question 4.
151 × 9
Answer:

Write the sentence as a numerical expression.

Question 5.
Add 5 and 8, then multiply by 4.
Answer:

Question 6.
Subtract 7 from 11, then divide by 2.
Answer:

Round the number to the indicated place value.

Question 7.
4.03785 to the tenths
Answer:

Question 8.
12.89503 to the hundredths
Answer:

Complete the sentence.

Question 9.
\(\frac{1}{10}\) of 30 is Big Ideas Math Answers 6th Grade 6 Advanced Chapter 1 Numerical Expressions and Factors 7.
Answer:

Question 10.
\(\frac{4}{5}\) of 25 is Big Ideas Math Answers 6th Grade 6 Advanced Chapter 1 Numerical Expressions and Factors 7.
Answer:

Concepts, Skills, & Problem Solving

WRITING EXPRESSIONS USING EXPONENTS
Copy and complete the table. (See Exploration 1 Page.No 3)

Repeated FactorsUsing an ExponentValue
11. 8 × 8
12. 4 × 4 × 4
13. 9 × 9 × 9 × 9
14. 12 × 12 × 12 × 12 × 12

Answer:

WRITING EXPRESSIONS AS POWERS
Write the product as a power.

Question 15.
9 × 9
Answer:

Question 16.
13 × 13
Answer:

Question 17.
15 × 15 × 15
Answer:

Question 18.
2 • 2 • 2 • 2 • 2
Answer:

Question 19.
14 × 14 × 14
Answer:

Question 20.
8 • 8 • 8 • 8
Answer:

Question 21.
11 × 11 × 11 × 11 × 11
Answer:

Question 22.
7 • 7 • 7 • 7 • 7 • 7
Answer:

Question 23.
16 • 16 • 16 • 16
Answer:

Question 24.
43 × 43 × 43 × 43 × 43
Answer:

Question 25.
167 • 167 • 167
Answer:

Question 26.
245 • 245 • 245 • 245
Answer:

FINDING VALUES OF POWERS
Find the value of the power.

Question 27.
52
Answer:

Question 28.
43
Answer:

Question 29.
62
Answer:

Question 30.
17
Answer:

Question 31.
03
Answer:

Question 32.
84
Answer:

Question 33.
24
Answer:

Question 34.
122
Answer:

Question 35.
73
Answer:

Question 36.
54
Answer:

Question 37.
25
Answer:

Question 38.
142
Answer:

USING TOOLS
Use a calculator to find the value of the power.

Question 39.
76
Answer:

Question 40.
48
Answer:

Question 41.
124
Answer:

Question 42.
175
Answer:

Question 43.
YOU BE THE TEACLIER
Your friend finds the value of 8. Is your friend correct? Explain your reasoning.
Big Ideas Math Answers 6th Grade 6 Advanced Chapter 1 Numerical Expressions and Factors 8
Answer:

IDENTIFYING PERFECT SQUARES
Determine whether the number is a perfect square.

Question 44.
8
Answer:

Question 45.
4
Answer:

Question 46.
81
Answer:

Question 47.
44
Answer:

Question 48.
49
Answer:

Question 49.
125
Answer:

Question 50.
150
Answer:

Question 51.
144
Answer:

Question 52.
MODELING REAL LIFE
On each square centimeter of a person’s skin, there are about 392 bacteria. How many bacteria does this expression represent?
Answer:

Question 53.
MP REPEATED REASONING
The smallest figurine in a gift shop is 2 inches tall. The height of each figurine is twice the height of the previous figurine. What is the height of the tallest figurine?
Answer:

Question 54.
MODELING REAL LIFE
A square painting measures 2 meters on each side. What is the area of the painting in square centimeters?
Big Ideas Math Answers 6th Grade 6 Advanced Chapter 1 Numerical Expressions and Factors 9
Answer:

Question 55.
MP NUMBER SENSE
Write three powers that have values greater than 120 and less than 130.
Answer:

Question 56.
DIG DEEPER!
A landscaper has 125 tiles to build a square patio. the patio must have an area of at least 80 square feet.
Big Ideas Math Answers 6th Grade 6 Advanced Chapter 1 Numerical Expressions and Factors 10
a. What arc the Possible arrangements for the patio?
b. How many tiles are not used in each arrangement?
Answer:

Question 57.
MP PATTERNS
Copy and complete the table. Describe what happens to the value of the power as the exponent decreases.
Use this pattern to find the value 0f 40.
Big Ideas Math Answers 6th Grade 6 Advanced Chapter 1 Numerical Expressions and Factors 11
Answer:

Question 58.
MP REPEATED REASONING
How many blocks do you need to add to Square 6 to get Square 7? to Square 9 to get Square 10? to Square 19 to get Square 20? Explain.
Big Ideas Math Answers 6th Grade 6 Advanced Chapter 1 Numerical Expressions and Factors 12
Answer:

Section 1.2 Order of Operations

Exploration 1

Comparing Different Orders

Work with a partner. Find the value of each expression by using different orders of operations. Are your answers the same?

a. Add, then multiply.                  Multiply, then add.
3 + 2 × 2                                      3 + 2 × 2

b. Subtract then multiply.             Multiply, then subtract.
18 – 3 • 3                                       18 – 3 • 3

c. Multiply, then subtract              Subtract, then multiply.
8 × 8 – 2                                        8 × 8 – 2

d. Multiply, then add.                    Add, then multiply.
6 • 6 + 2                                         6 • 6 + 2

Exploration 2

Determining Order of Operations

Work with a partner.
Big Ideas Math Answers Grade 6 Advanced Chapter 1 Numerical Expressions and Factors 13
a. Scientific calculators use a standard order of operations when evaluating expressions. Why is a standard order of operations needed?
Answer:

b. Lise a scientific calculator to evaluate each expression in Exploration 1. Enter each expression exactly as written. For each expression, which order of operations is correct?
Answer:

c. What order of operations should be used to evaluate 3 + 22, 18 – 32, 82 – 2, and 62 + 2?
Answer:

d. Do 18 ÷ 3 • 3 and 18 ÷ 32 have the same value? Justify your answer.
Answer:

e. How does evaluating powers fit into the order of operations?
Answer:

Math Practice
Use Technology to Explore
How does a scientific calculator help you explore order of operations?
Answer:

Lesson 1.2 Order of Operations

Try It Evaluate the expression.

Question 1.
7 • 5 + 3
Answer:

Question 2.
(28 – 20) ÷ 4
Answer:

Question 3.
[6 + (15 – 10)] × 5
Answer:

Try It Evaluate the expression.

Question 4.
6 + 24 – 1
Answer:

Question 5.
4 • 32 + 18 – 9
Answer:

Question 6.
16 + (52 – 7) ÷ 3
Answer:

Try It Evaluate the expression.

Question 7.
50 + 6(12 ÷ 4) – 82
Answer:

Question 8.
52 – \(\frac{1}{5}\)(10 – 5)
Answer:

Question 9.
\(\frac{8(2+5)}{7}\)
Answer:

USING ORDER OF OPERATIONS
Evaluate the expression.

Question 10.
7 + 2 • 4
Answer:

Question 11.
8 ÷ 4 × 2
Answer:

Question 12.
3(5 + 1) ÷ 32
Answer:

Question 13.
WRITING
Why does 12 – 8 ÷ 2 = 8, hut(12 – 8) ÷ 2 = 2?
Answer:

Question 14.
MP REASONING
Describe the steps ¡n evaluating the expression
8 ÷ (6 – 1) + 32.
Answer:

Question 15.
WHICH ONE DOESN’T BELONG?
Which expression does not belong with the other three? Explain your reasoning.
52 – 8 × 2
52 – (8 × 2)
52 – 2 × 8
(52 – 8) × 2
Answer:

Self-Assessment for problem solving

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

Question 16.
A square plot of land has side lengths of 10 meters. An archaeologist divides the land into 64 equal parts. What is the area of each part?
Big Ideas Math Answers Grade 6 Advanced Chapter 1 Numerical Expressions and Factors 14
Answer:

Question 17.
A glass block window is made of two different-sized glass squares. The window has side lengths of 40 inches. The large glass squares have side lengths of 10 inches. Find the total area of the small glass squares.
Big Ideas Math Answers Grade 6 Advanced Chapter 1 Numerical Expressions and Factors 14
Answer:

Question 18.
DIG DEEPER!
A square vegetable garden has side lengths of 12 feel You plant flowers in the center portion as shown. You divide the remaining space into 1 equal sections and plant tomatoes, onions, zucchini, and peppers. What is the area of the onion section?
Big Ideas Math Answers Grade 6 Advanced Chapter 1 Numerical Expressions and Factors 15
Answer:

Order of Operations Practice 1.2

Review & Refresh

Write the product as a power.

Question 1.
11 × 11 × 11 × 11
Answer:

Question 2.
13 × 13 × 13 × 13 × 13
Answer:

Find the missing dimension of the rectangular prism.

Question 3.
Big Ideas Math Answers Grade 6 Advanced Chapter 1 Numerical Expressions and Factors 16
Answer:

Question 4.
Big Ideas Math Answers Grade 6 Advanced Chapter 1 Numerical Expressions and Factors 17
Answer:

Tell whether the number is prime or composite.

Question 5.
9
Answer:

Question 6.
11
Answer:

Question 7.
23
Answer:

Concepts, skills, & Problem Solving

COMPARING DIFFERENT ORDERS
Find the value of the expression by using different orders of operations. Are your answers the saine? (See Exploration 1 Page.No 9)

Question 8.
Add, then multiply. Multiply, then add.
4 + 6 × 6 4 + 6 × 6
Answer:

Question 9.
Subtract, then multiply. MuItiply then subtract.
5 × 5 – 3 5 × 5 – 3
Answer:

USING ORDER OF OPERATIONS
Evaluate the expression.

Question 10.
5 + 8 ÷ 6
Answer:

Question 11.
(11 – 3) ÷ 2 + 1
Answer:

Question 12.
45 ÷ 9 × 12
Answer:

Question 13.
62 – 3 • 4
Answer:

Question 14.
42 ÷ (15 – 23)
Answer:

Question 15.
42 • 2 + 8 • 7
Answer:

Question 16.
(52 – 2) × 15 + 4
Answer:

Question 17.
4 + 2 × 32 – 9
Answer:

Question 18.
8 ÷ 2 × 3 + 42 ÷ 4
Answer:

Question 19.
32 + 12 ÷ (6 – 3) × 8
Answer:

Question 20.
(10 + 4) ÷ (26 – 19)
Answer:

Question 21.
(52 – 4) • 2 – 18
Answer:

Question 22.
2 × [(16 – 8) × 2]
Answer:

Question 23.
12 + 8 × 32 – 24
Answer:

Question 24.
62 ÷ [(2 + 4) × 23]
Answer:

YOU BE THE TEACHER
Your friend evaluates the expression. Is your friend correct? Explain your reasoning.
Question 25.
Big Ideas Math Answers Grade 6 Advanced Chapter 1 Numerical Expressions and Factors 18
Answer:

Question 26.
Big Ideas Math Answers Grade 6 Advanced Chapter 1 Numerical Expressions and Factors 19
Answer:

Question 27.
MP PROBLEM SOLVING
You need to read 20 poems in 5 days for an English project. Each poem is 2 pages long. Evaluate the expression 20 × 2 ÷ 5 to find how many pages you need to read each day.
Answer:

USING ORDER OF OPERATIONS
Evaluate the expression.

Question 28.
12 – 2(7 – 4)
Answer:

Question 29.
4(3 + 5) – 3(6 – 2)
Answer:

Question 30.
6 + \(\frac{1}{4}\) (12 – 8)
Answer:

Question 31.
92 – 8(6 + 2)
Answer:

Question 32.
4(3 – 1)3 + 7(6) – 52
Answer:

Question 33.
\(8\left[\left(1 \frac{1}{6}+\frac{5}{6}\right) \div 4\right]\)
Answer:

Question 34.
72 – 2\(\left(\frac{11}{8}-\frac{3}{8}\right)\)
Answer:

Question 35.
8(7.3 + 3.7 – 8) ÷ 2
Answer:

Question 36.
24(5.2 – 3.2) ÷ 4
Answer:

Question 37.
\(\frac{6^{2}(3+5)}{4}\)
Answer:

Question 38.
\(\frac{12^{2}-4(6)+1}{11^{2}}\)
Answer:

Question 39.
\(\frac{26 \div 2+5}{3^{2}-3}\)
Answer:

Question 40.
MP PROBLEM SOLVING
Before a show, there arc 8 people in a theater. Five groups of 4 people enter; and then three groups of 2 PeoPle leave. Evaluate the expression 8 5(4) – 3(2) to find how many people arc in the theater.
Answer:

Question 41.
MODELING REAL LIFE
The front door of a house is painted white and blue. Each window is a square with a side length of 7 inches. What is the area of the door that is painted blue?
Big Ideas Math Answers Grade 6 Advanced Chapter 1 Numerical Expressions and Factors 21
Answer:

Question 42.
MP PROBLEM SOLVING
You buy 6 notebooks, 10 folders, 1 pack of pencils, and 1 lunch box for school. After using a $10 gift card, how much do you owe? Explain how you solved the problem.
Big Ideas Math Answers Grade 6 Advanced Chapter 1 Numerical Expressions and Factors 21
Answer:

Question 43.
OPEN-ENDED
Use all four operations and at least one exponent to write an expression that has a value of 100.
Answer:

Question 44.
MP REPEATED REASONING
A Petri dish contains 35 cells. Every day, each cell in the Petri dish divides into 2 cells in a process called mitosis. How many cells are there after 11 days? Justify your answer.
Answer:

Question 45.
MP REASONING
Two groups collect litter along the
side of a road. It takes each group 5 minutes to clean up a 200-yard section. 1 low long does it take both groups working together to clean up 2 miles? Explain how you solved the problem.
Big Ideas Math Answers Grade 6 Advanced Chapter 1 Numerical Expressions and Factors 22
Answer:

Question 46.
MP NUMBER SENSE
Copy each statement Insert +, -, ×, or ÷ symbols to make each statement true.
Big Ideas Math Answers Grade 6 Advanced Chapter 1 Numerical Expressions and Factors 23
Answer:

Section 1.3 Prime Factorization

Exploration 1

Rewriting Numbers as Products of Factors

Work with a partner. Two students use factor trees to write 108 as a product of factors, as shown below.
Big Ideas Math Solutions Grade 6 Advanced Chapter 1 Numerical Expressions and Factors 24

a. Without using 1 as a factor, can you write 108 as a product with more factors than each student used? Justify your answer.
Answer:

b. Use factor trees to Write 80, 162, and 300 as products of as many factors as possible. Do not use 1 as a factor.
Big Ideas Math Solutions Grade 6 Advanced Chapter 1 Numerical Expressions and Factors 25
Answer:

c. Compare your results in parts (a) and (b) with other groups. For each number, identify the product with the greatest number of factors. What do these factors have in common?
Answer:

Math Practice
Interpret Results
How do you know your answer makes sense?
Answer:

Lesson 1.3 Prime Factorization

Try It List the factor pairs of the number.

Question 1.
18
Answer:

Question 2.
24
Answer:

Question 3.
51
Answer:

Question 4.
WHAT IF?
The woodwinds section of the marching hand has 38 members. Which has more possible arrangements, the brass section or the woodwinds section? Explain.
Answer:

Try It Write the prime factorization of the number.

Question 5.
20
Answer:

Question 6.
88
Answer:

Question 7.
90
Answer:

Question 8.
462
Answer:

Self-Assessment for Concepts & Skills

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

WRITING A PRIME FACTORIZATION
Write the prime factorization of the number.
Question 9.
14
Answer:

Question 10.
86
Answer:

Question 11.
10
Answer:

Question 12.
516
Answer:

Question 13.
WRITING
Explain the difference between prime numbers and composite numbers.
Answer:

Question 14.
MP STRUCTURE
Your friend lists the following factor pairs and concludes that there arc 6 factor pairs of 12. Explain why your friend incorrect.
1,12   2,6     3,4
12,1   6,2     4,3
Answer:

Question 15.
WHICH ONE DOESN’T BELONG?
Which factor pair does not belong with the other three? Explain your reasoning.
2, 28   4, 14    6, 9   7, 8
Answer:

Self-Assessment for Problem Solving

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

Question 16.
A group of 20 friends plays a card game. ‘The game can be played with 2 or more teams of equal size. Each team must have at least 2 members. List the possible numbers and sizes of teams.
Big Ideas Math Solutions Grade 6 Advanced Chapter 1 Numerical Expressions and Factors 26
Answer:

Question 17.
You arrange 150 chairs in rows for a school play. You want each row to have the same number of chairs. how many possible arrangements arc there? Are all of the possible arrangements appropriate for the play? Explain.
Answer:

Question 18.
What is the least perfect square that is a factor of 4536? What is the greatest perfect square that is a factor of 4536?
Answer:

Question 19.
DIG DEEPER!
The prime factorization of a number is 24 × 34 × 54 × 72 Is the number a Perfect square? Explain your reasoning.
Answer:

Prime Factorization Practice 1.3

Review & Refresh

Evaluate the expression.

Question 1.
2 + 42 (5 – 3)
Answer:

Question 2.
23 + 4 × 32
Answer:

Question 3.
9 × 5 – 24\(\left(\frac{5}{2}-\frac{1}{2}\right)\)
Answer:

Plot the points in a coordinate plane. Draw a line segment connecting the points.

Question 4.
(1, 1) and (4, 3)
Answer:

Question 5.
(2, 3) and (5, 9)
Answer:

Question 6.
(2, 5) and (4, 8)
Answer:

Use the Distributive Property to find the quotient. Justify your answer.

Question 7.
408 ÷ 4
Answer:

Question 8.
628 ÷ 2
Answer:

Question 9.
969 ÷ 3
Answer:

Classify the triangle in as many ways as possible.

Question 10.
Big Ideas Math Solutions Grade 6 Advanced Chapter 1 Numerical Expressions and Factors 27
Answer:

Question 11.
Big Ideas Math Solutions Grade 6 Advanced Chapter 1 Numerical Expressions and Factors 28
Answer:

Question 12.
Big Ideas Math Solutions Grade 6 Advanced Chapter 1 Numerical Expressions and Factors 29
Answer:

Concepts, Skills, & Problem Solving

REWRITING A NUMBER
Write the number as a product of as many factors as possible. (See Exploration 1 Page.No 15)
Question 13.
60
Answer:

Question 14.
63
Answer:

Question 15.
120
Answer:

Question 16.
150
Answer:

FINDING FACTOR PAIRS
List the factor pairs of the number.
Question 17.
15
Answer:

Question 18.
22
Answer:

Question 19.
31
Answer:

Question 20.
39
Answer:

Question 21.
45
Answer:

Question 22.
54
Answer:

Question 23.
59
Answer:

Question 24.
61
Answer:

Question 25.
100
Answer:

Question 26.
58
Answer:

Question 27.
25
Answer:

Question 28.
76
Answer:

Question 29.
52
Answer:

Question 30.
88
Answer:

Question 31.
71
Answer:

Question 32.
91
Answer:

WRITING A PRIME FACTORIZATION
Write the prime factorization of the number.
Question 33.
16
Answer:

Question 34.
25
Answer:

Question 35.
30
Answer:

Question 36.
26
Answer:

Question 37.
84
Answer:

Question 38.
54
Answer:

Question 39.
65
Answer:

Question 40.
77
Answer:

Question 41.
46
Answer:

Question 42.
39
Answer:

Question 43.
99
Answer:

Question 44.
Answer:

Question 45.
315
Answer:

Question 46.
190
Answer:

Question 47.
110
Answer:

Question 48.
640
Answer:

USING A PRIME FACTORIZATION
Find the number represented by the prime factorization.
Question 49.
22 • 32 • 5
Answer:

Question 50.
32 • 52 • 7
Answer:

Question 51.
23 • 112 • 13
Answer:

Question 52.
YOU BE THE TEACHER
Your friend finds the prime factorization of 72. Is your friend correct? Explain your reasoning.
Big Ideas Math Solutions Grade 6 Advanced Chapter 1 Numerical Expressions and Factors 30
Answer:

USING A PRIME FACTORIZATION
Find the greatest perfect square that is a factor of the number.
Question 53.
250
Answer:

Question 54.
275
Answer:

Question 55.
392
Answer:

Question 56.
338
Answer:

Question 57.
244
Answer:

Question 58.
650
Answer:

Question 59.
756
Answer:

Question 60.
1290
Answer:

Question 61.
2205
Answer:

Question 62.
1890
Answer:

Question 63.
495
Answer:

Question 64.
4725
Answer:

Question 65.
VOCABULARY
A botanist separates plants into equal groups of 5 for an experiment. Is the total number of plants in the experiment prime or composite? Explain.
Big Ideas Math Solutions Grade 6 Advanced Chapter 1 Numerical Expressions and Factors 31
Answer:

Question 66.
MP REASONING
A teacher divides 36 students into equal groups for a scavenger hunt. Each group should have at least 1 students hut no more than 8 students. What are the possible group sizes?
Answer:

Question 67.
CRITICAL THINKING
Is 2 the only even prime number? Explain.
Answer:

Question 68.
MP LOGIC
One topic at a back saic has 75 cookies. Another table has 60 cupcakes. Which table allows for more rectangular arrangements? Explain.
Answer:

Question 69.
PERFECT NUMBERS
A perfect number is a number that equals the sum of its factors, not including itself. Ëor example, the factors of 28 arc 1, 2, 1, 7, 14, and 28. Because 1 + 2 + 4 + 7 + 14 = 28, 28 is a perfect number. What are the perfect numbers between 1 and 27?
Answer:

Question 70.
MP REPEATED REASONING
Choose any two perfect squares and find their product. ‘Then multiply your answer by another perfect square. Continue this process. Are any of the products perfect squares? What can you conclude?
Answer:

Question 71.
MP PROBLEM SOLVING
The stage manager of a school play creates a rectangular stage that has whole number dimensions and an area of 42 square yards. String lights will outline the stage. What is the least number of yards of string lights needed to enclose the stage?
Answer:

Question 72.
DIG DEEPER!
Consider the rectangular prism shown. Using only whole number dimensions, how many different prisms are possible? Explain.
Big Ideas Math Solutions Grade 6 Advanced Chapter 1 Numerical Expressions and Factors 32
Answer:

Section 1.4 Greatest Common Factor

Exploration 1

Identifying Common Factors

Work with a partner. in parts (a) – (d), create a Venn diagram that represents the factors of each number and identify any common factors.
a. 36 and 18
b. 16 and 56
c. 30 and 75
d. 54 and 90
e. Look at the Venn diagrams in parts (a) – (d). Explain how to identify the greatest common factor of each pair of numbers. Then circle it in each diagram.
Answer:

Exploration 2

Using Prime Factors

Work with a partner.

a. Each Venn diagram represents the prime factorizations of two numbers. Identify each pair of numbers. Explain your reasoning.
Big Ideas Math Answer Key Grade 6 Advanced Chapter 1 Numerical Expressions and Factors 33
Answer:

b. Create a Venn diagram that represents the prime factorizations of 36 and 18.
Answer:

c. Repeat part (b) for the remaining number pairs in Exploration 1.
Answer:

d. MP STRUCTURE
Make a conjecture about the relationship between the greatest common factors you found in Exploration 1 and the numbers in the overlaps of the Venn diagrams you just created.
Answer:

Math Practice
Interpret a Solution
What does the diagram representing the prime factorizations mean?
Answer:

Lesson 1.4 Greatest Common Factor

Try It Find the GCF of the numbers using lists of factors.

Question 1.
8, 36
Answer:

Question 2.
18, 72
Answer:

Question 3.
14, 28, 49
Answer:

Try It Find the GCF of the numbers using prime factorizations.

Question 4.
20, 15
Answer:

Question 5.
32, 90
Answer:

Question 6.
45, 75, 120
Answer:

Try It

Question 7.
Write a pair of numbers whose greatest common factor is 10.
Answer:

Self-Assessment for concept & Skills

Solve each exercise. ‘Then rate your understanding of the success criteria in your journal.
FINDING THE GCF
Find the GCF of the numbers.
Question 8.
16, 40
Answer:

Question 9.
35, 63
Answer:

Question 10.
18, 72, 141
Answer:

Question 11.
MULTIPLE CHOICE
Which number is not a factor of 10? Explain.
A. 1
B. 2
C. 4
D. 5
Answer:

Question 12.
DIFFERENT WORDS, SAME QUESTION
Which is different? Find “both” answers.
What is the greatest common factor of 24 and 32?
What is the greatest common divisor of 24 and 32?
What is the greatest common prime factor of 24 and 32?
What is the product of the common prime factors of 24 and 32?
Answer:

Self-Assessment for Problem Solving

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

Question 13.
You usc 30 sandwiches and 42 granola bars to make identical picnic baskets. You make thc greatest number of picnic baskets with no food left over. How many sandwiches and how many granola bars are in each basket?
Big Ideas Math Answer Key Grade 6 Advanced Chapter 1 Numerical Expressions and Factors 34
Answer:

Question 14.
You fill bags with cookies to give to your friends. You bake 45 chocolate chip cookies, 30 peanut butter cookies, and
15 oatmeal cookies. You want identical groups of cookies in each hag with no cookies left over. What is the greatest number of bags you can make?
Answer:

Greatest Common Factor Practice 1.4

Review & Refresh

List the factor pairs of the number.
Question 1.
20
Answer:

Question 2.
16
Answer:

Question 3.
56
Answer:

Question 4.
87
Answer:

Tell whether the statement is always, sometimes, or sever true.

Question 5.
A rectangle is a rhombus.
Answer:

Question 6.
A rhombus is a square.
Answer:

Question 7.
A square is a rectangle.
Answer:

Question 8.
A trapezoid is a parallelogram.
Answer:

Concepts, Skills, & Problem Solving

USING A VENN DIAGRAM
Use a Venn diagram to find the greatest common factor of the numbers. (See Exploration 1 Page.No 21)

Question 9.
12, 30
Answer:

Question 10.
32, 54
Answer:

Question 11.
24, 108
Answer:

FINDING THE GCF
Find the GCF of the numbers using lists of factors
Question 12.
6, 15
Answer:

Question 13.
14, 84
Answer:

Question 14.
45, 76
Answer:

Question 15.
39, 65
Answer:

Question 16.
51, 85
Answer:

Question 17.
40, 63
Answer:

Question 18.
12, 48
Answer:

Question 19.
24, 52
Answer:

Question 20.
30, 58
Answer:

FINDING THE GCF
Find the GCF of the numbers using lists of factors prime factorizations.

Question 21.
45, 60
Answer:

Question 22.
27, 63
Answer:

Question 23.
36, 81
Answer:

Question 24.
72, 84
Answer:

Question 25.
61, 73
Answer:

Question 26.
38, 95
Answer:

Question 27.
60, 75
Answer:

Question 28.
42, 60
Answer:

Question 29.
42, 63
Answer:

Question 30.
24, 96
Answer:

Question 31.
189, 200
Answer:

Question 32.
90, 108
Answer:

OPEN-ENDED
write a pair of numbers with the indicated GCF.

Question 33.
5
Answer:

Question 34.
12
Answer:

Question 35.
37
Answer:

Question 36.
MODELING REAL LIFE
A teacher is making identical activity packets using 92 crayons and 23 sheets of paper. What is the greatest number of packets the teacher can make with no items left over?
Answer:

Question 37.
MODELING REAL LIFE
You arc making balloon arrangements for a birthday party. ‘there are 16 white balloons and 24 red balloons Each arrangement must he identical. What is the greatest number of arrangements you can make using every balloon?
Big Ideas Math Answer Key Grade 6 Advanced Chapter 1 Numerical Expressions and Factors 35
Answer:

YOU BE THE TEACHER
Your friend finds the CCI of the two numbers. Is your friend correct? Explain your reasoning.

Question 38.
Big Ideas Math Answer Key Grade 6 Advanced Chapter 1 Numerical Expressions and Factors 36
Answer:

Question 39.
Big Ideas Math Answer Key Grade 6 Advanced Chapter 1 Numerical Expressions and Factors 37
Answer:

FINDING THE GCF
Find the GCF of the numbers.
Question 40.
35, 56, 63
Answer:

Question 41.
30, 60, 78
Answer:

Question 42.
42, 70, 84
Answer:

Question 43.
40, 55, 72
Answer:

Question 44.
18, 54, 90
Answer:

Question 45.
16, 18, 88
Answer:

Question 46.
52, 78, 101
Answer:

Question 47.
96, 120, 156
Answer:

Question 48.
280, 300, 380
Answer:

Question 49.
OPEN-ENDED
Write three numbers that have a GCF of 16. What method did you use to find your answer?
Answer:

CRITICAL THINKING
Tell whether the statement ¡s always, someUmes, or never true. Explain your reasoning.
Question 50.
The GCF of two even numbers is 2.
Answer:

Question 51.
The GCF of two prime numbers is 1.
Answer:

Question 52.
When one number is a multiple of another, the GCF of the numbers is the greater of the numbers.
Answer:

Question 53.
PROBLEM SOLVING
A science museum makes gift bags for students using 168 magnets, 48 robot figurines, and 24 packs of freeze-dried ice cream. What is the greatest number of gift bags that can be made using all of the items? How many of each item are in each gift bag?
Big Ideas Math Answer Key Grade 6 Advanced Chapter 1 Numerical Expressions and Factors 38
Answer:

Question 54.
VENN DIAGRAM
Consider the numbers 252, 270, and 300.
a. Create a Venn diagram using the prime factors of the numbers.
Answer:

b. Use the Venn diagram to find the GCF of 252, 270, and 300.
Answer:

c. What is the GCF of 252 and 270? 252 and 300? 270 and 300? Explain how you found your answers.
Answer:

Question 55.
MP REASONING
You are making fruit baskets using 54 apples, 36 oranges, and 73 bananas.
a. Explain why you cannot make identical fruit baskets without leftover fruit.
Answer:

b. What is the greatest number of identical fruit baskets you can make with the least amount of fruit left over? Explain how you found your answer
Answer:

Question 56.
DIG DEEPER!
Two rectangular, adjacent rooms share a wall. One-foot-by-one-foot tiles cover the floor of each room. Describe how the greatest possible length of the adjoining wall is related to the total number of tiles in each room. Draw a diagram that represents one possibility.
Big Ideas Math Answer Key Grade 6 Advanced Chapter 1 Numerical Expressions and Factors 39
Answer:

Section 1.5 Least Common Multiple

Exploration 1

Identifying Common Multiples

Work with a partner. In paris (a) – (d), create a Venn diagram that represents the first several multiples of each number and identify any common multiples.
a. 8 and 12
Answer:

b. 4 and 14
Answer:

c. 10 and 15
Answer:

d. 20 and 35
Answer:

e. Look at the Venn diagrams in parts (a) – (d). Explain how to identify the least common multiple of each pair of numbers. Then circle it in each diagram.
Answer:

Exploration 2

Using Prime Factors

Work with a partner.

a. Create a Venn diagram that represents the prime factorizations of 8 and 12.

b. Repeat part (a) for the remaining number pairs in Exploration 1.

c. MP STRUCTURE
Make a conjecture about the relationship between the least common multiples you found in Exploration 1 and the numbers in the Venn diagrams you just created.

d. The Venn diagram shows the prime factors of two numbers.
Big Ideas Math Answers 6th Grade 6 Advanced Chapter 1 Numerical Expressions and Factors 40
Use the diagram to complete the following tasks.

  • Identify the two numbers.
  • Find the greatest common factor.
  • Find the least common multiple.

Answer:

Math Practice.
Analyze Conjectures
How can you test your conjecture in part (c)?
Answer:

Lesson 1.5 Least Common Multiple

Try It Find the LCM of the numbers using lists of multiples.
Question 1.
3, 8
Answer:

Question 2.
9, 12
Answer:

Question 3.
6, 10
Answer:

Try It Find the LCM of the numbers using prime factorizations.

Question 4.
14, 18
Answer:

Question 5.
28, 36
Answer:

Question 6.
24, 90
Answer:

Try It Find the LCM of the numbers.
Question 7.
2, 5, 8
Answer:

Question 8.
6, 10, 12
Answer:

Question 9.
Write three numbers that have a least common multiple of 100.
Answer:

Self-Assessment for Concepts & Skills

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

FINDING THE LCM
Find the LCM of the numbers.

Question 10.
6, 9
Answer:

Question 11.
30, 10
Answer:

Question 12.
5, 11
Answer:

Question 13.
MP REASONING
Write two numbers such that 18 and 30 are multiples of the numbers. justify your answer.
Answer:

Question 14.
MP REASONING
You need to find the LCM of 13 and 14. Would you rather list their multiples or use their prime factorizations? Explain.
Answer:

Question 15.
MP CHOOSE TOOLS
A student writes the prime factorizations of 8 and 12 in a table as shown. She claims she can use the table to find the greatest common factor and the least common multiple 8 and 12. How is this possible?
Big Ideas Math Answers 6th Grade 6 Advanced Chapter 1 Numerical Expressions and Factors 42
Big Ideas Math Answers 6th Grade 6 Advanced Chapter 1 Numerical Expressions and Factors 41
Answer:

Question 16.
CRITICAL THINKING
How can you use least common multiples to add or subtract fractions with different denominators?
Answer:

Self-Assessment for Problem Solving

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

Question 17.
A geyser erupts every fourth day. Another geyser erupts every sixth day. Today both geysers erupted. In how many days will b0th geysers erupt on the same day again?
Big Ideas Math Answers 6th Grade 6 Advanced Chapter 1 Numerical Expressions and Factors 43
Answer:

Question 18.
A water park has two large buckets that slowly fill with water. One bucket dumps water every 12 minutes. The other bucket dumps water every 10 minutes. Five minutes ago, b0th buckets dumped water. When will both buckets dump water at the same time again?
Big Ideas Math Answers 6th Grade 6 Advanced Chapter 1 Numerical Expressions and Factors 43
Answer:

Question 19.
DIG DEEPER!
You purchase disposable plates, cups, and forks for a cookout. Plates are sold in packages of 24, cups in packages of 32, and forks in packages of 18. What are the least numbers of packages you should buy in order to have the same number of plates, cups, and forks?
Answer:

Least Common Multiple Practice 1.5

Review & Refresh

Find the GCF of the numbers.

Question 1.
18, 42
Answer:

Question 2.
72, 96
Answer:

Question 3.
38, 76, 111
Answer:

Divide.
Question 4.
900 ÷ 6
Answer:

Question 5.
1941 ÷ 9
Answer:

Question 6.
672 ÷ 12
Answer:

Write an ordered pair that corresponds to the point.
Big Ideas Math Answers 6th Grade 6 Advanced Chapter 1 Numerical Expressions and Factors 44

Question 7.
Point A
Answer:

Question 8.
Point B
Answer:

Question 9.
Point C
Answer:

Question 10.
Point D
Answer:

Concepts, Skills, & Problem Solving

USING A VENN DIAGRAM
Use a Venn diagram to find the least common multiple of the numbers. (See Exploration 1 Page.No 27)
Question 11.
3, 7
Answer:

Question 12.
6, 8
Answer:

Question 13.
1, 5
Answer:

FINDING THE LCM
Find the LCM of the numbers using lists of multiplies.
Question 14.
1, 5
Answer:

Question 15.
2, 6
Answer:

Question 16.
2, 3
Answer:

Question 17.
2, 9
Answer:

Question 18.
3, 4
Answer:

Question 19.
8, 9
Answer:

20.
5, 8
Answer:

Question 21.
11, 12
Answer:

Question 22.
12, 18
Answer:

FINDING THE LCM
Find the LCM of the numbers using prime factorizations.
Question 23.
7, 12
Answer:

Question 24.
5, 9
Answer:

Question 25.
4, 11
Answer:

Question 26.
9, 10
Answer:

Question 27.
12, 27
Answer:

Question 28.
18, 45
Answer:

Question 29.
22, 33
Answer:

Question 30.
36, 60
Answer:

Question 31.
35, 50
Answer:

Question 32.
YOU BE THE TEACHER
Your friend finds the LCM of 6 and 9. Is your friend correct? Explain your reasoning.
Big Ideas Math Answers 6th Grade 6 Advanced Chapter 1 Numerical Expressions and Factors 45
Answer:

Question 33.
MODELING REAL LIFE
You have diving lessons every fifth day and swimming lessons every third day. Today you have both lessons. In how many days will you have both lessons on the same day again?
Answer:

Question 34.
MP REASONING
Which model represents an LCM that is different from the other three? Explain your reasoning.
Big Ideas Math Answers 6th Grade 6 Advanced Chapter 1 Numerical Expressions and Factors 46
Answer:

FINDING THE LCM
Find the LCM of the numbers.
Question 35.
2, 3, 7
Answer:

Question 36.
3, 5, 11
Answer:

Question 37.
4, 9, 12
Answer:

Question 38.
6, 8, 15
Answer:

Question 39.
7, 18, 21
Answer:

Question 40.
9, 10, 28
Answer:

Question 41.
MP PROBLEM SOLVING
At Union Station, you notice that three subway lines just arrived at the same time. How long must you wait until all three lines arrive at Union Station at the same time again?

Subway LineArrival Time
AEvery 10 min
BEvery 12 min
CEvery 15 min

Big Ideas Math Answers 6th Grade 6 Advanced Chapter 1 Numerical Expressions and Factors 47
Answer:

Question 42.
DIG DEEPER!
A radio station gives away $15 to every 15th caller, $25 to every 25th caller, and a free concert ticket to every 100th caller. When will the station first give away all three prizes to one caller? When this happens, how much money and how many tickets are given away?
Answer:

Question 43.
MP LOGIC
You and a friend are running on treadmills. You run 0.5 mile every 3 minutes, and your friend runs 2 miles every 14 minutes. You both start and stop running at the same time and run a whole number of miles. What are the least possible numbers of miles you and your friend can run?
Answer:

Question 44.
VENN DIAGRAM
Refer to the Venn diagram.
Big Ideas Math Answers 6th Grade 6 Advanced Chapter 1 Numerical Expressions and Factors 48
a. Copy and complete the Venn diagram.
Answer:

b. What is the LCM of 16, 24, and 40?
Answer:

c. What is the LCM of 16 and 40? 24 and 40? 16 and 21? Explain how you found your answers.
Answer:

CRITICAL THINKING
Tell whether the statement is always, sometimes, or never true. Explain your reasoning.
Question 45.
The LCM of two different prime numbers is their Product.
Answer:

Question 46.
The LCM of a set of numbers is equal to one of the numbers in the set.
Answer:

Question 47.
The GCF of two different numbers is the LCM of the numbers.
Answer:

Numerical Expressions and Factors Connecting Concepts

Using the Problem-Solving Plan

Question 1.
A sports team gives away shirts at the stadium. There arc 60 large shirts, 1.6 times as many small shirts as large shirts, and 1.5 times as many medium shirts as small shirts. The team wants to divide the shirts into identical groups to he distributed throughout the stadium. What is the greatest number of groups that can be formed using every shirt?

Understand the problem: You know the number of large shirts and two relationships among the numbers of small, medium, and large shirts. You are asked to find the greatest number of identical groups that can be formed using every shirt.

Make a plan: Break the problem into parts. First use multiplication to find the number of each size shirt. Then find the GCF of these numbers.

Solve and check: Lise the plan to solve the problem. Then check your solution.
Answer:

Question 2.
An escape artist fills the tank shown with water. Find the number of cubic feet of water needed to fill the tank. Then find the number of cubic yards of water that are needed to fill the tank. justify your answer.
Big Ideas Math Answers 6th Grade 6 Advanced Chapter 1 Numerical Expressions and Factors 49
Answer:

Numerical Expressions and Factors Chapter Review

Powers and Exponents Homework & Practice 1.1

Write the product as a power.

Question 1.
3 × 3 × 3 × 3 × 3 × 3
Answer:

Question 2.
5 × 5 × 5
Answer:

Question 3.
17 • 17 • 17 • 17 • 17
Answer:

Find the value of the power.
Question 4.
33
Answer:

Question 5.
26
Answer:

Question 6.
44
Answer:

Question 7.
Write a power that has a value greater than 23 and less than 33.
Answer:

Question 8.
Without evaluating, determine whether 25 or 22 is greater. Explain.
Answer:

Question 9.
The bases on a softball field arc square. What is the area of each base?
Big Ideas Math Answers Grade 6 Advanced Chapter 1 Numerical Expressions and Factors 50
Big Ideas Math Answers Grade 6 Advanced Chapter 1 Numerical Expressions and Factors 51
Answer:

Order of Operations Homework & Practice 1.2

Evaluate the expression.

Question 10.
3 × 6 – 12 ÷ 6
Answer:

Question 11.
30 ÷(14 – 22) × 5
Answer:

Question 12.
\(\frac{5(2.3+3.7)}{2}\)
Answer:

Question 13.
43 – \(\frac{1}{2}\)(72 + 5)
Answer:

Question 14.
20 × (32 – 4) ÷ 50
Answer:

Question 15.
5 + 3 (42 – 2) ÷ 6
Answer:

Question 16.
Use grouping symbols and at least one exponent to write a numerical expression that has a value of 80.
Answer:

Prime Factorization Homework & Practice 1.3

List the factor Pairs of the number.

Question 17.
28
Answer:

Question 18.
44
Answer:

Question 19.
96
Answer:

Question 20.
There arc 36 graduated cylinders to put away on a shelf after science class. The shelf can fit a maximum of 20 cylinders across and 4 cylinders deep. The teacher wants each row to have the same number of cylinders. List the possible arrangements of the graduated cylinders on the shelf.
Big Ideas Math Answers Grade 6 Advanced Chapter 1 Numerical Expressions and Factors 52
Answer:

Write the prime factorization of the number.

Question 21.
12
Answer:

Question 22.
50
Answer:

Question 23.
66
Answer:

Greatest Common Factor Homework & Practice 1.4

Find the GCF of the numbers using lists of factors.
Question 24.
27, 45
Answer:

Question 25.
30, 48
Answer:

Question 26.
28, 48
Answer:

Find the GCF of the numbers using prime factorizations.
Question 27.
24, 80
Answer:

Question 28.
52, 68
Answer:

Question 29.
32, 56
Answer:

Question 30.
Write a pair of numbers that have a GCF of 20.
Answer:

Question 31.
What is the greatest number of friends you can invite to an arcade using the coupon such that the tokens and slices of pizza are equally split between you and your friends with none left over? How many slices of pizza and tokens will each person receive?
Big Ideas Math Answers Grade 6 Advanced Chapter 1 Numerical Expressions and Factors 53
Answer:

Least Common Multiple Homework & Practice 1.5

Find the LCM of the numbers using lists of multiples.
Question 32.
4, 14
Answer:

Question 33.
6, 20
Answer:

Question 34.
12, 28
Answer:

Find the LCM of the numbers using prime factorbations.
Question 35.
6, 15
Answer:

Question 36.
10, 12
Answer:

Question 37.
18, 27
Answer:

Question 38.
Find the LCM of 8, 12, and 18.
Answer:

Question 39.
Write a pair of numbers that have an LCM of 84.
Answer:

Question 40.
Write three numbers that have an LCM of 45.
Answer:

Question 41.
You water your roses every sixth day and your hydrangeas every fifth day. Today you water both plants. In how man days will you water both plants on the same day again
Big Ideas Math Answers Grade 6 Advanced Chapter 1 Numerical Expressions and Factors 54
Answer:

Question 42.
Hamburgers are sold in packages of 20, while buns are sold in packages of 12. What are the least numbers of packages you should buy in order to have the same number of hamburgers and buns?
Answer:

Question 43.
A science museum is giving away a magnetic liquid kit to every 50th guest and a plasma ball to every 35th guest until someone receives both prizes.
Big Ideas Math Answers Grade 6 Advanced Chapter 1 Numerical Expressions and Factors 55
a. Which numbered guest will receive both a magnetic liquid kit and a plasma ball?
Answer:

b. How many people will receive a plasma ball?
Answer:

Numerical Expressions and Factors Practice Test

Question 1.
Find the value of 23.
Answer:

Question 2.
Evaluate \(\frac{5+4(12-2)}{3^{2}}\)
Answer:

Question 3.
Write 261 • 264 • 264 as a power.
Answer:

Question 4.
List the factor pairs of 66.
Answer:

Question 5.
Write the prime factorization of 56.
Answer:

Find the GCF of the numbers.
Question 6.
24, 54
Answer:

Question 7.
16, 32, 72
Answer:

Question 8.
52, 65
Answer:

Find the LCM of the numbers.
Question 9.
9, 24
Answer:

Question 10.
26, 39
Answer:

Question 11.
6, 12, 11
Answer:

Question 12.
You have 16 yellow beads, 20 red heads, and 21 orange beads to make identical bracelets. What is the greatest number of bracelets that you can make using all of the beads?
Answer:

Question 13.
A bag contains equal numbers of green marbles and blue marbles. You can divide all of the green marbles into groups of 12 and all the blue marbles into groups of 16. What is the least number of each color of marble that can be in the bag?
Answer:

Question 14.
The ages of the members of a family are 65, 58, 27, 25, 5, and 2 years old. What is the total admission price for the family to visit the zoo?
Big Ideas Math Answers Grade 6 Advanced Chapter 1 Numerical Expressions and Factors 56
Answer:

Question 15.
A competition awards prizes for fourth, third, second, and first place. The fourth place winner receives $5. Each place above that receives a prize that is five times the amount of the previous prize. How much prize money is awarded?
Answer:

Question 16.
You buy tealight candles and mints as party favors for a baby shower. The tealight candles come in packs of 12 for $3.50. The mints come in packs of 50 for $6.25. What is the least amount of money you can spend to buy the same number of candles and mints?
Big Ideas Math Answers Grade 6 Advanced Chapter 1 Numerical Expressions and Factors 57
Answer:

Numerical Expressions and Factors Cumulative Practice

Question 1.
What is the value of 8 × 135?
Answer:

Question 2.
Which number is equivalent to the expression below?
3 • 23 – 8 ÷ 4
A. 0
B. 4
C. 22
D. 214
Answer:

Question 3.
The top of an end table is a square with a side length of 16 inches. What is the area of the tabletop?
Big Ideas Math Solutions Grade 6 Advanced Chapter 1 Numerical Expressions and Factors 59
F. 16 in.2
G. 32 in.2
H. 64 in.2
I. 256 in.2
Answer:

Question 4.
You are filling baskets using 18 green eggs, 36 red eggs, and 54 blue eggs. What is the greatest number of baskets that you can fill so that the baskets are identical and there are no eggs left over?
A. 3
B. 6
C. 9
D. 18
Answer:

Question 5.
What is the value of 22 • 32 • 5?
Answer:

Question 6.
You hang the two strands of decorative lights shown below.
Big Ideas Math Solutions Grade 6 Advanced Chapter 1 Numerical Expressions and Factors 60
Strand 1: changes between red and blue every 15 seconds

Big Ideas Math Solutions Grade 6 Advanced Chapter 1 Numerical Expressions and Factors 61
Strand 2: changes between green and gold every 18 seconds
Both strands just changed color. After how many seconds will the strands change color at the same time again?
F. 3 seconds
G. 30 seconds
H. 90 seconds
I. 270 seconds
Answer:

Question 7.
Point p is plotted in the coordinate plane below.
Big Ideas Math Solutions Grade 6 Advanced Chapter 1 Numerical Expressions and Factors 62
What arc the coordinates of Point p?
A. (5, 3)
B. (4, 3)
C. (3, 5)
D. (3, 4)
Answer:

Question 8.
What is the prime factorization 1100?
F. 2 × 5 × 11
G. 22 × 52 × 11
H. 4 × 52 × 11
I. 22 × 5 × 55
Answer:

Question 9.
What is the least common multiple of 3, 8, and 10?
A. 21
B. 30
C. 80
D. 120
Answer:

Question 10.
What is the area of the shaded region of the figure below?
Big Ideas Math Solutions Grade 6 Advanced Chapter 1 Numerical Expressions and Factors 63
F. 16yd2
G. 65yd2
H. 81 yd2
I. 97 yd2
Answer:

Question 11.
Which expression represents a prime factorization?
A. 4 × 4 × 7
B. 22 × 21 × 23
C. 34 × 5 × 7
D. 5 × 5 × 9 × 11
Answer:

Question 12.
Find the greatest common factor for each pair of numbers.
Think
Solve
Explain
10 and 15      10 and 21     15 and 21
What can you conclude about the greatest common factor of 10, 15, and 21?
Explain your reasoning.
Answer:

Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions

Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions

If you stuck at solving complex problems on Quadratic Functions then stop calculating the problem and start practicing the concepts of Chapter 2 from Big Ideas Math Algebra 2 Answers. It holds all chapter’s answer keys in pdf format. Here, in this article, you will collect the details about Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions. This material is the complete guide for high school students to learn the concepts of the Quadratic functions. Hence, download the topic-wise BIM Algebra 2 Ch 1 Textbook Solutions from the below available links and start your practice sessions before any examination.

Big Ideas Math Book Algebra 2 Answer Key Chapter 2 Quadratic Functions

Students can access these Topicwise Big Ideas Math Algebra 2 Ch 1 Answers online or offline whenever required and kickstart their preparation. You can easily clear all your subject-related queries using the BIM Algebra 2 Ch 1 Answer key. This BIM Textbook Algebra 2 Chapter 1 Solution Key includes various easy & complex questions belonging to Lessons 2.1 to 2.4, Assessment Tests, Chapter Tests, Cumulative Assessments, etc. Apart from the Quadratic functions exercises, you can also find the exercise on the Lesson Focus of a Parabola. Excel in mathematics examinations by practicing more and more using the BigIdeas Math Algebra 2 Ch 2 Answer key.

Quadratic Functions Maintaining Mathematical Proficiency

Find the x-intercept of the graph of the linear equation.

Question 1.
y = 2x + 7

Question 2.
y = -6x + 8

Question 3.
y = -10x – 36

Question 4.
y = 3(x – 5)

Question 5.
y = -4(x + 10)

Question 6.
3x + 6y = 24

Find the distance between the two points.

Question 7.
(2, 5), (-4, 7)

Question 8.
(-1, 0), (-8, 4)

Question 9.
(3, 10), (5, 9)

Question 10.
(7, -4), (-5, 0)

Question 11.
(4, -8), (4, 2)

Question 12.
(0, 9), (-3, -6)

Question 13.
ABSTRACT REASONING Use the Distance Formula to write an expression for the distance between the two points (a, c) and (b, c). Is there an easier way to find the distance when the x-coordinates are equal? Explain your reasoning

Quadratic Functions Mathematical Practices

Monitoring Progress

Decide whether the syllogism represents correct or flawed reasoning. If flawed, explain why the conclusion is not valid.

Question 1.
All mammals are warm-blooded.
All dogs are mammals.
Therefore, all dogs are warm-blooded.

Question 2.
All mammals are warm-blooded.
My pet is warm-blooded.
Therefore, my pet is a mammal.

Question 3.
If I am sick, then I will miss school.
I missed school.
Therefore, I am sick.

Question 4.
If I am sick, then I will miss school.
I did not miss school.
Therefore, I am not sick.

Lesson 2.1 Transformations of Quadratic Functions

Essential Question

How do the constants a, h, and k affect the graph of the quadratic function g(x) = a(x – h)2 + k?
The parent function of the quadratic family is f(x) = x2. A transformation of the graph of the parent function is represented by the function g(x) = a(x – h)2 + k, where a ≠ 0.

EXPLORATION 1
Identifying Graphs of Quadratic Functions
Work with a partner.
Match each quadratic function with its graph. Explain your reasoning. Then use a graphing calculator to verify that your answer is correct.
a. g(x) = -(x – 2)2
b. g(x) = (x – 2)2 + 2
c. g(x) = -(x + 2)2 – 2
d. g(x) = 0.5(x – 2)2 + 2
e. g(x) = 2(x – 2)2
f. g(x) = -(x + 2)2 + 2
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 1

Communicate Your Answer

Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2
Question 2.
How do the constants a, h, and k affect the graph of the quadratic function g(x) =a(x – h)2 + k?

Question 3.
Write the equation of the quadratic function whose graph is shown at the right. Explain your reasoning. Then use a graphing calculator to verify that your equation is correct.
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 3

2.1 Lesson

Monitoring Progress

Describe the transformation of f(x) = x2 represented by g. Then graph each function.

Question 1.
g(x) = (x – 3)2

Question 2.
g(x) = (x + 2)2 – 2

Question 3.
g(x) = (x + 5)2 + 1

Describe the transformation of f(x) = x2 represented by g. Then graph each function.

Question 4.
g(x) = (\(\frac{1}{3} x\))2

Question 5.
g(x) = 3(x – 1)2

Question 6.
g(x) = -(x + 3)2 + 2

Question 7.
Let the graph of g be a vertical shrink by a factor of \(\frac{1}{2}\) followed by a translation 2 units up of the graph of f(x) = x2. Write a rule for g and identify the vertex.

Question 8.
Let the graph of g be a translation 4 units left followed by a horizontal shrink by a factor of \(\frac{1}{3}\) of the graph of f(x) = x2 + x. Write a rule for g.

Question 9.
WHAT IF? In Example 5, the water hits the ground 10 feet closer to the fire truck after lowering the ladder. Write a function that models the new path of the water.

Transformations of Quadratic Functions 2.1 Exercises

Vocabulary and Core Concept Check

Question 1.
COMPLETE THE SENTENCE The graph of a quadratic function is called a(n) ________.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.1 Question 1

Question 2.
VOCABULARY Identify the vertex of the parabola given by f(x) = (x + 2)2 – 4.
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3–12, describe the transformation of f(x) = x2 represented by g. Then graph each function.

Question 3.
g(x) = x2 – 3
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.1 Question 3

Question 4.
g(x) = x2 + 1
Answer:

Question 5.
g(x) = (x + 2)2

Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.1 Question 5

Question 6.
g(x) = (x – 4)2

Answer:

Question 7.
g(x) = (x – 1)2

Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.1 Question 7

Question 8.
g(x) = (x + 3)2

Answer:

Question 9.
g(x) = (x + 6)2 – 2
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.1 Question 9

Question 10.
g(x) = (x – 9)2 + 5
Answer:

Question 11.
g(x) = (x – 7)2 + 1
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.1 Question 11

Question 12.
g(x) = (x + 10)2 – 3
Answer:

ANALYZING RELATIONSHIPS In Exercises 13–16, match the function with the correct transformation of the graph of f. Explain your reasoning.
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 4

Question 13.
y = f(x – 1)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.1 Question 13

Question 14.
y = f(x) + 1
Answer:

Question 15.
y = f(x – 1) + 1
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.1 Question 15

Question 16.
y = f(x + 1)
Answer:

Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 5

In Exercises 17–24, describe the transformation of f(x) = x2 represented by g. Then graph each function.

Question 17.
g(x) = -x2

Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.1 Question 17

Question 18.
g(x) = (-x)2

Answer:

Question 19.
g(x) = 3x2

Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.1 Question 19

Question 20.
g(x) = \(\frac{1}{3}\)x2

Answer:

Question 21.
g(x) = (2x)2
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.1 Question 21

Question 22.
g(x) = -(2x)2

Answer:

Question 23.
g(x) = \(\frac{1}{5}\)x2 – 4
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.1 Question 23

Question 24.
g(x) = \(\frac{1}{2}\)(x – 1)2

Answer:

ERROR ANALYSIS In Exercises 25 and 26, describe and correct the error in analyzing the graph of f(x) = −6x2 + 4.

Question 25.
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 6
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.1 Question 25

Question 26.
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 7
Answer:

USING STRUCTURE In Exercises 27–30, describe the transformation of the graph of the parent quadratic function. Then identify the vertex.

Question 27.
f(x) = 3(x + 2)2 + 1
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.1 Question 27

Question 28.
f(x) = -4(x + 1)2 – 5
Answer:

Question 29.
f(x) = -2x2 + 5
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.1 Question 29

Question 30.
f(x) = \(\frac{1}{2}\)(x – 1)2
Answer:

In Exercises 31–34, write a rule for g described by the transformations of the graph of f. Then identify the vertex.

Question 31.
f(x) = x2 vertical stretch by a factor of 4 and a reflection in the x-axis, followed by a translation 2 units up
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.1 Question 31

Question 32.
f(x) = x2; vertical shrink by a factor of \(\frac{1}{3}\) and a reflection in the y-axis, followed by a translation 3 units right
Answer:

Question 33.
f(x) = 8x2 – 6; horizontal stretch by a factor of 2 and a translation 2 units up, followed by a reflection in the y-axis
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.1 Question 33

Question 34.
f(x) = (x + 6)2 + 3; horizontal shrink by a factor of \(\frac{1}{2}\) and a translation 1 unit down, followed by a reflection in the x-axis
Answer:

USING TOOLS In Exercises 35–40, match the function with its graph. Explain your reasoning.

Question 35.
g(x) = 2(x – 1)2 – 2
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.1 Question 35

Question 36.
g(x) = \(\frac{1}{2}\)(x + 1)2 – 2
Answer:

Question 37.
g(x) = -2(x – 1)2 + 2
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.1 Question 37

Question 38.
g(x) = 2(x + 1)2 + 2
Answer:

Question 39.
g(x) = -2(x + 1)2 – 2
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.1 Question 39

Question 40.
g(x) = 2(x – 1)2 + 2
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 8

JUSTIFYING STEPS In Exercises 41 and 42, justify eachstep in writing a function g based on the transformationsof f(x) = 2x2 + 6x.

Question 41.
translation 6 units down followed by a reflection in the x-axis
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 9
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.1 Question 41

Question 42.
reflection in the y-axis followed by a translation 4 units right
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 10
Answer:

Question 43.
MODELING WITH MATHEMATICS The function h(x) = -0.03(x – 14)2 + 6 models the jump of a red kangaroo, where x is the horizontal distance traveled (in feet) and h(x) is the height (in feet). When the kangaroo jumps from a higher location, it lands 5 feet farther away. Write a function that models the second jump.
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 11
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.1 Question 43

Question 44.
MODELING WITH MATHEMATICS The function f(t) = -16t2 + 10 models the height (in feet) of an object t seconds after it is dropped from a height of 10 feet on Earth. The same object dropped from the same height on the moon is modeled by g(t) = –\(\frac{8}{3}\)t2 + 10. Describe the transformation of the graph of f to obtain g. From what height must the object be dropped on the moon so it hits the ground at the same time as on Earth?
Answer:

Question 45.
MODELING WITH MATHEMATICS Flying fish use their pectoral fins like airplane wings to glide through the air.
a. Write an equation of the form y = a(x – h)2 + k with vertex (33, 5) that models the flight path, assuming the fish leaves the water at (0, 0).
b. What are the domain and range of the function? What do they represent in this situation?
c. Does the value of a change when the flight path has vertex (30, 4)? Justify your answer.
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 12
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.1 Question 45.1
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.1 Question 45.2
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.1 Question 45.3

Question 46.
HOW DO YOU SEE IT? Describe the graph of g as a transformation of the graph of f(x) = x2.
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 13
Answer:

Question 47.
COMPARING METHODS Let the graph of g be a translation 3 units up and 1 unit right followed by a vertical stretch by a factor of 2 of the graph of f(x) = x2.
a. Identify the values of a, h, and k and use vertex form to write the transformed function.
b. Use function notation to write the transformed function. Compare this function with your function in part (a).
c. Suppose the vertical stretch was performed first, followed by the translations. Repeat parts (a) and (b).
d. Which method do you prefer when writing a transformed function? Explain.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.1 Question 47.1
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.1 Question 47.2

Question 48.
THOUGHT PROVOKING A jump on a pogo stick with a conventional spring can be modeled by f(x) = -0.5(x – 6)2 + 18, where x is the horizontal distance (in inches) and f(x) is the vertical distance (in inches). Write at least one transformation of the function and provide a possible reason for your transformation.
Answer:

Question 49.
MATHEMATICAL CONNECTIONS The area of a circle depends on the radius, as shown in the graph. A circular earring with a radius of r millimeters has a circular hole with a radius of \(\frac{3 r}{4}\) millimeters. Describe a transformation of the graph below that models the area of the blue portion of the earring.
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 13.1
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.1 Question 49

Maintaining Mathematical Proficiency
A line of symmetry for the figure is shown in red. Find the coordinates of point A. (Skills Review Handbook)

Question 50.
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 14
Answer:

Question 51.
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 15
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.1 Question 51

Question 52.
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 16
Answer:

Lesson 2.2 Characteristics of Quadratic Functions

Essential Question
What type of symmetry does the graph of f(x) = a(x – h)2 + k have and how can you describe this symmetry?

EXPLORATION 1
Parabolas and Symmetry
Work with a partner.

a. Complete the table. Then use the values in the table to sketch the graph of the function
f(x) = \(\frac{1}{2}\)x2 – 2x – 2 on graph paper.
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 17
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 18
b. Use the results in part (a) to identify the vertex of the parabola.
c. Find a vertical line on your graph paper so that when you fold the paper, the left portion of the graph coincides with the right portion of the graph. What is the equation of this line? How does it relate to the vertex?
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 19
d. Show that the vertex form f(x) = \(\frac{1}{2}\)(x – 2)2 – 4 is equivalent to the function given in part (a).

EXPLORATION 2
Parabolas and Symmetry
Work with a partner.
Repeat Exploration 1 for the function given by f(x) = –\(\frac{1}{3}\)x2 + 2x + 3 = –\(\frac{1}{3}\)(x – 3),sup>2 + 6.

Communicate Your Answer

Question 3.
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 20
What type of symmetry does the graph of f(x) = a(x – h)2 + k have and how can you describe this symmetry?

Question 4.
Describe the symmetry of each graph. Then use a graphing calculator to verify your answer.
a. f(x) = -(x – 1)2 + 4
b. f(x) = (x + 1)2 – 2
c. f(x) = 2(x – 3)2 + 1
d. f(x) = \(\frac{1}{2}\)(x + 2)2
e. f(x) = -2x2 + 3
f. f(x) = 3(x – 5)2 + 2

2.2 Lesson

Monitoring Progress

Graph the function. Label the vertex and axis of symmetry.

Question 1.
f(x) = -3(x + 1)2

Question 2.
g(x) = 2(x – 2)2 + 5

Question 3.
h(x) = x2 + 2x – 1

Question 4.
p(x) = -2x2 – 8x + 1

Question 5.
Find the minimum value or maximum value of
(a) f(x) = 4x2 + 16x – 3 and
(b) h(x) = -x2 + 5x + 9. Describe the domain and range of each function, and where each function is increasing and decreasing.

Graph the function. Label the x-intercepts, vertex, and axis of symmetry.

Question 6.
f(x) = -(x + 1)(x + 5)

Question 7.
g(x) = \(\frac{1}{4}\)(x – 6)(x – 2)

Question 8.
WHAT IF? The graph of your third shot is a parabola through the origin that reaches a maximum height of 28 yards when x = 45. Compare the distance it travels before it hits the ground with the distances of the first two shots.

Characteristics of Quadratic Functions 2.2 Exercises

Vocabulary and Core Concept and Check

Question 1.
WRITING Explain how to determine whether a quadratic function will have a minimum value or a maximum value.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 1

Question 2.
WHICH ONE DOESN’T BELONG? The graph of which function does not belong with the other three? Explain.
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 21
Answer:

Question 3.
f(x) = (x – 3)2
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 3

Question 4.
h(x) = (x + 4)2
Answer:

Question 5.
g(x) = (x + 3)2 + 5
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 5

Question 6.
y = (x – 7)2 – 1
Answer:

Question 7.
y = -4(x – 2)2 + 4
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 7

Question 8.
g(x) = 2(x + 1)2 – 3
Answer:

Question 9.
f(x) = -2(x – 1)2 – 5
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 9

Question 10.
h(x) = 4(x + 4)2 + 6
Answer:

Question 11.
y = –\(\frac{1}{4}\)(x + 2)2 + 1
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 11

Question 12.
y = \(\frac{1}{2}\)(x – 3)2 + 2
Answer:

Question 13.
f(x) = 0.4(x – 1)2
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 13

Question 14.
g(x) = 0.75x2 – 5
Answer:

ANALYZING RELATIONSHIPS In Exercises 15–18, use the axis of symmetry to match the equation with its graph.

Question 15.
y = 2(x – 3)2 + 1
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 15

Question 16.
y = (x + 4)2 – 2
Answer:

Question 17.
y = \(\frac{1}{2}\)(x + 1)2 + 3
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 17

Question 18.
y = (x – 2)2 – 1
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 22
Answer:

REASONING In Exercises 19 and 20, use the axis of symmetry to plot the reflection of each point and complete the parabola.

Question 19.
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 23
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 19

Question 20.
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 24
Answer:

In Exercises 21–30, graph the function. Label the vertex and axis of symmetry.

Question 21.
y = x2 + 2x + 1
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 21.1
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 21.2

Question 22.
y = 3x2 – 6x + 4
Answer:

Question 23.
y = -4x2 + 8x + 2
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 23.1
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 23.2

Question 24.
f(x) = -x2 – 6x + 3
Answer:

Question 25.
g(x) = -x2 – 1
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 25.1
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 25.2

Question 26.
f(x) = 6x2 – 5
Answer:

Question 27.
g(x) = -1.5x2 + 3x + 2
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 27.1
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 27.2

Question 28.
f(x) = 0.5x2 + x – 3
Answer:

Question 29.
y = \(\frac{3}{2}\)x2 – 3x + 6
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 29.1
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 29.2

Question 30.
y = –\(\frac{5}{2}\)x2 – 4x – 1
Answer:

Question 31.
WRITING Two quadratic functions have graphs with vertices (2, 4) and (2, -3). Explain why you can not use the axes of symmetry to distinguish between the two functions.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 31

Question 32.
WRITING A quadratic function is increasing to the left of x = 2 and decreasing to the right of x = 2. Will the vertex be the highest or lowest point on the graph of the parabola? Explain.
Answer:

ERROR ANALYSIS In Exercises 33 and 34, describe and correct the error in analyzing the graph of y = 4x2 + 24x − 7.

Question 33.
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 25
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 33

Question 34.
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 26
Answer:

MODELING WITH MATHEMATICS In Exercises 35 and 36, x is the horizontal distance (in feet) and y is the vertical distance (in feet). Find and interpret the coordinates of the vertex.

Question 35.
The path of a basketball thrown at an angle of 45° can be modeled by y = -0.02x2 + x + 6.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 35

Question 36.
The path of a shot put released at an angle of 35° can be modeled by y = -0.01x2 + 0.7x + 6.
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 27
Answer:

Question 37.
ANALYZING EQUATIONS The graph of which function has the same axis of symmetry as the graph of y = x2 + 2x + 2?
A. y = 2x2 + 2x + 2
B. y = -3x2 – 6x + 2
C. y = x2 – 2x + 2
D. y = -5x2 + 10x + 23
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 37

Question 38.
USING STRUCTURE Which function represents the widest parabola? Explain your reasoning.
A. y = 2(x + 3)2
B. y = x2 – 5
C. y = 0.5(x – 1)2 + 1
D. y = -x2 + 6
Answer:

In Exercises 39–48, find the minimum or maximum value of the function. Describe the domain and range of the function, and where the function is increasing and decreasing.

Question 39.
y = 6x2 – 1
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 39

Question 40.
y = 9x2 + 7
Answer:

Question 41.
y = -x2 – 4x – 2
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 41

Question 42.
g(x) = -3x2 – 6x + 5
Answer:

Question 43.
f(x) = -2x2 + 8x + 7
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 43

Question 44.
g(x) = 3x2 + 18x – 5
Answer:

Question 45.
h(x) = 2x2 – 12x
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 45

Question 46.
h(x) = x2 – 4x
Answer:

Question 47.
y = \(\frac{1}{4}\)x2 – 3x + 2
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 47

Question 48.
f(x) = \(\frac{3}{2}\)x2 + 6x + 4
Answer:

Question 49.
PROBLEM SOLVING The path of a diver is modeled by the function f(x) = -9x2 + 9x + 1, where f(x) is the height of the diver (in meters) above the water and x is the horizontal distance (in meters) from the end of the diving board.
a. What is the height of the diving board?
b. What is the maximum height of the diver?
c. Describe where the diver is ascending and where the diver is descending.
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 28
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 49

Question 50.
PROBLEM SOLVING The engine torque y (in foot-pounds) of one model of car is given by y = -3.75x2 + 23.2x + 38.8, where x is the speed (in thousands of revolutions per minute) of the engine.
a. Find the engine speed that maximizes torque. What is the maximum torque?
b. Explain what happens to the engine torque as the speed of the engine increases.
Answer:

MATHEMATICAL CONNECTIONS In Exercises 51 and 52, write an equation for the area of the figure. Then determine the maximum possible area of the figure.

Question 51.
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 29
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 51

Question 52.
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 30
Answer:

In Exercises 53–60, graph the function. Label the x-intercept(s), vertex, and axis of symmetry.

Question 53.
y = (x + 3)(x – 3)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 53.1
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 53.2

Question 54.
y = (x + 1)(x – 3)
Answer:

Question 55.
y = 3(x + 2)(x + 6)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 55.1
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 55.2

Question 56.
f(x) = 2(x – 5)(x – 1)
Answer:

Question 57.
g(x) = -x(x + 6)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 57.1
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 57.2

Question 58.
y = -4x(x + 7)
Answer:

Question 59.
f(x) = -2(x – 3)2
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 59.1

Question 60.
y = 4(x – 7)2
Answer:

USING TOOLS In Exercises 61–64, identify the x-intercepts of the function and describe where the graph is increasing and decreasing. Use a graphing calculator to verify your answer.

Question 61.
f(x) = \(\frac{1}{2}\)(x – 2)(x + 6)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 61

Question 62.
y = \(\frac{3}{4}\)(x + 1)(x – 3)
Answer:

Question 63.
g(x) = -4(x – 4)(x – 2)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 63.1

Question 64.
h(x) = -5(x + 5)(x + 1)
Answer:

Question 65.
MODELING WITH MATHEMATICS A soccer player kicks a ball downfield. The height of the ball increases until it reaches a maximum height of 8 yards, 20 yards away from the player. A second kick is modeled by y = x(0.4 – 0.008x). Which kick travels farther before hitting the ground? Which kick travels higher?
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 31
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 65.1

Question 66.
MODELING WITH MATHEMATICS Although a football field appears to be flat, some are actually shaped like a parabola so that rain runs off to both sides. The cross section of a field can be modeled by y = -0.000234x(x – 160), where x and y are measured in feet. What is the width of the field? What is the maximum height of the surface of the field?
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 32
Answer:

Question 67.
REASONING The points (2, 3) and (-4, 2) lie on the graph of a quadratic function. Determine whether you can use these points to find the axis of symmetry. If not, explain. If so, write the equation of the axis of symmetry.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 67

Question 68.
OPEN-ENDED Write two different quadratic functions in intercept form whose graphs have the axis of symmetry x= 3.
Answer:

Question 69.
PROBLEM SOLVING An online music store sells about 4000 songs each day when it charges $1 per song. For each $0.05 increase in price, about 80 fewer songs per day are sold. Use the verbal model and quadratic function to determine how much the store should charge per song to maximize daily revenue.
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 33
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 69

Question 70.
PROBLEM SOLVING An electronics store sells 70 digital cameras per month at a price of $320 each. For each $20 decrease in price, about 5 more cameras per month are sold. Use the verbal model and quadratic function to determine how much the store should charge per camera to maximize monthly revenue.
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 34
Answer:

Question 71.
DRAWING CONCLUSIONS Compare the graphs of the three quadratic functions. What do you notice? Rewrite the functions f and g in standard form to justify your answer.
f(x) = (x + 3)(x + 1)
g(x) = (x + 2)2 – 1
h(x) = x2 + 4x + 3
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 71

Question 72.
USING STRUCTURE Write the quadratic function f(x) = x2 + x – 12 in intercept form. Graph the function. Label the x-intercepts, y-intercept, vertex, and axis of symmetry.
Answer:

Question 73.
PROBLEM SOLVING A woodland jumping mouse hops along a parabolic path given by y = -0.2x2 + 1.3x, where x is the mouse’s horizontal distance traveled (in feet) and y is the corresponding height (in feet). Can the mouse jump over a fence that is 3 feet high? Justify your answer.
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 35
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 73

Question 74.
HOW DO YOU SEE IT? Consider the graph of the function f(x) = a(x – p)(x – q).
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 36
a. What does f(\(\frac{p+q}{2}\)) represent in the graph?
b. If a < 0, how does your answer in part (a) change? Explain.
Answer:

Question 75.
MODELING WITH MATHEMATICS The Gateshead Millennium Bridge spans the River Tyne. The arch of the bridge can be modeled by a parabola. The arch reaches a maximum height of 50 meters at a point roughly 63 meters across the river. Graph the curve of the arch. What are the domain and range? What do they represent in this situation?
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 37
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 75

Quadratic 76.
THOUGHT PROVOKING
You have 100 feet of fencing to enclose a rectangular garden. Draw three possible designs for the garden. Of these, which has the greatest area? Make a conjecture about the dimensions of the rectangular garden with the greatest possible area. Explain your reasoning.
Answer:

Question 77.
MAKING AN ARGUMENT The point (1, 5) lies on the graph of a quadratic function with axis of symmetry x = -1. Your friend says the vertex could be the point (0, 5). Is your friend correct? Explain.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 77

Question 78.
CRITICAL THINKING Find the y-intercept in terms of a, p, and q for the quadratic function f(x) = a(x – p)(x – q).
Answer:

Question 79.
MODELING WITH MATHEMATICS A kernel of popcorn contains water that expands when the kernel is heated, causing it to pop. The equations below represent the “popping volume” y (in cubic centimeters per gram) of popcorn with moisture content x (as a percent of the popcorn’s weight).
Hot-air popping: y = -0.761(x – 5.52)(x – 22.6)
Hot-oil popping:y = -0.652(x – 5.35)(x – 21.8)
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 38
a. For hot-air popping, what moisture content maximizes popping volume? What is the maximum volume?
b. For hot-oil popping, what moisture content maximizes popping volume? What is the maximum volume?
c. Use a graphing calculator to graph both functions in the same coordinate plane. What are the domain and range of each function in this situation? Explain.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 79.1
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 79.2

Question 80.
ABSTRACT REASONING A function is written in intercept form with a > 0. What happens to the vertex of the graph as a increases? as a approaches 0?
Answer:

Maintaining Mathematical Proficiency

Solve the equation. Check for extraneous solutions. (Skills Review Handbook)

Question 81.
3\(\sqrt{x}\) – 6 = 0
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 81

Question 82.
2\(\sqrt{x-4}\) – 2 = 2
Answer:

Question 83.
\(\sqrt{5x}\) + 5 = 0
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 83

Question 84.
\(\sqrt{3x+8}\) = \(\sqrt{x+4}\)
Answer:

Solve the proportion. (Skills Review Handbook)

Question 85.
\(\frac{1}{2}\) = \(\frac{x}{4}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 85

Question 86.
\(\frac{2}{3}\) = \(\frac{x}{9}\)
Answer:

Question 87.
\(\frac{-1}{4}\) = \(\frac{3}{x}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 87

Question 88.
\(\frac{5}{2}\) =-\(\frac{20}{x}\)
Answer:

Quadratic Functions Study Skills Using the Features of Your Textbook to Prepare for Quizzes and Tests

Core Vocabulary
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 39

Core Concepts

Section 2.1
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 40
Section 2.2
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 40.1

Mathematical Practices

Question 1.
Why does the height you found in Exercise 44 on page 53 make sense in the context of the situation?

Question 2.
How can you effectively communicate your preference in methods to others in Exercise 47 on page 54?

Question 3.
How can you use technology to deepen your understanding of the concepts in Exercise 79 on page 64?

Study Skills
Using the Features of Your Textbook to Prepare for Quizzes and Tests

  • Read and understand the core vocabulary and the contents of the Core Concept boxes.
  • Review the Examples and the Monitoring Progress questions. Use the tutorials at BigIdeasMath.com for additional help.
  • Review previously completed homework assignments.

Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 41

Quadratic Functions 2.1 – 2.2 Quiz

2.1 – 2.2 Quiz

Describe the transformation of f(x) = x2 represented by g. (Section 2.1)

Question 1.
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 42

Question 2.
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 43

Question 3.
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 44

Write a rule for g and identify the vertex. (Section 2.1)

Question 4.
Let g be a translation 2 units up followed by a reflection in the x-axis and a vertical stretch by a factor of 6 of the graph of f(x) = x2.

Question 5.
Let g be a translation 1 unit left and 6 units down, followed by a vertical shrink by a factor of \(\frac{1}{2}\) of the graph of f(x) = 3(x + 2)2.

Question 6.
Let g be a horizontal shrink by a factor of \(\frac{1}{4}\), followed by a translation 1 unit up and 3 units right of the graph of f(x) = (2x + 1)2 – 11.

Graph the function. Label the vertex and axis of symmetry. (Section 2.2)

Question 7.
f(x) = 2(x – 1)2 – 5

Question 8.
h(x) = 3x2 + 6x – 2

Question 9.
f(x) = 7 – 8x – x2

Find the x-intercepts of the graph of the function. Then describe where the function is increasing and decreasing.(Section 2.2)

Question 10.
g(x) = -3(x + 2)(x + 4)

Question 11.
g(x) = \(\frac{1}{2}\)(x – 5)(x + 1)

Question 12.
f(x) = 0.4x(x – 6)

Question 13.
A grasshopper can jump incredible distances, up to 20 times its length. The height (in inches) of the jump above the ground of a 1-inch-long grasshopper is given by h(x) = –\(\frac{1}{20}\)x2 + x, where x is the horizontal distance (in inches) of the jump. When the grasshopper jumps off a rock, it lands on the ground 2 inches farther. Write a function that models the new path of the jump. (Section 2.1)
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 45

Question 14.
A passenger on a stranded lifeboat shoots a distress flare into the air. The height (in feet) of the flare above the water is given by f(t) = -16t(t – 8), where t is time (in seconds) since the flare was shot. The passenger shoots a second flare, whose path is modeled in the graph. Which flare travels higher? Which remains in the air longer? Justify your answer. (Section 2.2)
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 46

Lesson 2.3 Focus of a Parabola

Essential Question
What is the focus of a parabola?
EXPLORATION 1
Analyzing Satellite Dishes
Work with a partner
. Vertical rays enter a satellite dish whose cross section is a parabola. When the rays hit the parabola, they reflect at the same angle at which they entered. (See Ray 1 in the figure.)
a. Draw the reflected rays so that they intersect the y-axis.
b. What do the reflected rays have in common?
c. The optimal location for the receiver of the satellite dish is at a point called the focus of the parabola. Determine the location of the focus. Explain why this makes sense in this situation.
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 47

EXPLORATION 2
Analyzing Spotlights

Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 48
Work with a partner. Beams of light are coming from the bulb in a spotlight, located at the focus of the parabola. When the beams hit the parabola, they reflect at the same angle at which they hit. (See Beam 1 in the figure.) Draw the reflected beams. What do they have in common? Would you consider this to be the optimal result? Explain.
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 49

Communicate Your Answer

Question 3.
What is the focus of a parabola?

Question 4.
Describe some of the properties of the focus of a parabola.

2.3 Lesson

Monitoring Progress

Question 1.
Use the Distance Formula to write an equation of the parabola with focus F(0, -3) and directrix y = 3.
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 50

Identify the focus, directrix, and axis of symmetry of the parabola. Then graph the equation.

Question 2.
y = 0.5x2

Question 3.
-y = x2

Question 4.
y2 = 6x

Write an equation of the parabola with vertex at (0, 0) and the given directrix or focus.

Question 5.
directrix: x = -3

Question 6.
focus: (-2, 0)

Question 7.
focus: (0, \(\frac{3}{2}\))

Monitoring Progress

Question 8.
Write an equation of a parabola with vertex (-1, 4) and focus (-1, 2).

Question 9.
A parabolic microwave antenna is 16 feet in diameter. Write an equation that represents the cross section of the antenna with its vertex at (0, 0) and its focus 10 feet to the right of the vertex. What is the depth of the antenna?

Focus of a Parabola 2.3 Exercises

Vocabulary and Core Concept Check

Question 1.
COMPLETE THE SENTENCE A parabola is the set of all points in a plane equidistant from a fixed point called the ______ and a fixed line called the __________ .
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.3 Question 1

Question 2.
WRITING Explain how to find the coordinates of the focus of a parabola with vertex (0, 0)and directrix y = 5.
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3–10, use the Distance Formula to write an equation of the parabola.

Question 3.
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 51
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.3 Question 3

Question 4.
Big Ideas Math Answer Key Algebra 2 Chapter 2 Quadratic Functions 52
Answer:

Question 5.
focus: (0, -2)
directrix: y = 2
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.3 Question 5

Question 6.
directrix: y = 7
focus: (0, -7)
Answer:

Question 7.
vertex: (0, 0)
directrix: y = -6
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.3 Question 7

Question 8.
vertex: (0, 0)
focus: (0, 5)
Answer:

Question 9.
vertex: (0, 0)
focus: (0, -10)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.3 Question 9

Question 10.
vertex: (0, 0)
directrix: y = -9
Answer:

Question 11.
ANALYZING RELATIONSHIPS Which of the given characteristics describe parabolas that open down? Explain your reasoning.
A. focus: (0, -6)
directrix: y = 6
B. focus: (0, -2)
directrix: y = 2
C.focus: (0, 6)
directrix: y = -6
D. focus: (0, -1)
directrix: y = 1
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.3 Question 11

Question 12.
REASONING Which of the following are possible coordinates of the point P in the graph shown? Explain.
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 53
A. (-6, -1)
B. (3, –\(\frac{1}{4}\))
C. (4, –\(\frac{4}{9}\))
D. (1, –\(\frac{1}{36}\))
E. (6, -1)
F. (2, –\(\frac{1}{18}\))
Answer:

In Exercises 13–20, identify the focus, directrix, and axis of symmetry of the parabola. Graph the equation.

Question 13.
y = \(\frac{1}{8}\)x2
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.3 Question 13

Question 14.
y = –\(\frac{1}{12}\)x2
Answer:

Question 15.
x = –\(\frac{1}{20}\)y2
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.3 Question 15

Question 16.
x= \(\frac{1}{24}\)y2
Answer:

Question 17.
y2 = 16x
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.3 Question 17.1
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.3 Question 17.2

Question 18.
-x2 = 48y
Answer:

Question 19.
6x2 + 3y = 0
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.3 Question 19.1
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.3 Question 19.2

Question 20.
8x2 – y = 0
Answer:

ERROR ANALYSIS In Exercises 21 and 22, describe and correct the error in graphing the parabola.

Question 21.
Big Ideas Math Answer Key Algebra 2 Chapter 2 Quadratic Functions 54
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.3 Question 21

Question 22.
Big Ideas Math Answer Key Algebra 2 Chapter 2 Quadratic Functions 55
Answer:

Question 23.
ANALYZING EQUATIONS The cross section (with units in inches) of a parabolic satellite dish can be modeled by the equation y = \(\frac{1}{38}\)x2. How far is the receiver from the vertex of the cross section? Explain.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.3 Question 23

Question 24.
ANALYZING EQUATIONS The cross section (with units in inches) of a parabolic spotlight can be modeled by the equation x = \(\frac{1}{20}\)y2. How far is the bulb from the vertex of the cross section? Explain.
Big Ideas Math Answer Key Algebra 2 Chapter 2 Quadratic Functions 56
Answer:

In Exercises 25–28, write an equation of the parabola shown.

Question 25.
Big Ideas Math Answer Key Algebra 2 Chapter 2 Quadratic Functions 57
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.3 Question 25

Question 26.
Big Ideas Math Answer Key Algebra 2 Chapter 2 Quadratic Functions 58
Answer:

Question 27.
Big Ideas Math Answer Key Algebra 2 Chapter 2 Quadratic Functions 59
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.3 Question 27

Question 28.
Big Ideas Math Answer Key Algebra 2 Chapter 2 Quadratic Functions 60
Answer:

In Exercises 29–36, write an equation of the parabola with the given characteristics.

Question 29.
focus: (3, 0)
directrix: x = -3
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.3 Question 29

Question 30.
focus: (\(\frac{2}{3}\), 0)
directrix: x = –\(\frac{2}{3}\)
Answer:

Question 31.
directrix: x = -10
vertex: (0, 0)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.3 Question 31

Question 32.
directrix: y = \(\frac{8}{3}\)
vertex: (0, 0)
Answer:

Question 33.
focus: (0, –\(\frac{5}{3}\))
directrix: y = \(\frac{5}{3}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.3 Question 33

Question 34.
focus: (0, \(\frac{5}{4}\))
directrix: y = –\(\frac{5}{4}\)
Answer:

Question 35.
focus: (0, \(\frac{6}{7}\))
vertex: (0, 0)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.3 Question 35

Question 36.
focus: (-\(\frac{4}{5}\), 0)
vertex: (0, 0)
Answer:

In Exercises 37–40, write an equation of the parabola shown.

Question 37.
Big Ideas Math Answer Key Algebra 2 Chapter 2 Quadratic Functions 61
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.3 Question 37

Question 38.
Big Ideas Math Answer Key Algebra 2 Chapter 2 Quadratic Functions 62
Answer:

Question 39.
Big Ideas Math Answer Key Algebra 2 Chapter 2 Quadratic Functions 63
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.3 Question 39

Question 40.
Big Ideas Math Answer Key Algebra 2 Chapter 2 Quadratic Functions 64
Answer:

In Exercises 41–46, identify the vertex, focus, directrix, and axis of symmetry of the parabola. Describe the transformations of the graph of the standard equation with p = 1 and vertex (0, 0).

Question 41.
y = \(\frac{1}{8}\)(x – 3)2 + 2
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.3 Question 41

Question 42.
y = –\(\frac{1}{4}\)(x + 2)2 + 1
Answer:

Question 43.
x = \(\frac{1}{16}\)(y – 3)2 + 1
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.3 Question 43

Question 44.
y = (x + 3)2 – 5
Answer:

Question 45.
x = -3(y + 4)2 + 2
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.3 Question 45

Question 46.
x = 4(y + 5)2 – 1
Answer:

Question 47.
MODELING WITH MATHEMATICS Scientists studying dolphin echolocation simulate the projection of a bottlenose dolphin’s clicking sounds using computer models. The models originate the sounds at the focus of a parabolic reflector. The parabola in the graph shows the cross section of the reflector with focal length of 1.3 inches and aperture width of 8 inches. Write an equation to represent the cross section of the reflector. What is the depth of the reflector?
Big Ideas Math Answer Key Algebra 2 Chapter 2 Quadratic Functions 65
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.3 Question 47

Question 48.
MODELING WITH MATHEMATICS Solar energy can be concentrated using long troughs that have a parabolic cross section as shown in the figure. Write an equation to represent the cross section of the trough. What are the domain and range in this situation? What do they represent?
Big Ideas Math Answer Key Algebra 2 Chapter 2 Quadratic Functions 66
Answer:

Question 49.
ABSTRACT REASONING As | p | increases, how does the width of the graph of the equation y = \(\frac{1}{4 p}\)x2 change? Explain your reasoning.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.3 Question 49

Question 50.
HOW DO YOU SEE IT? The graph shows the path of a volleyball served from an initial height of 6 feet as it travels over a net.
Big Ideas Math Answer Key Algebra 2 Chapter 2 Quadratic Functions 67
a. Label the vertex, focus, and a point on the directrix.
b. An underhand serve follows the same parabolic path but is hit from a height of 3 feet. How does this affect the focus? the directrix?
Answer:

Question 51.
CRITICAL THINKING The distance from point P to the directrix is 2 units. Write an equation of the parabola.
Big Ideas Math Answer Key Algebra 2 Chapter 2 Quadratic Functions 68
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.3 Question 51

Question 52.
THOUGHT PROVOKING Two parabolas have the same focus (a, b) and focal length of 2 units. Write an equation of each parabola. Identify the directrix of each parabola.
Answer:

Question 53.
REPEATED REASONING Use the Distance Formula to derive the equation of a parabola that opens to the right with vertex (0, 0), focus (p, 0), and directrix x = -p.
Big Ideas Math Answer Key Algebra 2 Chapter 2 Quadratic Functions 69
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.3 Question 53

Question 54.
PROBLEM SOLVING The latus rectum of a parabola is the line segment that is parallel to the directrix, passes through the focus, and has endpoints that lie on the parabola. Find the length of the latus rectum of the parabola shown.
Big Ideas Math Answer Key Algebra 2 Chapter 2 Quadratic Functions 70
Answer:

Maintaining Mathematical Proficiency

Write an equation of the line that passes through the points.(Section 1.3)

Question 55.
(1, -4), (2, -1)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.3 Question 55

Question 56.
(-3, 12), (0, 6)
Answer:

Question 57.
(3, 1), (-5, 5)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.3 Question 57

Question 58.
(2, -1), (0, 1)
Answer:

Use a graphing calculator to find an equation for the line of best fit.

Question 59.
Big Ideas Math Answer Key Algebra 2 Chapter 2 Quadratic Functions 71
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.3 Question 59

Question 60.
Big Ideas Math Answer Key Algebra 2 Chapter 2 Quadratic Functions 72
Answer:

Lesson 2.4 Modeling with Quadratic Functions

Essential Question
How can you use a quadratic function to model a real-life situation?

EXPLORATION 1
Modeling with a Quadratic Function

Work with a partner. The graph shows a quadratic function of the form
P(t) = at2 + bt + c
Big Ideas Math Answer Key Algebra 2 Chapter 2 Quadratic Functions 73
which approximates the yearly profits for a company, where P(t) is the profit in year t.
a. Is the value of a positive, negative, or zero? Explain.
b. Write an expression in terms of a and b that represents the year t when the company made the least profit.
c. The company made the same yearly profits in 2004 and 2012. Estimate the year in which the company made the least profit.
d. Assume that the model is still valid today. Are the yearly profits currently increasing, decreasing, or constant? Explain.

EXPLORATION 2
Modeling with a Graphing Calculator
Work with a partner.
The table shows the heights h (in feet) of a wrench t seconds after it has been dropped from a building under construction.
Big Ideas Math Answer Key Algebra 2 Chapter 2 Quadratic Functions 74
a. Use a graphing calculator to create a scatter plot of the data, as shown at the right. Explain why the data appear to fit a quadratic model.
b. Use the quadratic regression feature to find a quadratic model for the data.
c. Graph the quadratic function on the same screen as the scatter plot to verify that it fits the data.
Big Ideas Math Answer Key Algebra 2 Chapter 2 Quadratic Functions 75
Big Ideas Math Answer Key Algebra 2 Chapter 2 Quadratic Functions 76
d. When does the wrench hit the ground? Explain.

Communicate Your Answer

Question 3.
How can you use a quadratic function to model a real-life situation?

Question 4.
Use the Internet or some other reference to find examples of real-life situations that can be modeled by quadratic functions.

2.4 Lesson

Monitoring Progress

Question 1.
WHAT IF? The vertex of the parabola is (50, 37.5). What is the height of the net?

Question 2.
Write an equation of the parabola that passes through the point (-1, 2) and has vertex (4, -9).

Question 3.
WHAT IF? The y-intercept is 4.8. How does this change your answers in parts (a) and (b)?

Question 4.
Write an equation of the parabola that passes through the point (2, 5) and has x-intercepts -2 and 4.

Question 5.
Write an equation of the parabola that passes through the points (-1, 4), (0, 1), and (2, 7).

Question 6.
The table shows the estimated profits y (in dollars) for a concert when the charge is x dollars per ticket. Write and evaluate a function to determine what the charge per ticket should be to maximize the profit.
Big Ideas Math Answer Key Algebra 2 Chapter 2 Quadratic Functions 77

Question 7.
The table shows the results of an experiment testing the maximum weights y (in tons) supported by ice x inches thick. Write a function that models the data. How much weight can be supported by ice that is 22 inches thick?
Big Ideas Math Answer Key Algebra 2 Chapter 2 Quadratic Functions 78

Modeling with Quadratic Functions 2.4 Exercises

Vocabulary and Core Concept Check

Question 1.
WRITING Explain when it is appropriate to use a quadratic model for a set of data.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.4 Question 1

Question 2.
DIFFERENT WORDS, SAME QUESTION
Which is different? Find “both” answers.
Big Ideas Math Answer Key Algebra 2 Chapter 2 Quadratic Functions 79
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3–8, write an equation of the parabola in vertex form.

Question 3.
Big Ideas Math Answer Key Algebra 2 Chapter 2 Quadratic Functions 80
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.4 Question 3

Question 4.
Big Ideas Math Answer Key Algebra 2 Chapter 2 Quadratic Functions 81
Answer:

Question 5.
passes through (13, 8) and has vertex (3, 2)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.4 Question 5

Question 6.
passes through (-7, -15) and has vertex (-5, 9)
Answer:

Question 7.
passes through (0, -24) and has vertex (-6, -12)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.4 Question 7

Question 8.
passes through (6, 35) and has vertex (-1, 14)
Answer:

In Exercises 9–14, write an equation of the parabola in intercept form.

Question 9.
Big Ideas Math Answer Key Algebra 2 Chapter 2 Quadratic Functions 82
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.4 Question 9.1

Question 10.
Big Ideas Math Answer Key Algebra 2 Chapter 2 Quadratic Functions 83
Answer:

Question 11.
x-intercepts of 12 and -6; passes through (14, 4)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.4 Question 11

Question 12.
x-intercepts of 9 and 1; passes through (0, -18)
Answer:

Question 13.
x-intercepts of -16 and -2; passes through (-18, 72)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.4 Question 13

Question 14.
x-intercepts of -7 and -3; passes through (-2, 0.05)
Answer:

Question 15.
WRITING Explain when to use intercept form and when to use vertex form when writing an equation of a parabola.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.4 Question 15

Question 16.
ANALYZING EQUATIONS Which of the following equations represent the parabola?
Big Ideas Math Answer Key Algebra 2 Chapter 2 Quadratic Functions 84
A. y = 2(x – 2)(x + 1)
B. y = 2(x + 0.5)2 – 4.5
C. y = 2(x – 0.5)2 – 4.5
D. y = 2(x + 2)(x – 1)
Answer:

In Exercises 17–20, write an equation of the parabola in vertex form or intercept form.

Question 17.
Big Ideas Math Answer Key Algebra 2 Chapter 2 Quadratic Functions 85
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.4 Question 17

Question 18.
Big Ideas Math Answer Key Algebra 2 Chapter 2 Quadratic Functions 86
Answer:

Question 19.
Big Ideas Math Answer Key Algebra 2 Chapter 2 Quadratic Functions 87
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.4 Question 19

Question 20.
Big Ideas Math Algebra 2 Solutions Chapter 2 Quadratic Functions 88
Answer:

Question 21.
ERROR ANALYSIS Describe and correct the error in writing an equation of the parabola.
Big Ideas Math Algebra 2 Solutions Chapter 2 Quadratic Functions 89
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.4 Question 21

Question 22.
MATHEMATICAL CONNECTIONS The area of a rectangle is modeled by the graph where y is the area (in square meters) and x is the width (in meters). Write an equation of the parabola. Find the dimensions and corresponding area of one possible rectangle. What dimensions result in the maximum area?
Big Ideas Math Algebra 2 Solutions Chapter 2 Quadratic Functions 90
Answer:

Question 23.
MODELING WITH MATHEMATICS Every rope has a safe working load. A rope should not be used to lift a weight greater than its safe working load. The table shows the safe working loads S (in pounds) for ropes with circumference C (in inches). Write an equation for the safe working load for a rope. Find the safe working load for a rope that has a circumference of 10 inches.
Big Ideas Math Algebra 2 Solutions Chapter 2 Quadratic Functions 91
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.4 Question 23.1
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.4 Question 23.2

Question 24.
MODELING WITH MATHEMATICS A baseball is thrown up in the air. The table shows the heights y (in feet) of the baseball after x seconds. Write an equation for the path of the baseball. Find the height of the baseball after 1.7 seconds.
Big Ideas Math Algebra 2 Solutions Chapter 2 Quadratic Functions 92
Answer:

Question 25.
COMPARING METHODS You use a system with three variables to find the equation of a parabola that passes through the points (−8, 0), (2, −20), and (1, 0). Your friend uses intercept form to find the equation. Whose method is easier? Justify your answer.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.4 Question 25

Question 26.
MODELING WITH MATHEMATICS The table shows the distances y a motorcyclist is from home after x hours.
Big Ideas Math Algebra 2 Solutions Chapter 2 Quadratic Functions 93
a. Determine what type of function you can use to model the data. Explain your reasoning.
b. Write and evaluate a function to determine the distance the motorcyclist is from home after 6 hours.
Answer:

Question 27.
USING TOOLS The table shows the heights h (in feet) of a sponge t seconds after it was dropped by a window cleaner on top of a skyscraper.
Big Ideas Math Algebra 2 Solutions Chapter 2 Quadratic Functions 94
a. Use a graphing calculator to create a scatter plot. Which better represents the data, a line or a parabola? Explain.
b. Use the regression feature of your calculator to find the model that best fits the data.
c. Use the model in part (b) to predict when the sponge will hit the ground.
d. Identify and interpret the domain and range in this situation.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.4 Question 27

Question 28.
MAKING AN ARGUMENT Your friend states that quadratic functions with the same x-intercepts have the same equations, vertex, and axis of symmetry. Is your friend correct? Explain your reasoning.
Answer:

In Exercises 29–32, analyze the differences in the outputs to determine whether the data are linear, quadratic, or neither. Explain. If linear or quadratic, write an equation that fits the data.

Question 29.
Big Ideas Math Algebra 2 Answer Key Chapter 2 Quadratic Functions 95
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.4 Question 29.1
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.4 Question 29.2

Question 30.
Big Ideas Math Algebra 2 Answer Key Chapter 2 Quadratic Functions 96
Answer:

Question 31.
Big Ideas Math Algebra 2 Answer Key Chapter 2 Quadratic Functions 97
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.4 Question 31

Question 32.
Big Ideas Math Algebra 2 Answer Key Chapter 2 Quadratic Functions 98
Answer:

Question 33.
PROBLEM SOLVING The graph shows the number y of students absent from school due to the flu each day x.
Big Ideas Math Algebra 2 Answer Key Chapter 2 Quadratic Functions 99
a. Interpret the meaning of the vertex in this situation.
b. Write an equation for the parabola to predict the number of students absent on day 10.
c. Compare the average rates of change in the students with the flu from 0 to 6 days and 6 to 11 days.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.4 Question 33.1
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.4 Question 33.2

Question 34.
THOUGHT PROVOKING Describe a real-life situation that can be modeled by a quadratic equation. Justify your answer.
Answer:

Question 35.
PROBLEM SOLVING The table shows the heights y of a competitive water-skier x seconds after jumping off a ramp. Write a function that models the height of the water-skier over time. When is the water-skier 5 feet above the water? How long is the skier in the air?
Big Ideas Math Algebra 2 Answer Key Chapter 2 Quadratic Functions 100
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.4 Question 35.1

Question 36.
HOW DO YOU SEE IT? Use the graph to determine whether the average rate of change over each interval is positive, negative, or zero.
Big Ideas Math Algebra 2 Answer Key Chapter 2 Quadratic Functions 101
a. 0 ≤ x ≤ 2
b. 2 ≤ x ≤ 5
c. 2 ≤ x ≤ 4
d. 0 ≤ x ≤ 4
Answer:

Question 37.
REPEATED REASONING The table shows the number of tiles in each figure. Verify that the data show a quadratic relationship. Predict the number of tiles in the 12th figure.
Big Ideas Math Algebra 2 Answer Key Chapter 2 Quadratic Functions 102
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.4 Question 37.1
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.4 Question 37.2

Maintaining Mathematical Proficiency

Factor the trinomial. (Skills Review Handbook)

Question 38.
x2 + 4x + 3
Answer:

Question 39.
x2 – 3x + 2
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.4 Question 39

Question 40.
3x2 – 15x + 12
Answer:

Question 41.
5x2 + 5x – 30
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.4 Question 41

Quadratic Functions Performance Task: Accident Reconstruction

2.3–2.4 What Did You Learn?

Core Vocabulary
focus, p. 68
directrix, p. 68

Core Concepts

Section 2.3
Standard Equations of a Parabola with Vertex at the Origin, p. 69
Standard Equations of a Parabola with Vertex at (h, k), p. 70

Section 2.4
Writing Quadratic Equations, p. 76
Writing Quadratic Equations to Model Data, p. 78

Mathematical Practices

Question 1.
Explain the solution pathway you used to solve Exercise 47 on page 73.

Question 2.
Explain how you used definitions to derive the equation in Exercise 53 on page 74.

Question 3.
Explain the shortcut you found to write the equation in Exercise 25 on page 81.

Question 4.
Describe how you were able to construct a viable argument in Exercise 28 on page 81.

Performance Task

Accident Reconstruction

Was the driver of a car speeding when the brakes were applied? What do skid marks at the scene of an accident reveal about the moments before the collision?
Big Ideas Math Algebra 2 Answer Key Chapter 2 Quadratic Functions 103
To explore the answers to these questions and more, go to BigIdeasMath.com.

Quadratic Functions Chapter Review

Describe the transformation of f(x) = x2 represented by g. Then graph each function.

Question 1.
g(x) = (x + 4)2

Question 2.
g(x) = (x – 7)2 + 2

Question 3.
g(x) = -3(x + 2)2 – 1

Question 4.
Let the graph of g be a horizontal shrink by a factor of \(\frac{2}{3}\), followed by a translation 5 units left and 2 units down of the graph of f(x) = x2.

Question 5.
Let the graph of g be a translation 2 units left and 3 units up, followed by a reflection in the y-axis of the graph of f(x) = x2 – 2x.

Graph the function. Label the vertex and axis of symmetry. Find the minimum or maximum value of f. Describe where the function is increasing and decreasing.

Question 6.
f(x) = 3(x – 1)2 – 4

Question 7.
g(x) = -2x2 + 16x + 3

Question 8.
h(x) = (x – 3)(x + 7)

Question 9.
You can make a solar hot-dog cooker by shaping foil-lined cardboard into a parabolic trough and passing a wire through the focus of each end piece. For the trough shown, how far from the bottom should the wire be placed?
Big Ideas Math Answers Algebra 2 Chapter 2 Quadratic Functions 105

Question 10.
Graph the equation 36y = x2. Identify the focus, directrix, and axis of symmetry.

Write an equation of the parabola with the given characteristics.

Question 11.
vertex: (0, 0)
directrix: x = 2

Question 12.
focus: (2, 2)
vertex: (2, 6)

Write an equation for the parabola with the given characteristics.

Question 13.
passes through (1, 12) and has vertex (10, -4)

Question 14.
passes through (4, 3) and has x-intercepts of -1 and 5

Question 15.
passes through (-2, 7), (1, 10), and (2, 27)

Question 16.
The table shows the heights y of a dropped object after x seconds. Verify that the data show a quadratic relationship. Write a function that models the data. How long is the object in the air?
Big Ideas Math Answers Algebra 2 Chapter 2 Quadratic Functions 106

Quadratic Functions Chapter Test

Question 1.
A parabola has an axis of symmetry y= 3 and passes through the point (2, 1). Find another point that lies on the graph of the parabola. Explain your reasoning.

Question 2.
Let the graph of g be a translation 2 units left and 1 unit down, followed by a reflection in the y-axis of the graph of f(x) = (2x + 1)2 – 4. Write a rule for g.

Question 3.
Identify the focus, directrix, and axis of symmetry of x = 2y2. Graph the equation.

Question 4.
Explain why a quadratic function models the data. Then use a linear system to find the model.
Big Ideas Math Answers Algebra 2 Chapter 2 Quadratic Functions 107

Write an equation of the parabola. Justify your answer.

Question 5.
Big Ideas Math Answers Algebra 2 Chapter 2 Quadratic Functions 108

Question 6.
Big Ideas Math Answers Algebra 2 Chapter 2 Quadratic Functions 109

Question 7.
Big Ideas Math Answers Algebra 2 Chapter 2 Quadratic Functions 110

Question 8.
A surfboard shop sells 40 surfboards per month when it charges $500 per surfboard. Each time the shop decreases the price by $10, it sells 1 additional surfboard per month. How much should the shop charge per surfboard to maximize the amount of money earned? What is the maximum amount the shop can earn per month? Explain.

Question 9.
Graph f(x) = 8x2 – 4x+ 3. Label the vertex and axis of symmetry. Describe where the function is increasing and decreasing.

Question 10.
Sunfire is a machine with a parabolic cross section used to collect solar energy. The Sun’s rays are reflected from the mirrors toward two boilers located at the focus of the parabola. The boilers produce steam that powers an alternator to produce electricity.
a. Write an equation that represents the cross section of the dish shown with its vertex at (0, 0).
b. What is the depth of Sunfire? Justify your answer.
Big Ideas Math Answers Algebra 2 Chapter 2 Quadratic Functions 111

Question 11.
In 2011, the price of gold reached an all-time high. The table shows the prices (in dollars per troy ounce) of gold each year since 2006 (t = 0 represents 2006). Find a quadratic function that best models the data. Use the model to predict the price of gold in the year 2016.
Big Ideas Math Answers Algebra 2 Chapter 2 Quadratic Functions 112

Quadratic Functions Cumulative Assessment

Question 1.
You and your friend are throwing a football. The parabola shows the path of your friend’s throw, where x is the horizontal distance (in feet) and y is the corresponding height (in feet). The path of your throw can be modeled by h(x) = −16x2 + 65x + 5. Choose the correct inequality symbol to indicate whose throw travels higher. Explain your reasoning.
Big Ideas Math Answers Algebra 2 Chapter 2 Quadratic Functions 113

Question 2.
The function g(x) = \(\frac{1}{2}\)∣x − 4 ∣ + 4 is a combination of transformations of f(x) = | x|. Which combinations describe the transformation from the graph of f to the graph of g?
A. translation 4 units right and vertical shrink by a factor of \(\frac{1}{2}\), followed by a translation 4 units up
B. translation 4 units right and 4 units up, followed by a vertical shrink by a factor of \(\frac{1}{2}\)
C. vertical shrink by a factor of \(\frac{1}{2}\) , followed by a translation 4 units up and 4 units right
D. translation 4 units right and 8 units up, followed by a vertical shrink by a factor of \(\frac{1}{2}\)

Question 3.
Your school decides to sell tickets to a dance in the school cafeteria to raise money. There is no fee to use the cafeteria, but the DJ charges a fee of $750. The table shows the profits (in dollars) when x students attend the dance.
Big Ideas Math Answers Algebra 2 Chapter 2 Quadratic Functions 114
a. What is the cost of a ticket?
b. Your school expects 400 students to attend and finds another DJ who only charges $650. How much should your school charge per ticket to still make the same profit?
c. Your school decides to charge the amount in part (a) and use the less expensive DJ. How much more money will the school raise?

Question 4.
Order the following parabolas from widest to narrowest.
A. focus: (0, −3); directrix: y = 3
B. y = \(\frac{1}{16}\)x2 + 4
C. x = \(\frac{1}{8}\)y2
D. y = \(\frac{1}{4}\)(x − 2)2 + 3

Question 5.
Your friend claims that for g(x) = b, where b is a real number, there is a transformation in the graph that is impossible to notice. Is your friend correct? Explain your reasoning.

Question 6.
Let the graph of g represent a vertical stretch and a reflection in the x-axis, followed by a translation left and down of the graph of f(x) = x2. Use the tiles to write a rule for g.
Big Ideas Math Answers Algebra 2 Chapter 2 Quadratic Functions 115

Question 7.
Two balls are thrown in the air. The path of the first ball is represented in the graph. The second ball is released 1.5 feet higher than the first ball and after 3 seconds reaches its maximum height 5 feet lower than the first ball.
Big Ideas Math Answers Algebra 2 Chapter 2 Quadratic Functions 116
a. Write an equation for the path of the second ball.
b. Do the balls hit the ground at the same time? If so, how long are the balls in the air? If not, which ball hits the ground first? Explain your reasoning.

Question 8.
Let the graph of g be a translation 3 units right of the graph of f. The points (−1, 6), (3, 14), and (6, 41) lie on the graph of f. Which points lie on the graph of g?
A. (2, 6)
B. (2, 11)
C. (6, 14)
D. (6, 19)
E. (9, 41)
F. (9, 46)

Question 9.
Gym A charges $10 per month plus an initiation fee of $100. Gym B charges $30 per month, but due to a special promotion, is not currently charging an initiation fee.
a. Write an equation for each gym modeling the total cost y for a membership lasting x months.
b. When is it more economical for a person to choose Gym A over Gym B?
c. Gym A lowers its initiation fee to $25. Describe the transformation this change represents and how it affects your decision in part (b).