Question

Carry out the divisions of rational numbers.

(i) $\frac{40}{12}÷\frac{10}{4}$

(ii) $\frac{-10}{11}÷\frac{-11}{10}$

(iii) $\frac{-7}{8}÷\frac{-3}{6}$

(iv) $\frac{2}{3}÷\left(-4\right)$

(v) $2\frac{1}{5}÷5\frac{3}{6}$

(vi) $\frac{-5}{13}÷\frac{7}{26}$

(vii) $\frac{9}{11}÷\left(-8\right)$

(viii) $5÷\frac{2}{5}$

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right)\frac{40}{12}÷\frac{10}{4} \\ =\frac{40}{12}\times \frac{4}{10} \\ =\frac{40\times 4}{12\times 10} \\ =\frac{160}{120} \\ =\frac{160÷40}{120÷40}\left(\mathrm{Since},\mathrm{HCF}\mathrm{of}160\mathrm{and}120\mathrm{is}40\right) \\ =\frac{40}{3} \end{gathered}$$
$$\begin{gathered} \left(\mathrm{ii}\right)\frac{-10}{11}÷\frac{-11}{10} \\ =\left(-\frac{10}{11}\right)\times \left(-\frac{10}{11}\right) \\ =\frac{\left(-10\right)\times \left(-10\right)}{11\times 11} \\ =\frac{100}{121} \end{gathered}$$
$$\begin{gathered} \left(\mathrm{iii}\right)\frac{-7}{8}÷\frac{-3}{6} \\ =\left(-\frac{7}{8}\right)\times \left(-\frac{6}{3}\right) \\ =\frac{\left(-7\right)\times \left(-6\right)}{8\times 3} \\ =\frac{42}{24} \\ =\frac{42÷6}{24÷6}\left(\mathrm{Since},\mathrm{HCF}\mathrm{of}42\mathrm{and}24\mathrm{is}6\right) \\ =\frac{7}{4} \end{gathered}$$
$$\begin{gathered} \left(\mathrm{iv}\right)\frac{2}{3}÷\left(-4\right) \\ =\frac{2}{3}\times \left(-\frac{1}{4}\right) \\ =\frac{2\times \left(-1\right)}{3\times 4} \\ =\frac{-2}{12} \\ =\frac{-2÷2}{12÷2}\left(\mathrm{Since},\mathrm{HCF}\mathrm{of}2\mathrm{and}12\mathrm{is}2\right) \\ =\frac{-1}{6} \end{gathered}$$
$$\begin{gathered} \left(\mathrm{v}\right)2\frac{1}{5}÷5\frac{3}{6} \\ =\frac{2\times 5+1}{5}÷\frac{5\times 6+3}{6} \\ =\frac{11}{5}÷\frac{33}{6} \\ =\frac{11}{5}\times \frac{6}{33} \\ =\frac{11\times 6}{5\times 33} \\ =\frac{66}{165} \\ =\frac{66÷33}{165÷33}\left(\mathrm{Since},\mathrm{HCF}\mathrm{of}66\mathrm{and}165\mathrm{is}33\right) \\ =\frac{2}{5} \end{gathered}$$