20 litres of mixture of acid and water contain 10% water. How much water…

Concept: When a pure substance (here, water) is added to a mixture, both the amount of that substance and the total volume increase, so the new percentage must be found against the new total volume, not the original one. If x is the quantity of pure water added to a mixture of total volume V containing a quantity w of water, the new water percentage is (w + x)/(V + x) times 100.

1. Water in the original 20 L mixture = 10% of 20 = 2 L, so acid = 18 L.

2. Let x L of pure water be added. New total volume = (20 + x) L; new water quantity = (2 + x) L.

3. Since the new water percentage must be 20%: (2 + x)/(20 + x) = 20/100.

4. Cross-multiplying: 100(2 + x) = 20(20 + x), so 200 + 100x = 400 + 20x.

5. Simplifying: 80x = 200, so x = 2.5.

Cross-check: The acid quantity never changes -- it stays at 18 L throughout. If water is 20% of the new mixture, acid must be the remaining 80%: 18/(20 + x) = 0.8, so 20 + x = 22.5, so x = 2.5 L, confirming the same value independently.

Answer: 2.5 L of water must be added.