To solve the problem, we need to determine how long it takes to fill the tank when all three pipes are used but Pipe C is closed after 4 hours. Let's break it down step-by-step:

Step 1: Determine individual rates of the pipes.

• Pipe A fills the tank in 9 hours, so it fills \frac{1}{9} of the tank in one hour.
• Pipe B fills the tank in 12 hours, so it fills \frac{1}{12} of the tank in one hour.
• Pipe C empties the tank in 18 hours, so it empties \frac{1}{18} of the tank in one hour.

Step 2: Calculate the net rate of the three pipes working together for the first 4 hours.

• The combined rate of Pipes A, B, and C is:
• \frac{1}{9} + \frac{1}{12} - \frac{1}{18}

To compute this, we find a common denominator:

• LCM of 9, 12, and 18 is 36.
• Convert each fraction:
• \frac{1}{9} = \frac{4}{36}, \frac{1}{12} = \frac{3}{36}, \frac{1}{18} = \frac{2}{36}
• Adding these gives: \frac{4}{36} + \frac{3}{36} - \frac{2}{36} = \frac{5}{36}

Therefore, in 1 hour, the three pipes together fill \frac{5}{36} of the tank.

Step 3: Calculate the amount of tank filled in the first 4 hours.

• In 4 hours, they will fill: 4 \times \frac{5}{36} = \frac{20}{36} = \frac{5}{9} of the tank.

Step 4: Calculate the remaining part of the tank to be filled and time required.

• The part of the tank remaining after 4 hours: 1 - \frac{5}{9} = \frac{4}{9}
• Only Pipes A and B are available now, so their combined rate is:
• \frac{1}{9} + \frac{1}{12} = \frac{4}{36} + \frac{3}{36} = \frac{7}{36} per hour.
• Time required to fill the remaining \frac{4}{9} of the tank:
• Time = \frac{\frac{4}{9}}{\frac{7}{36}} = \frac{4}{9} \times \frac{36}{7} = \frac{16}{7} hours.

Step 5: Calculate total time to fill the tank completely.

• Total time = 4 hours (initial) + \frac{16}{7} hours (remaining).
• Total time = 4 + \frac{16}{7} = \frac{28}{7} + \frac{16}{7} = \frac{44}{7} hours.

Therefore, the total time taken to fill the tank completely is \frac{44}{7} hours. Thus, the correct answer is \frac{44}{7} hours.