Question

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

Answer 1

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

By converting these into like fractions,

$\frac{2}{3}=\frac{2\times 2}{3\times 2}=\frac{4}{6}$

$\frac{5}{2}=\frac{5\times 3}{2\times 3}=\frac{15}{6}$

As $\frac{15}{6}>\frac{4}{6}$, therefore, $\frac{5}{2}$ is greater.

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

$\frac{-5}{6}=\frac{-5}{6}$

$\frac{-4}{3}=\frac{-4\times 2}{3\times 2}=\frac{-8}{6}$

As $\frac{-5}{6}>\frac{-8}{6}$, therefore $\frac{-5}{6}$ is greater.

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

or, $\frac{-3}{4},\frac{-2}{3}$

By converting these into like fractions,

$\frac{-3}{4}=\frac{-3\times 3}{4\times 3}=\frac{-9}{12}$

$\frac{-2}{3}=\frac{-2\times 4}{3\times 4}=\frac{-8}{12}$

As $\frac{-8}{12}>\frac{-9}{12}$, therefore $\frac{2}{-3}$ is greater.

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

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

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

$-3\frac{2}{7}=\frac{-23}{7}$

$-3\frac{4}{5}=\frac{-19}{5}$

By converting these into like fractions,

$\frac{-23}{7}=\frac{-23\times 5}{7\times 5}=\frac{-115}{35}$

$\frac{-19}{5}=\frac{-19\times 7}{5\times 7}=\frac{-133}{35}$

As, $\frac{-115}{35}>\frac{-133}{35},$

therefore $-3\frac{2}{7}$ is greater.