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(A) Let the concentration of wine in the first mixture be $\frac{3}{5}$ and in the second mixture be $\frac{4}{9}$.We want the final mixture to have e

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$5 \frac{2}{5}$

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(A) Let the concentration of wine in the first mixture be $\frac{3}{5}$ and in the second mixture be $\frac{4}{9}$.We want the final mixture to have equal quantities of wine and water,so the concentration of wine in the final mixture should be $\frac{1}{2}$.Using the rule of alligation:Ratio of first mixture to second mixture $= |\frac{4}{9} - \frac{1}{2}| : |\frac{3}{5} - \frac{1}{2}|$$= |\frac{8-9}{18}| : |\frac{6-5}{10}| = \frac{1}{18} : \frac{1}{10} = 10 : 18 = 5 : 9$.Given that $3 \, L$ of the first mixture is used,let $x$ be the quantity of the second mixture.$\frac{5}{9} = \frac{3}{x}$$5x = 27$$x = \frac{27}{5} = 5 \frac{2}{5} \, L$.

$A$ mixture contains wine and water in the ratio $3:2$ and another mixture contains them in the ratio $4:5$. How many litres of the latter must be mixed with $3 \, L$ of the former so that the resulting mixture may contain equal quantities of the wine and water? (in $L$)

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