Management Aptitude Test Sep 2020 Paper

Let amount of solution be $100\mathit{m}\mathit{l}$.
$\because$ Amount of water $=40\mathit{m}\mathit{l}$
$\therefore$ Amount of milk $=60\mathit{m}\mathit{l}$ Let amount of solution replaced be $\mathit{x}\mathit{m}\mathit{l}$.
New solution that is added has $81\%$ milk and $19\%$ water.
$\therefore$ Amount of water in new solutionobtained
$=26\%=26\mathit{m}\mathit{l}$
$\therefore 40-\frac{40\mathit{x}}{100}+\frac{19\mathit{x}}{100}=26$
$\Rightarrow 40-26=\frac{21\mathit{x}}{100}\Rightarrow 14=\frac{21\mathit{x}}{100}$
$\Rightarrow \mathit{x}=\frac{14}{21}\times 100=\frac{2}{3}\times 100$
$\therefore$ Quantity of milk replaced $=\frac{\frac{2}{3}\times 100}{100}=\frac{2}{3}$