Question

Suppose $x$ and $y$ are inversely proportional and positive. If $x$ increases $p\%$ then $y$ decreases by

• A. $p\%$
• B. $\frac{p}{1+p}\%$
• C. $\frac{100}{p}\%$
• D. $\frac{100p}{100+p}\%$

Answer 1

The correct option is D $\frac{100p}{100+p}\%$

x and y are inversely proportional

$\Rightarrow x\alpha \frac{1}{y}$

$\Rightarrow xy=$ constant

$x_1{y}_1=x_2{y}_2$

x has 'p' percentage increase :-

$x_2=x_1\times (1+\frac{P}{100})$

$x_1{y}_1=x_2{y}_2$

$x_1{y}_1=x_1(1+\frac{1}{100})y_2$

$y_2=\frac{100x_1}{100+p}$

$y_2=\left(\frac{100+p}{100+p}\right)y_1=\frac{py}{100+p}$

$y_2-y_1=\frac{p{y}_1}{100+p}$

$\frac{y_2-y_1}{y_1}=-\frac{p}{100+p}$

$\frac{y_1-y_2}{y_1}=\frac{p}{100+p}$

Percentage decrease $=\frac{y_1-y_2}{y_1}\times 100=\frac{100p}{100+p}\%$