A vessel contains a mixture of milk and water, where milk constitutes 85\% of the total volume. If 20 liters of this mixture are removed and then 30 liters of water are added to the vessel, the milk content reduces to 65\%. What was the initial volume of the mixture in the vessel (in liters)?

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**Solution:**

1. Let the initial quantity of the mixture be *x* liters.

2. Initially, milk is 85\% of *x*, so milk quantity = $0.85x$ liters.

3. When 20 liters of mixture is removed, the ratio of milk to water remains the same (85\% milk, 15\% water).
   - Milk removed = $20 	imes 0.85 = 17$ liters
   - Water removed = $20 	imes 0.15 = 3$ liters

4. After removal, remaining milk = $0.85x - 17$ liters
   Remaining water = $0.15x - 3$ liters

5. Then, 30 liters of water is added, so new water quantity = $(0.15x - 3) + 30 = 0.15x + 27$ liters

6. Total volume after these operations = $(x - 20 + 30) = x + 10$ liters

7. According to the problem, milk now forms 65\% of the total volume:
   $$\frac{0.85x - 17}{x + 10} = 0.65$$

8. Multiply both sides by $(x + 10)$:
   $$0.85x - 17 = 0.65x + 6.5$$

9. Rearranging terms:
   $$0.85x - 0.65x = 6.5 + 17$$
   $$0.20x = 23.5$$

10. Solving for *x*:
    $$x = \frac{23.5}{0.20} = 117.5$$ liters

**Answer:** $117.5$ liters

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