A container contains 40 litres of milk. From this container 4 litres of milk was taken out and replaced by water. This process was repeated further two times. How much milk is now contained by the container?

Answer

To solve the problem, we will use the formula for the remaining quantity of milk after repeated operations of removing a certain amount and replacing it with water.

### Step-by-Step Solution:

1. **Identify Initial Values**:
- Initial amount of milk in the container = 40 liters
- Amount of milk taken out each time = 4 liters
- Number of times the operation is performed = 3 (once initially and two more times)

2. **Apply the Formula**:
The formula to calculate the remaining quantity of milk after n operations is:
\[ \text{Final value} = \text{Initial value} \times \left(1 - \frac{a}{\text{Initial value}}\right)^n \]
where \( a \) is the amount taken out each time.

3. **Substitute the Values**:
Here, we substitute the values into the formula:
\[ \text{Final value} = 40 \times \left(1 - \frac{4}{40}\right)^3 \]

4. **Simplify the Fraction**:
Calculate \( \frac{4}{40} \):
\[ \frac{4}{40} = \frac{1}{10} \]
Therefore, the expression becomes:
\[ \text{Final value} = 40 \times \left(1 - \frac{1}{10}\right)^3 \]

5. **Calculate \( 1 - \frac{1}{10} \)**:
\[ 1 - \frac{1}{10} = \frac{9}{10} \]
Now the expression is:
\[ \text{Final value} = 40 \times \left(\frac{9}{10}\right)^3 \]

6. **Calculate \( \left(\frac{9}{10}\right)^3 \)**:
\[ \left(\frac{9}{10}\right)^3 = \frac{729}{1000} \]
Now substitute this back into the expression:
\[ \text{Final value} = 40 \times \frac{729}{1000} \]

7. **Multiply and Simplify**:
\[ \text{Final value} = \frac{40 \times 729}{1000} = \frac{29160}{1000} = 29.16 \text{ liters} \]

### Final Answer:
The amount of milk now contained in the container is **29.16 liters**.

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