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(D) Let the volume of each glass be $1$ unit.In the first glass,the amount of acid is $\frac{2}{5}$ and water is $\frac{3}{5}$.In the second glass,the

Vedclass · Accepted Answer

$401:544$

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(D) Let the volume of each glass be $1$ unit.In the first glass,the amount of acid is $\frac{2}{5}$ and water is $\frac{3}{5}$.In the second glass,the amount of acid is $\frac{3}{7}$ and water is $\frac{4}{7}$.In the third glass,the amount of acid is $\frac{4}{9}$ and water is $\frac{5}{9}$.Total acid in the large vessel $= \frac{2}{5} + \frac{3}{7} + \frac{4}{9} = \frac{126 + 135 + 140}{315} = \frac{401}{315}$.Total water in the large vessel $= \frac{3}{5} + \frac{4}{7} + \frac{5}{9} = \frac{189 + 180 + 175}{315} = \frac{544}{315}$.Therefore,the required ratio of acid to water $= \frac{401}{315} : \frac{544}{315} = 401:544$.

Three glasses of equal volume contain acid mixed with water. The ratios of acid to water are $2:3$,$3:4$,and $4:5$ respectively. If the contents of these glasses are poured into a large vessel,what is the ratio of acid to water in the large vessel?

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