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From a container having pure milk, 20% is replaced by water and the process is repeated thrice. At the end of the third operation, the milk is :
Answer & Solution
Correct Answer:
Option
C
Let total quantity of original milk = 1000 gm
Milk after first operation = 80% of 1000 = 800 gm
Milk after second operation = 80% of 800 = 640 gm
Milk after third operation = 80% of 640 = 512 gm
∴ Strength of final mixture = 51.2%
Milk after first operation = 80% of 1000 = 800 gm
Milk after second operation = 80% of 800 = 640 gm
Milk after third operation = 80% of 640 = 512 gm
∴ Strength of final mixture = 51.2%
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LoginLet the initial quantity of milk be 100 liters.
Since 20% is replaced each time, 80% of the mixture remains after each operation.
Therefore, the percentage of milk remaining after each operation can be calculated using the formula:
Remaining milk percentage = Initial milk percentage × (0.8)^n
where n is the number of times the operation is repeated.
After three operations:
Remaining milk percentage = 100% × (0.8)^3 = 100% × 0.512 = 51.2%
Thus, at the end of the third operation, the milk is 51.2% pure.