A vessel has 200 litre of milk and 40 litre of water. If ________ litres of mixture is taken from the vessel and ________ litres of water is added to the remaining mixture, then the final amount of milk in the vessel becomes 125 litre more than the amount of water in it. Which of the following integral values given in the options are possible in the blanks in same order? (A) (36, 11) (B) (30, 15) (C) (42, 12) (D) (24, 19) (E) (18, 24)

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Accepted Answer

**Solution:**

1. Initial quantities:
   - Milk = 200 litres
   - Water = 40 litres

2. Initial ratio of milk to water is given as 5:1, which matches 200:40.

3. Let *A* litres of mixture be removed and *B* litres of water be added.

4. Amount of milk removed when *A* litres of mixture is taken out:
   - Milk fraction in mixture = $ \frac{5}{6} $
   - Milk removed = $ \frac{5}{6} A $

5. Amount of water removed when *A* litres of mixture is taken out:
   - Water fraction in mixture = $ \frac{1}{6} $
   - Water removed = $ \frac{1}{6} A $

6. Remaining milk after removal:
   $$ 200 - \frac{5}{6} A $$

7. Remaining water after removal and addition:
   $$ 40 - \frac{1}{6} A + B $$

8. According to the problem, the final amount of milk is 125 litres more than the amount of water:
   $$
   200 - \frac{5}{6} A = 40 - \frac{1}{6} A + B + 125
   $$

9. Simplify the equation:
   $$
   200 - \frac{5}{6} A = 165 - \frac{1}{6} A + B
   $$

$$
   200 - 165 = \frac{5}{6} A - \frac{1}{6} A + B
   $$

$$
   35 = \frac{4}{6} A + B = \frac{2}{3} A + B
   $$

10. Multiply both sides by 3:
    $$
    105 = 2A + 3B
    $$

11. Check which options satisfy $ 105 = 2A + 3B $:
    - (36, 11): $ 2 \times 36 + 3 \times 11 = 72 + 33 = 105 $ ✔
    - (30, 15): $ 2 \times 30 + 3 \times 15 = 60 + 45 = 105 $ ✔
    - (42, 12): $ 2 \times 42 + 3 \times 12 = 84 + 36 = 120 $ ✘
    - (24, 19): $ 2 \times 24 + 3 \times 19 = 48 + 57 = 105 $ ✔
    - (18, 24): $ 2 \times 18 + 3 \times 24 = 36 + 72 = 108 $ ✘

**Answer:** only A, B and D

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