A mixture contains 80 % acid and the rest wat | vedclass

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(D) Initially,the mixture contains $80 \%$ acid and $20 \%$ water,so the ratio of Acid : Water $= 80:20 = 4:1$.Let the total volume of the mixture be

Vedclass · Accepted Answer

$\frac{2}{7}$

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(D) Initially,the mixture contains $80 \%$ acid and $20 \%$ water,so the ratio of Acid : Water $= 80:20 = 4:1$.Let the total volume of the mixture be $1$ unit. Let $x$ be the part of the mixture removed and replaced by water.After removing $x$ part of the mixture,the remaining mixture is $(1-x)$.The amount of acid remaining is $4(1-x)/5$ and the amount of water remaining is $(1-x)/5$.After adding $x$ amount of water,the new amount of water is $(1-x)/5 + x$.According to the problem,the new ratio of acid to water is $4:3$:$\frac{4(1-x)/5}{(1-x)/5 + x} = \frac{4}{3}$$\frac{4(1-x)}{1-x+5x} = \frac{4}{3}$$\frac{1-x}{1+4x} = \frac{1}{3}$$3 - 3x = 1 + 4x$$7x = 2$$x = \frac{2}{7}$

$A$ mixture contains $80 \%$ acid and the rest water. The part of the mixture that should be removed and replaced by the same amount of water to make the ratio of acid and water $4:3$ is

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