To solve this problem, we will first determine the rates at which pipes A, B, and C fill or empty the tank.
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Pipe A's filling rate: Pipe A can fill the tank in 3 hours. Thus, its rate is: \[ \text{Rate of A} = \frac{1 \text{ tank}}{3 \text{ hours}} = \frac{1}{3} \text{ tank/hour} \]
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Pipe B's filling rate: Pipe B can fill the tank in 6 hours. Thus, its rate is: \[ \text{Rate of B} = \frac{1 \text{ tank}}{6 \text{ hours}} = \frac{1}{6} \text{ tank/hour} \]
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Pipe C's emptying rate: Pipe C can empty the tank in 8 hours. Thus, its rate is: \[ \text{Rate of C} = -\frac{1 \text{ tank}}{8 \text{ hours}} = -\frac{1}{8} \text{ tank/hour} \] (It's negative because it empties the tank.)
Step 1: Combined rate of A and B
When both pipes A and B are open, we can calculate the combined rate of filling the tank: \[ \text{Combined rate of A and B} = \frac{1}{3} + \frac{1}{6} \] To add these fractions, we need a common denominator, which is 6: \[ \text{Combined rate of A and B} = \frac{2}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2} \text{ tank/hour} \]
Step 2: Filling the tank with A and B alone for 1 hour
If pipes A and B are working together for 1 hour, the amount of the tank filled in that hour is: \[ \text{Amount filled in 1 hour} = \text{Combined rate of A and B} \times 1 \text{ hour} = \frac{1}{2} \text{ tank} \]
So after 1 hour, \(\frac{1}{2}\) tank is filled.
Step 3: Remaining filling after 1 hour
After 1 hour, the remaining capacity of the tank is: \[ \text{Remaining tank to fill} = 1 - \frac{1}{2} = \frac{1}{2} \text{ tank} \]
Step 4: Adding Pipe C after 1 hour
After 1 hour, pipe C is opened along with pipes A and B. Now the combined rate becomes: \[ \text{Combined rate with C} = \frac{1}{3} + \frac{1}{6} - \frac{1}{8} \]
Let's find a common denominator for \(\frac{1}{3}\), \(\frac{1}{6}\), and \(-\frac{1}{8}\). The least common multiple of 3, 6, and 8 is 24. Let's convert each rate:
- \(\frac{1}{3} = \frac{8}{24}\)
- \(\frac{1}{6} = \frac{4}{24}\)
- \(-\frac{1}{8} = -\frac{3}{24}\)
Now add these rates: \[ \text{Combined rate with C} = \frac{8}{24} + \frac{4}{24} - \frac{3}{24} = \frac{9}{24} = \frac{3}{8} \text{ tank/hour} \]
Step 5: Time to fill remaining \(\frac{1}{2}\) tank
Now, we need to fill the remaining \(\frac{1}{2}\) tank with the rate of \(\frac{3}{8}\) tank/hour: \[ \text{Time} = \frac{\text{Remaining tank}}{\text{Combined rate}} = \frac{\frac{1}{2}}{\frac{3}{8}} = \frac{1}{2} \times \frac{8}{3} = \frac{8}{6} = \frac{4}{3} \text{ hours} \approx 1.33 \text{ hours} \]
Step 6: Total time to fill the tank
The total time taken to fill the tank is the initial hour plus \(\frac{4}{3}\) hours: \[ \text{Total time} = 1 + \frac{4}{3} = \frac{3}{3} + \frac{4}{3} = \frac{7}{3} \text{ hours} \approx 2.33 \text{ hours} \] This is approximately 2 hours and 20 minutes.
Therefore, the total time taken to fill the tank is: \[ \boxed{\frac{7}{3}} \text{ hours} \text{ or approximately } 2 \text{ hours and } 20 \text{ minutes.} \]