Eureka Math Grade 1 Module 1 Lesson 18 Answer Key

Eureka Math Grade 1 Module 1 Lesson 18 Answer Key

Engage NY Eureka Math 1st Grade Module 1 Lesson 18 Answer Key

Eureka Math Grade 1 Module 1 Lesson 18 Problem Set Answer Key

Question 1.
Add. Color the balloons that match the number in the boy’s mind. Find expressions that are equal. Connect them below with = to make true number sentences.
a. Eureka Math Grade 1 Module 1 Lesson 18 Problem Set Answer Key 1

Answer:
Eureka-Math-Grade-1-Module-1-Lesson-18-Problem-Set-Answer-Key-1

b. Eureka Math Grade 1 Module 1 Lesson 18 Problem Set Answer Key 2

Answer:
Eureka-Math-Grade-1-Module-1-Lesson-18-Problem-Set-Answer-Key-2

Question 2.
Are these number sentences true? Eureka Math Grade 1 Module 1 Lesson 18 Problem Set Answer Key 3

If it is false, rewrite the number sentence to make it true.

a. 3 + 1 = 2 + 2 Eureka Math Grade 1 Module 1 Lesson 18 Problem Set Answer Key 4
____________

Answer:
3 + 1 = 2 + 2  Eureka-Math-Grade-1-Module-1-Lesson-18-Problem-Set-Answer-Key-4
4 = 4

b. 9 + 1 = 1 + 2 Eureka Math Grade 1 Module 1 Lesson 18 Problem Set Answer Key 4
____________

Answer:
9 + 1 = 1 + 2 Eureka-Math-Grade-1-Module-1-Lesson-18-Problem-Set-Answer-Key-4 (1)
9 + 1 = 1 + 2
10 not equal to 3
correct expression is
9 + 1 = 1 + 9 Eureka-Math-Grade-1-Module-1-Lesson-18-Problem-Set-Answer-Key-4
10 = 10

c. 2 + 3 = 1 + 4 Eureka Math Grade 1 Module 1 Lesson 18 Problem Set Answer Key 4
____________

Answer:
2 + 3 = 1 + 4 Eureka-Math-Grade-1-Module-1-Lesson-18-Problem-Set-Answer-Key-4
5 = 5

d. 5 + 1 = 4 + 2 Eureka Math Grade 1 Module 1 Lesson 18 Problem Set Answer Key 4
____________

Answer:
5 + 1 = 4 + 2 Eureka-Math-Grade-1-Module-1-Lesson-18-Problem-Set-Answer-Key-4
6 = 6

e. 4 + 3 = 3 + 5 Eureka Math Grade 1 Module 1 Lesson 18 Problem Set Answer Key 4
____________

Answer:
4 + 3 = 3 + 5 Eureka-Math-Grade-1-Module-1-Lesson-18-Problem-Set-Answer-Key-4 (1)
7 not equal to 8
correct expression is
4 + 3 = 3 + 4 Eureka-Math-Grade-1-Module-1-Lesson-18-Problem-Set-Answer-Key-4
7 = 7

f. 0 + 10 = 2 + 8 Eureka Math Grade 1 Module 1 Lesson 18 Problem Set Answer Key 4
____________

Answer:
0 + 10 = 2 + 8 Eureka-Math-Grade-1-Module-1-Lesson-18-Problem-Set-Answer-Key-4
10 = 10

g. 6 + 3 = 4 + 5 Eureka Math Grade 1 Module 1 Lesson 18 Problem Set Answer Key 4
____________
Answer:
6 + 3 = 4 + 5 Eureka-Math-Grade-1-Module-1-Lesson-18-Problem-Set-Answer-Key-4
9 = 9

h. 3 + 7 = 2 + 6 Eureka Math Grade 1 Module 1 Lesson 18 Problem Set Answer Key 4
____________

Answer:
3 + 7 = 2 + 6 Eureka-Math-Grade-1-Module-1-Lesson-18-Problem-Set-Answer-Key-4 (1)
10 not equal to 8
correct expression is
3 + 7 = 7 + 3 Eureka-Math-Grade-1-Module-1-Lesson-18-Problem-Set-Answer-Key-4
10 = 10

Question 3.
Write a number in the expression and solve. Eureka Math Grade 1 Module 1 Lesson 18 Problem Set Answer Key 5
a. 1 + __ = 3 + 2 Eureka Math Grade 1 Module 1 Lesson 18 Problem Set Answer Key 8

Answer:
1 + 4 = 3 + 2  Eureka-Math-Grade-1-Module-1-Lesson-18-Problem-Set-Answer-Key-4

b. __ + 4 = 2 + 5 Eureka Math Grade 1 Module 1 Lesson 18 Problem Set Answer Key 8

Answer:
1 + 4 = 2 + 5 Eureka-Math-Grade-1-Module-1-Lesson-18-Problem-Set-Answer-Key-4 (1)

c. __ + 5 = 6 + __ Eureka Math Grade 1 Module 1 Lesson 18 Problem Set Answer Key 8

Answer:
2+ 5 = 6 + 1 Eureka-Math-Grade-1-Module-1-Lesson-18-Problem-Set-Answer-Key-4

d. 7 + __ = 8 + __ Eureka Math Grade 1 Module 1 Lesson 18 Problem Set Answer Key 8

Answer:
7 + 1= 8 + 0 Eureka-Math-Grade-1-Module-1-Lesson-18-Problem-Set-Answer-Key-4

Eureka Math Grade 1 Module 1 Lesson 18 Exit Ticket Answer Key

Find two ways to fix each number sentence to make it true.

a. Engage NY Math 1st Grade Module 1 Lesson 18 Exit Ticket Answer Key 9

Answer:
Engage-NY-Math-1st-Grade-Module-1-Lesson-18-Exit-Ticket-Answer-Key-9

b. Engage NY Math 1st Grade Module 1 Lesson 18 Exit Ticket Answer Key 10

Answer:
Engage-NY-Math-1st-Grade-Module-1-Lesson-18-Exit-Ticket-Answer-Key-10

Eureka Math Grade 1 Module 1 Lesson 18 Homework Answer Key

Question 1.
The pictures below are not equal. Make the pictures equal, and write a true number sentence.
Eureka Math 1st Grade Module 1 Lesson 18 Homework Answer Key 11

Answer:
Eureka-Math-1st-Grade-Module-1-Lesson-18-Homework-Answer-Key-11

Question 2.
Circle the true number sentences, and rewrite the false sentences to make them true.
a. Eureka Math 1st Grade Module 1 Lesson 18 Homework Answer Key 12

Answer:
Eureka-Math-1st-Grade-Module-1-Lesson-18-Homework-Answer-Key-12

b. Eureka Math 1st Grade Module 1 Lesson 18 Homework Answer Key 13

Answer:
Eureka-Math-1st-Grade-Module-1-Lesson-18-Homework-Answer-Key-13

c. Eureka Math 1st Grade Module 1 Lesson 18 Homework Answer Key 14

Answer:
Eureka-Math-1st-Grade-Module-1-Lesson-18-Homework-Answer-Key-14

d. Eureka Math 1st Grade Module 1 Lesson 18 Homework Answer Key 15

Answer:
Eureka-Math-1st-Grade-Module-1-Lesson-18-Homework-Answer-Key-15

e. Eureka Math 1st Grade Module 1 Lesson 18 Homework Answer Key 16

Answer:
Eureka-Math-1st-Grade-Module-1-Lesson-18-Homework-Answer-Key-16

f. Eureka Math 1st Grade Module 1 Lesson 18 Homework Answer Key 17

Answer:
Eureka-Math-1st-Grade-Module-1-Lesson-18-Homework-Answer-Key-17

g. Eureka Math 1st Grade Module 1 Lesson 18 Homework Answer Key 18

Answer:
Eureka-Math-1st-Grade-Module-1-Lesson-18-Homework-Answer-Key-18

h. Eureka Math 1st Grade Module 1 Lesson 18 Homework Answer Key 19

Answer:
Eureka-Math-1st-Grade-Module-1-Lesson-18-Homework-Answer-Key-19

i. Eureka Math 1st Grade Module 1 Lesson 18 Homework Answer Key 20

Answer:
Eureka-Math-1st-Grade-Module-1-Lesson-18-Homework-Answer-Key-20

Question 3.
Find the missing part to make the number of sentences true.

a. 8 + 0 = __ + 4

Answer:
8 + 0 = 4 + 4

b. 7 + 2 = 9 + __

Answer:
7 + 2 = 9 + 0

c. 5 + 2 = 4 + __

Answer:
5 + 2 = 4 + 3

d. 5 + __ = 6 + 0

Answer:
5 + 1 = 6 + 0

e. 5 + 4 = __ + 3

Answer:
5 + 4 = 6 + 3

Eureka Math Grade 4 Module 3 Lesson 32 Answer Key

Eureka Math Grade 4 Module 3 Lesson 32 Answer Key

Engage NY Eureka Math 4th Grade Module 3 Lesson 32 Answer Key

Eureka Math Grade 4 Module 3 Lesson 32 Problem Set Answer Key

Solve the following problems. Draw tape diagrams to help you solve. If there is a remainder, shade in a small portion of the tape diagram to represent that portion of the whole.

Question 1.
A concert hall contains 8 sections of seats with the same number of seats in each section. If there are 248 seats, how many seats are in each section?
Answer:
31 seats are there in each section,
Eureka Math Grade 4 Module 3 Lesson 32 Answer Key-1
Explanation:
Given a concert hall of 248 seats has 8 sections,
So the number of seats in each section are 31.

Question 2.
In one day, the bakery made 719 bagels. The bagels were divided into 9 equal shipments. A few bagels were left over and given to the baker. How many bagels did the baker get?
Answer:
The baker got 8 bagels,
Eureka Math Grade 4 Module 3 Lesson 32 Answer Key-2

Explanation:
Given in one day, the bakery made 719 bagels.
The bagels were divided into 9 equal shipments.
There are 9 shipments each to be loaded with equal number of bagels which is 79. After all the bagels were loaded 8 bagels were left which were given to the baker.

Question 3.
The sweet shop has 614 pieces of candy. They packed the candy into bags with 7 pieces in each bag. How many bags of candy did they fill? How many pieces of candy were left?
Answer:
The number of candies left are 5,
Eureka Math Grade 4 Module 3 Lesson 32 Answer Key-3
Explanation:
According to the problem number of candies in each bag are 7,
So number of bags filled are 87. After filling those 87 bags 5 candies are left.

Question 4.
There were 904 children signed up for the relay race. If there were 6 children on each team, how many teams were made?
The remaining children served as referees. How many children served as referees?
Answer:
4 children served as referees,
Eureka Math Grade 4 Module 3 Lesson 32 Answer Key-4

Explanation:
According to the problem number of children in each team is 6.
Number of teams formed are 904 ÷ 6 = 150 teams.
But after formation of those 150 teams 4 children were left who were made referees.

Question 5.
1,188 kilograms of rice are divided into 7 sacks. How many kilograms of rice are in 6 sacks of rice? How many kilograms of rice remain?
Answer:
Each bag has 169 kilograms of rice. So, there are 1,014 kilograms of rice in 6 bags of rice,
Eureka Math Grade 4 Module 3 Lesson 32 Answer Key-5
Explanation:
According to the problem 7 sacs are to be divided with 1,188 kilograms of rice, So each sack of rice gets 169 kilograms of rice and 5 kilograms of rice is left.

Eureka Math Grade 4 Module 3 Lesson 32 Exit Ticket Answer Key

Eureka Math Grade 4 Module 3 Lesson 32 Exit Ticket Answer Key

Solve the following problems. Draw tape diagrams to help you solve. If there is a remainder, shade in a small portion of the tape diagram to represent that portion of the whole.

Question 1.
Mr. Foote needs exactly 6 folders for each fourth-grade student at Hoover Elementary School. If he bought 726 folders, to how many students can he supply folders?
Answer:
Mr. Foote can provide 121 students with folders,
Eureka Math Grade 4 Module 3 Lesson 32 Answer Key-6

Explanation:
Mr. Foote should provide 6 folders to each child since he has 726 folders he can provide 121 students with folders.

Question 2.
Mrs. Terrance has a large bin of 236 crayons. She divides them equally among four containers. How many crayons does Mrs. Terrance have in each container?
Answer:
Number of crayons in each container are 59,
Eureka Math Grade 4 Module 3 Lesson 32 Answer Key-7

Explanation:
According to the problem number of crayons are 236 which have to be distibuted into 4 containers, So number of crayons in each container  are 59.

Eureka Math Grade 4 Module 3 Lesson 32 Homework Answer Key

Solve the following problems. Draw tape diagrams to help you solve. If there is a remainder, shade in a small portion of the tape diagram to represent that portion of the whole.

Question 1.
Meneca bought a package of 435 party favors to give to the guests at her birthday party. She calculated that she could give 9 party favors to each guest. How many guests is she expecting?
Answer:
Number of expected guests are 48 or 49 guests {approx},
Eureka Math Grade 4 Module 3 Lesson 32 Answer Key-8

Explanation:
according to the problem meneca brought 435 party favours.
she expcted she could give each guest 9 party favours. so, we can say she approximately expected 48 to 49 guests.

Question 2.
4,000 pencils were donated to an elementary school. If 8 classrooms shared the pencils equally, how many pencils did each class receive?
Answer:
Each classroom will receive 500 pencils,
Eureka Math Grade 4 Module 3 Lesson 32 Answer Key-9
Explanation:
according to the problem there are 4,000 pecils which have to be shared among 8 classroom, So each classroom will receive 500 pencils.

Question 3.
2,008 kilograms of potatoes were packed into sacks weighing 8 kilograms each. How many sacks were packed?
Answer:
251 sacks are packed weighing 8 kilograms of potatoes each,
Eureka Math Grade 4 Module 3 Lesson 32 Answer Key-10
Explanation:
2,008 kilograms of potatoes should be divided among sacks each having 8 kilograms of potates. So the number sacks packed are 251.

Question 4.
A baker made 7 batches of muffins. There was a total of 252 muffins. If there was the same number of muffins in each batch, how many muffins were in a batch?
Answer:
There are 36 muffins in each batch,
Eureka Math Grade 4 Module 3 Lesson 32 Answer Key-11
Explanation:
According to the problem the baker baked 252 muffins in total,
In each batch he baked equal number of muffins, that is 7 batches.
So each batch had 36 muffins in total.

Question 5.
Samantha ran 3,003 meters in 7 days. If she ran the same distance each day, how far did Samantha run in 3 days?
Answer:
She ran 1,287 meters in 3 days,
Eureka Math Grade 4 Module 3 Lesson 32 Answer Key-12
Explanation:
Samantha ran 3,003 meters in 7 days. So, she ran 429 meters
in 1 day. Number of meters she ran in three days is 1,287.

Eureka Math Grade 4 Module 3 Lesson 32 Fluency Template Answer Key

Eureka Math Grade 4 Module 3 Lesson 32 Fluency Template Answer Key 1
__12______________
shapes
Answer:
A. Trapezoid,
B. Square,
C. Parallelogram,
D. Rectangle,
E. Prism,
F. Tetrahedron,
G. Rhombus,
H. Parallelogram,
I. Rectangle,
J. Trapezium,
K. Parallelogram,
L. Trapezium,

Eureka Math Grade 4 Module 3 Lesson 31 Answer Key

Eureka Math Grade 4 Module 3 Lesson 31 Answer Key

Engage NY Eureka Math 4th Grade Module 3 Lesson 31 Answer Key

Eureka Math Grade 4 Module 3 Lesson 31 Sprint Answer Key

Divide.
Eureka Math Grade 4 Module 3 Lesson 31 Sprint Answer Key 1

Answer:

Eureka Math Grade 4 Module 3 Lesson 31 Answer Key-1

Question 1.
6 ÷ 2 =
Answer:
6 ÷ 2 = 3,

Explanation:
Divided 6 ÷ 2 we get quotient 3.

Question 2.
60 ÷ 2 =
Answer:
60 ÷ 2 = 30,

Explanation:
Divided 60 ÷ 2 we get quotient 30.

Question 3.
600 ÷ 2 =
Answer:
600 ÷ 2 = 300,

Explanation:
Divided 600 ÷ 2 we get quotient 300.

Question 4.
6,000 ÷ 2 =
Answer:
6,000 ÷ 2 = 3,000,

Explanation:
Divided 6,000 ÷ 2 we get quotient 3,000.

Question 5.
9 ÷ 3 =
Answer:
9 ÷ 3 = 3,

Explanation:
Divided 9 ÷ 3 we get quotient 3.

Question 6.
90 ÷ 3 =
Answer:
90 ÷ 3 = 30,

Explanation:
Divided 90 ÷ 3 we get quotient 30.

Question 7.
900 ÷ 3 =
Answer:
900 ÷ 3 = 300,

Explanation:
Divided 900 ÷ 3 we get quotient 3.

Question 8.
9,000 ÷ 3 =
Answer:
9,000 ÷ 3 = 3,000,

Explanation:
Divided 9,000 ÷ 3 we get quotient 3,000.

Question 9.
10 ÷ 5 =
Answer:
10 ÷ 5 = 2,

Explanation:
Divided 10 ÷ 5 we get quotient 2.

Question 10.
15 ÷ 5 =
Answer:
15 ÷ 5 = 3,

Explanation:
Divided 15 ÷ 5 we get quotient 3.

Question 11.
150 ÷ 5 =
Answer:
150 ÷ 5 = 3,

Explanation:
Divided 150 ÷ 5 we get quotient 30.

Question 12.
1,500 ÷ 5 =
Answer:
1,500 ÷ 5 = 300,

Explanation:
Divided 1,500 ÷ 5 we get quotient 300.

Question 13.
2,500 ÷ 5 =
Answer:
2,500 ÷ 5 = 500,

Explanation:
Divided 2,500 ÷ 5 we get quotient 500.

Question 14.
3,500 ÷ 5 =
Answer:
3,500 ÷ 5 = 700,

Explanation:
Divided 3,500 ÷ 5 we get quotient 700.

Question 15.
4,500 ÷ 5 =
Answer:
4,500 ÷ 5 = 900,

Explanation:
Divided 4,500 ÷ 5 we get quotient 900.

Question 16.
450 ÷ 5 =
Answer:
450 ÷ 5 = 90,

Explanation:
Divided 450 ÷ 5 we get quotient 90.

Question 17.
8 ÷ 4 =
Answer:
8 ÷ 4 = 2,

Explanation:
Divided 8 ÷ 4 we get quotient 2.

Question 18.
12 ÷ 4 =
Answer:
12 ÷ 4 = 3,

Explanation:
Divided 12 ÷ 4 we get quotient 3.

Question 19.
120 ÷ 4 =
Answer:
120 ÷ 4 = 30,

Explanation:
Divided 120 ÷ 4 we get quotient 30.

Question 20.
1,200 ÷ 4 =
Answer:
1,200 ÷ 4 = 300,

Explanation:
Divided 1,200 ÷ 4 we get quotient 300.

Question 21.
25 ÷ 5 =
Answer:
25 ÷ 5 = 5,

Explanation:
Divided 25 ÷ 5 we get quotient 5.

Question 22.
30 ÷ 5 =
Answer:
30 ÷ 5 = 6,

Explanation:
Divided 30 ÷ 5 we get quotient 6.

Question 23.
300 ÷ 5 =
Answer:
300 ÷ 5 = 60,

Explanation:
Divided 300 ÷ 5 we get quotient 60.

Question 24.
3,000 ÷ 5 =
Answer:
3,000 ÷ 5 = 600,

Explanation:
Divided 3,000 ÷ 5 we get quotient 600.

Question 25.
16 ÷ 4 =
Answer:
16 ÷ 4 = 4,

Explanation:
Divided 16 ÷ 4 we get quotient 4.

Question 26.
160 ÷ 4 =
Answer:
160 ÷ 4 = 40,

Explanation:
Divided 160 ÷ 4 we get quotient 40.

Question 27.
18 ÷ 6 =
Answer:
18 ÷ 6 = 3,

Explanation:
Divided 18 ÷ 6 we get quotient 3.

Question 28.
1,800 ÷ 6 =
Answer:
1,800 ÷ 6 = 300,

Explanation:
Divided 1,800 ÷ 6 we get quotient 300.

Question 29.
28 ÷ 7 =
Answer:
28 ÷ 7 = 4,

Explanation:
Divided 28 ÷ 7 we get quotient 4.

Question 30.
280 ÷ 7 =
Answer:
280 ÷ 7 = 40,

Explanation:
Divided 280 ÷ 7 we get quotient 40.

Question 31.
48 ÷ 8 =
Answer:
48 ÷ 8 = 6,

Explanation:
Divided 48 ÷ 8 we get quotient 6.

Question 32.
4,800 ÷ 8 =
Answer:
4,800 ÷ 8 = 600,

Explanation:
Divided 4,800 ÷ 8 we get quotient 600.

Question 33.
6,300 ÷ 9 =
Answer:
6,300 ÷ 9 = 7,

Explanation:
Divided 6,300 ÷ 9 we get quotient 7.

Question 34.
200 ÷ 5 =
Answer:
200 ÷ 5 = 40,

Explanation:
Divided 200 ÷ 5 we get quotient 40.

Question 35.
560 ÷ 7 =
Answer:
560 ÷ 7 = 80,

Explanation:
Divided 560 ÷ 7 we get quotient 80.

Question 36.
7,200 ÷ 9 =
Answer:
7,200 ÷ 9 = 800,

Explanation:
Divided 7,200 ÷ 9 we get quotient 800.

Question 37.
480 ÷ 6 =
Answer:
480 ÷ 6 = 80,

Explanation:
Divided 480 ÷ 6 we get quotient 80.

Question 38.
5,600 ÷ 8 =
Answer:
5,600 ÷ 8 = 700,

Explanation:
Divided 5,600 ÷ 8 we get quotient 700.

Question 39.
400 ÷ 5 =
Answer:
400 ÷ 5 = 80,

Explanation:
Divided 400 ÷ 5 we get quotient 80.

Question 40.
6,300 ÷ 7 =
Answer:
6,300 ÷ 7 = 900,

Explanation:
Divided 6,300 ÷ 7 we get quotient 900.

Question 41.
810 ÷ 9 =
Answer:
810 ÷ 9 = 90,

Explanation:
Divided 810 ÷ 9 we get quotient 90.

Question 42.
640 ÷ 8 =
Answer:
640 ÷ 8 = 80,

Explanation:
Divided 640 ÷ 8 we get quotient 80.

Question 43.
5,400 ÷ 6 =
Answer:
5,400 ÷ 6 = 900,

Explanation:
Divided 5,400 ÷ 6 we get quotient 900.

Question 44.
4,000 ÷ 5 =
Answer:
4,000 ÷ 5 = 800,

Explanation:
Divided 4,000 ÷ 5 we get quotient 800.

Divide.
Eureka Math Grade 4 Module 3 Lesson 31 Sprint Answer Key 2

Answer:

Eureka Math Grade 4 Module 3 Lesson 31 Answer Key-2

Question 1.
4 ÷ 2 =
Answer:
4 ÷ 2 = 2,

Explanation:
Divided 4 ÷ 2 we get quotient 2.

Question 2.
40 ÷ 2 =
Answer:
40 ÷ 2 = 20,

Explanation:
Divided 40 ÷ 2 we get quotient 20.

Question 3.
400 ÷ 2 =
Answer:
400 ÷ 2 = 200,

Explanation:
Divided 400 ÷ 2 we get quotient 200.

Question 4.
4,000 ÷ 2 =
Answer:
4,000 ÷ 2 = 2,000,

Explanation:
Divided 4,000 ÷ 2 we get quotient 2,000.

Question 5.
6 ÷ 3 =
Answer:
6 ÷ 3 = 2,

Explanation:
Divided 6 ÷ 3 we get quotient 2.

Question 6.
60 ÷ 3 =
Answer:
60 ÷ 3 = 20,

Explanation:
Divided 60 ÷ 3 we get quotient 20.

Question 7.
600 ÷ 3 =
Answer:
600 ÷ 3 = 200,

Explanation:
Divided 600 ÷ 3 we get quotient 200.

Question 8.
6,000 ÷ 3 =
Answer:
6,000 ÷ 3 = 2,000,

Explanation:
Divided 6,000 ÷ 3 we get quotient 2,000.

Question 9.
10 ÷ 5 =
Answer:
10 ÷ 5 = 2,

Explanation:
Divided 10 ÷ 5 we get quotient 2.

Question 10.
15 ÷ 5 =
Answer:
15 ÷ 5 = 3,

Explanation:
Divided 15 ÷ 5 we get quotient 3.

Question 11.
150 ÷ 5 =
Answer:
150 ÷ 5 = 30,

Explanation:
Divided 150 ÷ 5 we get quotient 30.

Question 12.
250 ÷ 5 =
Answer:
250 ÷ 5 = 50,

Explanation:
Divided 250 ÷ 5 we get quotient 50.

Question 13.
350 ÷ 5 =
Answer:
350 ÷ 5 = 70,

Explanation:
Divided 350 ÷ 5 we get quotient 70.

Question 14.
3,500 ÷ 5 =
Answer:
3,500 ÷ 5 = 700,

Explanation:
Divided 3,500 ÷ 5 we get quotient 700.

Question 15.
4,500 ÷ 5 =
Answer:
4,500 ÷ 5 = 900,

Explanation:
Divided 4,500 ÷ 5 we get quotient 900.

Question 16.
450 ÷ 5 =
Answer:
450 ÷ 5 = 90,

Explanation:
Divided 450 ÷ 5 we get quotient 90.

Question 17.
9 ÷ 3 =
Answer:
9 ÷ 3 = 3,

Explanation:
Divided 9 ÷ 3 we get quotient 3.

Question 18.
12 ÷ 3 =
Answer:
12 ÷ 3 = 4,

Explanation:
Divided 12 ÷ 3 we get quotient 4.

Question 19.
120 ÷ 3 =
Answer:
120 ÷ 3 = 40,

Explanation:
Divided 120 ÷ 3 we get quotient 40.

Question 20.
1,200 ÷ 3 =
Answer:
1,200 ÷ 3 = 400,

Explanation:
Divided 1,200 ÷ 3 we get quotient 400.

Question 21.
25 ÷ 5 =
Answer:
25 ÷ 5 = 5,

Explanation:
Divided 25 ÷ 5 we get quotient 5.

Question 22.
20 ÷ 5 =
Answer:
20 ÷ 5 = 4,

Explanation:
Divided 20 ÷ 5 we get quotient 4.

Question 23.
200 ÷ 5 =
Answer:
200 ÷ 5 = 40,

Explanation:
Divided 200 ÷ 5 we get quotient 40.

Question 24.
2,000 ÷ 5 =
Answer:
2,000 ÷ 5 = 400,

Explanation:
Divided 2,000 ÷ 5 we get quotient 400.

Question 25.
12 ÷ 4 =
Answer:
12 ÷ 4 = 3,

Explanation:
Divided 12 ÷ 4 we get quotient 3.

Question 26.
120 ÷ 4 =
Answer:
120 ÷ 4 = 30,

Explanation:
Divided 120 ÷ 4 we get quotient 30.

Question 27.
21 ÷ 7 =
Answer:
21 ÷ 7 = 3,

Explanation:
Divided 21 ÷ 7 we get quotient 3.

Question 28.
2,100 ÷ 7 =
Answer:
2,100 ÷ 7 = 300,

Explanation:
Divided 2,100 ÷ 7 we get quotient 300.

Question 29.
18 ÷ 6 =
Answer:
18 ÷ 6 = 3,

Explanation:
Divided 18 ÷ 6 we get quotient 3.

Question 30.
180 ÷ 6 =
Answer:
180 ÷ 6 = 30,

Explanation:
Divided 180 ÷ 6 we get quotient 30.

Question 31.
54 ÷ 9 =
Answer:
54 ÷ 9 = 6,

Explanation:
Divided 54 ÷ 9 we get quotient 6.

Question 32.
5,400 ÷ 9 =
Answer:
5,400 ÷ 9 = 600,

Explanation:
Divided 5,400 ÷ 9 we get quotient 600.

Question 33.
5,600 ÷ 8 =
Answer:
5,600 ÷ 8 = 700,

Explanation:
Divided 5,600 ÷ 8 we get quotient 700.

Question 34.
300 ÷ 5 =
Answer:
300 ÷ 5 = 60,

Explanation:
Divided 300 ÷ 5 we get quotient 60.

Question 35.
490 ÷ 7 =
Answer:
490 ÷ 7 = 70,

Explanation:
Divided 490 ÷ 7 we get quotient 70.

Question 36.
6,300 ÷ 9 =
Answer:
6,300 ÷ 9 = 700,

Explanation:
Divided 6,300 ÷ 9 we get quotient 700.

Question 37.
420 ÷ 6 =
Answer:
420 ÷ 6 = 70,

Explanation:
Divided 420 ÷ 6 we get quotient 70.

Question 38.
4,800 ÷ 8 =
Answer:
4,800 ÷ 8 = 600,

Explanation:
Divided 4,800 ÷ 8 we get quotient 600.

Question 39.
4,000 ÷ 5 =
Answer:
4,000 ÷ 5 = 800,

Explanation:
Divided 4,000 ÷ 5 we get quotient 800.

Question 40.
560 ÷ 8 =
Answer:
560 ÷ 8 = 70,

Explanation:
Divided 560 ÷ 8 we get quotient 70.

Question 41.
6,400 ÷ 8 =
Answer:
6,400 ÷ 8 = 800,

Explanation:
Divided 6,400 ÷ 8 we get quotient 800.

Question 42.
720 ÷ 8 =
Answer:
720 ÷ 8 = 90,

Explanation:
Divided 720 ÷ 8 we get quotient 90.

Question 43.
4,800 ÷ 6 =
Answer:
4,800 ÷ 6 = 800,

Explanation:
Divided 4,800 ÷ 6 we get quotient 800.

Question 44.
400 ÷ 5 =
Answer:
400 ÷ 5 = 80,

Explanation:
Divided 400 ÷ 5 we get quotient 80.

Eureka Math Grade 4 Module 3 Lesson 31 Problem Set Answer Key

Draw a tape diagram and solve. The first two tape diagrams have been drawn for you. Identify if the group size or the number of groups is unknown.

Question 1.
Monique needs exactly 4 plates on each table for the banquet. If she has 312 plates, how many tables is she able to prepare?
Eureka Math Grade 4 Module 3 Lesson 31 Problem Set Answer Key 3
Answer:
Monique is able to prepare 78 tables,
The number of groups is unknown,

Explanation:
Given Monique needs exactly 4 plates on each table for the banquet.
If she has 312 plates, number of tables is she able to prepare is
312 plates ÷ 4=
   78   
4|312
   28
   032
032
0
Therefore, Monique is able to prepare 78 tables, the number of groups is unknown.

Question 2.
2,365 books were donated to an elementary school. If 5 classrooms shared the books equally, how many books did each class receive?
Eureka Math Grade 4 Module 3 Lesson 31 Problem Set Answer Key 4
Answer:
Each class will receive 473 books,
The group size is unknown,

Explanation:
Given 2,365 books were donated to an elementary school.
If 5 classrooms shared the books equally,
Number of books did each class will receive is 2,365 ÷ 5 =
   473   
5|2,365
   20
     36
     35
       15
       15
 0
Therefore, Each class will receive 473 books,
The group size is unknown.

Question 3.
If 1,503 kilograms of rice was packed in sacks weighing 3 kilograms each, how many sacks were packed?
Answer:
501 sacks were packed,
The number of groups is unknown,

Explanation:
Eureka Math Grade 4 Module 3 Lesson 31 Answer Key-3
Given
If 1,503 kilograms of rice was packed in sacks weighing 3 kilograms each, number of sacks were packed are 1,503 ÷ 3 = 501
as shown above, The number of groups is unknown.

Question 4.
Rita made 5 batches of cookies. There was a total of 2,400 cookies. If each batch contained the same number of cookies, how many cookies were in 4 batches?
Answer:
There were 1,920 cookies in 4 batches,
the group size is unknown,

Explanation:
Eureka Math Grade 4 Module 3 Lesson 31 Answer Key-4
Given Rita made 5 batches of cookies.
There was a total of 2,400 cookies.
If each batch contained the same number of cookies,
So number of cookies were in 4 batches are 1,920,
the group size is unknown.

Question 5.
Every day, Sarah drives the same distance to work and back home. If Sarah drove 1,005 miles in 5 days, how far did Sarah drive in 3 days?
Answer:
Sarah drove 603 miles in 3 days,
the group size is unknown,

Explanation:
Eureka Math Grade 4 Module 3 Lesson 31 Answer Key-5
Given every day, Sarah drives the same distance to work and back home.
If Sarah drove 1,005 miles in 5 days, So Sarah drive in 3 days is 603 miles,
the group size is unknown.

Eureka Math Grade 4 Module 3 Lesson Exit Ticket Answer Key

Solve the following problems. Draw tape diagrams to help you solve. Identify if the group size or the number of groups is unknown.

Question 1.
572 cars were parked in a parking garage. The same number of cars was parked on each floor. If there were 4 floors, how many cars were parked on each floor?
Answer:
Number of cars were parked on each floor is 143 cars,
the group size is unknown,

Explanation:
Eureka Math Grade 4 Module 3 Lesson 31 Answer Key-6
Given 572 cars were parked in a parking garage.
The same number of carscwas parked on each floor.
If there were 4 floors, number of cars were parked on each floor are 143 cars as shown above,the group size is unknown.

Question 2.
356 kilograms of flour were packed into sacks holding 2 kilograms each. How many sacks were packed?
Answer:
178 sacks were packed, the number of groups is unknown,

Explanation:
Eureka Math Grade 4 Module 3 Lesson 31 Answer Key-7
Given 356 kilograms of flour were packed into sacks holding 2 kilograms each. Number of sacks were packed are 356 ÷ 2 = 178 sacks,
the number of groups is unknown.

Eureka Math Grade 4 Module 3 Lesson Homework Answer Key

Solve the following problems. Draw tape diagrams to help you solve. Identify if the group size or the number of groups is unknown.

Question 1.
500 milliliters of juice was shared equally by 4 children. How many milliliters of juice did each child get?
Answer:
Each child will get 125 milliliters of juice, the group size is unknown,
Eureka Math Grade 4 Module 3 Lesson 31 Answer Key-8

Explanation:
Given 500 milliliters of juice was shared equally by 4 children.
So number of milliliters of juice did each child will get is
125 milliliters as shown above, the group size is unknown.

Question 2.
Kelly separated 618 cookies into baggies. Each baggie contained 3 cookies. How many baggies of cookies did Kelly make?
Answer:
Kelly will make 206 baggies of cookies, the number of groups is unknown,
Eureka Math Grade 4 Module 3 Lesson 31 Answer Key-9

Explanation:
Given Kelly separated 618 cookies into baggies.
Each baggie contained 3 cookies.
Number of baggies of cookies did Kelly make are 206 baggies of cookies, the number of groups is unknown.

Question 3.
Jeff biked the same distance each day for 5 days. If he traveled 350 miles altogether, how many miles did he travel each day?
Answer:
Each day Jeff traveled 70 miles, the group size is unknown,
Eureka Math Grade 4 Module 3 Lesson 31 Answer Key-10

Explanation:
Given Jeff biked the same distance each day for 5 days. If he traveled 350 miles altogether, Number of miles did Jeff traveled each day is 70 miles, the group size is unknown.

Question 4.
A piece of ribbon 876 inches long was cut by a machine into 4-inch long strips to be made into bows. How many strips were cut?
Answer:
219 strips were cut, the number of groups is unknown,
Eureka Math Grade 4 Module 3 Lesson 31 Answer Key-11
Explanation:
Given a piece of ribbon 876 inches long was cut by a machine into 4-inch long strips to be made into bows. Number of strips were cut are 219 the number of groups is unknown.

Question 5.
Five Martians equally share 1,940 Groblarx fruits. How many Groblarx fruits will 3 of the Martians receive?
Answer:
1,164 Groblarx fruits will 3 of the Martians receive,
the group size is unknown,
Eureka Math Grade 4 Module 3 Lesson 31 Answer Key-12
Explanation:
Given five Martians equally share 1,940 Groblarx fruits.
So Groblarx fruits will 3 of the Martians receive are 1,164 as shown above, the group size is unknown.

Eureka Math Grade 2 Module 4 Lesson 11 Answer Key

Eureka Math Grade 2 Module 4 Lesson 11 Answer Key

Place value charts are the important topics in maths. This will be helpful in the real-time environment. Know-how and where to apply the formulas from this page. Get a detailed explanation for all the questions here. Download Eureka Math Answers Grade 2 chapter 11 pdf for free of cost. As per your convenience, we have provided the solutions in pdf format so that you can prepare offline.

Engage NY Eureka Math 2nd Grade Module 4 Lesson 11 Answer Key

Get the guided notes for the chapter 11 Answer Key from here. This will be the best resource to enhance your math skills. The topics covered in this chapter are Place value charts. test yourself by solving questions given at the end of the chapter.

Eureka Math Grade 2 Module 4 Lesson 11 Problem Set Answer Key

Question 1.
Solve using mental math.
a. 8 – 7 = ___1__                 38 – 7 = __31___                 38 – 8 = __30____                 38 – 9 = ___29___

Answer:
8 – 7 = 1.
38 – 7 = 31.
38 – 8 = 30.
38 – 9 = 29.

Explanation:
In the above-given question,
given that,
solve using mental math.
8 – 7 = 1.
38 – 7 = 31.
38 – 8 = 30.
38 – 9 = 29.

b. 7 – 6 = _1____                 87 – 6 = __81___                 87 – 7 = ___80___                 87 – 8 = __79____

Answer:
7 – 6 = 1.
87 – 6 = 81.
87 – 7 = 80.
87 – 8 = 79.

Explanation:
In the above-given question,
given that,
solve using mental math.
7 – 6 = 1.
87 – 6 = 81.
87 – 7 = 80.
87 – 8 = 79.

Question 2.
Solve using your place value chart and place value disks. Unbundle a ten if needed. Think about which problems you can solve mentally, too!
a. 28 – 7 = __21___                  28 – 9 = ___19___

Answer:
28 – 7 = 21.
28 – 9 = 19.

Explanation:
In the above-given question,
given that,
solve using place value chart and place value disks.
28 – 7 = 21.
28 – 9 = 19.
Eureka-Math-Grade-2-Module-4-Lesson -11- Answer Key-1

b. 25 – 5 = __20___                  25 – 6 = ___19___

Answer:
25 – 5 = 20.
25 – 6 = 19.

Explanation:
In the above-given question,
given that,
solve using place value chart and place value disks.
25 – 5 = 20.
25 – 6 = 19.
Eureka-Math-Grade-2-Module-4-Lesson -11- Answer Key-2

c. 30 – 5 = __25___                  33 – 5 = __28____

Answer:
30 – 5 = 25.
33 – 5 = 28.

Explanation:
In the above-given question,
given that,
solve using place value chart and place value disks.
30 – 5 = 25.
33 – 5 = 28.

d. 47 – 22 = _25____                41 – 22 = __19___

Answer:
47 – 22 = 25.
41 – 22 = 19.

Explanation:
In the above-given question,
given that,
solve using place value chart and place value disks.
47 – 22 = 25.
41 – 22 = 19.
Eureka-Math-Grade-2-Module-4-Lesson -11- Answer Key-3

e. 44 – 16 = __28___                44 – 26 = __18___

Answer:
44 – 16 = 28.
44 – 26 = 18.

Explanation:
In the above-given question,
given that,
solve using place value chart and place value disks.
44 – 16 = 28.
44 – 26 = 18.
Eureka-Math-Grade-2-Module-4-Lesson -11- Answer Key-4

f. 70 – 28 = __42___                80 – 28 = __52___

Answer:
70 – 28 = 42.
80 – 28 = 52.

Explanation:
In the above-given question,
given that,
solve using place value chart and place value disks.
70 – 28 = 42.
80 – 28 = 52.
Eureka-Math-Grade-2-Module-4-Lesson -11- Answer Key-5

Question 3.
Solve 56 – 28, and explain your strategy.
Eureka Math Grade 2 Module 4 Lesson 11 Problem Set Answer Key 1

Answer:
56 – 28 = 28.

Explanation:
In the above-given question,
given that,
solve using place value chart and place value disks.
56 – 28 = 28.
Eureka-Math-Grade-2-Module-4-Lesson -11- Answer Key-6

For early finishers:

Question 4.
There are 63 problems on the math test. Tamara answered 48 problems correctly, but the rest were incorrect. How many problems did she answer incorrectly?

Answer:
The number of problems did she answer incorrectly = 15.

Explanation:
In the above-given question,
given that,
There are 63 problems on the math test.
Tamara answered 48 problems correctly, but the rest were incorrect.
63 – 48 = 15.
so the number of problems did she answer incorrectly = 15.
Question 5.
Mr. Ross has 7 fewer students than Mrs. Jordan. Mr. Ross has 35 students. How many students does Mrs. Jordan have?

Answer:
The number of students Mrs. Jordan have = 42.

Explanation:
In the above-given question,
given that,
Mr. Ross has 7 fewer students than Mrs. Jordan.
Mr. Ross has 35 students.
35 + 7 = 42.
so the number of students Mrs. Jordan have = 42.

Eureka Math Grade 2 Module 4 Lesson 11 Exit Ticket Answer Key

Eureka Math Grade 2 Module 4 Lesson 11 Exit Ticket Answer Key

Solve for the missing part. Use your place value chart and place value disks.

Question 1.
Eureka Math Grade 2 Module 4 Lesson 11 Exit Ticket Answer Key 2

Answer:
71 – 56 = 15.

Explanation:
In the above-given question,
given that,
solve using place value chart and place value disks.
56 + 15 = 71.
71 – 15 = 56.
Eureka-Math-Grade-2-Module-4-Lesson -11- Answer Key-7

Question 2.
Eureka Math Grade 2 Module 4 Lesson 11 Exit Ticket Answer Key 3

Answer:
84 – 38 = 46.

Explanation:
In the above-given question,
given that,
solve using place value chart and place value disks.
46 + 38 = 84.
84 – 38 = 46.
Eureka-Math-Grade-2-Module-4-Lesson -11- Answer Key-8

Eureka Math Grade 2 Module 4 Lesson 11 Homework Answer Key

Question 1.
Solve using mental math.
a. 6 – 5 = ___1__                 26 – 5 = _21____                 26 – 6 = ___20___                 26 – 7 = __19____

Answer:
6 – 5 = 1.
26 – 5 = 21.
26 – 6 = 20.
26 – 7 = 19.

Explanation:
In the above-given question,
given that,
solve using mental math.
6 – 5 = 1.
26 – 5 = 21.
26 – 6 = 20.
26 – 7 = 19.

b. 8 – 7 = ___1__                 58 – 7 = __51___                 58 – 8 = __50____                 58 – 9 = __49____

Answer:
8 – 7 = 1.
58 – 7 = 51.
58 – 8 = 50.
58 – 9 = 49.

Explanation:
In the above-given question,
given that,
solve using mental math.
8 – 7 = 1.
58 – 7 = 51.
58 – 8 = 50.
58 – 9 = 49.

Question 2.
Solve using your place value chart and place value disks. Unbundle a ten, if needed. Think about which problems you can solve mentally, too!
a. 36 – 5 = __31___                 36 – 7 = __29___

Answer:
36 – 5 = 31.
36 – 7 = 29.

Explanation:
In the above-given question,
given that,
solve using place value chart and place value disks.
36 – 5 = 31.
36 – 7 = 29.
Eureka-Math-Grade-2-Module-4-Lesson -11- Answer Key-9

b. 37 – 6 = ___31__                 37 – 8 = ___29__

Answer:
37 – 6 = 31.
37 – 8 = 29.

Explanation:
In the above-given question,
given that,
solve using place value chart and place value disks.
37 – 6 = 31.
37 – 8 = 29.
Eureka-Math-Grade-2-Module-4-Lesson -11- Answer Key-10

c. 40 – 5 = __35___                 41 – 5 = __36___

Answer:
40 – 5 = 35.
41 – 5 = 36.

Explanation:
In the above-given question,
given that,
solve using place value chart and place value disks.
40 – 5 = 35.
41 – 5 = 36.
Eureka-Math-Grade-2-Module-4-Lesson -11- Answer Key-11

d. 58 – 32 = __26___               58 – 29 = __29___

Answer:
58 – 32 = 26.
58 – 29 = 29.

Explanation:
In the above-given question,
given that,
solve using place value chart and place value disks.
58 – 32 = 26.
58 – 29 = 29.

e. 60 – 26 = __34___                62 – 26 = _36____

Answer:
60 – 26 = 34.
62 – 26 = 36.

Explanation:
In the above-given question,
given that,
solve using place value chart and place value disks.
60 – 26 = 34.
62 – 26 = 36.

f. 70 – 41 = __29___ 80 – 41 = __39___

Answer:
70 – 41 = 29.
80 – 41 = 39.

Explanation:
In the above-given question,
given that,
solve using place value chart and place value disks.
70 – 41 = 29.
80 – 41 = 39.

Question 3.
Solve and explain your strategy.
a. 41 – 27 = __14___

Answer:
41 – 27 = 14.

Explanation:
In the above-given question,
given that,
solve using mental math.
41 – 27 = 14.
14 + 27 = 41.

b. 67 – 28 = __39___

Answer:
67 – 28 = 39.

Explanation:
In the above-given question,
given that,
solve using mental math.
67 – 28 = 39.
39 + 28 = 67.

Question 4.
The number of marbles in each jar is marked on the front. Miss Clark took 37 marbles out of each jar. How many marbles are left in each jar? Complete the number sentence to find out.
a.
Eureka Math Grade 2 Module 4 Lesson 11 Homework Answer Key 4
___45__ – _37____ = __8___

Answer:
45 – 37 = 8.

Explanation:
In the above-given question,
given that,
The number of marbles in each jar is marked on the front.
Miss Clark took 37 marbles out of each jar.
45 – 37 = 8.

b.
Eureka Math Grade 2 Module 4 Lesson 11 Homework Answer Key 5
___52__ – __37___ = __15___

Answer:
52 -37 = 15.

Explanation:
In the above-given question,
given that,
The number of marbles in each jar is marked on the front.
Miss Clark took 37 marbles out of each jar.
52 – 37 = 15.

c.
Eureka Math Grade 2 Module 4 Lesson 11 Homework Answer Key 6
__48___ – __37___ = ___11__

Answer:
48 – 37 = 11.

Explanation:
In the above-given question,
given that,
The number of marbles in each jar is marked on the front.
Miss Clark took 37 marbles out of each jar.
48 – 37 = 11.

d.
Eureka Math Grade 2 Module 4 Lesson 11 Homework Answer Key 7
___55__ – __37___ = __18___

Answer:
55 – 37 = 18.

Explanation:
In the above-given question,
given that,
The number of marbles in each jar is marked on the front.
Miss Clark took 37 marbles out of each jar.
55 – 37 = 18.

Big Ideas Math Answers Grade 5 Chapter 10 Divide Fractions

Big Ideas Math Answers Grade 5 Chapter 10 Divide Fractions

Download the Big Ideas Math Book Answer Key Grade 5 Chapter 10 Divide Fractions free of cost and kick start your preparation immediately. You will get the necessary skill set needed to solve the problems related to fractions here. Access Detailed Solutions to all the problems and learn how to solve related problems when you encounter them during your exams. Seek Homework Help needed by accessing the Big Ideas Math Answers Grade 5 Chapter 10 Divide Fractions.

Big Ideas Math Book Answer Key Grade 5 Chapter 10 Divide Fractions

Cross Check the Solutions from our Big Ideas Math Answers Grade 5 Chapter 10 Divide Fractions and understand the areas you are facing difficulty. Score higher grades in your exams and refer to Big Ideas Math Book Solution Key Grade 5 Chapter 10 Divide Fractions to have strong command over fractions.

Lesson: 1 Interpret Fractions as Division

Lesson: 1 Interpret Fractions as Division

Lesson: 2 Mixed Numbers as Quotients

Lesson: 2 Mixed Numbers as Quotients

Lesson: 3 Divide Whole Numbers by Unit Fractions

Divide Whole Numbers by Unit Fractions

Lesson: 4 Divide Unit Fractions by Whole Numbers

Lesson: 5 Problem Solving: Fraction Division

Chapter: 10 – Divide Fractions

Lesson 10.1 Interpret Fractions as Division

Explore and Grow`

You share 4 sheets of construction paper equally among 8 people. Write a division expression that represents the situation. What fraction of a sheet of paper does each person get? Use a model to support your answer?
Answer:
The division expression that represents the fraction of a sheet of paper does each person get is:
4 ÷ 8 = \(\frac{1}{2}\)

Explanation:
It is given that you have 4 sheets of construction paper equally among 8 people.
Hence,
The division expression that represents the fraction of a sheet of paper is:
( The number of sheets of construction paper ) ÷ ( The number of people )
= 4 ÷ 8
= \(\frac{1}{2}\)
Hence, from the above,
We can conclude that the fraction of a sheet of paper does each person get is: \(\frac{1}{2}\)

Structure
How can you check your answer using multiplication?
Answer: We can check the answer by using the partial products method or by using the simplification method.

Think and Grow: Divide Whole Numbers
You can use models to divide whole numbers that have a fraction as the quotient.
Answer: 
From the above model,
The number of colored parts is: 4
The total number of parts are: 8
So,
The fraction of the colored part out of the total number of parts = 4 ÷ 8
= \(\frac{4}{8}\) = \(\frac{1}{2}\)
In \(\frac{1}{2}\),
1 represents the quotient
2 represents the remainder
Example
Find 2 ÷ 3.
One Way: Use a tape diagram. Show 2 wholes. Divide each whole into 3 equal parts.

Another Way: Use an area model. Show 2 wholes. Divide each whole into 3 equal parts. Then separate the parts into 3 equal groups.

Show and Grow

Divide. Use a model to help

Question 1.
2 ÷ 4 =0.5
Answer:
From the above model,
The number of colored parts is: 2
The number of total parts is: 4
So,
The fraction of the colored parts out of the total number of parts = 2 ÷ 4
= \(\frac{2}{4}\)
= \(\frac{1}{2}\)
So,
The fraction of the colored parts out of the total number of parts in the decimal form is: 0.5

Question 2.
1 ÷ 3 = 0.33
Answer:

From the above model,
The number of colored parts is: 1
The number of total parts is: 3
So,
The fraction of the colored parts out of the total number of parts = 1 ÷ 3
= \(\frac{1}{3}\)
So,
The fraction of the colored parts out of the total number of parts in the decimal form is: 0.33

Apply and Grow: Practice

Divide. Use a model to help.

Question 3.
1 ÷ 8 =0.018
Answer:
From the above model,
The number of colored parts is: 1
The number of total parts is: 8
So,
The fraction of the colored parts out of the total number of parts = 1 ÷ 8
= \(\frac{1}{8}\)
So,
The fraction of the colored parts out of the total number of parts in the decimal form is: 0.018

Question 4.
1 ÷ 4 =0.25
Answer:
From the above model,
The number of colored parts is: 1
The number of total parts is: 4
So,
The fraction of the colored parts out of the total number of parts = 1 ÷ 4
= \(\frac{1}{4}\)
So,
The fraction of the colored parts out of the total number of parts in the decimal form is: 0.25

Question 5.
2 ÷ 6 =0.33
Answer:
From the above model,
The number of colored parts is: 2
The number of total parts is: 6
So,
The fraction of the colored parts out of the total number of parts = 2 ÷ 6
= \(\frac{2}{6}\)
= \(\frac{1}{3}\)
So,
The fraction of the colored parts out of the total number of parts in the decimal form is: 0.33

Question 6.
2 ÷ 5 = 0.4
Answer:
From the above model,
The number of colored parts is: 2
The number of total parts is: 5
So,
The fraction of the colored parts out of the total number of parts = 2 ÷ 5
= \(\frac{2}{5}\)
So,
The fraction of the colored parts out of the total number of parts in the decimal form is: 0.4

Question 7.
3 ÷ 7 = 0.42
Answer: 
From the above model,
The number of colored parts is: 3
The number of total parts is: 7
So,
The fraction of the colored parts out of the total number of parts = 3 ÷ 7
= \(\frac{3}{7}\)
So,
The fraction of the colored parts out of the total number of parts in the decimal form is: 0.42

Question 8.
5 ÷ 6 = 0.83
Answer:
From the above model,
The number of colored parts is: 5
The number of total parts is: 6
So,
The fraction of the colored parts out of the total number of parts = 5 ÷ 6
= \(\frac{5}{6}\)
So,
The fraction of the colored parts out of the total number of parts in the decimal form is: 0.83

Question 9.
How many 6s are in 1?
Answer: There are six \(\frac{1}{6}\)s in 1

Explanation:
The number of 6s in 1 can be obtained by dividing 1 into 6 equal parts.
So,
The figure obtained will be like;

From the above model,
The number of colored parts is: 1
The number of total parts is: 6
So,
The fraction of the colored parts out of the total number of parts = 1 ÷ 6
= \(\frac{1}{6}\)
Hence, from the above,
We can conclude that there are six 6s in 1

Question 10.
How many 10s are in 9?
Answer: There are 9 \(\frac{9}{10}\)s in 9

Explanation:
The model for the number of 10s in 9 are:

From the above model,
The number of colored parts is: 9
The number of total parts is: 10
So,
The fraction of the colored parts out of the total number of parts = 9 ÷ 10
= \(\frac{9}{10}\)
Hence, from the above,
We can conclude that there are nine 9s in 10

Question 11.
Number Sense
For which equations does k = 8?
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 3
Answer: Let the equations named A), B), C), and D)
So,
The four equations are:
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 3
From the above equations,
The value ‘k’ must be in the numerator.
So,
In A), the value of the numerator is: 3
In B), the value of the numerator is: k
In C), the value of the numerator is: 2
In D) the value of the numerator is: 8
So,
From the above numerator values,
We can say that “k=8” holds good for Equation B)

Question 12.
Writing
Write and solve a real-life problem for 7 ÷ 12.
Answer:
From the above model,
The number of colored parts is: 7
The number of total parts is: 12
So,
The fraction of the colored parts out of the total number of parts = 7 ÷ 12
= \(\frac{7}{12}\)
Hence,
The fraction of the colored parts out of the total number of parts in the decimal form is: 0.58

Think and Grow: Modeling Real Life

Example
Three fruit bars are shared equally among 4 friends. What fraction of a fruit bar does each friend get?
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 4
Divide 3 by 4 to find what fraction of a fruit bar each friend gets.
Use an area model to find 3 ÷ 4. Show 3 whole fruit bars. Divide each fruit bar into 4 equal parts. Then separate the parts into 4 equal groups.

Show and Grow

Question 13.
You cut a 5-foot streamer into 6 pieces of equal size. What is the length of each piece in feet? in inches?
Answer: The length of each piece in feet is: \(\frac{5}{6}\)

Explanation:
It is given that you cut a 5-foot streamer into 6 equal pieces of equal size.
So,
The model representing the 6 equal pieces of the 5-foot streamer is:

From the above model,
We can see that each part in the model represents \(\frac{5}{6}\) of each part.
Hence, from the above,
We can conclude that the length of each piece of a 5-foot streamer in feet is: \(\frac{5}{6}\)

Question 14.
Four circular lemon slices are shared equally among 8 glasses of water. What fraction of a lemon slice does each glass get?
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 7
Answer: The fraction of a lemon slice does each glass get is: \(\frac{1}{2}\)

Explanation:
It is given that the four circular lemon slices are shared equally among 8 glasses of water.
So,
The model representing the portion that each glass get is:

From the above model,
We can say that each part represents \(\frac{1}{2}\) part
Hence, from the above,
We can conclude that the portion of a lemon slice does glass get is: \(\frac{1}{2}\)

Question 15.
You cut a 5-foot streamer into 6 pieces of equal size. What is the length of each piece in feet? in inches?
Answer: The length of each piece in feet is: \(\frac{5}{6}\)

Explanation:
It is given that you cut a 5-foot streamer into 6 equal pieces of equal size.
So,
The model representing the 6 equal pieces of the 5-foot streamer is:

From the above model,
We can see that each part in the model represents \(\frac{5}{6}\) of each part.
Hence, from the above,
We can conclude that the length of each piece of a 5-foot streamer in feet is: \(\frac{5}{6}\)

Question 16.
DIG DEEPER!
A fruit drink is made using \(\frac{7}{4}\) quarts of orange juice and \(\frac{5}{4}\) quarts of pineapple juice. The drink is shared equally among 12 guests. What fraction of a quart does each guest get?
Answer: The fraction of a quart does each guest get is: \(\frac{1}{4}\)

Explanation:
It is given that a fruit drink is made using \(\frac{7}{4}\) quarts of orange juice and \(\frac{5}{4}\) quarts of pineapple juice.
So,
The total amount of fruit juice= \(\frac{7}{4}\) + \(\frac{5}{4}\)
= \(\frac{ 7 + 5}{4}\)
= \(\frac{12}{4}\)
It is also given that the drink is shared equally among 12 guests
So,
The fraction of a quart does each gust get = \(\frac{12}{4}\) ÷ 12
= \(\frac{12}{4}\) ÷ \(\frac{12}{1}\)
= \(\frac{12}{4}\) × \(\frac{1}{12}\)
= \(\frac{1}{4}\)
Hence, from the above,
We can conclude that the fraction of a quart does each person get is: \(\frac{1}{4}\)

Interpret Fractions as Division Homework & Practice 10.1

Divide. Use a model to help.

Question 1.
1 ÷ 6 =0.16
Answer:


From the above model,
The number of colored parts is: 1
The number of total parts is: 6
So,
The fraction of the colored parts out of the total number of parts = 1 ÷ 6
= \(\frac{1}{6}\)
So,
The fraction of the colored parts out of the total number of parts in the decimal form is: 0.16

Question 2.
1 ÷ 7 =0.14
Answer:
From the above model,
The number of colored parts is: 1
The number of total parts is: 7
So,
The fraction of the colored parts out of the total number of parts = 1 ÷ 7
= \(\frac{1}{7}\)
So,
The fraction of the colored parts out of the total number of parts in the decimal form is: 0.14

Question 3.
1 ÷ 5 = 0.20
Answer:
From the above model,
The number of colored parts is: 1
The number of total parts is: 5
So,
The fraction of the colored parts out of the total number of parts = 1 ÷ 5
= \(\frac{1}{5}\)
So,
The fraction of the colored parts out of the total number of parts in the decimal form is: 0.20

Question 4.
3 ÷ 4 = 0.75
Answer:
From the above model,
The number of colored parts is: 3
The number of total parts is: 4
So,
The fraction of the colored parts out of the total number of parts = 3 ÷ 4
= \(\frac{3}{4}\)
So,
The fraction of the colored parts out of the total number of parts in the decimal form is: 0.75

Question 5.
6 ÷ 7 = 0.85
Answer:
From the above model,
The number of colored parts is: 6
The number of total parts is: 7
So,
The fraction of the colored parts out of the total number of parts = 6 ÷ 7
= \(\frac{6}{7}\)
So,
The fraction of the colored parts out of the total number of parts in the decimal form is: 0.85

Question 6.
5 ÷ 9 = 0.55
Answer:
From the above model,
The number of colored parts is: 5
The number of total parts is: 9
So,
The fraction of the colored parts out of the total number of parts = 5 ÷ 9
= \(\frac{5}{9}\)
So,
The fraction of the colored parts out of the total number of parts in the decimal form is: 0.55

Question 7.
YOU BE THE TEACHER
Your friend says \(\frac{5}{12}\) is equivalent to 12 ÷ 5. Is your friend correct? Explain.
Answer: No, your friend s not correct.

Explanation:
The given fraction is: \(\frac{5}{12}\)
From the given fraction,
The numerator is: 5
The denominator is: 12
We can write a fraction in the following form:
Fraction = \(\frac{Numerator}{Denominator}\)
So,
\(\frac{5}{12}\) is equivalent to 5 ÷ 12
But, according to your friend,
\(\frac{5}{12}\) is equivalent to 12 ÷ 5
Hence, from the above,
we can conclude that your friend is not correct.

Question 8.
Writing
Explain how fractions and division are related.

Question 9.
Structure
Write a division equation represented by the model.
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 8
Answer:
The division equation represented by the model is: 1 ÷ 4

Explanation:
The given model is:
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 8
From the given model,
The number of shaded parts is: 1
The total number of parts are: 4
So,
The division equation can be represented as:
Division equation = (The number of shaded parts) ÷ ( The total number of parts )
= 1 ÷ 4
= \(\frac{1}{4}\)

Question 10.
Number Sense
Eight friends share multiple vegetable pizzas, and each gets \(\frac{3}{8}\) of a pizza. How many pizzas do they share?
Answer: The total number of pizzas the eight friends shared are: 3 pizzas

Explanation:
It is given that the eight friends share multiple vegetable pizzas and each gets \(\frac{3}{8}\) of a pizza.
So,
The total number of pizzas shared by the eight friends = \(\frac{3}{8}\) × 8
= \(\frac{3}{8}\) × \(\frac{8}{1}\)
= \(\frac{3 × 8}{8 × 1}\)
= \(\frac{3}{1}\)
= 3
Hence, from the above,
We can conclude that the total number of pizzas shared by the eight friends is: 3 pizzas

Question 11.
Modeling Real Life
Seven friends each run an equal part of a 5-kilometer relay race. What fraction of a kilometer does each friend complete?
Answer: The fraction of a kilometer does each friend complete is: \(\frac{5}{7}\) kilometer

Explanation:
It is given that there are seven friends each run an equal part of a 5-kilometer relay race.
So,
The fraction that each friend run = \(\frac{The total distance} {The number of friends}\)
= \(\frac{5}{7}\)
Hence, from the above,
We can conclude that the fraction of a kilometer does each friend complete is: \(\frac{5}{7}\) kilometer

Question 12.
Modeling Real Life
A group of friends equally share 3 bags of pretzels. Each friend gets \(\frac{3}{5}\) of a bag of pretzels. How many friends are in the group?
Answer: The total number of friends in the group are: 5

Explanation:
It is given that a group of friends equally share 3 bags of pretzels and each friend gets \(\frac{3}{5}\) of a bag of pretzels.
So,
The total number of friends = \(\frac{The total number of bags}{The amount each friend gets}\)
= \(\frac{3}{1}\) × \(\frac{5}{3}\)
= \(\frac{5}{1}\)
= 5
Hence, from the above,
We can conclude that the total number of friends are: 5

Review & Refresh

Multiply.

Question 13.
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 9
Answer: 9 × \(\frac{2}{3}\) = 6

Explanation:
The given fractions are: \(\frac{9}{1}\) and \(\frac{2}{3}\)
So,
\(\frac{9}{1}\) × \(\frac{2}{3}\)
= \(\frac{9 × 2}{1 × 3}\)
= \(\frac{6}{1}\)
= 6
Hence,
9 × \(\frac{2}{3}\) = 6

Question 14.
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 10
Answer: 5 × \(\frac{7}{10}\) = \(\frac{7}{2}\)

Explanation:
The given fractions are: \(\frac{5}{1}\) and \(\frac{7}{10}\)
So,
\(\frac{5}{1}\) × \(\frac{7}{10}\)
= \(\frac{5 × 7}{1 × 10}\)
= \(\frac{7}{2}\)
Hence,
5 × \(\frac{7}{10}\) = \(\frac{7}{2}\)

Question 15.
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 11
Answer: 3 × \(\frac{5}{12}\) = \(\frac{5}{4}\)

Explanation:
The given fractions are: \(\frac{3}{1}\) and \(\frac{5}{12}\)
So,
\(\frac{3}{1}\) × \(\frac{5}{12}\)
= \(\frac{3 × 5}{1 × 12}\)
= \(\frac{5}{4}\)
Hence,
3 × \(\frac{5}{12}\) = \(\frac{5}{4}\)

Lesson 10.2 Mixed Numbers as Quotients

Explore and Grow

You share 6 sheets of construction paper equally among 4 people. Write a division expression that represents the situation. How much paper does each person get? Use a model to support your answer.
Answer:
The division expression representing the situation is: 6 ÷ 4

Explanation:
It s given that you have shared 6 sheets of construction paper equally among 4 people
So,
The division equation representing the sharing of construction papers is: 6 ÷ 4
Now,
6 ÷ 4 = \(\frac{6}{4}\)
So,
The above equation represents that 4 is divided into 6 parts.
So,
The model representing the situation is:

From the above model,
We can say that the amount of does each person get is: 1\(\frac{1}{2}\) or 1.5 or \(\frac{3}{2}\)

Precision
Does each person get less than or more than 1 sheet of paper? Use the dividend and divisor to explain why your answer makes sense.
Answer:
From the above problem,
We can say that each person gets more than 1 paper.
So,
The division equation of the above problem is: 6 ÷ 4
The equivalent form of 6 ÷ 4 is: \(\frac{6}{4}\)
Now,
The simplest form of \(\frac{6}{4}\) is: \(\frac{3}{2}\) ( The simplest form is the division of the numerator and the denominator with the common multiple if we can divide)
The mixed form of \(\frac{3}{2}\) is: 1\(\frac{1}{2}\)

Think and Grow: Divide Whole Numbers

You can use models to divide whole numbers that have a mixed number as the quotient.
Example
Find 3 ÷ 2.
One Way:
Use a tape diagram. Show 3 wholes. Divide each whole into 2 equal parts.

Another Way: Use an area model. Show 3 wholes. Divide each whole into 2 equal parts. Then separate the parts into 2 equal groups.

Show and Grow

Divide. Use a model to help

Question 1.
5 ÷ 3 = ___
Answer: 5 ÷ 3 = 1\(\frac{2}{3}\)

Explanation;
The given division equation is: 5 ÷ 3
The model representing the division equation is:

From the above model,
5 ÷ 3 = 3 ÷ 3
= 1 R 2
Hence,
We can say that each part is divided into 1\(\frac{2}{3}\) or \(\frac{5}{3}\)

Question 2.
7 ÷ 2 = ___

Answer: 7 ÷ 2 = 3\(\frac{1}{2}\)

Explanation;
The given division equation is: 7 ÷ 2
The model representing the division equation is:

From the above model,
7 ÷ 2 = 6 ÷ 2
= 3 R 1
Hence,
We can say that each part is divided into 3\(\frac{1}{2}\) or \(\frac{7}{2}\) or 3.5

Apply and Grow: Practice

Divide. Use a model to help.

Question 3.
12 ÷ 7 = ___

Answer: 12 ÷ 7 = 1\(\frac{5}{7}\)

Explanation;
The given division equation is: 12 ÷ 7
The model representing the division equation is:

From the above model,
12 ÷ 7 = 7 ÷ 7
= 1 R 5
Hence,
We can say that each part is divided into 1\(\frac{5}{7}\) or \(\frac{12}{7}\)

Question 4.
25 ÷ 20 = ___

Answer: 25 ÷ 20 = 1\(\frac{5}{20}\) = \(\frac{5}{4}\)

Explanation;
The given division equation is: 25 ÷ 20
The model representing the division equation is:

From the above model,
25 ÷ 20 = 20 ÷ 20
= 1 R 5
Hence,
We can say that each part is divided into 1\(\frac{5}{20}\) or \(\frac{5}{4}\)

Question 5.
15 ÷ 4 = ___

Answer: 15 ÷ 4 = 3\(\frac{3}{4}\)

Explanation;
The given division equation is: 15 ÷ 4
The model representing the division equation is:

From the above model,
15 ÷ 4 = 12 ÷ 4
= 3 R 3
Hence,
We can say that each part is divided into 3\(\frac{3}{4}\) or \(\frac{15}{4}\)

Question 6.
13 ÷ 6 = ___

Answer: 13 ÷ 6 = 2\(\frac{1}{6}\)

Explanation;
The given division equation is: 13÷ 6
The model representing the division equation is:

From the above model,
13 ÷ 6 = 12 ÷ 6
= 2 R 1
Hence,
We can say that each part is divided into 2\(\frac{1}{6}\) or \(\frac{13}{6}\)

Question 7.
16 ÷ 8 = ___

Answer: 16 ÷ 8 = 2

Explanation;
The given division equation is: 16÷ 8
The model representing the division equation is:

From the above model,
16 ÷ 8
= 2 R 0
Hence,
We can say that each part is divided into 2 equal parts

Question 8.
92 ÷ 50 = ___

Answer: 92 ÷ 50 = 1\(\frac{21}{25}\)

Explanation;
The given division equation is: 92÷ 50
So,
92 ÷ 50 = 50 ÷ 50
= 1 R 42
Hence,
We can say that each part is divided into 1\(\frac{42}{50}\) or 1\(\frac{21}{25}\)

Question 9.
How many 3s are in 7?
Answer: The number of 3 in 7 are: \(\frac{7}{3}\) or 2\(\frac{1}{3}\)

Explanation:
The division equation is: 7 ÷ 3
So,
The model for the given division equation is:

From the above model,
7 ÷ 3 = 6 ÷ 3
= 2 R 1
Hence, from the above,
We can conclude that there are 2\(\frac{1}{3}\) 3s in 7

Question 10.
How many 6s are in 21?
Answer: The number of 6s in 21 are: \(\frac{21}{6}\) or 3\(\frac{3}{6}\)

Explanation:
The division equation is: 21 ÷ 6
So,
The model for the given division equation is:

From the above model,
21 ÷ 6 = 18 ÷ 6
= 3 R 3
Hence, from the above,
We can conclude that there are 3\(\frac{3}{6}\) 3s in 21

Question 11.
YOU BE THE TEACHER
Your friend says that \(\frac{35}{6}\) is equivalent to 35 ÷ 6. Is your friend correct? Explain.
Answer: Yes, your friend is correct

Explanation:
It is given that \(\frac{35}{6}\)
We know that,
The decimal equation can be converted into a fraction as \(\frac{Numerator}{Denominator}\)
So,
\(\frac{35}{6}\) = 35 ÷ 6
Hence, from the above,
We can conclude that your friend is correct

Question 12.
Writing
Write and solve a real-life problem for 24 ÷ 5.
Answer: 24 ÷ 5 = 4\(\frac{4}{5}\)

Explanation;
The given division equation is: 24÷ 5
The model for the above division equation is:

From the above model,
24 ÷ 5 = 20 ÷ 5
= 4 R 4
Hence,
We can say that each part is divided into 4\(\frac{4}{5}\)

Think and Grow: Modeling Real Life

Example
You share 7 bales of hay equally among 3 horse stalls. How many whole bales are in each stall? What fractional amount of a bale is in each stall?
Divide 7 by 3 to find how many bales of hay are in each stall. Use an area model to help.

Show and Grow

Question 13.
Six muffins are shared equally among 4 friends. How many whole muffins does each friend get? What fractional amount of a muffin does each friend get?
Answer: Each friend will get 1 muffin and 2 muffins are leftovers
The fractional part of a muffin does each friend get is: \(\frac{1}{2}\)

Explanation:
It is given that there are six muffins are shared equally among 4 friends.
So,
The number of muffins each friend get = 6 ÷ 4
= 4 ÷ 4
= 1 R 2
Hence, from the above,
We can conclude that each friend gets 1 muffin each and the fraction of each muffin get is: \(\frac{1}{2}\)

Question 14.
A cyclist bikes 44 miles in 5 days. She bikes the same distance each day. Does she bike more than 8\(\frac{1}{2}\) miles each day? Explain.
Answer: She bikes more than 8\(\frac{1}{2}\) miles each day.

Explanation:
It is given that a cyclist bikes 44 miles in 5 days.
So,
The distance that she bikes each day = 44 ÷ 5
So,
44 ÷ 5 = 40 ÷ 5
= 8 R 4
= 8\(\frac{4}{5}\) miles
But, it is given that she bikes 8\(\frac{1}{2}\) miles each day
Hence, from the above,
We can conclude that she bikes more than 8\(\frac{1}{2}\) miles each day.

Question 15.
DIG DEEPER!
At Table A, 4 students share 7 packs of clay equally. At Table B, 5 students share 8 packs of clay equally. At which table does each student get a greater amount of clay? Explain.
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 15
Answer: At Table A, each student gets a greater amount of clay.

Explanation:
It is given that at Table A, 4 students share 7 packs of clay equally.
So,
The representation of clay at table A is: \(\frac{7}{4}\)
It is also given that at Table B, 5 students share 8 packs of clay equally.
So,
The representation of clay at table B is: \(\frac{8}{5}\)
So,
For comparison, equate the denominators.
So,
Multiply the first fraction at table A by \(\frac{5}{5}\) and the fraction at table B by \(\frac{4}{4}\)
So,
\(\frac{7}{4}\) × \(\frac{5}{5}\)
= \(\frac{35}{20}\)
So,
\(\frac{8}{5}\) × \(\frac{4}{4}\)
= \(\frac{32}{20}\)
Hence, from the above,
We can conclude that at table A, the students will get more amount of clay.

Mixed Numbers as Quotients Homework & Practice 10.2

Divide. Use a model to help.

Question 1.
5 ÷ 2 = ___
Answer: 5 ÷ 2 = 2\(\frac{1}{2}\)

Explanation;
The given division equation is: 5 ÷ 2
The model representing the division equation is:

From the above model,
5 ÷ 2 = 4 ÷ 2
= 2 R 1
Hence,
We can say that 5 ÷ 2 = 2\(\frac{1}{2}\) or 2.5 or \(\frac{5}{2}\)

Question 2.
10 ÷ 7 = ___
Answer: 10 ÷ 7 = 1\(\frac{3}{7}\) = \(\frac{10}{7}\)

Explanation;
The given division equation is: 10 ÷ 7
The model representing the division equation is:

From the above model,
10 ÷ 7 = 7 ÷ 7
= 1 R 3
Hence,
We can say that 10 ÷ 7 = 1\(\frac{3}{7}\) or \(\frac{10}{7}\)

Question 3.
3 ÷ 9 = ___
Answer: 3 ÷ 9 = \(\frac{1}{3}\)

Explanation;
The given division equation is: 3 ÷ 9
The model representing the division equation is:

From the above model,
3 and 9 are the multiples of 3.
So,
3 ÷ 9 = \(\frac{1}{3}\)
Hence,
We can say that 3 ÷ 9 = \(\frac{1}{3}\)

Question 4.
11 ÷ 4 = ___
Answer: 11 ÷ 4 = 2\(\frac{3}{4}\)

Explanation;
The given division equation is: 11 ÷ 4
The model representing the division equation is:

From the above model,
11 ÷ 4 = 8 ÷ 4
= 2 R 3
Hence,
We can say that 11 ÷ 4 = \(\frac{11}{4}\) or 2\(\frac{3}{4}\)

Question 5.
13 ÷ 6 = ___
Answer: 13 ÷ 6 = 2\(\frac{1}{6}\)

Explanation;
The given division equation is: 13 ÷ 6
The model representing the division equation is:

From the above model,
13 ÷ 6 = 12 ÷ 6
= 2 R 1
Hence,
We can say that 13 ÷ 6 = \(\frac{13}{6}\) or 2\(\frac{1}{6}\)

Question 6.
45 ÷ 8 = ___
Answer: 45 ÷ 8 = 5\(\frac{5}{8}\)

Explanation;
The given division equation is: 45 ÷ 8
The model representing the division equation is:

From the above model,
45 ÷ 8 = 40 ÷ 8
= 5 R 5
Hence,
We can say that 45 ÷ 8 = \(\frac{45}{8}\) or 5\(\frac{5}{8}\)

Question 7.
Number Sense
Between which two whole numbers is the quotient of 74 and 9?
Answer: The quotient of 74 and 9 is between 8 and 9

Explanation:
The given two numbers are 7 and 9
So,
By using the partial quotients method,
74 ÷ 9= 72 ÷ 9
= 8 R 2
So,
74 ÷ 9 = \(\frac{74}{9}\) or 8\(\frac{2}{9}\) or 8.3
Hence, from the above,
We can conclude that the quotient of 74 and 9 is between 8 and 9

Question 8.
Reasoning
Three friends want to share 22 baseball cards. For this situation, why does the quotient 7 R1 make more sense than the quotient 7\(\frac{1}{3}\)?
Answer:
It is given that three friends want to share 22 baseball cards.
So,
We have to find the number of baseball cards each friend possesses.
So,
It is sufficient to write the number of baseball cards possessed by each friend in the remainder form rather than the fraction form.
So,
The number of baseball cards possessed by each friend = \(\frac{The total number of baseball cards}{The number of friends}\)
= 22 ÷ 3
= 21 ÷ 3
= 7 R 1
Hence, from the above,
We can conclude that the remainder form is sufficient to find the number of baseball cars possessed by each friend rather than the fraction form.

Question 9.
DIG DEEPER!
Is \(\frac{2}{5}\) × 3 equivalent to 2 × 3 ÷ 5? Explain.
Answer: Yes, \(\frac{2}{5}\) × 3 equivalent to 2 × 3 ÷ 5

Explanation:
The given fraction and the number is: \(\frac{2}{5}\) and 3
So,
\(\frac{2}{5}\) × 3 = \(\frac{2}{5}\) × \(\frac{3}{1}\)
= \(\frac{2 × 3}{5}\)
= 2 × 3 ÷ 5
Hence, from the above,
We can conclude that \(\frac{2}{5}\) × 3 equivalent to 2 × 3 ÷ 5

Question 10.
Modeling Real Life
A bag of 4 balls weighs 6 pounds. Each ball weighs the same amount. What is the weight of each ball?
Answer: The weight of each ball is: \(\frac{3}{2}\) pounds or 1.5 pounds

Explanation:
It is given that a bag of 4 balls weighs 6 pounds
So,
The weight of each ball = \(\frac{The total weight of the balls}{The number of balls}\)
= 6 ÷ 4
Since 6 and 4 are the multiples of 2, divide the two numbers by 2
So,
6 ÷ 4 = 3 ÷ 2
So,
3 ÷ 2 = 2 ÷ 2
= 1 R 1
= 1\(\frac{1}{2}\) pounds
Hence, from the above,
We can conclude that the weight of each ball is: 1\(\frac{1}{2}\) pounds or 1.5 pounds

Question 11.
Modeling Real Life
Zookeepers order 600 pounds of bamboo for the pandas. The bamboo lasts 7 days. How many whole pounds of bamboo do the pandas eat each day? What fractional amount of a pound do the pandas eat each day?
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 17
Answer:
The amount of bamboos the pandas eat each day is around 85 pounds
The amount of bamboos the pandas eat each day in the fraction form is: 85\(\frac{5}{7}\)

Explanation:
It is given that zookeepers order 600 pounds of bamboo for the pandas and the bamboos last 7 days for the pandas
So,
The number of bamboos the pandas eat each day = 600 ÷ 7
So,
By using the partial quotients method,
600 ÷ 7 = ( 560 + 35 ) ÷ 7
= ( 560 ÷ 7 ) + ( 35 ÷ 7 )
= 80 + 5
= 85 R 5
Hence, from the above,
We can conclude that
The amount of bamboos the pandas eat each day is around 85 pounds
The amount of bamboos the pandas eat each day in the fraction form is: 85\(\frac{5}{7}\)

Question 12.
Modeling Real Life
A plumber has 20 feet of piping. He cuts the piping into 6 equal pieces. Is each piece greater than, less than, or equal to 3\(\frac{1}{2}\) feet?
Answer: Each piece is less than 3\(\frac{1}{2}\) feet

Explanation:
It is given that a plumber has 20 feet of piping and he cuts the piping into 6 equal pieces.
So,
The length of each piece = 20 ÷ 6
By using the partial quotients method,
20 ÷ 6 = 18 ÷ 6
= 3 R 2
So,
20 ÷ 6 = 3\(\frac{2}{6}\)
Now,
3\(\frac{1}{2}\) = \(\frac{7}{2}\)
3\(\frac{2}{6}\) = \(\frac{20}{6}\)
For comparison, we have to equate whether the denominators or the numerators.
So,
Multiply 3\(\frac{1}{2}\) with \(\frac{3}{3}\)
So,
3\(\frac{1}{2}\) = \(\frac{21}{6}\)
Hence, from the above,
We can conclude that each piece is less than 3\(\frac{1}{2}\) feet

Review & Refresh

Add.

Question 13.
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 18
Answer: \(\frac{2}{9}\) + \(\frac{2}{3}\) = \(\frac{8}{9}\)

Explanation:
The two given fractions are: \(\frac{2}{9}\) and \(\frac{2}{3}\)
So, in addition, we have to make either the numerators or the denominators equal
So,
Multiply \(\frac{2}{3}\)  with \(\frac{3}{3}\)
So,
\(\frac{2}{3}\)  = \(\frac{6}{9}\)
Hence, from the above,
We can conclude that \(\frac{2}{9}\) + \(\frac{2}{3}\) = \(\frac{8}{9}\)

Question 14.
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 19
Answer: \(\frac{1}{10}\) + \(\frac{3}{4}\) = \(\frac{34}{40}\)

Explanation:
The two given fractions are: \(\frac{1}{10}\) and \(\frac{3}{4}\)
So, in addition, we have to make either the numerators or the denominators equal
So,
Multiply \(\frac{1}{10}\)  with \(\frac{4}{4}\)
Multiply \(\frac{3}{4}\)  with \(\frac{10}{10}\)
So,
\(\frac{1}{10}\)  = \(\frac{4}{40}\)
\(\frac{3}{4}\)  = \(\frac{30}{40}\)
Hence, from the above,
We can conclude that \(\frac{1}{10}\) + \(\frac{3}{4}\) = \(\frac{34}{40}\)

Question 15.
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 20
Answer: \(\frac{3}{5}\) + \(\frac{5}{6}\) + \(\frac{1}{5}\) = \(\frac{49}{30}\)

Explanation:
The three given fractions are: \(\frac{3}{5}\) , \(\frac{5}{6}\) and \(\frac{1}{5}\)
So, in addition, we have to make either the numerators or the denominators equal
So,
Multiply \(\frac{3}{5}\)  with \(\frac{6}{6}\)
Multiply \(\frac{5}{6}\)  with \(\frac{5}{5}\)
Multiply \(\frac{1}{5}\)  with \(\frac{6}{6}\)
So,
\(\frac{3}{5}\)  = \(\frac{18}{30}\)
\(\frac{5}{6}\)  = \(\frac{25}{30}\)
\(\frac{1}{5}\)  = \(\frac{6}{30}\)
Hence, from the above,
We can conclude that \(\frac{3}{5}\) + \(\frac{5}{6}\) +\(\frac{1}{5}\)  = \(\frac{49}{30}\)

Lesson 10.3 Divide Whole Numbers by Unit Fractions

Explore and Grow

Write a real-life problem that can be represented by 6 ÷ \(\frac{1}{2}\)?
Answer:
Suppose, we have an apple and there are 6 children and we are giving each child half of the piece.
So,
Each child receives 6 ÷ \(\frac{1}{2}\) piece of the apple

What is the solution to the problem? Use a model to support your answer?
Answer:
The above problem is the division of an apple among the six children
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
So,
The amount each child receive from an apple = 6 ÷ \(\frac{1}{2}\)
= 6 × \(\frac{2}{1}\)
= \(\frac{6}{1}\) × \(\frac{2}{1}\)
= \(\frac{ 6 × 2}{1 × 1}\)
= 12

Structure
How can you check your answer using multiplication?
Answer:
We can check the answer using multiplication by the two rules regarding division and multiplication. They are:
A) a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
B) a= \(\frac{a}{1}\)

Think and Grow: Divide Whole Numbers by Unit Fractions

You can use models to divide whole numbers by unit fractions.
Example
Find 4 ÷ \(\frac{1}{3}\)
One Way:
Use a tape diagram to find how many \(\frac{1}{3}\)s are in 4. There are 4 wholes.
Divide each whole into 3 equal parts. Each part is \(\frac{1}{3}\).
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 21
Because there are 3 one-thirds in 1 whole, there are
4 × 3 equal parts = 12 one-thirds in 4 wholes.

Show and Grow

Divide. Use a model to help

Question 1.
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 23
Answer: 3 ÷ \(\frac{1}{2}\) = 6

Explanation:
The given numbers are: 3 and \(\frac{1}{2}\)
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
So,
3 ÷ \(\frac{1}{2}\)  = 3 × \(\frac{2}{1}\)
= \(\frac{3}{1}\) × \(\frac{2}{1}\)
= \(\frac{ 3 × 2}{1 × 1}\)
= 6
Hence,
3÷ \(\frac{1}{2}\) = 6

Question 2.
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 24
Answer: 2 ÷ \(\frac{1}{5}\) = 10

Explanation:
The given numbers are: 2 and \(\frac{1}{5}\)
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
So,
2 ÷ \(\frac{1}{5}\)  = 2 × \(\frac{5}{1}\)
= \(\frac{5}{1}\) × \(\frac{2}{1}\)
= \(\frac{ 5 × 2}{1 × 1}\)
= 10
Hence,
2÷ \(\frac{1}{5}\) = 10

Apply and Grow: Practice

Divide. Use a model to help.

Question 3.

Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 25
Answer: 1 ÷ \(\frac{1}{3}\) = 3

Explanation:
The given numbers are: 1 and \(\frac{1}{3}\)
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
So,
1 ÷ \(\frac{1}{3}\)  = 1 × \(\frac{3}{1}\)
= \(\frac{3}{1}\) × \(\frac{1}{1}\)
= \(\frac{ 3 × 1}{1 × 1}\)
= 3
Hence,
1÷ \(\frac{1}{3}\) = 3

Question 4.
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 26
Answer: 3 ÷ \(\frac{1}{5}\) = 15

Explanation:
The given numbers are: 3 and \(\frac{1}{5}\)
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
So,
3 ÷ \(\frac{1}{5}\)  = 3 × \(\frac{5}{1}\)
= \(\frac{3}{1}\) × \(\frac{5}{1}\)
= \(\frac{ 3 × 5}{1 × 1}\)
= 15
Hence,
3÷ \(\frac{1}{5}\) = 15

Question 5.
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 27
Answer: 5 ÷ \(\frac{1}{3}\) = 15

Explanation:
The given numbers are: 5 and \(\frac{1}{3}\)
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
So,
5 ÷ \(\frac{1}{3}\)  = 5 × \(\frac{3}{1}\)
= \(\frac{3}{1}\) × \(\frac{5}{1}\)
= \(\frac{ 3 × 5}{1 × 1}\)
= 15
Hence,
5÷ \(\frac{1}{3}\) = 15

Question 6.
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 28
Answer: 4 ÷ \(\frac{1}{4}\) = 16

Explanation:
The given numbers are: 4 and \(\frac{1}{4}\)
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
So,
4 ÷ \(\frac{1}{4}\)  = 4 × \(\frac{4}{1}\)
= \(\frac{4}{1}\) × \(\frac{4}{1}\)
= \(\frac{ 4 × 4}{1 × 1}\)
= 16
Hence,
4÷ \(\frac{1}{4}\) = 16

Question 7.
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 29
Answer: 7 ÷ \(\frac{1}{2}\) = 14

Explanation:
The given numbers are: 7 and \(\frac{1}{2}\)
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
So,
7 ÷ \(\frac{1}{2}\)  = 7 × \(\frac{2}{1}\)
= \(\frac{7}{1}\) × \(\frac{2}{1}\)
= \(\frac{ 7 × 2}{1 × 1}\)
= 14
Hence,
7÷ \(\frac{1}{2}\) = 14

Question 8.
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 30
Answer: 2 ÷ \(\frac{1}{7}\) = 14

Explanation:
The given numbers are: 2 and \(\frac{1}{7}\)
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
So,
2 ÷ \(\frac{1}{7}\)  = 2 × \(\frac{7}{1}\)
= \(\frac{2}{1}\) × \(\frac{7}{1}\)
= \(\frac{ 7 × 2}{1 × 1}\)
= 14
Hence,
2÷ \(\frac{1}{7}\) = 14

Question 9.
How many \(\frac{1}{4}\)s are in 5?
Answer: There are 20 \(\frac{1}{4}\)s in 5

Explanation:
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
Now,
We have to find the number of \(\frac{1}{4}\)s in 5
So,
5 ÷ \(\frac{1}{4}\)  = 5 × \(\frac{4}{1}\)
= \(\frac{5}{1}\) × \(\frac{4}{1}\)
= \(\frac{ 5 × 4}{1 × 1}\)
= 20
Hence, from the above,
We can conclude that there are 20 \(\frac{1}{4}\)s in 5.

Question 10.
How many \(\frac{1}{6}\)s are in 2?
Answer: There are 12 \(\frac{1}{6}\)s in 2

Explanation:
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
Now,
We have to find the number of \(\frac{1}{6}\)s in 2
So,
2 ÷ \(\frac{1}{6}\)  = 2 × \(\frac{6}{1}\)
= \(\frac{2}{1}\) × \(\frac{6}{1}\)
= \(\frac{ 2 × 6}{1 × 1}\)
= 12
Hence, from the above,
We can conclude that there are 12 \(\frac{1}{6}\)s in 2.

Question 11.
YOU BE THE TEACHER
Newton finds 6 ÷ \(\frac{1}{3}\). Is he correct? Explain.
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 31
Answer: No, Newton is not correct

Explanation:
The given division equation is: 6 ÷ \(\frac{1}{3}\)
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
So,
6 ÷ \(\frac{1}{3}\)  = 6 × \(\frac{3}{1}\)
= \(\frac{6}{1}\) × \(\frac{3}{1}\)
= \(\frac{ 3 × 6}{1 × 1}\)
= 18
But, according to Newton,
6 ÷ \(\frac{1}{3}\) = 2
Hence, from the above,
We can conclude that Newton is not correct.

Question 12.
Writing
Write and solve a real-life problem for 4 ÷ \(\frac{1}{2}\).
Answer:
Suppose we have 4 bags of wheat and we have to distribute the 4 bags by dividing each bag of wheat in half
So,
Each person receives 4 ÷ \(\frac{1}{2}\) bag of wheat
Now,
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
So,
4 ÷ \(\frac{1}{2}\)  = 4 × \(\frac{2}{1}\)
= \(\frac{4}{1}\) × \(\frac{2}{1}\)
= \(\frac{ 4 × 2}{1 × 1}\)
= 8
Hence, from the above,
We can conclude that there are 8 bags of wheat when divide the 4 bags of wheat in half.

Think and Grow: Modeling Real Life

Example
A chef makes 3 cups of salsa. A serving of salsa is \(\frac{1}{8}\) cup. How many servings does the chef make?
To find the number of servings, find the number of \(\frac{1}{8}\) cups in 3 cups.
Use an area model to find 3 ÷ \(\frac{1}{8}\). Divide each cup into 8 equal parts.
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 32

Show and Grow

Question 13.
A litter of kittens weighs a total of 2 pounds. Each newborn kitten weighs \(\frac{1}{4}\) pound. How many kittens are in the litter?
Answer: The number of kittens in the litter are: 8 kittens

Explanation:
It is given that a litter of kittens weighs a total of 2 pounds and each newborn kitten weighs \(\frac{1}{4}\) pound.
So,
The number of kittens in the litter = \(\frac{The total weight of litter}{The weight of each newborn kitten}\)
= 2 ÷ \(\frac{1}{4}\)
Now,
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
So,
2 ÷ \(\frac{1}{4}\)  = 2 × \(\frac{4}{1}\)
= \(\frac{4}{1}\) × \(\frac{2}{1}\)
= \(\frac{ 4 × 2}{1 × 1}\)
= 8
Hence, from the above,
We can conclude that the number of kittens in the litter are: 8 kittens

Question 14.
You put signs on a walking trail that is 7 miles long. You put a sign at the start and at the end of the trail. You also put a sign every \(\frac{1}{10}\) mile. How many signs do you put on the trail?
Answer: The total number of signs you put on the trail is: 72

Explanation:
It is given that you put signs on a walking trail that is 7 miles long and you put a sign at the start and at the end of the trail.
It is also given that you put a sign every \(\frac{1}{10}\) mile.
So,
The total number of signs you put on the trail = The sign at the start of the trail + The sign at the end of the trail + The total number of signs for \(\frac{1}{10}\) mile
Now,
The total number of signs for \(\frac{1}{10}\) mile = 7 ÷ \(\frac{1}{10}\)
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
So,
7 ÷ \(\frac{1}{10}\)  = 7 × \(\frac{10}{1}\)
= \(\frac{7}{1}\) × \(\frac{10}{1}\)
= \(\frac{ 7 × 10}{1 × 1}\)
= 70
So,
The total number of signs you put on the trail = 1 + 1 + 70
= 72
hence, from the above,
We can conclude that there are 72 signs that you put on the trail

Question 15.
DIG DEEPER!
You have 2 boards that are each 8 feet long. You cut \(\frac{1}{2}\)– foot pieces to make square picture frames. How many picture frames can you make?
Answer: The number of picture frames you can make is: 32

Explanation:
It is given that you have 2 boards that are each 8 feet long.
So,
The total length of 2 boards = 2 × 8 = 16 feet
It is also given that you cut \(\frac{1}{2}\)– foot pieces to make square picture frames.
So,
The total number of picture frames = \(\frac{The total length of 2 boards}{The length of each square frame}\)
= 16 ÷ \(\frac{1}{2}\)
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
So,
16 ÷ \(\frac{1}{2}\)  = 16 × \(\frac{2}{1}\)
= \(\frac{16}{1}\) × \(\frac{2}{1}\)
= \(\frac{ 16 × 2}{1 × 1}\)
= 32
Hence, from the above,
We can conclude that we can make 32 picture frames.

Divide Whole Numbers by Unit Fractions Homework & Practice 10.3

Divide. Use a model to help.

Question 1.
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 33
Answer: 1 ÷ \(\frac{1}{9}\) = 9

Explanation:
The given numbers are: 1 and \(\frac{1}{9}\)
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
So,
1 ÷ \(\frac{1}{9}\)  = 1 × \(\frac{9}{1}\)
= \(\frac{1}{1}\) × \(\frac{9}{1}\)
= \(\frac{ 1 × 9}{1 × 1}\)
= 9
Hence,
1÷ \(\frac{1}{9}\) = 9

Question 2.
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 34
Answer: 2 ÷ \(\frac{1}{3}\) = 6

Explanation:
The given numbers are: 2 and \(\frac{1}{3}\)
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
So,
2 ÷ \(\frac{1}{3}\)  = 2 × \(\frac{3}{1}\)
= \(\frac{3}{1}\) × \(\frac{2}{1}\)
= \(\frac{ 3 × 2}{1 × 1}\)
= 6
Hence,
2÷ \(\frac{1}{3}\) = 6

Question 3.
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 35
Answer: 5 ÷ \(\frac{1}{2}\) = 10

Explanation:
The given numbers are: 5 and \(\frac{1}{2}\)
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
So,
5 ÷ \(\frac{1}{2}\)  = 5 × \(\frac{2}{1}\)
= \(\frac{5}{1}\) × \(\frac{2}{1}\)
= \(\frac{ 5 × 2}{1 × 1}\)
= 10
Hence,
5÷ \(\frac{1}{2}\) = 10

Divide. Use a model to help.

Question 4.
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 36
Answer: 9 ÷ \(\frac{1}{4}\) = 36

Explanation:
The given numbers are: 9 and \(\frac{1}{4}\)
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
So,
9 ÷ \(\frac{1}{4}\)  = 9 × \(\frac{4}{1}\)
= \(\frac{9}{1}\) × \(\frac{4}{1}\)
= \(\frac{ 9 × 4}{1 × 1}\)
= 36
Hence,
9÷ \(\frac{1}{4}\) = 36

Question 5.
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 37
Answer: 7 ÷ \(\frac{1}{3}\) = 21

Explanation:
The given numbers are: 7 and \(\frac{1}{3}\)
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
So,
7 ÷ \(\frac{1}{3}\)  = 7 × \(\frac{3}{1}\)
= \(\frac{3}{1}\) × \(\frac{7}{1}\)
= \(\frac{ 3 × 7}{1 × 1}\)
= 21
Hence,
7÷ \(\frac{1}{3}\) = 21

Question 6.
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 38
Answer: 8 ÷ \(\frac{1}{5}\) = 40

Explanation:
The given numbers are: 8 and \(\frac{1}{5}\)
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
So,
8 ÷ \(\frac{1}{5}\)  = 8 × \(\frac{5}{1}\)
= \(\frac{8}{1}\) × \(\frac{5}{1}\)
= \(\frac{ 8 × 5}{1 × 1}\)
= 40
Hence,
8÷ \(\frac{1}{5}\) = 40

Question 7.
Number Sense
Explain how you can check your answer for Exercise 6.
Answer:
We can check the answer for exercise 6 by using the below model:

From the above model,
Each part represents \(\frac{8}{5}\)
So,
The total value of the 5 parts is: \(\frac{40}{5}\)
Hence,
In the above way, we can say that we check the answer

Question 8.
YOU BE THE TEACHER
Descartes finds 5 ÷ \(\frac{1}{4}\). Is he correct? Explain.
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 38.1
Answer: Yes, he is correct

Explanation:
We can write 5 as \(\frac{20}{4}\) or \(\frac{5}{1}\)
But, we only take \(\frac{20}{4}\) because the divided number given is 4
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
So,
\(\frac{20}{4}\) ÷ \(\frac{1}{4}\)
= \(\frac{20}{4}\) × \(\frac{4}{1}\)
= \(\frac{ 20 × 4}{4 × 1}\)
= 20
Hence, from the above,
We can conclude that Descartes is correct.

Question 9.
Modeling Real Life
You need \(\frac{1}{2}\) pound of clay to make a pinch pot. How many pinch pots can you make with 12 pounds of clay?
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 42
Answer: You can make 24 pinch pots with 12 pounds of clay

Explanation:
It is given that you need \(\frac{1}{2}\) pound of clay to make a pinch pot.
It is also given that you have 12 pounds of clay
So,
The number of pinch pots you can make by using 12 pounds of clay = \(\frac{The total amount of clay}{The amount of clay used to make each pinch pot}\)
= 12 ÷ \(\frac{1}{2}\)
Now,
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
So,
12 ÷ \(\frac{1}{2}\)
= \(\frac{12}{1}\) × \(\frac{2}{1}\)
= \(\frac{ 12 × 2}{4 × 1}\)
= 24
Hence, from the above,
We can conclude that we can make 24 pinch pots by using 12 pounds of clay.

Question 10.
Modeling Real Life
Your art teacher has 5 yards of yellow string and 4 yards of green string. She cuts both colors \(\frac{1}{3}\)-yard pieces to hang of string into student artwork. How many pieces of student artwork can she hang?
Answer: The number of pieces of student artwork she can hang is: 27

Explanation:
It is given that your art teacher has 5 yards of yellow string and 4 yards of green string.
So,
The total number of yards of string = 5 + 4 = 9 yards of string
It is also given that she cuts both colors \(\frac{1}{3}\)-yard pieces to hang of string into student artwork.
So,
The number of pieces of student artwork she can hang = \(\frac{The total number of yards of strings}{The length of each yard f string}\)
= 9 ÷ \(\frac{1}{3}\)
Now,
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
So,
9 ÷ \(\frac{1}{3}\)
= \(\frac{9}{1}\) × \(\frac{3}{1}\)
= \(\frac{ 9 × 3}{1 × 1}\)
= 27
Hence, from the above,
We can conclude that there are 27 pieces of student artwork that she can hang.

Review & Refresh

Question 11.
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 43
Answer: \(\frac{2}{5}\) × \(\frac{3}{4}\) = \(\frac{6}{20}\)

Explanation:
The given fractions are: \(\frac{3}{4}\) and \(\frac{2}{5}\)
So,
\(\frac{2}{5}\) × \(\frac{3}{4}\)
= \(\frac{2 × 3}{5 × 4}\)
= \(\frac{6}{20}\)
Hence,
\(\frac{2}{5}\) × \(\frac{3}{4}\) = \(\frac{6}{20}\)

Question 12.
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 44
Answer: \(\frac{1}{8}\) × \(\frac{5}{8}\) = \(\frac{5}{64}\)

Explanation:
The given fractions are: \(\frac{1}{8}\) and \(\frac{5}{8}\)
So,
\(\frac{1}{8}\) × \(\frac{5}{8}\)
= \(\frac{1 × 5}{8 × 8}\)
= \(\frac{5}{64}\)
Hence,
\(\frac{1}{8}\) × \(\frac{5}{8}\) = \(\frac{5}{64}\)

Question 13.
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 45
Answer: \(\frac{4}{9}\) × \(\frac{2}{7}\) = \(\frac{8}{63}\)

Explanation:
The given fractions are: \(\frac{4}{9}\) and \(\frac{2}{7}\)
So,
\(\frac{4}{9}\) × \(\frac{2}{7}\)
= \(\frac{2 × 4}{7 × 9}\)
= \(\frac{8}{63}\)
Hence,
\(\frac{4}{9}\) × \(\frac{2}{7}\) = \(\frac{8}{63}\)

Lesson 10.4 Divide Unit Fractions by Whole Numbers

Write a real-life problem that can be represented by \(\frac{1}{2}\) ÷ 3?
Answer:
Suppose we have 3 people and those 3 people each has to share \(\frac{1}{2}\) of the apple

What is the solution to the problem? Use a model to support your answer?
Answer:
The above problem is: We have to share \(\frac{1}{2}\) each for the 3 people
Now,
We know that,
\(\frac{a}{b}\) ÷ a  = \(\frac{a}{b}\) × \(\frac{1}{a}\)
a= \(\frac{a}{1}\)
So,
\(\frac{1}{2}\) ÷ 3
= \(\frac{1}{2}\) × \(\frac{1}{3}\)
= \(\frac{ 1 × 1}{2 × 3}\)
= \(\frac{1}{6}\)
Hence,
\(\frac{1}{6}\) is the solution to the above problem.

Precision
Is the answer greater than or less than 1? Explain?
Answer: The answer is less than 1

Explanation:
The answer for the problem is: \(\frac{1}{6}\)
So,
For the comparison of \(\frac{1}{6}\) with 1, we have to see whether the numerators or the denominators are equal or not
So, in this case, the numerators are equal
So, compare the denominators
So,
1 < 6
Hence, from the above,
We can conclude that \(\frac{1}{6}\) is less than 1

Think and Grow: Divide Unit Fractions by Whole Numbers

You can use models to divide unit fractions by whole numbers.

Show and Grow

Divide. Use a model to help.

Question 1.
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 48
Answer: \(\frac{1}{4}\) ÷ 2 = \(\frac{1}{8}\)

Explanation:
The given numbers are: \(\frac{1}{4}\) and 2
We know that,
\(\frac{a}{b}\) ÷ a  = \(\frac{a}{b}\) × \(\frac{1}{a}\)
a= \(\frac{a}{1}\)
So,
\(\frac{1}{4}\) ÷ 2
= \(\frac{1}{4}\) × \(\frac{1}{2}\)
= \(\frac{ 1 × 1}{2 × 4}\)
= \(\frac{1}{8}\)
Hence,
\(\frac{1}{4}\) ÷ 2 = \(\frac{1}{8}\)

Question 2.
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 49
Answer: \(\frac{1}{2}\) ÷ 5 = \(\frac{1}{10}\)

Explanation:
The given numbers are: \(\frac{1}{2}\) and 5
We know that,
\(\frac{a}{b}\) ÷ a  = \(\frac{a}{b}\) × \(\frac{1}{a}\)
a= \(\frac{a}{1}\)
So,
\(\frac{1}{2}\) ÷ 5
= \(\frac{1}{2}\) × \(\frac{1}{5}\)
= \(\frac{ 1 × 1}{2 × 5}\)
= \(\frac{1}{10}\)
Hence,
\(\frac{1}{2}\) ÷ 5 = \(\frac{1}{10}\)

Apply and Grow: Practice

Divide. Use a model to help.

Question 3.
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 50
Answer: \(\frac{1}{5}\) ÷ 3 = \(\frac{1}{15}\)

Explanation:
The given numbers are: \(\frac{1}{5}\) and 3
We know that,
\(\frac{a}{b}\) ÷ a  = \(\frac{a}{b}\) × \(\frac{1}{a}\)
a= \(\frac{a}{1}\)
So,
\(\frac{1}{5}\) ÷ 3
= \(\frac{1}{5}\) × \(\frac{1}{3}\)
= \(\frac{ 1 × 1}{5 × 3}\)
= \(\frac{1}{15}\)
Hence,
\(\frac{1}{5}\) ÷ 3 = \(\frac{1}{15}\)

Question 4.
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 51
Answer: \(\frac{1}{6}\) ÷ 2 = \(\frac{1}{12}\)

Explanation:
The given numbers are: \(\frac{1}{6}\) and 2
We know that,
\(\frac{a}{b}\) ÷ a  = \(\frac{a}{b}\) × \(\frac{1}{a}\)
a= \(\frac{a}{1}\)
So,
\(\frac{1}{6}\) ÷ 2
= \(\frac{1}{6}\) × \(\frac{1}{2}\)
= \(\frac{ 1 × 1}{2 × 6}\)
= \(\frac{1}{12}\)
Hence,
\(\frac{1}{6}\) ÷ 2 = \(\frac{1}{12}\)

Question 5.
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 52
Answer: \(\frac{1}{3}\) ÷ 5 = \(\frac{1}{15}\)

Explanation:
The given numbers are: \(\frac{1}{3}\) and 5
We know that,
\(\frac{a}{b}\) ÷ a  = \(\frac{a}{b}\) × \(\frac{1}{a}\)
a= \(\frac{a}{1}\)
So,
\(\frac{1}{3}\) ÷ 5
= \(\frac{1}{3}\) × \(\frac{1}{5}\)
= \(\frac{ 1 × 1}{3 × 5}\)
= \(\frac{1}{15}\)
Hence,
\(\frac{1}{3}\) ÷ 5 = \(\frac{1}{15}\)

Question 6.
Big Ideas Math Answers 5th Grade Chapter 10 Divide Fractions 53
Answer: \(\frac{1}{5}\) ÷ 4 = \(\frac{1}{20}\)

Explanation:
The given numbers are: \(\frac{1}{5}\) and 4
We know that,
\(\frac{a}{b}\) ÷ a  = \(\frac{a}{b}\) × \(\frac{1}{a}\)
a= \(\frac{a}{1}\)
So,
\(\frac{1}{5}\) ÷ 4
= \(\frac{1}{5}\) × \(\frac{1}{4}\)
= \(\frac{ 1 × 1}{5 × 4}\)
= \(\frac{1}{20}\)
Hence,
\(\frac{1}{5}\) ÷ 4 = \(\frac{1}{20}\)

Question 7.
Big Ideas Math Answers 5th Grade Chapter 10 Divide Fractions 54
Answer: \(\frac{1}{3}\) ÷ 3 = \(\frac{1}{9}\)

Explanation:
The given numbers are: \(\frac{1}{3}\) and 3
We know that,
\(\frac{a}{b}\) ÷ a  = \(\frac{a}{b}\) × \(\frac{1}{a}\)
a= \(\frac{a}{1}\)
So,
\(\frac{1}{3}\) ÷ 3
= \(\frac{1}{3}\) × \(\frac{1}{3}\)
= \(\frac{ 1 × 1}{3 × 3}\)
= \(\frac{1}{9}\)
Hence,
\(\frac{1}{3}\) ÷ 3 = \(\frac{1}{9}\)

Question 8.
Big Ideas Math Answers 5th Grade Chapter 10 Divide Fractions 55
Answer: \(\frac{1}{8}\) ÷ 2 = \(\frac{1}{16}\)

Explanation:
The given numbers are: \(\frac{1}{8}\) and 2
We know that,
\(\frac{a}{b}\) ÷ a  = \(\frac{a}{b}\) × \(\frac{1}{a}\)
a= \(\frac{a}{1}\)
So,
\(\frac{1}{8}\) ÷ 2
= \(\frac{1}{8}\) × \(\frac{1}{2}\)
= \(\frac{ 1 × 1}{2 × 8}\)
= \(\frac{1}{16}\)
Hence,
\(\frac{1}{8}\) ÷ 2 = \(\frac{1}{16}\)

Question 9.
How many 6s are in \(\frac{1}{2}\)?
Answer: There are \(\frac{1}{12}\) 6s in \(\frac{1}{2}\)

Explanation:
We know that,
\(\frac{a}{b}\) ÷ a  = \(\frac{a}{b}\) × \(\frac{1}{a}\)
a= \(\frac{a}{1}\)
Now,
We have to find the number of 6s in \(\frac{1}{2}\)
So,
\(\frac{1}{2}\) ÷ 6
= \(\frac{1}{2}\) × \(\frac{1}{6}\)
= \(\frac{ 1 × 1}{2 × 6}\)
= \(\frac{1}{12}\)
Hence, from the above,
We can conclude that there are \(\frac{1}{12}\) 6s in \(\frac{1}{2}\)

Question 10.
How many 2s are in \(\frac{1}{3}\) ?
Answer: There are \(\frac{1}{6}\) 2s in \(\frac{1}{3}\)

Explanation:
We know that,
\(\frac{a}{b}\) ÷ a  = \(\frac{a}{b}\) × \(\frac{1}{a}\)
a= \(\frac{a}{1}\)
Now,
We have to find the number of 2s in \(\frac{1}{3}\)
So,
\(\frac{1}{3}\) ÷ 2
= \(\frac{1}{3}\) × \(\frac{1}{2}\)
= \(\frac{ 1 × 1}{2 × 3}\)
= \(\frac{1}{6}\)
Hence, from the above,
We can conclude that there are \(\frac{1}{6}\) 2s in \(\frac{1}{2}\)

Question 11.
Writing
Write and solve a real-life problem for
Big Ideas Math Answers 5th Grade Chapter 10 Divide Fractions 56
Answer:
Suppose a box has 7 chocolates. We have to divide these seven chocolates into further \(\frac{1}{2}\) parts so that the chocolates can be distributed to more people
So,
The each part of chocolate we can get = \(\frac{1}{2}\) ÷ 7
Now,
We know that,
\(\frac{a}{b}\) ÷ a  = \(\frac{a}{b}\) × \(\frac{1}{a}\)
a= \(\frac{a}{1}\)
So,
\(\frac{1}{2}\) ÷ 7
= \(\frac{1}{2}\) × \(\frac{1}{7}\)
= \(\frac{ 1 × 1}{2 × 7}\)
= \(\frac{1}{14}\)
Hence, from the above,
We can conclude that we can get \(\frac{1}{14}\) part of each chocolate.

Question 12.
Reasoning
Complete the statements.

Think and Grow: Modeling Real Life

You melt \(\frac{1}{4}\) quart of soap. You pour the soap into 4 of the same-sized molds. What fraction of a quart of soap does each mold hold?
You are dividing \(\frac{1}{4}\) quart into 4 equal parts, so you need to find \(\frac{1}{4}\) ÷ 4.

Show and Grow

Question 13.
You buy \(\frac{1}{2}\) pound of grapes. You equally divide the grapes into 2 bags. What fraction of a pound of grapes do you put into each bag?
Answer: The fraction of a pound of grapes you put into each bag is: \(\frac{1}{8}\) pound

Explanation:
It is given that you buy \(\frac{1}{2}\) pound of grapes.
It is also given that you equally divide the grapes into 2 bags.
So,
The number of grapes in each bag = \(\frac{1}{2}\) ÷ 2
Now,
The fraction of pound of grapes you put into each bag = \(\frac{The number of grapes in each bag}{2}\)
=  ( \(\frac{1}{2}\) ÷ 2 ) ÷ 2
Now,
We know that,
\(\frac{a}{b}\) ÷ a  = \(\frac{a}{b}\) × \(\frac{1}{a}\)
a= \(\frac{a}{1}\)
So,
( \(\frac{1}{2}\) ÷ 2 ) ÷ 2
= ( \(\frac{1}{2}\) × \(\frac{1}{2}\) ) × \(\frac{1}{2}\)
= \(\frac{ 1 × 1}{2 × 2}\) × \(\frac{1}{2}\)
= \(\frac{1}{4}\) × \(\frac{1}{2}\)
= \(\frac{ 1 × 1}{2 × 4}\)
= \(\frac{1}{8}\)
Hence, from the above
We can conclude that the fraction of pound of grapes in each bag is: \(\frac{1}{8}\) pound

Question 14.
You have \(\frac{1}{8}\) cup of red sand, \(\frac{1}{4}\) cup of blue sand, and \(\frac{1}{2}\) cup of white sand. You equally divide the sand into 3 containers. What fraction of a cup of sand do you pour into each container?
Answer: The fraction of a cup of sand you pour into each container is: \(\frac{7}{24}\)

Explanation:
It is given that you have \(\frac{1}{8}\) cup of red sand, \(\frac{1}{4}\) cup of blue sand, and \(\frac{1}{2}\) cup of white sand.
So,
The total amount of sand = \(\frac{1}{8}\) cup of red sand + \(\frac{1}{4}\) cup of blue sand + \(\frac{1}{2}\) cup of white sand
In addition, we have to see either the numerators are equal or the denominators are equal.
If the numerators are equal we have to ake the denominators also equal.
So,
\(\frac{1}{4}\) is multplied by \(\frac{2}{2}\)
\(\frac{1}{2}\) is multiplied by \(\frac{4}{4}\)
So,
\(\frac{1}{4}\) = \(\frac{2}{8}\)
\(\frac{1}{2}\) = \(\frac{4}{8}\)
So,
\(\frac{1}{8}\) + \(\frac{2}{8}\) + \(\frac{4}{8}\) = \(\frac{7}{8}\)
It is also given that all the sand is equally distributed into 3 containers
So,
The amount of sand in each container = \(\frac{7}{8}\) ÷ 3
Now,
We know that,
\(\frac{a}{b}\) ÷ a  = \(\frac{a}{b}\) × \(\frac{1}{a}\)
a= \(\frac{a}{1}\)
So,
\(\frac{7}{8}\) ÷ 3
= \(\frac{7}{8}\) × \(\frac{1}{3}\)
= \(\frac{ 7 × 1}{8 × 3}\)
= \(\frac{7}{24}\)
Hence, from the above,
We can conclude that the amount of sand in each container is: \(\frac{7}{24}\) cup.

Question 15.
DIG DEEPER!
You, your friend, and your cousin share \(\frac{1}{2}\) of a vegetable pizza and \(\frac{1}{4}\) of a cheese share pizza. The pizzas are the same size. What fraction of a pizza do you get in all?
Big Ideas Math Answers 5th Grade Chapter 10 Divide Fractions 58

Divide. Use a model to help
Answer: The fraction of a pizza you got is: \(\frac{3}{12}\)

Explanation:
It is given that you, your friend, and your cousin share \(\frac{1}{2}\) of a vegetable pizza and \(\frac{1}{4}\) of a cheese share pizza.
So,
The total amount of pizza = \(\frac{1}{2}\) of a vegetable pizza + \(\frac{1}{4}\) of a cheese share pizza
In addition, we have to see either the numerators are equal or the denominators are equal.
If the numerators are equal we have to ake the denominators also equal.
So,
\(\frac{1}{2}\) is multplied by \(\frac{2}{2}\)
So,
\(\frac{1}{2}\) = \(\frac{2}{4}\)
So,
\(\frac{2}{4}\) + \(\frac{1}{4}\) = \(\frac{3}{4}\)
So,
The fraction of pizza each get = \(\frac{The total amount of pizza}{3}\)
= \(\frac{3}{4}\) ÷ 3
Now,
We know that,
\(\frac{a}{b}\) ÷ a  = \(\frac{a}{b}\) × \(\frac{1}{a}\)
a= \(\frac{a}{1}\)
So,
\(\frac{3}{4}\) ÷ 3
= \(\frac{3}{4}\) × \(\frac{1}{3}\)
= \(\frac{ 3 × 1}{4 × 3}\)
= \(\frac{3}{12}\)
Hence, from the above,
We can conclude that the fraction of pizza each get is: \(\frac{3}{12}\)

Divide Unit Fractions by Whole Numbers Homework & Practice 10.4

Question 1.
Big Ideas Math Answers 5th Grade Chapter 10 Divide Fractions 59
Answer: \(\frac{1}{3}\) ÷ 4 = \(\frac{1}{12}\)

Explanation:
The given numbers are: \(\frac{1}{3}\) and 4
We know that,
\(\frac{a}{b}\) ÷ a  = \(\frac{a}{b}\) × \(\frac{1}{a}\)
a= \(\frac{a}{1}\)
So,
\(\frac{1}{3}\) ÷ 4
= \(\frac{1}{3}\) × \(\frac{1}{4}\)
= \(\frac{ 1 × 1}{3 × 4}\)
= \(\frac{1}{12}\)
Hence,
\(\frac{1}{3}\) ÷ 4 = \(\frac{1}{12}\)

Question 2.
Big Ideas Math Answers 5th Grade Chapter 10 Divide Fractions 60
Answer: \(\frac{1}{6}\) ÷ 3 = \(\frac{1}{18}\)

Explanation:
The given numbers are: \(\frac{1}{6}\) and 3
We know that,
\(\frac{a}{b}\) ÷ a  = \(\frac{a}{b}\) × \(\frac{1}{a}\)
a= \(\frac{a}{1}\)
So,
\(\frac{1}{6}\) ÷ 3
= \(\frac{1}{6}\) × \(\frac{1}{3}\)
= \(\frac{ 1 × 1}{6 × 3}\)
= \(\frac{1}{18}\)
Hence,
\(\frac{1}{6}\) ÷ 3 = \(\frac{1}{18}\)

Question 3.
Big Ideas Math Answers 5th Grade Chapter 10 Divide Fractions 61
Answer: \(\frac{1}{4}\) ÷ 5 = \(\frac{1}{20}\)

Explanation:
The given numbers are: \(\frac{1}{4}\) and 5
We know that,
\(\frac{a}{b}\) ÷ a  = \(\frac{a}{b}\) × \(\frac{1}{a}\)
a= \(\frac{a}{1}\)
So,
\(\frac{1}{4}\) ÷ 5
= \(\frac{1}{4}\) × \(\frac{1}{5}\)
= \(\frac{ 1 × 1}{5 × 4}\)
= \(\frac{1}{20}\)
Hence,
\(\frac{1}{4}\) ÷ 5 = \(\frac{1}{20}\)

Divide. Use a model to help.

Question 4.
Big Ideas Math Answers 5th Grade Chapter 10 Divide Fractions 62
Answer: \(\frac{1}{5}\) ÷ 9 = \(\frac{1}{45}\)

Explanation:
The given numbers are: \(\frac{1}{5}\) and 9
We know that,
\(\frac{a}{b}\) ÷ a  = \(\frac{a}{b}\) × \(\frac{1}{a}\)
a= \(\frac{a}{1}\)
So,
\(\frac{1}{5}\) ÷ 9
= \(\frac{1}{5}\) × \(\frac{1}{9}\)
= \(\frac{ 1 × 1}{5 × 9}\)
= \(\frac{1}{45}\)
Hence,
\(\frac{1}{5}\) ÷ 9 = \(\frac{1}{45}\)

Question 5.
Big Ideas Math Answers 5th Grade Chapter 10 Divide Fractions 63
Answer: \(\frac{1}{8}\) ÷ 6 = \(\frac{1}{48}\)

Explanation:
The given numbers are: \(\frac{1}{8}\) and 6
We know that,
\(\frac{a}{b}\) ÷ a  = \(\frac{a}{b}\) × \(\frac{1}{a}\)
a= \(\frac{a}{1}\)
So,
\(\frac{1}{8}\) ÷ 6
= \(\frac{1}{8}\) × \(\frac{1}{6}\)
= \(\frac{ 1 × 1}{8 × 6}\)
= \(\frac{1}{48}\)
Hence,
\(\frac{1}{8}\) ÷ 6 = \(\frac{1}{48}\)

Question 6.
Big Ideas Math Answers 5th Grade Chapter 10 Divide Fractions 64
Answer: \(\frac{1}{7}\) ÷ 4 = \(\frac{1}{28}\)

Explanation:
The given numbers are: \(\frac{1}{7}\) and 4
We know that,
\(\frac{a}{b}\) ÷ a  = \(\frac{a}{b}\) × \(\frac{1}{a}\)
a= \(\frac{a}{1}\)
So,
\(\frac{1}{7}\) ÷ 4
= \(\frac{1}{7}\) × \(\frac{1}{4}\)
= \(\frac{ 1 × 1}{7 × 4}\)
= \(\frac{1}{28}\)
Hence,
\(\frac{1}{7}\) ÷ 4 = \(\frac{1}{28}\)

Question 7.
YOU BE THE TEACHER
Your friend divides \(\frac{1}{3}\) by 7 to get \(\frac{1}{21}\). He checks his answer by multiplying \(\frac{1}{21}\) × \(\frac{1}{3}\). Does your friend check his answer correctly? Explain.
Answer: No, your friend does not check his answer correctly

Explanation:
It is given that your friend divides \(\frac{1}{3}\) by 7 to get \(\frac{1}{21}\).
We know that,
\(\frac{a}{b}\) ÷ a  = \(\frac{a}{b}\) × \(\frac{1}{a}\)
a= \(\frac{a}{1}\)
So,
\(\frac{1}{3}\) ÷ 7
= \(\frac{1}{3}\) × \(\frac{1}{7}\)
= \(\frac{ 1 × 1}{7 × 3}\)
= \(\frac{1}{21}\)
It is also given that your friend checks his answer by multiplying \(\frac{1}{21}\) × \(\frac{1}{3}\).
Now,
\(\frac{1}{21}\) × \(\frac{1}{3}\)
= \(\frac{1 × 1}{21 × 3}\)
= \(\frac{1}{63}\)
But, your friend wanted to check whether \(\frac{1}{21}\) × \(\frac{1}{3}\) = \(\frac{1}{7}\)
But, the value becomes \(\frac{1}{63}\)
Hence, from the above,
We can conclude that your friend does not check the answer correctly.

Question 8.
Logic
Find the missing numbers.

Question 9.
Modeling Real Life
You win tickets that you can exchange for prizes. You exchange \(\frac{1}{5}\) of your tickets and then divide them equally among 3 prizes. What fraction of your tickets do you spend on each prize?
Answer: The fraction of your tickets you spend on each prize is: \(\frac{1}{15}\)

Explanation:
It is given that you win tickets that you can exchange for prizes.
It is also given that you exchange \(\frac{1}{5}\) of your tickets and then divide them equally among 3 prizes
So,
The fraction of the tickets spent on each prize = \(\frac{The value of Exchange}{The number of prizes}\)
= \(\frac{1}{5}\) ÷ 3
Now,
We know that,
\(\frac{a}{b}\) ÷ a  = \(\frac{a}{b}\) × \(\frac{1}{a}\)
a= \(\frac{a}{1}\)
So,
\(\frac{1}{5}\) ÷ 3
= \(\frac{1}{5}\) × \(\frac{1}{3}\)
= \(\frac{ 1 × 1}{5 × 3}\)
= \(\frac{1}{15}\)
Hence, from the above,
We can conlude that the fraction of tickets you spend on each prize is: \(\frac{1}{15}\)

Question 10.
DIG DEEPER!
You have \(\frac{1}{8}\) gallon of melted crayon wax. You pour the wax equally into 8 different molds to make new crayons. What fraction of a cup of melted wax is in each mold? Think: 1 gallon is 16 cups.
Big Ideas Math Answers 5th Grade Chapter 10 Divide Fractions 66
Answer: The fraction of a cup of melted wax in each mold is: \(\frac{1}{4}\)

Explanation:
It is given that you have \(\frac{1}{8}\) gallon of melted crayon wax.
It is also given that you pour the wax equally into 8 different molds to make new crayons.
So,
The fraction of melted crayon wax in each mold in gallons = \(\frac{The total amount of melted crayon wax }{The number of molds}\)
= \(\frac{1}{8}\) ÷ 8
Now,
We know that,
\(\frac{a}{b}\) ÷ a  = \(\frac{a}{b}\) × \(\frac{1}{a}\)
a= \(\frac{a}{1}\)
So,
\(\frac{1}{8}\) ÷ 8
= \(\frac{1}{8}\) × \(\frac{1}{8}\)
= \(\frac{ 1 × 1}{8 × 8}\)
= \(\frac{1}{64}\) gallons
But, it is given that
1 gallon = 16 cups
So,
The total number of cups that the melted crayon wax contained = \(\frac{1}{64}\) × \(\frac{16}{1}\)
= \(\frac{1 × 16 }{64 × 1}\)
= \(\frac{1}{4}\)
Hence, from the above,
We can conclude that there are \(\frac{1}{4}\) cups of melted crayon wax in each mold.

Review & Refresh

Question 11.
0.9 ÷ 0.1 = ___
Answer: 0.9 ÷ 0.1 = 9

Explanation:
The given decimal numbers are: 0.9 and 0.1
The representation of the decimal numbers in the fraction form is: \(\frac{9}{10}\) and \(\frac{1}{10}\)
Now,
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
So,
\(\frac{9}{10}\) ÷ \(\frac{1}{10}\)  = \(\frac{9}{10}\) × \(\frac{10}{1}\)
= \(\frac{ 9 × 10}{10 × 1}\)
= 9
Hence, 0.9 ÷ 0.1 = 9

Question 12.
38.6 ÷ 100 = ___

Answer: 38.6 ÷ 100 = 0.386

Explanation:
The given numbers are: 38.6 and 100
The representation of the numbers in the fraction form is: \(\frac{386}{10}\) and \(\frac{100}{1}\)
Now,
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
So,
\(\frac{386}{10}\) ÷ \(\frac{100}{1}\)  = \(\frac{386}{10}\) × \(\frac{1}{100}\)
= \(\frac{ 386 × 1}{100 × 10}\)
= \(\frac{386}{1000}\)
= 0.386
Hence, 38.6 ÷ 100 = 0.386

Question 13.
2.57 ÷ 0.01 = ___
Answer: 2.57 ÷ 0.01 = 257

Explanation:
The given decimal numbers are: 2.57 and 0.01
The representation of the decimal numbers in the fraction form is: \(\frac{257}{100}\) and \(\frac{1}{100}\)
Now,
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
So,
\(\frac{257}{100}\) ÷ \(\frac{1}{100}\)  = \(\frac{257}{100}\) × \(\frac{100}{1}\)
= \(\frac{ 257 × 100}{100 × 1}\)
= 257
Hence, 2.57 ÷ 0.01 = 257

Lesson 10.5 Problem Solving: Fraction Division

Explore and Grow

You want to make a \(\frac{1}{3}\) batch of the recipe. How you can use division to find the amount of each ingredient you need?
Answer:
It is given that you want to make a \(\frac{1}{3}\) batch of the recipe.
So,
From \(\frac{1}{3}\),
1 represents a batch of the recipe
3 represents the total number of ingredients in a batch
So,
The amount of each ingredient you need = \(\frac{The amount of the batch of the recipe }{The total number of ingredients}\)
= \(\frac{1}{3}\) ÷ 3
Now,
We know that,
\(\frac{a}{b}\) ÷ a  = \(\frac{a}{b}\) × \(\frac{1}{a}\)
a= \(\frac{a}{1}\)
So,
\(\frac{1}{3}\) ÷ 3
= \(\frac{1}{3}\) × \(\frac{1}{3}\)
= \(\frac{ 1 × 1}{3 × 3}\)
= \(\frac{1}{9}\)
Hence, from the above,
We can conclude that the amount of each ingredient you need is: \(\frac{1}{9}\)

Reasoning
Without calculating, explain how you can tell whether you need more than or less than 1 tablespoon of olive oil?
Answer: You need less than 1 tablespoon of olive oil

Explanation:
From the above problem,
The amount of each ingredient is: \(\frac{1}{9}\)
Since the amount of each ingredient is less than 1, you need less than 1 tablespoon of olive oil

Think and Grow: Problem Solving: Fraction Division

Example
You have 4 cups of yellow paint and 3 cups of blue paint. How many batches of green paint can you make?
Big Ideas Math Answers 5th Grade Chapter 10 Divide Fractions 67

Understand the Problem

What do you know?

  • You have 4 cups of yellow paint and 3 cups of blue paint.
  • One batch of green paint is made of \(\frac{1}{2}\) cup of yellow and \(\frac{1}{3}\) cup of blue.

What do you need to find?

  • You need to find how many batches of green paint you can make.

Make a Plan
How will you solve?

  • Find how many batches are possible from yellow, and how many from blue.
  • Choose the lesser number of batches.

Solve

So, you can make 8 batches of green paint.

Show and Grow

Question 1.
In the example, explain why you choose the fewer number of batches.
Answer: In the above example, the yellow paint has the less number of batches as the amount of each batch of yellow paint-filled is more than the batch of green paint
Hence,
We choose the fewer number of batches of yellow paint

Apply and Grow: Practice

Understand the problem. What do you know? What do you need to find? Explain.

Question 2.
A landowner donates 3 acres of land to a city. The mayor of the city uses 1 acre of the land for a playground and the rest of the land for community garden plots. Each garden plot is \(\frac{1}{3}\) acre. How many plots are there?
Understand the problem. Then make a plan. How will you solve it? Explain?
Answer: The number of plots in the community is: 6

Explanation:
It is given that a landowner donates 3 acres of land to a city and the mayor of the city uses 1 acre of the land for a playground and the rest of the land for community garden plots.
So,
The portion of the land used for community garden plots is: 2 acres
It is also given that each garden plot is \(\frac{1}{3}\) acre.
So,
The number of plots = \(\frac{The portion of the land used for community garden plots}{The area of each garden plot}\)
= 2 ÷ \(\frac{1}{3}\)
= 2 × \(\frac{3}{1}\)
= \(\frac{2}{1}\) × \(\frac{3}{1}\)
= 6
Hence, from the above,
We can conclude that there are 6 plots

Question 3.
A craftsman uses \(\frac{3}{4}\) gallon of paint to paint 4 identical dressers. He uses the same amount of paint on each dresser. How much paint does he use to paint 7 of the same dressers?
Big Ideas Math Answers 5th Grade Chapter 10 Divide Fractions 69
Answer: The paint used by the craftsman to paint 7 of the same dressers is: \(\frac{21}{16}\)

Explanation:
It is given that a craftsman uses \(\frac{3}{4}\) gallon of paint to paint 4 identical dressers.
So,
The paint used to paint each dresser = \(\frac{3}{4}\) ÷ 4
= \(\frac{3}{4}\) × \(\frac{1}{4}\)
= \(\frac{3}{16}\) gallon
So,
The amount of paint used to paint the 7 identical dressers = \(\frac{The paint used to paint each dresser}{1}\) × 7
= \(\frac{3}{16}\) × \(\frac{7}{1}\)
= \(\frac{3 × 7}{16 × 1}\)
= \(\frac{21}{16}\) gallon
Hence, from the above,
We can conclude that the paint used to paint 7 identical dressers is: \(\frac{21}{16}\) gallon

Question 4.
An airplane travels 125 miles in \(\frac{1}{4}\) hour. It travels the same number of miles each hour. How many miles does the plane travel in 5 hours?
Answer: The number of miles the plane travel in 5 hours is: 2,500 miles

Explanation:
It is given that an airplane travels 125 miles in \(\frac{1}{4}\) hour
So,
The number of miles traveled by plane in 1 hour = 125 ÷ \(\frac{1}{4}\)
= 125 × \(\frac{4}{1}\)
= 125 × 4
= 500 miles
So,
The number of miles traveled by plane in 5 hours = ( The number of miles traveled by plane in 1 hour ) × 5
= 500 × 5
= 2,500 miles
Hence, from the above,
We can conclude that the number of miles traveled by plane in 5 hours is: 2,500 miles

Question 5.
You make bows for gifts using \(\frac{2}{3}\) yard of ribbon for each bow. You have 4 feet of red ribbon and 5 feet of green ribbon. How many bows can you make?
Answer: The number of bows you can make is: 2 bows

Explanation:
It is given that you make bows for gifts using \(\frac{2}{3}\) yard of ribbon for each bow.
It is also given that you have 4 feet of red ribbon and 5 feet of green ribbon
So,
The total length of ribbon = 5 + 4 = 9 feet
we know that,
1 foot = \(\frac{1}{3}\) yards
So,
9 feet = 9 × \(\frac{1}{3}\) yards
= \(\frac{9}{1}\) yards × \(\frac{1}{3}\) yards
= 3 yards
So,
The number of bows you can make = \(\frac{2}{3}\) yards × 3
= 2 bows
Hence, from the above,
We can conclude that the number of bows we can make is: 2

Question 6.
A landscaper buys 1 gallon of plant fertilizer. He uses \(\frac{1}{5}\) of the fertilizer, and then divides the rest into 3 smaller bottles. How many gallons does he put into each bottle?
Answer: The number of gallons he put into each bottle is: \(\frac{4}{15}\)

Explanation:
It is given that a landscaper buys 1 gallon of plant fertilizer and he uses \(\frac{1}{5}\) of the fertilizer
So,
The remaining amount of the fertilizer = 1 – \(\frac{1}{5}\)
= \(\frac{4}{5}\) gallons
It is also given that he divided the remaining amount of fertilizer into 3 smaller bottles.
So,
The amount of fertilizer put into each bottle = \(\frac{The remaining amount of the fertilizer}{The total number of bottles}\)
= \(\frac{4}{5}\) ÷ 3
= \(\frac{4}{5}\) × \(\frac{1}{3}\)
= \(\frac{4 × 1}{5 × 3}\)
= \(\frac{4}{15}\) gallons
hence, from the above,
We can conclude that the amount of remaining fertilizer put into each bottle is: \(\frac{4}{15}\) gallons

Think and Grow: Modeling Real Life

Example
A sponsor donates $0.10 to a charity for every \(\frac{1}{4}\) kilometer of the triathlon an athlete completes. The athlete completes the entire triathlon. How much money does the sponsor donate?
Big Ideas Math Answers 5th Grade Chapter 10 Divide Fractions 70
Think: What do you know? What do you need to find? How will you solve?
Write and solve an equation.
Add 1.9, 90, and 21.1 to find how many kilometers the athlete completes.
Divide the sum by \(\frac{1}{4}\) to find how many \(\frac{1}{4}\) kilometers the athlete completes.
Multiply the quotient by $0.10 to find how much money the sponsor donates.
Big Ideas Math Answers 5th Grade Chapter 10 Divide Fractions 71
Let m represent the total amount of money donated.

Show and Grow

Question 7.
You earn $5 for every \(\frac{1}{2}\) hour you do yard work. How much money do you earn in 1 week?
Big Ideas Math Answers 5th Grade Chapter 10 Divide Fractions 73
Answer: The amount you earn in 1 week is: $700

Explanation:
The given table is:
Big Ideas Math Answers 5th Grade Chapter 10 Divide Fractions 73
From the above table,
The total amount of time = 5\(\frac{1}{2}\) + 3 + 1\(\frac{1}{2}\)
= \(\frac{11}{2}\) + 3 + \(\frac{3}{2}\)
= \(\frac{11 + 3}{2}\) + 3
= 7 + 3
= 10 hours
It is given that you earn $5 for every \(\frac{1}{2}\) hour you do yard work
So,
The amount earned in 10 hours in a day = 10 ÷\(\frac{1}{2}\) × 5 ( Since we have the time in hours but the money earned is given in half an hour basis )
= 20 × 5
= $100
We know that 1 week = 7 days
So,
The amount earned in 1 week = 100 × 7 = $700
hence, from the above,
We can conclude that we can earn $700 in a week

Problem Solving: Fraction Division Homework &  10.5

Understand the problem. Then make a plan. How will you solve? Explain.

Question 1.
A train travels 75 miles in \(\frac{1}{2}\) hour. How many miles does the train travel in 8 hours?
Answer: The number of miles the train travel in  hours is: 1,200 miles

Explanation:
It is given that a train travels 75 miles in \(\frac{1}{2}\) hour.
So,
The number of miles the train travel in 1 hour = 75 ÷ \(\frac{1}{2}\)
= 75 × 2
= 150 miles
So,
The number of miles the train travel in 8 hours = The number of miles traveled by train in 1 hour × 8
= 150 × 8
= 1,200 miles
Hence, from the above,
We can conclude that the train travels 1,200 miles in 8 hours.

Question 2.
You need \(\frac{2}{3}\) yard of fabric to create a headband. You have 12 feet of blue fabric and 4 feet of yellow fabric. How many headbands can you make with all of the fabric?
Answer: The number of headbands you can make with all of the fabric is: 8 headbands

Explanation:
It is given that you need \(\frac{2}{3}\) yard of fabric to create a headband.
It is also given that you have 12 feet of blue fabric and 4 feet of yellow fabric.
So,
The total length of the fabric = 12 + 4 = 16 feet
We know that
1 foot = \(\frac{1}{3}\) yards
So,
16 feet = \(\frac{16}{3}\) yards
So,
The number of headbands you can create with all the fabric = \(\frac{The total length of the fabric}{The length of each fabric}\)
= \(\frac{16}{3}\) ÷ \(\frac{2}{3}\)
= \(\frac{16}{3}\) × \(\frac{3}{2}\)
= \(\frac{16 × 3}{3 × 2}\)
= 8 headbands
Hence, from the above,
We can conclude that we can create 8 headbands with all the fabric.

Question 3.
An art teacher has 8 gallons of paint. Her class uses \(\frac{3}{4}\) of the paint. The teacher divides the rest of the paint into 4 bottles. How much paint is in each bottle?
Answer: The amount of paint in each bottle is: \(\frac{1}{2}\)

Explanation:
It is given that an art teacher has 8 gallons of paint and her class uses \(\frac{3}{4}\) of the paint.
So,
The remaining amount of paint = \(\frac{1}{4}\) × 8
= \(\frac{1}{4}\) × \(\frac{8}{1}\)
=\(\frac{1 × 8}{4 × 1}\)
= 2 gallons
It is also given that the remaining amount of the paint divided into 4 bottles by the teacher
So,
The amount of paint present in each bottle = 2 ÷ 4
= \(\frac{1}{2}\) gallons
Hence, from the above,
We can conclude that the amout of paint present in each bottle is: \(\frac{1}{2}\) gallons

Question 4.
You mix 3\(\frac{1}{4}\) cups of frozen strawberries and 4\(\frac{1}{2}\) cups of frozen blueberries in a bowl. A smoothie requires \(\frac{1}{2}\) cup of your berry mix. How many smoothies can you make?
Answer: The number of smoothies you can make is:

Explanation:
It is given that you mix 3\(\frac{1}{4}\) cups of frozen strawberries and 4\(\frac{1}{2}\) cups of frozen blueberries in a bowl.
So,
The amount of berry mix = 3\(\frac{1}{4}\) cups of frozen strawberries + 4\(\frac{1}{2}\) cups of frozen blueberries
= 3\(\frac{1}{4}\) + 4\(\frac{1}{2}\)
= \(\frac{13}{4}\) + \(\frac{9}{2}\)
In addition, equate the denominators
So,
Multiply \(\frac{9}{2}\) with \(\frac{2}{2}\)
So,
\(\frac{9}{2}\) = \(\frac{18}{4}\)
So,
The amount of berry mix = \(\frac{13}{4}\) + \(\frac{18}{4}\)
= \(\frac{31}{4}\)
Now,
It is also given that the smoothie requires \(\frac{1}{2}\) cup of your berry mix.
So,
The number of smoothies = \(\frac{31}{4}\) ÷ \(\frac{1}{2}\)
= \(\frac{31}{4}\) × \(\frac{2}{1}\)
= \(\frac{31 × 2}{4 × 1}\)
= \(\frac{31}{2}\)
Hence, from the above,
We can conclude that the number of smoothies we can make are: \(\frac{31}{2}\)

Question 5.
Modeling Real Life
A sponsor donates $0.10 for every \(\frac{1}{4}\) dollar donated at the locations shown. How much money does the sponsor donate?
Big Ideas Math Answers 5th Grade Chapter 10 Divide Fractions 74
Answer: The amount of money the sponsor donates is: $40.4

Explanation:
The given table is:
Big Ideas Math Answers 5th Grade Chapter 10 Divide Fractions 74
From the above table,
The total amount of money collected = 25.25 + 12.50 + 63.25
= $101
It is given that a sponsor donates $0.10 for every \(\frac{1}{4}\) dollar
So,
the total amount of donated = The total amount of money collected ÷ \(\frac{1}{4}\) × 0.10
= 101 ÷ \(\frac{1}{4}\) × 0.10
= 101 × 4 × 0.10
= 04 × 0.10
= $40.4
Hence, from the above,
We can conclude that the amount of money donated by a sponsor is: $40.4

Question 6.
DIG DEEPER!
A nurse earns $16 for every \(\frac{1}{2}\) hour at work. How much money does she earn in 5 days?
Big Ideas Math Answers 5th Grade Chapter 10 Divide Fractions 75
Answer: The money she earns in 5 days is: $1,280

Explanation:
The given table is:
Big Ideas Math Answers 5th Grade Chapter 10 Divide Fractions 75
From the above table,
The total amount of time = 6\(\frac{3}{4}\) + 1 + \(\frac{1}{4}\)
= \(\frac{27}{4}\) + 1 + \(\frac{1}{4}\)
= \(\frac{27 + 1}{4}\) + 1
= 7 + 1
= 8 hours
It is given that a nurse earns $16 for every \(\frac{1}{2}\) hour at work.
So,
The money she earned for 1 hour = 16 ÷ \(\frac{1}{2}\)
= 16 × 2 = $32
So,
The money earned for 8 hours = The money earned in 1 hour × 8
= 32 × 8 = $256
So,
The money earned in 5 days = The money earned in 1 day × 5
= 256 × 5 = $1,280
hence, from the above,
we can conclude that she can earn $1,280 in 5 days.

Review & Refresh

Find the quotient. Then check your answer.

Question 7.
Big Ideas Math Answers 5th Grade Chapter 10 Divide Fractions 76
Answer:  186 ÷ 12 = 1 R 4

Explanation:
Let 185.88 be rounded to 186
So,
By using the partial quotients method,
186 ÷ 12 = ( 120 + 36 + 24 ) ÷ 12
= ( 120 ÷ 12 ) + ( 36 ÷ 12 ) + ( 24 ÷ 12 )
= 10 + 3 + 2
= 17 R 4
Hence, 186 ÷ 12 = 17 R 4

Question 8.
Big Ideas Math Answers 5th Grade Chapter 10 Divide Fractions 77
Answer: 74 ÷ 24 = 3 R 2

Explanation:
Let 74.4 be rounded to 74
So,
By using the partial quotients method,
74 ÷ 24 = 72 ÷ 24
= 3 R 2
Hence, 74 ÷ 24 = 3 R 2

Question 9.
Big Ideas Math Answers 5th Grade Chapter 10 Divide Fractions 78
Answer: 42 ÷ 46 = 0.9

Explanation:
Let 42.32 be rounded to 42
So,
By using the partial quotients method,
42 ÷ 46 = 0.9
Hence,
42 ÷ 46 = 0.9

Divide Fractions Performance Task 10

Your city has a robotics competition. Each team makes a robot that travels through a maze. The time each robot spends in the maze is used to find the team’s score.
1. One-third of the students in your grade participate in the competition. The number of participating students is divided into 12 teams.
Big Ideas Math Answers 5th Grade Chapter 10 Divide Fractions 79
a. What fraction of the total number of students in your grade is on each team?
Answer: The fraction of the total number of students in your grade is: \(\frac{1}{36}\)

Explanation:
It is given that there are \(\frac{1}{3}\) of the students in your grade are participating in the competition.
It is also given that the participating students are divided into 12 teams.
So,
The fraction of the total number of students in each team = \(\frac{The number of participating students}{Th total number of teams}\)
= \(\frac{1}{3}\) ÷ 12
= \(\frac{1}{3}\) ÷ \(\frac{12}{1}\)
= \(\frac{1}{3}\) × \(\frac{1}{12}\)
= \(\frac{1 × 1}{12 × 3}\)
= \(\frac{1}{36}\)
Hence, from the above,
We can conclude that the fraction of students that are in each team is: \(\frac{1}{36}\)

b. There are 3 students on each team. How many students are in your grade?
Answer: The number of students in your grade is: 36

Explanation:
It is given that the number of students is divided into 12 teams
It is also given that there are 3 students on each team
So,
The total number of students = The number of teams × The number of students in each team
= 12 × 3 = 36 students
Hence, from the above,
We can conclude that there are 36 students in your grade

Question 2.
The maze for the competition is shown.
a. Write the length of the maze in feet.
Big Ideas Math Answers 5th Grade Chapter 10 Divide Fractions 80
Answer:
The given maze for the competition is:
Big Ideas Math Answers 5th Grade Chapter 10 Divide Fractions 80
From the above maze,
The total length of the maze is: 8 feet 6 inches
We know that,
1 foot = 12 inches
Hence,
1 inch = \(\frac{1}{12}\) feet
So,
6 inches = 6 × \(\frac{1}{12}\)
= \(\frac{1}{12}\) × \(\frac{6}{1}\)
= \(\frac{1}{2}\) feet
So,
The total length of the maze in feet = 8 feet + \(\frac{1}{2}\) feet
= 8.5 feet or 8\(\frac{1}{2}\) feet

b. The length of the maze is divided into 6 equal sections. What is the length of each section of the maze?
Answer: The length of each section of the maze is: \(\frac{17}{12}\) feet

Explanation:
From the above Exercise,
The total length of the maze in feet is: 8.5 feet or 8\(\frac{1}{2}\) feet
It is given that the length of the maze is divided into 6 equal sections
So,
The length of each section of the maze = 8\(\frac{1}{2}\) ÷ 6
= 8\(\frac{1}{2}\) ÷ \(\frac{6}{1}\)
= 8\(\frac{1}{2}\) × \(\frac{1}{6}\)
= \(\frac{17}{2}\) × \(\frac{1}{6}\)
= \(\frac{17}{12}\) feet
Hence, from the above,
We can conclude that the length of each section in a maze is: \(\frac{17}{12}\) feet

Question 3.
Each team has 200 seconds to complete the maze. The rules require judges to use the expression (200 – x) ÷ \(\frac{1}{5}\), where x is the total number of seconds, to find a team’s total score.
a. Your robot completes the maze in 3 minutes 5 seconds. How many points does your team earn?
Answer: The number of points your team earn is: 75 points

Explanation:
It is given that each team has 200 seconds to complete the maze and the rules require judges to use the expression (200 – x) ÷ \(\frac{1}{5}\) where x is the total number of seconds
It is also given that your robot completes the maze in 3 minutes 5 seconds
We know that,
1 minute = 60 seconds
So,
The time taken by the robot to complete the maze in seconds = ( 3 × 60 ) + 5
= 185 seconds
So,
x= 185
So,
200 – x = 200 – 185 = 15
So,
The number of points the team earned = ( 200 – x ) ÷ \(\frac{1}{5}\)
= 15 ÷ \(\frac{1}{5}\)
= 15 × 5
= 75 points
Hence, from the above,
We can conclude that the number of points earned by the team is: 75 points

b. Do you think the team with the most points or the fewest points wins? Use an example to justify your answer.
Answer: The team with the most points wins the competition because
Reason:
Suppose team A takes 2 minutes and team B takes 3 minutes to complete the competition
So,
The time is taken by team A in seconds = 120 seconds
So,
x= 120
So,
200 – x = 200 – 120 = 80
Now,
The time is taken by team B in seconds = 180 seconds
So,
x= 180
So,
200 – x = 200 – 180 = 20
Now,
The number of points earned by team A = 80 ÷ \(\frac{1}{5}\)
= 400 points
The number of points earned by team B = 20 ÷ \(\frac{1}{5}\)
= 100 points
Hence, from the above,
We can conclude that the team with more points wins the competition

Divide Fractions Activity

Fraction Connection: Division

Directions:

  1. Players take turns rolling three dice.
  2. On your turn, evaluate the expression indicated by your roll and cover the answer.
  3. The first player to get four in a row, horizontally, vertically, or diagonally, wins!

Big Ideas Math Answers 5th Grade Chapter 10 Divide Fractions 81
Big Ideas Math Answers 5th Grade Chapter 10 Divide Fractions 82

Divide Fractions Chapter Practice 10

10.1 Interpret Fractions as Division

Divide. Use a model to help.

Question 1.
1 ÷ 2 = ___
Answer: 1 ÷ 2 = \(\frac{1}{2}\)

Explanation:
We know that,
a ÷ b = \(\frac{a}{b}\)
Hence,
1 ÷ 2 = \(\frac{1}{2}\)

Question 2.
3 ÷ 10 = __
Answer: 3 ÷ 10 = \(\frac{3}{10}\)

Explanation:
We know that,
a ÷ b = \(\frac{a}{b}\)
Hence,
3 ÷ 10 = \(\frac{3}{10}\)

Question 3.
4 ÷ 7 = __
Answer: 4 ÷ 7 = \(\frac{4}{7}\)

Explanation:
We know that,
a ÷ b = \(\frac{a}{b}\)
Hence,
4 ÷ 7 = \(\frac{4}{7}\)

Question 4.
11 ÷ 15 = ___
Answer: 11 ÷ 15 = \(\frac{11}{15}\)

Explanation:
We know that,
a ÷ b = \(\frac{a}{b}\)
Hence,
11 ÷ 15 = \(\frac{11}{15}\)

Question 5.
8 ÷ 9 = ___
Answer: 8 ÷ 9 = \(\frac{8}{9}\)

Explanation:
We know that,
a ÷ b = \(\frac{a}{b}\)
Hence,
8 ÷ 9 = \(\frac{8}{9}\)

Question 6.
13 ÷ 20 = ___
Answer: 13 ÷ 20 = \(\frac{13}{20}\)

Explanation:
We know that,
a ÷ b = \(\frac{a}{b}\)
Hence,
13 ÷ 20 = \(\frac{13}{20}\)

Question 7.
Modeling Real Life
Nine friends equally share 12 apples. What fraction of an apple does each friend get?
Answer: The fraction of an apple each friend get is: \(\frac{9}{12}\)

Explanation:
It is given that nine friends equally share 12 apples.
So,
The fraction of an apple each friend get = \(\frac{The number of friends}{The number of apples}\)
= \(\frac{9}{12}\)
Hence, from the above,
We can conclude that the fraction of an apple each friend get is: \(\frac{9}{12}\)

10.2 Mixed Numbers as Quotients

Divide. Use a model to help

Question 8.
8 ÷ 3 = ___

Answer: 8 ÷ 3 = 2\(\frac{2}{3}\)

Explanation:
We know that,
a ÷ b = \(\frac{a}{b}\)
a\(\frac{b}{c}\) = \(\frac{ac + b}{c}\)
\(\frac{a}{b}\) = Remainder\(\frac{Quotient}{Divisor}\)
So,
8 ÷ 3 = \(\frac{8}{3}\)
By using the partial quotients method,
8 ÷ 3 = 6 ÷ 3
= 2 R 2
Hence,
8 ÷ 3 = 2\(\frac{2}{3}\)

Question 9.
6 ÷ 5 = ___
Answer: 6 ÷ 5 = 1\(\frac{1}{5}\)

Explanation:
We know that,
a ÷ b = \(\frac{a}{b}\)
a\(\frac{b}{c}\) = \(\frac{ac + b}{c}\)
\(\frac{a}{b}\) = Remainder\(\frac{Quotient}{Divisor}\)
So,
6 ÷ 5 = \(\frac{6}{5}\)
By using the partial quotients method,
6 ÷ 5 = 5 ÷ 5
= 1 R 1
Hence,
6 ÷ 5 = 1\(\frac{1}{5}\)

Question 10.

10 ÷ 4 = ___
Answer: 10 ÷ 4 = 2\(\frac{2}{4}\)

Explanation:
We know that,
a ÷ b = \(\frac{a}{b}\)
a\(\frac{b}{c}\) = \(\frac{ac + b}{c}\)
\(\frac{a}{b}\) = Remainder\(\frac{Quotient}{Divisor}\)
So,
10 ÷ 4 = \(\frac{10}{4}\)
By using the partial quotients method,
10 ÷ 4 = 8 ÷ 4
= 2 R 2
Hence,
10 ÷ 4 = 2\(\frac{2}{4}\)

Question 11.

20 ÷ 11 = __
Answer: 20 ÷ 11 = 1\(\frac{9}{11}\)

Explanation:
We know that,
a ÷ b = \(\frac{a}{b}\)
a\(\frac{b}{c}\) = \(\frac{ac + b}{c}\)
\(\frac{a}{b}\) = Remainder\(\frac{Quotient}{Divisor}\)
So,
20 ÷ 11 = \(\frac{20}{11}\)
By using the partial quotients method,
20 ÷ 11 = 11 ÷ 11
= 1 R 9
Hence,
20 ÷ 11 = 1\(\frac{9}{11}\)

Question 12.

25 ÷ 2 = ___
Answer: 25 ÷ 2 = 12\(\frac{1}{2}\)

Explanation:
We know that,
a ÷ b = \(\frac{a}{b}\)
a\(\frac{b}{c}\) = \(\frac{ac + b}{c}\)
\(\frac{a}{b}\) = Remainder\(\frac{Quotient}{Divisor}\)
So,
25 ÷ 2 = \(\frac{25}{2}\)
By using the partial quotients method,
25 ÷ 2 = 24 ÷ 2
= 12 R 1
Hence,
25 ÷ 2 = 12\(\frac{1}{2}\)

Question 13.

64 ÷ 9 = ___
Answer: 64 ÷ 9 = 7\(\frac{1}{9}\)

Explanation:
We know that,
a ÷ b = \(\frac{a}{b}\)
a\(\frac{b}{c}\) = \(\frac{ac + b}{c}\)
\(\frac{a}{b}\) = Remainder\(\frac{Quotient}{Divisor}\)
So,
64 ÷ 9 = \(\frac{64}{9}\)
By using the partial quotients method,
64 ÷ 9 = 63 ÷ 9
= 7 R 1
Hence,
64 ÷ 9 = 7\(\frac{1}{9}\)

10.3 Divide Whole Numbers by Unit Fractions

Divide. Use a model to help.

Question 14.
Big Ideas Math Answers Grade 5 Chapter 10 Divide Fractions 83
Answer: 4 ÷ \(\frac{1}{2}\) = 8

Explanation:
The given numbers are: 4 and \(\frac{1}{2}\)
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
So,
4 ÷ \(\frac{1}{2}\)  = 4 × \(\frac{2}{1}\)
= \(\frac{4}{1}\) × \(\frac{2}{1}\)
= \(\frac{ 2 × 4}{1 × 1}\)
= 8
Hence,
4÷ \(\frac{1}{2}\) = 8

Question 15.
Big Ideas Math Answers Grade 5 Chapter 10 Divide Fractions 84
Answer: 6 ÷ \(\frac{1}{5}\) = 30

Explanation:
The given numbers are: 6 and \(\frac{1}{5}\)
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
So,
6 ÷ \(\frac{1}{5}\)  = 6 × \(\frac{5}{1}\)
= \(\frac{6}{1}\) × \(\frac{5}{1}\)
= \(\frac{ 6 × 5}{1 × 1}\)
= 30
Hence,
6÷ \(\frac{1}{5}\) = 30

Question 16.
Big Ideas Math Answers Grade 5 Chapter 10 Divide Fractions 85
Answer: 7 ÷ \(\frac{1}{4}\) = 28

Explanation:
The given numbers are: 7 and \(\frac{1}{4}\)
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
So,
7 ÷ \(\frac{1}{4}\)  = 7 × \(\frac{4}{1}\)
= \(\frac{7}{1}\) × \(\frac{4}{1}\)
= \(\frac{ 7 × 4}{1 × 1}\)
= 28
Hence,
7÷ \(\frac{1}{4}\) = 36

Question 17.
Big Ideas Math Answers Grade 5 Chapter 10 Divide Fractions 86
Answer: 8 ÷ \(\frac{1}{3}\) = 24

Explanation:
The given numbers are: 8 and \(\frac{1}{3}\)
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
So,
8 ÷ \(\frac{1}{3}\)  = 8 × \(\frac{3}{1}\)
= \(\frac{8}{1}\) × \(\frac{3}{1}\)
= \(\frac{ 8 × 3}{1 × 1}\)
= 24
Hence,
8÷ \(\frac{1}{3}\) = 24

Question 18.
Big Ideas Math Answers Grade 5 Chapter 10 Divide Fractions 87
Answer: 9 ÷ \(\frac{1}{2}\) = 18

Explanation:
The given numbers are: 9 and \(\frac{1}{2}\)
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
So,
9 ÷ \(\frac{1}{2}\)  = 9 × \(\frac{2}{1}\)
= \(\frac{9}{1}\) × \(\frac{2}{1}\)
= \(\frac{ 9 × 2}{1 × 1}\)
= 18
Hence,
9÷ \(\frac{1}{2}\) = 18

Question 19.
Big Ideas Math Answers Grade 5 Chapter 10 Divide Fractions 88
Answer: 2 ÷ \(\frac{1}{10}\) = 20

Explanation:
The given numbers are: 2 and \(\frac{1}{10}\)
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
So,
2 ÷ \(\frac{1}{10}\)  = 2 × \(\frac{10}{1}\)
= \(\frac{2}{1}\) × \(\frac{10}{1}\)
= \(\frac{ 2 × 10}{1 × 1}\)
= 20
Hence,
2÷ \(\frac{1}{10}\) = 20

10.4 Divide Unit Fractions by Whole Numbers

Divide. Use a model to help.

Question 20.
Big Ideas Math Answers Grade 5 Chapter 10 Divide Fractions 89
Answer: \(\frac{1}{7}\) ÷ 2 = \(\frac{1}{14}\)

Explanation:
The given numbers are: \(\frac{1}{7}\) and 2
We know that,
\(\frac{a}{b}\) ÷ a  = \(\frac{a}{b}\) × \(\frac{1}{a}\)
a= \(\frac{a}{1}\)
So,
\(\frac{1}{7}\) ÷ 2
= \(\frac{1}{7}\) × \(\frac{1}{2}\)
= \(\frac{ 1 × 1}{7 × 2}\)
= \(\frac{1}{14}\)
Hence,
\(\frac{1}{7}\) ÷ 2 = \(\frac{1}{14}\)

Question 21.
Big Ideas Math Answers Grade 5 Chapter 10 Divide Fractions 90
Answer: \(\frac{1}{2}\) ÷ 9 = \(\frac{1}{18}\)

Explanation:
The given numbers are: \(\frac{1}{2}\) and 9
We know that,
\(\frac{a}{b}\) ÷ a  = \(\frac{a}{b}\) × \(\frac{1}{a}\)
a= \(\frac{a}{1}\)
So,
\(\frac{1}{2}\) ÷ 9
= \(\frac{1}{2}\) × \(\frac{1}{9}\)
= \(\frac{ 1 × 1}{2 × 9}\)
= \(\frac{1}{18}\)
Hence,
\(\frac{1}{2}\) ÷ 9 = \(\frac{1}{18}\)

Question 22.
Big Ideas Math Solutions Grade 5 Chapter 10 Divide Fractions 91
Answer: \(\frac{1}{3}\) ÷ 7 = \(\frac{1}{21}\)

Explanation:
The given numbers are: \(\frac{1}{3}\) and 7
We know that,
\(\frac{a}{b}\) ÷ a  = \(\frac{a}{b}\) × \(\frac{1}{a}\)
a= \(\frac{a}{1}\)
So,
\(\frac{1}{3}\) ÷ 7
= \(\frac{1}{3}\) × \(\frac{1}{7}\)
= \(\frac{ 1 × 1}{7 × 3}\)
= \(\frac{1}{21}\)
Hence,
\(\frac{1}{3}\) ÷ 7 = \(\frac{1}{21}\)

Question 23.
Big Ideas Math Solutions Grade 5 Chapter 10 Divide Fractions 92
Answer: \(\frac{1}{6}\) ÷ 5 = \(\frac{1}{30}\)

Explanation:
The given numbers are: \(\frac{1}{6}\) and 5
We know that,
\(\frac{a}{b}\) ÷ a  = \(\frac{a}{b}\) × \(\frac{1}{a}\)
a= \(\frac{a}{1}\)
So,
\(\frac{1}{6}\) ÷ 5
= \(\frac{1}{6}\) × \(\frac{1}{5}\)
= \(\frac{ 1 × 1}{6 × 5}\)
= \(\frac{1}{30}\)
Hence,
\(\frac{1}{6}\) ÷ 5 = \(\frac{1}{30}\)

Question 24.
Big Ideas Math Solutions Grade 5 Chapter 10 Divide Fractions 93
Answer: \(\frac{1}{7}\) ÷ 3 = \(\frac{1}{21}\)

Explanation:
The given numbers are: \(\frac{1}{7}\) and 3
We know that,
\(\frac{a}{b}\) ÷ a  = \(\frac{a}{b}\) × \(\frac{1}{a}\)
a= \(\frac{a}{1}\)
So,
\(\frac{1}{7}\) ÷ 3
= \(\frac{1}{7}\) × \(\frac{1}{3}\)
= \(\frac{ 1 × 1}{7 × 3}\)
= \(\frac{1}{21}\)
Hence,
\(\frac{1}{7}\) ÷ 3 = \(\frac{1}{21}\)

Question 25.
Big Ideas Math Solutions Grade 5 Chapter 10 Divide Fractions 94

Answer: \(\frac{1}{8}\) ÷ 4 = \(\frac{1}{32}\)

Explanation:
The given numbers are: \(\frac{1}{8}\) and 4
We know that,
\(\frac{a}{b}\) ÷ a  = \(\frac{a}{b}\) × \(\frac{1}{a}\)
a= \(\frac{a}{1}\)
So,
\(\frac{1}{8}\) ÷ 4
= \(\frac{1}{8}\) × \(\frac{1}{4}\)
= \(\frac{ 1 × 1}{8 × 4}\)
= \(\frac{1}{32}\)
Hence,
\(\frac{1}{8}\) ÷ 4 = \(\frac{1}{32}\)

10.5 Problem Solving: Fraction Division

Question 26.
A mechanic buys 1 gallon of oil. She uses \(\frac{1}{6}\) of the oil, and then divides the rest into 4 smaller bottles. How much does she put into each bottle?
Big Ideas Math Solutions Grade 5 Chapter 10 Divide Fractions 95
Answer: The amount of oil she put into each bottle is: \(\frac{5}{24}\)

Explanation:
It is given that a mechanic buys 1 gallon of oil and she uses \(\frac{1}{6}\) of the oil
So,
the remaining part of the oil = 1 –  \(\frac{1}{6}\)
= \(\frac{5}{6}\)
It is also given that she divides the rest of the oil into 4 smaller bottles.
So,
The amount of oil in each bottle = \(\frac{The remaining part of the oil}{The number of bottles}\)
= \(\frac{5}{6}\) ÷ 4
= \(\frac{5}{6}\) × \(\frac{1}{4}\)
= \(\frac{5 × 1}{6 × 4}\)
= \(\frac{5}{24}\)
Hence, from the above,
We can conclude that the amount of oil in each bottle is: \(\frac{5}{24}\)

Conclusion:

Make use of the quick links and try to solve the problems in a simple manner. Redefine your true self with the BIM Answer Key for Grade 5 curated by subject experts. Test your knowledge by solving the questions which are given at the end of the chapter.

Eureka Math Grade 1 Module 2 Lesson 2 Answer Key

Eureka Math Grade 1 Module 2 Lesson 2 Answer Key

Engage NY Eureka Math 1st Grade Module 2 Lesson 2 Answer Key

Eureka Math Grade 1 Module 2 Lesson 2 Problem Set Answer Key

Eureka Math Grade 1 Module 2 Lesson 2 Problem Set Answer Key 1 the numbers that make ten. Draw a picture. Complete the number sentence.

Question 1.
Eureka Math Grade 1 Module 2 Lesson 2 Problem Set Answer Key 2
Answer:
Eureka-Math-Grade-1-Module-2-Lesson-2-Problem-Set-Answer-Key-2
10 + 4 = 14
Explanation:
An addition sentence is a mathematical expression that shows two or more values added together and their sum. The numbers that makes ten are seven and three as we can observe in the above image. Draw a circles for seven and three. By adding seven with three results ten. ADD ten with four which results fourteen.

Question 2.
9 + 1 + 4 = ☐
Eureka Math Grade 1 Module 2 Lesson 2 Problem Set Answer Key 3
Answer:
Eureka-Math-Grade-1-Module-2-Lesson-2-Problem-Set-Answer-Key-3
9 + 1 + 4 = 14
Explanation:
An addition sentence is a mathematical expression that shows two or more values added together and their sum. The numbers that makes ten are nine and one as we can observe in the above image. Draw a circles for nine and one. By adding nine with one results ten. ADD ten with four which results fourteen.

Question 3.
5 + 6 + 5 = ☐
Eureka Math Grade 1 Module 2 Lesson 2 Problem Set Answer Key 4
Answer:
Eureka-Math-Grade-1-Module-2-Lesson-2-Problem-Set-Answer-Key-4
5 + 5+ 6 = 16
Explanation:
An addition sentence is a mathematical expression that shows two or more values added together and their sum. The numbers that makes ten are five and five as we can observe in the above image. Draw a circles for five and five . By adding five with five results ten. ADD ten with six which results sixteen.

Question 4.
4 + 3 + 7 = ☐
Eureka Math Grade 1 Module 2 Lesson 2 Problem Set Answer Key 5
Answer:
Eureka-Math-Grade-1-Module-2-Lesson-2-Problem-Set-Answer-Key-5
4 + 3+ 7 = 14
Explanation:
An addition sentence is a mathematical expression that shows two or more values added together and their sum. The numbers that makes ten are three and seven as we can observe in the above image. Draw a circles for three and seven. By adding three with seven results ten. ADD ten with four which results fourteen.

Question 5.
2 + 7 + 8 = ☐
Eureka Math Grade 1 Module 2 Lesson 2 Problem Set Answer Key 5.1
Answer:
Eureka-Math-Grade-1-Module-2-Lesson-2-Problem-Set-Answer-Key-5.1
2 + 7 + 8 = 17
Explanation:
An addition sentence is a mathematical expression that shows two or more values added together and their sum. The numbers that makes ten are two and eight as we can observe in the above image. Draw a circles for two and eight. By adding two with eight results ten. ADD ten with seven which results seventeen.

Eureka Math Grade 1 Module 2 Lesson 2 Problem Set Answer Key 6 the numbers that make ten. Put them into a number bond, and solve.

Question 6.
Eureka Math Grade 1 Module 2 Lesson 2 Problem Set Answer Key 7
Answer:
Eureka-Math-Grade-1-Module-2-Lesson-2-Problem-Set-Answer-Key-7
Explanation:
A number bond is a simple addition of two numbers that add up to give the sum. The numbers that makes ten are nine and one as we can observe in the above image. Draw a circles for nine and one. By adding nine with one results ten. ADD ten with five which results fifteen.

Question 7.
8 + 2 + 4 = ____
Answer:
Eureka-Math-Grade-1-Module-2-Lesson-2-Problem-Set-Answer-Key-7(1)

8  +  2  +  4  =  14
Explanation:
A number bond is a simple addition of two numbers that add up to give the sum. The numbers that makes ten are eight and two as we can observe in the above image. Draw a circles for eight and two. By adding eight with two results ten. ADD ten with four which results fourteen.

Question 8.
3 + 5 + 5 = ____
Answer:
Eureka-Math-Grade-1-Module-2-Lesson-2-Problem-Set-Answer-Key-7(2)
3   +   5   +   5   =   13
Explanation:
A number bond is a simple addition of two numbers that add up to give the sum. The numbers that makes ten are five and five as we can observe in the above image. Draw a circles for five and five. By adding five with five results ten. ADD ten with three which results thirteen.

Question 9.
3 + 6 + 7 = ____
Answer:
Eureka-Math-Grade-1-Module-2-Lesson-2-Problem-Set-Answer-Key-7(3)
3  +  6  +  7  =  16
Explanation:
A number bond is a simple addition of two numbers that add up to give the sum. The numbers that makes ten are three and seven as we can observe in the above image. Draw a circles for three and seven. By adding three with seven results ten. ADD ten with six which results sixteen.

Eureka Math Grade 1 Module 2 Lesson 2 Exit Ticket Answer Key

Engage NY Math 1st Grade Module 2 Lesson 2 Exit Ticket Answer Key 8 the numbers that make ten.
Draw a picture, and complete the number sentences to solve.

a. 8 + 2 + 3 = ____
____ + ____ = ____
10 + ____ = ____
Answer:
Eureka-Math-Grade-1-Module-2-Lesson-2-Problem-Set-Answer-Key-7(4)
8  +   2  +  3  =  13
Explanation:
An addition sentence is a mathematical expression that shows two or more values added together and their sum. The numbers that makes ten are eight and two as we can observe in the above image. Draw a circles for eight and two. By adding eight with two results ten. ADD ten with three which results thirteen.

b. 7 + 4 + 3 = ____
____ + ____ = ____
10 + ____ = ____
Answer:
Eureka-Math-Grade-1-Module-2-Lesson-2-Problem-Set-Answer-Key-7(5)
7  +  4  +  3  =  14
Explanation:
An addition sentence is a mathematical expression that shows two or more values added together and their sum. The numbers that makes ten are seven and three as we can observe in the above image. Draw a circles for seven and three. By adding seven with three results ten. ADD ten with four which results fourteen.

Eureka Math Grade 1 Module 2 Lesson 2 Homework Answer Key

Eureka Math 1st Grade Module 2 Lesson 2 Homework Answer Key 25 the numbers that make ten. Draw a picture. Complete the number sentence.

Question 1.
Eureka Math 1st Grade Module 2 Lesson 2 Homework Answer Key 26
Answer:
Eureka-Math-1st-Grade-Module-2-Lesson-2-Homework-Answer-Key-26
6  +  2  +  4  =  12
Explanation:
An addition sentence is a mathematical expression that shows two or more values added together and their sum. The numbers that makes ten are six and four as we can observe in the above image. Draw a circles for six and four. By adding six with four results ten. ADD ten with two which results twelve.

Question 2.
5 + 3 + 5 = ☐
Eureka Math 1st Grade Module 2 Lesson 2 Homework Answer Key 27
Answer:
Eureka-Math-1st-Grade-Module-2-Lesson-2-Homework-Answer-Key-27
5 + 3 + 5 = 13
Explanation:
An addition sentence is a mathematical expression that shows two or more values added together and their sum. The numbers that makes ten are five and five as we can observe in the above image. Draw a circles for five and five. By adding five with five results ten. ADD ten with three which results thirteen.

Question 3.
5 + 2 + 8 = ☐
Eureka Math 1st Grade Module 2 Lesson 2 Homework Answer Key 28
Answer:
Eureka-Math-1st-Grade-Module-2-Lesson-2-Homework-Answer-Key-28
5 + 2 + 8 = 15
Explanation:
An addition sentence is a mathematical expression that shows two or more values added together and their sum. The numbers that makes ten are two and eight as we can observe in the above image. Draw a circles for two and eight. By adding two with eight results ten. ADD five with ten which results fifteen.

Question 4.
2 + 7 + 3 = ☐
Eureka Math 1st Grade Module 2 Lesson 2 Homework Answer Key 29
Answer:
Eureka-Math-1st-Grade-Module-2-Lesson-2-Homework-Answer-Key-29
2 + 7 + 3 = 12
Explanation:
An addition sentence is a mathematical expression that shows two or more values added together and their sum. The numbers that makes ten are seven and three as we can observe in the above image. Draw a circles for seven and three. By adding seven with three results ten. ADD two with ten which results twelve.

Eureka Math 1st Grade Module 2 Lesson 2 Homework Answer Key 30 the numbers that make ten, and put them into a number bond. Write a new number sentence.

Question 5.
Eureka Math 1st Grade Module 2 Lesson 2 Homework Answer Key 31
Answer:
Eureka-Math-1st-Grade-Module-2-Lesson-2-Homework-Answer-Key-31
A new number sentence is 10 + 5 = 15
Explanation:
A number bond is a simple addition of two numbers that add up to give the sum. The numbers that makes ten are three and seven as we can observe in the above image. Draw a circles for three and seven. By adding three with seven results ten. ADD ten with five which results fifteen. The new number sentence is 10 + 5 = 15.

Question 6.
Eureka Math 1st Grade Module 2 Lesson 2 Homework Answer Key 32
Answer:
Eureka-Math-1st-Grade-Module-2-Lesson-2-Homework-Answer-Key-32
A new number sentence is 4 + 10 = 14.
Explanation:
A number bond is a simple addition of two numbers that add up to give the sum. The numbers that makes ten are eight and two as we can observe in the above image. Draw a circles for eight and two. By adding eight with two results ten. ADD four with ten which results fourteen. The new number sentence is 4 + 10 = 14.

Challenge: Eureka Math 1st Grade Module 2 Lesson 2 Homework Answer Key 33 the addends that make ten. Eureka Math 1st Grade Module 2 Lesson 2 Homework Answer Key 33 the true number sentences.

a. Eureka Math 1st Grade Module 2 Lesson 2 Homework Answer Key 34
Answer:
Eureka-Math-1st-Grade-Module-2-Lesson-2-Homework-Answer-Key-34
Explanation:
The addends that make ten are five and five. Draw a circle for two addends that make ten. The number sentence is true as we can observe in the above image. The true number sentence is 10 + 3.

b. 4 + 6 + 6 = 10 + 6
Answer:
Eureka-Math-1st-Grade-Module-2-Lesson-2-Homework-Answer-Key-35
Explanation:
The addends that make ten are four and six. Draw a circle for two addends that make ten. The number sentence is true as we can observe in the above image. The true number sentence is 10 + 6. So make a circle for 10 + 6 number sentence.

c. 3 + 8 + 7 = 10 + 6
Answer:
Eureka-Math-1st-Grade-Module-2-Lesson-2-Homework-Answer-Key-36
Explanation:
The addends that make ten are three and seven. Draw a circle for two addends that make ten. The number sentence is false as we can observe in the above image. The number sentence 10 + 6 is not a correct number sentence. The correct number sentence is
10 + 8. So don’t make circle for the number sentence as shown in the above image.

d. 8 + 9 + 2 = 9 + 10
Answer:
Eureka-Math-1st-Grade-Module-2-Lesson-2-Homework-Answer-Key-37
Explanation:
The addends that make ten are eight and two. Draw a circle for two addends that make ten. The number sentence is true as we can observe in the above image. The true number sentence is 9 + 10. So make a circle for 9 + 10 number sentence.

 

Eureka Math Geometry Module 5 Lesson 5 Answer Key

Eureka Math Geometry Module 5 Lesson 5 Answer Key

Engage NY Eureka Math Geometry Module 5 Lesson 5 Answer Key

Eureka Math Geometry Module 5 Lesson 5 Example Answer Key

Example 1.
A and C are points on a circle with center O.
Engage NY Math Geometry Module 5 Lesson 5 Example Answer Key 1
a. What is the intercepted arc of ∠COA? Color it red.
Answer:
\(\widehat{A C}\)

b. Draw triangle AOC. What type of triangle is it? Why?
Answer:
An isosceles triangle because OC = OA (they are radii of the same circle).

c. What can you conclude about m∠OCA and m∠OAC? Why?
Answer:
They are equal because base angles of an isosceles triangle are equal in measure.

d. Draw a point B on the circle so that O is in the interior of the inscribed angle ABC.
Answer:
The diagram should resemble the inside case of the discussion diagrams.
Engage NY Math Geometry Module 5 Lesson 5 Example Answer Key 2

e. What is the intercepted arc of ∠ABC? Color it green.
Answer:
\(\widehat{A C}\)

f. What do you notice about \(\widehat{A C}\)?
Answer:
It is the same arc that was intercepted by the central angle.

g. Let the measure of ∠ABC be x and the measure of ∠AOC be y. Can you prove that y = 2x? (Hint: Draw the diameter that contains point B.)
Answer:
Let \(\overline{B D}\) be a diameter. Let x1, y1, x2, and y2 be the measures of ∠CBD, ∠COD, ∠ABD, and ∠AOD, respectively. We can express x and y in terms of these measures: x = x1 + x2, and y = y1 + y2. By the Opening Exercise, y1 = 2x1, and y2 = 2x2. Thus, y = 2x.

h. Does your conclusion support the inscribed angle theorem?
Answer:
Yes, even when the center of the circle is in the interior of the inscribed angle, the measure of the inscribed angle is equal to half the measure of the central angle that intercepts the same arc.

i. If we combine the Opening Exercise and this proof, have we finished proving the inscribed angle theorem?
Answer:
No. We still have to prove the case where the center is outside the inscribed angle.

Example 2.
A and C are points on a circle with center O.
Engage NY Math Geometry Module 5 Lesson 5 Example Answer Key 3
a. Draw a point B on the circle so that O is in the exterior of the inscribed angle ABC.
Answer:
The diagram should resemble the outside case of the discussion diagrams.

b. What is the intercepted arc of ∠ABC? Color it yellow.
Answer:
\(\widehat{A C}\)

c. Let the measure of ∠ABC be x and the measure of ∠AOC be y. Can you prove that y = 2x? (Hint: Draw the diameter that contains point B.)
Answer:
Let \(\overline{B D}\) be a diameter. Let x1, y1, x2, and y2 be the measures of ∠CBD, ∠COD, ∠ABD, and ∠AOD, respectively. We can express x and y in terms of these measures: x = x2 – x1 and y = y2 – y1. By the Opening Exercise, y1 = 2x1 and y2 = 2x2. Thus, y = 2x.
Engage NY Math Geometry Module 5 Lesson 5 Example Answer Key 4

d. Does your conclusion support the inscribed angle theorem?
Answer:
Yes, even when the center of the circle is in the exterior of the inscribed angle, the measure of the inscribed angle is equal to half the measure of the central angle that intercepts the same arc.

e. Have we finished proving the inscribed angle theorem?
Answer:
We have shown all cases of the inscribed angle theorem (central angle version). We do have one more case to study in Lesson 7, but it is ok not to mention it here. The last case is when the location of B is on the minor arc between A and C.

Eureka Math Geometry Module 5 Lesson 5 Exercise Answer Key

Eureka Math Geometry Module 5 Lesson 5 Exercise Answer Key

Opening Exercise
a. A and C are points on a circle with center O.
Engage NY Math Geometry Module 5 Lesson 5 Exercise Answer Key 1
i. Draw a point B on the circle so that \(\overline{A B}\) is a diameter. Then draw the angle ABC.

ii. What angle in your diagram is an inscribed angle?
Answer:
∠ABC

iii. What angle in your diagram is a central angle?
Answer:
∠AOC

iv. What is the intercepted arc of ∠ABC?
Answer:
\(\widehat{A C}\)

v. What is the intercepted arc of ∠AOC ?
Answer:
\(\widehat{A C}\)

b. The measure of the inscribed angle ACD is x, and the measure of the central angle CAB is y. Find m∠CAB in terms of x.
Engage NY Math Geometry Module 5 Lesson 5 Exercise Answer Key 2
Answer:
We are given that m∠ACD is x. We know that AB = AC = AD, so △CAD is an isosceles triangle, which means that m∠ADC is also x. The sum of the angles of a triangle is 180˚, so m∠CAD = 180˚ – 2x. ∠CAD and ∠CAB are supplementary meaning that m∠CAB = 180˚ – (180˚ – 2x) = 2x; therefore, y = 2x.

Exercises

Exercise 1.
Find the measure of the angle with measure x. Diagrams are not drawn to scale.
a. m∠D = 25°
Engage NY Math Geometry Module 5 Lesson 5 Exercise Answer Key 3
Answer:
50°

b. m∠D = 15°
Engage NY Math Geometry Module 5 Lesson 5 Exercise Answer Key 4
Answer:
15°

c. m∠BAC = 90°
Engage NY Math Geometry Module 5 Lesson 5 Exercise Answer Key 5
Answer:
45°

d. m∠B = 32°
Engage NY Math Geometry Module 5 Lesson 5 Exercise Answer Key 6
Answer:
58°

e. BD = AB
Engage NY Math Geometry Module 5 Lesson 5 Exercise Answer Key 7
Answer:
x = w = 60°

f. m∠D = 19°
Engage NY Math Geometry Module 5 Lesson 5 Exercise Answer Key 8
Answer:
71°

Exercise 2.
Toby says △BEA is a right triangle because m∠BEA = 90°. Is he correct? Justify your answer.
Engage NY Math Geometry Module 5 Lesson 5 Exercise Answer Key 9
Answer:
Toby is not correct. m∠BEA = 95°. ∠BCD is inscribed in the same arc as the central angle, so it has a measure of 35°. This means that m∠DEC = 95° because the sum of the angles of a triangle is 180°. m∠BEA = m∠DEC since they are vertical angles, so the triangle is not right.

Exercise 3.
Let’s look at relationships between inscribed angles.
a. Examine the inscribed polygon below. Express x in terms of y and y in terms of x. Are the opposite angles in any quadrilateral inscribed in a circle supplementary? Explain.
Engage NY Math Geometry Module 5 Lesson 5 Exercise Answer Key 10
Answer:
x = 180˚ – y; y = 180˚ – x. The angles are supplementary.

b. Examine the diagram below. How many angles have the same measure, and what are their measures in terms of x˚?
Engage NY Math Geometry Module 5 Lesson 5 Exercise Answer Key 11
Answer:
Let C and D be the points on the circle that the original angles contain. All the angles intercepting the minor arc between C and D have measure x˚, and the angles intercepting the major arc between C and D measure 180° – x.

Exercise 4.
Find the measures of the labeled angles.
a.
Engage NY Math Geometry Module 5 Lesson 5 Exercise Answer Key 12
Answer:
x = 28, y = 50

b.
Engage NY Math Geometry Module 5 Lesson 5 Exercise Answer Key 13
Answer:
y = 48

c.
Engage NY Math Geometry Module 5 Lesson 5 Exercise Answer Key 14
Answer:
x = 32

d.
Engage NY Math Geometry Module 5 Lesson 5 Exercise Answer Key 15
Answer:
x = 36, y = 120

e.
Engage NY Math Geometry Module 5 Lesson 5 Exercise Answer Key 16
Answer:
x = 40

f.
Engage NY Math Geometry Module 5 Lesson 5 Exercise Answer Key 17
Answer:
x = 30

Eureka Math Geometry Module 5 Lesson 5 Problem Set Answer Key

For Problems 1–8, find the value of x. Diagrams are not drawn to scale.
Question 1.
Eureka Math Geometry Module 5 Lesson 5 Problem Set Answer Key 1
Answer:
x = 34

Question 2.
Eureka Math Geometry Module 5 Lesson 5 Problem Set Answer Key 2
Answer:
x = 94

Question 3.
Eureka Math Geometry Module 5 Lesson 5 Problem Set Answer Key 3
Answer:
x = 30

Question 4.
Eureka Math Geometry Module 5 Lesson 5 Problem Set Answer Key 4
Answer:
x = 70

Question 5.
Eureka Math Geometry Module 5 Lesson 5 Problem Set Answer Key 5
Answer:
x = 60

Question 6.
Eureka Math Geometry Module 5 Lesson 5 Problem Set Answer Key 6
Answer:
x = 60

Question 7.
Eureka Math Geometry Module 5 Lesson 5 Problem Set Answer Key 7
Answer:
x = 20

Question 8.
Eureka Math Geometry Module 5 Lesson 5 Problem Set Answer Key 8
Answer:
x = 46

Question 9.
a. The two circles shown intersect at E and F. The center of the larger circle, D, lies on the circumference of the smaller circle. If a chord of the larger circle, (FG) ̅, cuts the smaller circle at H, find x and y.
Eureka Math Geometry Module 5 Lesson 5 Problem Set Answer Key 9
Answer:
x = 100; y = 50

b. How does this problem confirm the inscribed angle theorem?
Answer:
∠FDE is a central angle of the larger circle and is double ∠FGE, the inscribed angle of the larger circle. ∠FDE is inscribed in the smaller circle and equal in measure to ∠FHE, which is also inscribed in the smaller circle.

Question 10.
In the figure below, \(\overline{E D}\) and \(\overline{B C}\) intersect at point F.
Prove: m∠DAB + m∠EAC = 2(m∠BFD)
Eureka Math Geometry Module 5 Lesson 5 Problem Set Answer Key 10
Proof:
Join \(\overline{B E}\).

m∠BED = \(\frac{1}{2}\)(m∠___________)

m∠EBC = \(\frac{1}{2}\)(m∠___________)

In △EBF,
m∠BEF + m∠EBF = m∠___________

\(\frac{1}{2}\)(m∠___________) + \(\frac{1}{2}\)(m∠___________) = m∠___________
∴ m∠DAB + m∠EAC = 2(m∠BFD)
Answer:
BAD; EAC; BFD; DAB, EAC, BFD

Eureka Math Geometry Module 5 Lesson 5 Exit Ticket Answer Key

Question 1.
The center of the circle below is O. If angle B has a measure of 15 degrees, find the values of x and y. Explain how you know.
Eureka Math Geometry Module 5 Lesson 5 Exit Ticket Answer Key 1
Answer:
y = 15. Triangle COB is isosceles, so base angles ∠OCB and ∠OBC are congruent. m∠OBC = 15˚ = m∠OCB.
x = 30. ∠COA is a central angle inscribed in the same arc as inscribed ∠CBA. So, m∠COA = 2(m∠CBA).

Big Ideas Math Answers Grade 5 Chapter 14 Classify Two-Dimensional Shapes

Big Ideas Math Answers Grade 5 Chapter 14 Classify Two-Dimensional Shapes

Big Ideas Math Answers Grade 5 Chapter 14 Classify Two-Dimensional Shapes pdf is available here. So, the students who are in search of Big Ideas Math Book 5th Grade Answer Key Chapter 14 Classify Two-Dimensional Shapes can get them on this page. Go through the Big Ideas Math Answers Grade 5 Chapter 14 Classify Two-Dimensional Shapes and improve your performance skills. Test your skills by solving the Performance Task, Activity, Chapter Practice which is provided at the end of the chapter.

Big Ideas Math Book 5th Grade Answer Key Chapter 14 Classify Two-Dimensional Shapes

Big Ideas Math Answers 5th Grade Chapter 14 Classify Two-Dimensional Shapes are very easy to understand. You can quickly grasp the mathematical practices with the help of Big Ideas Math Answer Key Grade 5 Chapter 14 Classify Two-Dimensional Shapes. The topics covered in this chapter are Classify Triangles, Quadrilaterals, Relate Quadrilaterals. Just hit the below-provided links and Download Big Ideas Math Book 5th Grade Answer Key Chapter 14 Classify Two-Dimensional Shapes.

Lesson: 1 Classify Triangles

Lesson: 1 Classify Triangles

Lesson: 2 Classify Quadrilaterals

Lesson: 2 Classify Quadrilaterals

Lesson: 3 Relate Quadrilaterals

Classify Two-Dimensional Shapes

Classify Two-Dimensional Shapes

Lesson 14.1 Classify Triangles

Explore and Grow

Draw and label a triangle for each description. If a triangle cannot be drawn, explain why.
Big Ideas Math Answers Grade 5 Chapter 14 Classify Two-Dimensional Shapes 1

Precision
Draw a triangle that meets two of the descriptions above.

Answer:
Big-Ideas-Math-Answers-Grade-5-Chapter-14-Classify-Two-Dimensional-Shapes-1

Think and Grow: Classify Triangles

Key Idea
Triangles can be classified by their sides.

Big Ideas Math Answer Key Grade 5 Chapter 14 Classify Two-Dimensional Shapes 2
An equilateral triangle has three sides with the same length.
Big Ideas Math Answer Key Grade 5 Chapter 14 Classify Two-Dimensional Shapes 3
An isosceles triangle has two sides with the same length.
Big Ideas Math Answer Key Grade 5 Chapter 14 Classify Two-Dimensional Shapes 4
A scalene triangle has no sides with the same length.
Big Ideas Math Answer Key Grade 5 Chapter 14 Classify Two-Dimensional Shapes 5

Key Idea
Triangles can be classified by their angles.
Big Ideas Math Answer Key Grade 5 Chapter 14 Classify Two-Dimensional Shapes 6
An acute triangle has three acute angles.
Big Ideas Math Answer Key Grade 5 Chapter 14 Classify Two-Dimensional Shapes 7
An obtuse triangle has one obtuse angle.
Big Ideas Math Answer Key Grade 5 Chapter 14 Classify Two-Dimensional Shapes 8
A right triangle has one right angle.
Big Ideas Math Answer Key Grade 5 Chapter 14 Classify Two-Dimensional Shapes 9
An equiangular triangle has three angles with the same measure.
Example
Classify the triangle by its angles and its sides.
Big Ideas Math Answer Key Grade 5 Chapter 14 Classify Two-Dimensional Shapes 10
The triangle has one ___ angle
and ___ sides with the same length.
So, it is a ___ triangle.

Answer:
The triangle has one right angle
and no sides with the same length.
So, it is a right triangle.

Show and Grow

Classify the triangle by its angles and its sides

Question 1.
Big Ideas Math Answer Key Grade 5 Chapter 14 Classify Two-Dimensional Shapes 11

Answer:  Equilateral triangle.

Explanation: An equilateral triangle has three sides of the same length.

Question 2.
Big Ideas Math Answer Key Grade 5 Chapter 14 Classify Two-Dimensional Shapes 12

Answer: Isosceles triangle

Explanation: An Isosceles triangle has two sides of the same size. Two of its angle also measure equal.

Question 3.
Big Ideas Math Answer Key Grade 5 Chapter 14 Classify Two-Dimensional Shapes 13

Answer:  Scalene triangle.

Explanation: A Scalene triangle has no sides are congruent (same size)

Apply and Grow: Practice

Classify the triangle by its angles and its sides.

Question 4.
Big Ideas Math Answer Key Grade 5 Chapter 14 Classify Two-Dimensional Shapes 14

Answer:  Right triangle.

Explanation: In a triangle one of the angle is a right angle (90 Deg ) called as Right triangle.

Question 5.
Big Ideas Math Answer Key Grade 5 Chapter 14 Classify Two-Dimensional Shapes 15

Answer: Isosceles triangle.

Explanation: An Isosceles triangle has two sides of the same size. Two of its angle also measure equal.

Question 6.
Big Ideas Math Answer Key Grade 5 Chapter 14 Classify Two-Dimensional Shapes 16

Answer: Equiangular triangle

Explanation: In an equilateral triangle, all the lengths of the sides are equal. In such a case, each of the interior angles will have a measure of 60 degrees. Since the angles of an equilateral triangle are the same, it is also known as an equiangular triangle. The figure given below illustrates an equilateral triangle.

Question 7.
Big Ideas Math Answer Key Grade 5 Chapter 14 Classify Two-Dimensional Shapes 17

Answer:  Isosceles triangle.

Explanation: An Isosceles triangle has two sides of the same length. Two of its angle also measure equal.

Question 8.
Big Ideas Math Answer Key Grade 5 Chapter 14 Classify Two-Dimensional Shapes 18

Answer:  Scalene triangle.

Explanation: A Scalene triangle has no sides are congruent (Same size) and angles also all different.

Question 9.
Big Ideas Math Answer Key Grade 5 Chapter 14 Classify Two-Dimensional Shapes 19

Answer:  Right triangle.

Explanation: In a triangle one of the angle is a right angle (90 deg ) called a Right triangle.

Question 10.
A triangular sign has a 40° angle, a 55° angle, and an 85° angle. None of its sides have the same length. Classify the triangle by its angles and its sides.

Answer:  Scalene triangle.

Explanation: A Scalene triangle has no sides that are congruent (Same size) and angles also all different.

Question 11.
YOU BE THE TEACHER
Your friend says the triangle is an acute triangle because it has two acute angles. Is your friend correct? Explain.
Big Ideas Math Answer Key Grade 5 Chapter 14 Classify Two-Dimensional Shapes 20

Answer:  Above is no acute triangle and it is called a scalene triangle.

Explanation: A Scalene triangle has only no sides that are congruent (Same size) and angles also all different. So it is called a scalene triangle.

Question 12.
DIG DEEPER!
Draw one triangle for each category. Which is the appropriate category for an equiangular triangle? Explain your reasoning.
Big Ideas Math Answer Key Grade 5 Chapter 14 Classify Two-Dimensional Shapes 21

Answer:
Big-Ideas-Math-Answer-Key-Grade-5-Chapter-14-Classify-Two-Dimensional-Shapes-21
From the figure, we can say that acute triangles have the same length. So, the first triangle is the equiangular triangle.

Think and Grow: Modeling Real Life

Example
A bridge contains several identical triangles. Classify each triangle by its angles and its sides. What is the length of the bridge?
Big Ideas Math Answer Key Grade 5 Chapter 14 Classify Two-Dimensional Shapes 22
Each triangle has ___ angles with the same measure and ___ sides with the same length.
So, each triangle is ___ and ___.
The side lengths of 6 identical triangles meet to form the length of the bridge. So, multiply the side length by 6 to find the length of the bridge.
27 × 6 = ___
So, the bridge is ___ long.

Answer:
Each triangle has 3 angles with the same measure and 3 sides with the same length.
The side lengths of 6 identical triangles meet to form the length of the bridge. So, multiply the side length by 6 to find the length of the bridge.
27 × 6 = 162
So, the bridge is 162 ft long.

Show and Grow

Question 13.
The window is made using identical triangular panes of glass. Classify each triangle by its angles and its sides. What is the height of the window?
Big Ideas Math Answer Key Grade 5 Chapter 14 Classify Two-Dimensional Shapes 23

Answer:
Big-Ideas-Math-Answer-Key-Grade-5-Chapter-14-Classify-Two-Dimensional-Shapes-23
The length of the two sides of the triangle is the same.
18 in + 18 in = 36 inches
Thus the height of the window is 36 inches

Question 14.
DIG DEEPER!
You connect four triangular pieces of fabric to make the kite. Classify the triangles by their angles and their sides. Use a ruler and a protractor to verify your answer.
Big Ideas Math Answer Key Grade 5 Chapter 14 Classify Two-Dimensional Shapes 24

Answer:
The name of the blue triangle is isosceles right angle triangle.
The two sides of the triangle are the same.
The name of the red triangle is isosceles right-angle triangle.
The two sides of the triangle are the same.
The name of the green triangle is isosceles right angle triangle.
The two sides of the triangle are the same.
The name of the yellow triangle is isosceles right angle triangle.
The two sides of the triangle are the same.

Classify Triangles Homework & Practice 14.1

Classify the triangle by its angles and its sides.

Question 1.
Big Ideas Math Answer Key Grade 5 Chapter 14 Classify Two-Dimensional Shapes 25

Answer:  Scalene triangle.

Explanation: A Scalene triangle has only no sides that are congruent (Same size) and angles also all different. So it is called a scalene triangle

Question 2.
Big Ideas Math Answer Key Grade 5 Chapter 14 Classify Two-Dimensional Shapes 26

Answer: Isosceles triangle.

Explanation: An Isosceles triangle has two sides of the same size. Two of its angle also measure equal.

Question 3.
Big Ideas Math Answer Key Grade 5 Chapter 14 Classify Two-Dimensional Shapes 27

Answer: Scalene triangle.

Explanation: A Scalene triangle has no sides that are congruent (Same size) and angles also all different.

Question 4.
Big Ideas Math Answer Key Grade 5 Chapter 14 Classify Two-Dimensional Shapes 28

Answer: Isosceles triangle.

Explanation: An Isosceles triangle has two sides of the same length. Two of its angle also measure equal.

Question 5.

Big Ideas Math Answer Key Grade 5 Chapter 14 Classify Two-Dimensional Shapes 29

Answer:  Scalene triangle.

Explanation: A Scalene triangle has no sides that are congruent (Same size) and angles also all different.

Question 6.

Big Ideas Math Answer Key Grade 5 Chapter 14 Classify Two-Dimensional Shapes 30

Answer:  Equiangular triangle.

Explanation:
In an equilateral triangle, all the lengths of the sides are equal. In such a case, each of the interior angles will have a measure of 60 degrees. Since the angles of an equilateral triangle are the same, it is also known as an equiangular triangle. The figure given below illustrates an equilateral triangle.

Question 7.
A triangular race flag has two 65° angles and a 50° angle. Two of its sides have the same length. Classify the triangle by its angles and its sides.

Answer: Isosceles triangle.

Explanation:  An Isosceles triangle has two sides of the same size. Two of its angle also measure equal.

Question 8.
A triangular measuring tool has a 90° angle and no sides of the same length. Classify the triangle by its angles and its sides.

Answer: Right triangle.

Explanation: In a triangle one of the angle is a right angle (90 Deg ) called a Right triangle.

Question 9.

Structure

Draw a triangle with vertices A(2, 2), B(2, 6), and C(6, 2) in the coordinate plane. Classify the triangle by its angles and its sides. Explain your reasoning.

Big Ideas Math Answer Key Grade 5 Chapter 14 Classify Two-Dimensional Shapes 31

Answer:
Big-Ideas-Math-Answer-Key-Grade-5-Chapter-14-Classify-Two-Dimensional-Shapes-31

Question 10.

YOU BE THE TEACHER

Your friend says that both Newton and Descartes are correct. Is your friend correct? Explain.

Big Ideas Math Answer Key Grade 5 Chapter 14 Classify Two-Dimensional Shapes 32

Answer: Yes

Explanation:  An acute triangle is a triangle in which each angle is an acute angle. Any triangle which is not acute is either a right triangle or an obtuse triangle. All acute triangle angles are less than 90 degrees. For example, an equilateral triangle is always acute, since all angles (which are 60) are all less than 90.

Question 11.
DIG DEEPER!
The sum of all the angle measures in a triangle is 180°. A triangle has a 34° angle and a 26° angle. Is the triangle acute, right, or obtuse? Explain.

Answer: Scalene triangle.

Explanation: A Scalene triangle has no sides that are congruent (Same size) and angles also all different.

Question 12.
Modeling Real Life
A designer creates the logo using identical triangles. Classify each triangle by its angles and its sides. What is the perimeter of the logo?

Big Ideas Math Answer Key Grade 5 Chapter 14 Classify Two-Dimensional Shapes 33

Answer: Equiangular triangle

Explanation: In an equilateral triangle, all the lengths of the sides are equal. In such a case, each of the interior angles will have a measure of 60 degrees. Since the angles of an equilateral triangle are the same, it is also known as an equiangular triangle. The figure given below illustrates an equilateral triangle.

Question 13.
DIG DEEPER!
The window is made using identical triangular panes of glass. Classify each triangle by its angles and its sides. What are the perimeter and the area of the window? Explain your reasoning.
Big Ideas Math Answer Key Grade 5 Chapter 14 Classify Two-Dimensional Shapes 34

Answer: Right triangle.

Explanation: A right triangle is a triangle in which one of the angles is 90 degrees. In a right triangle, the side opposite to the right angle (90-degree angle) will be the longest side and is called the hypotenuse. You may come across triangle types with combined names like right isosceles triangle and such, but this only implies that the triangle has two equal sides with one of the interior angles being 90 degrees. The figure given below illustrates a right triangle

Review & Refresh

Question 14.
Big Ideas Math Answer Key Grade 5 Chapter 14 Classify Two-Dimensional Shapes 35

Answer : \(\frac{1}{4}\) =0.25

Explanation: 2 divides by 8  with 1/4 times, So the answer is 1/4.

Question 15.
Big Ideas Math Answer Key Grade 5 Chapter 14 Classify Two-Dimensional Shapes 36

Answer : \(\frac{15}{4}\) = 3.75

Explanation: 15 divides by 4 with \(\frac{15}{4}\) times, So the answer is \(\frac{15}{4}\) or 3.75.

Question 16.
Big Ideas Math Answer Key Grade 5 Chapter 14 Classify Two-Dimensional Shapes 37

Answer : \(\frac{15}{12}\) = \(\frac{1}{4}\) = 1.25

Explanation: 15 divides by 12 with \(\frac{1}{4}\) times, So the answer is \(\frac{1}{4}\).

Lesson 14.2 Classify Quadrilaterals

Explore and Grow

Draw and label a quadrilateral for each description. If a quadrilateral cannot be drawn, explain why

Big Ideas Math Answer Key Grade 5 Chapter 14 Classify Two-Dimensional Shapes 38

Precision
Draw a quadrilateral that meets three of the descriptions above.

Answer:
Big-Ideas-Math-Answer-Key-Grade-5-Chapter-14-Classify-Two-Dimensional-Shapes-38

Think and Grow: Classify Quadrilaterals

Key Idea
Quadrilaterals can be classified by their angles and their sides.
A trapezoid is a quadrilateral that has exactly one pair of parallel sides.
Big Ideas Math Answer Key Grade 5 Chapter 14 Classify Two-Dimensional Shapes 39
A parallelogram is a quadrilateral that has two pairs of parallel sides. Opposite sides have the same length.
Big Ideas Math Answer Key Grade 5 Chapter 14 Classify Two-Dimensional Shapes 40
A rectangle is a parallelogram that has four right angles.
Big Ideas Math Answer Key Grade 5 Chapter 14 Classify Two-Dimensional Shapes 41
A rhombus is a parallelogram that has four sides with the same length.
Big Ideas Math Answer Key Grade 5 Chapter 14 Classify Two-Dimensional Shapes 42
A square is a parallelogram that has four right angles and four sides with the same length.
Big Ideas Math Answer Key Grade 5 Chapter 14 Classify Two-Dimensional Shapes 43

Example
Classify the quadrilateral in as many ways as possible.
Big Ideas Math Answer Key Grade 5 Chapter 14 Classify Two-Dimensional Shapes 44

Answer:
Big-Ideas-Math-Answer-Key-Grade-5-Chapter-14-Classify-Two-Dimensional-Shapes-44

Show and Grow

Classify the quadrilateral in as many ways as possible.

Question 1.
Big Ideas Math Answer Key Grade 5 Chapter 14 Classify Two-Dimensional Shapes 45

Answer:  Square

Explanation: A square is a parallelogram that has four right angles and four sides with the same length.

Question 2.
Big Ideas Math Answer Key Grade 5 Chapter 14 Classify Two-Dimensional Shapes 46

Answer: Trapezoid 

Explanation: A trapezoid is a quadrilateral that has exactly one pair of parallel sides

Apply and Grow: Practice

Classify the quadrilateral in as many ways as possible.

Question 3.
Big Ideas Math Answer Key Grade 5 Chapter 14 Classify Two-Dimensional Shapes 47

Answer: Parallelogram

Explanation: A parallelogram is a quadrilateral that has two pairs of parallel sides. The opposite sides have the same length.

Question 4.
Big Ideas Math Answer Key Grade 5 Chapter 14 Classify Two-Dimensional Shapes 48

Answer:  Square

Explanation: A square is a parallelogram that has four right angles and four sides with the same length.

Question 5.
Big Ideas Math Answer Key Grade 5 Chapter 14 Classify Two-Dimensional Shapes 49

Answer: Rectangle

Explanation: A rectangle is a parallelogram that has four right angles and diagonals are congruent.

Question 6.
Big Ideas Math Answer Key Grade 5 Chapter 14 Classify Two-Dimensional Shapes 50

Answer: Rhombus

Explanation: A rhombus is a parallelogram with four congruent sides and A rhombus has all the properties of a parallelogram. The diagonals intersect at right angles.

Question 7.
A sign has the shape of a quadrilateral that has two pairs of parallel sides, four sides with the same length, and no right angles

Answer: Parallelogram

Explanation:
Assume that a quadrilateral has parallel sides or equal sides unless that is stated. A parallelogram has two parallel pairs of opposite sides. A rectangle has two pairs of opposite sides parallel, and four right angles.

Question 8.
A tabletop has the shape of a quadrilateral with exactly one pair of parallel sides.

Answer: A trapezoid is a quadrilateral that has exactly one pair of parallel sides. A parallelogram is a quadrilateral that has two pairs of parallel sides.

Question 9.
YOU BE THE TEACHER
Your friend says that a quadrilateral with at least two right angles must be a parallelogram. Is your friend correct? Explain.

Explanation: A trapezoid is only required to have two parallel sides. However, a trapezoid could have one of the sides connecting the two parallel sides perpendicular to the parallel sides which would yield two right angles.

enter image source here

Question 10.
Which One Doesn’t Belong? Which cannot set of lengths be the side lengths of a parallelogram?
Big Ideas Math Answer Key Grade 5 Chapter 14 Classify Two-Dimensional Shapes 51

Answer: 9 yd, 5 yd, 5 yd, 3 yd

Explanation: A parallelogram is a quadrilateral that has two pairs of parallel sides. Opposite sides have the same length, So the above one is not a parallelogram.

Think and Grow: Modeling Real Life

Example
The dashed line shows how you cut the bottom of a rectangular door so it opens more easily. Classify the new shape of the door.
Big Ideas Math Answer Key Grade 5 Chapter 14 Classify Two-Dimensional Shapes 52
Draw the new shape of the door.
Big Ideas Math Answer Key Grade 5 Chapter 14 Classify Two-Dimensional Shapes 53

Answer:
The original shape of the door was a rectangle, so it had one pairs of parallel sides. The new shape of the door has exactly one pair of parallel sides. So, the new shape of the door is a trapezoid.

Show and Grow

Question 11.
DIG DEEPER!
The dashed line shows how you cut the corner of the trapezoidal piece of fabric. The line you cut is parallel to the opposite side. Classify the new shape of the four-sided piece of fabric.
Big Ideas Math Answer Key Grade 5 Chapter 14 Classify Two-Dimensional Shapes 54

Answer: Parallelogram

Explanation: A parallelogram is a quadrilateral that has two pairs of parallel sides. The opposite sides have the same length.

Question 12.
A farmer encloses a section of land using the four pieces of fencing. Name all of the four-sided shapes that the farmer can enclose with the fencing.
Big Ideas Math Answer Key Grade 5 Chapter 14 Classify Two-Dimensional Shapes 55

Answer: Parallelogram

Explanation: A parallelogram is a quadrilateral that has two pairs of parallel sides. Opposite sides have the same length, So four-sided shapes of fencing look like Parallelogram.

Classify Quadrilaterals Homework & Practice 14.2

Classify the quadrilateral in as many ways as possible.

Question 1.
Big Ideas Math Answer Key Grade 5 Chapter 14 Classify Two-Dimensional Shapes 56

Answer: Trapezoid 

Explanation: A trapezoid is a quadrilateral that has exactly one pair of parallel sides

Question 2.
Big Ideas Math Answer Key Grade 5 Chapter 14 Classify Two-Dimensional Shapes 57

Answer: Trapezoid

Explanation: A Trapezoid is a quadrilateral with exactly one pair of parallel sides. (There may be some confusion about this word depending on which country you’re in. In India and Britain, they say trapezium; in America, trapezium usually means a quadrilateral with no parallel sides.)

Question 3.
Big Ideas Math Answer Key Grade 5 Chapter 14 Classify Two-Dimensional Shapes 58

Answer: Rectangle

Explanation: A rectangle is a parallelogram that has four right angles and diagonals are congruent.

Question 4.
Big Ideas Math Answer Key Grade 5 Chapter 14 Classify Two-Dimensional Shapes 59

Answer: Square

Explanation: The diagonals of a square bisect each other and meet at 90°. The diagonals of a square bisect its angles. The opposite sides of a square are both parallel and equal in length. All four angles of a square are equal (each being 360°/4 = 90°, a right angle).

Question 5.
A name tag has the shape of a quadrilateral that has two pairs of parallel sides and four right angles. Opposite sides are the same length, but not all four sides are the same length.

Answer: Rectangle

Explanation: A rectangle is a parallelogram that has four right angles and diagonals are congruent.

Question 6.
A napkin has the shape of a quadrilateral that has two pairs of parallel sides, four sides with the same length, and four right angles.

Answer: Square

Explanation: A square is a parallelogram that has four right angles and four sides of the same length.

Question 7.
Reasoning
Can you draw a quadrilateral that is not a square, but has four right angles? Explain.

Answer: A rectangle is a parallelogram that has four right angles

Big Ideas Math Answer Key Grade 5 Chapter 14 Classify Two-Dimensional Shapes 58

Question 8.
Structure
Plot two more points in the coordinate plane to form a square. What two points can you plot to form a parallelogram? What two points can you plot to form a trapezoid? Do not use the same pair of points twice.
Big Ideas Math Answer Key Grade 5 Chapter 14 Classify Two-Dimensional Shapes 60

Answer:
Big-Ideas-Math-Answer-Key-Grade-5-Chapter-14-Classify-Two-Dimensional-Shapes-60

Question 9.
DIG DEEPER!
Which quadrilateral can be classified as a parallelogram, and rectangle, square, rhombus? Explain.

Answer: Square

Explanation: A square can be defined as a rhombus which is also a rectangle – in other words, a parallelogram with four congruent sides and four right angles. A trapezoid is a quadrilateral with exactly one pair of parallel sides.

Question 10.
Modeling Real Life
The dashed line shows how you fold the flap of the envelope so it closes. Classify the new shape of the envelope.
Big Ideas Math Answer Key Grade 5 Chapter 14 Classify Two-Dimensional Shapes 61

Answer: Rectangle

Explanation: A rectangle is a parallelogram that has four right angles and diagonals are congruent.

Question 11.
DIG DEEPER!
A construction worker tapes off a section of land using the four pieces of caution tape. Name all of the possible shapes that the worker can enclose with the tape.
Big Ideas Math Answer Key Grade 5 Chapter 14 Classify Two-Dimensional Shapes 62

Answer: Trapezoid

Explanation: A trapezoid is a quadrilateral that has exactly one pair of parallel sides.

Question 12.
Big Ideas Math Answer Key Grade 5 Chapter 14 Classify Two-Dimensional Shapes 62.1

Answer: \(\frac{1}{2}\)

Explanation: \(\frac{2}{3}\) –\(\frac{1}{6}\) equal to \(\frac{1}{2}\).

Question 13.
Big Ideas Math Answer Key Grade 5 Chapter 14 Classify Two-Dimensional Shapes 62.2

Answer: 0.112

Explanation: \(\frac{1}{2}\) is equal to 0.5 and 7/18 is equal to 0.3888.So subtraction from 0.5 to 0.3888 is 0.112.

Question 14.
Big Ideas Math Answer Key Grade 5 Chapter 14 Classify Two-Dimensional Shapes 62.3

Answer: 0.289

Explanation: \(\frac{2}{5}\) is equal to 0.4 and 1/9 is equal to 0.111,So subtraction from 0.4 to 0.111is 0.289.

Lesson 14.3 Relate Quadrilaterals

Explore and Grow

Label the Venn diagram to show the relationships among quadrilaterals. The first one has been done for you.
Big Ideas Math Answer Key Grade 5 Chapter 14 Classify Two-Dimensional Shapes 63

Reasoning
Explain how you decided where to place each quadrilateral.

Think and Grow: Relate Quadrilaterals

Key Idea
The Venn diagram shows the relationships among quadrilaterals.
Big Ideas Math Answer Key Grade 5 Chapter 14 Classify Two-Dimensional Shapes 64

Example
Tell whether the statement is true or false.
All rhombuses are rectangles.
Rhombuses do not always have four right angles.
So, the statement is ___.

Answer: So, the statement is true.

Example
Tell whether the statement is true or false.
All rectangles are parallelograms.
All rectangles have two pairs of parallel sides.
So, the statement is ___.

Show and Grow

Tell whether the statement is true or false. Explain.

Question 1.
Some rhombuses are squares.

Answer: true

Explanation: A rhombus is a quadrilateral (plane figure, closed shape, four sides) with four equal-length sides and opposite sides parallel to each other. All squares are rhombuses, but not all rhombuses are squares. The opposite interior angles of rhombuses are congruent.

Question 2.
All parallelograms are rectangles.

Answer: False

Explanation: A rectangle is a parallelogram with four right angles, so all rectangles are also parallelograms and quadrilaterals. On the other hand, not all quadrilaterals and parallelograms are rectangles. A rectangle has all the properties of a parallelogram

Apply and Grow: Practice

Tell whether the statement is true or false. Explain.

Question 3.
All rectangles are squares.

Answer: False

Explanation: All squares are rectangles, but not all rectangles are squares.

Question 4.
Some parallelograms are trapezoids.

Answer: True

Explanation: A trapezoid has one pair of parallel sides and a parallelogram has two pairs of parallel sides. So a parallelogram is also a trapezoid.

Question 5.
Some rhombuses are rectangles.

Answer: False

Explanation: A rhombus is defined as a parallelogram with four equal sides. Is a rhombus always a rectangle? No, because a rhombus does not have to have 4 right angles. Kites have two pairs of adjacent sides that are equal.

Question 6.
All trapezoids are quadrilaterals.

Answer: True

Explanation: Trapezoids have only one pair of parallel sides; parallelograms have two pairs of parallel sides. The correct answer is that all trapezoids are quadrilaterals. Trapezoids are four-sided polygons, so they are all quadrilaterals

Question 7.
All squares are rhombuses.

Answer: True

Explanation: All squares are rhombuses, but not all rhombuses are squares. The opposite interior angles of rhombuses are congruent

Question 8.
Some trapezoids are squares.

Answer: True

Explanation: A trapezoid is a quadrilateral with at least one pair of parallel sides. In a square, there are always two pairs of parallel sides, so every square is also a trapezoid. Conversely, only some trapezoids are squares

Question 9.
Reasoning
Use the word cards to complete the graphic organizer.
Big Ideas Math Answer Key Grade 5 Chapter 14 Classify Two-Dimensional Shapes 65

Answer: The first box to be filled with Square, 3d box to be filled with Rectangle,4th box to be filled with trapezoid and final box to be filled with Quadrilateral.

Explanation: A Square can be defined as a rhombus which is also a rectangle, in other words, a parallelogram with four congruent sides and four right angles. A trapezoid is a quadrilateral with exactly one pair of parallel sides.

Question 10.
Reasoning
All rectangles are parallelograms. Are all parallelograms rectangles? Explain.

Answer: True

Explanation: A rectangle is considered a special case of a parallelogram because, A parallelogram is a quadrilateral with 2 pairs of opposite, equal and parallel sides. A rectangle is a quadrilateral with 2 pairs of opposite, equal and parallel sides but also forms right angles between adjacent sides.

Question 11.
Precision
Newton says the figure is a square. Descartes says the figure is a parallelogram. Your friend says the figure is a rhombus. Are all three correct? Explain.
Big Ideas Math Answer Key Grade 5 Chapter 14 Classify Two-Dimensional Shapes 66

Answer: No

Explanation:  A square has two pairs of parallel sides, four right angles, and all four sides are equal. It is also a rectangle and a parallelogram. A rhombus is defined as a parallelogram with four equal sides. No, because a rhombus does not have to have 4 right angles.

Think and Grow: Modeling Real Life

Example
You use toothpicks to create several parallelograms. You notice that opposite angles of parallelograms have the same measure. For what other quadrilaterals is this also true?
Big Ideas Math Answers 5th Grade Chapter 14 Classify Two-Dimensional Shapes 66.1
Parallelograms have the property that opposite angles have the same measure. Subcategories of parallelograms must also have this property.
___, ___, and ___ are subcategories of parallelograms.
Big Ideas Math Answers 5th Grade Chapter 14 Classify Two-Dimensional Shapes 67
So, ___, ____, and ____ also have opposite angles with the same measure.

Answer: Rectangle, Rhombus and Square are subcategories of parallelograms.

Show and Grow

Question 12.
You use pencils to create several rhombuses. You notice that diagonals of rhombuses are perpendicular and divide each other into two equal parts. For what other quadrilateral is this also true? Explain your reasoning.
Big Ideas Math Answers 5th Grade Chapter 14 Classify Two-Dimensional Shapes 68

Answer: Square, Parallelogram, Rhombus are perpendicular and divided into the equal parts.

Question 13.
DIG DEEPER!
You place two identical parallelograms side by side. What can you conclude about the measures of adjacent angles in a parallelogram? For what other quadrilaterals is this also true? Explain your reasoning.
Big Ideas Math Answers 5th Grade Chapter 14 Classify Two-Dimensional Shapes 69

Answer:
The adjacent angles of the parallelogram is supplementary.
Opposite angles of the parallelogram are equal.

Relate Quadrilaterals Homework & Practice 14.3

Tell whether the statement is true or false. Explain.

Question 1.
All trapezoids are parallelograms.

Answer: False

Explanation: The pair of opposite sides of a parallelogram are equal and parallel but in the case of trapezium, this is not true in that only one pair of opposite sides are equal. Therefore every parallelogram is not a trapezium.
Question 2.

All rectangles are parallelograms.

Answer: True

Explanation: Each pair of co-interior angles are supplementary, because two right angles add to a straight angle, so the opposite sides of a rectangle are parallel. This means that a rectangle is a parallelogram, so, Its opposite sides are equal and parallel. Its diagonals bisect each other.

Question 3.
All squares are quadrilaterals.

Answer: True

Explanation: A closed figure with four sides. For example, kites, parallelograms, rectangles, rhombuses, squares, and trapezoids are all quadrilaterals. Kite: A quadrilateral with two pairs of adjacent sides that are equal in length; a kite is a rhombus if all side lengths are equal.
Question 4.

Some quadrilaterals are trapezoids.

Answer: True

Explanation: Trapezoids have only one pair of parallel sides, parallelograms have two pairs of parallel sides. A trapezoid can never be a parallelogram. The correct answer is that all trapezoids are quadrilaterals.

Question 5.
Some parallelograms are rectangles.

Answer: True

Explanation: Not all parallelograms are rectangles. A parallelogram is a rectangle if it has four right angles and two pairs of parallel and congruent sides.

Question 6.
All squares are rectangles and rhombuses.

Answer: False

Explanation: No, because all four sides of a rectangle don’t have to be equal. However, the sets of rectangles and rhombuses do intersect, and their intersection is the set of squares, all squares are both a rectangle and a rhombus.

Question 7.
YOU BE THE TEACHER
Newton says he can draw a quadrilateral that is not a trapezoid and not a parallelogram. Is Newton correct? Explain.

Answer: False

Explanation: Trapezoids have only one pair of parallel sides; parallelograms have two pairs of parallel sides. A trapezoid can never be a parallelogram. The correct answer is that all trapezoids are quadrilaterals.

Question 8.
Writing
Explain why a parallelogram is not a trapezoid.

Explanation: a square is a quadrilateral, a parallelogram, a rectangle, and a rhombus Is a trapezoid a parallelogram? No, because a trapezoid has only one pair of parallel sides.

Reasoning
Write always, sometimes, or never to make the statement true? Explain.

Question 9.
A rhombus is ___ a square.

Answer: A rhombus is  some times a square

Explanation: A rhombus is a square. This is sometimes true. Â It is true when a rhombus has 4 right angles. It is not true when a rhombus does not have any right angles.

Question 10.
A trapezoid is __ a rectangle.

Answer: A trapezoid is sometimes a rectangle.

Explanation: A rectangle has one pair of parallel sides.

Question 11.
A parallelogram is ___ a quadrilateral.

Answer: A parallelogram is always a quadrilateral.

Explanation: A parallelogram must have 4 sides, so they must always be quadrilaterals.

Question 12.
DIG DEEPER!
A quadrilateral has exactly three sides that have the same length. Why can the figure not be a rectangle?

Explanation: A rectangle is a parallelogram that has four right angles. opposite sides are in the same length, so the above one is not a rectangle.

Question 13.

Modeling Real Life
You fold the rectangular piece of paper. You notice that the line segments connecting the halfway points of opposite sides are perpendicular. For what other quadrilateral is this also true?
Big Ideas Math Answers 5th Grade Chapter 14 Classify Two-Dimensional Shapes 70

Explanation: A rectangle is a parallelogram that has four right angles. opposite sides are in the same length, so the above one is not a rectangle.

Question 14.
DIG DEEPER!
You tear off the four corners of the square and arrange them to form a circle. You notice that the sum of the angle measures of a square is equal to 360°. For what other quadrilaterals is this also true?
Big Ideas Math Answers 5th Grade Chapter 14 Classify Two-Dimensional Shapes 71

Answer: The sum of the angles in a parallelogram are 360°

Review & Refresh

Question 15.
5 pt = ___ c

Answer : 5 pt = 10 c

Explanation:
Convert from pints to cups.
1 pint = 2 cups
5 pints = 5 × 2 cups
5 pints = 10 cups

Question 16.
32 fl oz = ___ c
Answer: 32 fl oz =  4 c
Explanation:
Convert from fl oz to cups.
1 fl oz = 0.125 cups
32 fl oz are equal to 4 c.

Question 17.
20 qt = ___ c

Answer : 20 qt = 80 c

Explanation:
Convert from quarts to cups.
1 quart = 4 cups
20 qt = 20 × 4 cups = 80 cups

Classify Two-Dimensional Shapes Performance Task 14

A homeowner wants to install solar panels on her roof to generate electricity for her house. A solar panel is 65 inches long and 39 inches wide.

Question 1.
a. The shape of the panel has 4 right angles. Sketch and classify the shape of the solar panel.

Answer : Big Ideas Math Answer Key Grade 5 Chapter 14 Classify Two-Dimensional Shapes 41

Explanation: A rectangle is a parallelogram that has four right angles. So the shape of the solar panel is a rectangle.

b. There are 60 identical solar cells in a solar panel, arranged in an array. Ten cells meet to form the length of the panel, and six cells meet to form the width. Classify the shape of each solar cell. Explain your reasoning.
Big Ideas Math Answers 5th Grade Chapter 14 Classify Two-Dimensional Shapes 72

Answer: A rectangle is a parallelogram that has four right angles. So the shape of the solar panel is a rectangle.

Question 2.
The home owner measures three sections of her roof.
Big Ideas Math Answers 5th Grade Chapter 14 Classify Two-Dimensional Shapes 73
a. Classify the shape of each section in as many ways as possible.

Answer:  Right triangle.

Explanation: In a triangle one of the angle is a right angle (90 Deg ) called as Right triangle

Answer: Rectangle

Explanation: Rectangle is a parallelogram with four right angles, so all rectangles are also parallelograms and quadrilaterals. On the other hand, not all quadrilaterals and parallelograms are rectangles.

Answer: Isosceles trapezoid

Explanation: An isosceles trapezoid is a trapezoid whose non-parallel sides are congruent.

b. About how many solar panels can fit on the measured sections of the roof? Explain your reasoning.

Question 3.
One solar panel can produce about 30 kilowatt-hours of electricity each month. The homeowner uses her electric bills to determine that she uses about 1,200 kilowatt-hours of electricity each month.

a. How many solar panels should the homeowner install on her roof?

Answer: 40 Solar panels

Explanation: 40 Solar panels X 30 kilowatt-hours of electricity each month per one solar panel equal to 1,200 kilowatt-hours of electricity per month, So the answer is 40 solar panels.

b. Will all of the solar panels fit on the measured sections of the roof? Explain.

Classify Two-Dimensional Shapes Activity

Quadrilateral Lineup

Directions:

  1. Players take turns spinning the spinner.
  2. On your turn, cover a quadrilateral that matches your spin.
  3. If you land on, Lose a turn, then do not cover a quadrilateral.
  4. The first player to get four in a row twice, horizontally, vertically, Recor diagonally, wins!

Big Ideas Math Answers 5th Grade Chapter 14 Classify Two-Dimensional Shapes 74
Big Ideas Math Answers 5th Grade Chapter 14 Classify Two-Dimensional Shapes 75

Classify Two-Dimensional Shapes Chapter Practice 14

14.1 Classify Triangles

Classify the triangle by its angles and its sides.

Question 1.
Big Ideas Math Answers 5th Grade Chapter 14 Classify Two-Dimensional Shapes 76

Answer: Scalene triangle.
Explanation: A Scalene triangle has no sides that are congruent (Same size) and angles also all different.

Question 2.
Big Ideas Math Answers 5th Grade Chapter 14 Classify Two-Dimensional Shapes 77

Answer: Right triangle.
Explanation: In a triangle one of the angle is a right angle (90 deg ) called as Right triangle.

Question 3.
Big Ideas Math Answers 5th Grade Chapter 14 Classify Two-Dimensional Shapes 78

Answer: Isosceles triangle.

Explanation: An Isosceles triangle has two sides of the same length. Two of its angle also measure equal.

Question 4.
Big Ideas Math Answers 5th Grade Chapter 14 Classify Two-Dimensional Shapes 79

Answer: Equiangular triangle

Explanation: In an equilateral triangle, all the lengths of the sides are equal. In such a case, each of the interior angles

Question 5.
Big Ideas Math Answers 5th Grade Chapter 14 Classify Two-Dimensional Shapes 80

Answer: Right triangle.

Explanation: In a triangle one of the angle is a right angle (90° ) called as Right triangle.

Question 6.
Big Ideas Math Answers 5th Grade Chapter 14 Classify Two-Dimensional Shapes 81

Answer: Scalene triangle.

Explanation: A Scalene triangle has no sides that are congruent (Same size) and angles also all different.

14.2 Classify Quadrilaterals

Classify the quadrilateral in as many ways as possible.

Question 7.
Big Ideas Math Answers 5th Grade Chapter 14 Classify Two-Dimensional Shapes 82

Answer: Square

Explanation: A square is a parallelogram that has four right angles and four sides of the same length.

Question 8.
Big Ideas Math Answers 5th Grade Chapter 14 Classify Two-Dimensional Shapes 83

Answer: Rectangle

Explanation: A rectangle is a parallelogram that has four right angle, Opposite sides are the same length

Question 9.
Big Ideas Math Solutions Grade 5 Chapter 14 Classify Two-Dimensional Shapes 84

Answer: Square

Explanation: The diagonals of a square bisect each other and meet at 90°. The diagonals of a square bisect its angles. The opposite sides of a square are both parallel and equal in length. All four angles of a square are equal (each being 360°/4 = 90°, a right angle).

Question 10.
Big Ideas Math Solutions Grade 5 Chapter 14 Classify Two-Dimensional Shapes 85

Answer: Isosceles trapezoid

Explanation: An isosceles trapezoid is a trapezoid whose non-parallel sides are congruent.

Question 11.
Big Ideas Math Solutions Grade 5 Chapter 14 Classify Two-Dimensional Shapes 86

Answer: Rectangle

Explanation: A rectangle is a parallelogram that has four right angle, Opposite sides are the same length

Question 12.
Big Ideas Math Solutions Grade 5 Chapter 14 Classify Two-Dimensional Shapes 87

Answer: Trapezoid

Explanation: A trapezoid is a quadrilateral that has exactly one pair of parallel sides.

Question 13.
Structure
Plot two more points in the coordinate plane to form a quadrilateral that has exactly two a rectangle. What two points can you plot to form a trapezoid? What two points can you plot to form a rhombus? Do not use the same pair of points twice.
Big Ideas Math Solutions Grade 5 Chapter 14 Classify Two-Dimensional Shapes 88

Answer:
Big-Ideas-Math-Solutions-Grade-5-Chapter-14-Classify-Two-Dimensional-Shapes-88

Question 14.
Reasoning
Can you draw a quadrilateral that has exactly two right angles? Explain.

Explanation: A quadrilateral with only 2 right angles and it is called a trapezoid .

enter image source here

Question 15.
Modeling Real Life
The dashed line shows how you break apart the graham cracker. Classify the new shape of each piece of the graham cracker.
Big Ideas Math Solutions Grade 5 Chapter 14 Classify Two-Dimensional Shapes 89

Answer: Square

Explanation: The diagonals of a square bisect each other and meet at 90°.

14.3 Relate Quadrilaterals

Tell whether the statement is true or false.

Question 16.
All rectangles are quadrilaterals.

Answer: True

Explanation: A closed figure with four sides. For example, kites, parallelograms, rectangles, rhombuses, squares, and trapezoids are all quadrilaterals

Question 17.
Some parallelograms are squares.

Answer: True

Explanation: Squares fulfill all criteria of being a rectangle because all angles are right angle and opposite sides are equal. Similarly, they fulfill all criteria of a rhombus, as all sides are equal and their diagonals bisect each other.

Question 18.

All trapezoids are rectangles.

Answer: False

Explanation: Rectangles are defined as a four-sided polygon with two pairs of parallel sides. On the other hand, a trapezoid is defined as a quadrilateral with only one pair of parallel sides.

Question 19.
Some rectangles are rhombuses.

Answer: True

Explanation: A rectangle is a parallelogram with all its interior angles being 90 degrees. A rhombus is a parallelogram with all its sides equal. This means that for a rectangle to be a rhombus, its sides must be equal. A rectangle can be a rhombus only if has extra properties which would make it a square

Question 20.
Some squares are trapezoids.

Answer: True

Explanation: The definition of a trapezoid is that it is a quadrilateral that has at least one pair of parallel sides. A square, therefore, would be considered a trapezoid.

Question 21.
All quadrilaterals are squares.

Answer: False

Explanation : Quadrilateral: A closed figure with four sides. For example, kites, parallelograms, rectangles, Square: A rectangle with four sides of equal length. Trapezoid: A quadrilateral with at least one pair of parallel sides So, All quadrilaterals are not squares.

Classify Two-Dimensional Shapes Cumulative Practice 1-14

Question 1.
Which model shows 0.4 × 0.2?
Big Ideas Math Solutions Grade 5 Chapter 14 Classify Two-Dimensional Shapes 90

Answer:

Question 2.
A triangle has angle measures of 82°, 53°, and 45°. Classify the triangle by its angles.
Big Ideas Math Solutions Grade 5 Chapter 14 Classify Two-Dimensional Shapes 91

Answer: C

Explanation: An obtuse triangle has one angle measuring more than 90º but less than 180º (an obtuse angle). It is not possible to draw a triangle with more than one obtuse angle

Question 3.
Which expressions have an estimated difference of \(\frac{1}{2}\) ?
Big Ideas Math Solutions Grade 5 Chapter 14 Classify Two-Dimensional Shapes 92

Answer:

Question 4.
A rectangular prism has a volume of 288 cubic centimeters. The height of the prism is 8 centimeters. The base is a square. What is a side length of the base?
Big Ideas Math Solutions Grade 5 Chapter 14 Classify Two-Dimensional Shapes 93

Answer: A

Explanation: volume of a rectangular prism, multiply its 3 dimensions: length x width x height. The volume is expressed in cubic units. So,6 X 6 X 8 is equal to 288 cubic centimeters, Therefore the side length of the base is 6 cm.

Question 5.
A sandwich at a food stand costs $3.00. Each additional topping costs the same extra amount. The coordinate plan shows the costs, in dollars, of sandwiches with different numbers of additional toppings. What is the cost of a sandwich with 3 additional toppings?
Big Ideas Math Solutions Grade 5 Chapter 14 Classify Two-Dimensional Shapes 94

Answer:

Question 6.
Which statements are true?
Big Ideas Math Solutions Grade 5 Chapter 14 Classify Two-Dimensional Shapes 95

Answer:  The following statements are true
Option 2,option 3 and option 4 .

Explanation :
Option 2 :
All squares are rectangles are parallelograms is true, why because squares fulfill all criteria of being a rectangle because all angles are right angle and opposite sides are equal. Similarly, they fulfill all criteria of a rhombus, as all sides are equal and their diagonals bisect each other.
Option 3: All squares are rhombuses is true, why because All squares are rhombuses, but not all rhombuses are squares. The opposite interior angles of rhombuses are congruent.
Option 4:  Every trapezoid is a quadrilateral is true, why because Trapezoids have only one pair of parallel sides; parallelograms have two pairs of parallel sides.

Question 7.
Your friend makes a volcano for a science project. She uses 10 cups of vinegar. How many pints of vinegar does he use?
Big Ideas Math Solutions Grade 5 Chapter 14 Classify Two-Dimensional Shapes 96

Answer: Option B

Explanation: 1 cup is equal to  0.5 pints, therefore 10 cups are equal to 5 pints.

Question 8.
The volume of the rectangular prism is 432 cubic centimeters. What is the length of the prism?
Big Ideas Math Answer Key Grade 5 Chapter 14 Classify Two-Dimensional Shapes 97

Answer: Option A

Explanation: volume of a rectangular prism, multiply its 3 dimensions: length x width x height. The volume is expressed in cubic units.
So, 6 cm X 9 cm X 8 cm is equal to 432 cubic centimeters.
Therefore the length of prim is 9 cm.

Question 9.
Descartes draws a pentagon by plotting another point in the coordinate plane and connecting the points. Which coordinates could he use?
Big Ideas Math Answer Key Grade 5 Chapter 14 Classify Two-Dimensional Shapes 98

Answer:

Question 10.
Newton rides to the dog park in a taxi. He owes the driver $12. He calculates the driver’s tip by multiplying $12 by 0.15. How much does he pay the driver, including the tip?
Big Ideas Math Answer Key Grade 5 Chapter 14 Classify Two-Dimensional Shapes 99

Answer: Option  C

Explanation: Driver cost $12 + ($12 X 0.15 )= 12+1.8 =13.8
Therefore answer is $13.8.

Question 11.
A quadrilateral has four sides with the same length, two pairs of parallel sides, and four 90° angles. Classify the quadrilateral in as many ways as possible.
Big Ideas Math Answer Key Grade 5 Chapter 14 Classify Two-Dimensional Shapes 100

Answer: Square, Parallelogram

Explanation: A quadrilateral has four sides with the same length, two pairs of parallel sides and four 90° angles is called as square. All squares are parallelograms.

Question 12.
Which ordered pair represents the location of a point shown in the coordinate plane?
Big Ideas Math Answer Key Grade 5 Chapter 14 Classify Two-Dimensional Shapes 101

Answer:

Question 13.
What is the product of 5,602 and 17?
Big Ideas Math Answer Key Grade 5 Chapter 14 Classify Two-Dimensional Shapes 102

Answer: 95234

Explanation: 5602 X 17 is equal to 95234.

Question 14.
Which pair of points do not lie on a line that is perpendicular to the x-axis?
Big Ideas Math Answer Key Grade 5 Chapter 14 Classify Two-Dimensional Shapes 103

Answer:

Question 15.
Newton has a gift in the shape of a rectangular prism that has a volume of 10,500 cubic inches. The box he uses to ship the gift is shown.
Big Ideas Math Answer Key Grade 5 Chapter 14 Classify Two-Dimensional Shapes 104
Part A What is the volume of the box?
Part B What is the volume, in cubic inches, of the space inside the box that is not taken up by gift? Explain.?

Answer:

Question 16.
Which expressions have a product greater than \(\frac{5}{6}=\)?
Big Ideas Math Answer Key Grade 5 Chapter 14 Classify Two-Dimensional Shapes 105

Answer:
Big-Ideas-Math-Answer-Key-Grade-5-Chapter-14-Classify-Two-Dimensional-Shapes-105

Question 17.
Newton is thinking of a shape that has 4 sides, only one pair of parallel sides, and angle measures of 90°, 40°, 140°, and 90°. Which is Newton’s shape?
Big Ideas Math Answer Key Grade 5 Chapter 14 Classify Two-Dimensional Shapes 106

Answer: D

Explanation: Trapezoid Only one pair of opposite sides is parallel.

Question 18.
Which rectangular prisms have a volume of 150 cubic feet?
Big Ideas Math Answer Key Grade 5 Chapter 14 Classify Two-Dimensional Shapes 107

Answer: Option 1

Explanation: volume of a rectangular prism, multiply its 3 dimensions: length x width x height. The volume is expressed in cubic feet.
So,2 ft X 25 ft X 3 ft is equal to 150 cubic ft, Therefore the right answer is option 1.

Classify Two-Dimensional Shapes Steam Performance Task 1-14

Each student in your grade makes a constellation display by making holes for the stars of a constellation on each side of the display. Each display is a rectangular prism with a square base.
Big Ideas Math Answer Key Grade 5 Chapter 14 Classify Two-Dimensional Shapes 108

Question 1.
Your science teacher orders a display for each student. The diagram shows the number of packages that can fit in a shipping box.
a. How many displays come in one box?
Big Ideas Math Answer Key Grade 5 Chapter 14 Classify Two-Dimensional Shapes 109
b. There are 108 students in your grade. How many boxes of displays does your teacher order? Explain.
c. The volume of the shipping box is 48,000 cubic inches. What is the volume of each display?
d. The height of each display is 15 inches. What are the dimensions of the square base?
e. Estimate the dimensions of the shipping box.
f. You paint every side of the display except the bottom. What is the total area you will paint?
g. You need a lantern to light up your display. Does the lantern fit inside of your display? Explain.
Big Ideas Math Answer Key Grade 5 Chapter 14 Classify Two-Dimensional Shapes 111

Question 2.
On one side of your display, you create an image of the constellation Libra. Each square on the grid is 1 square inch.
a. Classify the triangle formed by the points of the constellation.
Big Ideas Math Answer Key Grade 5 Chapter 14 Classify Two-Dimensional Shapes 112

Answer: Equiangular triangle
Explanation: In an equilateral triangle, all the lengths of the sides are equal. In such a case, each of the interior angles

b. What are the coordinates of the points of the constellation?
c. What is the height of the constellation on your display?

Question 3.
You use the coordinate plane to create the image of the Big Dipper.
a. Plot the points A(6, 2), B(8, 2), C(7, 6), D(5, 5), E(7, 9), F(6, 12), and G(4, 14).
b. Draw lines connecting the points of quadrilateral ABCD. Draw \(\overline{C E}\), \(\overline{E F}\) and \(\overline{F G}\).
Big Ideas Math Answer Key Grade 5 Chapter 14 Classify Two-Dimensional Shapes 113

Answer:
Big-Ideas-Math-Answer-Key-Grade-5-Chapter-14-Classify-Two-Dimensional-Shapes-113
c. Is quadrilateral ABCD a trapezoid? How do you know?

Answer: Yes ABCD is a trapezoid because all sides are not equal and only one pair has parallel sides.

Question 4.
Use the Internet or some other resource to learn more about constellations. Write one interesting thing you learn.

Answer: A constellation is a group of stars that appears to form a pattern or picture like Orion the Great Hunter, Leo the Lion, or Taurus the Bull. Constellations are easily recognizable patterns that help people orient themselves using the night sky. There are 88 “official” constellations.

Conclusion:

Sharpen your math skills by practicing the problems from Big Ideas Math Book 5th Grade Answer Key Chapter 14 Classify Two-Dimensional Shapes. All the solutions of Grade 5 Chapter 14 Classify Two-Dimensional Shapes are prepared by the math professionals. Thus you can prepare effectively and score good marks in the exams.

Eureka Math Grade 8 Module 4 Lesson 3 Answer Key

Eureka Math Grade 8 Module 4 Lesson 3 Answer Key

Engage NY Eureka Math 8th Grade Module 4 Lesson 3 Answer Key

Eureka Math Grade 8 Module 4 Lesson 3 Exercise Answer Key

Exercise 1.
Is the equation a true statement when x=-3? In other words, is -3 a solution to the equation 6x+5=5x+8+2x? Explain.
Answer:
If we replace x with the number -3, then the left side of the equation is
6∙(-3)+5=-18+5
=-13″,”
and the right side of the equation is
5∙(-3)+8+2∙(-3)=-15+8-6
=-7-6
=-13.
Since -13=-13, then x=-3 is a solution to the equation 6x+5=5x+8+2x.
Note: Some students may have transformed the equation.

Exercise 2.
Does x=12 satisfy the equation 16-\(\frac{1}{2}\) x=\(\frac{3}{4}\) x+1? Explain.
Answer:
If we replace x with the number 12, then the left side of the equation is
16-\(\frac{1}{2}\) x=16-\(\frac{1}{2}\)∙(12)
=16-6
=10,
and the right side of the equation is
\(\frac{3}{4}\) x+1=\(\frac{3}{4}\)∙(12)+1
=9+1
=10.
Since 10=10, then x=12 is a solution to the equation 16-\(\frac{1}{2}\) x=\(\frac{3}{4}\) x+1.

Exercise 3.
Chad solved the equation 24x+4+2x=3(10x-1) and is claiming that x=2 makes the equation true. Is Chad correct? Explain.
Answer:
If we replace x with the number 2, then the left side of the equation is
24x+4+2x=24∙2+4+2∙2
=48+4+4
=56,
and the right side of the equation is
3(10x-1)=3(10∙2-1)
=3(20-1)
=3(19)
=57.
Since 56≠57, then x=2 is not a solution to the equation 24x+4+2x=3(10x-1), and Chad is not correct.

Exercise 4.
Lisa solved the equation x+6=8+7x and claimed that the solution is x=-\(\frac{1}{3}\). Is she correct? Explain.
Answer:
If we replace x with the number –\(\frac{1}{3}\), then the left side of the equation is
x+6=-\(\frac{1}{3}\)+6
=5 \(\frac{2}{3}\),
and the right side of the equation is
8+7x=8+7∙(-\(\frac{1}{3}\))
=8-\(\frac{7}{3}\)
=\(\frac{24}{3}\)–\(\frac{7}{3}\)
=\(\frac{17}{3}\).
Since 5 \(\frac{2}{3}\)=\(\frac{17}{3}\), then x=-\(\frac{1}{3}\) is a solution to the equation x+6=8+7x, and Lisa is correct.

Exercise 5.
Angel transformed the following equation from 6x+4-x=2(x+1) to 10=2(x+1). He then stated that the solution to the equation is x=4. Is he correct? Explain.
Answer:
No, Angel is not correct. He did not transform the equation correctly. The expression on the left side of the equation 6x+4-x=2(x+1) would transform to
6x+4-x=6x-x+4
=(6-1)x+4
=5x+4.
If we replace x with the number 4, then the left side of the equation is
5x+4=5∙4+4
=20+4
=24,
and the right side of the equation is
2(x+1)=2(4+1)
=2(5)
=10.
Since 24≠10, then x=4 is not a solution to the equation 6x+4-x=2(x+1), and Angel is not correct.

Exercise 6.
Claire was able to verify that x=3 was a solution to her teacher’s linear equation, but the equation got erased from the board. What might the equation have been? Identify as many equations as you can with a solution of x=3.
Answer;
Answers will vary. Ask students to share their equations and justifications as to how they knew x=3 would make a true number sentence.

Exercise 7.
Does an equation always have a solution? Could you come up with an equation that does not have a solution?
Answer:
Answers will vary. Expect students to write equations that are false. Ask students to share their equations and justifications as to how they knew the equation they wrote did not have a solution. The concept of “no solution” is introduced in Lesson 6 and solidified in Lesson 7.

Eureka Math Grade 8 Module 4 Lesson 3 Problem Set Answer Key

Eureka Math Grade 8 Module 4 Lesson 3 Problem Set Answer Key

Students practice determining whether or not a given number is a solution to the linear equation.

Question 1.
Given that 2x+7=27 and 3x+1=28, does 2x+7=3x+1? Explain.
Answer:
No, because a linear equation is a statement about equality. We are given that 2x+7=27, but 3x+1=28. Since each linear expression is equal to a different number, 2x+7≠3x+1.

Question 2.
Is -5 a solution to the equation 6x+5=5x+8+2x? Explain.
Answer:
If we replace x with the number -5, then the left side of the equation is
6∙(-5)+5=-30+5
=-25,
and the right side of the equation is
5∙(-5)+8+2∙(-5)=-25+8-10
=-17-10
=-27.
Since -25≠-27, then -5 is not a solution of the equation 6x+5=5x+8+2x.
Note: Some students may have transformed the equation.

Question 3.
Does x=1.6 satisfy the equation 6-4x=-\(\frac{x}{4}\)? Explain.
Answer:
If we replace x with the number 1.6, then the left side of the equation is
6-4∙1.6=6-6.4
=-0.4,
and the right side of the equation is
–\(\frac{-1.6}{4}\)=-0.4.
Since -0.4=-0.4, then x=1.6 is a solution of the equation 6-4x=-\(\frac{x}{4}\).

Question 4.
Use the linear equation 3(x+1)=3x+3 to answer parts (a)–(d).
a. Does x=5 satisfy the equation above? Explain.
Answer:
If we replace x with the number 5, then the left side of the equation is
3(5+1)=3(6)
=18,
and the right side of the equation is
3x+3=3∙5+3
=15+3
=18.
Since 18=18, then x=5 is a solution of the equation 3(x+1)=3x+3.

b. Is x=-8 a solution of the equation above? Explain.
Answer:
If we replace x with the number -8, then the left side of the equation is
3(-8+1)=3(-7)
=-21,
and the right side of the equation is
3x+3=3∙(-8)+3
=-24+3
=-21.
Since -21=-21, then x=-8 is a solution of the equation 3(x+1)=3x+3.

c. Is x=\(\frac{1}{2}\) a solution of the equation above? Explain.
Answer:
If we replace x with the number \(\frac{1}{2}\), then the left side of the equation is
3(\(\frac{1}{2}\)+1)=3(\(\frac{1}{2}\)+\(\frac{2}{2}\))
=3(\(\frac{3}{2}\))
=\(\frac{9}{2}\),
and the right side of the equation is
3x+3=3∙(\(\frac{1}{2}\))+3
= \(\frac{3}{2}\)+3
=\(\frac{3}{2}\)+\(\frac{6}{2}\)
=\(\frac{9}{2}\).
Since \(\frac{9}{2}\)=\(\frac{9}{2}\), then x=\(\frac{1}{2}\) is a solution of the equation 3(x+1)=3x+3.

d. What interesting fact about the equation 3(x+1)=3x+3 is illuminated by the answers to parts (a), (b), and (c)? Why do you think this is true?
Answer:
Note to teacher: Ideally, students will notice that the equation 3(x+1)=3x+3 is an identity under the distributive law. The purpose of this problem is to prepare students for the idea that linear equations can have more than one solution, which is a topic of Lesson 7.

Eureka Math Grade 8 Module 4 Lesson 3 Exit Ticket Answer Key

Question 1.
Is 8 a solution to \(\frac{1}{2}\) x+9=13? Explain.
Answer:
If we replace x with the number 8, then the left side is \(\frac{1}{2}\) (8)+9=4+9=13, and the right side is 13. Since 13=13, then x=8 is a solution.

Question 2.
Write three different equations that have x=5 as a solution.
Answer:
Answers will vary. Accept equations where x=5 makes a true number sentence.

Question 3.
Is -3 a solution to the equation 3x-5=4+2x? Explain.
Answer:
If we replace x with the number -3, then the left side is 3(-3)-5=-9-5=-14. The right side is 4+2(-3)=4-6=-2. Since -14≠-2, then -3 is not a solution of the equation.

Eureka Math Grade 6 Module 1 Lesson 4 Answer Key

Eureka Math Grade 6 Module 1 Lesson 4 Answer Key

Engage NY Eureka Math 6th Grade Module 1 Lesson 4 Answer Key

Eureka Math Grade 6 Module 1 Lesson 4 Example Answer Key

Example 1.
The morning announcements said that two out of every seven sixth-grade students In the school have an overdue library book. Jasmine said, “That would mean 24 of us have overdue books!” Grace argued, “No way. That is way too high.” How can you determine who is right?
Answer:
You would have to know the total number of sixth-grade students, and then see if the ratio 24: total is equivalent to 2: 7.
Eureka Math Grade 6 Module 1 Lesson 4 Example Answer Key 1
Answer:

Eureka Math Grade 6 Module 1 Lesson 4 Exercise Answer Key

Eureka Math Grade 6 Module 1 Lesson 4 Exercise Answer Key

Exercise 1.
Decide whether or not each of the following pairs of ratios is equivalent.
→ If the ratios are not equivalent, find a ratio that is equivalent to the first ratio.
→ If the ratios are equivalent, identify the nonzero number, c, that could be used to multiply each number of the first ratio by in order to get the numbers for the second ratio.
a. 6: 11 and 42: 88
________ Yes, the value, c, is ________
________ No, an equivalent ratio would be ________
Answer:
Eureka Math Grade 6 Module 1 Lesson 4 Exercise Answer Key 2
________ Yes, the value, c, is ________
  x    No, an equivalent ratio would be    42: 77   
Answer:

b. 0: 5 and 0: 20
________ Yes, the value, c, is ________
________ No, an equivalent ratio would be ________
Answer:
Eureka Math Grade 6 Module 1 Lesson 4 Exercise Answer Key 3
  x _  Yes, the value, c, is   4   
________ No, an equivalent ratio would be ________

Exercise 2.
In a bag of mixed walnuts and cashews, the ratio of the number of walnuts to the number of cashews is 5: 6. Determine the number of walnuts that are in the bag if there are 54 cashews. Use a tape diagram to support your work. Justify your answer by showing that the new ratio you created of the number of walnuts to the number of cashews is equivalent to 5: 6.
Answer:
Eureka Math Grade 6 Module 1 Lesson 4 Exercise Answer Key 4
54 divided by 6 equals 9.
5 times 9 equals 45.
There are 45 walnuts in the bag.
The ratio of the number of walnuts to the number of cashews is 45: 54. That ratio is equivalent to 5: 6.
Eureka Math Grade 6 Module 1 Lesson 4 Exercise Answer Key 5
Answer:

Eureka Math Grade 6 Module 1 Lesson 4 Problem Set Answer Key

Question 1.
Use diagrams or the description of equivalent ratios to show that the ratios 2: 3, 4: 6, and 8: 12 are equivalent.
Answer:
Eureka Math Grade 6 Module 1 Lesson 4 Problem Set Answer Key 6

Question 2.
Prove that 3: 8 is equivalent to 12: 32.
a. Use diagrams to support your answer.
Answer:
Eureka Math Grade 6 Module 1 Lesson 4 Problem Set Answer Key 7

b. Use the description of equivalent ratios to support your answer.
Answer:
Answers will vary. Descriptions should include multiplicative comparisons, such as 12 is 3 times 4 and 32 is 8 times 4. The constant number, c, is 4.

Question 3.
The ratio of Isabella’s money to Shane’s money is 3: 11. If Isabella has $33, how much money do Shane and Isabella have together? Use diagrams to illustrate your answer.
Answer:
Isabella has $33, and Shane has $121. $33 + $121 = $154. Together, Isabella and Shane have $154. 00.
Eureka Math Grade 6 Module 1 Lesson 4 Problem Set Answer Key 8

Eureka Math Grade 6 Module 1 Lesson 4 Exit Ticket Answer Key

Eureka Math Grade 6 Module 1 Lesson 4 Exit Ticket Answer Key

Question 1.
There are 35 boys in the sixth grade. The number of girls in the sixth grade is 42. Lonnie says that means the ratio of the number of boys in the sixth grade to the number of girls in sixth grade is 5: 7. Is Lonnie correct? Show why or why not.
Answer:
No, Lonnie is not correct. The ratios 5:7 and 35:42 are not equivalent. They are not equivalent because 5 × 7 = 35, but 7 × 7 = 49, not 42.

Eureka Math Grade 6 Module 1 Lesson 3 Answer Key

Eureka Math Grade 6 Module 1 Lesson 3 Answer Key

Engage NY Eureka Math 6th Grade Module 1 Lesson 3 Answer Key

Eureka Math Grade 6 Module 1 Lesson 3 Exercise Answer Key

Exercise 1.
Write a one-sentence story problem about a ratio.
Answer:
Answers will vary. The ratio of the number of sunny days to the number of cloudy days in this town is 3: 1.

Write the ratio in two different forms.
Answer:
3: 1 and 3 to 1

Exercise 2.
Shanni and Mel are using ribbon to decorate a project in their art class. The ratio of the length of Shanni’s ribbon to the length of Mel’s ribbon is 7: 3.
Draw a tape diagram to represent this ratio.
Answer:
Eureka Math Grade 6 Module 1 Lesson 3 Exercise Answer Key 1

Exercise 3.
Mason and Laney ran laps to train for the long-distance running team. The ratio of the number of laps Mason ran to the number of laps Laney ran was 2 to 3.
a. If Mason ran 4 miles, how far did Laney run? Draw a tape diagram to demonstrate how you found the answer.
Answer:
Eureka Math Grade 6 Module 1 Lesson 3 Exercise Answer Key 2

b. If Laney ran 930 meters, how far did Mason run? Draw a tape diagram to determine how you found the answer.
Answer:
Eureka Math Grade 6 Module 1 Lesson 3 Exercise Answer Key 3

c. What ratios can we say are equivalent to 2: 3?
Answer:
4: 6 and 620: 930

Exercise 4.
Josie took a long multiple-choice, end-of-year vocabulary test. The ratio of the number of problems Josie got incorrect to the number of problems she got correct is 2: 9.
a. If Josie missed 8 questions, how many did she get correct? Draw a tape diagram to demonstrate how you found the answer.
Answer:
Eureka Math Grade 6 Module 1 Lesson 3 Exercise Answer Key 4

b. If Josie missed 20 questions, how many did she get correct? Draw a tape diagram to demonstrate how you found the answer.
Answer:
Eureka Math Grade 6 Module 1 Lesson 3 Exercise Answer Key 5

c. What ratios can we say are equivalent to 2: 9?
Answer:
8: 36 and 20: 90

d. Come up with another possible ratio of the number Josie got incorrect to the number she got correct.
Answer:
Eureka Math Grade 6 Module 1 Lesson 3 Exercise Answer Key 6

e. How did you find the numbers?
Answer:
Multiplied 5 × 2 and 5 × 9

f. Describe how to create equivalent ratios.
Answer:
Multiply both numbers of the ratio by the same number (any number you choose).

Eureka Math Grade 6 Module 1 Lesson 3 Problem Set Answer Key

Eureka Math Grade 6 Module 1 Lesson 3 Problem Set Answer Key

Question 1.
Write two ratios that are equivalent to 1: 1.
Answer:
Answers will vary. 2:2, 50: 50, etc.

Question 2.
Write two ratios that are equivalent to 3: 11.
Answer:
Answers will vary. 6:22, 9:33, etc.

Question 3.
a. The ratio of the width of the rectangle to the height of the rectangle is _______ to _______.
Eureka Math Grade 6 Module 1 Lesson 3 Problem Set Answer Key 7
Answer:
The ratio of the width of the rectangle to the height of the rectangle is   9     to   4    .

b. If each square in the grid has a side length of 8 mm, what is the width and height of the rectangle?
Answer:
72 mm wide and 32 mm high

Question 4.
For a project in their health class, Jasmine and Brenda recorded the amount of milk they drank every day. Jasmine drank 2 pints of milk each day, and Brenda drank 3 pints of milk each day.
a. Write a ratio of the number of pints of milk Jasmine drank to the number of pints of milk Brenda drank each day.
Answer:
2: 3

b. Represent this scenario with tape diagrams.
Answer:
Eureka Math Grade 6 Module 1 Lesson 3 Problem Set Answer Key 8

c. If one pint of milk Is equivalent to 2 cups of milk, how many cups of milk did Jasmine and Brenda each drink? How do you know?
Answer:
Jasmine drank 4 cups of milk, and Brenda drank 6 cups of milk. Since each pint represents 2 cups, I multiplied
Jasmine’s 2 pints by 2 and multiplied Brenda’s 3 pints by 2.

d. Write a ratio of the number of cups of milk Jasmine drank to the number of cups of milk Brenda drank.
Answer:
4: 6

e. Are the two ratios you determined equivalent? Explain why or why not.
Answer:
2: 3 and 4: 6 are equivalent because they represent the same value. The diagrams never changed, only the value of each unit in the diagram.

Eureka Math Grade 6 Module 1 Lesson 3 Exit Ticket Answer Key

Pam and her brother both open savings accounts. Each begin with a balance of zero dollars. For every two dollars that Pam saves in her account, her brother saves five dollars in his account.

Question 1.
Determine a ratio to describe the money in Pam’s account to the money in her brother’s account.
Answer:
2: 5

Question 2.
If Pam has 40 dollars in her account, how much money does her brother have in his account? Use a tape diagram to support your answer.
Answer:
Eureka Math Grade 6 Module 1 Lesson 3 Exit Ticket Answer 9

Question 3.
Record the equivalent ratio.
Answer:
40: 100

Question 4.
Create another possible ratio that describes the relationship between the amount of money in Pam’s account and the amount of money in her brother’s account.
Answer:
Answers will vary. 4: 10, 8: 20, etc.