A tank is filled by 3 pipes, second pipe take 10 hours

Solution

Let the second pipe can fill tank in x hours So, first pipe fill tank in (x-10) hours and  Third pipe fill tank in (x + 10) hours According to the question, Second pipe and third pipe together take 2 hour more time to fill the tank from first pipe. So time taken by second pipe third pipe together = (x – 10 + 2) hours = (x – 8) hours  1/x + 1/(x +10) = 1/(x -8) (x +10 + x )/(x²+ 10 x) = 1/(x - 8) (x – 8) (2x + 10) = x²+ 10x 2x²+ 10x - 16x – 80 = x²+ 10x x² - 16x – 80 = 0 x² - 20x + 4x – 80 = 0 x(x – 20) + 4(x – 20) = 0 (x – 20) (x + 4) = 0 x = - 4, 20 So second pipe can fill the tank in 20 hours alone.