Management Aptitude Test Dec 2013 Paper

(a ) Let amount of solution be 100 ml
∵ Amount of water = 40 ml
∴ Amount of milk = 60 ml
Let amount of solution replaced be x ml.
New solution that is added has 81% milk and 19% water.
∴ Amount of water in new solution obtained = 26% = 26 ml
40 - $\frac{40}{100}\mathit{x}+\frac{19}{100}\mathit{x}$ = 26
⇒40 - 26 = $\frac{21}{00}\mathit{x}$
⇒ 14 = $\frac{21}{00}\mathit{x}$
⇒ x = $\frac{14}{21}$ × 100 = $\frac{2}{3}$ × 100
∴ Quantity of milk replaced = $\frac{\frac{2}{3}\times 100}{100}=\frac{2}{3}$