A drum contains 80 litres of ethanol. 20 litres of this liquid is removed and replaced with water. 20 liters of this mixture is again removed and replaced with water. How much water (in litres) is present in this drum now?

After replacing 20 litre of liquid for first time, ratio of
\[\begin{array}{*{20}{c}} {{\text{Ethanol}}}&:&{{\text{Water}}} \\ {60}&:&{20} \end{array}\]
Now, again removing 20 litre of the mixture in the given proportion.
\[\begin{array}{*{20}{c}} {{\text{Ethanol}}}&:&{{\text{Water}}} \\ {45}&:&{15} \end{array}\]
Now, an adding 20 litre water
Total water in the drum = 15 + 20 = 35 litre

$$\eqalign{ & {\bf{Alternate}}\,{\bf{Solution:}} \cr & {\text{Final quantity}} = {\text{Initial quantity}} \times {\left( {1 - \frac{x}{c}} \right)^n} \cr & x \to {\text{ quantity taken out}} \cr & c \to {\text{ total capacity}} \cr & n \to {\text{ number of process}} \cr & = 80 \times {\left( {1 - \frac{{20}}{{80}}} \right)^2} \cr & = 80 \times {\left( {\frac{3}{4}} \right)^2} \cr & = 80 \times \frac{3}{4} \times \frac{3}{4} \cr & {\text{Final quantity of external}} = 45{\text{ litre}} \cr & {\text{and Water}} = 80 - 45 = 35{\text{ litre}} \cr} $$