Question

Two positive numbers x and y are inversely proportional. If x increases by 20%, then percentage decrease in y is _____.

• A. $17\frac{3}{4}$
• B. $16\frac{2}{3}$
• C. 15
• D. $1\frac{9}{11}$

Answer 1

The correct option is B $16\frac{2}{3}$

$x\propto \frac{1}{y}$
$⟹x=\frac{k}{y}$, where k is a constant
$⟹y=\frac{k}{x}$
When $x$ increases by 20%,
$x^{\prime }=x+\frac{20\left(x\right)}{100}=\frac{6x}{5}{y}^{\prime }=\frac{k}{\left(\frac{6x}{5}\right)}=\frac{5k}{6x}$
Change in $y=\frac{5k}{6x}-\frac{k}{x}=\frac{-k}{6x}$

$\therefore$ Value of y decreases.
Percentage decrease in $y$ is given by,
$\frac{\frac{k}{6x}}{\frac{k}{x}}\times 100$
= $\frac{1}{6}\times 100$
= $\frac{50}{3}$
= $16\frac{2}{3}$