4.5 Division of fractions  (Page 3/1)

$\begin{array}{c}\underbrace{\frac{3}{4}\text{and}\frac{4}{3}}\\ \frac{3}{4}\cdot \frac{4}{3}=1\end{array}$

$\begin{array}{c}\underbrace{\frac{7}{16}\text{and}\frac{16}{7}}\\ \frac{7}{16}\cdot \frac{16}{7}=1\end{array}$

$\begin{array}{c}\underbrace{\frac{1}{6}\text{and}\frac{6}{1}}\\ \frac{1}{6}\cdot \frac{6}{1}=1\end{array}$

$\frac{1}{3}÷\frac{3}{4}$ . The divisor is $\frac{3}{4}$ . Its reciprocal is $\frac{4}{3}$ . Multiply $\frac{1}{3}$ by $\frac{4}{3}$ .

$\frac{1}{3}\cdot \frac{4}{3}=\frac{1\cdot 4}{3\cdot 3}=\frac{4}{9}$

$\frac{1}{3}÷\frac{3}{4}=\frac{4}{9}$

$\frac{3}{8}÷\frac{5}{4}$ The divisor is $\frac{5}{4}$ . Its reciprocal is $\frac{4}{5}$ . Multiply $\frac{3}{8}$ by $\frac{4}{5}$ .

$\frac{3}{\underset{2}{\cancel{3}}}\cdot \frac{\stackrel{1}{\cancel{4}}}{5}=\frac{3\cdot 1}{2\cdot 5}=\frac{3}{\text{10}}$

$\frac{3}{8}÷\frac{5}{4}=\frac{3}{\text{10}}$

$\frac{5}{6}÷\frac{5}{\text{12}}$ . The divisor is $\frac{5}{\text{12}}$ . Its reciprocal is $\frac{\text{12}}{5}$ . Multiply $\frac{5}{6}$ by $\frac{\text{12}}{5}$ .

$\frac{\stackrel{1}{\cancel{5}}}{\underset{1}{\cancel{6}}}\cdot \frac{\stackrel{2}{\cancel{\text{12}}}}{\underset{1}{\cancel{5}}}=\frac{1\cdot 2}{1\cdot 1}=\frac{2}{1}=2$

$\frac{5}{6}÷\frac{5}{12}=2$

$2\frac{2}{9}÷3\frac{1}{3}$ . Convert each mixed number to an improper fraction.

$2\frac{2}{9}=\frac{9\cdot 2+2}{9}=\frac{\text{20}}{9}$ .

$3\frac{1}{3}=\frac{3\cdot 3+1}{3}=\frac{10}{3}$ .

$\frac{\text{20}}{9}÷\frac{\text{10}}{3}$ The divisor is $\frac{\text{10}}{3}$ . Its reciprocal is $\frac{3}{\text{10}}$ . Multiply $\frac{\text{20}}{9}$ by $\frac{3}{\text{10}}$ .

$\frac{\stackrel{2}{\cancel{20}}}{\underset{3}{\cancel{9}}}\cdot \frac{\stackrel{1}{\cancel{3}}}{\underset{1}{\cancel{10}}}=\frac{2\cdot 1}{3\cdot 1}=\frac{2}{3}$

$2\frac{2}{9}÷3\frac{1}{3}=\frac{2}{3}$

$\frac{\text{12}}{\text{11}}÷8$ . First conveniently write 8 as $\frac{8}{1}$ .

$\frac{\text{12}}{\text{11}}÷\frac{8}{1}$ The divisor is $\frac{8}{1}$ . Its reciprocal is $\frac{1}{8}$ . Multiply $\frac{\text{12}}{\text{11}}$ by $\frac{1}{8}$ .

$\frac{\stackrel{3}{\cancel{12}}}{\text{11}}\cdot \frac{1}{\underset{2}{\cancel{8}}}=\frac{3\cdot 1}{\text{11}\cdot 2}=\frac{3}{\text{22}}$

$\frac{\text{12}}{\text{11}}÷8=\frac{3}{\text{22}}$

$\frac{7}{8}÷\frac{\text{21}}{\text{20}}\cdot \frac{3}{\text{35}}$ . The divisor is $\frac{\text{21}}{\text{20}}$ . Its reciprocal is $\frac{\text{20}}{\text{21}}$ .

$\frac{\stackrel{1}{\cancel{7}}}{\underset{2}{\cancel{8}}}\cdot \frac{\stackrel{\stackrel{1}{\cancel{5}}}{\cancel{\text{20}}}}{\underset{\underset{1}{\cancel{3}}}{\cancel{\text{21}}}}\frac{\stackrel{1}{\cancel{3}}}{\underset{7}{\cancel{\text{35}}}}=\frac{1\cdot 1\cdot 1}{2\cdot 1\cdot 7}=\frac{1}{\text{14}}$

$\frac{7}{8}÷\frac{\text{21}}{\text{20}}\cdot \frac{3}{\text{25}}=\frac{1}{\text{14}}$

How many $2\frac{3}{8}$ -inch-wide packages can be placed in a box 19 inches wide?

The problem is to determine how many two and three eighths are contained in 19, that is, what is $\text{19}÷2\frac{3}{8}$ ?

$2\frac{3}{8}=\frac{\text{19}}{8}$ Convert the divisor $2\frac{3}{8}$ to an improper fraction.

$\text{19}=\frac{\text{19}}{1}$ Write the dividend 19 as $\frac{\text{19}}{1}$ .

$\frac{\text{19}}{1}÷\frac{\text{19}}{8}$ The divisor is $\frac{\text{19}}{8}$ . Its reciprocal is $\frac{8}{\text{19}}$ .

$\frac{\stackrel{1}{\cancel{\text{19}}}}{1}\cdot \frac{8}{\underset{1}{\cancel{\text{19}}}}=\frac{1\cdot 8}{1\cdot 1}=\frac{8}{1}=8$

Thus, 8 packages will fit into the box.