## Engage NY Eureka Math 5th Grade Module 4 Lesson 20 Answer Key

### Eureka Math Grade 5 Module 4 Lesson 20 Problem Set Answer Key

Question 1.

Convert. Show your work. Express your answer as a mixed number. (Draw a tape diagram if it helps you.) The first one is done for you.

a. 2 \(\frac{2}{3}\) yd = __8__ ft

2 \(\frac{2}{3}\) yd = 2 \(\frac{2}{3}\) Ã— 1 yd

= 2 \(\frac{2}{3}\) Ã— 3 ft

= \(\frac{8}{3}\) Ã— 3 ft

= \(\frac{24}{3}\) ft

= 8 ft

b. 1\(\frac{1}{2}\) qt = \(\frac{3}{8}\) gal

1\(\frac{1}{2}\) Ã— 1 qt

= 1 \(\frac{1}{2}\) Ã— \(\frac{1}{4}\) gal

= \(\frac{3}{2}\) Ã— \(\frac{1}{4}\) gal

= \(\frac{3}{8}\) gal.

c. 4 \(\frac{2}{3}\) ft = ______________ in

4 \(\frac{2}{3}\) Ã— 1 ft

= \(\frac{14}{3}\) Ã— 12 in

= \(\frac{168}{3}\) in

= 56 in.

d. 9 \(\frac{1}{2}\) pt = ______________ qt

9 \(\frac{1}{2}\) Ã— 1 pt

= Â \(\frac{19}{2}\) Ã— \(\frac{1}{2}\) qt

= Â \(\frac{19}{4}\) qt

= 4 \(\frac{3}{4}\) qt.

e. 3 \(\frac{3}{5}\) hr = ______________ min

3 \(\frac{3}{5}\) Ã— 1 hr

= \(\frac{18}{5}\) Ã— 60 min

= \(\frac{1080}{5}\) min

= 216 mins.

f. 3 \(\frac{2}{3}\) ft = ______________ yd

3 \(\frac{2}{3}\) Ã— 1 ft

= \(\frac{11}{3}\) Ã— \(\frac{1}{3}\) yd

= \(\frac{11}{9}\)

= 1 \(\frac{2}{9}\) yd.

Question 2.

Three dump trucks are carrying topsoil to a construction site. Truck A carries 3,545 lb, Truck B carries 1,758 lb, and Truck C carries 3,697 lb. How many tons of topsoil are the 3 trucks carrying altogether?

Answer:

The 3 trucks carrying altogether are 4.5 tons.

Explanation:

Given that there are three dump trucks are carrying topsoil to a construction site and Truck A carries 3,545 lb, Truck B carries 1,758 lb, and Truck C carries 3,697 lb, so the total weight carried altogether is 3,545 + 1,758 + 3,697 = 9,000 lb. As each ton is 2000 pounds, so altogether the trucks are carrying is 9000 Ã— \(\frac{1}{2000}\) which is 4.5 tons.

Question 3.

Melissa buys 3\(\frac{3}{4}\) gallons of iced tea. Denita buys 7 quarts more than Melissa. How much tea do they buy altogether? Express your answer in quarts.

Answer:

The total tea they bought is 37 quarts.

Explanation:

Given that Melissa buys 3\(\frac{3}{4}\) gallons of iced tea, so total iced tea for Melissa is \(\frac{15}{4}\) which is 3.75. And Denita buys 7 quarts more than Melissa, so the total iced tea for Denita is, as 1 quart is 0.25 gallon and for 7 quarts it will be 7 Ã— 0.25 which is 1.75 gallon. so the total iced tea for Denita is 1.75 + 3.75 which is 5.5. Then the total tea they bought is 3.75 + 5.5 = 9.25 gallon which is 9.25 Ã— 4 = 37 quarts.

Question 4.

Marvin buys a hose that is 27\(\frac{3}{4}\) feet long. He already owns a hose at home that is \(\frac{2}{3}\) the length of the new hose. How many total yards of hose does Marvin have now?

Answer:

The total yards of hose does Marvin have now is 15 \(\frac{5}{12}\) yd.

Explanation:

Given that Marvin buys a hose that is 27\(\frac{3}{4}\) feet long and he owns a hose at home that is \(\frac{2}{3}\) the length of the new hose, so \(\frac{2}{3}\) of 27\(\frac{3}{4}\) which is

\(\frac{2}{3}\) Ã— \(\frac{111}{4}\)

= \(\frac{222}{12}\)

= 18 \(\frac{1}{2}\),

So the total yards of hose does Marvin have now is

27\(\frac{3}{4}\) + 18\(\frac{1}{2}\)

= \(\frac{111}{4}\) + \(\frac{37}{2}\)

= \(\frac{185}{4}\)

= 46 \(\frac{1}{4}\).

So total in yards, it will be 46 \(\frac{1}{4}\) Ã— 1 yd

= \(\frac{185}{4}\) Ã— \(\frac{1}{3}\) yd

= \(\frac{185}{12}\)

=15 \(\frac{5}{12}\) yd.

### Eureka Math Grade 5 Module 4 Lesson 20 Exit Ticket Answer Key

Convert. Express your answer as a mixed number.

a. 2\(\frac{1}{6}\) ft = ______________ in

Answer:

26 in.

Explanation:

2\(\frac{1}{6}\) ft = \(\frac{13}{6}\) Ã— 1 ft

= \(\frac{13}{6}\) Ã— 12 in

= \(\frac{156}{6}\) in

= 26 in.

b. 3\(\frac{3}{4}\) ft = ______________ yd

Answer:

45 in.

Explanation:

3\(\frac{3}{4}\) ft = \(\frac{15}{4}\) ft Ã— 1 ft

= \(\frac{15}{4}\) Ã— 12 in

= \(\frac{180}{4}\) in

= 45 in.

c. 2\(\frac{1}{2}\)c = ______________ pt

Answer:

1 \(\frac{1}{4}\) pt.

Explanation:

2\(\frac{1}{2}\)c = \(\frac{5}{2}\) Ã— 1 c

= \(\frac{5}{2}\) Ã— \(\frac{1}{2}\) pt

= \(\frac{5}{4}\) pt

= 1 \(\frac{1}{4}\) pt.

d. 3\(\frac{2}{3}\) years = ______________ months

Answer:

44 months.

Explanation:

3\(\frac{2}{3}\) years = \(\frac{11}{3}\) Ã— 1 year

= \(\frac{11}{3}\) Ã— 12 months

= 44 months.

### Eureka Math Grade 5 Module 4 Lesson 20 Homework Answer Key

Question 1.

Convert. Show your work. Express your answer as a mixed number. The first one is done for you.

2 \(\frac{2}{3}\) yd = __8__ ft

2 \(\frac{2}{3}\) yd = 2 \(\frac{2}{3}\) Ã— 1 yd

= 2 \(\frac{2}{3}\) Ã— 3 ft

= \(\frac{8}{3}\) Ã— 3 ft

= \(\frac{24}{3}\) ft

= 8 ft

b. 1 \(\frac{1}{4}\) ft = \(\frac{5}{12}\) yd

1 \(\frac{1}{4}\) ft = 1 \(\frac{1}{4}\) Ã— 1 ft

= 1 \(\frac{1}{4}\) Ã— \(\frac{1}{3}\) yd

= \(\frac{5}{4}\) Ã— \(\frac{1}{3}\) yd

= \(\frac{5}{12}\) yd.

c. 3\(\frac{5}{6}\) ft = ______________ in

Answer:

46 in.

Explanation:

3\(\frac{5}{6}\) ft = 3\(\frac{5}{6}\) ft Ã— 1 ft

= \(\frac{23}{6}\) Ã— 12 in

= 46 in.

d. 7 \(\frac{1}{2}\) pt = ______________ qt

Answer:

3 \(\frac{3}{4}\) qt.

Explanation:

7 \(\frac{1}{2}\) pt = 7 \(\frac{1}{2}\) pt Ã— 1 pt

= \(\frac{15}{2}\) Ã— \(\frac{1}{2}\) qt

= \(\frac{15}{4}\) qt

= 3 \(\frac{3}{4}\) qt.

e. 4\(\frac{3}{10}\) hr = ______________ min

Answer:

258 mins.

Explanation:

4\(\frac{3}{10}\) hr = 4\(\frac{3}{10}\) Ã— 1 hr

= \(\frac{43}{10}\) Ã— 60 mins

= 258 mins

f. 33 months = ______________ years

Answer:

2 \(\frac{3}{4}\) years.

Explanation:

33 Ã— \(\frac{1}{12}\)

= \(\frac{33}{12}\)

= \(\frac{11}{4}\)

= 2 \(\frac{3}{4}\) years.

Question 2.

Four members of a track team run a relay race in 165 seconds. How many minutes did it take them to run the race?

Answer:

The number of minutes did it take them to run the race is 2 \(\frac{3}{4}\) mins.

Explanation:

Given that there are four members of a track team run a relay race in 165 seconds, so the number of minutes did it take them to run the race is 165 Ã— \(\frac{1}{60}\)

= \(\frac{11}{4}\)

= 2 \(\frac{3}{4}\) mins.

Question 3.

Horace buys 2\(\frac{3}{4}\) pounds of blueberries for a pie. He needs 48 ounces of blueberries for the pie. How many more pounds of blueberries does he need to buy?

Answer:

Horace need more blueberries to buy is 0.25 pounds.

Explanation:

Given that Horace buys 2\(\frac{3}{4}\) pounds of blueberries for a pie and he needs 48 ounces of blueberries for the pie, so the number of pounds of blueberries does he need to buy is, as 1 pound is 16 ounces and 1 ton is 2,200 pounds which is 32,000 ounces. As Horace needs 48 ounces of blueberries for the pie and we need to convert into ounce, so Horace need more blueberries to buy is 3 – 2\(\frac{3}{4}\) which is 3 – \(\frac{11}{4}\) = \(\frac{1}{4}\) = 0.25 pounds.

Question 4.

Tiffany is sending a package that may not exceed 16 pounds. The package contains books that weigh a total of 9\(\frac{3}{8}\) pounds. The other items to be sent weigh \(\frac{3}{5}\) the weight of the books. Will Tiffany be able to send the package?

Answer:

The total package is 15 pounds.

Explanation:

Given that Tiffany is sending a package that may not exceed 16 pounds and the package contains books that weigh a total of 9\(\frac{3}{8}\) pounds and the other items to be sent weigh \(\frac{3}{5}\) the weight of the books. Let the book package be X and let the other package be Y. So from the above problem,

X= 9\(\frac{3}{8}\) which is \(\frac{75}{8}\) and

Y = \(\frac{3}{5}\) X

= \(\frac{3}{5}\) Ã— \(\frac{75}{8}\)

= \(\frac{225}{40}\)

= 5 \(\frac{5}{8}\).

So the total package is

X + Y = 9\(\frac{3}{8}\) + 5 \(\frac{5}{8}\)

= \(\frac{75}{8}\) + \(\frac{225}{40}\)

= \(\frac{600}{40}\)

= 15 pounds.