A cask contains 25 liters of wine. First 5 liters are withdrawn and replaced by water; this is repeated twice more (three operations in total). Find the quantity of wine left and the ratio of wine to water in the final mixture.

Introduction / Context:
Repeated replacement problems use the exponential decay formula for the retained fraction after each withdrawal. Each time 5 liters are removed from 25 liters, the same fraction of wine is taken out and replaced by water.

Given Data / Assumptions:

• Initial wine = 25 L.
• Each operation: remove 5 L, then add 5 L water.
• Number of operations = 3.

Concept / Approach:
After each operation, the fraction of wine remaining equals (1 − 5/25) = 4/5 of the previous amount. After n operations: wine_left = 25*(4/5)^n.

Step-by-Step Solution:
wine_left = 25*(4/5)^3 = 25*(64/125) = 12.8 LWater = total − wine = 25 − 12.8 = 12.2 LWine : Water = 12.8 : 12.2 = 128 : 122 = 64 : 61

Verification / Alternative check:
Successive amounts: after 1st = 20 L, 2nd = 16 L, 3rd = 12.8 L; consistent with (4/5)^n scaling.

Why Other Options Are Wrong:
46 : 16 reverses values; 61 : 64 is water : wine, not wine : water; 1 : 1 is inaccurate.

Common Pitfalls:
Subtracting 5 liters of wine each time (linear) instead of multiplying by 4/5 (geometric).

Final Answer:
64 : 61