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1. 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. $\frac{1}{3}$

B. $\frac{1}{4}$

C. $\frac{1}{5}$

D. $\frac{1}{7}$

Solution:

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 =

(3 - \frac{3 x}{8} + x)

litres
Quantity of syrup in new mixture =

(5 - \frac{5 x}{8})

litres

\begin{aligned} ∴ 3 - \frac{3 x}{8} + x = 5 - \frac{5 x}{8} \\ \Rightarrow 5 x + 24 = 40 - 5 x \\ \Rightarrow 10 x = 16 \\ \Rightarrow x = \frac{8}{5} \end{aligned}

So, part of the mixture replaced

\begin{aligned} = \frac{8}{5} \times \frac{1}{8} \\ = \frac{1}{5} \end{aligned}

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