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1. 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. $\frac{2}{3}$

D. $\frac{1}{2}$

Solution:

Let the quantity of liquid drawn out = x

\begin{aligned} \Rightarrow \frac{3 - \frac{3}{4} x}{1 - \frac{1}{4} x + x} = \frac{1}{1} \\ \Rightarrow 12 - 3 x = 4 - x + 4 x \\ \Rightarrow 8 = 6 x \\ \Rightarrow x = \frac{4}{3} \end{aligned}

Hence, required part of quantity

\begin{aligned} = \frac{\frac{4}{3}}{4} \\ = \frac{1}{3} \end{aligned}

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