Delhi Police Constable 2012 Paper

Let price of 1 litre milk is ₹ x.
Increase in rate 25%,
new rate $=\mathit{x}+\frac{\mathit{x}\times 25}{100}=\frac{5\mathit{x}}{4}$
Quantity of milk in $\mathit{₹}\frac{5\mathit{x}}{4}$ is 1 litre
So, quantity of milk in $\mathit{₹}\mathit{x}=\frac{4\mathit{x}}{5\mathit{x}}$
In terms of percentage $=\left(\frac{4\mathit{x}}{5\mathit{x}}\times 100\right)\%$
$=80\%$
Hence, reduction in uses $=100-80\%=20\%$