A mixture of acid and water contains 20 percent acid. When 10 litres of water is added to the mixture, then the percentage of acid becomes 15 percent. What is the original quantity of mixture?

Let original mixture quantity be x liters. Acid content=0.2x. After adding 10L water, total mixture=(x+10)L with 15% acid, so acid=0.15(x+10). Since acid amount is unchanged: 0.2x=0.15(x+10). Solving: 0.2x=0.15x+1.5, 0.05x=1.5, x=30L. Verification: Original mixture=30L with 20% acid=6L acid. After adding 10L water, total=40L with 6L acid=15%. Concentration problems require tracking the solute amount.