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

Vocabulary

Associative Property of
benchmark
common denominator
Commutative Property of
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.
$$\frac{4}{10}$$ + $$\frac{7}{12}$$ _____
$$\frac{4}{10}$$ + $$\frac{7}{12}$$
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.
$$\frac{7}{8}$$ – $$\frac{2}{5}$$ _____
$$\frac{7}{8}$$ – $$\frac{2}{5}$$
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$$\frac{3}{4}$$ _____
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$$\frac{2}{3}$$ _____
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.
$$\frac{2}{3}$$ + $$\frac{5}{6}$$
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?
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?
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?
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
Then she works 1$$\frac{1}{5}$$ hours on her science homework.
She works a total of 4$$\frac{1}{4}$$ hours on homework.
m +  1$$\frac{1}{5}$$ = 4$$\frac{1}{4}$$
m = 4$$\frac{1}{4}$$ – 1$$\frac{1}{5}$$
m – 4$$\frac{1}{4}$$ = 1$$\frac{1}{5}$$