Eureka Math Grade 5 Module 3 End of Module Assessment Answer Key

Engage NY Eureka Math 5th Grade Module 3 End of Module Assessment Answer Key

Eureka Math Grade 5 Module 3 End of Module Assessment Task Answer Key

Question 1.
On Sunday, Sheldon bought 4\(\frac{1}{2}\) kg of plant food. He used 1\(\frac{2}{3}\) kg on his strawberry plants and used \(\frac{1}{4}\) kg for his tomato plants.
a. How many kilograms of plant food did Sheldon have left? Write one or more equations to show how you reached your answer.

b. Sheldon wants to feed his strawberry plants 2 more times and his tomato plants one more time.
He will use the same amounts of plant food as before. How much plant food will he need? Does he have enough left to do so? Explain your answer using words, pictures, or numbers.
Answer:
a.
Fraction of plant food bought by Sheldon = 4\(\frac{1}{2}\) kg = \(\frac{9}{2}\) kg
Fraction of plant food used on strawberry plant = 1\(\frac{2}{3}\) kg = \(\frac{5}{3}\) kg
Fraction of plant food used on tomato plants = \(\frac{1}{4}\) kg
Fraction of plant food left with Sheldon = \(\frac{9}{2}\) – \(\frac{5}{3}\) – \(\frac{1}{4}\)
= \(\frac{54}{12}\) – \(\frac{20}{12}\) – \(\frac{3}{12}\)
= \(\frac{31}{12}\) = 2\(\frac{7}{12}\) kg .
Therefore, Fraction of plant food left with Sheldon = \(\frac{31}{12}\) = 2\(\frac{7}{12}\) kg .
b.
Fraction of plant food to feed stawberry plant = 2 more times = 2 × \(\frac{5}{3}\) kg = \(\frac{10}{3}\) kg
Fraction of plant food to feed tomato plant = \(\frac{1}{4}\) kg
Total Plant food to feed plants = \(\frac{10}{3}\) kg + \(\frac{1}{4}\) kg = \(\frac{40}{12}\)  + \(\frac{3}{12}\) = \(\frac{43}{12}\) = 3\(\frac{7}{12}\)
3\(\frac{7}{12}\) is greater than 2\(\frac{7}{12}\)
Therefore the plant is not enough .
The plant required = \(\frac{43}{12}\) – \(\frac{31}{12}\) = \(\frac{12}{12}\) = 1 kg .

Question 2.
Sheldon harvests the strawberries and tomatoes in his garden.
a. He picks 1\(\frac{2}{5}\) kg less strawberries in the morning than in the afternoon. If Sheldon picks 2\(\frac{1}{4}\) kg in the morning, how many kilograms of strawberries does he pick in the afternoon? Explain your answer using words, pictures, or equations.

b. Sheldon also picks tomatoes from his garden. He picked 5\(\frac{3}{10}\) kg, but 1.5 kg were rotten and had to be thrown away. How many kilograms of tomatoes were not rotten? Write an equation that shows how you reached your answer.

c. After throwing away the rotten tomatoes, did Sheldon get more kilograms of strawberries or tomatoes? How many more kilograms? Explain your answer using an equation.
Answer:
a .
Fraction of strawberries picked in the morning = 2\(\frac{1}{4}\) kg = \(\frac{9}{4}\) kg
Fraction of strawberries picked in the afternoon = 1\(\frac{2}{5}\) kg less strawberries in the morning than in the afternoon = \(\frac{9}{4}\) + 1\(\frac{2}{5}\) = \(\frac{45}{20}\) + \(\frac{28}{20}\) = \(\frac{73}{20}\) = 3 \(\frac{13}{20}\) kg .

b .
Quantity of Tomatoes picked = 5\(\frac{3}{10}\) = \(\frac{53}{10}\) = 5.3 kg
Quantity of Tomatoes rotten = 1.5 kg
Quantity of tomatoes not rotten = 5.3 – 1.5 = 3.8 kg

c.
Fraction of Strawberries Morning + Evening = \(\frac{73}{20}\) kg + \(\frac{9}{4}\) kg =  \(\frac{73}{20}\) kg + \(\frac{45}{20}\) kg = \(\frac{59}{10}\) = 5 \(\frac{9}{10}\) kg
Fraction of Tomatoes = 3.8 kg = \(\frac{38}{10}\) =3\(\frac{8}{10}\) kg .
Sheldon got more strawberries than Tomatoes .
Quantity of more strawberries than Tomatoes = \(\frac{59}{10}\) – \(\frac{38}{10}\) kg = \(\frac{21}{10}\) = 2.1 kg

Eureka Math Grade 5 Module 3 Mid Module Assessment Answer Key

Engage NY Eureka Math 5th Grade Module 3 Mid Module Assessment Answer Key

Eureka Math Grade 5 Module 3 Mid Module Assessment Task Answer Key

Question 1.
Lila collected the honey from 3 of her beehives. From the first hive, she collected \(\frac{2}{3}\) gallon of honey. The last two hives yielded \(\frac{1}{4}\) gallon each.
a. How many gallons of honey did Lila collect in all? Draw a diagram to support your answer.

b. After using some of the honey she collected for baking, Lila found that she only had \(\frac{3}{4}\) gallon of honey left. How much honey did she use for baking? Support your answer using a diagram, numbers, and words.

c. With the remaining \(\frac{3}{4}\) gallon of honey, Lila decided to bake some loaves of bread and several batches of cookies for her school bake sale. The bread needed \(\frac{1}{6}\) gallon of honey and the cookies needed \(\frac{1}{4}\) gallon. How much honey was left over? Support your answer using a diagram, numbers, and words.

d. Lila decided to make more baked goods for the bake sale. She used \(\frac{1}{8}\) lb less flour to make bread than to make cookies. She used \(\frac{1}{4}\) lb more flour to make cookies than to make brownies. If she used \(\frac{1}{2}\) lb of flour to make the bread, how much flour did she use to make the brownies? Explain your answer using a diagram, numbers, and words.
Answer:
a.
Fraction of honey collected from first Hive = \(\frac{2}{3}\) gallon
Fraction of honey collected second and third hives = 2 × \(\frac{1}{4}\) = \(\frac{1}{2}\) .
Total Honey collected = \(\frac{2}{3}\) + \(\frac{1}{4}\) + \(\frac{1}{4}\) = \(\frac{2}{3}\) + \(\frac{1}{2}\) = \(\frac{4}{6}\) + \(\frac{3}{6}\) = \(\frac{7}{6}\) = 1 \(\frac{1}{6}\)
Therefore, The total Honey collected = 1 \(\frac{1}{6}\) gallon.
Engage-NY-Eureka-Math-5th-Grade-Module-3-Mid-Module-Assessment-Answer-Key-Eureka-Math-Grade-5-Module-3-Mid-Module-Assessment-Task-Answer-Key-1-a
b.
Fraction of Honey left after baking = \(\frac{3}{4}\) gallon
Fraction of Honey used for baking = \(\frac{7}{6}\)  – \(\frac{3}{4}\) = \(\frac{14}{12}\)  – \(\frac{9}{12}\) = \(\frac{5}{12}\).
Therefore, Fraction of Honey used for baking = \(\frac{5}{12}\) gallon .
Engage-NY-Eureka-Math-5th-Grade-Module-3-Mid-Module-Assessment-Answer-Key-Eureka-Math-Grade-5-Module-3-Mid-Module-Assessment-Task-Answer-Key-1-b
c.
Fraction of honey needed for bread = \(\frac{1}{6}\) gallon
Fraction of honey needed for cookies = \(\frac{1}{4}\) gallon
Fraction of honey left = \(\frac{3}{4}\) – \(\frac{1}{4}\) – \(\frac{1}{6}\) = \(\frac{1}{2}\) – \(\frac{1}{6}\) = \(\frac{3}{6}\) – \(\frac{1}{6}\) = \(\frac{2}{6}\) = \(\frac{1}{3}\) .
Engage-NY-Eureka-Math-5th-Grade-Module-3-Mid-Module-Assessment-Answer-Key-Eureka-Math-Grade-5-Module-3-Mid-Module-Assessment-Task-Answer-Key-1-c
d.
Fraction of flour used to bake bread = \(\frac{1}{2}\) lb
Fraction of  flour used to make cookies = \(\frac{1}{8}\) lb + \(\frac{1}{2}\) lb = \(\frac{1+4}{8}\) lb = \(\frac{5}{8}\) lb
Fraction of  flour used to make brownies =\(\frac{1}{4}\) – \(\frac{1}{8}\)  = \(\frac{2- 1 }{8}\) = \(\frac{1}{8}\)  lb .
Engage-NY-Eureka-Math-5th-Grade-Module-3-Mid-Module-Assessment-Answer-Key-Eureka-Math-Grade-5-Module-3-Mid-Module-Assessment-Task-Answer-Key-1-d

Eureka Math Grade 5 Module 3 Lesson 16 Answer Key

Engage NY Eureka Math 5th Grade Module 3 Lesson 16 Answer Key

Eureka Math Grade 5 Module 3 Lesson 16 Problem Set Answer Key

Question 1.
Draw the following ribbons. When finished, compare your work to your partner’s.
a. 1 ribbon. The piece shown below is only \(\frac{1}{3}\) of the whole. Complete the drawing to show the whole ribbon.
Engage NY Math Grade 5 Module 3 Lesson 16 Problem Set Answer Key 1
b. 1 ribbon. The piece shown below is \(\frac{4}{5}\) of the whole. Complete the drawing to show the whole ribbon.
Engage NY Math Grade 5 Module 3 Lesson 16 Problem Set Answer Key 2
c. 2 ribbons, A and B. One third of A is equal to all of B. Draw a picture of the ribbons.
d. 3 ribbons, C, D, and E. C is half the length of D. E is twice as long as D. Draw a picture of the ribbons.
Answer:
a.Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-16-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-16-Problem-Set-Answer-Key-Question-1-a
Explanation :
The below ribbon into 3 parts each part is \(\frac{1}{3}\) of the whole .
b.
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-16-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-16-Problem-Set-Answer-Key-Question-1-b
Explanation :
The below ribbon into 5 parts each part is \(\frac{1}{5}\) of the whole and  \(\frac{4}{5}\)  is shaded and shown in the above figure .
c.
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-16-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-16-Problem-Set-Answer-Key-Question-1-c
Explanation :
Ribbon A is into 3 parts and Each part is \(\frac{1}{3}\) of the whole . and Ribbon B is \(\frac{1}{3}\)of A which is 1 Whole .
d.
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-16-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-16-Problem-Set-Answer-Key-Question-1-d
Explanation :
C is half the length of D.
E is twice as long as D

Question 2.
Half of Robert’s piece of wire is equal to \(\frac{2}{3}\) of Maria’s wire. The total length of their wires is 10 feet. How much longer is Robert’s wire than Maria’s?
Answer:
length of Robert’s wire = R
Length of Maria’s wire = M
Half Length of Robert’s piece of wire = \(\frac{2}{3}\) Maria’s wire .
\(\frac{R}{2}\) = \(\frac{2}{3}\) M
\(\frac{R}{M}\) = \(\frac{4}{3}\)
3R = 4M
R = \(\frac{4M}{3}\)
R + M = 10 feet .
\(\frac{4M}{3}\) + M = 10 feet .
lcm is 3
4M +3M = 30
7M = 30
M = \(\frac{30}{7}\)
R = \(\frac{4(M)}{3}\) = \(\frac{4}{3}\) × \(\frac{30}{7}\) = \(\frac{40}{7}\) .
Length of Robert wire = \(\frac{40}{7}\)
Length of  Maria wire = \(\frac{30}{7}\)
Length of Robert wire more than Maria wire = \(\frac{40}{7}\) – \(\frac{30}{7}\) = \(\frac{10}{7}\) .
Therefore, Robert’s wire is \(\frac{10}{7}\) than Maria’s wire .

Question 3.
Half of Sarah’s wire is equal to \(\frac{2}{5}\) of Daniel’s. Chris has 3 times as much as Sarah. In all, their wire measures 6 ft. How long is Sarah’s wire in feet?
Answer:
Length of Sarah’s wire = S
Length of Daniel’s wire = D
Length of Chris wire = C
Half of Sarah’s wire is equal to \(\frac{2}{5}\) of Daniel’s
\(\frac{S}{2}\) = \(\frac{2D}{5}\)
D = \(\frac{S}{2}\) × \(\frac{5}{2}\) = 5S.
Chris has 3 times as much as Sarah
C = 3S
Total length of wire = 6 ft .
C + D + S = 6
3S + 5S + S = 6
9S = 6
S = \(\frac{6}{9}\) = \(\frac{2}{3}\) feet .
Therefore, Length of Sarah’s wire in feet = \(\frac{2}{3}\) feet .

Eureka Math Grade 5 Module 3 Lesson 16 Exit Ticket Answer Key

Draw the following ribbons.
a. 1 ribbon. The piece shown below is only \(\frac{2}{3}\) of the whole. Complete the drawing to show the whole ribbon.
Eureka Math 5th Grade Module 3 Lesson 16 Exit Ticket Answer Key 1
b. 1 ribbon. The piece shown below is \(\frac{1}{4}\) of the whole. Complete the drawing to show the whole ribbon.
Eureka Math 5th Grade Module 3 Lesson 16 Exit Ticket Answer Key 2
c. 3 ribbons, A, B, and C. 1 third of A is the same length as B. C is half as long as B. Draw a picture of the ribbons.
Answer:
a. Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-16-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-16-Exit-Ticket-Answer-Key-Question-1-a
Explanation :
The given Ribbon is divided into 3 parts each part is \(\frac{1}{3}\) and \(\frac{2}{3}\) is marked as shown in above figure .
b.
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-16-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-16-Exit-Ticket-Answer-Key-Question-1-b
Explanation :
The given Ribbon is divided into 4 parts each part is \(\frac{1}{4}\) .
c.
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-16-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-16-Exit-Ticket-Answer-Key-Question-1-c
Explanation :
Length of A = 3B.
Length of C = \(\frac{B}{2}\) .

Eureka Math Grade 5 Module 3 Lesson 16 Homework Answer Key

Draw the following roads.
a. 1 road. The piece shown below is only \(\frac{3}{7}\) of the whole. Complete the drawing to show the whole road.
Eureka Math Grade 5 Module 3 Lesson 16 Homework Answer Key 1
b. 1 road. The piece shown below is \(\frac{1}{6}\) of the whole. Complete the drawing to show the whole road.
Eureka Math Grade 5 Module 3 Lesson 16 Homework Answer Key 2
c. 3 roads, A, B, and C. B is three times longer than A. C is twice as long as B. Draw the roads. What fraction of the total length of the roads is the length of A? If Road B is 7 miles longer than Road A, what is the length of Road C?
d. Write your own road problem with 2 or 3 lengths.
Answer:
a.
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-16-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-16-Home-Work-Answer-Key-Question-1-a
Explanation :
Ribbon is divided into 7 parts and each part is \(\frac{1}{7}\) and \(\frac{3}{7}\) part is shaded with yellow as shown in above figure .
b.
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-16-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-16-Home-Work-Answer-Key-Question-1-b
Explanation :
Ribbon is divided into 6 parts and each part is \(\frac{1}{6}\) .
c.
B is three times longer than A. C is twice as long as B.
Length of Road A = \(\frac{1}{3}\) B
Length of Road B = B
Length of Road C = 2B
Road A = \(\frac{1}{3}\) B
Road B = A + 7 miles . that means \(\frac{2}{3}\) is 7 miles .
Road C = 2B = 2 (3) = \(\frac{6}{3}\) Each \(\frac{2}{3}\) is 7 miles then 3 (7) = 21 miles.
Total parts = 10
Length of Road C = 21 miles .

d.
3 roads P, Q and R. P is twice longer than R . Q is half of P . Total length of the roads are 12 miles . which is the shortest road and its length .

Eureka Math Grade 5 Module 3 Lesson 15 Answer Key

Engage NY Eureka Math 5th Grade Module 3 Lesson 15 Answer Key

Eureka Math Grade 5 Module 3 Lesson 15 Sprint Answer Key

A
Circle the Smaller Fraction
Engage NY Math 5th Grade Module 3 Lesson 15 Sprint Answer Key 1

Question 1.
\(\frac{1}{2}\) \(\frac{1}{4}\)
Answer:
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-15-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-15-Sprint-Answer-Key-A-Question-1

Question 2.
\(\frac{1}{2}\) \(\frac{3}{4}\)
Answer:
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-15-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-15-Sprint-Answer-Key-A-Question-2

Question 3.
\(\frac{1}{2}\) \(\frac{5}{8}\)
Answer:
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-15-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-15-Sprint-Answer-Key-A-Question-3

Question 4.
\(\frac{1}{2}\) \(\frac{7}{8}\)
Answer:
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-15-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-15-Sprint-Answer-Key-A-Question-4

Question 5.
\(\frac{1}{2}\) \(\frac{1}{10}\)
Answer:
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-15-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-15-Sprint-Answer-Key-A-Question-5

Question 6.
\(\frac{1}{2}\) \(\frac{3}{10}\)
Answer:
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-15-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-15-Sprint-Answer-Key-A-Question-6

Question 7.
\(\frac{1}{2}\) \(\frac{5}{12}\)
Answer:
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-15-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-15-Sprint-Answer-Key-A-Question-7

Question 8.
\(\frac{1}{2}\) \(\frac{11}{12}\)
Answer:
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-15-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-15-Sprint-Answer-Key-A-Question-8

Question 9.
\(\frac{1}{2}\) \(\frac{7}{10}\)
Answer:
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-15-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-15-Sprint-Answer-Key-A-Question-9

Question 10.
\(\frac{1}{5}\) \(\frac{9}{10}\)
Answer:
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-15-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-15-Sprint-Answer-Key-A-Question-10

Question 11.
\(\frac{2}{5}\) \(\frac{1}{10}\)
Answer:
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-15-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-15-Sprint-Answer-Key-A-Question-11

Question 12.
\(\frac{2}{5}\) \(\frac{3}{10}\)
Answer:
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-15-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-15-Sprint-Answer-Key-A-Question-11

Question 13.
\(\frac{3}{5}\) \(\frac{3}{10}\)
Answer:
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-15-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-15-Sprint-Answer-Key-A-Question-13

Question 14.
\(\frac{3}{5}\) \(\frac{7}{10}\)
Answer:
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-15-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-15-Sprint-Answer-Key-A-Question-14

Question 15.
\(\frac{4}{5}\) \(\frac{1}{10}\)
Answer:
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-15-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-15-Sprint-Answer-Key-A-Question-15

Question 16.
\(\frac{4}{5}\) \(\frac{9}{10}\)
Answer:
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-15-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-15-Sprint-Answer-Key-A-Question-16

Question 17.
\(\frac{1}{3}\) \(\frac{1}{9}\)
Answer:
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-15-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-15-Sprint-Answer-Key-A-Question-17

Question 18.
\(\frac{1}{3}\) \(\frac{2}{9}\)
Answer:
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-15-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-15-Sprint-Answer-Key-A-Question-18

Question 19.
\(\frac{1}{3}\) \(\frac{4}{9}\)
Answer:
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-15-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-15-Sprint-Answer-Key-A-Question-19

Question 20.
\(\frac{1}{3}\) \(\frac{8}{9}\)
Answer:
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-15-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-15-Sprint-Answer-Key-A-Question-20

Question 21.
\(\frac{1}{3}\) \(\frac{1}{12}\)
Answer:
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-15-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-15-Sprint-Answer-Key-A-Question-21

Question 22.
\(\frac{1}{3}\) \(\frac{5}{12}\)
Answer:
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-15-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-15-Sprint-Answer-Key-A-Question-22

Question 23.
\(\frac{1}{4}\) \(\frac{1}{8}\)
Answer:
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-15-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-15-Sprint-Answer-Key-A-Question-23

Question 24.
\(\frac{1}{4}\) \(\frac{3}{8}\)
Answer:
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-15-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-15-Sprint-Answer-Key-A-Question-24

Question 25.
\(\frac{1}{4}\) \(\frac{7}{12}\)
Answer:
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-15-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-15-Sprint-Answer-Key-A-Question-25

Question 26.
\(\frac{1}{4}\) \(\frac{11}{12}\)
Answer:
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-15-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-15-Sprint-Answer-Key-A-Question-26

Question 27.
\(\frac{1}{6}\) \(\frac{7}{12}\)
Answer:
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-15-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-15-Sprint-Answer-Key-A-Question-27

Question 28.
\(\frac{1}{6}\) \(\frac{11}{12}\)
Answer:
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-15-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-15-Sprint-Answer-Key-A-Question-28

Question 29.
\(\frac{2}{3}\) \(\frac{1}{6}\)
Answer:
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-15-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-15-Sprint-Answer-Key-A-Question-29

Question 30.
\(\frac{2}{3}\) \(\frac{5}{6}\)
Answer:
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-15-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-15-Sprint-Answer-Key-A-Question-30

Question 31.
\(\frac{2}{3}\) \(\frac{2}{9}\)
Answer:
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-15-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-15-Sprint-Answer-Key-A-Question-31

Question 32.
\(\frac{2}{3}\) \(\frac{4}{9}\)
Answer:
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-15-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-15-Sprint-Answer-Key-A-Question-32

Question 33.
\(\frac{2}{3}\) \(\frac{1}{12}\)
Answer:
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-15-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-15-Sprint-Answer-Key-A-Question-33

Question 34.
\(\frac{2}{3}\) \(\frac{5}{12}\)
Answer:
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-15-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-15-Sprint-Answer-Key-A-Question-34

Question 35.
\(\frac{2}{3}\) \(\frac{11}{12}\)
Answer:
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-15-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-15-Sprint-Answer-Key-A-Question-35

Question 36.
\(\frac{2}{3}\) \(\frac{7}{12}\)
Answer:
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-15-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-15-Sprint-Answer-Key-A-Question-36

Question 37.
\(\frac{3}{4}\) \(\frac{1}{8}\)
Answer:
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-15-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-15-Sprint-Answer-Key-A-Question-37

Question 38.
\(\frac{3}{4}\) \(\frac{1}{8}\)
Answer:
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-15-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-15-Sprint-Answer-Key-A-Question-37

Question 39.
\(\frac{5}{6}\) \(\frac{7}{12}\)
Answer:
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-15-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-15-Sprint-Answer-Key-A-Question-39

Question 40.
\(\frac{5}{6}\) \(\frac{5}{12}\)
Answer:
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-15-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-15-Sprint-Answer-Key-A-Question-40

Question 41.
\(\frac{6}{7}\) \(\frac{38}{42}\)
Answer:
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-15-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-15-Sprint-Answer-Key-A-Question-41

Question 42.
\(\frac{7}{8}\) \(\frac{62}{72}\)
Answer:
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-15-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-15-Sprint-Answer-Key-A-Question-42

Question 43.
\(\frac{49}{54}\) \(\frac{8}{9}\)
Answer:
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-15-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-15-Sprint-Answer-Key-A-Question-43

Question 44.
\(\frac{67}{72}\) \(\frac{11}{12}\)
Answer:
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-15-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-15-Sprint-Answer-Key-A-Question-44

B
Circle the Smaller Fraction
Engage NY Math 5th Grade Module 3 Lesson 15 Sprint Answer Key 2

Question 1.
\(\frac{1}{2}\) \(\frac{1}{6}\)
Answer:
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-15-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-15-Sprint-Answer-Key-B-Question-1

Question 2.
\(\frac{1}{2}\) \(\frac{5}{6}\)
Answer:
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-15-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-15-Sprint-Answer-Key-B-Question-2

Question 3.
\(\frac{1}{2}\) \(\frac{1}{8}\)
Answer:
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-15-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-15-Sprint-Answer-Key-B-Question-3

Question 4.
\(\frac{1}{2}\) \(\frac{3}{8}\)
Answer:
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-15-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-15-Sprint-Answer-Key-B-Question-4

Question 5.
\(\frac{1}{2}\) \(\frac{7}{10}\)
Answer:
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-15-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-15-Sprint-Answer-Key-B-Question-5

Question 6.
\(\frac{1}{2}\) \(\frac{9}{10}\)
Answer:
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-15-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-15-Sprint-Answer-Key-B-Question-6

Question 7.
\(\frac{1}{2}\) \(\frac{1}{12}\)
Answer:
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-15-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-15-Sprint-Answer-Key-B-Question-7

Question 8.
\(\frac{1}{2}\) \(\frac{7}{12}\)
Answer:
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-15-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-15-Sprint-Answer-Key-B-Question-8

Question 9.
\(\frac{1}{5}\) \(\frac{1}{10}\)
Answer:
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-15-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-15-Sprint-Answer-Key-B-Question-9

Question 10.
\(\frac{1}{5}\) \(\frac{3}{10}\)
Answer:
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-15-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-15-Sprint-Answer-Key-B-Question-10

Question 11.
\(\frac{2}{5}\) \(\frac{1}{10}\)
Answer:
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-15-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-15-Sprint-Answer-Key-B-Question-11

Question 12.
\(\frac{2}{5}\) \(\frac{9}{10}\)
Answer:
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-15-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-15-Sprint-Answer-Key-B-Question-12

Question 13.
\(\frac{3}{5}\) \(\frac{1}{10}\)
Answer:
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-15-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-15-Sprint-Answer-Key-B-Question-13

Question 14.
\(\frac{3}{5}\) \(\frac{9}{10}\)
Answer:
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-15-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-15-Sprint-Answer-Key-B-Question-14

Question 15.
\(\frac{4}{5}\) \(\frac{3}{10}\)
Answer:
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-15-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-15-Sprint-Answer-Key-B-Question-15

Question 16.
\(\frac{4}{5}\) \(\frac{7}{10}\)
Answer:
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-15-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-15-Sprint-Answer-Key-B-Question-16

Question 17.
\(\frac{1}{3}\) \(\frac{1}{6}\)
Answer:
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-15-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-15-Sprint-Answer-Key-B-Question-17

Question 18.
\(\frac{1}{3}\) \(\frac{5}{6}\)
Answer:
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-15-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-15-Sprint-Answer-Key-B-Question-18

Question 19.
\(\frac{1}{3}\) \(\frac{5}{9}\)
Answer:
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-15-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-15-Sprint-Answer-Key-B-Question-19

Question 20.
\(\frac{1}{3}\) \(\frac{7}{9}\)
Answer:
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-15-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-15-Sprint-Answer-Key-B-Question-20

Question 21.
\(\frac{1}{3}\) \(\frac{7}{12}\)
Answer:
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-15-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-15-Sprint-Answer-Key-B-Question-21

Question 22.
\(\frac{1}{3}\) \(\frac{11}{12}\)
Answer:
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-15-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-15-Sprint-Answer-Key-B-Question-22

Question 23.
\(\frac{1}{4}\) \(\frac{5}{8}\)
Answer:
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-15-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-15-Sprint-Answer-Key-B-Question-23

Question 24.
\(\frac{1}{4}\) \(\frac{7}{8}\)
Answer:
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-15-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-15-Sprint-Answer-Key-B-Question-24

Question 25.
\(\frac{1}{4}\) \(\frac{1}{12}\)
Answer:
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-15-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-15-Sprint-Answer-Key-B-Question-25

Question 26.
\(\frac{1}{4}\) \(\frac{5}{12}\)
Answer:
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-15-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-15-Sprint-Answer-Key-B-Question-26

Question 27.
\(\frac{1}{6}\) \(\frac{1}{12}\)
Answer:
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-15-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-15-Sprint-Answer-Key-B-Question-27

Question 28.
\(\frac{1}{6}\) \(\frac{5}{12}\)
Answer:
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-15-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-15-Sprint-Answer-Key-B-Question-28

Question 29.
\(\frac{2}{3}\) \(\frac{1}{9}\)
Answer:
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-15-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-15-Sprint-Answer-Key-B-Question-29

Question 30.
\(\frac{2}{3}\) \(\frac{7}{9}\)
Answer:
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-15-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-15-Sprint-Answer-Key-B-Question-30

Question 31.
\(\frac{2}{3}\) \(\frac{5}{9}\)
Answer:
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-15-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-15-Sprint-Answer-Key-B-Question-31

Question 32.
\(\frac{2}{3}\) \(\frac{8}{9}\)
Answer:
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-15-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-15-Sprint-Answer-Key-B-Question-32

Question 33.
\(\frac{2}{3}\) \(\frac{1}{12}\)
Answer:
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-15-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-15-Sprint-Answer-Key-B-Question-33

Question 34.
\(\frac{3}{4}\) \(\frac{1}{2}\)
Answer:
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-15-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-15-Sprint-Answer-Key-B-Question-33

Question 35.
\(\frac{3}{4}\) \(\frac{5}{12}\)
Answer:
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-15-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-15-Sprint-Answer-Key-B-Question-35

Question 36.
\(\frac{3}{4}\) \(\frac{7}{12}\)
Answer:
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-15-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-15-Sprint-Answer-Key-B-Question-36

Question 37.
\(\frac{5}{6}\) \(\frac{1}{12}\)
Answer:
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-15-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-15-Sprint-Answer-Key-B-Question-37

Question 38.
\(\frac{5}{6}\) \(\frac{11}{12}\)
Answer:
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-15-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-15-Sprint-Answer-Key-B-Question-38

Question 39.
\(\frac{3}{4}\) \(\frac{5}{8}\)
Answer:
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-15-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-15-Sprint-Answer-Key-B-Question-39

Question 40.
\(\frac{3}{4}\) \(\frac{3}{8}\)
Answer:
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-15-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-15-Sprint-Answer-Key-B-Question-40

Question 41.
\(\frac{6}{7}\) \(\frac{34}{42}\)
Answer:
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-15-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-15-Sprint-Answer-Key-B-Question-41

Question 42.
\(\frac{7}{8}\) \(\frac{64}{72}\)
Answer:
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-15-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-15-Sprint-Answer-Key-B-Question-42

Question 43.
\(\frac{47}{54}\) \(\frac{8}{9}\)
Answer:
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-15-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-15-Sprint-Answer-Key-B-Question-43

Question 44.
\(\frac{65}{72}\) \(\frac{11}{12}\)
Answer:
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-15-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-15-Sprint-Answer-Key-B-Question-44

Eureka Math Grade 5 Module 3 Lesson 15 Problem Set Answer Key

Solve the word problems using the RDW strategy. Show all of your work.
Question 1.
In a race, the-second place finisher crossed the finish line 1\(\frac{1}{3}\) minutes after the winner. The third-place finisher was 1\(\frac{3}{4}\) minutes behind the second-place finisher. The third-place finisher took 34\(\frac{2}{3}\) minutes. How long did the winner take?
Answer:
Fraction of time of Second place finisher crossed the line after = 1\(\frac{1}{3}\) minutes = \(\frac{4}{3}\)
Fraction of time of Third place finisher is behind the second place  = 1\(\frac{3}{4}\) minutes  = \(\frac{7}{4}\)
Fraction of time the third place finisher took = 34\(\frac{2}{3}\) = \(\frac{105}{3}\)
Fraction of time the second place runner took = \(\frac{105}{3}\) – \(\frac{7}{4}\) = \(\frac{420}{12}\) – \(\frac{21}{12}\) = \(\frac{399}{12}\) = \(\frac{133}{4}\)
Fraction of time the First place runner took = \(\frac{133}{4}\) – \(\frac{4}{3}\) = \(\frac{399}{12}\) – \(\frac{16}{12}\) = \(\frac{383}{12}\) = 31 \(\frac{11}{12}\) .
Therefore the First Runner took = 31 \(\frac{11}{12}\) .minutes.

Question 2.
John used 1\(\frac{3}{4}\) kg of salt to melt the ice on his sidewalk. He then used another 3\(\frac{4}{5}\) kg on the driveway. If he originally bought 10 kg of salt, how much does he have left?
Answer:
Fraction of Salt used by John =1\(\frac{3}{4}\) kg = \(\frac{7}{4}\) kg
Fraction of Salt used again =3\(\frac{4}{5}\) kg = \(\frac{24}{5}\) kg
Fraction of salt used = \(\frac{7}{4}\)  + \(\frac{24}{5}\) = \(\frac{35}{20}\)  + \(\frac{96}{20}\) = \(\frac{131}{20}\)  = 6 \(\frac{11}{20}\) .
Total Salt = 10 kg.
Fraction of salt left = 10 – \(\frac{131}{20}\)  = \(\frac{200}{20}\)  – \(\frac{131}{20}\)  = \(\frac{69}{20}\)  = 3\(\frac{9}{20}\)  .
Therefore Fraction of salt left = 3\(\frac{9}{20}\)  .

Question 3.
Sinister Stan stole 3\(\frac{3}{4}\) oz of slime from Messy Molly, but his evil plans require 6\(\frac{3}{8}\) oz of slime. He stole another 2\(\frac{3}{5}\) oz of slime from Rude Ralph. How much more slime does Sinister Stan need for his evil plan?
Answer:
Fraction of slime stolen from Messy Molly = 3\(\frac{3}{4}\) = \(\frac{15}{4}\) oz
Fraction of slime stolen from Messy Molly again = 2\(\frac{3}{5}\) = \(\frac{13}{5}\) oz
Total Fraction Stolen = \(\frac{15}{4}\)  + \(\frac{13}{5}\) = \(\frac{75}{20}\) + \(\frac{52}{20}\) = \(\frac{127}{20}\) = 6\(\frac{7}{20}\) .
Fraction of more slime required = 6\(\frac{3}{8}\) – \(\frac{127}{20}\) = \(\frac{51}{8}\) – \(\frac{127}{20}\) = \(\frac{255}{40}\) – \(\frac{254}{40}\) = \(\frac{1}{40}\) .
Therefore, Fraction of more slime required = \(\frac{1}{40}\) oz.

Question 4.
Gavin had 20 minutes to do a three-problem quiz. He spent 9\(\frac{3}{4}\) minutes on Problem 1 and 3\(\frac{4}{5}\) minutes on Problem 2. How much time did he have left for Problem 3? Write the answer in minutes and seconds.
Answer:
Time given for 3 problems = 20 minutes
Fraction of time Spent on Problem 1 = 9\(\frac{3}{4}\) minutes = \(\frac{39}{4}\) .
Fraction of Time spent on Problem 2 = 3\(\frac{4}{5}\) = \(\frac{19}{5}\) .
Fraction of Time spent on Problem 3 = x
20 = \(\frac{39}{4}\) + \(\frac{19}{5}\) + x
x = 20 – \(\frac{39}{4}\) – \(\frac{19}{5}\)
x = \(\frac{400}{20}\) – \(\frac{195}{20}\) – \(\frac{76}{20}\)
x = \(\frac{129}{20}\) = 6\(\frac{9}{20}\) .
Therefore, Fraction of Time spent on Problem 3 = 6\(\frac{9}{20}\) .

Question 5.
Matt wants to shave 2\(\frac{1}{2}\) minutes off his 5K race time. After a month of hard training, he managed to lower his overall time from 21\(\frac{1}{5}\) minutes to 19\(\frac{1}{4}\) minutes. By how many more minutes does Matt need to lower his race time?
Answer:
Fraction of Time lowered = 21\(\frac{1}{5}\) minutes to 19\(\frac{1}{4}\) minutes. = \(\frac{106}{5}\) – \(\frac{77}{4}\) = \(\frac{424}{20}\) – \(\frac{385}{20}\) = \(\frac{39}{20}\) =1\(\frac{19}{20}\) .
Fraction of Time shaved = 2\(\frac{1}{2}\) =\(\frac{5}{2}\) .
Fraction of More Time Matt need to lower his race time = \(\frac{5}{2}\) – \(\frac{39}{20}\) = \(\frac{50}{20}\) – \(\frac{39}{20}\) = \(\frac{11}{20}\) = \(\frac{33}{60}\) = 33 minutes .

Eureka Math Grade 5 Module 3 Lesson 15 Exit Ticket Answer Key

Solve the word problem using the RDW strategy. Show all of your work.
Cheryl bought a sandwich for 5\(\frac{1}{2}\) dollars and a drink for $2.60. If she paid for her meal with a $10 bill, how much money did she have left? Write your answer as a fraction and in dollars and cents.
Answer:
Fraction of Cost of sandwich = 5\(\frac{1}{2}\) = \(\frac{11}{2}\) dollar = 5.5 dollar
Fraction of Cost of Drink = $2.60.
Total Cost = 5.5 +2.60 = 8.1 $.
Amount paid = 10$.
Money left = 10 – 8.1 = 1.9 $ .

Eureka Math Grade 5 Module 3 Lesson 15 Homework Answer Key

Solve the word problems using the RDW strategy. Show all of your work.
Question 1.
A baker buys a 5 lb bag of sugar. She uses 1\(\frac{2}{3}\) lb to make some muffins and 2\(\frac{3}{4}\) lb to make a cake. How much sugar does she have left?
Answer:
Total Quantity of Sugar = 5 lb
Fraction of Quantity of Suagr used for muffins = 1\(\frac{2}{3}\) lb = \(\frac{5}{3}\)
Fraction of Quantity of Suagr used cake = 2\(\frac{3}{4}\) lb = \(\frac{11}{4}\)
Fraction of Quantity of Sugar used = \(\frac{5}{3}\) + \(\frac{11}{4}\) = \(\frac{20}{12}\) + \(\frac{33}{12}\) = \(\frac{53}{12}\)
Fraction of Quantity of Sugar left = 5 – \(\frac{53}{12}\) = \(\frac{60}{12}\) – \(\frac{53}{12}\) =\(\frac{7}{12}\) .
Therefore, Fraction of Quantity of sugar left = \(\frac{7}{12}\) .

Question 2.
A boxer needs to lose 3\(\frac{1}{2}\) kg in a month to be able to compete as a flyweight. In three weeks, he lowers his weight from 55.5 kg to 53.8 kg. How many kilograms must the boxer lose in the final week to be able to compete as a flyweight?
Answer:
Fraction of weight need to lose in month = 3\(\frac{1}{2}\) = \(\frac{7}{2}\) = 3.5 kg
Weight lost in 3 weeks = 55.5 –  53.8 = 1.7 kg
Weight need to lose in final week = 3.5 – 1.7 = 1.8 kg.

Question 3.
A construction company builds a new rail line from Town A to Town B. They complete 1\(\frac{1}{4}\) miles in their first week of work and 1\(\frac{2}{3}\) miles in the second week. If they still have 25\(\frac{3}{4}\) miles left to build, what is the distance from Town A to Town B?
Answer:
Fraction of work completed in first week = 1\(\frac{1}{4}\) miles = \(\frac{5}{4}\)
Fraction of work completed in second week = 1\(\frac{2}{3}\) miles = \(\frac{5}{3}\)
Fraction of work left to built = 25\(\frac{3}{4}\) miles = \(\frac{103}{4}\)
Fraction of Distance from Town A to Town B = \(\frac{103}{4}\) + \(\frac{5}{4}\)  + \(\frac{5}{3}\) = \(\frac{108}{4}\) + \(\frac{5}{3}\) = \(\frac{324}{12}\) + \(\frac{20}{12}\) = \(\frac{344}{12}\)= 28\(\frac{2}{3}\) .
Therefore, Fraction of Distance from Town A to Town B = 28\(\frac{2}{3}\) miles.

Question 4.
A catering company needs 8.75 lb of shrimp for a small party. They buy 3\(\frac{2}{3}\) lb of jumbo shrimp, 2\(\frac{5}{8}\) lb of medium-sized shrimp, and some mini-shrimp. How many pounds of mini-shrimp do they buy?
Answer:
Quantity of shrimp needed = 8.75 lb =8\(\frac{3}{4}\) = \(\frac{27}{4}\)
Quantity of jumbo shrimp = 3\(\frac{2}{3}\) lb = \(\frac{11}{3}\)
Quantity of  medium – sized shrimp = 2\(\frac{5}{8}\) lb = \(\frac{21}{8}\)
Quantity of mini shrimp = x
\(\frac{35}{4}\)  = \(\frac{11}{3}\) + \(\frac{21}{8}\) + x
x = \(\frac{210}{24}\)  – \(\frac{88}{24}\) – \(\frac{63}{24}\)
x =  \(\frac{59}{24}\) = 2 \(\frac{11}{24}\)
Therefore, Quantity of mini shrimp = x = 2 \(\frac{11}{24}\) lb .

Question 5.
Mark breaks up a 9-hour drive into 3 segments. He drives 2\(\frac{1}{2}\) hours before stopping for lunch. After driving some more, he stops for gas. If the second segment of his drive was 1\(\frac{2}{3}\) hours longer than the first segment, how long did he drive after stopping for gas?
Answer:
Total time of the drive = 9 hours .
Fraction of Time drived for first segment = 2\(\frac{1}{2}\) hours  = \(\frac{5}{2}\)
Fraction of Time of second segment = 1\(\frac{2}{3}\) hours longer than the first segment = \(\frac{5}{3}\) + \(\frac{5}{2}\) = \(\frac{10}{6}\) + \(\frac{15}{6}\) = \(\frac{25}{6}\) =4\(\frac{1}{6}\)
Fraction of Time of first and second segment= \(\frac{5}{2}\) + \(\frac{25}{6}\) = \(\frac{15}{6}\) + \(\frac{25}{6}\) = \(\frac{40}{6}\) = 6\(\frac{4}{6}\)
Fraction of Time he drive after stopping gas = 9 – \(\frac{40}{6}\) = \(\frac{54}{6}\) –  \(\frac{40}{6}\) = \(\frac{14}{6}\) = 2\(\frac{2}{6}\) .
Therefore, Fraction of Time he drive after stopping gas = third segment = 2\(\frac{2}{6}\) hours .

 

Eureka Math Grade 5 Module 3 Lesson 14 Answer Key

Engage NY Eureka Math 5th Grade Module 3 Lesson 14 Answer Key

Eureka Math Grade 5 Module 3 Lesson 14 Sprint Answer Key

A
Make Larger Units
Engage NY Math 5th Grade Module 3 Lesson 14 Sprint Answer Key 1

Question 1.
\(\frac{2}{4}\) =
Answer:
\(\frac{2}{4}\) = \(\frac{2 × 2}{4 × 2}\) = \(\frac{4}{8}\)
Explanation :
Multiply numerator and denominator by 2 to make the given fraction into a bigger fraction .

Question 2.
\(\frac{2}{6}\) =
Answer:
\(\frac{2}{6}\) = \(\frac{2 × 2}{6 × 2}\) = \(\frac{4}{12}\)
Explanation :
Multiply numerator and denominator by 2 or another number to make the given fraction into a larger fraction unit .

Question 3.
\(\frac{2}{8}\) =
Answer:
\(\frac{2}{8}\) = \(\frac{2 × 2}{8 × 2}\) = \(\frac{4}{16}\)
Explanation :
Multiply numerator and denominator by 2 or another number to make the given fraction into a larger fraction unit .

Question 4.
\(\frac{5}{10}\) =
Answer:
\(\frac{5}{10}\) = \(\frac{5 × 2}{10 × 2}\) = \(\frac{10}{20}\)
Explanation :
Multiply numerator and denominator by 2 or another number to make the given fraction into a larger fraction unit .

Question 5.
\(\frac{5}{15}\) =
Answer:
\(\frac{5}{15}\) = \(\frac{5 × 2}{15 × 2}\) = \(\frac{10}{30}\)
Explanation :
Multiply numerator and denominator by 2 or another number to make the given fraction into a larger fraction unit .

Question 6.
\(\frac{5}{20}\) =
Answer:
\(\frac{5}{20}\) = \(\frac{5 × 2}{20 × 2}\) = \(\frac{10}{40}\)
Explanation :
Multiply numerator and denominator by 2 or another number to make the given fraction into a larger fraction unit .

Question 7.
\(\frac{4}{8}\) =
Answer:
\(\frac{4}{8}\) = \(\frac{4 × 2}{8 × 2}\) = \(\frac{8}{16}\)
Explanation :
Multiply numerator and denominator by 2 or another number to make the given fraction into a larger fraction unit .

Question 8.
\(\frac{4}{12}\) =
Answer:
\(\frac{4}{12}\) = \(\frac{4 × 2}{12 × 2}\) = \(\frac{8}{24}\)
Explanation :
Multiply numerator and denominator by 2 or another number to make the given fraction into a larger fraction unit .

Question 9.
\(\frac{4}{16}\) =
Answer:
\(\frac{4}{16}\) = \(\frac{4 × 2}{16 × 2}\) = \(\frac{8}{32}\)
Explanation :
Multiply numerator and denominator by 2 or another number to make the given fraction into a larger fraction unit .

Question 10.
\(\frac{3}{6}\) =
Answer:
\(\frac{3}{6}\) = \(\frac{3 × 2}{6 × 2}\) = \(\frac{6}{12}\)
Explanation :
Multiply numerator and denominator by 2 or another number to make the given fraction into a larger fraction unit .

Question 11.
\(\frac{3}{9}\) =
Answer:
\(\frac{3}{9}\) = \(\frac{3 × 2}{9 × 2}\) = \(\frac{6}{18}\)
Explanation :
Multiply numerator and denominator by 2 or another number to make the given fraction into a larger fraction unit .

Question 12.
\(\frac{3}{12}\) =
Answer:
\(\frac{3}{12}\) = \(\frac{3 × 2}{12 × 2}\) = \(\frac{6}{24}\)
Explanation :
Multiply numerator and denominator by 2 or another number to make the given fraction into a larger fraction unit .

Question 13.
\(\frac{4}{6}\) =
Answer:
\(\frac{4}{6}\) = \(\frac{4 × 2}{6 × 2}\) = \(\frac{8}{12}\)
Explanation :
Multiply numerator and denominator by 2 or another number to make the given fraction into a larger fraction unit .

Question 14.
\(\frac{6}{12}\) =
Answer:
\(\frac{6}{12}\) = \(\frac{6 × 2}{12 × 2}\) = \(\frac{12}{24}\)
Explanation :
Multiply numerator and denominator by 2 or another number to make the given fraction into a larger fraction unit .

Question 15.
\(\frac{6}{18}\) =
Answer:
\(\frac{6}{8}\) = \(\frac{6 × 2}{8 × 2}\) = \(\frac{12}{16}\)
Explanation :
Multiply numerator and denominator by 2 or another number to make the given fraction into a larger fraction unit .

Question 16.
\(\frac{6}{30}\) =
Answer:
\(\frac{6}{30}\) = \(\frac{6 × 2}{30 × 2}\) = \(\frac{12}{60}\)
Explanation :
Multiply numerator and denominator by 2 or another number to make the given fraction into a larger fraction unit .

Question 17.
\(\frac{6}{9}\) =
Answer:
\(\frac{6}{9}\) = \(\frac{6 × 2}{9 × 2}\) = \(\frac{12}{18}\)
Explanation :
Multiply numerator and denominator by 2 or another number to make the given fraction into a larger fraction unit .

Question 18.
\(\frac{7}{14}\) =
Answer:
\(\frac{7}{14}\) = \(\frac{7 × 2}{14 × 2}\) = \(\frac{14}{28}\)
Explanation :
Multiply numerator and denominator by 2 or another number to make the given fraction into a larger fraction unit .

Question 19.
\(\frac{7}{21}\) =
Answer:
\(\frac{7}{21}\) = \(\frac{7 × 2}{21 × 2}\) = \(\frac{14}{42}\)
Explanation :
Multiply numerator and denominator by 2 or another number to make the given fraction into a larger fraction unit .

Question 20.
\(\frac{7}{42}\) =
Answer:
\(\frac{7}{42}\) = \(\frac{7 × 2}{42 × 2}\) = \(\frac{14}{84}\)
Explanation :
Multiply numerator and denominator by 2 or another number to make the given fraction into a larger fraction unit .

Question 21.
\(\frac{8}{12}\) =
Answer:
\(\frac{8}{12}\) = \(\frac{8 × 2}{12 × 2}\) = \(\frac{16}{24}\)
Explanation :
Multiply numerator and denominator by 2 or another number to make the given fraction into a larger fraction unit .

Question 22.
\(\frac{9}{18}\) =
Answer:
\(\frac{9}{18}\) = \(\frac{9 × 2}{18 × 2}\) = \(\frac{18}{36}\)
Explanation :
Multiply numerator and denominator by 2 or another number to make the given fraction into a larger fraction unit .

Question 23.
\(\frac{9}{27}\) =
Answer:
\(\frac{9}{18}\) = \(\frac{9 × 2}{27 × 2}\) = \(\frac{18}{54}\)
Explanation :
Multiply numerator and denominator by 2 or another number to make the given fraction into a larger fraction unit .

Question 24.
\(\frac{2}{4}\) =
Answer:
\(\frac{2}{4}\) = \(\frac{2 × 2}{4 × 2}\) = \(\frac{4}{8}\)
Explanation :
Multiply numerator and denominator by 2 or another number to make the given fraction into a larger fraction unit .

Question 25.
\(\frac{8}{12}\) =
Answer:
\(\frac{8}{12}\) = \(\frac{8 × 2}{12 × 2}\) = \(\frac{16}{24}\)
Explanation :
Multiply numerator and denominator by 2 or another number to make the given fraction into a larger fraction unit .

Question 26.
\(\frac{8}{16}\) =
Answer:
\(\frac{8}{16}\) = \(\frac{8 × 2}{16 × 2}\) = \(\frac{16}{32}\)
Explanation :
Multiply numerator and denominator by 2 or another number to make the given fraction into a larger fraction unit .

Question 27.
\(\frac{8}{24}\) =
Answer:
\(\frac{8}{24}\) = \(\frac{8 × 2}{24 × 2}\) = \(\frac{16}{48}\)
Explanation :
Multiply numerator and denominator by 2 or another number to make the given fraction into a larger fraction unit .

Question 28.
\(\frac{8}{64}\) =
Answer:
\(\frac{8}{64}\) = \(\frac{8 × 2}{64 × 2}\) = \(\frac{16}{128}\)
Explanation :
Multiply numerator and denominator by 2 or another number to make the given fraction into a larger fraction unit .

Question 29.
\(\frac{12}{18}\) =
Answer:
\(\frac{12}{18}\) = \(\frac{12 × 2}{18 × 2}\) = \(\frac{24}{36}\)
Explanation :
Multiply numerator and denominator by 2 or another number to make the given fraction into a larger fraction unit .

Question 30.
\(\frac{12}{16}\) =
Answer:
\(\frac{12}{16}\) = \(\frac{12 × 2}{16 × 2}\) = \(\frac{24}{32}\)
Explanation :
Multiply numerator and denominator by 2 or another number to make the given fraction into a larger fraction unit .

Question 31.
\(\frac{9}{12}\) =
Answer:
\(\frac{9}{12}\) = \(\frac{9 × 2}{12 × 2}\) = \(\frac{18}{24}\)
Explanation :
Multiply numerator and denominator by 2 or another number to make the given fraction into a larger fraction unit .

Question 32.
\(\frac{6}{8}\) =
Answer:
\(\frac{6}{8}\) = \(\frac{6 × 2}{8 × 2}\) = \(\frac{12}{16}\)
Explanation :
Multiply numerator and denominator by 2 or another number to make the given fraction into a larger fraction unit .

Question 33.
\(\frac{10}{12}\) =
Answer:
\(\frac{10}{12}\) = \(\frac{10 × 2}{12 × 2}\) = \(\frac{20}{24}\)
Explanation :
Multiply numerator and denominator by 2 or another number to make the given fraction into a larger fraction unit .

Question 34.
\(\frac{15}{18}\) =
Answer:
\(\frac{15}{18}\) = \(\frac{15 × 2}{18 × 2}\) = \(\frac{30}{36}\)
Explanation :
Multiply numerator and denominator by 2 or another number to make the given fraction into a larger fraction unit .

Question 35.
\(\frac{8}{10}\) =
Answer:
\(\frac{8}{10}\) = \(\frac{8 × 2}{10 × 2}\) = \(\frac{16}{20}\)
Explanation :
Multiply numerator and denominator by 2 or another number to make the given fraction into a larger fraction unit .

Question 36.
\(\frac{16}{20}\) =
Answer:
\(\frac{16}{20}\) = \(\frac{16 × 2}{20 × 2}\) = \(\frac{32}{40}\)
Explanation :
Multiply numerator and denominator by 2 or another number to make the given fraction into a larger fraction unit .

Question 37.
\(\frac{12}{15}\) =
Answer:
\(\frac{12}{15}\) = \(\frac{12 × 2}{15 × 2}\) = \(\frac{24}{30}\)
Explanation :
Multiply numerator and denominator by 2 or another number to make the given fraction into a larger fraction unit .

Question 38.
\(\frac{18}{27}\) =
Answer:
\(\frac{18}{27}\) = \(\frac{18 × 2}{27 × 2}\) = \(\frac{36}{54}\)
Explanation :
Multiply numerator and denominator by 2 or another number to make the given fraction into a larger fraction unit .

Question 39.
\(\frac{27}{36}\) =
Answer:
\(\frac{27}{36}\) = \(\frac{27 × 2}{36 × 2}\) = \(\frac{54}{72}\)
Explanation :
Multiply numerator and denominator by 2 or another number to make the given fraction into a larger fraction unit .

Question 40.
\(\frac{32}{40}\) =
Answer:
\(\frac{32}{40}\) = \(\frac{32 × 2}{40 × 2}\) = \(\frac{64}{80}\)
Explanation :
Multiply numerator and denominator by 2 or another number to make the given fraction into a larger fraction unit .

Question 41.
\(\frac{45}{54}\) =
Answer:
\(\frac{45}{54}\) = \(\frac{45 × 2}{54 × 2}\) = \(\frac{90}{108}\)
Explanation :
Multiply numerator and denominator by 2 or another number to make the given fraction into a larger fraction unit .

Question 42.
\(\frac{24}{36}\) =
Answer:
\(\frac{24}{36}\) = \(\frac{24 × 2}{36 × 2}\) = \(\frac{48}{72}\)
Explanation :
Multiply numerator and denominator by 2 or another number to make the given fraction into a larger fraction unit .

Question 43.
\(\frac{60}{72}\) =
Answer:
\(\frac{60}{72}\) = \(\frac{60 × 2}{72 × 2}\) = \(\frac{120}{144}\)
Explanation :
Multiply numerator and denominator by 2 or another number to make the given fraction into a larger fraction unit .

Question 44.
\(\frac{48}{60}\) =
Answer:
\(\frac{48}{60}\) = \(\frac{48 × 2}{60 × 2}\) = \(\frac{96}{120}\)
Explanation :
Multiply numerator and denominator by 2 or another number to make the given fraction into a larger fraction unit .

B
Make Larger Units
Engage NY Math 5th Grade Module 3 Lesson 14 Sprint Answer Key 2

Question 1.
\(\frac{5}{10}\) =
Answer:
\(\frac{5}{10}\) = \(\frac{5 × 2}{10 × 2}\) = \(\frac{10}{20}\)
Explanation :
Multiply numerator and denominator by 2 or another number to make the given fraction into a larger fraction unit .

Question 2.
\(\frac{5}{15}\) =
Answer:
\(\frac{5}{15}\) = \(\frac{5 × 2}{15 × 2}\) = \(\frac{10}{30}\)
Explanation :
Multiply numerator and denominator by 2 or another number to make the given fraction into a larger fraction unit .

Question 3.
\(\frac{5}{20}\) =
Answer:
\(\frac{5}{20}\) = \(\frac{5 × 2}{20 × 2}\) = \(\frac{10}{40}\)
Explanation :
Multiply numerator and denominator by 2 or another number to make the given fraction into a larger fraction unit .

Question 4.
\(\frac{2}{4}\) =
Answer:
\(\frac{2}{4}\) = \(\frac{2 × 2}{4 × 2}\) = \(\frac{4}{8}\)
Explanation :
Multiply numerator and denominator by 2 or another number to make the given fraction into a larger fraction unit .

Question 5.
\(\frac{2}{6}\) =
Answer:
\(\frac{2}{6}\) = \(\frac{2 × 2}{6 × 2}\) = \(\frac{4}{12}\)
Explanation :
Multiply numerator and denominator by 2 or another number to make the given fraction into a larger fraction unit .

Question 6.
\(\frac{2}{8}\) =
Answer:
\(\frac{2}{8}\) = \(\frac{2 × 2}{8 × 2}\) = \(\frac{4}{16}\)
Explanation :
Multiply numerator and denominator by 2 or another number to make the given fraction into a larger fraction unit .

Question 7.
\(\frac{3}{6}\) =
Answer:
\(\frac{3}{6}\) = \(\frac{3 × 2}{6 × 2}\) = \(\frac{5}{12}\)
Explanation :
Multiply numerator and denominator by 2 or another number to make the given fraction into a larger fraction unit .

Question 8.
\(\frac{3}{9}\) =
Answer:
\(\frac{2}{4}\) = \(\frac{2 × 2}{4 × 2}\) = \(\frac{4}{8}\)
Explanation :
Multiply numerator and denominator by 2 or another number to make the given fraction into a larger fraction unit .

Question 9.
\(\frac{3}{12}\) =
Answer:
\(\frac{3}{12}\) = \(\frac{3 × 2}{12 × 2}\) = \(\frac{6}{24}\)
Explanation :
Multiply numerator and denominator by 2 or another number to make the given fraction into a larger fraction unit .

Question 10.
\(\frac{4}{8}\) =
Answer:
\(\frac{4}{8}\) = \(\frac{4 × 2}{8 × 2}\) = \(\frac{8}{16}\)
Explanation :
Multiply numerator and denominator by 2 or another number to make the given fraction into a larger fraction unit .

Question 11.
\(\frac{4}{12}\) =
Answer:
\(\frac{4}{12}\) = \(\frac{4 × 2}{12 × 2}\) = \(\frac{8}{24}\)
Explanation :
Multiply numerator and denominator by 2 or another number to make the given fraction into a larger fraction unit .

Question 12.
\(\frac{4}{16}\) =
Answer:
\(\frac{4}{16}\) = \(\frac{4 × 2}{16 × 2}\) = \(\frac{8}{32}\)
Explanation :
Multiply numerator and denominator by 2 or another number to make the given fraction into a larger fraction unit .

Question 13.
\(\frac{4}{6}\) =
Answer:
\(\frac{4}{6}\) = \(\frac{4 × 2}{6 × 2}\) = \(\frac{8}{12}\)
Explanation :
Multiply numerator and denominator by 2 or another number to make the given fraction into a larger fraction unit .

Question 14.
\(\frac{7}{14}\) =
Answer:
\(\frac{7}{14}\) = \(\frac{7 × 2}{14 × 2}\) = \(\frac{14}{28}\)
Explanation :
Multiply numerator and denominator by 2 or another number to make the given fraction into a larger fraction unit .

Question 15.
\(\frac{7}{21}\) =
Answer:
\(\frac{7}{21}\) = \(\frac{7 × 2}{21 × 2}\) = \(\frac{14}{42}\)
Explanation :
Multiply numerator and denominator by 2 or another number to make the given fraction into a larger fraction unit .

Question 16.
\(\frac{7}{35}\) =
Answer:
\(\frac{7}{35}\) = \(\frac{7 × 2}{35 × 2}\) = \(\frac{14}{70}\)
Explanation :
Multiply numerator and denominator by 2 or another number to make the given fraction into a larger fraction unit .

Question 17.
\(\frac{6}{9}\) =
Answer:
\(\frac{6}{9}\) = \(\frac{6 × 2}{9 × 2}\) = \(\frac{12}{18}\)
Explanation :
Multiply numerator and denominator by 2 or another number to make the given fraction into a larger fraction unit .

Question 18.
\(\frac{6}{12}\) =
Answer:
\(\frac{6}{12}\) = \(\frac{6 × 2}{12 × 2}\) = \(\frac{12}{24}\)
Explanation :
Multiply numerator and denominator by 2 or another number to make the given fraction into a larger fraction unit .

Question 19.
\(\frac{6}{18}\) =
Answer:
\(\frac{6}{18}\) = \(\frac{6 × 2}{18 × 2}\) = \(\frac{12}{36}\)
Explanation :
Multiply numerator and denominator by 2 or another number to make the given fraction into a larger fraction unit .

Question 20.
\(\frac{6}{36}\) =
Answer:
\(\frac{6}{36}\) = \(\frac{6 × 2}{36 × 2}\) = \(\frac{12}{72}\)
Explanation :
Multiply numerator and denominator by 2 or another number to make the given fraction into a larger fraction unit .

Question 21.
\(\frac{8}{12}\) =
Answer:
\(\frac{8}{12}\) = \(\frac{8 × 2}{12 × 2}\) = \(\frac{16}{24}\)
Explanation :
Multiply numerator and denominator by 2 or another number to make the given fraction into a larger fraction unit .

Question 22.
\(\frac{8}{16}\) =
Answer:
\(\frac{8}{16}\) = \(\frac{8 × 2}{16 × 2}\) = \(\frac{16}{32}\)
Explanation :
Multiply numerator and denominator by 2 or another number to make the given fraction into a larger fraction unit .

Question 23.
\(\frac{8}{24}\) =
Answer:
\(\frac{8}{24}\) = \(\frac{8 × 2}{24 × 2}\) = \(\frac{16}{48}\)
Explanation :
Multiply numerator and denominator by 2 or another number to make the given fraction into a larger fraction unit .

Question 24.
\(\frac{8}{56}\) =
Answer:
\(\frac{8}{56}\) = \(\frac{8 × 2}{56 × 2}\) = \(\frac{16}{112}\)
Explanation :
Multiply numerator and denominator by 2 or another number to make the given fraction into a larger fraction unit .

Question 25.
\(\frac{8}{12}\) =
Answer:
\(\frac{8}{12}\) = \(\frac{8 × 2}{12 × 2}\) = \(\frac{16}{24}\)
Explanation :
Multiply numerator and denominator by 2 or another number to make the given fraction into a larger fraction unit .

Question 26.
\(\frac{9}{18}\) =
Answer:
\(\frac{8}{12}\) = \(\frac{8 × 2}{12 × 2}\) = \(\frac{16}{24}\)
Explanation :
Multiply numerator and denominator by 2 or another number to make the given fraction into a larger fraction unit .

Question 27.
\(\frac{9}{27}\) =
Answer:
\(\frac{9}{27}\) = \(\frac{9 × 2}{27 × 2}\) = \(\frac{18}{54}\)
Explanation :
Multiply numerator and denominator by 2 or another number to make the given fraction into a larger fraction unit .

Question 28.
\(\frac{9}{72}\) =
Answer:
\(\frac{9}{72}\) = \(\frac{9 × 2}{72 × 2}\) = \(\frac{18}{144}\)
Explanation :
Multiply numerator and denominator by 2 or another number to make the given fraction into a larger fraction unit .

Question 29.
\(\frac{12}{18}\) =
Answer:
\(\frac{12}{18}\) = \(\frac{12 × 2}{18 × 2}\) = \(\frac{24}{36}\)
Explanation :
Multiply numerator and denominator by 2 or another number to make the given fraction into a larger fraction unit .

Question 30.
\(\frac{12}{16}\) =
Answer:
\(\frac{12}{16}\) = \(\frac{12 × 2}{16 × 2}\) = \(\frac{24}{32}\)
Explanation :
Multiply numerator and denominator by 2 or another number to make the given fraction into a larger fraction unit .

Question 31.
\(\frac{9}{12}\) =
Answer:
\(\frac{9}{12}\) = \(\frac{9 × 2}{12 × 2}\) = \(\frac{18}{24}\)
Explanation :
Multiply numerator and denominator by 2 or another number to make the given fraction into a larger fraction unit .

Question 32.
\(\frac{12}{16}\) =
Answer:
\(\frac{12}{16}\) = \(\frac{12 × 2}{16 × 2}\) = \(\frac{24}{32}\)
Explanation :
Multiply numerator and denominator by 2 or another number to make the given fraction into a larger fraction unit .

Question 33.
\(\frac{8}{10}\) =
Answer:
\(\frac{8}{10}\) = \(\frac{8 × 2}{10 × 2}\) = \(\frac{16}{20}\)
Explanation :
Multiply numerator and denominator by 2 or another number to make the given fraction into a larger fraction unit .

Question 34.
\(\frac{16}{20}\) =
Answer:
\(\frac{16}{20}\) = \(\frac{16 × 2}{20 × 2}\) = \(\frac{32}{40}\)
Explanation :
Multiply numerator and denominator by 2 or another number to make the given fraction into a larger fraction unit .

Question 35.
\(\frac{12}{15}\) =
Answer:
\(\frac{12}{15}\) = \(\frac{12 × 2}{15 × 2}\) = \(\frac{24}{30}\)
Explanation :
Multiply numerator and denominator by 2 or another number to make the given fraction into a larger fraction unit .

Question 36.
\(\frac{10}{12}\) =
Answer:
\(\frac{10}{12}\) = \(\frac{10 × 2}{12 × 2}\) = \(\frac{20}{24}\)
Explanation :
Multiply numerator and denominator by 2 or another number to make the given fraction into a larger fraction unit .

Question 37.
\(\frac{15}{18}\) =
Answer:
\(\frac{15}{18}\) = \(\frac{15 × 2}{18 × 2}\) = \(\frac{30}{36}\)
Explanation :
Multiply numerator and denominator by 2 or another number to make the given fraction into a larger fraction unit .

Question 38.
\(\frac{16}{24}\) =
Answer:
\(\frac{16}{24}\) = \(\frac{16 × 2}{24 × 2}\) = \(\frac{32}{48}\)
Explanation :
Multiply numerator and denominator by 2 or another number to make the given fraction into a larger fraction unit .

Question 39.
\(\frac{24}{32}\) =
Answer:
\(\frac{24}{32}\) = \(\frac{24 × 2}{32 × 2}\) = \(\frac{48}{64}\)
Explanation :
Multiply numerator and denominator by 2 or another number to make the given fraction into a larger fraction unit .

Question 40.
\(\frac{36}{45}\) =
Answer:
\(\frac{36}{45}\) = \(\frac{36 × 2}{45 × 2}\) = \(\frac{72}{90}\)
Explanation :
Multiply numerator and denominator by 2 or another number to make the given fraction into a larger fraction unit .

Question 41.
\(\frac{40}{48}\) =
Answer:
\(\frac{40}{48}\) = \(\frac{40 × 2}{48 × 2}\) = \(\frac{80}{96}\)
Explanation :
Multiply numerator and denominator by 2 or another number to make the given fraction into a larger fraction unit .

Question 42.
\(\frac{24}{36}\) =
Answer:
\(\frac{24}{36}\) = \(\frac{24 × 2}{36 × 2}\) = \(\frac{48}{72}\)
Explanation :
Multiply numerator and denominator by 2 or another number to make the given fraction into a larger fraction unit .

Question 43.
\(\frac{48}{60}\) =
Answer:
\(\frac{48}{60}\) = \(\frac{48 × 2}{60 × 2}\) = \(\frac{96}{120}\)
Explanation :
Multiply numerator and denominator by 2 or another number to make the given fraction into a larger fraction unit .

Question 44.
\(\frac{60}{72}\) =
Answer:
\(\frac{60}{72}\) = \(\frac{60 × 2}{72 × 2}\) = \(\frac{120}{144}\)
Explanation :
Multiply numerator and denominator by 2 or another number to make the given fraction into a larger fraction unit .

Eureka Math Grade 5 Module 3 Lesson 14 Problem Set Answer Key

Question 1.
Rearrange the terms so that you can add or subtract mentally. Then, solve.
a. \(\frac{1}{4}\) + 2\(\frac{2}{3}\) + \(\frac{7}{4}\) + \(\frac{1}{3}\)
b. 2\(\frac{3}{5}\) – \(\frac{3}{4}\) + \(\frac{2}{5}\)
c. 4\(\frac{3}{7}\) – \(\frac{3}{4}\) – 2\(\frac{1}{4}\) – \(\frac{3}{7}\)
d. \(\frac{5}{6}\) + \(\frac{1}{3}\) – \(\frac{4}{3}\) + \(\frac{1}{6}\)
Answer:
a. \(\frac{1}{4}\) + 2\(\frac{2}{3}\) + \(\frac{7}{4}\) + \(\frac{1}{3}\) = 5
b. 2\(\frac{3}{5}\) – \(\frac{3}{4}\) + \(\frac{2}{5}\) = 2 \(\frac{1}{4}\)
c. 4\(\frac{3}{7}\) – \(\frac{3}{4}\) – 2\(\frac{1}{4}\) – \(\frac{3}{7}\) = 1
d. \(\frac{5}{6}\) + \(\frac{1}{3}\) – \(\frac{4}{3}\) + \(\frac{1}{6}\) = 0
Explanation :
a. \(\frac{1}{4}\) + 2\(\frac{2}{3}\) + \(\frac{7}{4}\) + \(\frac{1}{3}\)
= \(\frac{1}{4}\) + \(\frac{7}{4}\)  + \(\frac{8}{3}\) + \(\frac{1}{3}\)
= \(\frac{8}{4}\) + \(\frac{9}{3}\) = 2 + 3 = 5

b. 2\(\frac{3}{5}\) – \(\frac{3}{4}\) + \(\frac{2}{5}\)
= \(\frac{13}{5}\) – \(\frac{3}{4}\) + \(\frac{2}{5}\)
= \(\frac{13}{5}\) + \(\frac{2}{5}\) – \(\frac{3}{4}\)
= \(\frac{15}{5}\) – \(\frac{3}{4}\) = 3 – \(\frac{3}{4}\) = 2 \(\frac{1}{4}\)

c. 4\(\frac{3}{7}\) – \(\frac{3}{4}\) – 2\(\frac{1}{4}\) – \(\frac{3}{7}\)
= \(\frac{31}{7}\) – \(\frac{3}{7}\) – \(\frac{3}{4}\) – \(\frac{9}{4}\)
= \(\frac{28}{7}\) – \(\frac{12}{4}\) = 4 – 3 = 1

d. \(\frac{5}{6}\) + \(\frac{1}{3}\) – \(\frac{4}{3}\) + \(\frac{1}{6}\)
=  \(\frac{5}{6}\) + \(\frac{1}{6}\) + \(\frac{1}{3}\) – \(\frac{4}{3}\)
= \(\frac{6}{6}\) – \(\frac{3}{3}\) = 1 – 1 = 0

Question 2.
Fill in the blank to make the statement true.
a. 11\(\frac{2}{5}\) – 3\(\frac{2}{3}\) – \(\frac{11}{3}\) = ________
b. 11\(\frac{7}{8}\) + 3\(\frac{1}{5}\) – ____________ = 15
c.\(\frac{5}{12}\) -________+ \(\frac{5}{4}\) = \(\frac{2}{3}\)
d. ________- 30 – 7\(\frac{1}{4}\) = 21\(\frac{2}{3}\)
e. \(\frac{24}{5}\) + ________ + \(\frac{8}{7}\) = 9
f. 11.1 + 3 \(\frac{1}{10}\) – ________= \(\frac{99}{10}\)
Answer:
a. 11\(\frac{2}{5}\) – 3\(\frac{2}{3}\) – \(\frac{11}{3}\) =4\(\frac{1}{15}\)
b. 11\(\frac{7}{8}\) + 3\(\frac{1}{5}\) – \(\frac{3}{40}\) = 15
c.\(\frac{5}{12}\) -1+ \(\frac{5}{4}\) = \(\frac{2}{3}\)
d. 58\(\frac{11}{12}\) . – 30 – 7\(\frac{1}{4}\) = 21\(\frac{2}{3}\)
e. \(\frac{24}{5}\) + 21\(\frac{2}{5}\) + \(\frac{8}{7}\) = 9
f. 11.1 + 3 \(\frac{1}{10}\) – 8 = \(\frac{99}{10}\)
Explanation :
a. 11\(\frac{2}{5}\) – 3\(\frac{2}{3}\) – \(\frac{11}{3}\)
= \(\frac{57}{5}\) – \(\frac{11}{3}\) – \(\frac{11}{3}\)
= \(\frac{57}{5}\) – \(\frac{22}{3}\)
lcm of 5 and 3 is 15 .
= \(\frac{171}{15}\) – \(\frac{110}{15}\)
= \(\frac{61}{15}\) = 4\(\frac{1}{15}\)

b. 11\(\frac{7}{8}\) + 3\(\frac{1}{5}\) – x = 15
= \(\frac{95}{8}\) + \(\frac{16}{5}\) – x = 15
x = \(\frac{95}{8}\) + \(\frac{16}{5}\) – 15
lcm is 40 .
x = \(\frac{475}{40}\) + \(\frac{128}{40}\) – \(\frac{600}{40}\)
x = \(\frac{3}{40}\)

c. \(\frac{5}{12}\) – y + \(\frac{5}{4}\) = \(\frac{2}{3}\)
= \(\frac{5}{12}\) – y + \(\frac{5}{4}\) = \(\frac{2}{3}\)
lcm of 12 , 4 and 3 is 12 .
y = \(\frac{5}{12}\)  + \(\frac{15}{12}\) – \(\frac{8}{12}\)
y = \(\frac{12}{12}\) = 1
y = 1.

d. x – 30 – 7\(\frac{1}{4}\) = 21\(\frac{2}{3}\) = x – 30 – \(\frac{29}{4}\) = \(\frac{65}{3}\)
lcm of 4 and 3 is 12.
x = \(\frac{260}{12}\) + \(\frac{360}{12}\) + \(\frac{87}{12}\)
x = \(\frac{707}{12}\) = 58\(\frac{11}{12}\) .

e. \(\frac{24}{5}\) + y + \(\frac{8}{7}\) = 9
y = 9 – \(\frac{24}{5}\)  – \(\frac{8}{7}\)
lcm of 5 and 7 is 35 .
y = \(\frac{315}{35}\)  – \(\frac{168}{35}\)  – \(\frac{40}{35}\)
y =  \(\frac{107}{5}\) = 21\(\frac{2}{5}\)

f. 11.1 + 3 \(\frac{1}{10}\) – x = \(\frac{99}{10}\)
11.1 + \(\frac{31}{10}\) –  \(\frac{99}{10}\) = x
11.1  – \(\frac{31}{10}\) = x
11.1  – 3.1 = x
8.0 = x

Question 3.
DeAngelo needs 100 lb of garden soil to landscape a building. In the company’s storage area, he finds 2 cases holding 24\(\frac{3}{4}\) lb of garden soil each, and a third case holding 19\(\frac{3}{8}\) lb. How much gardening soil does DeAngelo still need in order to do the job?
Answer:
Fraction of soil needed to landscape a building = 100lb .
Fraction of 2 cases holding soil = 2 × 24\(\frac{3}{4}\) =  \(\frac{99}{2}\) lb
Fraction of third case holding soil = 19\(\frac{3}{8}\) = \(\frac155}{8}\) lb
Fraction of cases holding capacity = \(\frac{99}{2}\) lb + \(\frac{155}{8}\) lb  = \(\frac{396}{8}\) + \(\frac{155}{8}\) = \(\frac{551}{8}\) = 68 \(\frac{7}{8}\) .
Fraction of gardening soil still need to do = 100 – 68 \(\frac{7}{8}\) = \(\frac{800}{8}\) – \(\frac{551}{8}\) = \(\frac{249}{8}\) = 31\(\frac{1}{8}\) .
Therefore Fraction of gardening soil still need to do = 31\(\frac{1}{8}\) .

Question 4.
Volunteers helped clean up 8.2 kg of trash in one neighborhood and 11\(\frac{1}{2}\) kg in another. They sent 1\(\frac{1}{4}\) kg to be recycled and threw the rest away. How many kilograms of trash did they throw away?
Answer:
Fraction of trash cleaned up in one neighborhood = 8.2 kg
Fraction of trash cleaned up in another neighborhood = 11\(\frac{1}{2}\) = \(\frac{23}{2}\) kg
Fraction of capacity of recycled = 1\(\frac{1}{4}\) = \(\frac{5}{4}\) = 1.25 kg
Total trash cleaned up = 8.2 + \(\frac{23}{2}\) = \(\frac{16.4}{2}\) + \(\frac{23}{2}\) = \(\frac{39.4}{2}\) = 19.7 kg .
Fraction of trash threw = 19.7 – 1.25kg = 18.45 kg
Therefore fraction of trash threw = 18.45 kg .

Eureka Math Grade 5 Module 3 Lesson 14 Exit Ticket Answer Key

Fill in the blank to make the statement true.
Question 1.
1\(\frac{3}{4}\) + \(\frac{1}{6}\) + __________ = 7\(\frac{1}{2}\)
Answer:
1\(\frac{3}{4}\) + \(\frac{1}{6}\) + \(\frac{67}{12}\)  = 7\(\frac{1}{2}\)
Explanation :
1\(\frac{3}{4}\) + \(\frac{1}{6}\) + x = 7\(\frac{1}{2}\)
x = \(\frac{15}{2}\) – \(\frac{7}{4}\) – \(\frac{1}{6}\)
lcm of 2 , 4 and 6 is 12 .
x = \(\frac{90}{12}\) – \(\frac{21}{12}\) – \(\frac{2}{12}\)
x = \(\frac{67}{12}\)

Question 2.
8\(\frac{4}{5}\) – \(\frac{2}{3}\) – _____________ = 3\(\frac{1}{10}\)
Answer:
8\(\frac{4}{5}\) – \(\frac{2}{3}\) – _____________ = 3\(\frac{1}{10}\)
Explanation :
8\(\frac{4}{5}\) – \(\frac{2}{3}\) – y = 3\(\frac{1}{10}\)
y = \(\frac{44}{5}\) – \(\frac{2}{3}\) – \(\frac{31}{10}\)
lcm of 5 , 3 and 10 is 30.
y = \(\frac{264}{30}\) – \(\frac{20}{30}\) – \(\frac{186}{30}\)
y = \(\frac{58}{30}\) = 1\(\frac{28}{30}\) .

Eureka Math Grade 5 Module 3 Lesson 14 Homework Answer Key

Question 1.
Rearrange the terms so that you can add or subtract mentally. Then, solve.
a. 1\(\frac{3}{4}\) + \(\frac{1}{2}\) + \(\frac{1}{4}\) +\(\frac{1}{2}\)
b. 3\(\frac{1}{6}\) – \(\frac{3}{4}\) + \(\frac{5}{6}\)
c. 5\(\frac{5}{8}\) – 2\(\frac{6}{7}\) – \(\frac{2}{7}\) – \(\frac{5}{8}\)
d. \(\frac{7}{9}\) + \(\frac{1}{2}\) – \(\frac{3}{2}\) + \(\frac{2}{9}\)
Answer:
a. 1\(\frac{3}{4}\) + \(\frac{1}{2}\) + \(\frac{1}{4}\) +\(\frac{1}{2}\) = 3
b. 3\(\frac{1}{6}\) – \(\frac{3}{4}\) + \(\frac{5}{6}\) = 3\(\frac{1}{4}\)
c. 5\(\frac{5}{8}\) – 2\(\frac{6}{7}\) – \(\frac{2}{7}\) – \(\frac{5}{8}\) =1\(\frac{6}{7}\)
d. \(\frac{7}{9}\) + \(\frac{1}{2}\) – \(\frac{3}{2}\) + \(\frac{2}{9}\) = 0
Explanation :
a. 1\(\frac{3}{4}\) + \(\frac{1}{2}\) + \(\frac{1}{4}\) +\(\frac{1}{2}\)
= \(\frac{7}{4}\) + \(\frac{1}{2}\) + \(\frac{1}{4}\) +\(\frac{1}{2}\)
= \(\frac{7}{4}\)+ \(\frac{1}{4}\) +\(\frac{1}{2}\) + \(\frac{1}{2}\)
= \(\frac{8}{4}\) + \(\frac{2}{2}\)
= 2 + 1
=3

b. 3\(\frac{1}{6}\) – \(\frac{3}{4}\) + \(\frac{5}{6}\)
= \(\frac{19}{6}\)+ \(\frac{5}{6}\) – \(\frac{3}{4}\)
= \(\frac{24}{6}\) – \(\frac{3}{4}\)
lcm of 6 and 4 is 24 .
= \(\frac{96}{24}\) – \(\frac{18}{24}\)
= \(\frac{78}{24}\)
= \(\frac{13}{4}\)
= 3\(\frac{1}{4}\)

c. 5\(\frac{5}{8}\) – 2\(\frac{6}{7}\) – \(\frac{2}{7}\) – \(\frac{5}{8}\)
= \(\frac{45}{8}\) – \(\frac{5}{8}\) – \(\frac{20}{7}\) – \(\frac{2}{7}\)
= \(\frac{40}{8}\) – \(\frac{22}{7}\)
= \(\frac{280}{56}\) – \(\frac{176}{56}\)
= \(\frac{104}{56}\)
= \(\frac{13}{7}\)
=1\(\frac{6}{7}\)

d. \(\frac{7}{9}\) + \(\frac{1}{2}\) – \(\frac{3}{2}\) + \(\frac{2}{9}\)
= \(\frac{7}{9}\) + \(\frac{2}{9}\) + \(\frac{1}{2}\) – \(\frac{3}{2}\)
= \(\frac{9}{9}\) – \(\frac{2}{2}\)
= 1- 1
= 0

Question 2.
Fill in the blank to make the statement true.
a. 7\(\frac{3}{4}\) – 1\(\frac{2}{7}\) – \(\frac{3}{2}\) = ________
b. 9\(\frac{5}{6}\) + 1\(\frac{1}{4}\) + ________ = 14
c. \(\frac{7}{10}\) – _______ + \(\frac{3}{2}\) = \(\frac{6}{5}\)
d. ________ – 20 – 3\(\frac{1}{4}\) = 14\(\frac{5}{8}\)
e. \(\frac{17}{3}\) + ________ + \(\frac{5}{2}\) = 10\(\frac{4}{5}\)
f. 23.1 + 1\(\frac{7}{10}\) – ________= \(\frac{66}{10}\)
Answer:
a. 7\(\frac{3}{4}\) – 1\(\frac{2}{7}\) – \(\frac{3}{2}\) = = 4\(\frac{27}{28}\)
b. 9\(\frac{5}{6}\) + 1\(\frac{1}{4}\) + 2\(\frac{11}{12}\) = 14
c. \(\frac{7}{10}\) – 1 + \(\frac{3}{2}\) = \(\frac{6}{5}\)
d. 34\(\frac{5}{8}\)  – 20 – 3\(\frac{1}{4}\) = 14\(\frac{5}{8}\)
e. \(\frac{17}{3}\) + 2 \(\frac{19}{30}\) + \(\frac{5}{2}\) = 10\(\frac{4}{5}\)
f. 23.1 + 1\(\frac{7}{10}\) – 18.2 = \(\frac{66}{10}\)

Explanation :
a. 7\(\frac{3}{4}\) – 1\(\frac{2}{7}\) – \(\frac{3}{2}\)
= \(\frac{31}{4}\) – \(\frac{9}{7}\) – \(\frac{3}{2}\)
lcm of 4 , 7 and 2 is 28
= \(\frac{217}{28}\) – \(\frac{36}{28}\) – \(\frac{42}{28}\)
= \(\frac{139}{28}\)
= 4\(\frac{27}{28}\)
b. 9\(\frac{5}{6}\) + 1\(\frac{1}{4}\) + x = 14
x = 14 – \(\frac{59}{6}\) – \(\frac{5}{4}\)
x = \(\frac{168}{12}\) – \(\frac{118}{12}\) – \(\frac{15}{12}\)
x = \(\frac{35}{12}\)
x = 2\(\frac{11}{12}\)

c. \(\frac{7}{10}\) – y + \(\frac{3}{2}\) = \(\frac{6}{5}\)
y = \(\frac{7}{10}\) + \(\frac{3}{2}\) – \(\frac{6}{5}\)
lcm of 10, 2 and 5 is 10 .
y = \(\frac{7}{10}\) + \(\frac{15}{10}\) – \(\frac{12}{10}\)
y = \(\frac{10}{10}\)
y = 1.

d. x – 20 – 3\(\frac{1}{4}\) = 14\(\frac{5}{8}\)
x = 14\(\frac{5}{8}\) + 20 + 3\(\frac{1}{4}\)
x = \(\frac{117}{8}\) + 20 + \(\frac{13}{4}\)
lcm is 8
x = \(\frac{117}{8}\) + \(\frac{26}{8}\) + \(\frac{160}{8}\)
x = \(\frac{277}{8}\)
x = 34\(\frac{5}{8}\)

e. \(\frac{17}{3}\) + y + \(\frac{5}{2}\) = 10\(\frac{4}{5}\)
y = \(\frac{54}{5}\) – \(\frac{17}{3}\) – \(\frac{5}{2}\)
lcm of 5 , 3 and 2 is 30 .
y = \(\frac{324}{30}\) – \(\frac{170}{30}\) – \(\frac{75}{30}\)
y = \(\frac{79}{30}\)
y = 2 \(\frac{19}{30}\) .

f. 23.1 + 1\(\frac{7}{10}\) – y = \(\frac{66}{10}\)
y = 23.1 + \(\frac{17}{10}\) – \(\frac{66}{10}\)
y = 23.1 – \(\frac{49}{10}\)
y = 23.1 – 4.9
y = 18.2

Question 3.
Laura bought 8\(\frac{3}{10}\) yd of ribbon. She used 1\(\frac{2}{5}\) yd to tie a package and 2\(\frac{1}{3}\) yd to make a bow. Joe later gave her 4\(\frac{3}{5}\) yd. How much ribbon does she now have?
Answer:
Fraction of Ribbon brought by Laura = 8\(\frac{3}{10}\) yd = \(\frac{83}{10}\)
Fraction of Ribbon used to tie a package = 1\(\frac{2}{5}\) yd
Fraction of Ribbon used to make a bow = 2\(\frac{1}{3}\) yd
Fraction of ribbon used by Laura = 1\(\frac{2}{5}\) yd + 2\(\frac{1}{3}\) yd = \(\frac{7}{5}\) + \(\frac{7}{3}\) = \(\frac{21}{15}\) + \(\frac{35}{15}\) = \(\frac{56}{15}\)
Fraction of ribbon left with laura = \(\frac{83}{10}\) – \(\frac{56}{15}\) = \(\frac{249}{10}\) + \(\frac{112}{15}\) = \(\frac{137}{15}\)= 9\(\frac{2}{15}\) .
Fraction of ribbon joe gave to laura = 4\(\frac{3}{5}\) = \(\frac{23}{5}\) yd.
Fraction of ribbon with laura now = \(\frac{137}{15}\) + \(\frac{23}{5}\) =\(\frac{137}{15}\) + \(\frac{69}{15}\) = \(\frac{206}{15}\) = 13\(\frac{11}{15}\) .
Therefore Fraction of ribbon with laura now = 13\(\frac{11}{15}\) .

Question 4.
Mia bought 10\(\frac{1}{9}\) lb of flour. She used 2\(\frac{3}{4}\) lb of flour to bake banana cakes and some to bake chocolate cakes. After baking all the cakes, she had 3\(\frac{5}{6}\) lb of flour left. How much flour did she use to bake the chocolate cakes?
Answer:
Fraction of Flour with Mia = 10\(\frac{1}{9}\) = \(\frac{91}{9}\) lb
Fraction of Flour used to bake bake banana cakes = 2\(\frac{3}{4}\) = \(\frac{11}{4}\) lb
Fraction of flour left = 3\(\frac{5}{6}\) = \(\frac{23}{6}\) lb
Fraction of flour used for chocolate cake = \(\frac{91}{9}\) – \(\frac{11}{4}\) – \(\frac{23}{6}\) =
\(\frac{364}{36}\) – \(\frac{99}{36}\) – \(\frac{138}{36}\) = \(\frac{127}{36}\) = 3\(\frac{17}{36}\).
Therefore, Fraction of flour used for chocolate cake = 3\(\frac{17}{36}\).

Eureka Math Grade 5 Module 3 Lesson 13 Answer Key

Engage NY Eureka Math 5th Grade Module 3 Lesson 13 Answer Key

Eureka Math Grade 5 Module 3 Lesson 13 Problem Set Answer Key

Question 1.
Are the following expressions greater than or less than 1? Circle the correct answer.
a. \(\frac{1}{2}\) + \(\frac{2}{7}\)    greater than 1    less than 1

b. \(\frac{5}{8}\) + \(\frac{3}{5}\)   greater than 1     less than 1

c. 1\(\frac{1}{4}\) – \(\frac{1}{3}\)   greater than 1     less than 1

d. 3\(\frac{5}{8}\) – 2\(\frac{5}{9}\)  greater than1     less than 1
Answer:
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-13-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-13-Problem-Set-Answer-Key-Question-1
Explanation :
a. \(\frac{1}{2}\) + \(\frac{2}{7}\) =
Lcm of 2 and 7 are 14 .
\(\frac{7}{14}\) + \(\frac{4}{14}\) = \(\frac{11}{14}\)

b. \(\frac{5}{8}\) + \(\frac{3}{5}\)
lcm of 8 and 5 is 40.
\(\frac{25}{40}\) + \(\frac{24}{40}\)   = \(\frac{49}{40}\) = 1\(\frac{9}{40}\)

c. 1\(\frac{1}{4}\) – \(\frac{1}{3}\)   = \(\frac{5}{4}\) – \(\frac{1}{3}\)
lcm of 4 and 3 is 12.
\(\frac{15}{12}\) – \(\frac{4}{12}\) = \(\frac{11}{12}\)

d. 3\(\frac{5}{8}\) – 2\(\frac{5}{9}\)  = \(\frac{29}{8}\) – \(\frac{23}{9}\)
lcm of 8 and 9 is 72 .
\(\frac{261}{72}\) – \(\frac{184}{72}\)  = \(\frac{77}{72}\) = 1\(\frac{5}{72}\) .

Question 2.
Are the following expressions greater than or less than \(\frac{1}{2}\) ? Circle the correct answer.
a. \(\frac{1}{4}\) + \(\frac{2}{3}\)      greater than \(\frac{1}{2}\)          less than \(\frac{1}{2}\)

b.\(\frac{3}{7}\) – \(\frac{1}{8}\)        greater than \(\frac{1}{2}\)          less than \(\frac{1}{2}\)

c. 1\(\frac{1}{7}\) – \(\frac{7}{8}\)       greater than \(\frac{1}{2}\)        less than \(\frac{1}{2}\)

d. \(\frac{3}{7}\) + \(\frac{2}{6}\)        greater than \(\frac{1}{2}\)        less than \(\frac{1}{2}\)
Answer:
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-13-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-13-Problem-Set-Answer-Key-Question-2
Explanation :
a. \(\frac{1}{4}\) + \(\frac{2}{3}\)
lcm of 4 and 3 is 12 .
\(\frac{3}{12}\) + \(\frac{8}{12}\) =  \(\frac{11}{12}\) greater than \(\frac{1}{2}\)

b.\(\frac{3}{7}\) – \(\frac{1}{8}\)
lcm of 7 and 8 is 56.
\(\frac{24}{56}\) – \(\frac{7}{56}\) =  \(\frac{17}{56}\) less than \(\frac{1}{2}\)

c. 1\(\frac{1}{7}\) – \(\frac{7}{8}\) = \(\frac{8}{7}\) – \(\frac{7}{8}\)
lcm of 7 and 8 is 56.
\(\frac{64}{56}\) – \(\frac{49}{56}\) = \(\frac{15}{56}\) less than \(\frac{1}{2}\)

d. \(\frac{3}{7}\) + \(\frac{2}{6}\)
lcm of 7 and 6 is 42.
\(\frac{18}{42}\) + \(\frac{14}{42}\)  = \(\frac{32}{42}\) = \(\frac{16}{21}\) greater than \(\frac{1}{2}\) .

Question 3.
Use > , < , or = to make the following statements true.
a. 5\(\frac{2}{3}\) + 3\(\frac{3}{4}\) _______ 8\(\frac{2}{3}\)
b. 4\(\frac{5}{8}\) – 3\(\frac{2}{5}\) _______ 1\(\frac{5}{8}\) + \(\frac{2}{5}\)
c. 5\(\frac{1}{2}\) + 1\(\frac{3}{7}\) _______ 6 + \(\frac{13}{14}\)
d. 15\(\frac{4}{7}\) – 11\(\frac{2}{5}\) _______ 4\(\frac{4}{7}\) + \(\frac{2}{5}\)
Answer:
a. 5\(\frac{2}{3}\) + 3\(\frac{3}{4}\) = 8\(\frac{2}{3}\)
b. 4\(\frac{5}{8}\) – 3\(\frac{2}{5}\) < 1\(\frac{5}{8}\) + \(\frac{2}{5}\)
c. 5\(\frac{1}{2}\) + 1\(\frac{3}{7}\) = 6 + \(\frac{13}{14}\)
d. 15\(\frac{4}{7}\) – 11\(\frac{2}{5}\) > 4\(\frac{4}{7}\) + \(\frac{2}{5}\)
Explanation :
a. 5\(\frac{2}{3}\) + 3\(\frac{3}{4}\) = \(\frac{17}{3}\) + \(\frac{12}{4}\)
lcm of 3 and 4 is 12.
\(\frac{68}{12}\) + \(\frac{36}{12}\) = \(\frac{104}{12}\) = \(\frac{26}{3}\) =8\(\frac{2}{3}
\)

b. 4\(\frac{5}{8}\) – 3\(\frac{2}{5}\) = \(\frac{37}{8}\) – \(\frac{17}{5}\)
lcm of 8 and 5 is 40 .
\(\frac{185}{40}\) – \(\frac{136}{40}\) = \(\frac{49}{40}\) =1\(\frac{9}{40}\)
1\(\frac{5}{8}\) + \(\frac{2}{5}\) = \(\frac{13}{8}\) + \(\frac{2}{5}\)
lcm of 8 and 5 is 40 .
\(\frac{65}{40}\) + \(\frac{16}{5}\) = \(\frac{81}{40}\) = 2\(\frac{1}{40}\)

c. 5\(\frac{1}{2}\) + 1\(\frac{3}{7}\) = \(\frac{11}{2}\) + \(\frac{10}{7}\)
lcm of 2 and 7 is 14.
\(\frac{77}{14}\) + \(\frac{20}{14}\) = \(\frac{97}{14}\) = 6 \(\frac{13}{14}\)
6 + \(\frac{13}{14}\) =\(\frac{84}{14}\) + \(\frac{13}{14}\) = \(\frac{97}{14}\) = 6 \(\frac{13}{14}\)

d. 15\(\frac{4}{7}\) – 11\(\frac{2}{5}\) =\(\frac{109}{7}\) – \(\frac{57}{5}\)
lcm of 7 and 5 is 35 .
\(\frac{545}{35}\) – \(\frac{399}{35}\) = \(\frac{944}{35}\) = 26\(\frac{34}{35}\) .
4\(\frac{4}{7}\) + \(\frac{2}{5}\) = \(\frac{32}{7}\) + \(\frac{2}{5}\)
lcm of 7 and 5 is 35 .
\(\frac{160}{35}\) + \(\frac{14}{35}\) = \(\frac{174}{35}\)= 4\(\frac{34}{35}\)

Question 4.
Is it true that 4\(\frac{3}{5}\) – 3\(\frac{2}{3}\) = 1 + \(\frac{3}{5}\) + \(\frac{2}{3}\)? Prove your answer.
Answer:
No it is wrong
4\(\frac{3}{5}\) – 3\(\frac{2}{3}\) < 1 + \(\frac{3}{5}\) + \(\frac{2}{3}\)
Explanation :
4\(\frac{3}{5}\) – 3\(\frac{2}{3}\) = \(\frac{23}{5}\) – \(\frac{11}{3}\)
lcm of 5 and 3 is 15 .
\(\frac{69}{15}\) – \(\frac{55}{15}\) = \(\frac{14}{15}\)

1 + \(\frac{3}{5}\) + \(\frac{2}{3}\)
lcm of 5 and 3 is 15 .
\(\frac{15}{15}\) + \(\frac{9}{15}\) + \(\frac{10}{15}\) = latex]\frac{34}{15}[/latex] = 2latex]\frac{4}{15}[/latex] .

Question 5.
Jackson needs to be 1\(\frac{3}{4}\) inches taller in order to ride the roller coaster. Since he can’t wait, he puts on a pair of boots that add 1\(\frac{1}{6}\) inches to his height and slips an insole inside to add another \(\frac{1}{8}\) inch to his height. Will this make Jackson appear tall enough to ride the roller coaster?
Answer:
Fraction of Height required to ride a roller coaster for Jackson = 1\(\frac{3}{4}\) inches.
Fraction of his height = 1\(\frac{1}{6}\) inches = \(\frac{7}{6}\)
Fraction of his boots length = \(\frac{1}{8}\) inches
Total fraction of his height with boots = \(\frac{7}{6}\) + \(\frac{1}{8}\) = \(\frac{28}{24}\) + \(\frac{3}{24}\) = \(\frac{31}{24}\) = 1\(\frac{7}{24}\) .
1\(\frac{3}{4}\) = multiply by 6 both numerator and denominator = 1\(\frac{18}{24}\)
therefore, 1\(\frac{18}{24}\) > is greater than 1\(\frac{7}{24}\)  So, he is not taller enough to ride roller coaster .
So, he cant ride the roller coaster .

Question 6.
A baker needs 5 lb of butter for a recipe. She found 2 portions that each weigh 1\(\frac{1}{6}\) lb and a portion that weighs 2\(\frac{2}{7}\) lb. Does she have enough butter for her recipe?
Answer:
Fraction of butter required for a recipe = 5 lb
Fraction of 2 portions that weigh = 2 × \(\frac{7}{6}\) lb = \(\frac{7}{3}\)
Fraction of portions that weighs = 2\(\frac{2}{7}\) lb. = \(\frac{16}{7}\) lb.
Fraction of butter of portions = \(\frac{7}{3}\) + \(\frac{16}{7}\) = \(\frac{49}{21}\) + \(\frac{48}{21}\) = \(\frac{97}{21}\) = 4\(\frac{13}{21}\)
Therefore, she doesnot have enough butter for the recipe =  4\(\frac{13}{21}\)

Eureka Math Grade 5 Module 3 Lesson 13 Exit Ticket Answer Key

Question 1.
Circle the correct answer.
a. \(\frac{1}{2}\) +\(\frac{5}{12}\)        greater than 1          less than 1

b. 2\(\frac{7}{8}\) – 1\(\frac{7}{9}\)      greater than 1         less than 1

c. 1\(\frac{1}{12}\) – \(\frac{7}{10}\)     greater than \(\frac{1}{2}\)      less than \(\frac{1}{2}\)

d. \(\frac{3}{7}\) + \(\frac{1}{8}\)         greater than \(\frac{1}{2}\)      less than \(\frac{1}{2}\)
Answer:
a. \(\frac{1}{2}\) +\(\frac{5}{12}\) = \(\frac{11}{12}\)
b.2\(\frac{7}{8}\) – 1\(\frac{7}{9}\)  = 8\(\frac{7}{9}\)
c. 1\(\frac{1}{12}\) – \(\frac{7}{10}\) = \(\frac{23}{60}\)
d. \(\frac{3}{7}\) + \(\frac{1}{8}\) = \(\frac{31}{56}\)
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-13-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-13-Exit-Ticket-Answer-Key-Question-1
Explanation :
a. \(\frac{1}{2}\) +\(\frac{5}{12}\)
lcm of 2 and 12 is 12.
\(\frac{6}{12}\) +\(\frac{5}{12}\) = \(\frac{11}{12}\) less than 1
b. 2\(\frac{7}{8}\) – 1\(\frac{7}{9}\)   = \(\frac{23}{8}\) – \(\frac{16}{9}\)
lcm of 8 and 9 is 72.
\(\frac{207}{72}\) – \(\frac{128}{9}\)  = \(\frac{79}{9}\) =  8\(\frac{7}{9}\) greater than1.
c. . 1\(\frac{1}{12}\) – \(\frac{7}{10}\)  = . \(\frac{13}{12}\) – \(\frac{7}{10}\)
lcm of 12 and 10 is 60.
. \(\frac{65}{60}\) – \(\frac{42}{60}\)  = \(\frac{23}{60}\) less than 1 .
d. \(\frac{3}{7}\) + \(\frac{1}{8}\)
lcm of 7 and 8 is 56
\(\frac{24}{56}\) + \(\frac{7}{56}\) = \(\frac{31}{56}\) = less than 1 .

Question 2.
Use > , < , or = to make the following statement true.
4\(\frac{4}{5}\) + 3\(\frac{2}{3}\) < 8\(\frac{1}{2}\)
Answer:
4\(\frac{4}{5}\) + 3\(\frac{2}{3}\) < 8\(\frac{1}{2}\)
Explanation :
4\(\frac{4}{5}\) + 3\(\frac{2}{3}\) = \(\frac{24}{5}\) + \(\frac{11}{3}\)
lcm of 5 and 3 is 15 .
\(\frac{72}{15}\) + \(\frac{55}{15}\) = \(\frac{127}{15}\) = 8 \(\frac{7}{15}\)

Eureka Math Grade 5 Module 3 Lesson 13 Homework Answer Key

Question 1.
Are the following expressions greater than or less than 1? Circle the correct answer.
a. \(\frac{1}{2}\) + \(\frac{4}{9}\)        greater than 1          less than 1

b. \(\frac{5}{8}\) + \(\frac{3}{5}\)        greater than 1          less than 1

c. 1\(\frac{1}{5}\) – \(\frac{1}{3}\)       greater than 1           less than 1

d. 4\(\frac{3}{5}\) – 3\(\frac{3}{4}\)     greater than 1           less than 1
Answer:
a. \(\frac{1}{2}\) + \(\frac{4}{9}\) = \(\frac{17}{18}\) less than 1.
b. \(\frac{5}{8}\) + \(\frac{3}{5}\)  = 1 \(\frac{9}{40}\) greater than 1 .
c. 1\(\frac{1}{5}\) – \(\frac{1}{3}\) = \(\frac{13}{15}\)  less than 1.
d. 4\(\frac{3}{5}\) – 3\(\frac{3}{4}\) = \(\frac{17}{20}\) less than 1 .
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-13-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-13-Home-Work-Answer-Key-Question-1
Explanation :
a. \(\frac{1}{2}\) + \(\frac{4}{9}\)
lcm of 2 and 9 is 18.
\(\frac{9}{18}\) + \(\frac{8}{18}\) = \(\frac{17}{18}\)
b. \(\frac{5}{8}\) + \(\frac{3}{5}\)
lcm of 8 and 5 is 40 .
\(\frac{25}{40}\) + \(\frac{24}{40}\) = \(\frac{49}{40}\) =1 \(\frac{9}{40}\)
c. 1\(\frac{1}{5}\) – \(\frac{1}{3}\) = \(\frac{6}{5}\) – \(\frac{1}{3}\)
lcm of 5 and 3 is 15 .
\(\frac{18}{15}\) – \(\frac{5}{15}\)  = \(\frac{13}{15}\)
d. 4\(\frac{3}{5}\) – 3\(\frac{3}{4}\) = \(\frac{23}{5}\) – \(\frac{15}{4}\)
lcm of 5 and 4 is 20 .
\(\frac{92}{20}\) – \(\frac{75}{20}\) = \(\frac{17}{20}\)

Question 2.
Are the following expressions greater than or less than \(\frac{1}{2}\)? Circle the correct answer.
a. \(\frac{1}{5}\) + \(\frac{1}{4}\)        greater than \(\frac{1}{2}\)       less than \(\frac{1}{2}\)

b. \(\frac{6}{7}\) – \(\frac{1}{6}\)         greater than \(\frac{1}{2}\)        less than \(\frac{1}{2}\)

c. 1\(\frac{1}{7}\) – \(\frac{5}{6}\)        greater than \(\frac{1}{2}\)       less than \(\frac{1}{2}\)

d. \(\frac{4}{7}\) + \(\frac{1}{8}\)         greater than \(\frac{1}{2}\)      less than \(\frac{1}{2}\)
Answer:
a. \(\frac{1}{5}\) + \(\frac{1}{4}\) = \(\frac{9}{20}\) less than \(\frac{1}{2}\)
b. \(\frac{6}{7}\) – \(\frac{1}{6}\)  = \(\frac{29}{42}\) greater than \(\frac{1}{2}\)
c. 1\(\frac{1}{7}\) – \(\frac{5}{6}\) = \(\frac{13}{42}\) less than \(\frac{1}{2}\)
d. \(\frac{4}{7}\) + \(\frac{1}{8}\)  =   \(\frac{39}{56}\) greater than \(\frac{1}{2}\)
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-13-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-13-Home-Work-Answer-Key-Question-2

Explanation :
a. \(\frac{1}{5}\) + \(\frac{1}{4}\)
lcm of 5 and 4 is 20 .
\(\frac{4}{20}\) + \(\frac{5}{20}\) = \(\frac{9}{20}\)
b. \(\frac{6}{7}\) – \(\frac{1}{6}\)
lcm of 7 and 6 is 42 .
\(\frac{36}{42}\) – \(\frac{7}{42}\) = \(\frac{29}{42}\)
c. 1\(\frac{1}{7}\) – \(\frac{5}{6}\) = \(\frac{8}{7}\) – \(\frac{5}{6}\)
lcm of 7 and 6 is 42 .
\(\frac{48}{42}\) – \(\frac{35}{42}\) = \(\frac{13}{42}\)
d. \(\frac{4}{7}\) + \(\frac{1}{8}\)
lcm of 7 and 8 is 56 .
\(\frac{32}{56}\) + \(\frac{7}{56}\) = \(\frac{39}{56}\)

Question 3.
Use > , < , or = to make the following statements true.
a. 5\(\frac{4}{5}\) + 2\(\frac{2}{3}\) _______ 8\(\frac{3}{4}\)
b. 3\(\frac{4}{7}\) – 2\(\frac{3}{5}\) _______ 1\(\frac{4}{7}\) + \(\frac{3}{5}\)
c. 4\(\frac{1}{2}\) + 1\(\frac{4}{9}\) _______ 5 + \(\frac{13}{18}\)
d. 10\(\frac{3}{8}\) – 7\(\frac{3}{5}\) _______ 3\(\frac{3}{8}\) + \(\frac{3}{5}\)
Answer:
a. 5\(\frac{4}{5}\) + 2\(\frac{2}{3}\) < 8\(\frac{3}{4}\)
b. 3\(\frac{4}{7}\) – 2\(\frac{3}{5}\) < 1\(\frac{4}{7}\) + \(\frac{3}{5}\)
c. 4\(\frac{1}{2}\) + 1\(\frac{4}{9}\) < 5 + \(\frac{13}{18}\)
d. 10\(\frac{3}{8}\) – 7\(\frac{3}{5}\) > 3\(\frac{3}{8}\) + \(\frac{3}{5}\)
Explanation :
a. 5\(\frac{4}{5}\) + 2\(\frac{2}{3}\) = \(\frac{29}{5}\) + \(\frac{8}{3}\)
lcm of 5 and 3 is 15 .
\(\frac{57}{15}\) + \(\frac{40}{15}\) = \(\frac{97}{15}\) = 6\(\frac{7}{15}\) .

b. 3\(\frac{4}{7}\) – 2\(\frac{3}{5}\) = \(\frac{25}{7}\) – \(\frac{13}{5}\) .
lcm of 7 and 5 is 35 .
\(\frac{125}{35}\) – \(\frac{91}{35}\) = \(\frac{34}{35}\)
1\(\frac{4}{7}\) + \(\frac{3}{5}\) = \(\frac{11}{7}\) + \(\frac{3}{5}\)
lcm of 5 and 7 is 35 .
\(\frac{55}{35}\) + \(\frac{21}{35}\) = \(\frac{76}{35}\) = 2 \(\frac{6}{35}\)

c. 4\(\frac{1}{2}\) + 1\(\frac{4}{9}\) = \(\frac{9}{2}\) + \(\frac{13}{9}\)
lcm of 2 and 9 is 18 .
4\(\frac{1}{2}\) + 1\(\frac{4}{9}\)
5 + \(\frac{13}{18}\) = \(\frac{90}{18}\) + \(\frac{13}{18}\) = \(\frac{103}{18}\) = 5\(\frac{13}{18}\)

d. 10\(\frac{3}{8}\) – 7\(\frac{3}{5}\) = \(\frac{83}{8}\) – \(\frac{38}{5}\)
lcm of 8 and 5 is 40 .
\(\frac{415}{40}\) – \(\frac{304}{40}\) = \(\frac{311}{40}\) = 7\(\frac{31}{40}\)
3\(\frac{3}{8}\) + \(\frac{3}{5}\) = \(\frac{27}{8}\) + \(\frac{3}{5}\)
lcm of 8 and 5 is 40 .
\(\frac{135}{40}\) + \(\frac{24}{40}\) = \(\frac{159}{40}\)= 3\(\frac{39}{40}\)

Question 4.
Is it true that 5\(\frac{2}{3}\) – 3\(\frac{3}{4}\) = 1 + \(\frac{2}{3}\) + \(\frac{3}{4}\)? Prove your answer.
Answer:
It is not true .
5\(\frac{2}{3}\) – 3\(\frac{3}{4}\) < 1 + \(\frac{2}{3}\) + \(\frac{3}{4}\)
Explanation :
5\(\frac{2}{3}\) – 3\(\frac{3}{4}\) = \(\frac{17}{3}\) – \(\frac{15}{4}\)
lcm of 3 and 4 is 12.
\(\frac{68}{12}\) – \(\frac{45}{12}\) = \(\frac{23}{12}\) = 1\(\frac{11}{12}\)

1 + \(\frac{2}{3}\) + \(\frac{3}{4}\)
lcm of 3 and 4 is 12 .
\(\frac{12}{12}\) + \(\frac{8}{12}\) + \(\frac{9}{12}\) = \(\frac{29}{12}\) = 2\(\frac{5}{12}\) .

Question 5.
A tree limb hangs 5\(\frac{1}{4}\) feet from a telephone wire. The city trims back the branch before it grows within 2 \(\frac{1}{2}\) feet of the wire. Will the city allow the tree to grow 2\(\frac{3}{4}\) more feet?
Answer:
Fraction of height at which telephone wire is hung = 5\(\frac{1}{4}\) =\(\frac{21}{4}\)  feet
Fraction of height city allow the tree to grow = 2\(\frac{3}{4}\) = \(\frac{11}{4}\) feet .
Fraction of height city trims back the branch before it grows = 2\(\frac{1}{2}\) = \(\frac{5}{2}\) feet
Fraction of height of telephone wire can be hang = \(\frac{21}{4}\)  – \(\frac{11}{4}\)  = \(\frac{10}{4}\) = \(\frac{5}{2}\)
both are equal that means the tree will be trim back .

Question 6.
Mr. Kreider wants to paint two doors and several shutters. It takes 2\(\frac{1}{8}\) gallons of paint to coat each door and 1\(\frac{3}{5}\) gallons of paint to coat all of his shutters. If Mr. Kreider buys three 2-gallon cans of paint, does he have enough to complete the job?
Answer:
Fraction of cost of paint to coat each door = 2\(\frac{1}{8}\) gallons = \(\frac{17}{8}\)
Fraction of cost of paint to coat all his shutters = 1\(\frac{3}{5}\) gallons = \(\frac{8}{5}\)
Fraction of cost to paint 2 doors and shutters = 2 × \(\frac{17}{8}\) + \(\frac{8}{5}\) = \(\frac{17}{4}\) + \(\frac{8}{5}\) = \(\frac{85}{20}\) + \(\frac{32}{20}\) = \(\frac{117}{20}\) = 5latex]\frac{17}{20}[/latex]
Total paint = three 2-gallon cans of paint = 3 × 2 = 6 gallons.
Therefore Kreider doesn’t have sufficient amount of paint .

Eureka Math Grade 5 Module 3 Lesson 12 Answer Key

Engage NY Eureka Math 5th Grade Module 3 Lesson 12 Answer Key

Eureka Math Grade 5 Module 3 Lesson 12 Sprint Answer Key

A
Subtract Fractions with Unlike Units
Engage NY Math 5th Grade Module 3 Lesson 12 Sprint Answer Key 1

Question 1.
\(\frac{2}{4}\) – \(\frac{1}{4}\) =
Answer:
\(\frac{2}{4}\) – \(\frac{1}{4}\) = \(\frac{1}{4}\)

Question 2.
\(\frac{1}{2}\) – \(\frac{1}{4}\) =
Answer:
\(\frac{1}{2}\) – \(\frac{1}{4}\) = \(\frac{1}{4}\)
Explanation :
\(\frac{1}{2}\) – \(\frac{1}{4}\) = \(\frac{2}{4}\) – \(\frac{1}{4}\) = \(\frac{1}{4}\)

Question 3.
\(\frac{2}{6}\) – \(\frac{1}{6}\) =
Answer:
\(\frac{2}{6}\) – \(\frac{1}{6}\) = \(\frac{1}{6}\)

Question 4.
\(\frac{1}{3}\) – \(\frac{1}{6}\) =
Answer:
\(\frac{1}{3}\) – \(\frac{1}{6}\) = \(\frac{1}{6}\)
Explanation :
\(\frac{1}{3}\) – \(\frac{1}{6}\) =\(\frac{2}{6}\) – \(\frac{1}{6}\) = \(\frac{1}{6}\)

Question 5.
\(\frac{2}{8}\) – \(\frac{1}{8}\) =
Answer:
\(\frac{2}{8}\) – \(\frac{1}{8}\) = \(\frac{1}{8}\)

Question 6.
\(\frac{1}{4}\) – \(\frac{1}{8}\) =
Answer:
\(\frac{1}{4}\) – \(\frac{1}{8}\) =\(\frac{1}{8}\)
Explanation :
\(\frac{1}{4}\) – \(\frac{1}{8}\) = \(\frac{2}{8}\) – \(\frac{1}{8}\) = \(\frac{1}{8}\)

Question 7.
\(\frac{6}{8}\) – \(\frac{1}{8}\) =
Answer:
\(\frac{6}{8}\) – \(\frac{1}{8}\) = \(\frac{5}{8}\)

Question 8.
\(\frac{3}{4}\) – \(\frac{1}{8}\) =
Answer:
\(\frac{3}{4}\) – \(\frac{1}{8}\) = \(\frac{5}{8}\)
Explanation :
\(\frac{3}{4}\) – \(\frac{1}{8}\) = \(\frac{6}{8}\) – \(\frac{1}{8}\) = \(\frac{5}{8}\)

Question 9.
\(\frac{3}{4}\) – \(\frac{3}{8}\) =
Answer:
\(\frac{3}{4}\) – \(\frac{3}{8}\) = \(\frac{3}{8}\)
Explanation :
\(\frac{3}{4}\) – \(\frac{3}{8}\) = \(\frac{6}{8}\) – \(\frac{3}{8}\) = \(\frac{3}{8}\)

Question 10.
\(\frac{5}{10}\) – \(\frac{2}{10}\) =
Answer:
\(\frac{5}{10}\) – \(\frac{2}{10}\) = \(\frac{3}{10}\)

Question 11.
\(\frac{1}{2}\) – \(\frac{2}{10}\) =
Answer:
\(\frac{1}{2}\) – \(\frac{2}{10}\) = \(\frac{3}{10}\)
Explanation :
\(\frac{1}{2}\) – \(\frac{2}{10}\) = \(\frac{5}{10}\) – \(\frac{2}{10}\) = \(\frac{3}{10}\)

Question 12.
\(\frac{1}{2}\) – \(\frac{2}{10}\) =
Answer:
\(\frac{1}{2}\) – \(\frac{2}{10}\) = \(\frac{3}{10}\)
Explanation :
\(\frac{1}{2}\) – \(\frac{2}{10}\) = \(\frac{5}{10}\) – \(\frac{2}{10}\) = \(\frac{3}{10}\)

Question 13.
\(\frac{4}{10}\) – \(\frac{1}{10}\) =
Answer:
\(\frac{4}{10}\) – \(\frac{1}{10}\) = \(\frac{3}{10}\)

Question 14.
\(\frac{2}{5}\) – \(\frac{1}{10}\) =
Answer:
\(\frac{2}{5}\) – \(\frac{1}{10}\) = \(\frac{3}{10}\)
Explanation :
\(\frac{2}{5}\) – \(\frac{1}{10}\) = \(\frac{4}{10}\) – \(\frac{1}{10}\) = \(\frac{3}{10}\)

Question 15.
\(\frac{2}{5}\) – \(\frac{3}{10}\) =
Answer:
\(\frac{2}{5}\) – \(\frac{3}{10}\) = \(\frac{1}{10}\)
Explanation :
\(\frac{2}{5}\) – \(\frac{3}{10}\) = \(\frac{4}{10}\) – \(\frac{3}{10}\) = \(\frac{1}{10}\)

Question 16.
\(\frac{6}{10}\) – \(\frac{3}{10}\) =
Answer:
\(\frac{6}{10}\) – \(\frac{3}{10}\) =\(\frac{3}{10}\)

Question 17.
\(\frac{3}{5}\) – \(\frac{3}{10}\) =
Answer:
\(\frac{3}{5}\) – \(\frac{3}{10}\) = \(\frac{3}{10}\)
Explanation :
\(\frac{3}{5}\) – \(\frac{3}{10}\) = \(\frac{6}{10}\) – \(\frac{3}{10}\) = \(\frac{3}{10}\)

Question 18.
\(\frac{3}{5}\) – \(\frac{5}{10}\) =
Answer:
\(\frac{3}{5}\) – \(\frac{5}{10}\) = \(\frac{1}{10}\)
Explanation :
\(\frac{3}{5}\) – \(\frac{5}{10}\) = \(\frac{6}{10}\) – \(\frac{5}{10}\) = \(\frac{1}{10}\)

Question 19.
\(\frac{8}{10}\) – \(\frac{1}{10}\) =
Answer:
\(\frac{8}{10}\) – \(\frac{1}{10}\) =  \(\frac{7}{10}\)

Question 20.
\(\frac{4}{5}\) – \(\frac{1}{10}\) =
Answer:
\(\frac{4}{5}\) – \(\frac{1}{10}\) = \(\frac{7}{10}\)
Explanation :
\(\frac{4}{5}\) – \(\frac{1}{10}\) = \(\frac{8}{10}\) – \(\frac{1}{10}\) =  \(\frac{7}{10}\)

Question 21.
\(\frac{4}{5}\) – \(\frac{5}{10}\) =
Answer:
\(\frac{4}{5}\) – \(\frac{5}{10}\) = \(\frac{3}{10}\)
Explanation :
\(\frac{4}{5}\) – \(\frac{5}{10}\) = \(\frac{8}{10}\) – \(\frac{5}{10}\) =  \(\frac{4}{10}\)

Question 22.
\(\frac{4}{5}\) – \(\frac{5}{10}\) =
Answer:
\(\frac{4}{5}\) – \(\frac{5}{10}\) = \(\frac{3}{10}\)
Explanation :
\(\frac{4}{5}\) – \(\frac{5}{10}\) = \(\frac{8}{10}\) – \(\frac{5}{10}\) =  \(\frac{4}{10}\)

Question 23.
\(\frac{4}{5}\) – \(\frac{7}{10}\) =
Answer:
\(\frac{4}{5}\) – \(\frac{7}{10}\) = \(\frac{1}{10}\)
Explanation :
\(\frac{4}{5}\) – \(\frac{7}{10}\) = \(\frac{8}{10}\) – \(\frac{7}{10}\) =  \(\frac{1}{10}\)

Question 24.
\(\frac{2}{12}\) – \(\frac{1}{12}\) =
Answer:
\(\frac{2}{12}\) – \(\frac{1}{12}\) = \(\frac{1}{12}\)

Question 25.
\(\frac{1}{6}\) – \(\frac{1}{12}\) =
Answer:
\(\frac{1}{6}\) – \(\frac{1}{12}\) =\(\frac{1}{12}\)
Explanation :
\(\frac{1}{6}\) – \(\frac{1}{12}\) = \(\frac{2}{12}\) – \(\frac{1}{12}\) = \(\frac{1}{12}\)

Question 26.
\(\frac{6}{12}\) – \(\frac{1}{12}\) =
Answer:
\(\frac{6}{12}\) – \(\frac{1}{12}\) = \(\frac{5}{12}\)

Question 27.
\(\frac{1}{2}\) – \(\frac{1}{12}\) =
Answer:
\(\frac{1}{2}\) – \(\frac{1}{12}\) =  \(\frac{5}{12}\)
Explanation :
\(\frac{1}{2}\) – \(\frac{1}{12}\) = \(\frac{6}{12}\) – \(\frac{1}{12}\) = \(\frac{5}{12}\)

Question 28.
\(\frac{1}{2}\) – \(\frac{5}{12}\) =
Answer:
\(\frac{1}{2}\) – \(\frac{5}{12}\) =  \(\frac{1}{12}\)
Explanation :
\(\frac{1}{2}\) – \(\frac{5}{12}\) = \(\frac{6}{12}\) – \(\frac{5}{12}\) = \(\frac{1}{12}\)

Question 29.
\(\frac{10}{12}\) – \(\frac{5}{12}\) =
Answer:
\(\frac{10}{12}\) – \(\frac{5}{12}\) = \(\frac{5}{12}\)

Question 30.
\(\frac{5}{6}\) – \(\frac{5}{12}\) =
Answer:
\(\frac{5}{6}\) – \(\frac{5}{12}\) = \(\frac{5}{12}\)
Explanation :
\(\frac{5}{6}\) – \(\frac{5}{12}\) = \(\frac{10}{12}\) – \(\frac{5}{12}\) = \(\frac{5}{12}\)

Question 31.
\(\frac{1}{3}\) – \(\frac{3}{12}\) =
Answer:
\(\frac{1}{3}\) – \(\frac{3}{12}\) = \(\frac{1}{12}\)
Explanation :
\(\frac{1}{3}\) – \(\frac{3}{12}\) = \(\frac{4}{12}\) – \(\frac{3}{12}\) = \(\frac{1}{12}\)

Question 32.
\(\frac{2}{3}\) – \(\frac{1}{12}\) =
Answer:
\(\frac{2}{3}\) – \(\frac{1}{12}\) = \(\frac{7}{12}\)
Explanation :
\(\frac{2}{3}\) – \(\frac{1}{12}\) = \(\frac{8}{12}\) – \(\frac{1}{12}\) = \(\frac{7}{12}\)

Question 33.
\(\frac{2}{3}\) – \(\frac{3}{12}\) =
Answer:
\(\frac{2}{3}\) – \(\frac{3}{12}\) = \(\frac{5}{12}\)
Explanation :
\(\frac{2}{3}\) – \(\frac{3}{12}\) = \(\frac{8}{12}\) – \(\frac{3}{12}\) = \(\frac{5}{12}\)

Question 34.
\(\frac{2}{3}\) – \(\frac{7}{12}\) =
Answer:
\(\frac{2}{3}\) – \(\frac{7}{12}\) = \(\frac{1}{12}\)
Explanation :
\(\frac{2}{3}\) – \(\frac{7}{12}\) = \(\frac{8}{12}\) – \(\frac{7}{12}\) = \(\frac{1}{12}\)

Question 35.
\(\frac{1}{4}\) – \(\frac{2}{12}\) =
Answer:
\(\frac{1}{4}\) – \(\frac{2}{12}\) =  \(\frac{1}{12}\)
Explanation :
\(\frac{1}{4}\) – \(\frac{2}{12}\) = \(\frac{3}{12}\) – \(\frac{2}{12}\) = \(\frac{1}{12}\)

Question 36.
\(\frac{1}{5}\) – \(\frac{1}{15}\) =
Answer:
\(\frac{1}{5}\) – \(\frac{1}{15}\) = \(\frac{2}{15}\)
Explanation :
\(\frac{1}{5}\) – \(\frac{1}{15}\) = \(\frac{3}{15}\) – \(\frac{1}{15}\) = \(\frac{2}{15}\)

Question 37.
\(\frac{1}{3}\) – \(\frac{1}{15}\) =
Answer:
\(\frac{1}{3}\) – \(\frac{1}{15}\) =  \(\frac{4}{15}\)
Explanation :
\(\frac{1}{3}\) – \(\frac{1}{15}\) = \(\frac{5}{15}\) – \(\frac{1}{15}\) = \(\frac{4}{15}\)

Question 38.
\(\frac{2}{3}\) – \(\frac{3}{15}\) =
Answer:
\(\frac{2}{3}\) – \(\frac{3}{15}\) =  \(\frac{7}{15}\)
Explanation :
\(\frac{2}{3}\) – \(\frac{3}{15}\) = \(\frac{10}{15}\) – \(\frac{3}{15}\) = \(\frac{7}{15}\)

Question 39.
\(\frac{2}{5}\) – \(\frac{4}{15}\) =
Answer:
\(\frac{2}{5}\) – \(\frac{4}{15}\) = \(\frac{2}{15}\)
Explanation :
\(\frac{2}{5}\) – \(\frac{4}{15}\) = \(\frac{6}{15}\) – \(\frac{4}{15}\) = \(\frac{2}{15}\)

Question 40.
\(\frac{3}{4}\) – \(\frac{2}{12}\) =
Answer:
\(\frac{3}{4}\) – \(\frac{2}{12}\) =  \(\frac{7}{12}\)
Explanation :
\(\frac{3}{4}\) – \(\frac{2}{12}\) = \(\frac{9}{12}\) – \(\frac{2}{12}\) = \(\frac{7}{12}\)

Question 41.
\(\frac{3}{4}\) – \(\frac{5}{16}\) =
Answer:
\(\frac{3}{4}\) – \(\frac{5}{16}\) = \(\frac{7}{16}\)
Explanation :
\(\frac{3}{4}\) – \(\frac{5}{16}\) = \(\frac{12}{16}\) – \(\frac{5}{16}\) = \(\frac{7}{16}\)

Question 42.
\(\frac{4}{5}\) – \(\frac{5}{15}\) =
Answer:
\(\frac{4}{5}\) – \(\frac{5}{15}\) = \(\frac{7}{15}\)
Explanation :
\(\frac{4}{5}\) – \(\frac{5}{15}\) = \(\frac{12}{15}\) – \(\frac{5}{15}\) = \(\frac{7}{15}\)

Question 43.
\(\frac{3}{4}\) – \(\frac{4}{12}\) =
Answer:
\(\frac{3}{4}\) – \(\frac{4}{12}\) =  \(\frac{5}{12}\)
Explanation :
\(\frac{3}{4}\) – \(\frac{4}{12}\) = \(\frac{9}{12}\) – \(\frac{4}{12}\) = \(\frac{5}{12}\)

Question 44.
\(\frac{3}{4}\) – \(\frac{7}{16}\) =
Answer:
\(\frac{3}{4}\) – \(\frac{7}{16}\) = \(\frac{5}{16}\)
Explanation :
\(\frac{3}{4}\) – \(\frac{5}{16}\) = \(\frac{12}{16}\) – \(\frac{7}{16}\) = \(\frac{5}{16}\)

B
Subtract Fractions with Unlike Units
Engage NY Math 5th Grade Module 3 Lesson 12 Sprint Answer Key 2

Question 1.
\(\frac{2}{10}\) – \(\frac{1}{10}\) =
Answer:
\(\frac{2}{10}\) – \(\frac{1}{10}\) = \(\frac{1}{10}\)

Question 2.
\(\frac{1}{5}\) – \(\frac{1}{10}\) =
Answer:
\(\frac{1}{5}\) – \(\frac{1}{10}\) = \(\frac{1}{10}\)
Explanation :
\(\frac{1}{5}\) – \(\frac{1}{10}\) = \(\frac{2}{10}\) – \(\frac{1}{10}\) = \(\frac{1}{10}\)

Question 3.
\(\frac{2}{4}\) – \(\frac{1}{4}\) =
Answer:
\(\frac{2}{4}\) – \(\frac{1}{4}\) = \(\frac{1}{4}\)

Question 4.
\(\frac{1}{2}\) – \(\frac{1}{4}\) =
Answer:
\(\frac{1}{2}\) – \(\frac{1}{4}\) = \(\frac{1}{4}\)
Explanation :
\(\frac{1}{2}\) – \(\frac{1}{4}\) = \(\frac{2}{4}\) – \(\frac{1}{4}\) = \(\frac{1}{4}\)

Question 5.
\(\frac{5}{10}\) – \(\frac{2}{10}\) =
Answer:
\(\frac{5}{10}\) – \(\frac{2}{10}\) = \(\frac{3}{10}\)

Question 6.
\(\frac{1}{2}\) – \(\frac{2}{10}\) =
Answer:
\(\frac{1}{2}\) – \(\frac{2}{10}\) = \(\frac{3}{10}\)
Explanation :
\(\frac{1}{2}\) – \(\frac{2}{10}\) = \(\frac{5}{10}\) – \(\frac{2}{10}\) = \(\frac{3}{10}\)

Question 7.
\(\frac{1}{2}\) – \(\frac{4}{10}\) =
Answer:
\(\frac{1}{2}\) – \(\frac{4}{10}\) = \(\frac{1}{10}\)
Explanation :
\(\frac{1}{2}\) – \(\frac{4}{10}\) = \(\frac{5}{10}\) – \(\frac{4}{10}\) = \(\frac{1}{10}\)

Question 8.
\(\frac{4}{10}\) – \(\frac{1}{10}\) =
Answer:
\(\frac{4}{10}\) – \(\frac{1}{10}\) = \(\frac{3}{10}\)

Question 9.
\(\frac{2}{5}\) – \(\frac{1}{10}\) =
Answer:
\(\frac{2}{5}\) – \(\frac{1}{10}\) = \(\frac{3}{10}\)
Explanation :
\(\frac{2}{5}\) – \(\frac{1}{10}\) = \(\frac{4}{10}\) – \(\frac{1}{10}\) = \(\frac{3}{10}\)

Question 10.
\(\frac{2}{5}\) – \(\frac{3}{10}\) =
Answer:
\(\frac{2}{5}\) – \(\frac{3}{10}\) = \(\frac{1}{10}\)
Explanation :
\(\frac{2}{5}\) – \(\frac{3}{10}\) = \(\frac{4}{10}\) – \(\frac{3}{10}\) = \(\frac{1}{10}\)

Question 11.
\(\frac{6}{10}\) – \(\frac{3}{10}\) =
Answer:
\(\frac{6}{10}\) – \(\frac{3}{10}\) =\(\frac{3}{10}\)

Question 12.
\(\frac{3}{5}\) – \(\frac{3}{10}\) =
Answer:
\(\frac{3}{5}\) – \(\frac{3}{10}\) = \(\frac{3}{10}\)
Explanation :
\(\frac{3}{5}\) – \(\frac{3}{10}\) = \(\frac{6}{10}\) – \(\frac{3}{10}\) = \(\frac{3}{10}\)

Question 13.
\(\frac{3}{5}\) – \(\frac{5}{10}\) =
Answer:
\(\frac{3}{5}\) – \(\frac{5}{10}\) = \(\frac{1}{10}\)
Explanation :
\(\frac{3}{5}\) – \(\frac{5}{10}\) = \(\frac{6}{10}\) – \(\frac{5}{10}\) = \(\frac{1}{10}\)

Question 14.
\(\frac{8}{10}\) – \(\frac{1}{10}\) =
Answer:
\(\frac{8}{10}\) – \(\frac{1}{10}\) =  \(\frac{7}{10}\)

Question 15.
\(\frac{4}{5}\) – \(\frac{1}{10}\) =
Answer:
\(\frac{4}{5}\) – \(\frac{1}{10}\) = \(\frac{7}{10}\)
Explanation :
\(\frac{4}{5}\) – \(\frac{1}{10}\) = \(\frac{8}{10}\) – \(\frac{1}{10}\) =  \(\frac{7}{10}\)

Question 16.
\(\frac{4}{5}\) – \(\frac{5}{10}\) =
Answer:
\(\frac{4}{5}\) – \(\frac{5}{10}\) = \(\frac{3}{10}\)
Explanation :
\(\frac{4}{5}\) – \(\frac{5}{10}\) = \(\frac{8}{10}\) – \(\frac{5}{10}\) =  \(\frac{4}{10}\)

Question 17.
\(\frac{4}{5}\) – \(\frac{5}{10}\) =
Answer:
\(\frac{4}{5}\) – \(\frac{5}{10}\) = \(\frac{3}{10}\)
Explanation :
\(\frac{4}{5}\) – \(\frac{5}{10}\) = \(\frac{8}{10}\) – \(\frac{5}{10}\) =  \(\frac{4}{10}\)

Question 18.
\(\frac{4}{5}\) – \(\frac{7}{10}\) =
Answer:
\(\frac{4}{5}\) – \(\frac{7}{10}\) = \(\frac{1}{10}\)
Explanation :
\(\frac{4}{5}\) – \(\frac{7}{10}\) = \(\frac{8}{10}\) – \(\frac{7}{10}\) =  \(\frac{1}{10}\)

Question 19.
\(\frac{2}{8}\) – \(\frac{1}{8}\) =
Answer:
\(\frac{2}{8}\) – \(\frac{1}{8}\) = \(\frac{1}{8}\)

Question 20.
\(\frac{1}{4}\) – \(\frac{1}{8}\) =
Answer:
\(\frac{1}{4}\) – \(\frac{1}{8}\) =\(\frac{1}{8}\)
Explanation :
\(\frac{1}{4}\) – \(\frac{1}{8}\) = \(\frac{2}{8}\) – \(\frac{1}{8}\) = \(\frac{1}{8}\)

Question 21.
\(\frac{6}{8}\) – \(\frac{1}{8}\) =
Answer:
\(\frac{6}{8}\) – \(\frac{1}{8}\) = \(\frac{5}{8}\)

Question 22.
\(\frac{3}{4}\) – \(\frac{1}{8}\) =
Answer:
\(\frac{3}{4}\) – \(\frac{1}{8}\) = \(\frac{5}{8}\)
Explanation :
\(\frac{3}{4}\) – \(\frac{1}{8}\) = \(\frac{6}{8}\) – \(\frac{1}{8}\) = \(\frac{5}{8}\)

Question 23.
\(\frac{3}{4}\) – \(\frac{3}{8}\) =
Answer:
\(\frac{3}{4}\) – \(\frac{3}{8}\) = \(\frac{3}{8}\)
Explanation :
\(\frac{3}{4}\) – \(\frac{3}{8}\) = \(\frac{6}{8}\) – \(\frac{3}{8}\) = \(\frac{3}{8}\)

Question 24.
\(\frac{5}{15}\) – \(\frac{1}{15}\) =
Answer:
\(\frac{5}{15}\) – \(\frac{1}{15}\) = \(\frac{4}{15}\)

Question 25.
\(\frac{1}{3}\) – \(\frac{1}{15}\) =
Answer:
\(\frac{1}{3}\) – \(\frac{1}{15}\) =  \(\frac{4}{15}\)
Explanation :
\(\frac{1}{3}\) – \(\frac{1}{15}\) = \(\frac{5}{15}\) – \(\frac{1}{15}\) = \(\frac{4}{15}\)

Question 26.
\(\frac{3}{15}\) – \(\frac{1}{15}\) =
Answer:
\(\frac{3}{15}\) – \(\frac{1}{15}\) = \(\frac{2}{15}\)

Question 27.
\(\frac{1}{5}\) – \(\frac{1}{15}\) =
Answer:
\(\frac{1}{5}\) – \(\frac{1}{15}\) = \(\frac{2}{15}\)
Explanation :
\(\frac{1}{5}\) – \(\frac{1}{15}\) = \(\frac{3}{15}\) – \(\frac{1}{15}\) = \(\frac{2}{15}\)

Question 28.
\(\frac{1}{5}\) – \(\frac{2}{15}\) =
Answer:
\(\frac{1}{5}\) – \(\frac{2}{15}\) = \(\frac{1}{15}\)
Explanation :
\(\frac{1}{5}\) – \(\frac{2}{15}\) = \(\frac{3}{15}\) – \(\frac{2}{15}\) = \(\frac{1}{15}\)

Question 29.
\(\frac{12}{15}\) – \(\frac{4}{15}\) =
Answer:
\(\frac{12}{15}\) – \(\frac{4}{15}\) = \(\frac{8}{15}\)

Question 30.
\(\frac{4}{5}\) – \(\frac{4}{15}\) =
Answer:
\(\frac{4}{5}\) – \(\frac{2}{15}\) = \(\frac{2}{3}\)
Explanation :
\(\frac{4}{5}\) – \(\frac{2}{15}\) = \(\frac{12}{15}\) – \(\frac{2}{15}\) = \(\frac{10}{15}\)= \(\frac{2}{3}\)

Question 31.
\(\frac{1}{4}\) – \(\frac{2}{12}\) =
Answer:
\(\frac{1}{4}\) – \(\frac{2}{12}\) =  \(\frac{1}{12}\)
Explanation :
\(\frac{1}{4}\) – \(\frac{2}{12}\) = \(\frac{3}{12}\) – \(\frac{2}{12}\) = \(\frac{1}{12}\)

Question 32.
\(\frac{3}{4}\) – \(\frac{2}{12}\) =
Answer:
\(\frac{3}{4}\) – \(\frac{2}{12}\) =  \(\frac{7}{12}\)
Explanation :
\(\frac{3}{4}\) – \(\frac{2}{12}\) = \(\frac{9}{12}\) – \(\frac{2}{12}\) = \(\frac{7}{12}\)

Question 33.
\(\frac{3}{4}\) – \(\frac{4}{12}\) =
Answer:
\(\frac{3}{4}\) – \(\frac{4}{12}\) =  \(\frac{5}{12}\)
Explanation :
\(\frac{3}{4}\) – \(\frac{4}{12}\) = \(\frac{9}{12}\) – \(\frac{4}{12}\) = \(\frac{5}{12}\)

Question 34.
\(\frac{3}{4}\) – \(\frac{8}{12}\) =
Answer:
\(\frac{3}{4}\) – \(\frac{8}{12}\) =  \(\frac{1}{12}\)
Explanation :
\(\frac{3}{4}\) – \(\frac{8}{12}\) = \(\frac{9}{12}\) – \(\frac{8}{12}\) = \(\frac{1}{12}\)

Question 35.
\(\frac{1}{3}\) – \(\frac{3}{12}\) =
Answer:
\(\frac{1}{3}\) – \(\frac{3}{12}\) = \(\frac{1}{12}\)
Explanation :
\(\frac{1}{3}\) – \(\frac{3}{12}\) = \(\frac{4}{12}\) – \(\frac{3}{12}\) = \(\frac{1}{12}\)

Question 36.
\(\frac{1}{6}\) – \(\frac{1}{12}\) =
Answer:
\(\frac{1}{6}\) – \(\frac{1}{12}\) =\(\frac{1}{12}\)
Explanation :
\(\frac{1}{6}\) – \(\frac{1}{12}\) = \(\frac{2}{12}\) – \(\frac{1}{12}\) = \(\frac{1}{12}\)

Question 37.
\(\frac{1}{3}\) – \(\frac{3}{15}\) =
Answer:
\(\frac{1}{3}\) – \(\frac{3}{15}\) =  \(\frac{2}{15}\)
Explanation :
\(\frac{1}{3}\) – \(\frac{3}{15}\) = \(\frac{5}{15}\) – \(\frac{3}{15}\) = \(\frac{2}{15}\)

Question 38.
\(\frac{2}{3}\) – \(\frac{2}{15}\) =
Answer:
\(\frac{2}{3}\) – \(\frac{2}{15}\) =  \(\frac{8}{15}\)
Explanation :
\(\frac{2}{3}\) – \(\frac{2}{15}\) = \(\frac{10}{15}\) – \(\frac{2}{15}\) = \(\frac{8}{15}\)

Question 39.
\(\frac{2}{5}\) – \(\frac{2}{15}\) =
Answer:
\(\frac{2}{5}\) – \(\frac{2}{15}\) = \(\frac{4}{15}\)
Explanation :
\(\frac{2}{5}\) – \(\frac{2}{15}\) = \(\frac{6}{15}\) – \(\frac{2}{15}\) = \(\frac{4}{15}\)

Question 40.
\(\frac{3}{4}\) – \(\frac{4}{12}\) =
Answer:
\(\frac{3}{4}\) – \(\frac{4}{12}\) =  \(\frac{5}{12}\)
Explanation :
\(\frac{3}{4}\) – \(\frac{4}{12}\) = \(\frac{9}{12}\) – \(\frac{4}{12}\) = \(\frac{5}{12}\)

Question 41.
\(\frac{3}{4}\) – \(\frac{7}{16}\) =
Answer:
\(\frac{3}{4}\) – \(\frac{7}{16}\) = \(\frac{5}{16}\)
Explanation :
\(\frac{3}{4}\) – \(\frac{5}{16}\) = \(\frac{12}{16}\) – \(\frac{7}{16}\) = \(\frac{5}{16}\)

Question 42.
\(\frac{4}{5}\) – \(\frac{4}{15}\) =
Answer:
\(\frac{4}{5}\) – \(\frac{4}{15}\) = \(\frac{8}{15}\)
Explanation :
\(\frac{4}{5}\) – \(\frac{4}{15}\) = \(\frac{12}{15}\) – \(\frac{4}{15}\) = \(\frac{8}{15}\)

Question 43.
\(\frac{3}{4}\) – \(\frac{2}{12}\) =
Answer:
\(\frac{3}{4}\) – \(\frac{2}{12}\) =  \(\frac{7}{12}\)
Explanation :
\(\frac{3}{4}\) – \(\frac{2}{12}\) = \(\frac{9}{12}\) – \(\frac{2}{12}\) = \(\frac{7}{12}\)

Question 44.
\(\frac{3}{4}\) – \(\frac{5}{16}\) =
Answer:
\(\frac{3}{4}\) – \(\frac{5}{16}\) = \(\frac{7}{16}\)
Explanation :
\(\frac{3}{4}\) – \(\frac{5}{16}\) = \(\frac{12}{16}\) – \(\frac{5}{16}\) = \(\frac{7}{16}\)

Eureka Math Grade 5 Module 3 Lesson 12 Problem Set Answer Key

Question 1.
Subtract.
a. 3\(\frac{1}{5}\) – 2\(\frac{1}{4}\) =
b. 4\(\frac{2}{5}\) – 3\(\frac{3}{4}\) =
c. 7\(\frac{1}{5}\) – 4\(\frac{1}{5}\) =
d. 7\(\frac{2}{5}\) – 5\(\frac{2}{3}\) =
e. 4\(\frac{2}{7}\) – 3\(\frac{1}{3}\) =
f. 9\(\frac{2}{3}\) – 2\(\frac{6}{7}\) =
g. 17\(\frac{2}{3}\) – 5\(\frac{5}{6}\) =
h. 18\(\frac{1}{3}\) – 3\(\frac{3}{8}\) =
Answer:
a. 3\(\frac{1}{5}\) – 2\(\frac{1}{4}\) = \(\frac{19}{20}\)
Explanation :
3\(\frac{1}{5}\) – 2\(\frac{1}{4}\) = \(\frac{16}{5}\) – \(\frac{9}{4}\)
lcm of 5 and 4 is 20 .
\(\frac{64}{20}\) – \(\frac{45}{20}\) = \(\frac{19}{20}\)

b. 4\(\frac{2}{5}\) – 3\(\frac{3}{4}\) = \(\frac{13}{20}\)
Explanation :
4\(\frac{2}{5}\) – 3\(\frac{3}{4}\) = \(\frac{22}{5}\) – \(\frac{15}{4}\)
lcm of 5 and 4 is 20 .
\(\frac{88}{20}\) – \(\frac{75}{20}\) = \(\frac{13}{20}\)

c. 7\(\frac{1}{5}\) – 4\(\frac{1}{5}\) =
Explanation :
7\(\frac{1}{5}\) – 4\(\frac{1}{5}\) = \(\frac{36}{5}\) – \(\frac{21}{5}\) = \(\frac{15}{5}\) = 3 .

d. 7\(\frac{2}{5}\) – 5\(\frac{2}{3}\) = 1\(\frac{11}{15}\) .
Explanation :
7\(\frac{2}{5}\) – 5\(\frac{2}{3}\) = \(\frac{37}{5}\) – \(\frac{17}{3}\)
lcm of 5 and 3 is 15 .
\(\frac{111}{15}\) – \(\frac{85}{15}\) = \(\frac{26}{15}\) = 1\(\frac{11}{15}\) .

e. 4\(\frac{2}{7}\) – 3\(\frac{1}{3}\) = \(\frac{20}{21}\)
Explanation :
4\(\frac{2}{7}\) – 3\(\frac{1}{3}\) = \(\frac{30}{7}\) – \(\frac{10}{3}\)
lcm of 7 and 3 is 21.
\(\frac{90}{21}\) – \(\frac{70}{21}\) = \(\frac{20}{21}\)

f. 9\(\frac{2}{3}\) – 2\(\frac{6}{7}\) = 6\(\frac{17}{21}\)
Explanation :
9\(\frac{2}{3}\) – 2\(\frac{6}{7}\) = \(\frac{29}{3}\) – \(\frac{20}{7}\)
lcm of 3 and 7 is 21.
\(\frac{203}{21}\) – \(\frac{60}{21}\) = \(\frac{143}{21}\) = 6\(\frac{17}{21}\)

g. 17\(\frac{2}{3}\) – 5\(\frac{5}{6}\) = 11\(\frac{5}{6}\)
Explanation :
17\(\frac{2}{3}\) – 5\(\frac{5}{6}\) = \(\frac{53}{3}\) – \(\frac{35}{6}\)
lcm of 3 and 6 is 6 .
\(\frac{106}{6}\) – \(\frac{35}{6}\) = \(\frac{71}{6}\) = 11\(\frac{5}{6}\)

h. 18\(\frac{1}{3}\) – 3\(\frac{3}{8}\) = 14\(\frac{23}{24}\)
Explanation :
18\(\frac{1}{3}\) – 3\(\frac{3}{8}\) = \(\frac{55}{3}\) – \(\frac{27}{8}\)
lcm of 3 and 8 is 24 .
\(\frac{440}{24}\) – \(\frac{81}{24}\) = \(\frac{359}{24}\) = 14\(\frac{23}{24}\)

Question 2.
Toby wrote the following:
7\(\frac{1}{4}\) – 3\(\frac{3}{4}\) = 4\(\frac{2}{4}\) = 4\(\frac{1}{2}\)
Is Toby’s calculation correct? Draw a number line to support your answer.
Answer:
Tody’s calculation is wrong .
7\(\frac{1}{4}\) – 3\(\frac{3}{4}\) = 3\(\frac{1}{2}\) .
Explanation :
7\(\frac{1}{4}\) – 3\(\frac{3}{4}\) = \(\frac{29}{4}\) – \(\frac{15}{4}\) = \(\frac{14}{4}\) = \(\frac{7}{2}\) = 3\(\frac{1}{2}\) .
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-12-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-12-Problem-Set-Answer-Key-Question-2

Question 3.
Mr. Neville Iceguy mixed up 12\(\frac{3}{5}\) gallons of chili for a party. If 7\(\frac{3}{4}\)gallons of chili was mild, and the rest was extra spicy, how much extra spicy chili did Mr. Iceguy make?
Answer:
Fraction of chili mixed up for a party = 12\(\frac{3}{5}\) gallons = \(\frac{63}{5}\) gallons
Fraction of chilli is mild = 7\(\frac{3}{4}\)gallons = \(\frac{31}{4}\)gallons
Fraction of chilli is extra spicy = x
\(\frac{63}{5}\) =  \(\frac{31}{4}\) + x
x = \(\frac{63}{5}\) – \(\frac{31}{4}\)
lcm of 5 and 4 is 20 .
x = \(\frac{252}{20}\) – \(\frac{155}{20}\) = \(\frac{97}{20}\) = 4\(\frac{17}{20}\)
Therefore, Fraction of chilli is extra spicy = x = 4\(\frac{17}{20}\) gallons .

Question 4.
Jazmyne decided to spend 6\(\frac{1}{2}\) hours studying over the weekend. She spent 1\(\frac{1}{4}\) hours studying on Friday evening and 2\(\frac{2}{3}\) hours on Saturday. How much longer does she need to spend studying on Sunday in order to reach her goal?
Answer:
Total Fraction of time should spend on Studying = 6\(\frac{1}{2}\) hours = \(\frac{13}{2}\) hours
Fraction of Time spent on Friday for studying = 1\(\frac{1}{4}\) hours = \(\frac{5}{4}\) hours
Fraction of Time spent on Saturday for studying = 2\(\frac{2}{3}\) hours = \(\frac{8}{3}\) hours
Fraction of Time should spend on Sunday for studying = x
\(\frac{13}{2}\) = \(\frac{5}{4}\) + \(\frac{8}{3}\) + x
lcm of 2 , 4 , and 3 is 12 .
x = \(\frac{13}{2}\) – \(\frac{5}{4}\) – \(\frac{8}{3}\)
x = \(\frac{78}{12}\) – \(\frac{15}{12}\) – \(\frac{32}{12}\)
x = \(\frac{78}{12}\) – \(\frac{47}{12}\)
x = \(\frac{31}{12}\)
x = 2\(\frac{7}{12}\)
Therefore, Fraction of Time should spend on Sunday for studying = x = 2\(\frac{7}{12}\) hours .

Eureka Math Grade 5 Module 3 Lesson 12 Exit Ticket Answer Key

Subtract.
Question 1.
5\(\frac{1}{2}\) – 1\(\frac{1}{3}\) =
Answer:
5\(\frac{1}{2}\) – 1\(\frac{1}{3}\) = 4\(\frac{1}{6}\) .
Explanation :
5\(\frac{1}{2}\) – 1\(\frac{1}{3}\) = \(\frac{11}{2}\) – \(\frac{4}{3}\)
lcm of 2 and 3 is 6
\(\frac{33}{6}\) – \(\frac{8}{6}\) = \(\frac{25}{6}\) = 4\(\frac{1}{6}\) .

Question 2.
8\(\frac{3}{4}\) – 5\(\frac{5}{6}\) =
Answer:
8\(\frac{3}{4}\) – 5\(\frac{5}{6}\) = 2\(\frac{11}{12}\)
Explanation :
8\(\frac{3}{4}\) – 5\(\frac{5}{6}\) = \(\frac{35}{4}\) – \(\frac{35}{6}\)
lcm of 4 and 6 is 12 .
\(\frac{105}{12}\) – \(\frac{70}{12}\) = \(\frac{35}{12}\) = 2\(\frac{11}{12}\)

Eureka Math Grade 5 Module 3 Lesson 12 Homework Answer Key

Question 1.
Subtract.
a. 3\(\frac{1}{4}\) – 2\(\frac{1}{3}\) =
b. 3\(\frac{2}{3}\) – 2\(\frac{3}{4}\) =
c. 6\(\frac{1}{5}\) – 4\(\frac{1}{4}\) =
d. 6\(\frac{3}{5}\) – 4\(\frac{3}{4}\) =
e. 5\(\frac{2}{7}\) – 4\(\frac{1}{3}\) =
f. 8\(\frac{2}{3}\) – 3\(\frac{5}{7}\) =
g. 18\(\frac{3}{4}\) – 5\(\frac{7}{8}\) =
h. 17\(\frac{1}{5}\) – 2\(\frac{5}{8}\) =
Answer:
a. 3\(\frac{1}{4}\) – 2\(\frac{1}{3}\) = \(\frac{11}{12}\)
Explanation :
3\(\frac{1}{4}\) – 2\(\frac{1}{3}\) = \(\frac{13}{4}\) – \(\frac{7}{3}\)
lcm of 4 and 3 is 12
\(\frac{39}{12}\) – \(\frac{28}{12}\) = \(\frac{11}{12}\)

b. 3\(\frac{2}{3}\) – 2\(\frac{3}{4}\) = \(\frac{11}{12}\)
Explanation :
3\(\frac{2}{3}\) – 2\(\frac{3}{4}\) = \(\frac{11}{3}\) – \(\frac{11}{4}\)
lcm of 3 and 4 is 12
\(\frac{44}{12}\) – \(\frac{33}{12}\) = \(\frac{11}{12}\)

c. 6\(\frac{1}{5}\) – 4\(\frac{1}{4}\) = 3\(\frac{13}{20}\)
Explanation :
6\(\frac{1}{5}\) – 4\(\frac{1}{4}\) = \(\frac{31}{5}\) – \(\frac{17}{4}\)
lcm of 4 and 5 is 20 .
\(\frac{124}{20}\) – \(\frac{51}{20}\) = \(\frac{73}{20}\) = 3\(\frac{13}{20}\) .

d. 6\(\frac{3}{5}\) – 4\(\frac{3}{4}\) = 1\(\frac{17}{20}\)
Explanation :
6\(\frac{3}{5}\) – 4\(\frac{3}{4}\) = \(\frac{33}{5}\) – \(\frac{19}{4}\)
lcm of 4 and 5 is 20 .
\(\frac{132}{20}\) – \(\frac{95}{20}\) = \(\frac{37}{20}\) = 1\(\frac{17}{20}\)

e. 5\(\frac{2}{7}\) – 4\(\frac{1}{3}\) = \(\frac{20}{21}\)
Explanation :
5\(\frac{2}{7}\) – 4\(\frac{1}{3}\) = \(\frac{37}{7}\) – \(\frac{13}{3}\)
lcm of 3 and 7 is 21.
\(\frac{111}{21}\) – \(\frac{91}{21}\) = \(\frac{20}{21}\)

f. 8\(\frac{2}{3}\) – 3\(\frac{5}{7}\) = 4\(\frac{20}{21}\)
Explanation :
8\(\frac{2}{3}\) – 3\(\frac{5}{7}\) = \(\frac{26}{3}\) – \(\frac{26}{7}\)
lcm of 3 and 7 is 21.
\(\frac{182}{21}\) – \(\frac{78}{21}\) = \(\frac{104}{21}\) = 4\(\frac{20}{21}\)

g. 18\(\frac{3}{4}\) – 5\(\frac{7}{8}\) = 11\(\frac{9}{8}\)
Explanation :
18\(\frac{3}{4}\) – 5\(\frac{7}{8}\) = \(\frac{75}{4}\) – \(\frac{47}{8}\)
lcm of 4 and 8 is 8 .
\(\frac{130}{8}\) – \(\frac{47}{8}\) = \(\frac{97}{8}\) = 11\(\frac{9}{8}\)

h. 17\(\frac{1}{5}\) – 2\(\frac{5}{8}\) = 14\(\frac{23}{40}\)
Explanation :
17\(\frac{1}{5}\) – 2\(\frac{5}{8}\) = \(\frac{86}{5}\) – \(\frac{21}{8}\)
lcm of 5 and 8 is 40 .
\(\frac{688}{40}\) – \(\frac{105}{40}\) = \(\frac{583}{40}\) = 14\(\frac{23}{40}\)

Question 2.
Tony wrote the following:
7\(\frac{1}{4}\) – 3\(\frac{3}{4}\) = 4\(\frac{1}{4}\) – \(\frac{3}{4}\)
Is Tony’s statement correct? Draw a number line to support your answer.
Answer:
Yes Tony’s statement is correct .
7\(\frac{1}{4}\) – 3\(\frac{3}{4}\) = 4\(\frac{1}{4}\) – \(\frac{3}{4}\) = 3\(\frac{1}{2}\) .
Explanation :
7\(\frac{1}{4}\) – 3\(\frac{3}{4}\) = \(\frac{29}{4}\) – \(\frac{15}{4}\) = \(\frac{14}{4}\) = \(\frac{7}{2}\) = 3\(\frac{1}{2}\) .
4\(\frac{1}{4}\) – \(\frac{3}{4}\) = \(\frac{17}{4}\) – \(\frac{3}{4}\) = \(\frac{14}{4}\) = \(\frac{7}{2}\) = 3\(\frac{1}{2}\) .
Therefore , 7\(\frac{1}{4}\) – 3\(\frac{3}{4}\) = 4\(\frac{1}{4}\) – \(\frac{3}{4}\) = 3\(\frac{1}{2}\) .
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-12-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-12-Homework-Answer-Key-Question-2

Question 3.
Ms. Sanger blended 8\(\frac{3}{4}\) gallons of iced tea with some lemonade for a picnic. If there were 13\(\frac{2}{5}\) gallons of the beverage, how many gallons of lemonade did she use?
Answer:
Fraction of gallons of iced tea with some lemonade for a picnic = 8\(\frac{3}{4}\) = \(\frac{35}{4}\) gallons .
Fraction of gallons of the beverage = 13\(\frac{2}{5}\) = \(\frac{67}{5}\) .
Fraction of gallons of the lemonade = x
\(\frac{67}{5}\) = \(\frac{35}{4}\) – x
x =  \(\frac{67}{5}\) – \(\frac{35}{4}\)
lcm of 4 and 5 is 20 .
x = \(\frac{268}{20}\) – \(\frac{175}{20}\)
x = \(\frac{93}{20}\) = 4\(\frac{13}{20}\)
Therefore, Fraction of gallons of the lemonade = x = 4\(\frac{13}{20}\) gallons .

Question 4.
A carpenter has 10\(\frac{1}{2}\) feet of wooden plank. He cuts off 4\(\frac{1}{4}\) feet to replace the slat of a deck and 3\(\frac{2}{3}\) feet to repair a bannister. He uses the rest of the plank to fix a stair. How many feet of wood does the carpenter use to fix the stair?
Answer:
Total Fraction of wooden plank = 10\(\frac{1}{2}\) feet = \(\frac{21}{2}\) feet
Fraction of wooden plank cut off = 4\(\frac{1}{4}\) feet = \(\frac{17}{4}\) feet
Fraction of wooden plank used to repair a bannister = 3\(\frac{2}{3}\) feet = \(\frac{11}{3}\) feet
Fraction of wooden plank used for stair = x feet
10\(\frac{1}{2}\) = \(\frac{17}{4}\) + \(\frac{11}{3}\) + x
\(\frac{21}{2}\) = \(\frac{17}{4}\) + \(\frac{11}{3}\) + x
lcm of 2 , 4 and 3 is 12 .
\(\frac{126}{12}\) = \(\frac{51}{4}\) + \(\frac{44}{3}\) + x
x = \(\frac{126}{12}\) – \(\frac{95}{4}\)
x = \(\frac{31}{4}\) = 7\(\frac{3}{4}\) .
Therefore, Fraction of wooden plank used for stair = 7\(\frac{3}{4}\) feet .

Eureka Math Grade 5 Module 3 Lesson 11 Answer Key

Engage NY Eureka Math 5th Grade Module 3 Lesson 11 Answer Key

Eureka Math Grade 5 Module 3 Lesson 11 Problem Set Answer Key

Question 1.
Generate equivalent fractions to get like units. Then, subtract.
a. \(\frac{1}{2}\) – \(\frac{1}{3}\) =
b. \(\frac{7}{10}\) – \(\frac{1}{3}\) =
c. \(\frac{7}{8}\) – \(\frac{3}{4}\) =
d. 1\(\frac{2}{5}\) – \(\frac{3}{8}\) =
e. 1\(\frac{3}{10}\) – \(\frac{1}{6}\) =
f. 2\(\frac{1}{3}\) – 1\(\frac{1}{5}\) =
g. 5\(\frac{6}{7}\) – 2\(\frac{2}{3}\) =
h. Draw a number line to show that your answer to (g) is reasonable.
Answer:
a. \(\frac{1}{2}\) – \(\frac{1}{3}\) = \(\frac{1}{6}\)
Explanation :
\(\frac{1}{2}\) – \(\frac{1}{3}\)
lcm of 2 and 3 is 6
\(\frac{3}{6}\) – \(\frac{2}{6}\) = \(\frac{1}{6}\)

b. \(\frac{7}{10}\) – \(\frac{1}{3}\) = \(\frac{11}{30}\)
Explanation :
\(\frac{7}{10}\) – \(\frac{1}{3}\)
lcm of 10 and 3 is 30 .
\(\frac{21}{30}\) – \(\frac{10}{30}\) = \(\frac{11}{30}\)

c. \(\frac{7}{8}\) – \(\frac{3}{4}\) = \(\frac{1}{8 }\)
Explanation :
\(\frac{7}{8}\) – \(\frac{3}{4}\)
lcm of 8 and 4 is 8 .
\(\frac{7}{8}\) – \(\frac{6}{8 }\) = \(\frac{1}{8 }\)

d. 1\(\frac{2}{5}\) – \(\frac{3}{8}\) = 1\(\frac{31}{40}\)
Explanation :
1\(\frac{2}{5}\) – \(\frac{3}{8}\) = \(\frac{7}{5}\) – \(\frac{3}{8}\)
lcm of 5 and 8 is 40 .
\(\frac{56}{40}\) – \(\frac{15}{40}\) = \(\frac{71}{40}\) = 1\(\frac{31}{40}\)

e. 1\(\frac{3}{10}\) – \(\frac{1}{6}\) = 1\(\frac{4}{30}\)
Explanation :
1\(\frac{3}{10}\) – \(\frac{1}{6}\) = \(\frac{13}{10}\) – \(\frac{1}{6}\)
lcm of 6 and 10 is 30.
\(\frac{39}{30}\) – \(\frac{5}{30}\) = \(\frac{34}{30}\) = 1\(\frac{4}{30}\)

f. 2\(\frac{1}{3}\) – 1\(\frac{1}{5}\) =1\(\frac{2}{15}\)
Explanation :
2\(\frac{1}{3}\) – 1\(\frac{1}{5}\) = \(\frac{7}{3}\) – \(\frac{6}{5}\)
lcm of 3 and 5 is 15 .
\(\frac{35}{15}\) – \(\frac{18}{15}\) = \(\frac{17}{15}\) = 1\(\frac{2}{15}\)

g. 5\(\frac{6}{7}\) – 2\(\frac{2}{3}\) = 3\(\frac{4}{21}\) .
Explanation :
5\(\frac{6}{7}\) – 2\(\frac{2}{3}\) = \(\frac{41}{7}\) – \(\frac{8}{3}\)
lcm of 7 and 3 is 21 .
\(\frac{123}{21}\) – \(\frac{56}{21}\) = \(\frac{67}{21}\) = 3\(\frac{4}{21}\) .

Question 2.
George says that, to subtract fractions with different denominators, you always have to multiply the denominators to find the common unit; for example:
\(\frac{3}{8}-\frac{1}{6}=\frac{18}{48}-\frac{8}{48}\)
Show George how he could have chosen a denominator smaller than 48, and solve the problem.
Answer:
\(\frac{3}{8}\) – \(\frac{1}{6}\) = \(\frac{3}{8}\) – \(\frac{1}{6}\)
lcm of 8 and 6 is 24 .
[late3x]\frac{9}{24}[/latex] – \(\frac{4}{24}\) = \(\frac{5}{24}\)
Multiplies of 8 and 6 are .
8 : 16, 24, 32, 40, 48 .
6 : 12, 18, 24, 30, 36, 48.
common multiple smaller than 48 is 24 .

Question 3.
Meiling has 1\(\frac{1}{4}\) liter of orange juice. She drinks \(\frac{1}{3}\) liter. How much orange juice does she have left? (Extension: If her brother then drinks twice as much as Meiling, how much is left?)
Answer:
Fraction of Quantity of Juice with Meiling = 1\(\frac{1}{4}\) = \(\frac{5}{4}\)
Fraction of Quantity of Juice drank by Meiling = \(\frac{1}{3}\)
Fraction of Quantity of Juice left = \(\frac{5}{4}\) – \(\frac{1}{3}\) = \(\frac{15}{12}\) – \(\frac{4}{12}\) = \(\frac{11}{12}\) .
Therefore , Fraction of Quantity of Juice left = \(\frac{11}{12}\) .

Question 4.
Harlan used 3\(\frac{1}{2}\) kg of sand to make a large hourglass. To make a smaller hourglass, he only used 1\(\frac{3}{7}\) kg of sand. How much more sand did it take to make the large hourglass than the smaller one?
Answer:
Fraction of Quantity of sand used for large hourglass = 3\(\frac{1}{2}\) kg = \(\frac{7}{2}\) kg
Fraction of Quantity of sand used for small hourglass = \(\frac{10}{7}\) kg
Fraction of Quantity of sand to make the large hourglass than the smaller one = \(\frac{7}{2}\) – \(\frac{10}{7}\) = \(\frac{49}{14}\) – \(\frac{20}{14}\) = \(\frac{29}{14}\) = 2\(\frac{1}{14}\) .
Therefore, Fraction of Quantity of sand to make the large hourglass than the smaller one = 2\(\frac{1}{14}\) .

Eureka Math Grade 5 Module 3 Lesson 11 Exit Ticket Answer Key

Generate equivalent fractions to get like units. Then, subtract.
a. \(\frac{3}{4}\) – \(\frac{3}{10}\) =
b. 3\(\frac{1}{2}\) – 1\(\frac{1}{3}\) =
Answer:
a. \(\frac{3}{4}\) – \(\frac{3}{10}\) = \(\frac{9}{20}\)
Explanation :
\(\frac{3}{4}\) – \(\frac{3}{10}\)
lcm of 4 and 10 are 20 .
\(\frac{15}{20}\) – \(\frac{6}{20}\) = \(\frac{9}{20}\)

b. 3\(\frac{1}{2}\) – 1\(\frac{1}{3}\) = 2\(\frac{1}{6}\)
Explanation :
3\(\frac{1}{2}\) – 1\(\frac{1}{3}\) = \(\frac{7}{2}\) – \(\frac{4}{3}\)
lcm of 2 and 3 is 6
\(\frac{21}{6}\) – \(\frac{8}{6}\) = \(\frac{13}{6}\) = 2\(\frac{1}{6}\)

Eureka Math Grade 5 Module 3 Lesson 11 Homework Answer Key

Question 1.
Generate equivalent fractions to get like units. Then, subtract.
a. \(\frac{1}{2}\) – \(\frac{1}{5}\) =
b. \(\frac{7}{8}\) – \(\frac{1}{3}\) =
c. \(\frac{7}{10}\) – \(\frac{3}{5}\) =
d. 1\(\frac{5}{6}\) – \(\frac{2}{3}\) =
e. 2\(\frac{1}{4}\) – 1\(\frac{1}{5}\) =
f. 5\(\frac{6}{7}\) – 3\(\frac{2}{3}\) =
g. 15\(\frac{7}{8}\) – 5\(\frac{3}{4}\) =
h. 15\(\frac{5}{8}\) – 3\(\frac{1}{3}\) =
Answer:
a. \(\frac{1}{2}\) – \(\frac{1}{5}\) = \(\frac{3}{10}\)
Explanation :
\(\frac{1}{2}\) – \(\frac{1}{5}\)
lcm of 2 and 5 is 10 .
\(\frac{5}{10}\) – \(\frac{2}{10}\) = \(\frac{3}{10}\)

b. \(\frac{7}{8}\) – \(\frac{1}{3}\) = \(\frac{13}{24}\)
Explanation :
\(\frac{7}{8}\) – \(\frac{1}{3}\)
lcm of 8 and 3 is 24 .
\(\frac{21}{24}\) – \(\frac{8}{24}\) = \(\frac{13}{24}\)

c. \(\frac{7}{10}\) – \(\frac{3}{5}\) = \(\frac{1}{10}\)
Explanation :
\(\frac{7}{10}\) – \(\frac{3}{5}\)
lcm of 10 and 5 is 10 .
\(\frac{7}{10}\) – \(\frac{6}{10}\) =  \(\frac{1}{10}\)

d. 1\(\frac{5}{6}\) – \(\frac{2}{3}\) = \(\frac{1}{2}\)
Explanation :
1\(\frac{5}{6}\) – \(\frac{2}{3}\) = \(\frac{11}{6}\) – \(\frac{2}{3}\)
lcm of 6 and 3 is 6
\(\frac{11}{6}\) – \(\frac{4}{6}\) = \(\frac{3}{6}\) = \(\frac{1}{2}\)

e. 2\(\frac{1}{4}\) – 1\(\frac{1}{5}\) = 1\(\frac{1}{20}\)
Explanation :
2\(\frac{1}{4}\) – 1\(\frac{1}{5}\) = \(\frac{9}{4}\) – \(\frac{6}{5}\)
lcm of 4 and 5 is 20 .
\(\frac{45}{20}\) – \(\frac{24}{20}\) = \(\frac{21}{20}\) = 1\(\frac{1}{20}\)

f. 5\(\frac{6}{7}\) – 3\(\frac{2}{3}\) = 2 \(\frac{4}{21}\)
Explanation :
5\(\frac{6}{7}\) – 3\(\frac{2}{3}\) = \(\frac{41}{7}\) – \(\frac{11}{3}\)
lcm of 7 and 3 is 21
\(\frac{123}{21}\) – \(\frac{77}{21}\) = \(\frac{46}{21}\) = 2 \(\frac{4}{21}\)

g. 15\(\frac{7}{8}\) – 5\(\frac{3}{4}\) = 10\(\frac{1}{8}\)
Explanation :
15\(\frac{7}{8}\) – 5\(\frac{3}{4}\) = \(\frac{127}{8}\) – \(\frac{23}{4}\)
lcm of 8 and 4 is 8 .
\(\frac{127}{8}\) – \(\frac{46}{8}\) = \(\frac{81}{8}\) = 10\(\frac{1}{8}\) .

h. 15\(\frac{5}{8}\) – 3\(\frac{1}{3}\) = 12 \(\frac{7}{24}\)
Explanation :
15\(\frac{5}{8}\) – 3\(\frac{1}{3}\) = \(\frac{125}{8}\) – \(\frac{10}{3}\)
lcm of 3 and 8 is 24 .
\(\frac{375}{24}\) – \(\frac{80}{24}\) = \(\frac{295}{24}\) =12 \(\frac{7}{24}\)

Question 2.
Sandy ate \(\frac{1}{6}\) of a candy bar. John ate \(\frac{3}{4}\) of it. How much more of the candy bar did John eat than Sandy?
Answer:
Fraction of candy ate by sandy = \(\frac{1}{6}\)
Fraction of candy ate by John = \(\frac{3}{4}\)
Fraction of the candy bar ate more by John eat than Sandy = \(\frac{3}{4}\) – \(\frac{1}{6}\) = \(\frac{9}{12}\) – \(\frac{2}{12}\) = \(\frac{7}{12}\)
Therefore, Fraction of the candy bar ate more by John eat than Sandy = \(\frac{7}{12}\) .

Question 3.
4\(\frac{1}{2}\) yards of cloth are needed to make a woman’s dress. 2\(\frac{2}{7}\) yards of cloth are needed to make a girl’s dress. How much more cloth is needed to make a woman’s dress than a girl’s dress?
Answer:
Fraction of cloth needed to make women’s dress = 4\(\frac{1}{2}\) yards = \(\frac{9}{2}\) yards
Fraction of cloth needed to make girl’s dress = 2\(\frac{2}{7}\) yards = \(\frac{16}{7}\) yards
Fraction of more cloth needed to make a woman’s dress than a girl’s dress = \(\frac{9}{2}\) – \(\frac{16}{7}\)
= \(\frac{63}{14}\) – \(\frac{32}{14}\) = \(\frac{31}{14}\) = 2\(\frac{3}{14}\) yards .
Therefore, Fraction of more cloth needed to make a woman’s dress than a girl’s dress = 2\(\frac{3}{14}\) yards

Question 4.
Bill reads \(\frac{1}{5}\) of a book on Monday. He reads \(\frac{2}{3}\) of the book on Tuesday. If he finishes reading the book on Wednesday, what fraction of the book did he read on Wednesday?
Answer:
Fraction of book read on Monday =\(\frac{1}{5}\)
Fraction of book read on Tuesday = \(\frac{2}{3}\)
Fraction of book read on both days = \(\frac{1}{5}\) + \(\frac{2}{3}\) = \(\frac{3}{15}\) + \(\frac{10}{15}\) = \(\frac{13}{15}\) .
Therefore, Fraction of book read on both days = \(\frac{13}{15}\)

Question 5.
Tank A has a capacity of 9.5 gallons. 6\(\frac{1}{3}\) gallons of the tank’s water are poured out. How many gallons of water are left in the tank?
Answer:
Fraction of Capacity of Tank A = 9.5 gallons
Fraction of Capacity of water poured out = 6\(\frac{1}{3}\) gallons = \(\frac{19}{3}\) gallons .
Fraction of Capacity of water left = 9.5 – \(\frac{19}{3}\) = \(\frac{95}{10}\) – \(\frac{19}{3}\) = \(\frac{285}{30}\) – \(\frac{190}{30}\) = \(\frac{95}{30}\) = \(\frac{19}{6}\) = 3\(\frac{1}{6}\)
Therefore, Fraction of Capacity of water left = 3\(\frac{1}{6}\) .

Eureka Math Grade 5 Module 3 Lesson 10 Answer Key

Engage NY Eureka Math 5th Grade Module 3 Lesson 10 Answer Key

Eureka Math Grade 5 Module 3 Lesson 10 Sprint Answer Key

A
Add and Subtract Whole Numbers and Ones with Fraction Units
Engage NY Math 5th Grade Module 3 Lesson 10 Sprint Answer Key 1

Question 1.
3 + 1 =
Answer:
3 + 1 = 4
Explanation :
Adding 1 to 3 we get 4 a s sum .

Question 2.
3 + \(\frac{1}{2}\) =
Answer:
3 + \(\frac{1}{2}\) = 3\(\frac{1}{2}\)
Explanation :
3 + \(\frac{1}{2}\) = \(\frac{6}{2}\) + \(\frac{1}{2}\) = \(\frac{7}{2}\) = 3\(\frac{1}{2}\)

Question 3.
3\(\frac{1}{2}\) + 1 =
Answer:
3\(\frac{1}{2}\) + 1 = 4\(\frac{1}{2}\)
Explanation :
3\(\frac{1}{2}\) + 1 = \(\frac{7}{2}\) + \(\frac{2}{2}\) = \(\frac{9}{2}\) =4\(\frac{1}{2}\)

Question 4.
3 – 1 =
Answer:
3 – 1 = 2
Explanation :
subtracting 1 from 3 we get 2 .

Question 5.
3\(\frac{1}{2}\) – 1 =
Answer:
3\(\frac{1}{2}\) – 1 = 2\(\frac{1}{2}\)
Explanation :
3\(\frac{1}{2}\) – 1 = \(\frac{7}{2}\) – \(\frac{2}{2}\) = \(\frac{5}{2}\) =2\(\frac{1}{2}\)

Question 6.
4 – 2 =
Answer:
4 – 2 = 2
Explanation :
Subtracting 2 from 4 we get difference as 2.

Question 7.
4\(\frac{1}{2}\) – 2 =
Answer:
4\(\frac{1}{2}\) – 2 = 2\(\frac{1}{2}\)
Explanation :
4\(\frac{1}{2}\) – 2 = \(\frac{9}{2}\) – \(\frac{4}{2}\) = \(\frac{5}{2}\) =2\(\frac{1}{2}\)

Question 8.
5 – 2 =
Answer:
5 – 2 = 3
Explanation :
Subtracting 2 from 5 we get difference as 3.

Question 9.
5\(\frac{1}{3}\) – 2 =
Answer:
5\(\frac{1}{3}\) – 2 = 3\(\frac{1}{3}\)
Explanation :
5\(\frac{1}{3}\) – 2 = \(\frac{16}{3}\) – \(\frac{6}{3}\) = \(\frac{10}{3}\) =3\(\frac{1}{3}\)

Question 10.
5\(\frac{2}{3}\) – 2 =
Answer:
5\(\frac{2}{3}\) – 2 = 3\(\frac{2}{3}\)
Explanation :
5\(\frac{2}{3}\) – 2 = \(\frac{17}{3}\) – \(\frac{6}{3}\) = \(\frac{11}{3}\) =3\(\frac{2}{3}\)

Question 11.
5\(\frac{2}{3}\) + 2 =
Answer:
5\(\frac{2}{3}\) + 2 = 7\(\frac{2}{3}\)
Explanation :
5\(\frac{2}{3}\) + 2 = \(\frac{17}{3}\) + \(\frac{6}{3}\) = \(\frac{23}{3}\) = 7\(\frac{2}{3}\)

Question 12.
6 + 2 =
Answer:
6 + 2 = 8
Explanation :
Adding 2 to 6 we get sum as 8 .

Question 13.
6 + \(\frac{3}{4}\) =
Answer:
6 + \(\frac{3}{4}\) = 6\(\frac{3}{4}\)
Explanation :
6 + \(\frac{3}{4}\) = \(\frac{24}{4}\) + \(\frac{3}{4}\) = \(\frac{27}{4}\) = 6\(\frac{3}{4}\)

Question 14.
6\(\frac{3}{4}\) + 2 =
Answer:
6\(\frac{3}{4}\) + 2 = 8\(\frac{3}{4}\)
Explanation :
6\(\frac{3}{4}\) + 2 = \(\frac{27}{4}\) + \(\frac{8}{4}\) = \(\frac{35}{4}\) = 8\(\frac{3}{4}\)

Question 15.
6\(\frac{3}{4}\) – 2 =
Answer:
6\(\frac{3}{4}\) – 2 = 4\(\frac{3}{4}\)
Explanation :
6\(\frac{3}{4}\) – 2 = \(\frac{27}{4}\) – \(\frac{8}{4}\) = \(\frac{19}{4}\) = 4\(\frac{3}{4}\)

Question 16.
6\(\frac{3}{4}\) – 3 =
Answer:
6\(\frac{3}{4}\) – 3 = 3\(\frac{3}{4}\)
Explanation :
6\(\frac{3}{4}\) – 3 = \(\frac{27}{4}\) – \(\frac{12}{4}\) = \(\frac{15}{4}\) = 3\(\frac{3}{4}\)

Question 17.
6\(\frac{3}{4}\) – 4 =
Answer:
6\(\frac{3}{4}\) – 4 = 2\(\frac{3}{4}\)
Explanation :
6\(\frac{3}{4}\) – 4 = \(\frac{27}{4}\) – \(\frac{16}{4}\) = \(\frac{11}{4}\) = 2\(\frac{3}{4}\)

Question 18.
6\(\frac{3}{4}\) – 6 =
Answer:
6\(\frac{3}{4}\) – 6 = \(\frac{3}{4}\)
Explanation :
6\(\frac{3}{4}\) – 6 = \(\frac{27}{4}\) – \(\frac{24}{4}\) = \(\frac{3}{4}\)

Question 19.
6\(\frac{3}{4}\) – \(\frac{3}{4}\) =
Answer:
6\(\frac{3}{4}\) – \(\frac{3}{4}\) = 6
Explanation :
6\(\frac{3}{4}\) – \(\frac{3}{4}\) = \(\frac{27}{4}\) – \(\frac{3{4}\) = \(\frac{24}{4}\) = 6

Question 20.
2\(\frac{5}{6}\) + 3 =
Answer:
2\(\frac{5}{6}\) + 3 = 5\(\frac{5}{6}\)
Explanation :
2\(\frac{5}{6}\) + 3 = \(\frac{17}{6}\) + \(\frac{18}{6}\) = \(\frac{35}{6}\) = 5\(\frac{5}{6}\)

Question 21.
2\(\frac{1}{6}\) + 3 =
Answer:
2\(\frac{1}{6}\) + 3 = 5\(\frac{1}{6}\)
Explanation :
2\(\frac{1}{6}\) + 3 = \(\frac{13}{6}\) + \(\frac{18}{6}\) = \(\frac{31}{6}\) = 5\(\frac{1}{6}\)

Question 22.
2\(\frac{5}{6}\) + 7 =
Answer:
2\(\frac{5}{6}\) + 7 = 9\(\frac{5}{6}\)
Explanation :
2\(\frac{5}{6}\) + 7 = \(\frac{17}{6}\) + \(\frac{42}{6}\) = \(\frac{59}{6}\) = 9\(\frac{5}{6}\)

Question 23.
3\(\frac{5}{6}\) + 7 =
Answer:
3\(\frac{5}{6}\) + 7 = 10\(\frac{5}{6}\)
Explanation :
3\(\frac{5}{6}\) + 7 = \(\frac{23}{6}\) + \(\frac{42}{6}\) = \(\frac{65}{6}\) = 10\(\frac{5}{6}\)

Question 24.
7\(\frac{5}{6}\) + 3 =
Answer:
7\(\frac{5}{6}\) + 3 = 10\(\frac{5}{6}\)
Explanation :
7\(\frac{5}{6}\) + 3 = \(\frac{47}{6}\) + \(\frac{18}{6}\) = \(\frac{65}{6}\) = 10\(\frac{5}{6}\)

Question 25.
10\(\frac{5}{6}\) – 3 =
Answer:
10\(\frac{5}{6}\) – 3 = 7\(\frac{5}{6}\)
Explanation :
10\(\frac{5}{6}\) – 3 = \(\frac{65}{6}\) – \(\frac{18}{6}\) = \(\frac{47}{6}\) = 7\(\frac{5}{6}\)

Question 26.
10\(\frac{5}{6}\) – 7 =
Answer:
10\(\frac{5}{6}\) – 7 = 3\(\frac{5}{6}\)
Explanation :
10\(\frac{5}{6}\) – 7 = \(\frac{65}{6}\) – \(\frac{42}{6}\) = \(\frac{23}{6}\) = 3\(\frac{5}{6}\)

Question 27.
3 + \(\frac{4}{5}\) + 2 =
Answer:
3 + \(\frac{4}{5}\) + 2 = 5\(\frac{4}{5}\)
Explanation :
3 + \(\frac{4}{5}\) + 2 = 5 + \(\frac{4}{5}\) = \(\frac{25}{5}\) + \(\frac{4}{5}\) = \(\frac{29}{5}\) = 5\(\frac{4}{5}\)

Question 28.
5 + \(\frac{7}{8}\) + 4 =
Answer:
5 + \(\frac{7}{8}\) + 4 = 9 \(\frac{7}{8}\)
Explanation :
5 + \(\frac{7}{8}\) + 4 = 9 + \(\frac{7}{8}\) = \(\frac{72}{8}\) + \(\frac{7}{8}\) = \(\frac{79}{8}\) = 9\(\frac{7}{8}\)

Question 29.
7 + \(\frac{4}{5}\) – 2 =
Answer:
7 + \(\frac{4}{5}\) – 2 = 5\(\frac{4}{5}\)
Explanation :
7 + \(\frac{4}{5}\) – 2 = 5 + \(\frac{4}{5}\) = \(\frac{29}{5}\) + \(\frac{4}{5}\) = \(\frac{33}{5}\) = 5\(\frac{4}{5}\)

Question 30.
9 + \(\frac{5}{12}\) – 5 =
Answer:
9 + \(\frac{5}{12}\) – 5 = 4\(\frac{5}{12}\)
Explanation :
9 + \(\frac{5}{12}\) – 5 = 4 + \(\frac{5}{12}\) = \(\frac{48}{12}\) + \(\frac{5}{12}\) + \(\frac{53}{12}\) = 4\(\frac{5}{12}\) .

Question 31.
7 + \(\frac{1}{5}\) + \(\frac{1}{5}\) + 2
Answer:
7 + \(\frac{1}{5}\) + \(\frac{1}{5}\) + 2 = 9\(\frac{2}{5}\)
Explanation :
7 + \(\frac{1}{5}\) + \(\frac{1}{5}\) + 2 = 9 + \(\frac{2}{5}\) = \(\frac{45}{5}\) + \(\frac{2}{5}\) = \(\frac{47}{5}\) = 9\(\frac{2}{5}\)

Question 32.
7 + \(\frac{2}{5}\) + 2 =
Answer:
7 + \(\frac{2}{5}\) + 2 =  9\(\frac{2}{5}\) 
Explanation :
7 + \(\frac{2}{5}\) + 2 = 9 + \(\frac{2}{5}\) = \(\frac{45}{5}\) + \(\frac{2}{5}\) = \(\frac{47}{5}\) = 9\(\frac{2}{5}\)

Question 33.
7 + \(\frac{2}{5}\) + 2 + \(\frac{2}{5}\) =
Answer:
7 + \(\frac{2}{5}\) + 2 + \(\frac{2}{5}\) = 9\(\frac{4}{5}\)
Explanation :
7 + \(\frac{2}{5}\) + 2 + \(\frac{2}{5}\) = 9 + \(\frac{4}{5}\) = \(\frac{45}{5}\)  + \(\frac{4}{5}\) = \(\frac{49}{5}\) = 9\(\frac{4}{5}\)

Question 34.
7\(\frac{2}{5}\) + 2\(\frac{2}{5}\) =
Answer:
7\(\frac{2}{5}\) + 2\(\frac{2}{5}\) = 9\(\frac{4}{5}\)
Explanation :
7\(\frac{2}{5}\) + 2\(\frac{2}{5}\) = \(\frac{37}{5}\)  + \(\frac{12}{5}\) = \(\frac{49}{5}\) = 9\(\frac{4}{5}\)

Question 35.
6 + \(\frac{1}{3}\) + 1 + \(\frac{1}{3}\) =
Answer:
6 + \(\frac{1}{3}\) + 1 + \(\frac{1}{3}\) = 7 \(\frac{2}{3}\)
Explanation :
6 + \(\frac{1}{3}\) + 1 + \(\frac{1}{3}\) = 7 + \(\frac{2}{3}\) = \(\frac{22}{3}\) + \(\frac{2}{3}\) = \(\frac{24}{3}\)

Question 36.
6\(\frac{1}{3}\) + 1\(\frac{1}{3}\) =
Answer:
6\(\frac{1}{3}\) + 1\(\frac{1}{3}\) = 7\(\frac{2}{3}\)
Explanation :
6\(\frac{1}{3}\) + 1\(\frac{1}{3}\) = \(\frac{19}{3}\) + \(\frac{4}{3}\) = \(\frac{23}{3}\) = 7\(\frac{2}{3}\)

Question 37.
6 + \(\frac{2}{3}\) – 1 =
Answer:
6 + \(\frac{2}{3}\) – 1 = 5\(\frac{2}{3}\)
Explanation :
6 + \(\frac{2}{3}\) – 1 = 5 + \(\frac{2}{3}\) = \(\frac{15}{3}\) + \(\frac{2}{3}\) = \(\frac{17}{3}\) = 5\(\frac{2}{3}\)

Question 38.
6\(\frac{2}{3}\) – 1\(\frac{1}{3}\) =
Answer:
6\(\frac{2}{3}\) – 1\(\frac{1}{3}\) = 5\(\frac{1}{3}\)
Explanation :
6\(\frac{2}{3}\) – 1\(\frac{1}{3}\) = \(\frac{20}{3}\) – \(\frac{4}{3}\) = \(\frac{16}{3}\) = 5\(\frac{1}{3}\)

Question 39.
6\(\frac{2}{3}\) – 1\(\frac{2}{3}\) =
Answer:
6\(\frac{2}{3}\) – 1\(\frac{2}{3}\) = 5
Explanation :
6\(\frac{2}{3}\) – 1\(\frac{2}{3}\) = \(\frac{20}{3}\) – \(\frac{5}{3}\) = 5

Question 40.
3 + \(\frac{4}{7}\) + 1 + \(\frac{2}{7}\) =
Answer:
3 + \(\frac{4}{7}\) + 1 + \(\frac{2}{7}\) = 4 \(\frac{6}{7}\)
Explanation :
3 + \(\frac{4}{7}\) + 1 + \(\frac{2}{7}\) = 4 + \(\frac{6}{7}\) = \(\frac{28}{7}\) + \(\frac{6}{7}\) = \(\frac{34}{7}\) = 4\(\frac{6}{7}\)

Question 41.
3\(\frac{4}{7}\) + 1\(\frac{2}{7}\) =
Answer:
3\(\frac{4}{7}\) + 1\(\frac{2}{7}\) = 4\(\frac{6}{7}\)
Explanation :
3\(\frac{4}{7}\) + 1\(\frac{2}{7}\) = \(\frac{25}{7}\) + \(\frac{9}{7}\) = \(\frac{34}{7}\) = 4\(\frac{6}{7}\)

Question 42.
7\(\frac{4}{5}\) – 2\(\frac{3}{5}\) =
Answer:
7\(\frac{4}{5}\) – 2\(\frac{3}{5}\) = 5\(\frac{1}{5}\)
Explanation :
7\(\frac{4}{5}\) – 2\(\frac{3}{5}\) = \(\frac{39}{5}\) – \(\frac{13}{5}\) = \(\frac{26}{5}\) = 5\(\frac{1}{5}\)

Question 43.
7\(\frac{4}{5}\) – 2\(\frac{2}{5}\) =
Answer:
7\(\frac{4}{5}\) – 2\(\frac{2}{5}\) = 5\(\frac{2}{5}\)
Explanation :
7\(\frac{4}{5}\) – 2\(\frac{2}{5}\) = \(\frac{39}{5}\) – \(\frac{12}{5}\) = \(\frac{27}{5}\) = 5\(\frac{2}{5}\)

Question 44.
13\(\frac{7}{9}\) – 7\(\frac{5}{9}\) =
Answer:
13\(\frac{7}{9}\) – 7\(\frac{5}{9}\) = 6\(\frac{2}{9}\)
Explanation :
13\(\frac{7}{9}\) – 7\(\frac{5}{9}\) = \(\frac{124}{9}\) – \(\frac{68}{9}\) = \(\frac{56}{9}\)= 6\(\frac{2}{9}\)

B
Add and Subtract Whole Numbers and Ones with Fraction Units
Engage NY Math 5th Grade Module 3 Lesson 10 Sprint Answer Key 2

Question 1.
2 + 1 =
Answer:
2 + 1 = 3
Explanation :
adding 1 to 2 we get sum as 3 .

Question 2.
2 + \(\frac{1}{2}\) =
Answer:
2 + \(\frac{1}{2}\) = 2\(\frac{1}{2}\)
Explanation :
2 + \(\frac{1}{2}\) = \(\frac{4}{2}\) + \(\frac{1}{2}\) = \(\frac{5}{2}\) =2\(\frac{1}{2}\)

Question 3.
2\(\frac{1}{2}\) + 1 =
Answer:
2\(\frac{1}{2}\) + 1 = 3\(\frac{1}{2}\)
Explanation :
2\(\frac{1}{2}\) + 1 = \(\frac{5}{2}\)  + \(\frac{2}{2}\) = \(\frac{7}{2}\) = 3\(\frac{1}{2}\)

Question 4.
2 – 1 =
Answer:
2 – 1 = 1
Explanation :
Subtracting 1 from 2 we get 1as difference .

Question 5.
2\(\frac{1}{2}\) – 1 =
Answer:
2\(\frac{1}{2}\) – 1 = 1\(\frac{1}{2}\)
Explanation :
2\(\frac{1}{2}\) – 1 = \(\frac{5}{2}\)  – \(\frac{2}{2}\) = \(\frac{3}{2}\) = 1\(\frac{1}{2}\)

Question 6.
5 – 2 =
Answer:
5 – 2 = 3
Explanation :
subtract 2 from 5 we get 3 as difference .

Question 7.
5\(\frac{1}{2}\) – 2 =
Answer:
5\(\frac{1}{2}\) – 2 = 3\(\frac{1}{2}\)
Explanation :
5\(\frac{1}{2}\) – 2 = \(\frac{11}{2}\)  – \(\frac{4}{2}\)  = \(\frac{7}{2}\) = 3\(\frac{1}{2}\)

Question 8.
6 – 2 =
Answer:
6 – 2 = 4
Explanation :
subtract 2 from 6 we get 4 as difference .

Question 9.
6\(\frac{1}{3}\) – 2 =
Answer:
6\(\frac{1}{3}\) – 2 = 4\(\frac{1}{3}\)
Explanation :
6\(\frac{1}{3}\) – 2 = \(\frac{19}{3}\) – \(\frac{6}{3}\)= \(\frac{13}{3}\)= 4\(\frac{1}{3}\)

Question 10.
6\(\frac{2}{3}\) – 2 =
Answer:
6\(\frac{2}{3}\) – 2 = 4\(\frac{2}{3}\)
Explanation :
6\(\frac{2}{3}\) – 2 = \(\frac{20}{3}\) – \(\frac{6}{3}\)= \(\frac{14}{3}\)= 4\(\frac{2}{3}\)

Question 11.
6\(\frac{2}{3}\) + 2 =
Answer:
6\(\frac{2}{3}\) + 2 = 8\(\frac{2}{3}\)
Explanation :
6\(\frac{2}{3}\) + 2 = \(\frac{20}{3}\) + \(\frac{6}{3}\)= \(\frac{26}{3}\)= 8\(\frac{2}{3}\)

Question 12.
7 + 2 =
Answer:
7 + 2 = 9
Explanation:
Adding 2 to 7 we get sum as 9

Question 13.
7 + \(\frac{3}{4}\) =
Answer:
7 + \(\frac{3}{4}\) = 7\(\frac{3}{4}\)
Explanation :
7 + \(\frac{3}{4}\) = \(\frac{28}{4}\) + \(\frac{3}{4}\) = \(\frac{31}{4}\)= 7\(\frac{3}{4}\)

Question 14.
7\(\frac{3}{4}\) + 2 =
Answer:
7\(\frac{3}{4}\) + 2= 9\(\frac{3}{4}\)
Explanation :
7\(\frac{3}{4}\) + 2 = \(\frac{31}{4}\) + \(\frac{8}{4}\) = \(\frac{39}{4}\)= 9\(\frac{3}{4}\)

Question 15.
7\(\frac{3}{4}\) – 2 =
Answer:
7\(\frac{3}{4}\) – 2= 5\(\frac{3}{4}\)
Explanation :
7\(\frac{3}{4}\) – 2 = \(\frac{31}{4}\) – \(\frac{8}{4}\) = \(\frac{23}{4}\)= 5\(\frac{3}{4}\)

Question 16.
7\(\frac{3}{4}\) – 3 =
Answer:
7\(\frac{3}{4}\) – 3= 4\(\frac{3}{4}\)
Explanation :
7\(\frac{3}{4}\) – 3 = \(\frac{31}{4}\) – \(\frac{12}{4}\) = \(\frac{19}{4}\)= 4\(\frac{3}{4}\)

Question 17.
7\(\frac{3}{4}\) – 4 =
Answer:
7\(\frac{3}{4}\) – 4= 3\(\frac{3}{4}\)
Explanation :
7\(\frac{3}{4}\) – 4 = \(\frac{31}{4}\) – \(\frac{16}{4}\) = \(\frac{15}{4}\)= 3\(\frac{3}{4}\)

Question 18.
7\(\frac{3}{4}\) – 7 =
Answer:
7\(\frac{3}{4}\) – 7= \(\frac{3}{4}\)
Explanation :
7\(\frac{3}{4}\) – 7 = \(\frac{31}{4}\) – \(\frac{28}{4}\) = \(\frac{3}{4}\)

Question 19.
7\(\frac{3}{4}\) – \(\frac{3}{4}\) =
Answer:
7\(\frac{3}{4}\) – \(\frac{3}{4}\) = 7
Explanation :
7\(\frac{3}{4}\) – \(\frac{3}{4}\) = \(\frac{31}{4}\) – \(\frac{3}{4}\) = \(\frac{28}{4}\) = 7

Question 20.
3\(\frac{5}{6}\) + 2 =
Answer:
3\(\frac{5}{6}\) + 2 = 5\(\frac{5}{6}\)
Explanation :
3\(\frac{5}{6}\) + 2 = \(\frac{23}{6}\) + \(\frac{12}{6}\) = \(\frac{35}{6}\) = 5\(\frac{5}{6}\)

Question 21.
3\(\frac{1}{6}\) + 2 =
Answer:
3\(\frac{5}{6}\) + 2 = 5\(\frac{5}{6}\)
Explanation :
3\(\frac{5}{6}\) + 2 = \(\frac{23}{6}\) + \(\frac{12}{6}\) = \(\frac{35}{6}\) = 5\(\frac{5}{6}\)

Question 22.
3\(\frac{5}{6}\) + 6 =
Answer:
3\(\frac{5}{6}\) + 6 = 9\(\frac{5}{6}\)
Explanation :
3\(\frac{5}{6}\) + 6 = \(\frac{23}{6}\) + \(\frac{36}{6}\) = \(\frac{59}{6}\) = 9\(\frac{5}{6}\)

Question 23.
4\(\frac{5}{6}\) + 6 =
Answer:
4\(\frac{5}{6}\) + 6 = 10\(\frac{5}{6}\)
Explanation :
4\(\frac{5}{6}\) + 6 = \(\frac{29}{6}\) + \(\frac{36}{6}\) = \(\frac{65}{6}\) = 10\(\frac{5}{6}\)

Question 24.
6\(\frac{5}{6}\) + 4 =
Answer:
6\(\frac{5}{6}\) + 4 = 10\(\frac{5}{6}\)
Explanation :
6\(\frac{5}{6}\) + 4 = \(\frac{41}{6}\) + \(\frac{24}{6}\) = \(\frac{65}{6}\) = 10\(\frac{5}{6}\)

Question 25.
10\(\frac{5}{6}\) – 4 =
Answer:
10\(\frac{5}{6}\) – 4 = 6\(\frac{5}{6}\)
Explanation :
10\(\frac{5}{6}\) – 4 = \(\frac{65}{6}\) – \(\frac{24}{6}\) = \(\frac{41}{6}\) = 6\(\frac{5}{6}\)

Question 26.
10\(\frac{5}{6}\) – 6 =
Answer:
10\(\frac{5}{6}\) – 6 = 4\(\frac{5}{6}\)
Explanation :
10\(\frac{5}{6}\) – 6 = \(\frac{65}{6}\) – \(\frac{36}{6}\) = \(\frac{29}{6}\) = 4\(\frac{5}{6}\)

Question 27.
4 + \(\frac{4}{5}\) + 2 =
Answer:
4 + \(\frac{4}{5}\) + 2 = 6\(\frac{4}{5}\)
Explanation :
4 + \(\frac{4}{5}\) + 2 = 6 + \(\frac{4}{5}\) = \(\frac{30}{5}\) + \(\frac{4}{5}\) = \(\frac{34}{5}\) = 6\(\frac{4}{5}\)

Question 28.
6 + \(\frac{7}{8}\) + 3 =
Answer:
6 + \(\frac{7}{8}\) + 3 = 9\(\frac{7}{8}\)
Explanation :
6 + \(\frac{7}{8}\) + 3 = 9 + \(\frac{7}{8}\) = \(\frac{72}{8}\) + \(\frac{7}{8}\) = \(\frac{79}{8}\) = 9\(\frac{7}{8}\)

Question 29.
6 + \(\frac{4}{5}\) – 2 =
Answer:
6 + \(\frac{4}{5}\) – 2 = 4\(\frac{4}{5}\)
Explanation :
6 + \(\frac{4}{5}\) – 2 = 4 + \(\frac{4}{5}\) = \(\frac{20}{5}\)  + \(\frac{4}{5}\)  = \(\frac{24}{5}\) = 4\(\frac{4}{5}\)

Question 30.
9 + \(\frac{5}{12}\) – 4 =
Answer:
9 + \(\frac{5}{12}\) – 4 = 5\(\frac{5}{12}\)
Explanation :
9 + \(\frac{5}{12}\) – 4 = 5 + \(\frac{5}{12}\) = \(\frac{60}{12}\) + \(\frac{5}{12}\) = \(\frac{65}{12}\) = 5\(\frac{5}{12}\)

Question 31.
6 + \(\frac{1}{5}\) + \(\frac{1}{5}\) + 2 =
Answer:
6 + \(\frac{1}{5}\) + \(\frac{1}{5}\) + 2 = 8 \(\frac{2}{5}\)
Explanation :
6 + \(\frac{1}{5}\) + \(\frac{1}{5}\) + 2 = 8 + \(\frac{2}{5}\) = \(\frac{40}{5}\) + \(\frac{2}{5}\) = \(\frac{42}{5}\) = 8\(\frac{2}{5}\)

Question 32.
6 + \(\frac{2}{5}\) + 2 =
Answer:
6 + \(\frac{2}{5}\) + 2 = 8\(\frac{2}{5}\)
Explanation :
6 + \(\frac{2}{5}\) + 2 = 8 + \(\frac{2}{5}\) = \(\frac{40}{5}\) + \(\frac{2}{5}\)  = \(\frac{42}{5}\) = 8\(\frac{2}{5}\)

Question 33.
6 + \(\frac{2}{5}\) + 2 + \(\frac{2}{5}\) =
Answer:
6 + \(\frac{2}{5}\) + 2 + \(\frac{2}{5}\) = 8\(\frac{4}{5}\)
Explanation :
6 + \(\frac{2}{5}\) + 2 + \(\frac{2}{5}\) = 8 + \(\frac{4}{5}\) = \(\frac{40}{5}\) + \(\frac{4}{5}\) = \(\frac{44}{5}\) = 8\(\frac{4}{5}\)

Question 34.
6\(\frac{2}{5}\) + 2\(\frac{2}{5}\) =
Answer:
6\(\frac{2}{5}\) + 2\(\frac{2}{5}\) = 8\(\frac{4}{5}\)
Explanation :
6\(\frac{2}{5}\) + 2\(\frac{2}{5}\) = \(\frac{32}{5}\) + \(\frac{12}{5}\) = \(\frac{44}{5}\)= 8\(\frac{4}{5}\)

Question 35.
5 + \(\frac{1}{3}\) + 1 + \(\frac{1}{3}\) =
Answer:
5 + \(\frac{1}{3}\) + 1 + \(\frac{1}{3}\) = 6\(\frac{2}{3}\)
Explanation :
5 + \(\frac{1}{3}\) + 1 + \(\frac{1}{3}\) = 6 + \(\frac{2}{3}\) = \(\frac{18}{3}\) + \(\frac{2}{3}\) = \(\frac{20}{3}\) = 6\(\frac{2}{3}\)

Question 36.
5\(\frac{1}{3}\) + 1\(\frac{1}{3}\) =
Answer:
5\(\frac{1}{3}\) + 1\(\frac{1}{3}\) = 6\(\frac{2}{3}\)
Explanation :
5\(\frac{1}{3}\) + 1\(\frac{1}{3}\) = \(\frac{16}{3}\) + \(\frac{4}{3}\) = \(\frac{20}{3}\)= 6\(\frac{2}{3}\)

Question 37.
7 + \(\frac{2}{3}\) – 1 =
Answer:
7 + \(\frac{2}{3}\) – 1 = 6\(\frac{2}{3}\)
Explanation :
7 + \(\frac{2}{3}\) – 1 = 6 + \(\frac{2}{3}\) = \(\frac{18}{3}\)  + \(\frac{2}{3}\)  = \(\frac{20}{3}\) = 6\(\frac{2}{3}\)

Question 38.
7\(\frac{2}{3}\) – 1\(\frac{1}{3}\) =
Answer:
7\(\frac{2}{3}\) – 1\(\frac{1}{3}\) = 6\(\frac{1}{3}\)
Explanation :
7\(\frac{2}{3}\) – 1\(\frac{1}{3}\) = \(\frac{23}{3}\) – \(\frac{4}{3}\) = \(\frac{19}{3}\) = 6\(\frac{1}{3}\)

Question 39.
7\(\frac{2}{3}\) – 1\(\frac{2}{3}\) =
Answer:
7\(\frac{2}{3}\) – 1\(\frac{2}{3}\) = 6
Explanation :
7\(\frac{2}{3}\) – 1\(\frac{2}{3}\) = \(\frac{23}{3}\) – \(\frac{5}{3}\) = \(\frac{18}{3}\) = 6

Question 40.
5 + \(\frac{4}{7}\) + 1 + \(\frac{2}{7}\) =
Answer:
5 + \(\frac{4}{7}\) + 1 + \(\frac{2}{7}\) = 6\(\frac{6}{7}\)
Explanation :
5 + \(\frac{4}{7}\) + 1 + \(\frac{2}{7}\) = 6 + \(\frac{6}{7}\) = \(\frac{42}{7}\) + \(\frac{6}{7}\)  = \(\frac{48}{7}\) =6\(\frac{6}{7}\)

Question 41.
5\(\frac{4}{7}\) + 1\(\frac{2}{7}\) =
Answer:
5\(\frac{4}{7}\) + 1\(\frac{2}{7}\) = 6\(\frac{6}{7}\)
Explanation :
5\(\frac{4}{7}\) + 1\(\frac{2}{7}\) = \(\frac{39}{7}\) + \(\frac{9}{7}\) = \(\frac{48}{7}\)
= 6\(\frac{6}{7}\)

Question 42.
6 + \(\frac{4}{5}\) – 2\(\frac{3}{5}\) =
Answer:
6 + \(\frac{4}{5}\) – 2\(\frac{3}{5}\) =  4\(\frac{1}{5}\)
Explanation :
6 + \(\frac{4}{5}\) – 2\(\frac{3}{5}\) = \(\frac{30}{5}\) + \(\frac{4}{5}\) – \(\frac{13}{5}\) = \(\frac{21}{5}\) = 4\(\frac{1}{5}\)

Question 43.
6\(\frac{4}{5}\) – 2\(\frac{3}{5}\) =
Answer:
6\(\frac{4}{5}\) – 2\(\frac{3}{5}\) = 4\(\frac{1}{5}\)
Explanation :
6\(\frac{4}{5}\) – 2\(\frac{3}{5}\) = \(\frac{34}{5}\) – \(\frac{13}{5}\) = \(\frac{21}{5}\)
= 4\(\frac{1}{5}\)

Question 44.
13\(\frac{7}{9}\) – 6\(\frac{5}{9}\) =
Answer:
13\(\frac{7}{9}\) – 6\(\frac{5}{9}\) = 7\(\frac{2}{9}\)
Explanation :
13\(\frac{7}{9}\) – 6\(\frac{5}{9}\) = \(\frac{124}{9}\) – \(\frac{59}{9}\) = \(\frac{65}{9}\) = 7\(\frac{2}{9}\)

Eureka Math Grade 5 Module 3 Lesson 9 Problem Set Answer Key

Question 1.
Add.
a. 2\(\frac{1}{4}\) + 1\(\frac{1}{5}\) =
b. 2\(\frac{3}{4}\) + 1\(\frac{2}{5}\) =
c. 1\(\frac{1}{5}\) + 2\(\frac{1}{3}\) =
d. 4\(\frac{2}{3}\) + 1\(\frac{2}{5}\) =
e. 3\(\frac{1}{3}\) + 4\(\frac{5}{7}\) =
f. 2\(\frac{6}{7}\) + 5\(\frac{2}{3}\) =
g. 15\(\frac{1}{5}\) + 3\(\frac{5}{8}\) =
h. 15\(\frac{5}{8}\) + 5\(\frac{2}{5}\) =
Answer:
a.
2\(\frac{1}{4}\) + 1\(\frac{1}{5}\) =\(\frac{9}{4}\) + \(\frac{6}{5}\)
lcm of 4 and 5 is 20 .
\(\frac{45}{20}\) + \(\frac{24}{20}\) = \(\frac{69}{20}\) = 3\(\frac{9}{20}\)

b.
2\(\frac{3}{4}\) + 1\(\frac{2}{5}\) = \(\frac{11}{4}\) + \(\frac{7}{5}\)
lcm of 4 and 5 is 20 .
\(\frac{55}{20}\) + \(\frac{28}{20}\) =\(\frac{88}{20}\) = \(\frac{22}{5}\) = 4\(\frac{2}{5}\)

c.
1\(\frac{1}{5}\) + 2\(\frac{1}{3}\) =\(\frac{6}{5}\) + \(\frac{7}{3}\)
lcm of 5 and 3 is 15 .
\(\frac{18}{15}\) + \(\frac{35}{15}\) = \(\frac{53}{15}\) =3\(\frac{8}{15}\)

d.
4\(\frac{2}{3}\) + 1\(\frac{2}{5}\) = \(\frac{14}{3}\) + \(\frac{7}{5}\)
lcm of 3 and 5 is 15 .
\(\frac{70}{15}\) + \(\frac{21}{15}\) = \(\frac{91}{15}\) = 6\(\frac{1}{15}\)

e.
3\(\frac{1}{3}\) + 4\(\frac{5}{7}\) = \(\frac{10}{3}\) + \(\frac{33}{7}\)
lcm of 3 and 7 is 21.
\(\frac{70}{21}\) + \(\frac{99}{21}\) = \(\frac{169}{21}\) =8\(\frac{1}{21}\)

f.
2\(\frac{6}{7}\) + 5\(\frac{2}{3}\) = \(\frac{20}{7}\) + \(\frac{17}{3}\)
lcm of 7 and 3 is 21 .
\(\frac{60}{21}\) + \(\frac{139}{21}\) = \(\frac{199}{21}\) = 9\(\frac{10}{21}\)

g.
15\(\frac{1}{5}\) + 3\(\frac{5}{8}\) =\(\frac{76}{5}\) + \(\frac{29}{8}\)
lcm of 5 and 8 is 40 .
\(\frac{608}{40}\) + \(\frac{145}{40}\) = \(\frac{753}{40}\) = 18 \(\frac{33}{40}\)

h.
15\(\frac{5}{8}\) + 5\(\frac{2}{5}\) = \(\frac{125}{8}\) + \(\frac{12}{5}\)
lcm of 8 and 5 is 40 .
\(\frac{625}{40}\) + \(\frac{96}{40}\) = \(\frac{721}{40}\) = 18 \(\frac{1}{40}\)

Question 2.
Erin jogged 2\(\frac{1}{4}\) miles on Monday. Wednesday, she jogged 3\(\frac{1}{3}\) miles, and on Friday, she jogged 2\(\frac{2}{3}\) miles. How far did Erin jog altogether?
Answer:
Fraction of miles jogged on Monday = 2\(\frac{1}{4}\) = \(\frac{9}{4}\)
Fraction of miles jogged on Wednesday = 3\(\frac{1}{3}\) = \(\frac{10}{3}\)
Fraction of miles jogged on Friday = 2\(\frac{2}{3}\) = \(\frac{8}{3}\)
Total Fraction of miles jogged altogether = \(\frac{9}{4}\) + \(\frac{10}{3}\) + \(\frac{8}{3}\) = \(\frac{27}{12}\) + \(\frac{40}{12}\) + \(\frac{32}{12}\) = \(\frac{99}{12}\) =\(\frac{33}{4}\) = 8\(\frac{1}{4}\) .
Therefore, Total Fraction of miles jogged altogether = 8\(\frac{1}{4}\)

Question 3.
Darren bought some paint. He used 2\(\frac{1}{4}\) gallons painting his living room. After that, he had 3\(\frac{5}{6}\) gallons left. How much paint did he buy?
Answer:
Fraction of Quantity of paint used for living room = 2\(\frac{1}{4}\) gallons
Fraction of Quantity of paint left = 3\(\frac{5}{6}\) gallons
Total Fraction of Quantity of Paint = 2\(\frac{1}{4}\) + 3\(\frac{5}{6}\) = \(\frac{9}{4}\) + \(\frac{23}{6}\) = \(\frac{27}{12}\) + \(\frac{46}{12}\) = \(\frac{73}{12}\) .
Therefore, Total Fraction of Quantity of Paint = \(\frac{73}{12}\) .

Question 4.
Clayton says that 2\(\frac{1}{2}\) + 3\(\frac{3}{5}\) will be more than 5 but less than 6 since 2 + 3 is 5. Is Clayton’s reasoning correct? Prove him right or wrong.
Answer:
2\(\frac{1}{2}\) + 3\(\frac{3}{5}\) = \(\frac{5}{2}\) + \(\frac{18}{5}\) = \(\frac{25}{10}\) + \(\frac{36}{10}\) = \(\frac{61}{10}\) = 6\(\frac{1}{10}\) .
Clyton is not correct \(\frac{1}{2}\) + \(\frac{3}{5}\) is more than 1 .

Eureka Math Grade 5 Module 3 Lesson 9 Exit Ticket Answer Key

Add.
Question 1.
3\(\frac{1}{2}\) + 1\(\frac{1}{3}\) =
Answer:
3\(\frac{1}{2}\) + 1\(\frac{1}{3}\) = \(\frac{7}{2}\) + \(\frac{4}{3}\)
lcm of 2 and 3 is 6
\(\frac{21}{6}\) + \(\frac{8}{6}\) = \(\frac{29}{6}\)

Question 2.
4\(\frac{5}{7}\) + 3\(\frac{3}{4}\) =
Answer:
4\(\frac{5}{7}\) + 3\(\frac{3}{4}\) = \(\frac{33}{7}\) + \(\frac{15}{4}\)
lcm of 7 and 4 is 28
\(\frac{132}{28}\) + \(\frac{105}{28}\) = \(\frac{237}{28}\) = 8\(\frac{13}{28}\)

Eureka Math Grade 5 Module 3 Lesson 9 Homework Answer Key

Question 1.
Add.
a.2\(\frac{1}{2}\) + 1\(\frac{1}{5}\) =
b. 2\(\frac{1}{2}\) + 1\(\frac{3}{5}\) =
c. 1\(\frac{1}{5}\) + 3\(\frac{1}{3}\) =
d. 3\(\frac{2}{3}\) + 1\(\frac{3}{5}\) =
e. 2\(\frac{1}{3}\) + 4\(\frac{4}{7}\) =
f. 3\(\frac{5}{7}\) + 4\(\frac{2}{3}\) =
g. 1\(\frac{1}{5}\) + 4\(\frac{3}{8}\) =
h. 18\(\frac{3}{8}\) + 2\(\frac{2}{5}\) =
Answer:
a. 2\(\frac{1}{2}\) + 1\(\frac{1}{5}\) = 3\(\frac{7}{10}\)
Explanation :
2\(\frac{1}{2}\) + 1\(\frac{1}{5}\) = \(\frac{5}{2}\) + \(\frac{6}{5}\)
lcm of 2 and 5 is 10 .
\(\frac{25}{10}\) + \(\frac{12}{10}\) = \(\frac{37}{10}\) = 3\(\frac{7}{10}\)

b. 2\(\frac{1}{2}\) + 1\(\frac{3}{5}\) = 4 \(\frac{1}{10}\)
Explanation :
2\(\frac{1}{2}\) + 1\(\frac{3}{5}\) = \(\frac{5}{2}\) + \(\frac{8}{5}\)
lcm of 2 and 5 is 10
\(\frac{25}{10}\) + \(\frac{16}{10}\) = \(\frac{41}{10}\) = 4 \(\frac{1}{10}\)

c. 1\(\frac{1}{5}\) + 3\(\frac{1}{3}\) =4\(\frac{8}{15}\)
Explanation :
1\(\frac{1}{5}\) + 3\(\frac{1}{3}\) = \(\frac{6}{5}\) + \(\frac{10}{3}\)
lcm of 5 and 3 is 15.
\(\frac{18}{15}\) + \(\frac{50}{15}\) = \(\frac{68}{15}\) = 4\(\frac{8}{15}\)

d. 3\(\frac{2}{3}\) + 1\(\frac{3}{5}\) = 5\(\frac{4}{15}\)
Explanation :
3\(\frac{2}{3}\) + 1\(\frac{3}{5}\) = \(\frac{11}{3}\) + \(\frac{8}{5}\)
lcm of 3 and 5 = 15 .
\(\frac{55}{15}\) + \(\frac{24}{15}\) = \(\frac{79}{15}\) = 5\(\frac{4}{15}\)

e. 2\(\frac{1}{3}\) + 4\(\frac{4}{7}\) = 6\(\frac{19}{21}\)
Explanation :
2\(\frac{1}{3}\) + 4\(\frac{4}{7}\) = \(\frac{7}{3}\) + \(\frac{32}{7}\)
lcm of 3 and 7 is 21
\(\frac{49}{21}\) + \(\frac{96}{21}\) = \(\frac{145}{21}\) = 6\(\frac{19}{21}\)

f. 3\(\frac{5}{7}\) + 4\(\frac{2}{3}\) = 8 \(\frac{8}{21}\)
Explanation :
3\(\frac{5}{7}\) + 4\(\frac{2}{3}\) = \(\frac{26}{7}\) + \(\frac{14}{3}\)
lcm of 7 and 3 is 21
\(\frac{78}{21}\) + \(\frac{98}{21}\) = \(\frac{176}{21}\) = 8 \(\frac{8}{21}\)

g. 1\(\frac{1}{5}\) + 4\(\frac{3}{8}\) = 4 \(\frac{23}{40}\)
Explanation :
1\(\frac{1}{5}\) + 4\(\frac{3}{8}\) = \(\frac{6}{5}\) + \(\frac{27}{8}\)
lcm of 5 and 8 is 40 .
\(\frac{48}{40}\) + \(\frac{135}{40}\) = \(\frac{183}{40}\) = 4 \(\frac{23}{40}\)

h. 18\(\frac{3}{8}\) + 2\(\frac{2}{5}\) =
Explanation :
18\(\frac{3}{8}\) + 2\(\frac{2}{5}\) = \(\frac{147}{8}\) + \(\frac{12}{5}\)
lcm of 8 and 5 is 40 .
\(\frac{735}{40}\) + \(\frac{96}{40}\) = \(\frac{831}{40}\) = 20\(\frac{31}{40}\)

Question 2.
Angela practiced piano for 2\(\frac{1}{2}\) hours on Friday, 2\(\frac{1}{23}\) hours on Saturday, and 3\(\frac{2}{3}\) hours on Sunday. How much time did Angela practice piano during the weekend?
Answer:
Fraction of hours practiced piano on Friday = 2\(\frac{1}{2}\) hours = \(\frac{5}{2}\)
Fraction of hours practiced piano on Saturday = 2\(\frac{1}{3}\) hours = \(\frac{7}{3}\)
Fraction of hours practiced piano on Sunday = 3\(\frac{2}{3}\) hours = \(\frac{11}{3}\)
Total Fraction of hours practice piano during the weekend = \(\frac{5}{2}\) + \(\frac{7}{3}\) + \(\frac{11}{3}\) = \(\frac{15}{6}\) + \(\frac{14}{6}\) + \(\frac{22}{6}\) = \(\frac{51}{6}\) = \(\frac{17}{2}\) = 8 \(\frac{1}{2}\)
Therefore, Total Fraction of hours practice piano during the weekend = 8 \(\frac{1}{2}\)  .

Question 3..
String A is 3\(\frac{5}{6}\) meters long. String B is 2\(\frac{1}{4}\) meters long. What’s the total length of both strings?
Answer:
Fraction of length of String A = 3\(\frac{5}{6}\) = \(\frac{23}{6}\) meters
Fraction of length of String B = 2\(\frac{1}{4}\) = \(\frac{9}{4}\) meters
Fraction of total length of both strings = \(\frac{23}{6}\) + \(\frac{9}{4}\) = \(\frac{46}{12}\) + \(\frac{27}{12}\) = \(\frac{73}{12}\) = 6\(\frac{1}{12}\) meters .
Therefore, Fraction of total length of both strings = 6\(\frac{1}{12}\) meters .

Question 4.
Matt says that 5 – 1\(\frac{1}{4}\) will be more than 4, since 5 – 1 is 4. Draw a picture to prove that Matt is wrong.
Answer:
Matt is wrong 5 – 1\(\frac{1}{4}\) = 3\(\frac{3}{4}\) which is less than 4 .
Explanation :
5 – 1\(\frac{1}{4}\) = 5 – \(\frac{5}{4}\)
lcm is 4
5 – \(\frac{5}{4}\) = \(\frac{20}{4}\) – \(\frac{5}{4}\) = \(\frac{15}{4}\) = 3\(\frac{3}{4}\)
Engage-NY-Eureka-Math-5th-Grade-Module-3-Lesson-9-Answer-Key-Eureka-Math-Grade-5-Module-3-Lesson-9-Home-Work-Answer-Key-Question-4

Eureka Math Grade 5 Module 3 Lesson 9 Answer Key

Engage NY Eureka Math 5th Grade Module 3 Lesson 9 Answer Key

Eureka Math Grade 5 Module 3 Lesson 9 Sprint Answer Key

A
Add and Subtract Fractions with Like Units
Engage NY Math 5th Grade Module 3 Lesson 9 Sprint Answer Key 1

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

Question 2.
\(\frac{1}{10}\) + \(\frac{5}{10}\) =
Answer:
\(\frac{1}{10}\) + \(\frac{5}{10}\) = \(\frac{6}{10}\) = \(\frac{3}{5}\)

Question 3.
\(\frac{1}{10}\) + \(\frac{7}{10}\) =
Answer:
\(\frac{1}{10}\) + \(\frac{7}{10}\) = \(\frac{8}{10}\) = \(\frac{4}{5}\)

Question 4.
\(\frac{2}{5}\) + \(\frac{2}{5}\) =
Answer:
\(\frac{2}{5}\) + \(\frac{2}{5}\) = \(\frac{4}{5}\)

Question 5.
\(\frac{5}{10}\) – \(\frac{4}{10}\) =
Answer:
\(\frac{5}{10}\) – \(\frac{4}{10}\) = \(\frac{1}{10}\)

Question 6.
\(\frac{3}{5}\) – \(\frac{1}{5}\) =
Answer:
\(\frac{3}{5}\) – \(\frac{1}{5}\) = \(\frac{2}{5}\)

Question 7.
\(\frac{3}{10}\) + \(\frac{3}{10}\) =
Answer:
\(\frac{3}{10}\) + \(\frac{3}{10}\) = \(\frac{6}{10}\) = \(\frac{3}{5}\)

Question 8.
\(\frac{4}{5}\) – \(\frac{1}{5}\) =
Answer:
\(\frac{4}{5}\) – \(\frac{1}{5}\) = \(\frac{3}{5}\)

Question 9.
\(\frac{1}{4}\) + \(\frac{1}{4}\) =
Answer:
\(\frac{1}{4}\) + \(\frac{1}{4}\) = \(\frac{2}{4}\) = \(\frac{1}{2}\)

Question 10.
\(\frac{1}{4}\) + \(\frac{2}{4}\) =
Answer:
\(\frac{1}{4}\) + \(\frac{2}{4}\) = \(\frac{3}{4}\)

Question 11.
\(\frac{3}{12}\) – \(\frac{2}{12}\) =
Answer:
\(\frac{3}{12}\) – \(\frac{2}{12}\) = \(\frac{1}{12}\)

Question 12.
\(\frac{1}{4}\) + \(\frac{3}{4}\) =
Answer:
\(\frac{1}{4}\) + \(\frac{3}{4}\) = \(\frac{4}{4}\) = 1

Question 13.
\(\frac{1}{12}\) + \(\frac{1}{12}\) =
Answer:
\(\frac{1}{12}\) + \(\frac{1}{12}\) = \(\frac{2}{12}\) = \(\frac{1}{6}\)

Question 14.
\(\frac{1}{3}\) + \(\frac{1}{3}\) =
Answer:
\(\frac{1}{3}\) + \(\frac{1}{3}\) = \(\frac{2}{3}\)

Question 15.
\(\frac{3}{12}\) – \(\frac{2}{12}\) =
Answer:
\(\frac{3}{12}\) – \(\frac{2}{12}\) = \(\frac{1}{12}\)

Question 16.
\(\frac{5}{12}\) + \(\frac{6}{12}\) =
Answer:
\(\frac{5}{12}\) + \(\frac{6}{12}\) = \(\frac{11}{12}\)

Question 17.
\(\frac{7}{12}\) + \(\frac{4}{12}\) =
Answer:
\(\frac{7}{12}\) + \(\frac{4}{12}\) = \(\frac{11}{12}\)

Question 18.
\(\frac{4}{6}\) – \(\frac{1}{6}\) =
Answer:
\(\frac{4}{6}\) – \(\frac{1}{6}\) = \(\frac{3}{6}\) = \(\frac{1}{2}\)

Question 19.
\(\frac{1}{6}\) + \(\frac{2}{6}\) =
Answer:
\(\frac{1}{6}\) + \(\frac{2}{6}\) = \(\frac{3}{6}\) = \(\frac{1}{2}\)

Question 20.
\(\frac{1}{6}\) + \(\frac{1}{6}\) + \(\frac{1}{6}\) =
Answer:
\(\frac{1}{6}\) + \(\frac{1}{6}\) + \(\frac{1}{6}\) = \(\frac{3}{6}\) =  \(\frac{1}{2}\)

Question 21.
\(\frac{1}{3}\) + \(\frac{1}{3}\) + \(\frac{1}{3}\) =
Answer:
\(\frac{1}{3}\) + \(\frac{1}{3}\) + \(\frac{1}{3}\) = \(\frac{3}{3}\) = 1

Question 22.
\(\frac{1}{12}\) + \(\frac{1}{12}\) + \(\frac{1}{12}\) =
Answer:
\(\frac{1}{12}\) + \(\frac{1}{12}\) + \(\frac{1}{12}\) = \(\frac{3}{12}\) = \(\frac{1}{4}\)

Question 23.
\(\frac{1}{9}\) + \(\frac{1}{9}\) + \(\frac{1}{9}\) =
Answer:
\(\frac{1}{9}\) + \(\frac{1}{9}\) + \(\frac{1}{9}\) = \(\frac{3}{9}\) =  \(\frac{1}{3}\)

Question 24.
\(\frac{1}{9}\) + \(\frac{3}{9}\) + \(\frac{1}{9}\) =
Answer:
\(\frac{1}{9}\) + \(\frac{3}{9}\) + \(\frac{1}{9}\) = \(\frac{5}{9}\)

Question 25.
\(\frac{4}{9}\) – \(\frac{1}{9}\) – \(\frac{3}{9}\) =
Answer:
\(\frac{4}{9}\) – \(\frac{1}{9}\) – \(\frac{3}{9}\) = \(\frac{8}{9}\)

Question 26.
\(\frac{1}{4}\) + \(\frac{2}{4}\) + \(\frac{1}{4}\) =
Answer:
\(\frac{1}{4}\) + \(\frac{2}{4}\) + \(\frac{1}{4}\) = \(\frac{4}{4}\) = 1

Question 27.
\(\frac{1}{8}\) + \(\frac{3}{8}\) + \(\frac{2}{8}\) =
Answer:
\(\frac{1}{8}\) + \(\frac{3}{8}\) + \(\frac{2}{8}\) = \(\frac{6}{8}\) = \(\frac{3}{4}\)

Question 28.
\(\frac{5}{12}\) + \(\frac{1}{12}\) + \(\frac{5}{12}\) =
Answer:
\(\frac{5}{12}\) + \(\frac{1}{12}\) + \(\frac{5}{12}\) = \(\frac{11}{12}\)

Question 29.
\(\frac{2}{9}\) + \(\frac{3}{9}\) + \(\frac{2}{9}\) =
Answer:
\(\frac{2}{9}\) + \(\frac{3}{9}\) + \(\frac{2}{9}\) = \(\frac{7}{9}\)

Question 30.
\(\frac{3}{10}\) – \(\frac{3}{10}\) + \(\frac{3}{10}\) =
Answer:
\(\frac{3}{10}\) – \(\frac{3}{10}\) + \(\frac{3}{10}\) = \(\frac{9}{10}\)

Question 31.
\(\frac{3}{5}\) – \(\frac{1}{5}\) – \(\frac{1}{5}\) =
Answer:
\(\frac{3}{5}\) – \(\frac{1}{5}\) – \(\frac{1}{5}\) = \(\frac{3}{5}\) – \(\frac{2}{5}\) = \(\frac{1}{5}\)

Question 32.
\(\frac{1}{6}\) + \(\frac{2}{6}\) =
Answer:
\(\frac{1}{6}\) + \(\frac{2}{6}\) = \(\frac{3}{6}\) = \(\frac{1}{2}\) =\

Question 33.
\(\frac{3}{12}\) + \(\frac{4}{12}\) =
Answer:
\(\frac{3}{12}\) + \(\frac{4}{12}\) = \(\frac{7}{12}\)

Question 34.
\(\frac{3}{12}\) + \(\frac{6}{12}\) =
Answer:
\(\frac{3}{12}\) + \(\frac{6}{12}\) = \(\frac{9}{12}\) = \(\frac{3}{4}\)

Question 35.
\(\frac{4}{8}\) + \(\frac{2}{8}\) =
Answer:
\(\frac{4}{8}\) + \(\frac{2}{8}\) = \(\frac{6}{8}\) = \(\frac{3}{4}\)

Question 36.
\(\frac{4}{12}\) + \(\frac{1}{12}\) =
Answer:
\(\frac{4}{12}\) + \(\frac{1}{12}\) = \(\frac{5}{12}\)

Question 37.
\(\frac{1}{5}\) + \(\frac{3}{5}\) =
Answer:
\(\frac{1}{5}\) + \(\frac{3}{5}\) = \(\frac{4}{5}\)

Question 38.
\(\frac{2}{5}\) + \(\frac{2}{5}\) =
Answer:
\(\frac{2}{5}\) + \(\frac{2}{5}\) = \(\frac{4}{5}\)

Question 39.
\(\frac{1}{6}\) + \(\frac{2}{6}\) =
Answer:
\(\frac{1}{6}\) + \(\frac{2}{6}\) = \(\frac{3}{6}\) = \(\frac{1}{2}\)

Question 40.
\(\frac{5}{12}\) – \(\frac{3}{12}\) =
Answer:
\(\frac{5}{12}\) – \(\frac{3}{12}\) = \(\frac{2}{12}\) = \(\frac{1}{6}\)

Question 41.
\(\frac{7}{15}\) – \(\frac{2}{15}\) =
Answer:
\(\frac{7}{15}\) – \(\frac{2}{15}\) = \(\frac{5}{15}\) = \(\frac{1}{3}\)

Question 42.
\(\frac{7}{15}\) – \(\frac{3}{15}\) =
Answer:
\(\frac{7}{15}\) – \(\frac{3}{15}\) = \(\frac{4}{15}\)

Question 43.
\(\frac{11}{15}\) – \(\frac{2}{15}\) =
Answer:
\(\frac{11}{15}\) – \(\frac{2}{15}\) = \(\frac{9}{15}\)

Question 44.
\(\frac{2}{15}\) + \(\frac{4}{15}\) =
Answer:
\(\frac{2}{15}\) + \(\frac{4}{15}\) = \(\frac{6}{15}\) = \(\frac{2}{5}\)

B
Add and Subtract Fractions with Like Units
Engage NY Math 5th Grade Module 3 Lesson 9 Sprint Answer Key 2

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

Question 2.
\(\frac{2}{8}\) + \(\frac{1}{8}\) =
Answer:
\(\frac{2}{8}\) + \(\frac{1}{8}\) = \(\frac{3}{8}\)

Question 3.
\(\frac{2}{8}\) + \(\frac{3}{8}\) =
Answer:
\(\frac{2}{8}\) + \(\frac{3}{8}\) = \(\frac{5}{8}\)

Question 4.
\(\frac{2}{12}\) – \(\frac{1}{12}\) =
Answer:
\(\frac{2}{12}\) – \(\frac{1}{12}\) = \(\frac{1}{12}\)

Question 5.
\(\frac{5}{12}\) + \(\frac{2}{12}\) =
Answer:
\(\frac{5}{12}\) + \(\frac{2}{12}\) = \(\frac{7}{12}\)

Question 6.
\(\frac{4}{8}\) – \(\frac{3}{8}\) =
Answer:
\(\frac{4}{8}\) – \(\frac{3}{8}\) = \(\frac{1}{8}\)

Question 7.
\(\frac{4}{8}\) – \(\frac{3}{8}\) =
Answer:
\(\frac{4}{8}\) – \(\frac{3}{8}\) = \(\frac{1}{8}\)

Question 8.
\(\frac{1}{8}\) + \(\frac{5}{8}\) =
Answer:
\(\frac{1}{8}\) + \(\frac{5}{8}\) = \(\frac{6}{8}\) = \(\frac{3}{4}\)

Question 9.
\(\frac{3}{4}\) – \(\frac{1}{4}\) =
Answer:
\(\frac{3}{4}\) – \(\frac{1}{4}\) = \(\frac{2}{4}\) = \(\frac{1}{2}\)

Question 10.
\(\frac{3}{6}\) – \(\frac{3}{6}\) =
Answer:
\(\frac{3}{6}\) – \(\frac{3}{6}\) = 0

Question 11.
\(\frac{3}{9}\) + \(\frac{3}{9}\) =
Answer:
\(\frac{3}{9}\) + \(\frac{3}{9}\) = \(\frac{6}{9}\) = \(\frac{2}{3}\)

Question 12.
\(\frac{2}{3}\) + \(\frac{1}{3}\) =
Answer:
\(\frac{2}{3}\) + \(\frac{1}{3}\) = \(\frac{3}{3}\) = 1

Question 13.
\(\frac{6}{9}\) – \(\frac{4}{9}\) =
Answer:
\(\frac{6}{9}\) – \(\frac{4}{9}\) = \(\frac{2}{9}\)

Question 14.
\(\frac{5}{9}\) – \(\frac{3}{9}\) =
Answer:
\(\frac{5}{9}\) – \(\frac{3}{9}\) = \(\frac{2}{9}\)

Question 15.
\(\frac{2}{9}\) + \(\frac{2}{9}\) =
Answer:
\(\frac{2}{9}\) + \(\frac{2}{9}\) = \(\frac{4}{9}\)

Question 16.
\(\frac{1}{12}\) + \(\frac{3}{12}\) =
Answer:
\(\frac{1}{12}\) + \(\frac{3}{12}\) = \(\frac{4}{12}\) = \(\frac{1}{4}\)

Question 17.
\(\frac{5}{12}\) – \(\frac{4}{12}\) =
Answer:
\(\frac{5}{12}\) – \(\frac{4}{12}\) = \(\frac{1}{12}\)

Question 18.
\(\frac{9}{12}\) – \(\frac{6}{12}\) =
Answer:
\(\frac{9}{12}\) – \(\frac{6}{12}\) = \(\frac{3}{12}\) = \(\frac{1}{4}\)

Question 19.
\(\frac{6}{10}\) – \(\frac{4}{10}\) =
Answer:
\(\frac{6}{10}\) – \(\frac{4}{10}\) = \(\frac{2}{10}\) = \(\frac{1}{5}\)

Question 20.
\(\frac{2}{8}\) + \(\frac{2}{8}\) + \(\frac{2}{8}\) =
Answer:
\(\frac{2}{8}\) + \(\frac{2}{8}\) + \(\frac{2}{8}\) = \(\frac{6}{8}\) = \(\frac{3}{4}\)

Question 21.
\(\frac{1}{10}\) + \(\frac{1}{10}\) + \(\frac{1}{10}\) =
Answer:
\(\frac{1}{10}\) + \(\frac{1}{10}\) + \(\frac{1}{10}\) = \(\frac{3}{10}\) +

Question 22.
\(\frac{7}{12}\) – \(\frac{2}{10}\) – \(\frac{4}{10}\) =
Answer:
\(\frac{7}{12}\) – \(\frac{2}{10}\) – \(\frac{4}{10}\) = \(\frac{7}{12}\) – \(\frac{6}{10}\) = \(\frac{1}{10}\)

Question 23.
\(\frac{1}{12}\) + \(\frac{6}{12}\) + \(\frac{2}{12}\) =
Answer:
\(\frac{1}{12}\) + \(\frac{6}{12}\) + \(\frac{2}{12}\) = \(\frac{9}{12}\) = \(\frac{3}{4}\)

Question 24.
\(\frac{4}{12}\) + \(\frac{3}{12}\) + \(\frac{3}{12}\) =
Answer:
\(\frac{4}{12}\) + \(\frac{3}{12}\) + \(\frac{3}{12}\) = \(\frac{10}{12}\) = \(\frac{5}{6}\)

Question 25.
\(\frac{8}{12}\) – \(\frac{4}{12}\) – \(\frac{4}{12}\) =
Answer:
\(\frac{8}{12}\) – \(\frac{4}{12}\) – \(\frac{4}{12}\) = \(\frac{8}{12}\) – \(\frac{8}{12}\) = 0

Question 26.
\(\frac{1}{10}\) + \(\frac{2}{10}\) + \(\frac{4}{10}\) =
Answer:
\(\frac{1}{10}\) + \(\frac{2}{10}\) + \(\frac{4}{10}\) = \(\frac{7}{10}\)

Question 27.
\(\frac{1}{10}\) + \(\frac{1}{10}\) + \(\frac{6}{10}\) =
Answer:
\(\frac{1}{10}\) + \(\frac{1}{10}\) + \(\frac{6}{10}\) = \(\frac{8}{10}\) = \(\frac{4}{5}\)

Question 28.
\(\frac{4}{6}\) + \(\frac{1}{6}\) + \(\frac{1}{6}\) =
Answer:
\(\frac{4}{6}\) + \(\frac{1}{6}\) + \(\frac{1}{6}\) = \(\frac{6}{6}\) = 1

Question 29.
\(\frac{2}{12}\) + \(\frac{3}{12}\) + \(\frac{4}{12}\) =
Answer:
\(\frac{2}{12}\) + \(\frac{3}{12}\) + \(\frac{4}{12}\) = \(\frac{9}{12}\) = \(\frac{3}{4}\)

Question 30.
\(\frac{2}{10}\) + \(\frac{4}{10}\) + \(\frac{4}{10}\) =
Answer:
\(\frac{2}{10}\) + \(\frac{4}{10}\) + \(\frac{4}{10}\) = \(\frac{10}{10}\) = 1

Question 31.
\(\frac{3}{10}\) + \(\frac{1}{10}\) + \(\frac{2}{10}\) =
Answer:
\(\frac{3}{10}\) + \(\frac{1}{10}\) + \(\frac{2}{10}\) = \(\frac{6}{10}\) = \(\frac{3}{5}\)

Question 32.
\(\frac{4}{6}\) – \(\frac{2}{6}\) =
Answer:
\(\frac{4}{6}\) – \(\frac{2}{6}\) = \(\frac{2}{6}\) = \(\frac{1}{3}\)

Question 33.
\(\frac{3}{12}\) – \(\frac{2}{12}\) =
Answer:
\(\frac{3}{12}\) – \(\frac{2}{12}\) = \(\frac{1}{12}\)

Question 34.
\(\frac{2}{3}\) + \(\frac{1}{3}\) =
Answer:
\(\frac{2}{3}\) + \(\frac{1}{3}\) = \(\frac{3}{3}\) = 1

Question 35.
\(\frac{2}{4}\) + \(\frac{1}{4}\) =
Answer:
\(\frac{2}{4}\) + \(\frac{1}{4}\) = \(\frac{3}{4}\)

Question 36.
\(\frac{3}{12}\) + \(\frac{2}{12}\) =
Answer:
\(\frac{3}{12}\) + \(\frac{2}{12}\) = \(\frac{5}{12}\)

Question 37.
\(\frac{1}{5}\) + \(\frac{2}{5}\) =
Answer:
\(\frac{1}{5}\) + \(\frac{2}{5}\) = \(\frac{3}{5}\)

Question 38.
\(\frac{4}{5}\) – \(\frac{4}{5}\) =
Answer:
\(\frac{4}{5}\) – \(\frac{4}{5}\) = 0

Question 39.
\(\frac{5}{12}\) – \(\frac{1}{12}\) =
Answer:
\(\frac{5}{12}\) – \(\frac{1}{12}\) = \(\frac{4}{12}\) = \(\frac{1}{3}\)

Question 40.
\(\frac{6}{8}\) + \(\frac{2}{8}\) =
Answer:
\(\frac{6}{8}\) + \(\frac{2}{8}\) = \(\frac{8}{8}\) = 1

Question 41.
\(\frac{2}{8}\) + \(\frac{2}{8}\) + \(\frac{2}{8}\) =
Answer:
\(\frac{2}{8}\) + \(\frac{2}{8}\) + \(\frac{2}{8}\) = \(\frac{6}{8}\) = \(\frac{3}{4}\)

Question 42.
\(\frac{9}{10}\) – \(\frac{7}{10}\) – \(\frac{1}{10}\) =
Answer:
\(\frac{9}{10}\) – \(\frac{7}{10}\) – \(\frac{1}{10}\) = \(\frac{9}{10}\) – \(\frac{8}{10}\) = \(\frac{1}{10}\) =

Question 43.
\(\frac{2}{10}\) + \(\frac{5}{10}\) + \(\frac{2}{10}\) =
Answer:
\(\frac{2}{10}\) + \(\frac{5}{10}\) + \(\frac{2}{10}\) = \(\frac{9}{10}\)

Question 44.
\(\frac{9}{12}\) – \(\frac{1}{12}\) – \(\frac{4}{12}\) =
Answer:
\(\frac{9}{12}\) – \(\frac{1}{12}\) – \(\frac{4}{12}\) = \(\frac{9}{12}\) – \(\frac{5}{12}\) = \(\frac{4}{12}\) = \(\frac{1}{3}\)  .

Eureka Math Grade 5 Module 3 Lesson 9 Problem Set Answer Key

Question 1.
First make like units, and then add.
a. \(\frac{3}{4}\) + \(\frac{1}{7}\) =
b. \(\frac{1}{4}\) + \(\frac{9}{8}\) =
c. \(\frac{3}{8}\) + \(\frac{3}{7}\) =
d. \(\frac{4}{9}\) + \(\frac{4}{7}\) =
e. \(\frac{1}{5}\) + \(\frac{2}{3}\) =
f. \(\frac{3}{4}\) + \(\frac{5}{6}\) =
g. \(\frac{2}{3}\) + \(\frac{1}{11}\) =
h. \(\frac{3}{4}\) + 1\(\frac{1}{10}\) =
Answer:
a.
\(\frac{3}{4}\) + \(\frac{1}{7}\)
lcm of 4 and 7 is 28
=\(\frac{21}{28}\) + \(\frac{4}{28}\) = \(\frac{25}{28}\)
b.
\(\frac{1}{4}\) + \(\frac{9}{8}\)
lcm of 4 and 8 is 8
\(\frac{2}{8}\) + \(\frac{9}{8}\) =\(\frac{11}{8}\) = 1\(\frac{3}{8}\)
c.
\(\frac{3}{8}\) + \(\frac{3}{7}\)
lcm of 8 and 7 is 56
\(\frac{21}{56}\) + \(\frac{24}{56}\) = \(\frac{45}{56}\)
d.
\(\frac{4}{9}\) + \(\frac{4}{7}\)
lcm of 9 and 7 is 63
\(\frac{28}{63}\) + \(\frac{36}{63}\) = \(\frac{64}{63}\) = 1\(\frac{1}{63}\)
e.
\(\frac{1}{5}\) + \(\frac{2}{3}\)
lcm of 5 and 3 is 15 .
\(\frac{3}{15}\) + \(\frac{10}{15}\) = \(\frac{13}{15}\)
f.
\(\frac{3}{4}\) + \(\frac{5}{6}\)
lcm of 4 and 6 is 12.
\(\frac{9}{12}\) + \(\frac{10}{12}\) = \(\frac{19}{12}\) =1 \(\frac{7}{12}\)
g.
\(\frac{2}{3}\) + \(\frac{1}{11}\)
lcm of 3 and 11 is 33
\(\frac{22}{33}\) + \(\frac{3}{33}\) = \(\frac{25}{33}\)
h.
\(\frac{3}{4}\) + 1\(\frac{1}{10}\) = \(\frac{3}{4}\) + \(\frac{11}{10}\)
lcm of 4 and 10 is 20.
\(\frac{15}{20}\) + \(\frac{22}{10}\) = \(\frac{37}{20}\) = 1\(\frac{17}{20}\)

Question 2.
Whitney says that to add fractions with different denominators, you always have to multiply the denominators to find the common unit; for example:
\(\frac{1}{4}+\frac{1}{6}=\frac{6}{24}+\frac{4}{24}\)
Show Whitney how she could have chosen a denominator smaller than 24, and solve the problem.
Answer:
multiples of 4 and 6 are
4 : 4, 8, 12, 16, 20, 24
6: 6, 12, 18, 24, 30 .
12 and 24 are the common multiplies of 4 and 6. smaller than 24 we get 12 multiple .
(\(\frac{1 × 3}{4 × 3}\)) + (\(\frac{1 × 2}{6 × 2}\)) = \(\frac{3}{12}\) + \(\frac{2}{12}\) = \(\frac{5}{12}\)

Question 3.
Jackie brought \(\frac{3}{4}\) of a gallon of iced tea to the party. Bill brought \(\frac{7}{8}\) of a gallon of iced tea to the same party. How much iced tea did Jackie and Bill bring to the party?
Answer:
Fraction of iced tea brought by Jackie = \(\frac{3}{4}\)
Fraction of iced tea brought by Bill = \(\frac{7}{8}\)
Total Fraction of iced tea brought to party = \(\frac{3}{4}\) + \(\frac{7}{8}\)  = \(\frac{6}{8}\) + \(\frac{7}{8}\) = \(\frac{13}{8}\) = 1\(\frac{5}{8}\)
Therefore, Total Fraction of iced tea brought to party = \(\frac{13}{8}\) = 1\(\frac{5}{8}\) .

Question 4.
Madame Curie made some radium in her lab. She used \(\frac{2}{5}\) kg of the radium in an experiment and had 1\(\frac{1}{4}\) kg left. How much radium did she have at first? (Extension: If she performed the experiment twice, how much radium would she have left?)
Answer:
Quantity of Radium made by Madam Curie = x kgs
Fraction of Quantity of Radium used by Experiment = \(\frac{2}{5}\) kg
Fraction of Quantity of Radium left = 1\(\frac{1}{4}\) kg = \(\frac{5}{4}\) kg
Quantity of Radium made by Madam Curie = \(\frac{2}{5}\) + \(\frac{5}{4}[/latex
lcm of 5 and 4 is 20 .
[latex]\frac{8}{20}\)  + \(\frac{25}{20}\) = \(\frac{33}{20}\) =1\(\frac{13}{20}\) .
Therefore if the experiment is done once the Total Quantity = \(\frac{33}{20}\) =1\(\frac{13}{20}\)
If the Experiment if done twiced .
Total Quantity – Quantity Used for Experiment twice = left Quantity .
\(\frac{33}{20}\) – 2 × \(\frac{2}{5}\) = \(\frac{33}{20}\) –  \(\frac{4}{5}\) = \(\frac{33}{20}\) – \(\frac{16}{20}\) = \(\frac{17}{20}\)
Therefore if the experiment is done once the Total Quantity = \(\frac{17}{20}\)

Eureka Math Grade 5 Module 3 Lesson 9 Exit Ticket Answer Key

Make like units, and then add.
a. \(\frac{1}{6}\) + \(\frac{3}{4}\) =
b. 1\(\frac{1}{2}\) + \(\frac{2}{5}\) =
Answer:
a.
\(\frac{1}{6}\) + \(\frac{3}{4}\)
lcm of 6 and 4 is 12
\(\frac{2}{12}\) + \(\frac{9}{12}\) = \(\frac{11}{12}\)
b.
1\(\frac{1}{2}\) + \(\frac{2}{5}\) = \(\frac{3}{2}\) + \(\frac{2}{5}\)
lcm of 2 and 5 is 10.
\(\frac{15}{10}\) + \(\frac{4}{10}\) =\(\frac{19}{10}\) = 1\(\frac{9}{10}\)

Eureka Math Grade 5 Module 3 Lesson 9 Homework Answer Key

Question 1.
Make like units, and then add.
a. \(\frac{3}{5}\) + \(\frac{1}{3}\) =
b. \(\frac{3}{5}\) + \(\frac{1}{11}\) =
c. \(\frac{2}{9}\) + \(\frac{5}{6}\) =
d. \(\frac{2}{5}\) + \(\frac{1}{4}\) + \(\frac{1}{10}\) =
e. \(\frac{1}{3}\) + \(\frac{7}{5}\) =
f. \(\frac{5}{8}\) + \(\frac{7}{12}\) =
g. 1\(\frac{1}{3}\) + \(\frac{3}{4}\) =
h. \(\frac{5}{6}\) + 1\(\frac{1}{4}\) =
Answer:
a.
\(\frac{3}{5}\) + \(\frac{1}{3}\)
lcm of 5 and 3 is 15
\(\frac{9}{15}\) + \(\frac{5}{15}\) = \(\frac{14}{15}\)
b.
\(\frac{3}{5}\) + \(\frac{1}{11}\)
lcm of 5 and 11 is 55
\(\frac{33}{55}\) + \(\frac{5}{55}\) = \(\frac{38}{55}\)
c.
\(\frac{2}{9}\) + \(\frac{5}{6}\)
lcm of 9 and 6 is 18 .
\(\frac{4}{18}\) + \(\frac{15}{18}\) = \(\frac{19}{18}\) = 1 \(\frac{1}{18}\)
d.
\(\frac{2}{5}\) + \(\frac{1}{4}\) + \(\frac{1}{10}\)
lcm of 5 , 4 and 10 is 20 .
\(\frac{8}{20}\) + \(\frac{5}{20}\) + \(\frac{2}{20}\) = \(\frac{15}{20}\)= \(\frac{3}{4}\)
e.
\(\frac{1}{3}\) + \(\frac{7}{5}\)
lcm of 3 and 5 is 15 .
\(\frac{5}{15}\) + \(\frac{21}{15}\) =\(\frac{26}{3}\) =1\(\frac{11}{15}\)
f.
\(\frac{5}{8}\) + \(\frac{7}{12}\)
lcm of 8 and 12 is 24.
\(\frac{15}{24}\) + \(\frac{14}{24}\) = \(\frac{29}{24}\) = 1\(\frac{5}{24}\)
g.
1\(\frac{1}{3}\) + \(\frac{3}{4}\) = \(\frac{4}{3}\) + \(\frac{3}{4}\)
lcm of 3 and 4  is 12
\(\frac{16}{12}\) + \(\frac{9}{12}\) = \(\frac{25}{12}\) = 2 \(\frac{1}{12}\)
h.
\(\frac{5}{6}\) + 1\(\frac{1}{4}\) =\(\frac{5}{6}\) + \(\frac{5}{4}\)
lcm of 4 and 6 is 12 .
\(\frac{10}{12}\) + \(\frac{15}{12}\) = \(\frac{25}{12}\) = 2\(\frac{1}{12}\)

Question 2.
On Monday, Ka practiced guitar for \(\frac{2}{3}\) of one hour. When she finished, she practiced piano for \(\frac{3}{4}\) of one hour. How much time did Ka spend practicing instruments on Monday?
Answer:
Fraction of Time spent in playing guitar of one hour = \(\frac{2}{3}\)
Fraction of Time spent in playing guitar when finished = \(\frac{3}{4}\)
Total Time taken for practicing = \(\frac{2}{3}\) + \(\frac{3}{4}\) = \(\frac{8}{12}\) + \(\frac{9}{12}\) = \(\frac{17}{12}\) = 1\(\frac{5}{12}\) hour .
Therefore, Total Time taken in practicing = \(\frac{17}{12}\) = 1\(\frac{5}{12}\) hour

Question 3.
Ms. How bought a bag of rice for dinner. She used \(\frac{3}{5}\) kg of the rice and still had 2\(\frac{1}{4}\) kg left. How heavy was the bag of rice that Ms. How bought?
Answer:
Fraction of Quantity of rice used = \(\frac{3}{5}\) kg
Fraction of Quantity of rice left = 2\(\frac{1}{4}\) kg
Total Quantity of rice = \(\frac{3}{5}\)  + 2\(\frac{1}{4}\)  = \(\frac{3}{5}\) + \(\frac{9}{4}\)
= \(\frac{12}{20}\) + \(\frac{45}{20}\) =\(\frac{57}{20}\) = 2\(\frac{17}{20}\)
Therefore, Total Quantity of rice = \(\frac{57}{20}\) = 2\(\frac{17}{20}\) .

Question 4.
Joe spends \(\frac{2}{5}\) of his money on a jacket and \(\frac{3}{8}\) of his money on a shirt. He spends the rest on a pair of pants. What fraction of his money does he use to buy the pants?
Answer:
Money spent on jacket = \(\frac{2}{5}\)
Money spent on a shirt = \(\frac{3}{8}\)
Money spent on pair of pants = x
1 = \(\frac{2}{5}\) + \(\frac{3}{8}\)  + x
lcm of 5 and 8 is 40.
\(\frac{40}{40}\)  = \(\frac{16}{40}\) + \(\frac{10}{40}\) +x
\(\frac{40}{40}\)  = \(\frac{26}{40}\) + x
x = \(\frac{40}{40}\)  – \(\frac{13}{40}\)
x = \(\frac{27}{40}\)