# 3.2.1 Fractions, PT3 Practice

Question 1:
$3\frac{2}{5}+1\frac{5}{6}-\frac{1}{5}=$

Solution:
$\begin{array}{l}3\frac{2}{5}+1\frac{5}{6}-\frac{1}{5}\\ =3\frac{12}{30}+1\frac{25}{30}-\frac{6}{30}\\ =4\frac{37}{30}-\frac{6}{30}\\ =4\frac{31}{30}\\ =5\frac{1}{30}\end{array}$

Question 2:
$6\frac{1}{12}-3\frac{2}{3}-1\frac{5}{6}=$

Solution:
$\begin{array}{l}6\frac{1}{12}-3\frac{2}{3}-1\frac{5}{6}\\ =6\frac{1}{12}-3\frac{8}{12}-1\frac{10}{12}\\ =\left(5\frac{13}{12}-3\frac{8}{12}\right)-1\frac{10}{12}\\ =2\frac{5}{12}-1\frac{10}{12}\\ =1\frac{17}{12}-1\frac{10}{12}\\ =\frac{7}{12}\end{array}$

Question 3:
$30-\left(\frac{2}{7}÷\frac{2}{21}\right)=$

Solution:
$\begin{array}{l}30-\left(\frac{2}{7}÷\frac{2}{21}\right)\\ =30-\left(\frac{\overline{)2}}{\overline{)7}}×\frac{{\overline{)21}}^{3}}{\overline{)2}}\right)\\ =30-3\\ =27\end{array}$

Question 4:
$\left(7\frac{1}{2}-3\frac{2}{3}\right)×\frac{12}{69}=$

Solution:
$\begin{array}{l}\left(7\frac{1}{2}-3\frac{2}{3}\right)×\frac{12}{69}\\ =\left(7\frac{3}{6}-3\frac{4}{6}\right)×\frac{12}{69}\\ =\left(6\frac{9}{6}-3\frac{4}{6}\right)×\frac{12}{69}\\ =3\frac{5}{6}×\frac{12}{69}\\ =\frac{\overline{)23}}{\overline{)6}}×\frac{{\overline{)12}}^{2}}{{\overline{)69}}^{3}}\\ =\frac{2}{3}\end{array}$

Question 5:
$1\frac{1}{5}÷\left(4-2\frac{1}{2}\right)=$

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