Question

Pipes A and C are fill pipes while Pipe B is a drain pipe of a tank. Pipe B empties the full tank in one hour less than the time taken by Pipe A to fill the empty tank. When pipes A, B and C are turned on together, the empty tank is filled in two hours. If pipes B and C are turned on together when the tank is empty and Pipe B is turned off after one hour, then Pipe C takes another one hour and 15 minutes to fill the remaining tank. If Pipe A can fill the empty tank in less than five hours, then

Accepted Answer

A. Let Pipe A fill the tank in $a$ hours ($a < 5$). Then Pipe B empties the tank in $a - 1$ hours. Let Pipe C fill the tank in $c$ hours. Given that A, B, C together fill the tank in 2 hours: $$\frac{1}{a} - \frac{1}{a-1} + \frac{1}{c} = \frac{1}{2} \implies \frac{1}{c} = \frac{1}{2} + \frac{1}{a(a-1)}$$ When B and C run together for 1 hour, and C runs alone for another $1.25$ hours ($2.25 = \frac{9}{4}$ hours total for C): $$-\frac{1}{a-1}(1) + \frac{1}{c}\left(\frac{9}{4}\right) = 1$$ Substitute $\frac{1}{c}$: $$\frac{9}{4}\left(\frac{1}{2} + \frac{1}{a(a-1)}\right) - \frac{1}{a-1} = 1$$ $$\frac{9}{8} + \frac{9}{4a(a-1)} - \frac{1}{a-1} = 1$$ $$\frac{9 - 4a}{4a(a-1)} = 1 - \frac{9}{8} = -\frac{1}{8}$$ $$18 - 8a = -a^2 + a \implies a^2 - 9a + 18 = 0$$ $$(a - 3)(a - 6) = 0$$ Since $a < 5$, $a = 3$ hours. Now, $\frac{1}{c} = \frac{1}{2} + \frac{1}{3(2)} = \frac{2}{3} \implies c = 1.5$ hours

Pipes and cisterns filling time

Explanation

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