Question

A bottle contains $\frac{3}{4}$ of milk and the rest water. How much of the mixture must be taken away and replaced by an equal quantity of water so that the mixture has half milk and half water?

Answer 1

Let the total amount of mixture be $X$

Then Initial amount of milk = $\frac{3}{4}\times X$

Initial amount of water = $\frac{1}{4}\times X$

Let $\frac{1}{y}th$ of the mixture is required to be taken away

Then According to question $\frac{3}{4}\times X$ $-\frac{1}{y}\times \frac{3}{4}X$ = $\frac{1}{4}\times X$ $-\frac{1}{y}\times \frac{1}{4}X$ $+\frac{1}{y}X$

Solving we get $y=3$

So $\frac{1}{3}rd$ of the mixture is required to be taken away