Question

The variable $x$ is inversely proportional to $y.$ If $x$ increase by $p\%$, then by what per cent will $y$ decrease?

Answer 1

The variable $x$ is inversely proportional $y$.
$\therefore$ $xy=k$ (constant)

$x=\frac{k}{y}\dots \dots \left(i\right)$

When $x$ is increased by $p\%$, then new value will be

$x^{\prime }=x+p\%$ of $x$

$x=x+\frac{p}{100}\times x$

$x^{\prime }=x\left(\frac{100+p}{100}\right)$

Since we know that two quantities $x$ and $y$ are said to be in inverse proportion, if an increase in $x$ causes a proportional decrease in $y$ and vice -versa.

So, $y$ will decrease when $x$ is increased, and the new decreased value of $y$ will be $y^{\prime }$

on putting the above value of $x^{\prime }$ in eq.(i), we get

$x\left(\frac{100+p}{100}\right)=\frac{k}{y^{\prime }}$

$\Rightarrow y^{\prime }=\frac{k}{x}\times \frac{100}{100+p}$

$\Rightarrow y^{\prime }=\frac{100}{100+p}\times y$ [from $eq.\left(i\right),y=\frac{k}{x}$]

$\Rightarrow y^{\prime }=\frac{(100+p)-p}{100+p}\times y$ [on adding and subtracting $p$ in the numerator]

$\Rightarrow y^{\prime }=\left(1-\frac{p}{100+p}\right)\times y$

$\Rightarrow y^{\prime }=\left(1\times y-\frac{p}{100+p}y\right)$

$\Rightarrow y^{\prime }=\left(y-\frac{p}{100+p}y\right)$

So, $y$ is decreased by $\frac{p}{100+p}\%$