Question 1

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 five maintaining half milk and half water?

Solution

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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