A vessel contains some litre of pure milk. 5 litre of milk taken out from it and replaced by water. This process is repeated once more. Then the ratio of milk and water in the result ant mixture becomes 64 : 17. Find the initial quantity of milk.

Answer

To solve the problem, we need to determine the initial quantity of pure milk in the vessel. Let's denote the initial quantity of milk as \( x \) liters.

### Step-by-Step Solution:

1. **Initial Condition**:
The vessel initially contains \( x \) liters of pure milk.

2. **First Operation**:
When 5 liters of milk is taken out, the amount of milk left in the vessel is:
\[ x - 5 \text{ liters} \]
After removing 5 liters of milk, we replace it with 5 liters of water. Therefore, the total volume remains \( x \) liters, but now the composition is:
- Milk: \( x - 5 \) liters
- Water: \( 5 \) liters

3. **Second Operation**:
We repeat the process of taking out 5 liters of the mixture. The concentration of milk in the mixture after the first operation is:
\[ \text{Concentration of milk} = \frac{x - 5}{x} \]
The amount of milk removed in the second operation is:
\[ 5 \times \frac{x - 5}{x} = \frac{5(x - 5)}{x} \]
Therefore, the amount of milk left after the second operation is:
\[ \left(x - 5\right) - \frac{5(x - 5)}{x} = \left(x - 5\right) \left(1 - \frac{5}{x}\right) = \left(x - 5\right) \left(\frac{x - 5}{x}\right) \]

4. **Final Amount of Milk**:
The amount of milk left after the second operation can be simplified to:
\[ \frac{(x - 5)^2}{x} \]

5. **Final Ratio of Milk to Water**:
After the second operation, the total amount of water in the vessel is:
\[ 5 + 5 = 10 \text{ liters} \]
The ratio of milk to water is given as \( 64:17 \). Therefore, we can set up the equation:
\[ \frac{\frac{(x - 5)^2}{x}}{10} = \frac{64}{17} \]

6. **Cross-Multiplying**:
Cross-multiplying gives us:
\[ 17 \cdot (x - 5)^2 = 640 \]

7. **Expanding and Rearranging**:
Expanding the left side:
\[ 17(x^2 - 10x + 25) = 640 \]
\[ 17x^2 - 170x + 425 = 640 \]
Rearranging gives:
\[ 17x^2 - 170x - 215 = 0 \]

8. **Dividing by 17**:
Dividing the entire equation by 17 simplifies it:
\[ x^2 - 10x - 12.647 = 0 \]

9. **Using the Quadratic Formula**:
We can use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ x = \frac{10 \pm \sqrt{(-10)^2 - 4 \cdot 1 \cdot (-12.647)}}{2 \cdot 1} \]
\[ x = \frac{10 \pm \sqrt{100 + 50.588}}{2} \]
\[ x = \frac{10 \pm \sqrt{150.588}}{2} \]

10. **Calculating the Value**:
Calculate the square root and solve for \( x \):
\[ x \approx \frac{10 \pm 12.29}{2} \]
Taking the positive root:
\[ x \approx \frac{22.29}{2} \approx 11.145 \]

### Final Answer:
The initial quantity of milk in the vessel is approximately \( 11.145 \) liters.