Two taps can fill a cistern in 3 hours and 32 hours, respectively. A third tap…

Concept: in a pipes-and-cisterns problem every pipe works at a constant rate equal to 1 divided by the time it needs on its own, an inlet counting as a positive rate and an outlet as a negative rate. When pipes run together their rates add algebraically, and the time needed for any required part of the tank equals that part divided by the combined rate.

Applying this to the given cistern:

1. The tap that fills the cistern in 3 hours works at 1/3 of the cistern per hour.

2. The tap that fills the cistern in 32 hours works at 1/32 of the cistern per hour.

3. The tap that empties the cistern in 3 hours works at −1/3 of the cistern per hour.

4. Combined rate = 1/3 + 1/32 − 1/3 = 1/32 of the cistern per hour, because the 3-hour inlet and the 3-hour outlet cancel each other exactly.

5. The part that has to be filled is three-fourth of the cistern, that is 3/4.

6. Time = (3/4) ÷ (1/32) = (3/4) × 32 = 24 hours.

Cross-check with the LCM (unit) method:

• Take the capacity of the cistern as 96 units, the LCM of 3 and 32.

• The two inlets then supply 32 units and 3 units per hour while the outlet removes 32 units per hour, so the net inflow is 3 units per hour.

• Three-fourth of 96 units is 72 units, and 72 ÷ 3 = 24 hours, the same value as before.

• Note that 96 and 72 are counts of volume units in this model and not durations; they still have to be divided by the net inflow of 3 units per hour.

So, with all three taps opened together, three-fourth of the empty cistern is filled in 24 hours.