# 1.3.2 Fractions Revision, PT3 Practice

Question 6:
$1\frac{1}{2}÷\frac{1}{4}×\frac{4}{5}=$

Solution:
$\begin{array}{l}1\frac{1}{2}÷\frac{1}{4}×\frac{4}{5}\\ =\frac{3}{\overline{)2}}×\frac{{\overline{)4}}^{2}}{1}×\frac{4}{5}\\ =\frac{24}{5}\\ =4\frac{4}{5}\end{array}$

Question 7:
$4\frac{1}{5}÷1\frac{1}{6}×2\frac{2}{9}=$

Solution:
$\begin{array}{l}4\frac{1}{5}÷1\frac{1}{6}×2\frac{2}{9}\\ =\frac{{\overline{)21}}^{\overline{)3}}}{\overline{)5}}×\frac{{\overline{)6}}^{2}}{\overline{)7}}×\frac{{\overline{)20}}^{4}}{{\overline{)9}}^{\overline{)3}}}\\ =8\end{array}$

Question 8:
$2\frac{3}{4}×1\frac{1}{3}÷7\frac{1}{3}=$

Solution:
$\begin{array}{l}2\frac{3}{4}×1\frac{1}{3}÷7\frac{1}{3}\\ =\frac{11}{4}×\frac{4}{3}÷\frac{22}{3}\\ =\frac{\overline{)11}}{\overline{)4}}×\frac{\overline{)4}}{\overline{)3}}×\frac{\overline{)3}}{{\overline{)22}}^{2}}\\ =\frac{1}{2}\end{array}$

Question 9:
$4\frac{5}{6}-2\frac{1}{3}÷\frac{5}{9}=$

Solution:
$\begin{array}{l}4\frac{5}{6}-2\frac{1}{3}÷\frac{5}{9}\\ =4\frac{5}{6}-\left(\frac{7}{3}÷\frac{5}{9}\right)\\ =4\frac{5}{6}-\left(\frac{7}{\overline{)3}}×\frac{{\overline{)9}}^{3}}{5}\right)\\ =4\frac{5}{6}-\frac{21}{5}\\ =4\frac{5}{6}-4\frac{1}{5}\\ =\frac{25}{30}-\frac{6}{30}\\ =\frac{19}{30}\end{array}$

Question 10:
$\left(\frac{7}{9}+\frac{11}{15}\right)÷\left(7\frac{8}{15}-4\frac{1}{5}\right)=$

Solution:
$\begin{array}{l}\left(\frac{7}{9}+\frac{11}{15}\right)÷\left(4\frac{8}{15}-3\frac{1}{5}\right)\\ =\left(\frac{35}{45}+\frac{33}{45}\right)÷\left(4\frac{8}{15}-3\frac{3}{15}\right)\\ =\frac{68}{45}÷1\frac{5}{15}\\ =\frac{68}{45}÷1\frac{1}{3}\\ =\frac{68}{45}÷\frac{4}{3}\\ =\frac{{\overline{)68}}^{17}}{{\overline{)45}}^{15}}×\frac{\overline{)3}}{\overline{)4}}\\ =\frac{17}{15}\\ =1\frac{2}{15}\end{array}$