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.