Two pipes A and B can alone fill a tank in 6 hours and 9 hours respectively. Pipe C can alone empty the same tank in 6 hours. If all three pipes are opened together, then in how many hours will the tank be filled?

Options

A. 9
B. 18
C. 8
D. 6

Answer

To solve the problem, we need to determine how long it will take to fill the tank when all three pipes (A, B, and C) are opened together.

### Step-by-Step Solution:

1. **Determine the rate of each pipe:**
- Pipe A fills the tank in 6 hours. Therefore, in one hour, Pipe A fills:
\[ \text{Rate of A} = \frac{1}{6} \text{ tank/hour} \]
- Pipe B fills the tank in 9 hours. Therefore, in one hour, Pipe B fills:
\[ \text{Rate of B} = \frac{1}{9} \text{ tank/hour} \]
- Pipe C empties the tank in 6 hours. Therefore, in one hour, Pipe C empties:
\[ \text{Rate of C} = -\frac{1}{6} \text{ tank/hour} \]

2. **Calculate the combined rate of all three pipes:**
- When all three pipes are opened together, their combined rate is:
\[ \text{Combined Rate} = \text{Rate of A} + \text{Rate of B} + \text{Rate of C} \]
- Substituting the rates we found:
\[ \text{Combined Rate} = \frac{1}{6} + \frac{1}{9} - \frac{1}{6} \]

3. **Finding a common denominator:**
- The least common multiple (LCM) of 6 and 9 is 18. We can express each rate with a denominator of 18:
\[ \frac{1}{6} = \frac{3}{18}, \quad \frac{1}{9} = \frac{2}{18}, \quad -\frac{1}{6} = -\frac{3}{18} \]
- Now substituting these into the combined rate:
\[ \text{Combined Rate} = \frac{3}{18} + \frac{2}{18} - \frac{3}{18} = \frac{2}{18} = \frac{1}{9} \text{ tank/hour} \]

4. **Calculate the time to fill the tank:**
- If the combined rate is \(\frac{1}{9}\) tank per hour, then the time taken to fill the entire tank (1 tank) is:
\[ \text{Time} = \frac{1 \text{ tank}}{\frac{1}{9} \text{ tank/hour}} = 9 \text{ hours} \]

### Final Answer:
The tank will be filled in **9 hours** when all three pipes are opened together.