8 litres are drawn from a cask filled with wine and is then filled with water. This operation is performed three more times. The ratio of the quantity of wine now left in cask to that of the total solution is 16 : 81. How much wine did the cask hold originally?

Let the quantity of the wine in the cask originally be x litres.
Using formula:
Final Amount of solute that is not replaced = Initial Amount × $${\left( {\frac{{{\text{Vol}}{\text{. after removal}}}}{{{\text{Vol}}{\text{. after replacing}}}}} \right)^{\text{N}}}$$
Where N = No. of operation done.
Then ratio of wine to total solution in cask after 4 operations,
$$\eqalign{ & 1 \times { {\left( {\frac{{x - 8}}{x}} \right)} ^4} = \frac{{16}}{{81}} \cr & \Rightarrow 1 \times {\left\{ {\frac{{ {x - 8} }}{x}} \right\}^4} = {\left( {\frac{2}{3}} \right)^4} \cr & \Rightarrow \frac{{ {x - 8} }}{x} = \frac{2}{3} \cr & \Rightarrow 3x - 24 = 2x \cr & \Rightarrow x = 24\,{\text{litres}} \cr} $$