To solve this problem, we need to calculate the time at which pipe Y was closed. We know the following:
- Pipe X fills the tank in 12 hours.
- Pipe Y empties the tank in 18 hours.
- Both pipes are opened at 8 am and the tank is fully filled by 10 pm.
The question is asking at what time pipe Y was closed. Let's break down the problem using step-by-step reasoning:
- Determine the rates of work of Pipe X and Pipe Y:
- Rate of filling by Pipe X = \(\frac{1}{12}\) of the tank per hour.
- Rate of emptying by Pipe Y = \(\frac{1}{18}\) of the tank per hour.
- Find the net rate when both pipes are open:
- Net rate = Rate of Pipe X - Rate of Pipe Y
- Net rate = \(\frac{1}{12} - \frac{1}{18} = \frac{1}{36}\) of the tank per hour.
- Calculate the total time when both pipes are open:
- Both pipes are open from 8 am to the time when Y is closed. Suppose Y is closed after 't' hours.
- So, the effective time with both pipes open is 't' hours.
- Therefore, after 't' hours, fraction of tank filled = \(\frac{t}{36}\).
- Calculate the remaining time for full tank:
- After pipe Y is closed, pipe X fills the tank alone.
- If total time from 8 am to 10 pm is 14 hours, then remaining time with only Pipe X open is 14 - t hours.
- During this time, Pipe X fills \(\frac{14-t}{12}\) of the tank.
- Equate the two expressions to find 't':
- For the tank to be full, the sum of the filled parts should be 1 (full tank).
- So, \(\frac{t}{36} + \frac{14-t}{12} = 1\)
- Solve for 't':
- \(\frac{t}{36} + \frac{14-t}{12} = 1\)
- \(\frac{t}{36} + \frac{14 \times 3 - t \times 3}{36} = 1\)
- Simplifying: \(\frac{t + 42 - 3t}{36} = 1\)
- \(\frac{42 - 2t}{36} = 1\)
- Simplify and solve for 't':
- \(42 - 2t = 36\)
- \(2t = 6\)
- \(t = 3\)
Therefore, pipe Y was closed after 3 hours from 8 am, which is 11 am.