Question

Arrange $\frac{3}{5}$, $\frac{2}{5}$, $\frac{1}{5}$, $\frac{7}{6}$, $\frac{1}{7}$, and $\frac{3}{8}$ in ascending order.

• A. $\frac{1}{7}$ < $\frac{1}{5}$ < $\frac{3}{8}$ < $\frac{2}{5}$ < $\frac{3}{5}$ < $\frac{7}{6}$
• B. $\frac{1}{7}$ < $\frac{1}{5}$ < $\frac{2}{5}$ < $\frac{3}{8}$ < $\frac{3}{5}$ < $\frac{7}{6}$
• C. $\frac{7}{6}$ < $\frac{1}{5}$ < $\frac{3}{8}$ < $\frac{2}{5}$ < $\frac{3}{5}$ < $\frac{1}{7}$
• D. $\frac{1}{5}$ < $\frac{2}{5}$ < $\frac{3}{8}$ < $\frac{3}{5}$ < $\frac{1}{7}$ < $\frac{7}{6}$

Answer 1

The correct option is A $\frac{1}{7}$ < $\frac{1}{5}$ < $\frac{3}{8}$ < $\frac{2}{5}$ < $\frac{3}{5}$ < $\frac{7}{6}$

$\frac{7}{6}$ is greater than other fractions as it is improper and rest being proper fractions.

For fractions with denominator 5:
$\frac{1}{5}$ < $\frac{2}{5}$ < $\frac{3}{5}$

Comparing fractions with denominator as 8 and 5 by equating the denominators to same value.

$\frac{3}{8}$ = $\frac{3\times 5}{8\times 5}$ = $\frac{15}{40}$

$\frac{1}{5}$ = $\frac{1\times 8}{5\times 8}$ = $\frac{8}{40}$

and $\frac{2}{5}$ = $\frac{2\times 8}{5\times 8}$ = $\frac{16}{40}$

$⟹$ $\frac{1}{5}$ < $\frac{3}{8}$ < $\frac{2}{5}$

Comparing fractions with denominator as 7 and 5 by equating the denominators to same value.

$\frac{1}{7}$ = $\frac{1\times 5}{7\times 5}$ = $\frac{5}{35}$

and $\frac{1}{5}$ = $\frac{1\times 7}{5\times 7}$ = $\frac{7}{35}$

$⟹$ $\frac{1}{7}$ < $\frac{1}{5}$

$\therefore$ $\frac{1}{7}$ < $\frac{1}{5}$ < $\frac{3}{8}$ < $\frac{2}{5}$ < $\frac{3}{5}$ < $\frac{7}{6}$