Three taps A, B, and C can fill a tank individually in 12, 15, and 20 hours respectively. If tap A runs continuously while taps B and C are opened alternately for one hour each, how long will it take to fill the tank completely?

Explanation

The combined work done by taps A and B in one hour is (1/12 + 1/15) = 3/20 of the tank. Similarly, taps A and C together fill (1/12 + 1/20) = 2/15 of the tank in one hour. Since B and C alternate every hour while A runs continuously, every two hours they fill (3/20 + 2/15) = 17/60 of the tank. After 6 hours (three such cycles), the tank is filled by 3 × 17/60 = 17/20. The remaining 3/20 of the tank is filled by taps A and B in the next hour. Therefore, the total time to fill the tank is 6 + 1 = 7 hours.