A mixture contains milk and water such that the quantity of milk is 20 liters less than that of water. When 40\% of this mixture is replaced by orange juice, the ratio of milk, water, and orange juice in the new mixture becomes 4 : 5 : 6 respectively. Then, a certain percentage of the resulting mixture is removed, after which the total quantity of milk and orange juice left in the final mixture is 66 liters. What percentage of the resulting mixture is taken out?

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Accepted Answer

**Solution:**

1. Let the initial quantities of milk and water be *M* and *W* liters respectively.
   Given, *M = W - 20*.

2. The total initial mixture quantity = *M + W*.

3. 40\% of the mixture is taken out and replaced by orange juice.
   - Quantity removed = 0.4 × (M + W)
   - Quantity remaining = 0.6 × (M + W)

4. After replacement, the mixture contains milk, water, and orange juice in the ratio 4 : 5 : 6.
   - Total parts = 4 + 5 + 6 = 15 parts

5. Since milk and water quantities remain only from the 60\% remaining portion (no milk or water in orange juice),
   - Milk in resulting mixture = 0.6 × M
   - Water in resulting mixture = 0.6 × W
   - Orange juice = 0.4 × (M + W)

6. According to the ratio,
   $$\frac{	ext{Milk}}{	ext{Water}} = \frac{4}{5} \implies \frac{0.6M}{0.6W} = \frac{4}{5} \implies \frac{M}{W} = \frac{4}{5}$$

7. From step 1, *M = W - 20* and from step 6, *M/W = 4/5*.
   So,
   $$\frac{W - 20}{W} = \frac{4}{5}$$
   $$5(W - 20) = 4W$$
   $$5W - 100 = 4W$$
   $$W = 100 	ext{ liters}$$

8. Then,
   $$M = W - 20 = 100 - 20 = 80 	ext{ liters}$$

9. Total initial mixture = 80 + 100 = 180 liters.

10. After replacement:
    - Milk = 0.6 × 80 = 48 liters
    - Water = 0.6 × 100 = 60 liters
    - Orange juice = 0.4 × 180 = 72 liters

11. Now, a certain percentage *x* of the resulting mixture is taken out, so the remaining mixture is (100 - *x*)\%.

12. The total quantity of milk and orange juice after removal is 66 liters:
    $$	ext{Milk} + 	ext{Orange juice} = 66$$
    $$ (100 - x)\% 	imes (48 + 72) = 66$$
    $$ \frac{100 - x}{100} 	imes 120 = 66$$

13. Solving for *x*:
    $$100 - x = \frac{66 	imes 100}{120} = 55$$
    $$x = 100 - 55 = 45$$

14. But 45\% is not matching the options correctly because the question asks for the percentage taken out, so re-check step 13:

Actually, the calculation in step 13 is correct, but the options include 45\% as option (d).

15. However, the question states the correct answer is 25\% (option b). Let's verify calculations carefully.

**Rechecking step 12 and 13:**

- Milk + Orange juice after removal = 66 liters
- Total milk + orange juice before removal = 48 + 72 = 120 liters

Let the percentage taken out be *x*\%, so remaining mixture is (100 - *x*)\%.

So,
$$\frac{100 - x}{100} 	imes 120 = 66$$
$$100 - x = \frac{66 	imes 100}{120} = 55$$
$$x = 45$$

This means 45\% of the mixture is taken out, which corresponds to option (d).

**Answer:** 45\%

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