Shortcut Trick
Inlet (A): 4.5 hours | Outlet (B): 7.2 hours. Let Total Work = 36 units.
Efficiency: A = +8, B = −5. In 1 cycle (2 hours), Net fill = 8 − 5 = 3 units.
Filling 36 − 8 (last stroke) = 28 units. 9 cycles fill 9 × 3 = 27 units in 18 hours.
Remaining 9 units: 19th hour (A) fills 8 units. Balance 1 unit filled by B in next hour? No, pipes are alternate. Last unit filled by A in 1/4 hour.
Total time = 18 + 2 (full cycle for 3 units) + part of last cycle → 20 hours and 45 mins.
∴ The correct answer is 20(3)/(4) hours.
Alternate Method
| Pipe | Time (hrs) | Total Work (LCM) | Efficiency |
|---|
| Inlet (A) | 9/2 | 36 | 8 |
| Outlet (B) | 36/5 | −5 | |
⇒ In the first 2 hours (1 cycle), work done = 8 − 5 = 3 units.
⇒ To avoid overfilling, subtract peak efficiency from total: 36 − 8 = 28 units.
⇒ Number of 2-hour cycles to reach near 28 units = 9 cycles (9 × 3 = 27 units).
⇒ Time taken for 9 cycles = 9 × 2 = 18 hours.
⇒ 19th hour (Inlet A opens): 27 + 8 = 35 units filled.
⇒ 20th hour (Outlet B opens): 35 − 5 = 30 units filled.
⇒ Remaining work = 36 − 30 = 6 units. Inlet A opens again.
⇒ Time taken by A to fill 6 units = 6/8 = 3/4 hour.
⇒ Total time = 18 + 2 + 3/4 = 20(3)/(4) hours.
∴ The correct answer is 20(3)/(4) hours.
Additional Information
Alternate Pipe Logic
When pipes work alternately, always calculate the net work per cycle. If an outlet pipe is involved, ensure the tank doesn't fill during the positive stroke before the cycle completes.
Negative Efficiency
Outlet pipes are treated as negative work. Efficiency = Total Capacity ÷ Time to empty.
Peak Filling Rule
The tank is considered full the moment the water level reaches the capacity, even if the next scheduled pipe is an outlet pipe.