The tank is filled by three pipes with different uniform flow rates. While the first two pipes are operating simultaneously, they fill the tank in the same duration that the third pipe takes to fill it alone. The second pipe can fill the tank 5 hours quicker than the first pipe, yet 4 hours slower than the third pipe. What is the time (in hours) needed for the first pipe to fill the tank?
15
Let first-pipe time be \(x\) hours. Then second-pipe time = \(x-5\) and third-pipe time = \((x-5)-4 = x-9\).
Combined rate of pipes 1 and 2 equals the rate of pipe 3:
\(\dfrac{1}{x} + \dfrac{1}{x-5} = \dfrac{1}{x-9}\)
Check \(x = 15\): LHS \(= \dfrac{1}{15} + \dfrac{1}{10} = \dfrac{2 + 3}{30} = \dfrac{1}{6}\); RHS \(= \dfrac{1}{6}\). ✓
Hence the first pipe takes 15 hours — option (4).
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