A container has 80 L of milk. From this container 8 L of milk was taken out and replaced by water. The process was further repeated twice. The volue of milk in the container after that is :

Options

A. 85.32L
B. 58.23 L
C. 85.23 L
D. 58.32 L

Answer

To solve the problem step by step, we will use the formula for the remaining quantity of liquid after repeated removals and replacements.

### Step-by-Step Solution:

1. **Initial Setup**:
- The initial volume of milk in the container is \( A = 80 \) liters.
- The volume of milk taken out in each operation is \( b = 8 \) liters.
- The number of operations performed is \( n = 3 \).

2. **Understanding the Formula**:
- The formula to calculate the remaining quantity of milk after \( n \) operations is:
\[ \text{Remaining Milk} = A \left(1 - \frac{b}{A}\right)^n \]
- Here, \( \frac{b}{A} \) represents the fraction of milk removed in each operation.

3. **Calculate the Fraction Removed**:
- Calculate \( \frac{b}{A} \):
\[ \frac{b}{A} = \frac{8}{80} = \frac{1}{10} \]

4. **Substituting Values into the Formula**:
- Substitute \( A \), \( b \), and \( n \) into the formula:
\[ \text{Remaining Milk} = 80 \left(1 - \frac{1}{10}\right)^3 \]

5. **Simplifying the Expression**:
- Simplify \( 1 - \frac{1}{10} \):
\[ 1 - \frac{1}{10} = \frac{9}{10} \]
- Now substitute this back into the equation:
\[ \text{Remaining Milk} = 80 \left(\frac{9}{10}\right)^3 \]

6. **Calculating \( \left(\frac{9}{10}\right)^3 \)**:
- Calculate \( \left(\frac{9}{10}\right)^3 \):
\[ \left(\frac{9}{10}\right)^3 = \frac{729}{1000} \]

7. **Final Calculation**:
- Now multiply by 80:
\[ \text{Remaining Milk} = 80 \times \frac{729}{1000} = \frac{58320}{1000} = 58.32 \text{ liters} \]

### Conclusion:
The volume of milk remaining in the container after three operations is **58.32 liters**.