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 :
A. 40% pure
B. 50% pure
C. 51.2% pure
D. 58.8% pure
Answer: Option C
Solution (By Examveda Team)
Let total quantity of original milk = 1000 gmMilk 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%
This Question Belongs to Arithmetic Ability >> Percentage
You're absolutely correct; after the first operation, the mixture will no longer contain pure milk, and each replacement operation will remove both milk and water. Here’s the corrected calculation:
Let 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.
The calculation is wrong. From the second operation certain amount of water will also be withdrawn with the milk as after the first operation is done the milk will no longer be pure.