A barrel contains a mixture of wine and water | vedclass

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(B) Let the total volume of the mixture in the barrel be $4\,L$.Since the ratio of wine to water is $3:1$,the quantity of wine is $3\,L$ and the quant

Vedclass · Accepted Answer

$\frac{1}{3}$

Vedclass · Answer

(B) Let the total volume of the mixture in the barrel be $4\,L$.Since the ratio of wine to water is $3:1$,the quantity of wine is $3\,L$ and the quantity of water is $1\,L$.Let $x$ be the fraction of the mixture drawn off.When $x$ fraction of the mixture is removed,the amount of wine removed is $3x$ and the amount of water removed is $x$.The remaining amount of wine is $3 - 3x$ and the remaining amount of water is $1 - x$.After adding $x$ amount of water (since the total volume removed is $x$ fraction of $4\,L$,i.e.,$4x$),the new amount of water becomes $(1 - x) + 4x = 1 + 3x$.According to the problem,the new ratio of wine to water is $1:1$,so:$\frac{3 - 3x}{1 + 3x} = \frac{1}{1}$$3 - 3x = 1 + 3x$$6x = 2$$x = \frac{2}{6} = \frac{1}{3}$Thus,the required fraction of the mixture to be drawn off is $\frac{1}{3}$.

$A$ 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 $1:1$?

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