Question

A barrel contains a mixture of wine and water in the ratio 3 : 1 How much fraction of the mixture must be drawn off and substituted by water so that the ratio of wine and water in the resultant mixture in the barrel becomes 1 : 1?

• A. $\frac{1}{4}$
• B. $\frac{1}{3}$
• C. $\frac{3}{4}$
• D. $\frac{2}{3}$

Answer 1

The correct option is B $\frac{1}{3}$

Let the barrel contains wine 3x litres and water x litres.
Then, total mixture=4x litres
Let the part of mixture drawn out be p litres.
$\therefore (4x-p)\times \frac{3}{4}:(4x-p)\times \frac{1}{4}+p=1:1\Rightarrow 3x-\frac{3p}{4}=x+\frac{3p}{4}\Rightarrow 2x=\frac{6p}{4}\Rightarrow p=\frac{2x\times 4}{6}=\frac{1}{3}\left(4x\right)\therefore \frac{1}{3}$ part of mixture is drawn out.