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Q.2 Arrange the following in descending order: (i)\( \frac{2}{9}, \frac{2}{3}, \frac{8}{21} \)
(ii) \(\frac{1}{5} ,\frac{3}{7}, \frac{7}{10} \)
Answer :

(i)\( \frac{2}{9}, \frac{2}{3}, \frac{8}{21} \)
LCM of 9, 3 ,21 = 63 then
$$ \frac{2}{9}=\frac{2×7}{9×7}=\frac{14}{63} $$ $$ \frac{2}{3}=\frac{2×21}{3×21}=\frac{42}{63} $$ $$ \frac{8}{21}=\frac{8×3}{21×3}=\frac{24}{63} $$ Hence \(\frac{42}{63} > \frac{24}{63} > \frac{14}{63} \)
\(\frac{2}{3} > \frac{8}{21} > \frac{2}{9} \)

(ii) \(\frac{1}{5} ,\frac{3}{7}, \frac{7}{10} \)
LCM of 5,7,10 = 70 then
$$ \frac{1}{5}=\frac{1×14}{5×14}=\frac{14}{70} $$ $$ \frac{3}{7}=\frac{3×10}{7×10}=\frac{21}{70} $$ $$ \frac{7}{10}=\frac{7×7}{10×7}=\frac{49}{70} $$ Hence \(\frac{49}{70} > \frac{21}{70} > \frac{14}{70} \)
\(\frac{7}{10} > \frac{3}{7} > \frac{1}{5} \)