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?

Answer: Option C

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 = $$\left(3-\frac{3x}{8}+x\right)$$   litres
Quantity of syrup in new mixture = $$\left(5-\frac{5x}{8}\right)$$   litres
$$\begin{array}{rl}& \therefore 3-\frac{3x}{8}+x=5-\frac{5x}{8}\\ & \Rightarrow 5x+24=40-5x\\ & \Rightarrow 10x=16\\ & \Rightarrow x=\frac{8}{5}\end{array}$$
So, part of the mixture replaced
$$\begin{array}{rl}& =\frac{8}{5}\times \frac{1}{8}\\ & =\frac{1}{5}\end{array}$$