A pot initially contains pure alcohol. In each operation, 20% of the mixture is removed and replaced with water. This replacement is done three times in total. What percentage of alcohol remains after the third operation?

Introduction / Context:
Successive replacement problems rely on the fact that each operation removes a fixed fraction of whatever is currently in the vessel. If a fraction r is replaced each time, the remaining fraction of the original solute after k operations is (1 − r)^k.

Given Data / Assumptions:

• Initially: 100% alcohol.
• Each time, 20% of the mixture is removed and replaced with water.
• Number of operations k = 3.
• No volume losses; perfect mixing after each operation.

Concept / Approach:
After one replacement, the fraction of original alcohol left is (1 − 0.20) = 0.80. After k operations, the fraction of alcohol remaining equals 0.80^k. Convert this fraction to a percentage for the final answer.

Step-by-Step Solution:

Verification / Alternative check:
Equivalent formula: Remaining% = (initial%) * (1 − r)^k = 100% * 0.8^3 = 51.2%.

Why Other Options Are Wrong:
49% and 38% are typical rounding or arithmetic slips; 29% corresponds to removing a larger fraction or more cycles.

Common Pitfalls:
Confusing ”removed fraction” with ”remaining fraction” or adding percentages instead of multiplying factors across steps.

Final Answer:
51.2%