Amarge the following in descending order.
(i) $\frac{2}{9}, \frac{2}{3} \cdot \frac{8}{21} $
(ii) $\frac{1}{5}, \frac{3}{7}+\frac{7}{10} $
In a magic square"
Given:
$(i)\ \frac{2}{9}, \frac{2}{3} \times \frac{8}{21}$
$(ii)\ \frac{1}{5}, \frac{3}{7},\ \frac{7}{10}$
Solution::
$(i)\ \frac{2}{9}, \frac{2}{3} \times \frac{8}{21}$
LCM of 3, 9, 21 is 63
Given fractions can be rewritten as
$\frac{14}{63}$, $\frac{42}{63}$, $\frac{24}{63}$
So descending order would be $\frac{42}{63}$, $\frac{24}{63}$, $\frac{14}{63}$
or $\frac{2}{3}$, $\frac{8}{21}$, $\frac{2}{9}$ or
$\frac{2}{3}> \frac{8}{21}> \frac{2}{9}$
$(ii)\ \frac{1}{5}, \frac{3}{7},\ \frac{7}{10} $
LCM of 5, 7, 10 is 70
Given fractions can be rewritten as $\frac{14}{70}$, $\frac{30}{70}$, \frac{49}{70}$
Descending order is $\frac{49}{70} > \frac{30}{70} > \frac{14}{70}$
or $\frac{7}{10} > \frac{3}{7} > \frac{1}{5}$
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