When taps close, which time are we finding?
6 identical taps fill an empty tank in 12 hours. After 3 hours, 3 taps close. The other 3 continue. How long does filling take?
Assume constant equal flow with no water lost. This is a simplified mathematical model.
Before the change
6 taps · 1/4 of the tank filled
Elapsed time: 3 hours
After the change
3 taps · 3/4 of the tank left to fill
Remaining time: 18 hours
Total time: 21 hours
Solve in two stages
- With 6 taps, the combined rate is 1/12 of a tank per hour. In the first 3 hours: 3 × 1/12 = 1/4.
- The remaining part is 1 − 1/4 = 3/4 of the tank.
- With only 3 taps, the rate halves: 1/24 of a tank per hour.
- The remaining time is (3/4) ÷ (1/24) = 18 h.
- Total elapsed time also includes the first 3 hours: 3 + 18 = 21 h.
Fewer taps, more time
For the same amount of water, the number of taps and the time are inversely proportional: halving the number of taps doubles the time. The 3 taps would fill a whole tank in 24 hours, but here they need to fill only the remaining 3/4.
Why not 24 hours?
24 hours would be needed if only 3 taps filled the whole tank from the start. But a quarter is already full. Nor is 18 hours the total: it is only the remaining time after the change. Check: 3/12 + 18/24 = 1.
Try a variation
What if the 3 taps close after 6 hours? The tank is half full; (1/2) ÷ (1/24) = 12 more hours are needed. Total: 6 + 12 = 18 hours. More time with all taps means less total time.