A tap fills a cistern in 16 hours. Another tap empties the full tank in 80…

Concept — A pipe works at a constant rate equal to the reciprocal of the time it needs on its own: a filling pipe contributes +1/T of the tank per hour and an emptying pipe contributes −1/T per hour. When pipes run together, the tank changes at the algebraic sum of their rates, and the time needed for any fraction of the job is that fraction divided by the net rate.

Applying it to this cistern

1. Filling tap: it alone fills the whole cistern in 16 hours, so its rate is 1/16 of the tank per hour.

2. Emptying tap: it alone empties the full tank in 80 hours, so its rate is −1/80 of the tank per hour.

3. Net rate with both taps open: 1/16 − 1/80 = 5/80 − 1/80 = 4/80 = 1/20 of the tank per hour (positive, so the tank does fill).

4. Work required: only one-fourth of the tank has to be filled, i.e. 1/4 of the job.

5. Time = work ÷ net rate = (1/4) ÷ (1/20) = (1/4) × 20 = 5 hours.

Cross-check — In 5 hours the filling tap puts in 5/16 of the tank while the emptying tap takes out 5/80 = 1/16, leaving 5/16 − 1/16 = 4/16 = 1/4 of the tank, exactly the level asked for. Equivalently, a net rate of 1/20 per hour fills the whole tank in 20 hours, so one-fourth of it takes 20 ÷ 4 = 5 hours.

Result — the quarter-fill takes 5 hours.