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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 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
1). $\frac{1}{4}$
2). $\frac{1}{3}$
3). $\frac{3}{4}$
4). $\frac{2}{3}$

This Question has 3 answers.

i think option 2 is correct

Let the barrel contain 4 litres
of mixture.
$\therefore$ Wine = 3 litres
Water = 1 litre
Let x litre mixture is taken out.
$\therefore$ Wine in (4 – x) litres mixture = $\frac{3}{4}(4-x)$

On adding x litres water, water in
mixture, $(4-x)\times\frac{1}{4}$+4 ,

$\therefore$ x=$\frac{4}{3}$

Required answer = $\frac{\frac{4}{3}}{4}$ = $\frac{1}{3}$

required ans is 1/3.........................................

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