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(A) Let the ratio of milk to water in the first container be $5:1$. The fraction of milk is $\frac{5}{6}$.Let the ratio of milk to water in the second

Vedclass · Accepted Answer

$2:3$

Vedclass · Answer

(A) Let the ratio of milk to water in the first container be $5:1$. The fraction of milk is $\frac{5}{6}$.Let the ratio of milk to water in the second container be $7:2$. The fraction of milk is $\frac{7}{9}$.Let the mixtures be mixed in the ratio $x:y$. The fraction of milk in the final mixture is $80\% = \frac{80}{100} = \frac{4}{5}$.Using the weighted average formula:$\frac{x(\frac{5}{6}) + y(\frac{7}{9})}{x+y} = \frac{4}{5}$Multiply by $90(x+y)$ to clear denominators:$15(5x) + 10(7y) = 72(x+y)$$75x + 70y = 72x + 72y$$3x = 2y$$\frac{x}{y} = \frac{2}{3}$Thus,the required ratio is $2:3$.

$A$ mixture of milk and water is kept in two separate containers. The ratio of milk to water in one container is $5:1$ and in the other container is $7:2$. In what ratio should the mixtures from these two containers be mixed so that the quantity of milk in the new mixture becomes $80\%$?

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