Pipe A can fill a tank in 12 hours, pipe B in 15

Solution

Rates in tank/day (take “unit tank”): A’s rate = 1/12 B’s rate = 1/15 C’s rate = −1/10 (emptying) Net rate (A + B + C) = 1/12 + 1/15 − 1/10 LCM of 12, 15, 10 is 60: = (5 + 4 − 6) / 60 = 3/60 = 1/20 tank/hour Work done in first 3 hours = 3 × (1/20) = 3/20 of tank. Remaining = 1 − 3/20 = 17/20 of tank. After C is closed, only A and B work: (A + B) rate = 1/12 + 1/15 = (5 + 4) / 60 = 9/60 = 3/20 tank/hour Time to finish remaining work = (17/20) ÷ (3/20) = 17/3 hours.