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 the time taken, in minutes, by Pipe C to fill the empty tank is

Question

Accepted Answer

Let $$A$$, $$B$$, and $$C$$ be the number of hours taken by pipes A, B, and C to completely fill or empty a tank. The fraction of the tank filled (or emptied) by them in one hour is $$\frac{1}{A}$$, $$\frac{1}{B}$$, and $$\frac{1}{C}$$, respectively.

1. **Given Relationships:**  
   - Pipe B empties the full tank in one hour less than the time taken by Pipe A to fill it: $$B = A - 1$$.  
   - When pipes A, B, and C are turned on together, they fill the tank in 2 hours:  
     $$
     \frac{1}{A} + \frac{1}{B} + \frac{1}{C} = \frac{1}{2}.
     $$

2. **When pipes B and C are turned on together for 1 hour:**  
   - After Pipe B is turned off, Pipe C takes another 1 hour and 15 minutes (or $$1\frac{1}{4}$$ hours) to fill the remaining tank.

3. **Solving for $$C$$:**  
   Using algebraic manipulation (as shown in the attached solution):  
   - The quadratic equation derived is:  
     $$
     A^2 - 6A + 9 = 0.
     $$  
   - Solving this gives $$A = 3$$.

4. **Finding $$C$$:**  
   Substituting $$A = 3$$ into the equations:  
   $$
   C = \frac{9}{4} 	imes \left(1 - \frac{1}{A}ight) = \frac{9}{4} 	imes \left(1 - \frac{1}{3}ight) = \frac{9}{4} 	imes \frac{2}{3} = \frac{3}{2}.
   $$  
   This means Pipe C takes $$1.5$$ hours (or $$90$$ minutes) to fill the empty tank.

Thus, the time taken by Pipe C to fill the empty tank is **90 minutes**.

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