Successive dilution of wine with water: A 64-L cask of wine has 8 L withdrawn and replaced with water. This process is repeated two more times. How many liters of wine remain after the third replacement?

Introduction / Context:
Repeatedly removing a fixed volume and refilling leads to a multiplicative retention factor each time. For volume V and withdrawal w, the fraction retained after one cycle is (1 − w/V). After n cycles, the retained fraction is (1 − w/V)^n of the original amount.

Given Data / Assumptions:

• Initial wine = 64 L.
• Each cycle: remove 8 L, refill with water.
• Cycles n = 3.

Concept / Approach:
Remaining wine = 64 * (1 − 8/64)^3 = 64 * (7/8)^3. Compute the value to get the exact liters remaining of wine (water makes up the rest).

Step-by-Step Solution:

Verification / Alternative check:
Numeric check: 42.875 L; options present 42 7/8 L which is identical.

Why Other Options Are Wrong:
Other fractional values correspond to incorrect powers or arithmetic and do not match the retention factor (7/8)^3.

Common Pitfalls:
Subtracting 8 L of wine each time (linear) instead of applying the retention fraction (multiplicative).

Final Answer:
42 7/8 L