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A swimming pool is filled by three pipes with uniform flow. The first two pipes operating simultaneously fill the pool in the same time during which the pool is filled by the third pipe alone. The second pipe fills the pool 5 hours faster than the first pipe and 4 hours slower than the third pipe. The time required by the first pipe is?

Correct Answer: 15 hours
Suppose first pipe alone takes x hours to fill the tank.
Then second and third pipes will takes (x - 5) and (x - 9) hours respectively to fill the tank.
$$\eqalign{ & \therefore \frac{1}{x} + \frac{1}{{\left( {x - 5} \right)}} = \frac{1}{{\left( {x - 9} \right)}} \cr & \Rightarrow \frac{{x - 5 + x}}{{x\left( {x - 5} \right)}} = \frac{1}{{\left( {x - 9} \right)}} \cr & \Rightarrow \left( {2x - 5} \right)\left( {x - 9} \right) = x\left( {x - 5} \right) \cr & \Rightarrow {x^2} - 18x + 45 = 0 \cr & \Rightarrow \left( {x - 15} \right)\left( {x - 3} \right) = 0 \cr & \Rightarrow x = 15\left \cr} $$
So, first pipe alone takes 15 hrs to fill the tank.

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