Successive dilution fraction removed after five replacements: From a vessel of alcohol, one-fifth of the contents is removed and replaced with water. This operation is repeated five times. What fraction of the original alcohol has been removed?

Introduction / Context: Each cycle retains a fixed fraction of the alcohol. If one-fifth is removed, four-fifths remain. After n cycles, the remaining fraction is (4/5)^n, so the removed fraction is 1 − (4/5)^n. Apply this with n = 5.

Given Data / Assumptions:

• Removal each time = 1/5 of whatever is present at that time.
• n = 5 successive replacements.

Concept / Approach: Remaining fraction = (4/5)^5 = 1024/3125. Therefore removed fraction = 1 − 1024/3125 = 2101/3125.

Step-by-Step Solution:

Verification / Alternative check: Decimal check: 1024/3125 ≈ 0.32768 ⇒ removed ≈ 0.67232; aligns with five successive 20% removals.

Why Other Options Are Wrong: 1024/3125 is the remaining fraction, not removed; 1024/2101 is not meaningful in this context; other numerators/denominators do not match (4/5)^5.

Common Pitfalls: Subtracting 1/5 each time linearly instead of compounding the retention factor.

Final Answer: 2101/3125