A sump is filled by three tankers with uniform flow. The first two tankers operating
simultaneously fill the sump in the same time during which the sump is filled by the third tanker alone. The second tanker fills the sump 5 hours faster than the first tanker and 4 hours slower than the third tanker. The time required bythefirst tanker is:

Suppose, first pipe alone takes x hours to fill the tank .

Then, second and third pipes will take (x -5) and (x - 9) hours respectively to fill the tank.

As per question, we get


$$\frac{1}{x} + \frac{1}{x-5} = \frac{1}{x-9}$$
=> $$\frac{x-5+x}{x(x-5)} = \frac{1}{x-9}$$
=> $$(2x - 5)(x - 9) = x(x - 5)$$
=> $$x^2 - 18x + 45 = 0$$



After solving this euation, we get
(x-15)(x+3) = 0,
As value can not be negative, so x = 15