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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