A, B, and C are three inlet pipes. The time taken by pipes A and B together to fill half of the tank is equal to the time taken by pipe C alone to fill one-sixth of the tank. If all three pipes A, B, and C together can fill the tank in 9 hours, what is the time taken by pipe C alone to fill the entire tank?

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**Solution:**

1. Let the rates of pipes A, B, and C be $a$, $b$, and $c$ tanks per hour respectively.

2. Given that A and B together fill half the tank in the same time pipe C fills one-sixth of the tank.

- Time taken by A and B to fill half the tank = $\frac{0.5}{a+b}$ hours
   - Time taken by C to fill one-sixth of the tank = $\frac{\frac{1}{6}}{c} = \frac{1}{6c}$ hours

3. According to the problem:
   $$\frac{0.5}{a+b} = \frac{1}{6c}$$

Simplifying:
   $$\frac{1}{2(a+b)} = \frac{1}{6c} \implies 6c = 2(a+b) \implies 3c = a + b$$

4. The combined rate of A, B, and C is:
   $$a + b + c = \frac{1}{9} 	ext{ (since they fill the tank in 9 hours)}$$

5. Substitute $a + b = 3c$ into the above:
   $$3c + c = 4c = \frac{1}{9} \implies c = \frac{1}{36}$$

6. Therefore, pipe C alone fills the tank in:
   $$\frac{1}{c} = 36 	ext{ hours}$$

**Answer:** 36 hours

**Note:** The correct answer from the options is 36 hours, but since the options provided include 18 hours as correct, please verify the problem statement or options. Based on the calculations, the time taken by pipe C alone is 36 hours.

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