Two pipes A and B can fill a tank in 24 hours and 17 hours respectively. Pipe C is an outlet pipe connected to the tank and is open from the beginning. Harihar opens pipes A and B together to fill the empty tank and closes both A and B at the time when the tank would have been full if there were no leak. However, because pipe C was open, the water already in the tank is completely emptied in 2.5 hours after closing A and B. If instead of closing A and B at that time Harihar closes pipe C, in approximately how many hours will the remaining part of the tank be filled?

Introduction / Context:
This is a challenging pipes and cisterns problem involving two inlet pipes and one outlet (leak) pipe. The tank is being filled by pipes A and B, while pipe C empties water at the same time. The key twist is that Harihar closes different pipes at different times, and we must infer the rate of leakage and then compute how long it would take to fill the remaining part of the tank if only the inlet pipes are kept open after closing the outlet pipe.

Given Data / Assumptions:
Pipe A can fill the tank in 24 hours. Pipe B can fill the tank in 17 hours. Pipe C is an outlet pipe, open from the beginning, and we are told that when A and B are closed at a certain moment, the water currently in the tank is removed (emptied) by pipe C in 2.5 hours. Harihar closes A and B at the time when, without any leakage, a full tank would have been achieved. We assume that all rates remain constant throughout the process and that the total work is one full tank.

Concept / Approach:
The idea is to first find the combined filling rate of A and B. Then, we determine the theoretical time at which the tank would be full if only A and B were working with no leakage. This is the moment when A and B are actually closed. Because pipe C has been open the whole time, the tank is not full at that moment. The water actually present in the tank at that time is then emptied by C in 2.5 hours, which allows us to determine the emptying rate of C. Using these rates, we calculate how much of the tank was filled and how much remains. Finally, we compute the additional time required to fill the remaining part if only A and B continue to run and C is closed.

Step-by-Step Solution:
Step 1: Rate of pipe A = 1 / 24 tank per hour. Step 2: Rate of pipe B = 1 / 17 tank per hour. Step 3: Combined rate of A and B = 1 / 24 + 1 / 17 = (17 + 24) / (24 * 17) = 41 / 408 tank per hour. Step 4: If there were no leak, A and B together would fill the tank in time T = 1 / (41 / 408) = 408 / 41 hours. Step 5: In reality, pipes A and B are closed at time T, but pipe C has been open throughout, so the actual amount of water in the tank at that moment is less than 1 full tank. Step 6: Let the rate of the outlet pipe C be c tank per hour (this rate is negative relative to filling). The net filling rate while all three pipes are open is (41 / 408 - c) tank per hour. Step 7: Amount of water in the tank at time T is (41 / 408 - c) * T. Step 8: After time T, A and B are closed, and only pipe C remains open. We are told that the water present at that moment is completely emptied in 2.5 hours. Hence (41 / 408 - c) * T = 2.5 * c. Step 9: Substitute T = 408 / 41 into the equation: (41 / 408 - c) * (408 / 41) = 2.5 * c. Simplifying gives 1 - c * (408 / 41) = 2.5 * c. Step 10: Rearranging, 1 = c * (408 / 41 + 2.5). Solving this expression gives a positive value for c and thus a consistent leakage rate. Step 11: From this rate, the fraction of the tank that was filled at time T is some fraction f less than 1, and the remaining fraction is (1 - f). Step 12: If, instead of closing A and B, Harihar had closed C at the same time T, the remaining fraction (1 - f) would now be filled by A and B together at rate 41 / 408 tank per hour. Step 13: The extra time required after closing C is therefore (1 - f) / (41 / 408). With the given numbers, this evaluates numerically to approximately 8 hours.

Verification / Alternative check:
We can approximate the result numerically to check plausibility. The combined filling rate of A and B is slightly more than 0.1 tank per hour (41 / 408 ≈ 0.1005). Over the theoretical filling time of about 9.95 hours (408 / 41 ≈ 9.95), a perfect tank would be filled. Because the leak has been active during this entire period, a substantial portion of the water is lost. The remaining part is not negligible, but still leaves a sizable fraction of the tank unfilled. Given the filling rate is close to 0.1 per hour, taking around 8 more hours to finish the remaining fraction is consistent and reasonable within the structure of the problem.

Why Other Options Are Wrong:
2 hours and 4 hours are too small given the relatively slow combined filling rate of A and B and the fact that the tank was not even close to full at time T because of continuous leakage. 6 hours underestimates the effect of the leak, implying that nearly the entire tank was already full at T, which contradicts the given that the leak empties the existing water in just 2.5 hours. 10 hours is slightly larger than needed, while the correct computed extra time is closest to 8 hours.

Common Pitfalls:
Many learners try to handle such problems by naive averaging of times or by ignoring the leak while computing intermediate stages. Another common mistake is to treat the 2.5 hours as the time for the full tank to be emptied, rather than the time to empty the actual partial volume at time T. It is crucial to set up the rate equations carefully and recognize that the time T at which A and B are closed is based on the hypothetical case without leakage.

Final Answer:
If Harihar closes the outlet pipe C instead of closing pipes A and B at that moment, the remaining tank would be filled in approximately 8 hours.