$8$ litres are drawn from a cask full of 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 water is $16 : 65$. How much wine did the cask hold originally ?

### Concept & Reverse Dilution Logic We use the standard replacement formula, but in reverse, by linking it to the final ratio of the mixture components. $$ \frac{\text{Final Wine}}{\text{Initial Capacity}} = \left(1 - \frac{\text{Replaced Volume}}{\text{Initial Capacity}}\right)^n $$ ### Step-by-Step Solution * **Given:** Replaced volume = $8$ L. Total operations $n = 1 + 3 = 4$. Final ratio of Wine : Water = $16 : 65$. * Let the original capacity of the cask be $x$ litres. * The ratio of Wine to Total Mixture (which equals the original capacity $x$) is $\frac{16}{16 + 65} = \frac{16}{81}$. * Using the formula: $\frac{\text{Final Wine}}{x} = \left(1 - \frac{8}{x}\right)^4$ * Substitute the known ratio: $\frac{16}{81} = \left(1 - \frac{8}{x}\right)^4$ * Take the fourth root of both sides: * $\left(\frac{2}{3}\right)^4 = \left(1 - \frac{8}{x}\right)^4$ * $\frac{2}{3} = 1 - \frac{8}{x}$ * Solve for $x$: * $\frac{8}{x} = 1 - \frac{2}{3} = \frac{1}{3}$ * $x = 8 \times 3 = 24$ litres. ### Exam Strategy & Shortcut Whenever a process is done 4 times and you see numbers like 16 and 81 (total), recognize them immediately as $2^4$ and $3^4$. The ratio of remaining wine to total capacity is $\left(\frac{2}{3}\right)^4$. The fraction removed is $1 - \frac{2}{3} = \frac{1}{3}$. Since $\frac{1}{3}$ of the volume is $8$L, the total volume is $24$L. ### Common Pitfall Using the ratio of Wine to Water ($\frac{16}{65}$) in the formula instead of Wine to Total Mixture ($\frac{16}{81}$). The formula gives the ratio of remaining pure liquid to the *total* initial volume. ### Final Answer Therefore, the correct answer is **24 litres**.