From a vessel containing pure milk, 20% is replaced by water and the process…
2025
From a vessel containing pure milk, 20% is replaced by water and the process is repeated 2 more times. At the end of the third operation, the percentage of pure milk in the mixture is :
- A.
40%
- B.
51.2%
- C.
60%
- D.
25%
Attempted by 12 students.
Show answer & explanation
Correct answer: B
Concept: When a fraction of a solution is removed and replaced by another liquid, the fraction of the original substance that survives one operation equals (1 − replaced fraction) of whatever was present just before that operation. Repeating the operation n times multiplies the retained fraction by itself n times, so the substance left after n operations = initial quantity × (1 − replaced fraction)n.
Application:
Take the vessel to initially hold 100 units of pure milk, so that the amount of milk left after each step reads directly as a percentage.
Each operation replaces 20% of the current mixture with water, so the fraction of the mixture retained after one operation is 1 − 0.20 = 0.80.
After the 1st operation: milk left = 100 × 0.80 = 80 units.
After the 2nd operation: milk left = 80 × 0.80 = 64 units.
After the 3rd operation: milk left = 64 × 0.80 = 51.2 units.
Because the base quantity was taken as 100, the milk left after the third operation equals the percentage of pure milk remaining in the mixture: 51.2%.
Cross-check: Applying the successive-replacement formula directly gives the same figure: retained fraction after 3 operations = 0.803 = 0.512, i.e. 51.2%, matching the step-by-step calculation above.