Question

Find the value of:

(i) $(-4)÷\frac{2}{3}$

(ii) $\frac{-3}{5}÷2$

(iii) $\frac{-4}{5}÷-3$

(iv) $\frac{-1}{8}÷\frac{3}{4}$

(v) $\frac{-2}{13}÷\frac{1}{7}$

(vi) $\frac{-7}{12}÷\frac{-2}{13}$

(vii) $\frac{3}{13}÷\frac{-4}{65}$

Answer 1

Recall the following properties of divisin of the fractions.

$a÷\frac{b}{c}=a\times \frac{c}{b}$ ---(1)

$\frac{a}{b}÷c=\frac{a}{b}\times \frac{1}{c}$----(2)

$\frac{a}{b}÷\frac{c}{d}=\frac{a}{b}\times \frac{d}{c}$ ----(3)

Now, lets evaluate given simplifications,

(i)

$(-4)÷\frac{2}{3}$

$=-4\times \frac{3}{2}$

$=-2\times 3$

$=-6$

$\therefore (-4)÷\frac{2}{3}=-6$

(ii)

$\frac{-3}{5}÷2$

$=\frac{-3}{5}\times \frac{1}{2}$

$=\frac{-3\times 1}{5\times 2}$

$=\frac{-3}{10}$

$\therefore \frac{-3}{5}÷2=\frac{-3}{10}$

(iii)

$\frac{-4}{5}÷-3$

$=\frac{-4}{5}\times \frac{1}{-3}$

$=\frac{4}{15}$

$\frac{-4}{5}÷-3=\frac{4}{15}$

(iv)

$\frac{-1}{8}÷\frac{3}{4}$

$=\frac{-1}{8}\times \frac{4}{3}$

$=\frac{-1}{2}\times \frac{1}{3}$

$=\frac{-1\times 1}{2\times 3}$

$=\frac{-1}{6}$

$\therefore \frac{-1}{8}÷\frac{3}{4}=\frac{-1}{6}$

(v)

$\frac{-2}{13}÷\frac{1}{7}$

$=\frac{-2}{13}\times \frac{7}{1}$

$=\frac{-2\times 7}{13\times 1}$

$=-\frac{14}{13}$

$=-1\frac{1}{13}$

$\therefore \frac{-2}{13}÷\frac{1}{7}=-1\frac{1}{13}$

(vi)

$\frac{-7}{12}÷\frac{-2}{13}$

$=\frac{-7}{12}\times \frac{13}{-2}$

$=\frac{-7\times 13}{12\times -2}$

$=\frac{91}{24}$

$=3\frac{19}{24}$

$\therefore \frac{-7}{12}÷\frac{-2}{13}=3\frac{19}{24}$

(vii)

$\frac{3}{13}÷\frac{-4}{65}$

$=\frac{3}{13}\times \frac{65}{-4}$

$=\frac{3}{1}\times \frac{5}{-4}$

$=-\frac{15}{4}$

$=-3\frac{3}{4}$

$\therefore \frac{3}{13}÷\frac{-4}{65}=-3\frac{3}{4}$​​​