A container is full of 75 litre milk. If 15 litre content of container is replaced by water and the same process is further repeated two times, then find the quantity of milk left in the final solution?

Options

A. A)36.4 litre
B. B)38.4 litre
C. C)40 litre
D. D)41.4 litre

Answer

To solve the problem of finding the quantity of milk left after replacing part of it with water three times, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Initial Condition**:
- The container is initially full of 75 liters of milk.

2. **Determine the Amount Replaced**:
- Each time, 15 liters of the mixture (milk + water) is removed and replaced with 15 liters of water.

3. **Use the Formula**:
- We will use the formula for the final quantity of milk after multiple replacements:
\[ \text{Final Milk} = \text{Initial Milk} \times \left(1 - \frac{a}{V}\right)^n \]
where:
- \( a \) = amount of mixture replaced (15 liters),
- \( V \) = initial volume of the mixture (75 liters),
- \( n \) = number of times the operation is performed (3 times).

4. **Plug in the Values**:
- Substitute the values into the formula:
\[ \text{Final Milk} = 75 \times \left(1 - \frac{15}{75}\right)^3 \]

5. **Calculate the Fraction**:
- Simplify the fraction:
\[ \frac{15}{75} = \frac{1}{5} \]
- Therefore:
\[ 1 - \frac{1}{5} = \frac{4}{5} \]

6. **Calculate the Power**:
- Now, substitute back into the formula:
\[ \text{Final Milk} = 75 \times \left(\frac{4}{5}\right)^3 \]

7. **Calculate \(\left(\frac{4}{5}\right)^3\)**:
- Calculate:
\[ \left(\frac{4}{5}\right)^3 = \frac{64}{125} \]

8. **Final Calculation**:
- Now multiply:
\[ \text{Final Milk} = 75 \times \frac{64}{125} \]
- Simplifying:
\[ = \frac{75 \times 64}{125} = \frac{4800}{125} = 38.4 \]

9. **Conclusion**:
- The quantity of milk left after three replacements is **38.4 liters**.