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Milk and water in two vessels A and B are in the ratio 4 : 3 and 2 : 3 respectively. In what ratio the liquids in both the plate vessels should be mixed to obtain a new mixture in vessel C, only maintaining half milk and half water?
The fraction of milk in A = 4/(4+3) = 4/7
The fraction of milk in B = 2/(2+3) = 2/5
Let them be mixed in the ratio x:y such that their fraction is 1/2
It can be written as
$$\frac{4x}{7}$$ + $$\frac{2y}{5}$$ = $$\frac{x+y}{2}$$
Upon solving,
$$\frac{4x}{7}$$ - $$\frac{x}{2}$$ = $$\frac{y}{2}$$ - $$\frac{2y}{5}$$
$$\frac{x}{14}$$ = $$\frac{y}{10}$$
$$\frac{x}{y}$$ = $$\frac{7}{5}$$
Required ratio is 7:5
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