Tag: Eureka Math Grade 5 Answer Key

Engage NY Eureka Math 5th Grade Module 4 Lesson 17 Answer Key

Eureka Math Grade 5 Module 4 Lesson 17 Problem Set Answer Key

Question 1.
Multiply and model. Rewrite each expression as a multiplication sentence with decimal factors. The first one is done for you.
\(\frac{1}{10}\) × \(\frac{1}{10}\)
= \(\frac{1 × 1}{10 × 10}\)
= \(\frac{1}{100}\)

b. \(\frac{4}{10}\) × \(\frac{3}{10}\)

Answer:

Explanation:
Given that \(\frac{4}{10}\) × \(\frac{3}{10}\)
= \(\frac{4 × 3}{10 × 10}\)
= \(\frac{12}{100}\)
= 0.12.

c. \(\frac{1}{10}\) × 1.4

d. \(\frac{6}{10}\) × 1.7

Question 2.
Multiply. The first few are started for you.
a. 5 × 0.7 = _______
= 5 × \(\frac{7}{10}\)
= \(\frac{5 × 7}{10}\)
= \(\frac{35}{10}\)
= 3.5

b. 0.5 × 0.7 = ___
= \(\frac{5}{10}\) × \(\frac{7}{10}\)
= \(\frac{5 × 7}{10 × 10}\)
= \(\frac{35}{100}\)
= 0.35

c. 0.05 × 0.7 = __
= \(\frac{5}{100}\) × \(\frac{7}{10}\)

= \(\frac{5}{7}\) × \(\frac{10}{100}\)
= \(\frac{35}{1000}\)
= 0.035

d. 6 × 0.3 = _______
= 6 × \(\frac{3}{10}\)
= \(\frac{6 × 3}{10}\)
= \(\frac{18}{10}\)
= 1.8

e. 0.6 × 0.3 = _______
= \(\frac{6}{10}\) × \(\frac{3}{10}\)
= \(\frac{6 × 3}{10 × 10}\)
= \(\frac{18}{100}\)
= 0.18

f. 0.06 × 0.3 = _______
= \(\frac{6}{100}\) × \(\frac{3}{10}\)
= \(\frac{6 × 3}{100 × 10}\)
= \(\frac{18}{1000}\)
= 0.018

g. 1.2 × 4 = _______
= \(\frac{12}{10}\) × 4
= \(\frac{12 × 4}{10}\)
= \(\frac{48}{10}\)
= 4.8.

h. 1.2 × 0.4 = _______
= \(\frac{12}{10}\) × \(\frac{4}{10}\)
= \(\frac{12 × 4}{10 × 10}\)
= \(\frac{48}{100}\)
= 0.48.

i. 0.12 × 0.4 = _______
= \(\frac{12}{100}\) × \(\frac{4}{10}\)
= \(\frac{12 × 4}{100 × 10}\)
= \(\frac{48}{1000}\)
= 0.048.

Question 3.
A Boy Scout has a length of rope measuring 0.7 meter. He uses 2 tenths of the rope to tie a knot at one end. How many meters of rope are in the knot?

Question 4.
After just 4 tenths of a 2.5-mile race was completed, Lenox took the lead and remained there until the end of the race.

a. How many miles did Lenox lead the race?

b. Reid, the second-place finisher, developed a cramp with 3 tenths of the race remaining. How many miles did Reid run without a cramp?

Eureka Math Grade 5 Module 4 Lesson 17 Exit Ticket Answer Key

Question 1.
Multiply and model. Rewrite the expression as a number sentence with decimal factors.
\(\frac{1}{10}\) × 1.2

Question 2.
Multiply.

a. 1.5 × 3 = _______

b. 1.5 × 0.3 = _______

c. 0.15 × 0.3 = _______

Eureka Math Grade 5 Module 4 Lesson 17 Homework Answer Key

Question 1.
Multiply and model. Rewrite each expression as a number sentence with decimal factors. The first one is done for you.
a. \(\frac{1}{10}\) × \(\frac{1}{10}\)
= \(\frac{1 × 1}{10 × 10}\)
= \(\frac{1}{100}\)
0.1 × 0.1 = 0.01

b. \(\frac{6}{10}\) × \(\frac{2}{10}\)

c. \(\frac{1}{10}\) × 1.6

d. \(\frac{6}{10}\) × 1.9

Question 2.
Multiply. The first few are started for you.
a. 4 × 0.6 = _______
= 4 × \(\frac{6}{10}\)
= \(\frac{4 × 6}{10}\)
= \(\frac{24}{10}\)
= 2.4

b. 0.4 × 0.6 = _______
= \(\frac{4}{10}\) × \(\frac{6}{10}\)
= \(\frac{4 × 6}{10 × 10}\)
=

c. 0.04 × 0.6 = _______

d. 7 × 0.3 = _______

e. 0.7 × 0.3 = _______

f. 0.07 × 0.3 = _______

g. 1.3 × 5 = _______

h. 1.3 × 0.5 = _______

i. 0.13 × 0.5 = _______

Question 3.
Jennifer makes 1.7 liters of lemonade. If she pours 3 tenths of the lemonade in the glass, how many liters of lemonade are in the glass?

Question 4.
Cassius walked 6 tenths of a 3.6-mile trail.
a. How many miles did Cassius have left to hike?

b. Cameron was 1.3 miles ahead of Cassius. How many miles did Cameron hike already?

Engage NY Eureka Math 5th Grade Module 4 Lesson 15 Answer Key

Eureka Math Grade 5 Module 4 Lesson 15 Problem Set Answer Key

Question 1.
Solve. Draw a rectangular fraction model to explain your thinking. Then, write a multiplication sentence. The first one is done for you.

a. \(\frac{2}{3}\) of \(\frac{3}{5}\)
\(\frac{2}{3}\) × \(\frac{3}{5}\) = \(\frac{6}{15}\) = \(\frac{2}{5}\)

b. \(\frac{3}{4}\) of \(\frac{4}{5}\) =

Answer:
latex]\frac{3}{4}[/latex] of \(\frac{4}{5}\) = \(\frac{3}{5}\).

Explanation:
Given that \(\frac{3}{4}\) of \(\frac{4}{5}\) which is
\(\frac{3}{4}\) × \(\frac{4}{5}\) = \(\frac{3}{5}\)

c. \(\frac{2}{5}\) of \(\frac{2}{3}\)=

Answer:
latex]\frac{2}{5}[/latex] of \(\frac{2}{3}\) = \(\frac{4}{15}\).

Explanation:
Given that \(\frac{2}{5}\) of \(\frac{2}{3}\) which is
\(\frac{2}{5}\) × \(\frac{2}{3}\) = \(\frac{4}{15}\)

d. \(\frac{4}{5}\) × \(\frac{2}{3}\) =

Answer:
latex]\frac{4}{5}[/latex] of \(\frac{2}{3}\) = \(\frac{8}{15}\).

Explanation:
Given that \(\frac{4}{5}\) of \(\frac{2}{3}\) which is
\(\frac{4}{5}\) × \(\frac{2}{3}\) = \(\frac{8}{15}\)

e. \(\frac{3}{4}\) × \(\frac{2}{3}\)=

Answer:
latex]\frac{3}{4}[/latex] of \(\frac{2}{3}\) = \(\frac{1}{2}\).

Explanation:
Given that \(\frac{3}{4}\) of \(\frac{2}{3}\) which is
\(\frac{3}{4}\) × \(\frac{2}{3}\) = \(\frac{1}{2}\)

Question 2.
Multiply. Draw a rectangular fraction model if it helps you, or use the method in the example.

a. \(\frac{3}{4}\) × \(\frac{5}{6}\)

Answer:
latex]\frac{3}{4}[/latex] of \(\frac{5}{6}\) = \(\frac{5}{8}\).

Explanation:
Given that \(\frac{3}{4}\) of \(\frac{5}{6}\) which is
\(\frac{3}{4}\) × \(\frac{5}{6}\) = \(\frac{5}{8}\).

b. \(\frac{4}{5}\) × \(\frac{5}{8}\)

Answer:
latex]\frac{4}{5}[/latex] of \(\frac{5}{8}\) = \(\frac{1}{2}\).

Explanation:
Given that \(\frac{4}{5}\) of \(\frac{5}{8}\) which is
\(\frac{4}{5}\) × \(\frac{5}{8}\) = \(\frac{1}{2}\)

c. \(\frac{2}{3}\) × \(\frac{6}{7}\)

Answer:
latex]\frac{2}{3}[/latex] of \(\frac{6}{7}\) = \(\frac{1}{7}\).

Explanation:
Given that \(\frac{2}{3}\) of \(\frac{6}{7}\) which is
\(\frac{2}{3}\) × \(\frac{6}{7}\) = \(\frac{1}{7}\)

d. \(\frac{4}{9}\) × \(\frac{3}{10}\)

Answer:
latex]\frac{4}{9}[/latex] of \(\frac{3}{10}\) = \(\frac{2}{15}\).

Explanation:
Given that \(\frac{4}{9}\) of \(\frac{3}{10}\) which is
\(\frac{4}{9}\) × \(\frac{3}{10}\) = \(\frac{2}{15}\).

Question 3.
Phillip’s family traveled \(\frac{3}{10}\) of the distance to his grandmother’s house on Saturday. They traveled \(\frac{4}{7}\) of the remaining distance on Sunday. What fraction of the total distance to his grandmother’s house was traveled on Sunday?

Answer:
Philip’s family traveled on Sunday is \(\frac{2}{5}\).

Explanation:
Given that Phillip’s family traveled \(\frac{3}{10}\) of the distance to his grandmother’s house on Saturday, so the remaining is 1 – \(\frac{3}{10}\) which is \(\frac{7}{10}\). So Philip’s family traveled on Sunday is \(\frac{4}{7}\) × \(\frac{7}{10}\) which is \(\frac{2}{5}\).

Question 4.
Santino bought a \(\frac{3}{4}\)-pound bag of chocolate chips. He used \(\frac{2}{3}\) of the bag while baking. How many pounds of chocolate chips did he use while baking?

Answer:
The number of pounds of chocolate chips did he use while baking is \(\frac{1}{2}\) lb.

Explanation:
Given that Santino bought a \(\frac{3}{4}\)-pound bag of chocolate chips and he used \(\frac{2}{3}\) of the bag while baking. So the number of pounds of chocolate chips did he use while baking is \(\frac{3}{4}\) × \(\frac{2}{3}\) which is \(\frac{1}{2}\) lb.

Question 5.
Farmer Dave harvested his corn. He stored \(\frac{5}{9}\) of his corn in one large silo and \(\frac{3}{4}\) of the remaining corn in a small silo. The rest was taken to market to be sold.
a. What fraction of the corn was stored in the small silo?

Answer:
The fraction of the corn was stored in the small silo \(\frac{1}{3}\).

Explanation:
Given that Dave has stored \(\frac{5}{9}\) of his corn in one large silo. Let the total corn be ‘X’, and the amount of corn stored in the silo is \(\frac{5}{9}\)X. The amount of corn remaining is X – \(\frac{5}{9}\)X which is \(\frac{9X – 5X}{9}\) = \(\frac{4X}{9}\). Thus the amount of corn stored in the small silo is \(\frac{3}{4}\) × \(\frac{4}{9}\)X which is \(\frac{1}{3}\)X. Thus the fraction of the corn was stored in the small silo \(\frac{1}{3}\).

b. If he harvested 18 tons of corn, how many tons did he take to market?

Answer:
The amount of corn taken to market is 9 tonnes.

Explanation:
The amount of corn solid in the market is \(\frac{4X}{9}\) – \(\frac{X}{3}\) which is \(\frac{X}{9}\). Thus the amount of corn taken to market is 18 × \(\frac{1}{9}\) which is 9 tonnes.

Eureka Math Grade 5 Module 4 Lesson 15 Exit Ticket Answer Key

Question 1.
Solve. Draw a rectangular fraction model to explain your thinking. Then, write a multiplication sentence.

a. \(\frac{2}{3}\) of \(\frac{3}{5}\) =

Answer:
latex]\frac{2}{3}[/latex] of \(\frac{3}{5}\) = \(\frac{2}{5}\).

Explanation:
Given that \(\frac{2}{3}\) of \(\frac{3}{5}\) which is
\(\frac{2}{3}\) × \(\frac{3}{5}\) = \(\frac{2}{5}\).

b. \(\frac{4}{9}\) × \(\frac{3}{8}\) =

Answer:
latex]\frac{4}{9}[/latex] of \(\frac{3}{8}\) = \(\frac{1}{6}\).

Explanation:
Given that \(\frac{4}{9}\) of \(\frac{3}{8}\) which is
\(\frac{4}{9}\) × \(\frac{3}{8}\) = \(\frac{1}{6}\).

Question 2.
A newspaper’s cover page is \(\frac{3}{8}\) text, and photographs fill the rest. If \(\frac{2}{5}\) of the text is an article about endangered species, what fraction of the cover page is the article about endangered species?

Answer:
The fraction of the cover page is the article about endangered species \(\frac{3}{20}\).

Explanation:
Given that a newspaper’s cover page is \(\frac{3}{8}\) text, and photographs fill the rest, and if \(\frac{2}{5}\) of the text is an article about endangered species. So the fraction of the cover page is the article about endangered species \(\frac{3}{8}\) × \(\frac{2}{5}\) which is \(\frac{3}{20}\).

Eureka Math Grade 5 Module 4 Lesson 15 Homework Answer Key

Question 1.
Solve. Draw a rectangular fraction model to explain your thinking. Then, write a multiplication sentence.
a. \(\frac{2}{3}\) of \(\frac{3}{4}\) =

Answer:
latex]\frac{2}{3}[/latex] of \(\frac{3}{4}\) = \(\frac{1}{2}\).

Explanation:
Given that \(\frac{2}{3}\) of \(\frac{3}{4}\) which is
\(\frac{2}{3}\) × \(\frac{3}{4}\) = \(\frac{1}{2}\).

b. \(\frac{2}{5}\) of \(\frac{3}{4}\) =

Answer:
latex]\frac{2}{5}[/latex] of \(\frac{3}{4}\) = \(\frac{3}{10}\).

Explanation:
Given that \(\frac{2}{5}\) of \(\frac{3}{4}\) which is
\(\frac{2}{5}\) × \(\frac{3}{4}\) = \(\frac{3}{10}\).

c. \(\frac{2}{5}\) of \(\frac{4}{5}\) =

Answer:
latex]\frac{2}{5}[/latex] of \(\frac{4}{5}\) = \(\frac{8}{25}\).

Explanation:
Given that \(\frac{2}{5}\) of \(\frac{4}{5}\) which is
\(\frac{2}{5}\) × \(\frac{4}{5}\) = \(\frac{8}{25}\).

d. \(\frac{4}{5}\) of \(\frac{3}{4}\) =

Answer:
latex]\frac{4}{5}[/latex] of \(\frac{3}{4}\) = \(\frac{3}{5}\).

Explanation:
Given that \(\frac{4}{5}\) of \(\frac{3}{4}\) which is
\(\frac{4}{5}\) × \(\frac{3}{4}\) = \(\frac{3}{5}\).

Question 2.
Multiply. Draw a rectangular fraction model if it helps you.
a. \(\frac{5}{6}\) × \(\frac{3}{10}\)

Answer:
latex]\frac{5}{6}[/latex] of \(\frac{3}{10}\) = \(\frac{1}{4}\).

Explanation:
Given that \(\frac{5}{6}\) of \(\frac{3}{10}\) which is
\(\frac{5}{6}\) × \(\frac{3}{10}\) = \(\frac{1}{4}\).

b. \(\frac{3}{4}\) × \(\frac{4}{5}\)

Answer:
latex]\frac{3}{4}[/latex] of \(\frac{4}{5}\) = \(\frac{3}{5}\).

Explanation:
Given that \(\frac{3}{4}\) of \(\frac{4}{5}\) which is
\(\frac{3}{4}\) × \(\frac{4}{5}\) = \(\frac{3}{5}\).

c. \(\frac{5}{6}\) × \(\frac{5}{8}\)

Answer:
latex]\frac{4}{9}[/latex] of \(\frac{3}{8}\) = \(\frac{1}{6}\).

Explanation:
Given that \(\frac{4}{9}\) of \(\frac{3}{8}\) which is
\(\frac{4}{9}\) × \(\frac{3}{8}\) = \(\frac{1}{6}\).

d. \(\frac{3}{4}\) × \(\frac{5}{12}\)

Answer:
latex]\frac{4}{9}[/latex] of \(\frac{3}{8}\) = \(\frac{1}{6}\).

Explanation:
Given that \(\frac{4}{9}\) of \(\frac{3}{8}\) which is
\(\frac{4}{9}\) × \(\frac{3}{8}\) = \(\frac{1}{6}\).

e. \(\frac{8}{9}\) × \(\frac{2}{3}\)

Answer:
latex]\frac{8}{9}[/latex] of \(\frac{2}{3}\) = \(\frac{16}{27}\).

Explanation:
Given that \(\frac{8}{9}\) of \(\frac{2}{3}\) which is
\(\frac{8}{9}\) × \(\frac{2}{3}\) = \(\frac{16}{27}\).

f. \(\frac{3}{7}\) × \(\frac{2}{9}\)

Answer:
latex]\frac{3}{7}[/latex] of \(\frac{2}{9}\) = \(\frac{2}{21}\).

Explanation:
Given that \(\frac{3}{7}\) of \(\frac{2}{9}\) which is
\(\frac{3}{7}\) × \(\frac{2}{9}\) = \(\frac{2}{21}\).

Question 3.
Every morning, Halle goes to school with a 1-liter bottle of water. She drinks \(\frac{1}{4}\) of the bottle before school starts and \(\frac{2}{3}\) of the rest before lunch.
a. What fraction of the bottle does Halle drink after school starts but before lunch?

Answer:
The fraction of the bottle does Halle drinks after school starts but before lunch is \(\frac{1}{2}\).

Explanation:
Given that Halle goes to school with a 1-liter bottle of water and she drinks \(\frac{1}{4}\) of the bottle before school starts and \(\frac{2}{3}\) of the rest before lunch and the amount left after drinking before school starts are 1 – \(\frac{1}{4}\) which is \(\frac{3}{4}\) and the fraction of the bottle does Halle drinks after school starts but before lunch is \(\frac{2}{3}\) of Amount left
= \(\frac{2}{3}\) × \(\frac{3}{4}\)
= \(\frac{1}{2}\).

b. How many milliliters are left in the bottle at lunch?

Answer:
The amount that left in the bottle at lunch is 250 milliliters.

Explanation:
The amount that left in the bottle at lunch is 1 – (\(\frac{3}{4}\) + \(\frac{1}{2}\))
= \(\frac{1}{4}\), as we know that 1 litre is 1000 milliliters, so \(\frac{1}{4}\) litre is \(\frac{1}{4}\) × 1000 which is 250 milliliters.

Question 4.
Moussa delivered \(\frac{3}{8}\) of the newspapers on his route in the first hour and \(\frac{4}{5}\) of the rest in the second hour. What fraction of the newspapers did Moussa deliver in the second hour?

Question 5.
Rose bought some spinach. She used \(\frac{3}{5}\) of the spinach on a pan of spinach pie for a party and \(\frac{3}{4}\) of the remaining spinach for a pan for her family. She used the rest of the spinach to make a salad.
a. What fraction of the spinach did she use to make the salad?

b. If Rose used 3 pounds of spinach to make the pan of spinach pie for the party, how many pounds of spinach did Rose use to make the salad?

Engage NY Eureka Math 5th Grade Module 4 Lesson 8 Answer Key

Eureka Math Grade 5 Module 4 Lesson 8 Problem Set Answer Key

Question 1.
Laura and Sean find the product of \(\frac{2}{3}\) × 4 using different methods.
Laura: It’s 2 thirds of 4.
\(\frac{2}{3}\) × 4 = \(\frac{4}{3}\) + \(\frac{4}{3}\) = 2 × \(\frac{4}{3}\) = \(\frac{8}{3}\)

Sean: It’s 4 groups of 2 thirds.
\(\frac{2}{3}\) + \(\frac{2}{3}\) + \(\frac{2}{3}\) + \(\frac{2}{3}\) = 4 × \(\frac{2}{3}\) = \(\frac{8}{3}\)
Use words, pictures, or numbers to compare their methods in the space below.
Answer:
Laura:
2/3 * 4 = 4/3 + 4/3 = 2*4/3 = 8/3
Sean:
2/3 + 2/3 + 2/3 + 2/3 = 4 * 2/3 = 8/3
Both methods are correct. 2/3 *4 is 2 thirds of 4, and it will also have the same product as the 4 groups of 2 thirds.

Question 2.
Rewrite the following addition expressions as fractions as shown in the example.
Example: \(\frac{2}{3}\) + \(\frac{2}{3}\) + \(\frac{2}{3}\) + \(\frac{2}{3}\) = (\(\frac{4 × 2}{3}\)) = \(\frac{8}{3}\)
a. \(\frac{7}{4}\) + \(\frac{7}{4}\) + \(\frac{7}{4}\) =
Answer:
21/4

Explanation:
The addition expression for the fraction given is :
7/4 + 7/4 + 7/4 = (3 × 7)/4 = 21/4

b. \(\frac{14}{5}\) + \(\frac{14}{5}\) =
Answer:
28/5

Explanation:
The addition expression for the fraction given is :
14/5 + 14/5 = (2×14)/5 = 28/5

c. \(\frac{4}{7}\) + \(\frac{4}{7}\) + \(\frac{4}{7}\) =
Answer:
12/7

Explanation:
4/7 +4/7 +4/7 = (3×4)/7 = 12/7

Question 3.
Solve and model each problem as a fraction of a set and as repeated addition.

a. \(\frac{1}{2}\) × 8 8 × \(\frac{1}{2}\)
Answer:
1 ×8/2 = 1×4 = 4
8 × 1/2 = (8×1)/2
= 8/2
= 4

b. \(\frac{3}{5}\) × 10 10 × \(\frac{3}{5}\)
Answer:
6

Explanation:
3 ×10/5 = 3 × 2 = 6
10 × 3/5 = (10×3)/5 = 30/5 = 6

Question 4.
Solve each problem in two different ways as modeled in the example.

a. 14 × \(\frac{3}{7}\) 14 × \(\frac{3}{7}\)
Answer:
6

Explanation:
(14×3)/7 = (7×2×3)/7 = (7×6)/7 = 6
(14 × 3)/7 = (2×3)/1 = 6

b. \(\frac{3}{4}\) × 36 \(\frac{3}{4}\) ×36
Answer:
27

Explanation:
(3×36)/4 = (3×4×9)/4 = (4×27)/4 = 27
(3×36)/4 = (3×9) = (3×9) = 27

c. 30 × \(\frac{13}{10}\) 30 × \(\frac{13}{10}\)
Answer:
39

Explanation:
(30×13)/10 = (10×3×13)/10 = (10×39)/10 = 39
(30 ×13)/10 = (3×13)/1 = 39

d. \(\frac{9}{8}\) ×32 \(\frac{9}{8}\) × 32
Answer:
36

Explanation:
(9×32)/8 = (9×4×8)/8 = (36×8)/8 = 36
(9×32)/8 = (9×32)/8 = (9×4)/1 = 36

Question 5.
Solve each problem any way you choose.
a. \(\frac{1}{2}\) × 60 \(\frac{1}{2}\) minute = __________ seconds
Answer:
30seconds

Explanation:
The answer and the procedure are explained clearly in the below steps.
1 minute = 60 seconds
(1×60)/2 = (1×30)/1 = 30

b. \(\frac{3}{4}\) × 60 \(\frac{3}{4}\) hour = __________ minutes
Answer:
45 minutes

Explanation:
The answer and the procedure are explained clearly in the below steps.
1 hour = 60 minutes
(3×60)/4 = (3×15)/1 = 45

c. \(\frac{3}{10}\) × 1,000 \(\frac{3}{10}\) kilogram = __________ grams”
Answer:
300 grams

Explanation:
The answer and the procedure are explained clearly in the below steps.
1 kilogram = 1000 grams
(3×1000)/1 = 300

d. \(\frac{4}{5}\) × 100 \(\frac{4}{5}\) meter = __________ centimeters
Answer:
80 centimeters

Explanation:
The answer and the procedure are explained clearly in the below steps.
1 meter
(4×100)/5 = (4×20)/1 = 80

Eureka Math Grade 5 Module 4 Lesson 8 Exit Ticket Answer Key

Solve each problem in two different ways as modeled in the example.

a. \(\frac{2}{3}\) × 15 \(\frac{2}{3}\) × 15
Answer:
10

Explanation:
By solving the given question in two methods we could get the same answer i.e 10. The Explanation is given below.
2/3 × 15 = (2 ×15)/3 = 30/3 = 10
2/3 ×15 = (2×15)/3 = (2×5) = 10

b. \(\frac{5}{4}\) × 12 \(\frac{5}{4}\) × 12
Answer:
15

Explanation:
By solving the given question in two methods we could get the same answer i.e 15. The process for the question is done below.
5/4 × 12 = (5×12)/4 = (60)/4 =15
5/4 ×12 = ( 5×12)/4 = (5×3) =15

Eureka Math Grade 5 Module 4 Lesson 8 Homework Answer Key

Question 1.
Rewrite the following expressions as shown in the example.
Example: \(\frac{2}{3}\) + \(\frac{2}{3}\) + \(\frac{2}{3}\) + \(\frac{2}{3}\) = (\(\frac{4 × 2}{3}\)) = \(\frac{8}{3}\)

a. \(\frac{5}{3}\) + \(\frac{5}{3}\) + \(\frac{5}{3}\)
Answer:
5

Explanation:
(\(\frac{3 × 5}{3}\)) = \(\frac{15}{3}\) =5

b. \(\frac{13}{5}\) + \(\frac{13}{5}\)
Answer:
\(\frac{26}{5}\)

Explanation:
(\(\frac{2 × 13}{5}\)) = \(\frac{26}{5}\)

c. \(\frac{9}{4}\) + \(\frac{9}{4}\) + \(\frac{9}{4}\)
Answer:
\(\frac{27}{4}\)

Expalanation:
(\(\frac{3 × 9}{4}\)) = \(\frac{27}{4}\)

Question 2.
Solve each problem in two different ways as modeled in the example.

a. \(\frac{3}{4}\) × 16 \(\frac{3}{4}\) × 16
Answer:
12

Explanation:
(\(\frac{3 × 16}{4}\)) = \(\frac{48}{4}\) = 12
(\(\frac{3 × 16}{4}\)) = \(\frac{3 × 4}{1}\) = 12

b. \(\frac{4}{3}\) × 12 \(\frac{4}{3}\) × 12
Answer:
16

Explanation:
(\(\frac{4 × 12}{3}\)) = \(\frac{48}{3}\) = 16
(\(\frac{4 × 12}{3}\)) = \(\frac{4 × 12}{1}\) = 16

c. 40 × \(\frac{11}{10}\) 40 × \(\frac{11}{10}\)
Answer:
44

Explanation:
(\(\frac{40 × 11}{10}\)) = \(\frac{440}{10}\) = 44
(\(\frac{40× 11}{10}\)) = \(\frac{4 × 11}{1}\) = \(\frac{44}{1}\) = 44

d. \(\frac{7}{6}\) × 36 \(\frac{7}{6}\)× 36
Answer:
42

Explanation:
(\(\frac{7 × 36}{6}\)) = \(\frac{252}{6}\) = 42
(\(\frac{7 × 36}{6}\)) = \(\frac{7 × 6}{1}\) = 42

e. 24 × \(\frac{5}{8}\) 24 × \(\frac{5}{8}\)
Answer:
15

Explanation:
(\(\frac{24 × 5}{8}\)) = \(\frac{126}{8}\) = 15
(\(\frac{24 × 5}{8}\)) = \(\frac{3 × 5}{1}\) = 15

f. 18 × \(\frac{5}{12}\) 18 × \(\frac{5}{12}\)
Answer:
7 1/2

Explanation:
(\(\frac{18 × 5}{12}\)) = \(\frac{90}{12}\) = 7 6/12 = 7 1/2
(\(\frac{18 × 5}{12}\)) = \(\frac{3 × 5 }{2}\) = \(\frac{15}{2}\)= 7 1/2

g. \(\frac{10}{9}\) × 21 \(\frac{10}{9}\) × 21
Answer:
23 3/9 = 23 1/3

Explanation:
(\(\frac{10 × 21}{9}\)) = \(\frac{210}{9}\) = 23 3/9 = 23 1/3
(\(\frac{10 × 21}{9}\)) = \(\frac{10 × 7}{3}\) = \(\frac{70}{3}\)  = 23 1/3

Question 3.
Solve each problem any way you choose.

a. \(\frac{1}{3}\) × 60 \(\frac{1}{3}\) minute = _________ seconds
Answer:
20 seconds

Explanation:
(\(\frac{1 × 60}{3}\)) = \(\frac{60}{3\) = 20

b. \(\frac{4}{5}\) × 60 \(\frac{4}{5}\) hour = _________ minutes
Answer:
48 minutes

Explanation:
(\(\frac{4 × 60}{5}\)) = \(\frac{48}{1\) = 48

c. \(\frac{7}{10}\) × 1000 \(\frac{7}{10}\) kilogram = _________ grams
Answer:
700 grams

Explanation:
(\(\frac{7 × 1000}{10}\)) = \(\frac{700}{1\) = 700

d. \(\frac{3}{5}\)× 100 \(\frac{3}{5}\) meter = _________ centimeters
Answer:
60 centimeres

Explanation:
(\(\frac{3 × 100}{5}\)) = \(\frac{60}{1\) = 60

Engage NY Eureka Math 5th Grade Module 4 Lesson 7 Answer Key

Eureka Math Grade 5 Module 4 Lesson 7 Problem Set Answer Key

Solve using a tape diagram.
a. \(\frac{1}{3}\) of 18
Answer:
6

Explanation:

b. \(\frac{1}{3}\) of 36
Answer:
11

Explanation:

c. \(\frac{3}{4}\) × 24
Answer:
18

Explanation:

d. \(\frac{3}{8}\) × 24
Answer:
9

Explanation:

e. \(\frac{4}{5}\) × 25
Answer:
20

Explanation:

f. \(\frac{1}{7}\) × 140
Answer:
20

Explanation:

g. \(\frac{1}{4}\) × 9
Answer:
2 1/4

Explanation:

h. \(\frac{2}{5}\) × 12

i. \(\frac{2}{3}\) of a number is 10. What’s the number?

j. \(\frac{3}{4}\) of a number is 24. What’s the number?

Question 2.
Solve using tape diagrams.
a. There are 48 students going on a field trip. One-fourth are girls. How many boys are going on the trip?
Answer:
36 boys are going on the field trip.

Explanation:
As 48 students are going on a field trip in this \(\frac{1}{4}\) are girls. Hence 36 boys are going on the field trip.
\(\frac{1}{4}\) × 48 = \(\frac{48}{4}\) = 12
12 × 3 = 36

b. Three angles are labeled below with arcs. The smallest angle is \(\frac{3}{8}\) as large as the 160° angle. Find the value of angle a.

Answer:
Angle a is 140°

Explanation:
The smallest angle given is \(\frac{3}{8}\) and the largest angle is 160°.
160 × \(\frac{3}{8}\) = \(\frac{480}{8}\) = 60
160 + 60 = 220
360-220= 140
Hence the value of angle a is 140°

c. Abbie spent \(\frac{5}{8}\) of her money and saved the rest. If she spent $45, how much money did she have at first?
Answer:
Abbie had $72.

Explanation:
45 ÷ \(\frac{5}{8}\) = 45 × \(\frac{8}{5}\) = \(\frac{360}{5}\) =72

d. Mrs. Harrison used 16 ounces of dark chocolate while baking. She used \(\frac{2}{5}\) of the chocolate to make some frosting and used the rest to make brownies. How much more chocolate did Mrs. Harrison use in the brownies than in the frosting?
Answer:
Mrs. Harrison use 3 × \(\frac{1}{5}\) chocolate for the brownie

Explanation:
Harrison used 16 ounces of dark chocolate. She used \(\frac{2}{5}\) of the chocolate to make some frosting and the remaining to make brownies.
16 × \(\frac{1}{5}\) = \(\frac{16}{5}\) = 3 × \(\frac{1}{5}\)
Hence, Mrs. Harrison use 3 × \(\frac{1}{5}\) chocolate for the brownie.

Eureka Math Grade 5 Module 4 Lesson 7 Exit Ticket Answer Key

Solve using a tape diagram.

a. \(\frac{3}{5}\) of 30

b. \(\frac{3}{5}\) of a number is 30. What’s the number?

c. Mrs. Johnson baked 2 dozen cookies. Two-thirds of the cookies were oatmeal. How many oatmeal cookies did Mrs. Johnson bake?

Eureka Math Grade 5 Module 4 Lesson 7 Homework Answer Key

Question 1.
Solve using a tape diagram.
a. \(\frac{1}{4}\) of 24

b. \(\frac{1}{4}\) of 48

c. \(\frac{2}{3}\) × 18

d. \(\frac{2}{6}\) × 18

e. \(\frac{3}{7}\) × 49

f. \(\frac{3}{10}\) × 120

g. \(\frac{1}{3}\) × 31

h. \(\frac{2}{5}\) × 20

i. \(\frac{1}{4}\) × 25

j. \(\frac{3}{4}\) × 25

k. \(\frac{3}{4}\) of a number is 27. What’s the number?

i. \(\frac{2}{5}\) of a number is 14. What’s the number?

Question 2.
Solve using tape diagrams.
a. A skating rink sold 66 tickets. Of these, \(\frac{2}{3}\) were children’s tickets, and the rest were adult tickets. What total number of adult tickets were sold?

b. A straight angle is split into two smaller angles as shown. The smaller angle’s measure is \(\frac{1}{6}\) that of a straight angle. What is the value of angle a?

c. Annabel and Eric made 17 ounces of pizza dough. They used \(\frac{5}{8}\) of the dough to make a pizza and used the rest to make calzones. What is the difference between the amount of dough they used to make pizza and the amount of dough they used to make calzones?

d. The New York Rangers hockey team won \(\frac{3}{4}\) of their games last season. If they lost 21 games, how many games did they play in the entire season?