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

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

Question 1.

Determine the following sums and differences. Show your work.

a. 1 ft + 2 ft = _______ yd

b. 3 yd 1 ft + 2 ft = _______ yd

c. 1 yd – 1 ft = _______ ft

d. 8 yd – 1 ft = _______ yd _______ ft

e. 3 in + 9 in = _______ ft

f. 6 in + 9 in = _______ ft _______ in

g. 1 ft – 8 in = _______ in

h. 5 ft – 8 in = _______ ft _______ in

Answer:

a. 1 ft + 2 ft = 1 yd

b. 3 yd 1 ft + 2 ft = 4 yd

c. 1 yd – 1 ft = 2 ft

d. 8 yd – 1 ft =7 yd 2 ft

e. 3 in + 9 in = 1 ft

f. 6 in + 9 in = 1 ft 3 in

g. 1 ft – 8 in = 4 in

h. 5 ft – 8 in = 4 ft 4 in

Question 2.

Find the following sums and differences. Show your work.

a. 5 yd 2 ft + 2 ft = _______ yd _______ ft

b. 7 yd 2 ft + 2 yd 2 ft = _______ yd _______ ft

c. 4 yd 1 ft – 2 ft = _______ yd _______ ft

d. 6 yd 1 ft – 2 yd 2 ft =_______ yd _______ ft

e. 6 ft 9 in + 4 in = _______ ft _______ in

f. 4 ft 4 in + 3 ft 11 in =_______ ft _______ in

g. 34 ft 4 in – 8 in = _______ ft _______ in

h. 7 ft 1 in – 5 ft 10 in = _______ ft _______ in

Answer:

a. 5 yd 2 ft + 2 ft = 6 yd 1 ft

b. 7 yd 2 ft + 2 yd 2 ft = 10 yd 1 ft

c. 4 yd 1 ft – 2 ft = 3 yd 2 ft

d. 6 yd 1 ft – 2 yd 2 ft =3 yd 2 ft

e. 6 ft 9 in + 4 in = 7 ft 1 in

f. 4 ft 4 in + 3 ft 11 in = 8 ft 3 in

g. 34 ft 4 in – 8 in = 33 ft 8 in

h. 7 ft 1 in – 5 ft 10 in = 1 ft 3 in

Question 3.

Matthew is 6 feet 2 inches tall. His little cousin Emma is 3 feet 6 inches tall. How much taller is Matthew than Emma?

Answer:

The height of Matthew = 6 feet 2 inches Which means, 74 inches

The height of his little cousin Emma = 3 feet 6 inches Which means, 42 inches

Now, the age difference = 6 ft 2 in – 3 ft 6 in = 2 ft 8 in

Therefore, the Mathew is 2 feet 8 inches than Emma.

Question 4.

In gym class, Jared climbed 10 feet 4 inches up a rope. Then, he continued to climb up another 3 feet 9 inches. How high did Jared climb?

Answer:

Given that,

Firstly,Jared climbed 10 feet 4 inches

The length he continued climbing = 3 feet 9 inches

Total length = 10 feet 4 inches + 3 feet 9 inches = 13 feet and 13 inches

Also written as 14 feet 1 inch

Therefore, the total length Jared climbed = 14 feet 1 inch.

Question 5.

A quadrilateral has a perimeter of 18 feet 2 inches. The sum of three of the sides is 12 feet 4 inches.

a. What is the length of the fourth side?

b. An equilateral triangle has a side length equal to the fourth side of the quadrilateral. What is the perimeter of the triangle?

Answer:

a.

Given , the parameter of quadrilateral = 18 feet 2 inches or 17 feet 14 inches

The sum of three of the sides = 12 feet 4 inches

The length of fourth side = 18 ft 2 in – 12 ft 4 in = 5 ft 10 in

Therefore, the length of forth side = 5 ft 10 in

b.

The perimeter of triangle =

3 x (5 ft 10 in)

15 ft 30 inches

30 inches can also be written as 24 inches 6 inches

24 inches = 2 ft

Total = 17 ft 6 in

Therefore, the perimeter of triangle = 17 ft 6 in.

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

Determine the following sums and differences. Show your work.

Question 1.

4 yd 1 ft + 2 ft ________ yd

Answer:

4 yd 1 ft + 2 ft = 5 yd

Question 2.

6 yd – 1 ft = ________ yd ________ ft

Answer:

6 yd – 1 ft = 5 yd 2 ft

Question 3.

4 yd 1 ft + 3 yd 2 ft = ________ yd

Answer:

4 yd 1 ft + 3 yd 2 ft = 8 yd

Question 4.

8 yd 1 ft – 3 yd 2 ft =________ yd ________ ft

Answer:

8 yd 1 ft – 3 yd 2 ft = 4 yd 2 ft

### Eureka Math Grade 4 Module 7 Lesson 7 Homework Answer Key

Question 1.

Determine the following sums and differences. Show your work.

a. 2 yd 2 ft + 1 ft = ________ yd

b. 2 yd – 1 ft = ________ yd ________ ft

c. 2 ft + 2 ft = ________ yd ________ ft

d. 5 yd – 1 ft = ________ yd ________ft

e. 7 in + 5 in = ________ ft

f. 7 in + 7 in =________ ft ________ in

g. 1 ft – 2 in = ________ in

h. 2 ft – 6 in = ________ ft ________ in

Answer:

a. 2 yd 2 ft + 1 ft = 3 yd

b. 2 yd – 1 ft = 1 yd 2 ft

c. 2 ft + 2 ft = 1 yd 1 ft

d. 5 yd – 1 ft = 4 yd 2 ft

e. 7 in + 5 in = 1 ft

f. 7 in + 7 in = 1 ft 2 in

g. 1 ft – 2 in = 10 in

h. 2 ft – 6 in = 1 ft 6 in

Question 2.

Find the following sums and differences. Show your work.

a. 4 yd 2 ft + 2 ft = ________ yd ________ ft

b. 6 yd 2 ft + 1 yd 1 ft = ________ yd ________ ft

c. 5 yd 1 ft – 2 ft = ________ yd ________ ft

d. 7 yd 1 ft – 5 yd 2 ft = ________ yd ________ ft

e. 7 ft 8 in + 5 in = ________ ft ________ in

f. 6 ft 5 in + 5 ft 9 in = ________ ft ________ in

g. 32 ft 3 in – 7 in = ________ ft ________ in

h. 8 ft 2 in – 3 ft 11 in = ________ ft ________ in

Answer:

a. 4 yd 2 ft + 2 ft = 5 yd 1 ft

b. 6 yd 2 ft + 1 yd 1 ft = 8 yd 0 ft

c. 5 yd 1 ft – 2 ft = 4 yd 2 ft

d. 7 yd 1 ft – 5 yd 2 ft = 1 yd 2 ft

e. 7 ft 8 in + 5 in = 8 ft 1 in

f. 6 ft 5 in + 5 ft 9 in = 12 ft 2 in

g. 32 ft 3 in – 7 in = 31 ft 8 in

h. 8 ft 2 in – 3 ft 11 in = 4 ft 3 in

Question 3.

Laurie bought 9 feet 5 inches of blue ribbon. She also bought 6 feet 4 inches of green ribbon. How much ribbon did she buy altogether?

Answer:

Given that,

The length of blue ribbon Laurie bought = 9 feet 5 inches

The length of green ribbon Laurie bought = 6 feet 4 inches

Total length : 9 ft 5 in + 6 ft 4 in

= 15 ft 9 in

Therefore, the total length of ribbons Laurie bought altogether = 15 feet 9 inches.

Question 4.

The length of the room is 11 feet 6 inches. The width of the room is 2 feet 9 inches shorter than the length. What is the width of the room?

Answer:

11 feet 6 inches can also be written as 10 feet 18 inches

Given that the width of the room is 2 feet 9 inches smaller than the length

Now, 10 feet 18 inches – 2 feet 9 inches

= 8 feet 9 inches

Therefore, the width of the room = 8 feet 9 inches.

Question 5.

Tim’s bedroom is 12 feet 6 inches wide. The perimeter of the rectangular-shaped bedroom is 50 feet.

a. What is the length of Tim’s bedroom?

b. How much longer is the length of Tim’s room than the width?

Answer:

a.

12 feet 6 inches + 12 feet 6 inches = 50

25 ft = 50

= 12 ft 6 in

Therefore, the length of Tim’s room is 12 feet 6 inches.

b.

The length and width are the same .