Question

A vessel has 500 litres of milk. 50 litres of milk is replaced with 50 litres of water in this vessel. If this operation is repeated another 2 more times, what is the percentage of milk in the vessel at the end?

• A. 70.0
• B. 80.0
• C. 72.9
• D. 81.0

Answer 1

The Correct Option is B

We need to determine the percentage of milk remaining in a vessel after repeatedly replacing part of it with water. Let's go through the step-by-step calculation:

1. Initially, the vessel contains 500 litres of pure milk.
2. In the first operation, 50 litres of milk is replaced with 50 litres of water. The remaining quantity of milk after the first replacement is calculated as follows: \(500 \, \text{litres} - 50 \, \text{litres} = 450 \, \text{litres of milk remaining}\). Thus, the fraction of milk after the first replacement is \(\frac{450}{500}\).
3. After the first replacement, the new concentration of milk is \(\frac{450}{500} \times 100\% = 90\%\) milk.
4. In the second operation, again 50 litres of the new solution is replaced with water. The amount of milk remaining is: \(450 \, \text{litres} \times \left(\frac{450}{500}\right)\) = \(450 \times 0.9 = 405 \, \text{litres of milk}\).
5. The concentration of milk after the second operation is \(\frac{405}{500} \times 100\% = 81\% \, \text{milk}\).
6. In the final (third) operation, 50 litres of the solution is again replaced with water. The remaining milk is: \(405 \, \text{litres} \times \left(\frac{450}{500}\right) = 405 \times 0.9 = 364.5 \, \text{litres of milk}\).
7. Thus, the final concentration of milk after the third replacement is \(\frac{364.5}{500} \times 100\% = 72.9\% \, \text{milk}\).

Based on this calculation, the final percentage of milk in the vessel is 72.9%. The apparent correct answer given is 80.0%, but this series of calculations shows that the mathematical result for the operations described is 72.9%.