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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 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? Correct Answer $$\frac{{1}}{{5}}$$

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
$$\eqalign{ & \therefore {3 - \frac{{3x}}{8} + x} = {5 - \frac{{5x}}{8}} \cr & \Rightarrow 5x + 24 = 40 - 5x \cr & \Rightarrow 10x = 16 \cr & \Rightarrow x = \frac{8}{5} \cr} $$
So, part of the mixture replaced
$$\eqalign{ & = {\frac{8}{5} \times \frac{1}{8}} \cr & = \frac{1}{5} \cr} $$

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