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 × \(\frac{3}{4}\)
Answer:
5 × \(\frac{3}{4}\) = 3\(\frac{3}{4}\)
Explanation:
5 × \(\frac{3}{4}\)
= \(\frac{15}{4}\)
= 3\(\frac{3}{4}\)
Question 2.
4 × \(\frac{2}{7}\)
Answer:
4 × \(\frac{2}{7}\) = 1\(\frac{1}{7}\)
Explanation:
4 × \(\frac{2}{7}\)
= \(\frac{8}{7}\)
= 1\(\frac{1}{7}\)
Question 3.
21 × \(\frac{5}{8}\)
Answer:
21 × \(\frac{5}{8}\) = 13\(\frac{1}{8}\)
Explanation:
21 × \(\frac{5}{8}\)
= \(\frac{105}{8}\)
= 13\(\frac{1}{8}\)
Question 4.
11 × \(\frac{2}{9}\)
Answer:
11 × \(\frac{2}{9}\) = 2\(\frac{4}{9}\)
Explanation:
11 × \(\frac{2}{9}\)
= \(\frac{22}{9}\)
= 2\(\frac{4}{9}\)
Question 5.
4 × \(\frac{3}{11}\)
Answer:
4 × \(\frac{3}{11}\) = 1\(\frac{1}{11}\)
Explanation:
4 × \(\frac{3}{11}\)
= \(\frac{12}{11}\)
= 1\(\frac{1}{11}\)
Question 6.
13 × \(\frac{13}{14}\)
Answer:
13 × \(\frac{13}{14}\) = 12\(\frac{1}{14}\)
Explanation:
13 × \(\frac{13}{14}\)
= \(\frac{169}{14}\)
= 12\(\frac{1}{14}\)
Question 7.
21 × \(\frac{10}{23}\)
Answer:
21 × \(\frac{10}{23}\) = 9 \(\frac{3}{23}\)
Explanation:
21 × \(\frac{10}{23}\)
= \(\frac{210}{23}\)
= 9 \(\frac{3}{23}\)
Question 8.
7 × \(\frac{2}{3}\)
Answer:
7 × \(\frac{2}{3}\) = 4 \(\frac{2}{3}\)
Explanation:
7 × \(\frac{2}{3}\)
= \(\frac{14}{3}\)
= 4 \(\frac{2}{3}\)
Question 9.
12 × \(\frac{7}{19}\)
Answer:
12 × \(\frac{7}{19}\) = 4\(\frac{8}{19}\)
Explanation:
12 × \(\frac{7}{19}\)
= \(\frac{84}{19}\)
= 4\(\frac{8}{19}\)
Question 10.
14 × \(\frac{3}{5}\)
Answer:
14 × \(\frac{3}{5}\) = 8\(\frac{2}{5}\)
Explanation:
14 × \(\frac{3}{5}\)
=\(\frac{42}{5}\)
= 8\(\frac{2}{5}\)
Question 11.
14 × \(\frac{11}{13}\)
Answer:
14 × \(\frac{11}{13}\) = 11\(\frac{11}{13}\)
Explanation:
14 × \(\frac{11}{13}\)
= \(\frac{154}{13}\)
= 11\(\frac{11}{13}\)
Question 12.
13 × \(\frac{4}{17}\)
Answer:
13 × \(\frac{4}{17}\) = 3\(\frac{1}{17}\)
Explanation:
13 × \(\frac{4}{17}\)
= \(\frac{52}{17}\)
= 3\(\frac{1}{17}\)
Question 13.
10 × \(\frac{9}{23}\)
Answer:
10 × \(\frac{9}{23}\) = 3\(\frac{21}{23}\)
Explanation:
10 × \(\frac{9}{23}\)
= \(\frac{90}{23}\)
= 3\(\frac{21}{23}\)
Question 14.
13 × \(\frac{21}{22}\)
Answer:
13 × \(\frac{21}{22}\) = 12\(\frac{9}{22}\)
Explanation:
13 × \(\frac{21}{22}\)
= \(\frac{273}{22}\)
= 12\(\frac{9}{22}\)
Question 15.
5 × \(\frac{14}{27}\)
Answer:
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?
Answer:
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\(\frac{1}{5}\)