Three taps A, B and C can fill a tank in 12, 15 and 20 hours respectively. If…

Concept: When several pipes/taps work in a repeating (cyclic) pattern, first find how much work one full cycle completes by adding the rates of whichever taps are open in each hour of that cycle. Then find how many complete cycles fit before the tank is full, and account for the leftover fraction using the next hour's rate — the tank finishes exactly at the end of that hour only when the leftover exactly equals a whole hour's work; otherwise it finishes partway through that hour, needing only a fraction of it.

Application: A, B and C fill the tank alone in 12, 15 and 20 hours, so their hourly rates are 1/12, 1/15 and 1/20 of the tank. A stays open every hour; B and C alternate hourly starting with B, giving the repeating pattern AB, AC, AB, AC, …

1. Work done by A and B together in 1 hour = 1/12 + 1/15 = 5/60 + 4/60 = 9/60.

2. Work done by A and C together in 1 hour = 1/12 + 1/20 = 5/60 + 3/60 = 8/60.

3. Work done in one AB+AC cycle (2 hours) = 9/60 + 8/60 = 17/60.

4. After 3 complete cycles (6 hours), work done = 3 × 17/60 = 51/60 of the tank.

5. Work still remaining = 1 − 51/60 = 9/60.

6. The 7th hour follows the pattern with A and B open again, and their hourly rate is exactly 9/60 — the remaining work — so the tank finishes exactly at the end of this hour, with no fractional hour needed.

7. Total time taken = 6 + 1 = 7 hours.

Cross-check: Adding the work across all 7 hours directly — 9/60, 8/60, 9/60, 8/60, 9/60, 8/60, 9/60 — gives 60/60, i.e. the full tank, confirming the tank is exactly filled after 7 hours with none left over.