Find:
(i) $\frac{1}{4}$ of (a) $\frac{1}{4}$(b) $\frac{3}{5}$ (c) $\frac{4}{3}$
(ii) $\frac{1}{7}$ of (a) $\frac{2}{9}$ (b) $\frac{6}{5}$ (c) $\frac{3}{10}$

To do: 

We have to find

(i) $\frac{1}{4}$ of (a) $\frac{1}{4}$ (b) $\frac{3}{5}$ (c) $\frac{4}{3}$

(ii) $\frac{1}{7}$ of (a) $\frac{2}{9}$ (b) $\frac{6}{5}$ (c) $\frac{3}{10}$

Solution:

 (i)

(a) $\frac{1}{4}$ of $\frac{1}{4}$

$\frac{1}{4}$ of $\frac{1}{4}=\frac{1}{4}\times\frac{1}{4}$

$=\frac{1\times1}{4\times4}$

$=\frac{1}{16}$

(b) $\frac{1}{4}$ of $\frac{3}{5}$

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

$=\frac{1\times3}{4\times5}$

$=\frac{3}{20}$

(c) $\frac{1}{4}$ of $\frac{4}{3}$

$\frac{1}{4}$ of $\frac{4}{3}=\frac{1}{4}\times\frac{4}{3}$

$=\frac{1\times4}{4\times3}$

$=\frac{1}{3}$

(ii)

(a) $\frac{1}{7}$ of $\frac{2}{9}$

$\frac{1}{7}$ of $\frac{2}{9}=\frac{1}{7}\times\frac{2}{9}$

$=\frac{1}{7}\times\frac{2}{9}$

$=\frac{1\times2}{7\times9}$

$=\frac{2}{63}$

(b) $\frac{1}{7}$ of $\frac{6}{5}$

$\frac{1}{7}$ of $\frac{6}{5}=\frac{1}{7}\times\frac{6}{5}$

$=\frac{1}{7}\times(\frac{6}{5})$

$=\frac{1\times6}{7\times5}$

$=\frac{6}{35}$

(c) $\frac{1}{7}$ of $\frac{6}{5}$

$\frac{1}{7}$ of $\frac{6}{5}=\frac{1}{7}\times(\frac{3}{10})$

$=\frac{1\times3}{7\times10}$

$=\frac{3}{70}$

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Updated on: 10-Oct-2022

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