$\frac{1}{2},\frac{3}{4},\frac{4}{6},\frac{3}{2},\frac{6}{3}$ Among the given fractions $\frac{6}{3}$ is greater than $\frac{1}{2}$ .State whether the statement is True or False .

The correct option is A True
Coneverting $\frac{1}{2},\frac{3}{4},\frac{4}{6},\frac{3}{2},\frac{6}{3}$ into like fractions .

$\frac{1}{2}$ = $\frac{1\times 12}{2\times 12}$ = $\frac{12}{24}$
$\frac{3}{4}$ = $\frac{3\times 6}{4\times 6}$ = $\frac{18}{24}$
$\frac{4}{6}$ = $\frac{2}{3}$ = $\frac{2\times 8}{3\times 8}$ = $\frac{16}{24}$
$\frac{3}{2}$ = $\frac{3\times 12}{2\times 12}$ = $\frac{36}{24}$
$\frac{6}{3}$ = $\frac{6\times 8}{3\times 8}$ = $\frac{48}{24}$

$\frac{48}{24}$ > $\frac{12}{24}$ . The value present in the numerator 48 > 12 as the denominator is 24 .