A container is filled with a mixture containing 40\% water and the rest milk. A certain portion of this mixture is removed and replaced with another mixture that contains 81\% milk and the remaining water. After this replacement, the new mixture has 74\% milk. What fraction of the original mixture was replaced?

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**Solution:**

1. Let the total quantity of the mixture be 1 unit.

2. The original mixture contains 40\% water, so milk percentage is 60\%.

3. Let the fraction of the mixture replaced be $ x $.

4. The quantity of milk in the original mixture before replacement is $ $0.6 \times 1 = 0.6$ $.

5. When $ x $ fraction of the original mixture is removed, the milk removed is $ 0.6x $.

6. The replaced mixture contains 81\% milk, so milk added is $ 0.81x $.

7. After replacement, the total milk quantity is:
   
   $$$ \text{Milk remaining} + \text{Milk added} = 0.6 - 0.6x + 0.81x = 0.6 + 0.21x $$$

8. The new mixture still has total quantity 1 unit, and milk percentage is 74\%, so milk quantity is $ 0.74 $.

9. Equate the milk quantity:

$$ 0.6 + 0.21x = 0.74 $$

10. Solve for $ x $:

$$ $0.21x = 0.14 \implies x = \frac{0.14}{0.21} = \frac{2}{3} $$$

**Answer:** $ $\frac{2}{3} $$

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