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.
Answer:
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?
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.
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.
Answer:
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
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.
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?
d. Write your own road problem with 2 or 3 lengths.
Answer:
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 .