A swimming pool is fitted with three pipes. The first two pipes working simultaneously, fill the pool in the same time as the third pipe alone. The second pipe alone fills the pool 5 hours faster than the first pipe and 4 hours slower than the third pipe. In what time will the second and third pipes together fill the pool?

Let the first pipe alone can fill the tank in $$x$$ hours

=> Second pipe can fill the tank in $$(x-5)$$ hours

and the third pipe can fill the tank in $$(x-5)-4 = (x-9)$$ hours

Since, part filled by first pipe and second pipe together in 1 hr = part filled by third pipe in 1 hr

=> $$\frac{1}{x} + \frac{1}{x-5} = \frac{1}{x-9}$$

=> $$(x-5)(x-9) + x(x-9) = x(x-5)$$

=> $$x^2 - 18x + 45 = 0$$

=> $$x = 15 , 3$$

When $$x = 3$$, the expression, $$(x-5)$$ will be negative, thus it's not possible.

=> $$x = 15$$

Time taken by second and third pipes to fill the tank = $$\frac{1}{\frac{1}{10} + \frac{1}{6}}$$

= $$\frac{60}{16}$$ = 3.75 hours