Question

If the numerator of a fraction is increased by $20\%$ and the denominator is decreased by $10\%$, then the value of the fraction is $\frac{16}{21}$. What is the original fraction?

• A. $\frac{3}{5}$
• B. $\frac{3}{10}$
• C. $\frac{4}{5}$
• D. $\frac{4}{7}$

Answer 1

Answer: (D)

$\frac{4}{7}$

Let the fraction be $\frac{x}{y}$.

Then from the given condition, $\frac{x+20}{y-10}=\frac{16}{21}$
$\Rightarrow \frac{x+\frac{x}{5}}{y-\frac{y}{10}}=\frac{16}{21}$

$\Rightarrow \frac{\frac{6x}{5}}{\frac{9y}{10}}=\frac{16}{21}$

$\Rightarrow \frac{x}{y}=\frac{16}{21}\times \frac{9}{10}\times \frac{5}{6}=\frac{4}{7}$