A container initially contains pure milk.
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A container initially contains pure milk. If 20% of the milk is replaced by water and then the same process is repeated twice (total 3 operations), what will be the percentage of milk remaining?
Show answer & explanation
When 20% of the liquid is replaced by water in each operation, 80% of the existing milk remains. After 3 successive operations, the milk percentage becomes 100% × (0.8)³ = 100% × 0.512 = 51.2%. Each replacement operation multiplies the milk concentration by the retention factor (1 - replacement ratio).
Step-by-step Derivation:
Step-by-step calculation:
Initial milk concentration: 100%
Operation 1: Remove 20% of milk, retain 80%
- Milk remaining = 100% × 0.8 = 80%
Operation 2: Remove 20% of the current mixture (80% milk), retain 80%
- Milk remaining = 80% × 0.8 = 64%
Operation 3: Remove 20% of the current mixture (64% milk), retain 80%
- Milk remaining = 64% × 0.8 = 51.2%
Alternative formula: After n operations with replacement ratio r:
Final concentration = Initial concentration × (1 - r)ⁿ
Final milk = 100% × (1 - 0.2)³ = 100% × (0.8)³ = 100% × 0.512 = 51.2%