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Q 9.     Which is greater in each of the following:

(i)      $\frac{2}{3}, \frac{5}{2}$ (ii)      $\frac{-5}{6}, \frac{-4}{3}$

(iii)    $\frac{-3}{4}, \frac{2}{-3}$ (iv)      $\frac{-1}{4}, \frac{1}{4}$

(v)   $-3\frac{2}{7}, -3\frac{4}{5}$

Answers (1)

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(i) $\frac{2}{3}, \frac{5}{2}$

$\Rightarrow \frac{2\times2}{3\times2}, \frac{5\times3}{2\times3}$

$\Rightarrow \frac{4}{6}, \frac{15}{6}$

Since, $\frac{15}{6}> \frac{4}{6}$

So, $\frac{5}{2}> \frac{2}{3}.$

(ii) $\frac{-5}{6}, \frac{-4}{3}$

$\Rightarrow \frac{-5\times3}{6\times3}, \frac{-4\times6}{3\times6}$

$\Rightarrow \frac{-15}{18}, \frac{-24}{18}$

Since, $\frac{-15}{18}> \frac{-24}{18}$

So, $\frac{-5}{6}> \frac{-4}{3}.$

(iii) $\frac{-3}{4}, \frac{2}{-3}$

$\Rightarrow \frac{-3\times-3}{4\times-3}, \frac{2\times4}{-3\times4}$

$\Rightarrow \frac{9}{-12}, \frac{8}{-12}$

Since, $\frac{8}{-12}> \frac{9}{-12}$

So, $\frac{2}{-3}> \frac{-3}{4}$

(iv) $\frac{-1}{4}, \frac{1}{4}$

$\Rightarrow \frac{1}{4}> \frac{-1}{4}$

As each positive number is greater than its negative.

(v) $-3\frac{2}{7}, -3\frac{4}{5}$

$\Rightarrow \frac{-23}{7}, \frac{-19}{5} =\frac{-23\times5}{7\times5}, \frac{-19\times7}{5\times7}$

$\Rightarrow \frac{-115}{35} > \frac{-133}{35}$

So, $-3\frac{2}{7}> -3\frac{4}{5}$

Posted by

Divya Prakash Singh

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