Pipe A can fill a tank in 20 minutes, Pipe B can fill it

Solution

ATQ, A's 1-minute work = 1/20 B's 1-minute work = 1/30 C's 1-minute work = -1/40 So, work done in 1 minute by all three together: 1/20 + 1/30 - 1/40 = 6/120 + 4/120 - 3/120 = 7/120 Let Pipe C remain open for x minutes. Then for x minutes, work done = 7x/120 For remaining (15 - x) minutes, only A and B work. A + B work in 1 minute: 1/20 + 1/30 = 5/60 = 1/12 So, work done in remaining time = (15 - x)/12 Total work = 1 Therefore, 7x/120 + (15 - x)/12 = 1 Multiply by 120: 7x + 10(15 - x) = 120 7x + 150 - 10x = 120 -3x = -30 x = 10 Hence, Pipe C was open for 10 minutes.