We included **HMH Into Math Grade 5 Answer Key**** PDF** **Module 7 Review **to make students experts in learning maths.

## HMH Into Math Grade 5 Module 7 Review Answer Key

**Vocabulary**

Associative Property of

Addition

benchmark

common denominator

Commutative Property of

Addition

equivalent fraction

**Complete each sentence by choosing a term from the Vocabulary Box.**

Question 1.

You can use a ___ to compare and estimate fractions.

Answer: You can use a **benchmark** to compare and estimate fractions.

Question 2.

An ___ uses a different numerator and denominator to name the same part of a whole.

Answer: An **equivalent fraction** uses a different numerator and denominator to name the same part of a whole.

Question 3.

The ___ of two fractions is a common multiple of their denominators.

Answer: The **common denominator** of two fractions is a common multiple of their denominators.

Question 4.

\(\frac{1}{3}\) + \(\frac{3}{4}\) = \(\frac{3}{4}\) + \(\frac{1}{3}\) is an example of the __________

Answer: \(\frac{1}{3}\) + \(\frac{3}{4}\) = \(\frac{3}{4}\) + \(\frac{1}{3}\) is an example of the **Commutative Property of addition**

Question 5.

(\(\frac{3}{8}\) + \(\frac{1}{6}\)) + \(\frac{5}{6}\) = \(\frac{3}{8}\) + (\(\frac{1}{6}\) + \(\frac{5}{6}\)) is an example of the

_______________________

Answer: (\(\frac{3}{8}\) + \(\frac{1}{6}\)) + \(\frac{5}{6}\) = \(\frac{3}{8}\) + (\(\frac{1}{6}\) + \(\frac{5}{6}\)) is an example of the **Associative Property of addition**

**Concepts and Skills**

**Estimate the sum or difference.**

Question 6.

\(\frac{4}{10}\) + \(\frac{7}{12}\) _____

Answer:

\(\frac{4}{10}\) + \(\frac{7}{12}\)

The denominators of the given fractions are not the same. So we have to find the LCD.

LCD of 10 and 12 is 60.

\(\frac{24}{60}\) + \(\frac{35}{60}\) = \(\frac{59}{60}\)

\(\frac{4}{10}\) + \(\frac{7}{12}\) = \(\frac{59}{60}\)

Question 7.

\(\frac{7}{8}\) – \(\frac{2}{5}\) _____

Answer:

\(\frac{7}{8}\) – \(\frac{2}{5}\)

The denominators of the given fractions are not the same. So we have to find the LCD.

LCD of 8 and 5 is 40.

\(\frac{7}{8}\) – \(\frac{2}{5}\) = \(\frac{35}{40}\) – \(\frac{16}{40}\) = \(\frac{19}{40}\)

Question 8.

14\(\frac{1}{3}\) – 3\(\frac{3}{4}\) _____

Answer:

14\(\frac{1}{3}\) – 3\(\frac{3}{4}\)

Rewriting our equation with parts separated

14 + \(\frac{1}{3}\) – 3 – \(\frac{3}{4}\)

14 – 3 = 11

The denominators of the given fractions are not the same. So we have to find the LCD.

\(\frac{1}{3}\) – \(\frac{3}{4}\)

LCD of 3 and 4 is 12.

\(\frac{3}{12}\) – \(\frac{9}{12}\) = –\(\frac{5}{12}\)

11 – \(\frac{5}{12}\) = 10\(\frac{7}{12}\)

Question 9.

5\(\frac{2}{12}\) + 6\(\frac{2}{3}\) _____

Answer:

5\(\frac{2}{12}\) + 6\(\frac{2}{3}\)

Rewriting our equation with parts separated

5 + \(\frac{2}{12}\) + 6 + \(\frac{2}{3}\)

5 + 6 = 11

\(\frac{2}{12}\) + \(\frac{2}{3}\)

The denominators of the given fractions are not the same. So we have to find the LCD.

LCD of 3 and 12 is 12.

\(\frac{2}{12}\) + \(\frac{8}{12}\) = \(\frac{10}{12}\) = \(\frac{5}{6}\)

11 + \(\frac{5}{6}\) = 11\(\frac{5}{6}\)

Question 10.

**Use Tools** Find \(\frac{2}{3}\) + \(\frac{5}{6}\). Name the strategy or tool you will use to solve the problem, explain your choice, and then find the answer.

Answer:

\(\frac{2}{3}\) + \(\frac{5}{6}\)

The denominators of the given fractions are not the same. So we have to find the LCD.

LCD of 3 and 6 is 6.

\(\frac{4}{6}\) + \(\frac{5}{6}\) = \(\frac{9}{6}\) = \(\frac{3}{2}\) = 1\(\frac{1}{2}\)

Question 11.

Which is equivalent to \(\frac{10}{12}\) – \(\frac{1}{4}\)?

A. \(\frac{9}{12}\)

B. \(\frac{7}{12}\)

C. \(\frac{1}{2}\)

D. \(\frac{1}{3}\)

Answer: B. \(\frac{7}{12}\)

Explanation:

\(\frac{10}{12}\) – \(\frac{1}{4}\)

The denominators of the given fractions are not the same. So we have to find the LCD.

LCD of 12 and 4 is 12.

\(\frac{10}{12}\) – \(\frac{1}{4}\)

\(\frac{10}{12}\) – \(\frac{3}{12}\) = \(\frac{7}{12}\)

So, option B is the correct answer.

Question 12.

The distance from Malik’s home to a grocery store is 6 miles. The distance from his home to a park is 3\(\frac{3}{5}\) miles. How much farther from Malik’s home is the grocery store than the park?

Answer:

Given,

The distance from Malik’s home to a grocery store is 6 miles.

The distance from his home to a park is 3\(\frac{3}{5}\) miles.

6 – 3\(\frac{3}{5}\)

Rewriting our equation with parts separated

6 – 3 – \(\frac{3}{5}\)

3 – \(\frac{3}{5}\) = 2\(\frac{2}{5}\)

Question 13.

Luisa writes a history report that is 2\(\frac{3}{4}\) pages long. Therese writes a report that is 1\(\frac{1}{2}\) pages longer than Luisa’s report. How long is Therese’s report?

Answer:

Given,

Luisa writes a history report that is 2\(\frac{3}{4}\) pages long.

Therese writes a report that is 1\(\frac{1}{2}\) pages longer than Luisa’s report.

2\(\frac{3}{4}\) – 1\(\frac{1}{2}\)

Rewriting our equation with parts separated

2 + \(\frac{3}{4}\) – 1 – \(\frac{1}{2}\)

2 – 1 = 1

\(\frac{3}{4}\) – \(\frac{1}{2}\) = \(\frac{1}{4}\)

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

Question 14.

A bag of sunflower seeds weighs 5\(\frac{1}{2}\) pounds. Mr. Clark puts 1\(\frac{11}{16}\) pounds of the seeds in his bird feeder. How many pounds of sunflower seeds are left in the bag?

Answer:

Given,

A bag of sunflower seeds weighs 5\(\frac{1}{2}\) pounds.

Mr. Clark puts 1\(\frac{11}{16}\) pounds of the seeds in his bird feeder.

5\(\frac{1}{2}\) – 1\(\frac{11}{16}\)

Rewriting our equation with parts separated

5 + \(\frac{1}{2}\) – 1 – \(\frac{11}{16}\)

5 – 1 = 4

\(\frac{1}{2}\) – \(\frac{11}{16}\)

The denominators of the given fractions are not the same. So we have to find the LCD.

LCD of 2 and 16 is 16.

\(\frac{8}{16}\) – \(\frac{11}{16}\) = –\(\frac{3}{16}\)

4 – \(\frac{3}{16}\) = 3\(\frac{13}{16}\)

Question 15.

Which is equivalent to the sum?

2\(\frac{3}{4}\) + 1\(\frac{1}{3}\) + 2\(\frac{1}{4}\)

A. 6\(\frac{4}{12}\)

B. 6

C. 5\(\frac{5}{11}\)

D. 5\(\frac{1}{3}\)

Answer: A. 6\(\frac{4}{12}\)

2\(\frac{3}{4}\) + 1\(\frac{1}{3}\) + 2\(\frac{1}{4}\)

Rewriting our equation with parts separated

2 + 1 + 2 = 5

\(\frac{3}{4}\) + \(\frac{1}{3}\) + \(\frac{1}{4}\) = 1\(\frac{1}{3}\)

5 + 1\(\frac{1}{3}\) = 6\(\frac{1}{3}\)

The equivalent fraction to 6\(\frac{1}{3}\) is 6\(\frac{4}{12}\)

So, option A is the correct answer.

Question 16.

Jessica works on her math homework. Then she works 1\(\frac{1}{5}\) hours on her science homework. She works a total of 4\(\frac{1}{4}\) hours on homework. Select all equations that Jessica could use to find m, the number of hours she works on her math homework.

A. m – 4\(\frac{1}{4}\) = 1\(\frac{1}{5}\)

B. 1\(\frac{1}{5}\) + 4\(\frac{1}{4}\) = m

C. m + 1\(\frac{1}{5}\) = 4\(\frac{1}{4}\)

D. m – 1\(\frac{1}{5}\) = 4\(\frac{1}{4}\)

E. 4\(\frac{1}{4}\) – 1\(\frac{1}{5}\) = m

Answer:

Given,

Jessica works on her math homework.

Then she works 1\(\frac{1}{5}\) hours on her science homework.

She works a total of 4\(\frac{1}{4}\) hours on homework.

Let m be the number of hours she works on her math.

m + 1\(\frac{1}{5}\) = 4\(\frac{1}{4}\)

m = 4\(\frac{1}{4}\) – 1\(\frac{1}{5}\)

m – 4\(\frac{1}{4}\) = 1\(\frac{1}{5}\)

Option A, C and D are the correct answers.