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

Given:

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

(ii) $\frac{1}{5}, \frac{3}{7},\ \frac{7}{10}$

To do:

We have to arrange the given fractions in descending order in each case.

Solution:

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

LCM of $3, 9, 21$ is $63$

Convert the given fractions into equivalent fractions having $63$ as the denominator.

Given fractions can be rewritten as,

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

$=\frac{2\times7}{9\times7}$

$=\frac{14}{63}$

$\frac{2}{3}=\frac{2}{3}\times\frac{21}{21}$

$=\frac{2\times21}{3\times21}$

$=\frac{42}{63}$

$\frac{8}{21}=\frac{8}{21}\times\frac{3}{3}$

$=\frac{8\times3}{21\times3}$

$=\frac{24}{63}$

Here,

$\frac{42}{63}>\frac{24}{63}>\frac{14}{63}$

This implies,

$\frac{2}{3}>\frac{8}{21}>\frac{2}{9}$

Therefore, the descending order of the given fractions is $\frac{2}{3}$, $\frac{8}{21}$, $\frac{2}{9}$.

(ii) $\frac{1}{5},\ \frac{3}{7},\ \frac{7}{10}$

LCM of $5, 7, 10$ is $70$

Convert the given fractions into equivalent fractions having $70$ as the denominator.

Given fractions can be rewritten as,

$\frac{1}{5}=\frac{1}{5}\times\frac{14}{14}$

$=\frac{1\times14}{5\times14}$

$=\frac{14}{70}$

$\frac{3}{7}=\frac{3}{7}\times\frac{10}{10}$

$=\frac{3\times10}{7\times10}$

$=\frac{30}{70}$

$\frac{7}{10}=\frac{7}{10}\times\frac{7}{7}$

$=\frac{7\times7}{10\times7}$

$=\frac{49}{70}$

Here,

$\frac{49}{70}>\frac{30}{70}>\frac{14}{70}$

This implies,

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

Therefore, the descending order of the given fractions is $\frac{7}{10}$, $\frac{3}{7}$, $\frac{1}{5}$.

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Updated on: 10-Oct-2022

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