Question

A tub contains 60 litres of milk. From this tub, 6 litres of milk was taken out and replaced with water. This whole process was repeated further two more times. How much milk is there in the tub now?

• A. 29.16 litre
• B. 43.74 litre
• C. 42.24 litre
• D. 38.74 litre

Hint:

In mixture replacement problems, focus on the fraction of the original liquid that remains after one operation. In this case, \(1 - 6/60 = 9/10\) of the milk remains. For three operations, you simply calculate \( \text{Initial} \times (\frac{9}{10}) \times (\frac{9}{10}) \times (\frac{9}{10}) \).

Answer 1

The Correct Option is B

Step 1: Conceptualization:
This is a mixture problem involving repeated removal and replacement of a substance. The objective is to determine the residual fraction of the original substance after each operation.
Step 2: Governing Formula:
The quantity of the original substance remaining after 'n' operations is calculated using the formula:\[ \text{Final Quantity} = \text{Initial Quantity} \times \left(1 - \frac{\text{Quantity Replaced}}{\text{Total Volume}}\right)^n \]'n' represents the total count of operations performed.
Step 3: Calculation Procedure:
Given parameters:
Initial milk volume = 60 litres.
Total mixture volume = 60 litres (constant).
Volume replaced per operation = 6 litres.
The operation was executed once, followed by two additional repetitions, totaling \(n = 1 + 2 = 3\) operations.
Applying the formula:\[ \text{Amount of milk left} = 60 \times \left(1 - \frac{6}{60}\right)^3 \]Simplify the fractional term within the parenthesis:\[ \frac{6}{60} = \frac{1}{10} = 0.1 \]The expression simplifies to:\[ \text{Amount of milk left} = 60 \times (1 - 0.1)^3 \]\[ = 60 \times (0.9)^3 \]Compute \( (0.9)^3 \):\[ (0.9)^3 = 0.9 \times 0.9 \times 0.9 = 0.81 \times 0.9 = 0.729 \]Multiply the result by the initial quantity:\[ \text{Amount of milk left} = 60 \times 0.729 \]\[ = 43.74 \]Step 4: Conclusion:
Following three operations, 43.74 litres of milk remain in the tub.