Two pipes A and B can fill a tank alone in 10 hours and 30 hours respectively.…

Concept: In a pipe-and-cistern problem, take the tank capacity as the LCM of the individual fill times so each pipe's rate becomes a whole number of units per hour. When pipes work together their efficiencies add; a leak works against filling, so its efficiency is subtracted from the combined filling efficiency to get the net rate actually observed.

Application:

1. Let the tank capacity be LCM(10, 30) = 30 units.

2. Efficiency of pipe A = 30 / 10 = 3 units/hour; efficiency of pipe B = 30 / 30 = 1 unit/hour.

3. Combined efficiency of A and B (with no leak) = 3 + 1 = 4 units/hour.

4. Time to fill the tank with no leak = 30 / 4 = 7.5 hours.

5. With the leak present, the tank actually takes 2.5 hours longer, so the actual fill time = 7.5 + 2.5 = 10 hours.

6. The extra water A and B together would supply in that additional 2.5 hours = 2.5 x 4 = 10 units; this is exactly the water the leak drains over the full 10-hour actual fill time.

7. So the leak's efficiency E satisfies 10 x E = 10, giving E = 1 unit/hour.

8. Time for the leak alone to empty the full 30-unit tank = 30 / 1 = 30 hours.

Cross-check: using the net-rate equation directly, net filling rate with the leak = combined efficiency - leak efficiency = 4 - 1 = 3 units/hour, so fill time = 30 / 3 = 10 hours - matching the actual 10-hour fill time given in the question.

Therefore, the leak alone would take 30 hours to empty the full tank.