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?

Explanation:-

$$\eqalign{ & {\text{Suppose the vessel initially contains 8 litres of liquid}}{\text{.}} \cr & {\text{Let}}\,x{\text{}}\,{\text{litres of this liquid be replaced with water}}{\text{.}} \cr & {\text{Quantity}}\,{\text{of}}\,{\text{water}}\,{\text{in}}\,{\text{new}}\,{\text{mixture}} \cr & = \left( {3 - \frac{{3x}}{8} + x} \right)\,{\text{litres}} \cr & {\text{Quantity}}\,{\text{of}}\,{\text{syrup}}\,{\text{in}}\,{\text{new}}\,{\text{mixture}} \cr & = \left( {5 - \frac{{5x}}{8}} \right)\,{\text{litres}} \cr & \therefore \left( {3 - \frac{{3x}}{8} + x} \right) = \left( {5 - \frac{{5x}}{8}} \right) \cr & \Rightarrow 5x + 24 = 40 - 5x \cr & \Rightarrow 10x = 16 \cr & \Rightarrow x = \frac{8}{5} \cr & {\text{So,}}\,{\text{part}}\,{\text{of}}\,{\text{the}}\,{\text{mixture}}\,{\text{replaced}} \cr & = \left( {\frac{8}{5} \times \frac{1}{8}} \right) = \frac{1}{5} \cr}$$