## 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. b. 1 ribbon. The piece shown below is $$\frac{4}{5}$$ of the whole. Complete the drawing to show the whole ribbon. 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.
a. Explanation :
The below ribbon into 3 parts each part is $$\frac{1}{3}$$ of the whole .
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. 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. 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?
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?
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. b. 1 ribbon. The piece shown below is $$\frac{1}{4}$$ of the whole. Complete the drawing to show the whole ribbon. 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.
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. Explanation :
The given Ribbon is divided into 4 parts each part is $$\frac{1}{4}$$ .
c. Explanation :
Length of A = 3B.
Length of C = $$\frac{B}{2}$$ .

### Eureka Math Grade 5 Module 3 Lesson 16 Homework Answer Key

a. 1 road. The piece shown below is only $$\frac{3}{7}$$ of the whole. Complete the drawing to show the whole road. b. 1 road. The piece shown below is $$\frac{1}{6}$$ of the whole. Complete the drawing to show the whole road. 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?
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. 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 .