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?

A. $\large\frac{1}{3}$ B. $\large\frac{1}{4}$ C. $\large\frac{1}{5}$ D. $\large\frac{1}{7}$ Answer: Option C
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Solution(By Apex Team)

Suppose the vessel initially contains 8 litres of liquid. Let x litres of this liquid be replaced with water. Quantity of water in new mixture = $\left(3-\Large\frac{3 x}{8}+x\right)$ litres Quantity of syrup in new mixture = $\left(5-\Large\frac{5 x}{8}\right)$ litres $\begin{aligned}&\therefore3-\frac{3x}{8}+x=5-\frac{5x}{8}\\ &\Rightarrow5x+24=40-5x\\ &\Rightarrow10x=16\\ &\Rightarrow x=\frac{8}{5}\end{aligned}$ So, part of the mixture replaced $\begin{array}{l} =\Large\frac{8}{5} \times \frac{1}{8} \\ =\Large\frac{1}{5} \end{array}$

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