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(B) Let the quantity of milk taken from the first can be $x$ litres and from the second can be $(12 - x)$ litres.In the first can,water is $25\%$,so m

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$6$ litres,$6$ litres

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(B) Let the quantity of milk taken from the first can be $x$ litres and from the second can be $(12 - x)$ litres.In the first can,water is $25\%$,so milk is $75\% = \frac{3}{4}$.In the second can,water is $50\%$,so milk is $50\% = \frac{1}{2}$.The final mixture of $12$ litres has a water to milk ratio of $3:5$,which means milk is $\frac{5}{3+5} = \frac{5}{8}$ of the total volume.Total milk in the final mixture $= 12 	imes \frac{5}{8} = 7.5$ litres.Now,set up the equation based on the milk content:$\frac{3}{4}x + \frac{1}{2}(12 - x) = 7.5$Multiply by $4$ to clear the denominators:$3x + 2(12 - x) = 30$$3x + 24 - 2x = 30$$x = 6$ litres.Therefore,the quantity from the first can is $6$ litres and from the second can is $12 - 6 = 6$ litres.

$A$ milk vendor has $2$ cans of milk. The first contains $25\%$ water and the rest milk. The second contains $50\%$ water. How much milk should he mix from each of the containers so as to get $12$ litres of milk such that the ratio of water to milk is $3:5$?

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