This handy Spectrum Math Grade 7 Answer Key Chapter 2 Lesson 2.2 Multiplying Fractions and Mixed Numbers provides detailed answers for the workbook questions.
Spectrum Math Grade 7 Chapter 2 Lesson 2.2 Multiplying Fractions and Mixed Numbers Answers Key
Reduce to simplest form if possible. Then, multiply the numerators and multiply the denominators.
Rename the numbers as improper fractions. Reduce to simplest form. Multiply the numerators and denominators. Simplify.
Multiply. Write each answer in simplest form.
Question 1.
a. \(\frac{1}{2}\) × \(\frac{3}{4}\)
Answer: \(\frac{3}{8}\)
\(\frac{1}{2}\) × \(\frac{3}{4}\)
First step is to reduce the above fractions into simplest form if possible. In this case, it is not possible.
Then, multiply the numerators and denominators separately.
= \(\frac{1 × 3}{2 × 4}\)
= \(\frac{3}{8}\)
Therefore, \(\frac{1}{2}\) × \(\frac{3}{4}\) = \(\frac{3}{8}\)
b. \(\frac{2}{3}\) × \(\frac{4}{5}\)
Answer: \(\frac{8}{15}\)
\(\frac{2}{3}\) × \(\frac{4}{5}\)
First step is to reduce the above fractions into simplest form if possible. In this case, it is not possible.
Then, multiply the numerators and denominators separately.
= \(\frac{2 × 4}{3 × 5}\)
= \(\frac{8}{15}\)
Therefore, \(\frac{2}{3}\) × \(\frac{4}{5}\) = \(\frac{8}{15}\)
c. \(\frac{3}{4}\) × \(\frac{3}{4}\)
Answer: \(\frac{9}{16}\)
\(\frac{3}{4}\) × \(\frac{3}{4}\)
First step is to reduce the above fractions into simplest form if possible. In this case, it is not possible.
Then, multiply the numerators and denominators separately.
= \(\frac{3 ×3}{4 × 4}\)
= \(\frac{9}{16}\)
Therefore, \(\frac{3}{4}\) × \(\frac{3}{4}\) = \(\frac{9}{16}\)
d. \(\frac{4}{5}\) × \(\frac{1}{8}\)
Answer: \(\frac{1}{10}\)
\(\frac{4}{5}\) × \(\frac{1}{8}\)
First step is to reduce the above fractions into simplest form if possible. In this case, it is possible.
Then, multiply the numerators and denominators separately.
= \(\frac{4 × 1}{5 × 8}\)
Divide the 4 in numerator and 8 in denominator by 4, which is a common factor.
= \(\frac{1 × 1}{5×2}\)
= \(\frac{1}{10}\)
Therefore, \(\frac{4}{5}\) × \(\frac{1}{8}\) = \(\frac{1}{10}\)
Question 2.
a. \(\frac{3}{5}\) × \(\frac{7}{8}\)
Answer: \(\frac{21}{40}\)
\(\frac{3}{5}\) × \(\frac{7}{8}\)
First step is to reduce the above fractions into simplest form if possible. In this case, it is not possible.
Then, multiply the numerators and denominators separately.
= \(\frac{3 × 7}{5 × 8}\)
= \(\frac{21}{40}\)
Therefore, \(\frac{3}{5}\) × \(\frac{7}{8}\) = \(\frac{21}{40}\)
b. \(\frac{1}{3}\) × \(\frac{3}{5}\)
Answer: \(\frac{1}{5}\)
\(\frac{1}{3}\) × \(\frac{3}{5}\)
First step is to reduce the above fractions into simplest form if possible. In this case, it is possible.
Then, multiply the numerators and denominators separately.
= \(\frac{1 × 3}{3× 5}\)
Divide the 3 in numerator and 3 in denominator by 3, which is a common factor.
= \(\frac{1 × 1}{1 × 5}\)
= \(\frac{1}{5}\)
Therefore, \(\frac{1}{3}\) × \(\frac{3}{5}\) = \(\frac{1}{5}\)
c. \(\frac{3}{7}\) × \(\frac{1}{5}\)
Answer: \(\frac{3}{35}\)
\(\frac{3}{7}\) × \(\frac{1}{5}\)
First step is to reduce the above fractions into simplest form if possible. In this case, it is not possible.
Then, multiply the numerators and denominators separately.
= \(\frac{3 × 1}{7×5}\)
= \(\frac{3}{35}\)
Therefore, \(\frac{3}{7}\) × \(\frac{1}{5}\) = \(\frac{3}{35}\)
d. \(\frac{3}{10}\) × \(\frac{4}{5}\)
Answer: \(\frac{6}{25}\)
\(\frac{3}{10}\) × \(\frac{4}{5}\)
First step is to reduce the above fractions into simplest form if possible. In this case, it is possible.
Then, multiply the numerators and denominators separately.
= \(\frac{3 × 4}{10 × 5}\)
Divide the 4 in numerator and 10 in denominator by 2, which is a common factor.
= \(\frac{3 × 2}{5 × 5}\)
= \(\frac{6}{25}\)
Therefore, \(\frac{3}{10}\) × \(\frac{4}{5}\) = \(\frac{6}{25}\)
Question 3.
a. \(\frac{5}{8}\) × \(\frac{3}{8}\)
Answer: \(\frac{15}{64}\)
\(\frac{5}{8}\) × \(\frac{3}{8}\)
First step is to reduce the above fractions into simplest form if possible. In this case, it is not possible.
Then, multiply the numerators and denominators separately.
= \(\frac{5 × 3}{8× 8}\)
= \(\frac{15}{64}\)
Therefore, \(\frac{5}{8}\) × \(\frac{3}{8}\) = \(\frac{15}{64}\)
b. \(\frac{2}{3}\) × \(\frac{1}{2}\)
Answer: \(\frac{1}{3}\)
\(\frac{2}{3}\) × \(\frac{1}{2}\)
First step is to reduce the above fractions into simplest form if possible. In this case, it is possible.
Then, multiply the numerators and denominators separately.
= \(\frac{2 × 1}{3 × 2}\)
Divide the 2 in numerator and 2 in denominator by 2, which is a common factor.
= \(\frac{1× 1}{3 × 1}\)
= \(\frac{1}{3}\)
Therefore, \(\frac{2}{3}\) × \(\frac{1}{2}\) = \(\frac{1}{3}\)
c. \(\frac{5}{6}\) × \(\frac{2}{3}\)
Answer: \(\frac{5}{9}\)
\(\frac{5}{6}\) × \(\frac{2}{3}\)
First step is to reduce the above fractions into simplest form if possible. In this case, it is possible.
Then, multiply the numerators and denominators separately.
= \(\frac{5 × 2}{6 × 3}\)
Divide the 2 in numerator and 6 in denominator by 2, which is a common factor.
= \(\frac{5 ×1}{3 × 3}\)
= \(\frac{5}{9}\)
Therefore, \(\frac{5}{6}\) × \(\frac{2}{3}\) = \(\frac{5}{9}\)
d. \(\frac{4}{7}\) × \(\frac{1}{3}\)
Answer: \(\frac{4}{21}\)
\(\frac{4}{7}\) × \(\frac{1}{3}\)
First step is to reduce the above fractions into simplest form if possible. In this case, it is not possible.
Then, multiply the numerators and denominators separately.
= \(\frac{4 × 1}{7 × 3}\)
= \(\frac{4}{21}\)
Therefore, \(\frac{4}{7}\) × \(\frac{1}{3}\) = \(\frac{4}{21}\)
Question 4.
a. 3 × 1\(\frac{2}{7}\)
Answer: \(\frac{27}{7}\) = 3\(\frac{6}{7}\)
3 × 1\(\frac{2}{7}\)
Convert the above numbers in improper fractions to make calculations easy.
= \(\frac{3}{1}\) × \(\frac{9}{7}\)
First step is to reduce the above fractions into simplest form if possible. In this case, it is not possible.
Then, multiply the numerators and denominators separately.
= \(\frac{3 × 9}{1 × 7}\)
= \(\frac{27}{7}\)
= 3\(\frac{6}{7}\)
Therefore, 3 × 1\(\frac{2}{7}\) = \(\frac{27}{7}\) = 3\(\frac{6}{7}\)
b. 2\(\frac{1}{4}\) × 3\(\frac{1}{3}\)
Answer: \(\frac{15}{2}\) = 7\(\frac{1}{2}\)
2\(\frac{1}{4}\) × 3\(\frac{1}{3}\)
Convert the above numbers in improper fractions to make calculations easy.
= \(\frac{9}{4}\) × \(\frac{10}{3}\)
First step is to reduce the above fractions into simplest form if possible. In this case, it is possible.
Then, multiply the numerators and denominators separately.
= \(\frac{9 × 10}{4 × 3}\)
Divide the 9 in numerator and 3 in denominator by 3 and 10 in numerator and 4 in denominator by 2 , which is a common factor.
= \(\frac{3× 5}{2 ×1}\)
= \(\frac{15}{2}\)
= 7\(\frac{1}{2}\)
Therefore, 2\(\frac{1}{4}\) × 3\(\frac{1}{3}\) = \(\frac{15}{2}\) = 7\(\frac{1}{2}\)
c. 1\(\frac{1}{9}\) × 3\(\frac{1}{4}\)
Answer: \(\frac{65}{18}\) = 3\(\frac{11}{18}\)
1\(\frac{1}{9}\) × 3\(\frac{1}{4}\)
Convert the above numbers in improper fractions to make calculations easy.
= \(\frac{10}{9}\) × \(\frac{13}{4}\)
First step is to reduce the above fractions into simplest form if possible. In this case, it is possible.
Then, multiply the numerators and denominators separately.
= \(\frac{10 × 13}{9 v 4}\)
Divide the 10 in numerator and 4 in denominator by 2, which is a common factor.
= \(\frac{5 ×13}{9 × 2}\)
= \(\frac{65}{18}\)
= 3\(\frac{11}{18}\)
Therefore, 1\(\frac{1}{9}\) × 3\(\frac{1}{4}\) = \(\frac{65}{18}\) = 3\(\frac{11}{18}\)
d. 2\(\frac{1}{4}\) × 6
Answer: \(\frac{27}{2}\) = 13\(\frac{1}{2}\)
2\(\frac{1}{4}\) × 6
Convert the above numbers in improper fractions to make calculations easy.
= \(\frac{9}{4}\)× \(\frac{6}{1}\)
First step is to reduce the above fractions into simplest form if possible. In this case, it is possible.
Then, multiply the numerators and denominators separately.
= \(\frac{9 × 6}{4 × 1}\)
Divide the 6 in numerator and 4 in denominator by 2, which is a common factor.
= \(\frac{9 × 3}{2 × 1}\)
= \(\frac{27}{2}\)
= 13\(\frac{1}{2}\)
Therefore, 2\(\frac{1}{4}\) × 6 = \(\frac{27}{2}\) = 13\(\frac{1}{2}\)
Question 5.
a. 1\(\frac{2}{3}\) × 3\(\frac{7}{8}\)
Answer: \(\frac{155}{24}\)= 6\(\frac{11}{24}\)
1\(\frac{2}{3}\) × 3\(\frac{7}{8}\)
Convert the above numbers in improper fractions to make calculations easy.
= \(\frac{5}{3}\) ×\(\frac{31}{8}\)
First step is to reduce the above fractions into simplest form if possible. In this case, it is not possible.
Then, multiply the numerators and denominators separately.
= \(\frac{5 × 31}{3 ×8}\)
= \(\frac{155}{24}\)
= 6\(\frac{11}{24}\)
Therefore, 1\(\frac{2}{3}\) × 3\(\frac{7}{8}\) = \(\frac{155}{24}\) = 6\(\frac{11}{24}\)
b. 2\(\frac{1}{7}\) × 1\(\frac{1}{3}\)
Answer: \(\frac{60}{21}\)= 2\(\frac{18}{21}\)
2\(\frac{1}{7}\) × 1\(\frac{1}{3}\)
Convert the above numbers in improper fractions to make calculations easy.
= \(\frac{15}{7}\) × \(\frac{4}{3}\)
First step is to reduce the above fractions into simplest form if possible. In this case, it is not possible.
Then, multiply the numerators and denominators separately.
= \(\frac{15× 4}{7× 3}\)
= \(\frac{60}{21}\)
= 2\(\frac{18}{21}\)
Therefore, 2\(\frac{1}{7}\) × 1\(\frac{1}{3}\) = \(\frac{60}{21}\) = 2\(\frac{18}{21}\)
c. 4\(\frac{1}{2}\) × 2\(\frac{1}{3}\) × 3
Answer: \(\frac{63}{2}\) = 31\(\frac{1}{2}\)
4\(\frac{1}{2}\) × 2\(\frac{1}{3}\) × 3
Convert the above numbers in improper fractions to make calculations easy.
= \(\frac{9}{2}\) × \(\frac{7}{3}\) × \(\frac{3}{1}\)
First step is to reduce the above fractions into simplest form if possible. In this case, it is possible.
Then, multiply the numerators and denominators separately.
= \(\frac{9 × 7× 3}{2 ×3 ×1}\)
Divide the 3 in numerator and 3 in denominator by 3, which is a common factor.
= \(\frac{9 × 7× 1}{2 × 1 × 1}\)
= \(\frac{63}{2}\)
= 31\(\frac{1}{2}\)
Therefore, 4\(\frac{1}{2}\) × 2\(\frac{1}{3}\) × 3 = \(\frac{63}{2}\)
= 31\(\frac{1}{2}\)
d. 5\(\frac{1}{4}\) × 2\(\frac{1}{2}\) × 1\(\frac{1}{3}\)
Answer: \(\frac{35}{2}\) = 17\(\frac{1}{2}\)
5\(\frac{1}{4}\) × 2\(\frac{1}{2}\) × 1\(\frac{1}{3}\)
Convert the above numbers in improper fractions to make calculations easy.
= \(\frac{21}{4}\) × \(\frac{5}{2}\) × \(\frac{4}{3}\)
First step is to reduce the above fractions into simplest form if possible. In this case, it is possible.
Then, multiply the numerators and denominators separately.
= \(\frac{21 × 5 × 4}{4 × 2 × 3}\)
Divide the 4 in numerator and 4 in denominator by 4, which is a common factor.
= \(\frac{21 × 5 × 1}{1 × 2 × 3}\)
Now, Divide the 21 in numerator and 3 in denominator by 3, which is a common factor.
= \(\frac{7 × 5 × 1}{1 × 2 × 1}\)
= \(\frac{35}{2}\)
= 17\(\frac{1}{2}\)
Therefore, 5\(\frac{1}{4}\) × 2\(\frac{1}{2}\) × 1\(\frac{1}{3}\) = \(\frac{35}{2}\) = 17\(\frac{1}{2}\)
Question 6.
a. 4\(\frac{1}{8}\) × 3\(\frac{2}{7}\) × 7
Answer: \(\frac{759}{8}\) = 94\(\frac{7}{8}\)
4\(\frac{1}{8}\) × 3\(\frac{2}{7}\) × 7
Convert the above numbers in improper fractions to make calculations easy.
= \(\frac{33}{8}\) × \(\frac{23}{7}\) × \(\frac{7}{1}\)
First step is to reduce the above fractions into simplest form if possible. In this case, it is possible.
Then, multiply the numerators and denominators separately.
= \(\frac{33 ×23 × 7}{8 × 7 × 1}\)
Divide the 4 in numerator and 4 in denominator by 4, which is a common factor.
= \(\frac{33 × 23 × 1}{8 × 1 × 1}\)
= \(\frac{759}{8}\)
= 94\(\frac{7}{8}\)
4\(\frac{1}{8}\) × 3\(\frac{2}{7}\) × 7 = \(\frac{759}{8}\) = 94\(\frac{7}{8}\)
b. \(\frac{5}{6}\) × 1\(\frac{1}{3}\) × 2
Answer: \(\frac{20}{9}\) = 3\(\frac{2}{9}\)
\(\frac{5}{6}\) × 1\(\frac{1}{3}\) × 2
Convert the above numbers in improper fractions to make calculations easy.
= \(\frac{5}{6}\) × \(\frac{4}{3}\) ×\(\frac{2}{1}\)
First step is to reduce the above fractions into simplest form if possible. In this case, it is possible.
Then, multiply the numerators and denominators separately.
= \(\frac{5 × 4 ×2}{6 × 3 ×1}\)
Divide the 2 in numerator and 6 in denominator by 2, which is a common factor.
= \(\frac{5 × 4 × 1}{3 × 3 × 1}\)
= \(\frac{20}{9}\)
= 3\(\frac{2}{9}\)
Therefore, \(\frac{5}{6}\) × 1\(\frac{1}{3}\) × 2 = \(\frac{20}{9}\) = 3\(\frac{2}{9}\)
c. \(\frac{2}{3}\) × 1\(\frac{5}{8}\) × 3\(\frac{1}{4}\)
Answer:\(\frac{169}{48}\) = 3\(\frac{25}{48}\)
\(\frac{2}{3}\) × 1\(\frac{5}{8}\) × 3\(\frac{1}{4}\)
Convert the above numbers in improper fractions to make calculations easy.
= \(\frac{2}{3}\) × \(\frac{13}{8}\) × \(\frac{13}{4}\)
First step is to reduce the above fractions into simplest form if possible. In this case, it is possible.
Then, multiply the numerators and denominators separately.
= \(\frac{2 ×13 ×13}{3 × 8 × 4}\)
Divide the 2 in numerator and 4 in denominator by 2, which is a common factor.
= \(\frac{1 × 13 × 13}{3 × 8 × 2}\)
= \(\frac{169}{48}\)
= 3\(\frac{25}{48}\)
Therefore, \(\frac{2}{3}\) × 1\(\frac{5}{8}\) × 3\(\frac{1}{4}\) = \(\frac{169}{48}\) = 3\(\frac{25}{48}\)
d. 1\(\frac{1}{2}\) × 2\(\frac{2}{3}\) × 1\(\frac{1}{8}\)
Answer: \(\frac{9}{2}\) = 4 \(\frac{1}{2}\)
1\(\frac{1}{2}\) × 2\(\frac{2}{3}\) × 1\(\frac{1}{8}\)
Convert the above numbers in improper fractions to make calculations easy.
= \(\frac{3}{2}\) ×\(\frac{8}{3}\)× \(\frac{9}{8}\)
First step is to reduce the above fractions into simplest form if possible. In this case, it is possible.
Then, multiply the numerators and denominators separately.
= \(\frac{3 ×8 ×9}{2× 3 × 8}\)
Divide the 8 in numerator and 8 in denominator by 8, which is a common factor.
= \(\frac{3× 1×9}{2×3× 1}\)
Now, Divide the 3 in numerator and 3 in denominator by 3, which is a common factor.
= \(\frac{1 × 1 ×9}{2 × 1× 1}\)
= \(\frac{9}{2}\)
= 4 \(\frac{1}{2}\)
Therefore, 1\(\frac{1}{2}\) × 2\(\frac{2}{3}\) × 1\(\frac{1}{8}\) = \(\frac{9}{2}\) = 4 \(\frac{1}{2}\)