Question : A vessel is filled with liquid, 3 parts of which are water and 5 parts syrup. How much of the mixture must be drawn off and replaced with water so that the mixture may be half water and half syrup?
Option 1: $\frac{1}{3}$
Option 2: $\frac{1}{4}$
Option 3: $\frac{1}{5}$
Option 4: $\frac{1}{7}$
Correct Answer: $\frac{1}{5}$
Solution : Given: Suppose the vessel initially contains 8 litres of liquid. In this liquid, water is 3 litres and syrup is 5 litres. So the ratio is 3 : 5. Let $x$ litres of this liquid be replaced with water. Quantity of water in $x$ litre of liquid $=\frac{3x}{8}$ Quantity of syrup in $x$ litre of liquid $=\frac{5x}{8}$ As per given condition, The quantity of water in the new mixture is $=3-\frac{3x}{8}+x$ The quantity of syrup in the new mixture is $=5-\frac{5x}{8}$ After replacement, the quantity should be the same. $3-\frac{3x}{8}+x=5-\frac{5x}{8}$ ⇒ $\frac{5x}{8}-\frac{3x}{8}+x=2$ ⇒ $\frac{10x}{8}=2$ $\therefore x=\frac{8}{5}$ So, part of the mixture replaced $=\frac{\frac{8}{5}}{8}=\frac{8}{5}×\frac{1}{8}=\frac{1}{5}$ Hence, the correct answer is $\frac{1}{5}$.
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