Milk and water in a solution of 56 ml is such that after removing 16 ml of the…

Concept: When a fraction of a mixture is removed, it takes away each component (milk and water) in exactly the same proportion in which they exist in the mixture at that moment — so a uniform removal only scales milk and water down by the same factor and never changes their ratio to each other. The ratio changes only when a pure ingredient (here, extra water) is added afterward.

Application:

1. Let the initial quantity of milk be x ml, so the initial quantity of water is (56 − x) ml.

2. Removing 16 ml from the 56 ml mixture removes 16/56 = 2/7 of it, leaving 5/7 of each component: milk left = (5/7)x ml, water left = (5/7)(56 − x) ml.

3. Adding 5 ml of water changes only the water quantity: water becomes (5/7)(56 − x) + 5 ml.

4. The final milk-to-water ratio is given as 5 ∶ 4, so: (5x/7) ÷ [(5/7)(56 − x) + 5] = 5/4.

5. Cross-multiplying: 4 × (5x/7) = 5 × [(5/7)(56 − x) + 5] ⟹ 20x/7 = 25(56 − x)/7 + 25 ⟹ 20x = 25(56 − x) + 175 ⟹ 20x = 1400 − 25x + 175 ⟹ 45x = 1575 ⟹ x = 35.

Cross-check: With x = 35, milk left after removal = (5/7) × 35 = 25 ml and water left after removal = (5/7) × 21 = 15 ml; adding 5 ml of water makes the water 20 ml, giving milk ∶ water = 25 ∶ 20 = 5 ∶ 4, exactly as required. Equivalently, the net final volume is 56 − 16 + 5 = 45 ml, and since the ratio 5 ∶ 4 has 9 total parts, the milk share is (5/9) × 45 = 25 ml, which matches.

So the initial quantity of milk in the mixture was 35 ml.