An electric pump can fill a tank in 4 hours. Because of a leak in the tank it…

Concept: Pipe-and-cistern problems apply the work-rate principle: rate = 1 ÷ time taken, and rates combine by addition when helping and by subtraction when opposing. A leak that drains a tank while a pump fills it behaves as a rate that subtracts from the pump's own fill-rate, so the leak's own emptying rate equals the pump's rate minus the slower, leak-affected combined rate.

1. The pump's own rate is 1/4 tank per hour, since it fills the tank alone in 4 hours.

2. With the leak working against it, the tank fills in 4½ hours (= 9/2 hours), so the combined rate is 1 ÷ (9/2) = 2/9 tank per hour.

3. The leak's own rate is the pump's rate minus the combined rate: 1/4 − 2/9. Using a common denominator of 36, this is 9/36 − 8/36 = 1/36 tank per hour.

4. Time for the leak alone to empty a full tank = 1 ÷ (1/36) = 36 hours.

Cross-check: If the leak alone drains 1/36 of the tank each hour while the pump alone fills 9/36 (= 1/4) each hour, the net rate when both act together is 9/36 − 1/36 = 8/36 = 2/9 tank per hour — exactly 1 tank in 4½ hours, matching the time given in the question.

Answer: The leak alone takes 36 hours to empty a full tank.