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

Solution

(i) $\frac{2}{9}, \frac{2}{3}, \frac{8}{21}$

Changing them to like fractions, we obtain

$\frac{2}{9} = \frac{2 \times 7}{9 \times 7} = \frac{14}{63} \frac{2}{3} = \frac{2 \times 21}{3 \times 21} = \frac{42}{63} \frac{8}{21} = \frac{8 \times 3}{21 \times 3} = \frac{24}{63}$

Since 42>24>14,

$\frac{2}{3} > \frac{8}{21} > \frac{2}{9} (ii) \frac{1}{5}, \frac{3}{7}, \frac{7}{10}$

Changing them to like fractions, we obtain

$\frac{1}{5} = \frac{1 \times 14}{5 \times 14} = \frac{14}{70} \frac{3}{7} = \frac{3 \times 10}{7 \times 10} = \frac{30}{70} \frac{7}{10} = \frac{7 \times 7}{10 \times 7} = \frac{49}{70}$

As 49 > 30>14,

$\frac{70}{10} > \frac{3}{7} > \frac{1}{5}$

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

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