# McGraw Hill Math Grade 6 Lesson 7.1 Answer Key Multiplying Fractions and Whole Numbers

Practice questions available in McGraw Hill Math Grade 6 Answer Key PDF Lesson 7.1 Multiplying Fractions and Whole Numbers will engage students and is a great way of informal assessment.

## McGraw-Hill Math Grade 6 Answer Key Lesson 7.1 Multiplying Fractions and Whole Numbers

Exercises Estimate

Question 1.
5 × $$\frac{3}{4}$$
5 × $$\frac{3}{4}$$ = 3$$\frac{3}{4}$$

Explanation:
5 × $$\frac{3}{4}$$
= $$\frac{15}{4}$$
= 3$$\frac{3}{4}$$

Question 2.
4 × $$\frac{2}{7}$$
4 × $$\frac{2}{7}$$ = 1$$\frac{1}{7}$$

Explanation:
4 × $$\frac{2}{7}$$
= $$\frac{8}{7}$$
= 1$$\frac{1}{7}$$

Question 3.
21 × $$\frac{5}{8}$$
21 × $$\frac{5}{8}$$ = 13$$\frac{1}{8}$$

Explanation:
21 × $$\frac{5}{8}$$
= $$\frac{105}{8}$$
= 13$$\frac{1}{8}$$

Question 4.
11 × $$\frac{2}{9}$$
11 × $$\frac{2}{9}$$ = 2$$\frac{4}{9}$$

Explanation:
11 × $$\frac{2}{9}$$
= $$\frac{22}{9}$$
= 2$$\frac{4}{9}$$

Question 5.
4 × $$\frac{3}{11}$$
4 × $$\frac{3}{11}$$ = 1$$\frac{1}{11}$$

Explanation:
4 × $$\frac{3}{11}$$
= $$\frac{12}{11}$$
= 1$$\frac{1}{11}$$

Question 6.
13 × $$\frac{13}{14}$$
13 × $$\frac{13}{14}$$ = 12$$\frac{1}{14}$$

Explanation:
13 × $$\frac{13}{14}$$
= $$\frac{169}{14}$$
= 12$$\frac{1}{14}$$

Question 7.
21 × $$\frac{10}{23}$$
21 × $$\frac{10}{23}$$ = 9 $$\frac{3}{23}$$

Explanation:
21 × $$\frac{10}{23}$$
= $$\frac{210}{23}$$
= 9 $$\frac{3}{23}$$

Question 8.
7 × $$\frac{2}{3}$$
7 × $$\frac{2}{3}$$ = 4 $$\frac{2}{3}$$

Explanation:
7 × $$\frac{2}{3}$$
= $$\frac{14}{3}$$
= 4 $$\frac{2}{3}$$

Question 9.
12 × $$\frac{7}{19}$$
12 × $$\frac{7}{19}$$ = 4$$\frac{8}{19}$$

Explanation:
12 × $$\frac{7}{19}$$
= $$\frac{84}{19}$$
= 4$$\frac{8}{19}$$

Question 10.
14 × $$\frac{3}{5}$$
14 × $$\frac{3}{5}$$ = 8$$\frac{2}{5}$$

Explanation:
14 × $$\frac{3}{5}$$
=$$\frac{42}{5}$$
= 8$$\frac{2}{5}$$

Question 11.
14 × $$\frac{11}{13}$$
14 × $$\frac{11}{13}$$ = 11$$\frac{11}{13}$$

Explanation:
14 × $$\frac{11}{13}$$
= $$\frac{154}{13}$$
= 11$$\frac{11}{13}$$

Question 12.
13 × $$\frac{4}{17}$$
13 × $$\frac{4}{17}$$ = 3$$\frac{1}{17}$$

Explanation:
13 × $$\frac{4}{17}$$
= $$\frac{52}{17}$$
= 3$$\frac{1}{17}$$

Question 13.
10 × $$\frac{9}{23}$$
10 × $$\frac{9}{23}$$ = 3$$\frac{21}{23}$$

Explanation:
10 × $$\frac{9}{23}$$
= $$\frac{90}{23}$$
= 3$$\frac{21}{23}$$

Question 14.
13 × $$\frac{21}{22}$$
13 × $$\frac{21}{22}$$ = 12$$\frac{9}{22}$$

Explanation:
13 × $$\frac{21}{22}$$
= $$\frac{273}{22}$$
= 12$$\frac{9}{22}$$

Question 15.
5 × $$\frac{14}{27}$$
5 × $$\frac{14}{27}$$ = 2$$\frac{16}{27}$$

Explanation:
5 × $$\frac{14}{27}$$
= $$\frac{70}{27}$$
= 2$$\frac{16}{27}$$

Question 16.
Chet found a pair of sunglasses that he would like to buy. They normally cost $43.00, but this week they are on sale for only $$\frac{7}{8}$$ of the usual price. How much money will Chet need to buy the sunglasses? Answer: Amount of money Chet needs to buy the sunglasses =$37$$\frac{5}{8}$$

Explanation:
Cost of a pair of sunglasses usually = $43.00 Cost of a pair of sunglasses this week on sale for only of the usual price = $$\frac{7}{8}$$ Amount of money Chet needs to buy the sunglasses = Cost of a pair of sunglasses usually × Cost of a pair of sunglasses this week on sale for only of the usual price = 43 × $$\frac{7}{8}$$ = $$\frac{301}{8}$$ =$37$$\frac{5}{8}$$

Question 17.
Ashlee is riding her bike to the beach. If the beach is 17 miles from her home, and she has already traveled $$\frac{3}{5}$$ of the way, how much farther does she have to cycle?
Number of miles more she needs to cycle = 10$$\frac{1}{5}$$

Explanation:
Number of miles from her home the beach = 17.
Number of miles already travelled of the way = $$\frac{3}{5}$$
Number of miles more she needs to cycle = Number of miles from her home the beach × Number of miles already travelled of the way
= 17 × $$\frac{3}{5}$$
= $$\frac{51}{5}$$
= 10$$\frac{1}{5}$$