## Engage NY Eureka Math 4th Grade Module 5 Lesson 37 Answer Key

### Eureka Math Grade 4 Module 5 Lesson 37 Problem Set Answer Key

Question 1.

Draw tape diagrams to show two ways to represent 2 units of 4\(\frac{2}{3}\).

Write a multiplication expression to match each tape diagram.

Answer:

2 × 4\(\frac{2}{3}\)

Explanation:

In the above-given question,

given that,

2 × 4\(\frac{2}{3}\)

2 × (4 + \(\frac{2}{3}\))

(2 × 4) + (2 × \(\frac{2}{3}\))

8 + \(\frac{4}{3}\)

8 + 1\(\frac{1}{3}\)

9\(\frac{1}{3}\)

Question 2.

Solve the following using the distributive property. The first one has been done for you. (As soon as you are ready, you may omit the step that is in line 2.)

a. 3 × 6\(\frac{4}{5}\) = 3 × (6 + \(\frac{4}{5}\))

= (3 × 6) + (3 × \(\frac{4}{5}\))

= 18 + \(\frac{12}{5}\)

= 18 + 2\(\frac{2}{5}\)

= 20\(\frac{2}{5}\)

b. 2 × 4\(\frac{2}{3}\)

Answer:

2 × 4\(\frac{2}{3}\)

Explanation:

In the above-given question,

given that,

2 × 4\(\frac{2}{3}\)

2 × (4 + \(\frac{2}{3}\))

(2 × 4) + (2 × \(\frac{2}{3}\))

8 + \(\frac{4}{3}\)

8 + 1\(\frac{1}{3}\)

9\(\frac{1}{3}\)

c. 3 × 2\(\frac{5}{8}\)

Answer:

3 × 2\(\frac{5}{8}\)

Explanation:

In the above-given question,

given that,

3 × 2\(\frac{5}{8}\)

3 × (2 + \(\frac{5}{8}\))

(3 × 2) + (3 × \(\frac{5}{8}\))

6 + \(\frac{15}{8}\)

6 + 1\(\frac{7}{8}\)

7\(\frac{7}{8}\)

d. 2 × 4\(\frac{7}{10}\)

Answer:

2 × 4\(\frac{7}{10}\)

Explanation:

In the above-given question,

given that,

2 × 4\(\frac{7}{10}\)

2 × (4 + \(\frac{7}{10}\))

(2 × 4) + (2 × \(\frac{7}{10}\))

8 + \(\frac{14}{10}\)

8 + 1\(\frac{4}{10}\)

9\(\frac{4}{10}\)

e. 3 7\(\frac{3}{4}\)

Answer:

3 × 7\(\frac{3}{4}\)

Explanation:

In the above-given question,

given that,

3 × 7\(\frac{3}{4}\)

3 × (7 + \(\frac{3}{4}\))

(3 × 7) + (3 × \(\frac{3}{4}\))

21 + \(\frac{9}{4}\)

21 + 2\(\frac{1}{4}\)

23\(\frac{1}{4}\)

f. 6 × 3\(\frac{1}{2}\)

Answer:

6 × 3\(\frac{1}{2}\)

Explanation:

In the above-given question,

given that,

6 × 3\(\frac{1}{2}\)

6 × (3 + \(\frac{1}{2}\))

(6 × 3) + (6 × \(\frac{1}{2}\))

18 + \(\frac{6}{2}\)

18 + 2\(\frac{2}{2}\)

20\(\frac{2}{2}\)

g. 4 × 9\(\frac{1}{5}\)

Answer:

4 × 9\(\frac{1}{5}\)

Explanation:

In the above-given question,

given that,

4 × 9\(\frac{1}{5}\)

4 × (9 + \(\frac{1}{5}\))

(4 × 9) + (4 × \(\frac{1}{5}\))

36 + \(\frac{4}{5}\)

36 + \(\frac{4}{5}\)

36\(\frac{4}{5}\)

h. 5\(\frac{6}{8}\) × 4

Answer:

5 × 4\(\frac{6}{8}\)

Explanation:

In the above-given question,

given that,

5 × 4\(\frac{6}{8}\)

5 × (4 + \(\frac{6}{8}\))

(5 × 4) + (5 × \(\frac{6}{8}\))

20 + \(\frac{30}{8}\)

20 + 3\(\frac{6}{8}\)

23\(\frac{6}{8}\)

Question 3.

For one dance costume, Saisha needs 4\(\frac{2}{3}\) feet of ribbon. How much ribbon does she need for 5 identical costumes?

Answer:

The ribbon does she need for 5 identical costumes = 13 feet.

Explanation:

In the above-given question,

given that,

For one dance costume, Saisha needs 4(2/3) feet of ribbon.

4(2/3) = 4 x 2/3.

4 x 2 = 8/3.

8/3 = 2.6.

2.6 x 5 = 13.

### Eureka Math Grade 4 Module 5 Lesson 37 Exit Ticket Answer Key

Multiply. Write each product as a mixed number.

Question 1.

4 × 5\(\frac{3}{8}\)

Answer:

4 × 5\(\frac{3}{8}\)

Explanation:

In the above-given question,

given that,

4 × 5\(\frac{3}{8}\)

4 × (5 + \(\frac{3}{8}\))

(4 × 5) + (4 × \(\frac{3}{8}\))

20 + \(\frac{12}{8}\)

20 + 1\(\frac{4}{8}\)

21\(\frac{4}{8}\)

Question 2.

4\(\frac{3}{10}\) × 3

Answer:

4 × 3\(\frac{3}{10}\)

Explanation:

In the above-given question,

given that,

4 × 3\(\frac{3}{10}\)

4 × (3 + \(\frac{3}{10}\))

(4 × 3) + (4 × \(\frac{3}{10}\))

12 + \(\frac{12}{10}\)

12 + 1\(\frac{2}{10}\)

13\(\frac{2}{10}\)

## Eureka Math Grade 4 Module 5 Lesson 37 Homework Answer Key

Question 1.

Draw tape diagrams to show two ways to represent 3 units of 5\(\frac{1}{12}\)” .

Write a multiplication expression to match each tape diagram.

Answer:

3 × 5\(\frac{1}{12}\)

Explanation:

In the above-given question,

given that,

3 × 5\(\frac{1}{12}\)

3 × (5 + \(\frac{1}{12}\))

(3 × 5) + (3 × \(\frac{1}{12}\))

15 + \(\frac{3}{12}\)

15 + \(\frac{1}{4}\)

15\(\frac{1}{4}\)

Question 2.

Solve the following using the distributive property. The first one has been done for you. (As soon as you are ready, you may omit the step that is in line 2.)

a. 3 × 6\(\frac{4}{5}\) = 3 × (6 + \(\frac{4}{5}\))

= (3 × 6) + (3 × \(\frac{4}{5}\))

= 18 + \(\frac{12}{5}\)

= 18 + 2\(\frac{2}{5}\)

= 20\(\frac{2}{5}\)

b. 5 × 4\(\frac{1}{6}\)

Answer:

5 × 4\(\frac{1}{6}\)

Explanation:

In the above-given question,

given that,

5 × 4\(\frac{1}{6}\)

5 × (4 + \(\frac{1}{6}\))

(5 × 4) + (5 × \(\frac{1}{6}\))

20 + \(\frac{5}{6}\)

20 + \(\frac{5}{6}\)

20\(\frac{5}{6}\)

c. 6 × 2\(\frac{3}{5}\)

Answer:

6 × 2\(\frac{3}{5}\)

Explanation:

In the above-given question,

given that,

6 × 2\(\frac{3}{5}\)

6 × (2 + \(\frac{3}{5}\))

(6 × 2) + (6 × \(\frac{3}{5}\))

12 + \(\frac{18}{5}\)

12 + 3\(\frac{3}{5}\)

15\(\frac{3}{5}\)

d. 2 × 7\(\frac{3}{10}\)

Answer:

2 × 7\(\frac{3}{10}\)

Explanation:

In the above-given question,

given that,

2 × 7\(\frac{3}{10}\)

2 × (7 + \(\frac{3}{10}\))

(2× 7) + (2 × \(\frac{3}{10}\))

14 + \(\frac{6}{10}\)

14 + \(\frac{3}{5}\)

14\(\frac{3}{5}\)

e. 8 × 7\(\frac{1}{4}\)

Answer:

8 × 7\(\frac{1}{4}\)

Explanation:

In the above-given question,

given that,

8 × 7\(\frac{1}{4}\)

8 × (7 + \(\frac{1}{4}\))

(8 × 7) + (8 × \(\frac{1}{4}\))

56 + \(\frac{8}{4}\)

56 + \(\frac{2}{0}\)

56\(\frac{2}{0}\)

f. 3\(\frac{3}{8}\) × 12

Answer:

3 × 12\(\frac{3}{8}\)

Explanation:

In the above-given question,

given that,

3 × 12\(\frac{3}{8}\)

3 × (12 + \(\frac{3}{8}\))

(3 × 12) + (3 × \(\frac{3}{8}\))

36 + \(\frac{9}{8}\)

36 + 1\(\frac{1}{8}\)

37\(\frac{1}{8}\)

Question 3.

Sara’s street is 2\(\frac{3}{10}\) miles long. She ran the length of the street 6 times. How far did she run?

Answer:

The far did she run = 16.8

Explanation:

In the above-given question,

given that,

Sara’s street is 2(3/10) miles long.

she ran the length of the street 6 times.

2(3/10).

6/10 = 0.6.

Question 4.

Kelly’s new puppy weighed 4\(\frac{7}{10}\) pounds when she brought him home. Now, he weighs six times as much. How much does he weigh now?

Answer:

The weight now = 16.8

Explanation:

In the above-given question,

given that,

Kelly’s new puppy weighed 4(7/10) pounds when she brought him home.

he weighs six times as much.

4(7/10) = 28/10.

28/10 = 2.8.

2.8 x 6 = 16.8.