Pipe A and B can fill a cistern in 10 hours and 15 hours Respectively. When a third pipe C which works as an outlet pipe is also open then the cistern can be filled in 18 hours. The outlet pipe can empty a full cistern in :

Options

A. 12 hours
B. 8 hours
C. 9 hours
D. 14 hours

Answer

To solve the problem, we need to determine how long the outlet pipe C takes to empty a full cistern. We will follow these steps:

### Step-by-Step Solution:

1. **Determine the filling rates of pipes A and B:**
- Pipe A can fill the cistern in 10 hours. Therefore, its rate of filling is:
\[ \text{Rate of A} = \frac{1 \text{ cistern}}{10 \text{ hours}} = \frac{1}{10} \text{ cistern per hour} \]
- Pipe B can fill the cistern in 15 hours. Therefore, its rate of filling is:
\[ \text{Rate of B} = \frac{1 \text{ cistern}}{15 \text{ hours}} = \frac{1}{15} \text{ cistern per hour} \]

2. **Calculate the combined filling rate of pipes A and B:**
- To find the combined rate of A and B, we add their individual rates:
\[ \text{Combined Rate of A and B} = \frac{1}{10} + \frac{1}{15} \]
- To add these fractions, we need a common denominator. The least common multiple of 10 and 15 is 30:
\[ \frac{1}{10} = \frac{3}{30}, \quad \frac{1}{15} = \frac{2}{30} \]
- Thus,
\[ \text{Combined Rate of A and B} = \frac{3}{30} + \frac{2}{30} = \frac{5}{30} = \frac{1}{6} \text{ cistern per hour} \]

3. **Incorporate the effect of the outlet pipe C:**
- When all three pipes (A, B, and C) are open, they can fill the cistern in 18 hours. Therefore, their combined rate is:
\[ \text{Combined Rate of A, B, and C} = \frac{1 \text{ cistern}}{18 \text{ hours}} = \frac{1}{18} \text{ cistern per hour} \]

4. **Set up the equation to find the rate of outlet pipe C:**
- Let the rate of outlet pipe C be \( R_C \). The combined rate of A, B, and C can be expressed as:
\[ \text{Rate of A} + \text{Rate of B} - \text{Rate of C} = \text{Combined Rate of A, B, and C} \]
- Substituting the known values:
\[ \frac{1}{10} + \frac{1}{15} - R_C = \frac{1}{18} \]
- We already calculated \(\frac{1}{10} + \frac{1}{15} = \frac{1}{6}\). Therefore:
\[ \frac{1}{6} - R_C = \frac{1}{18} \]

5. **Solve for \( R_C \):**
- Rearranging gives:
\[ R_C = \frac{1}{6} - \frac{1}{18} \]
- Finding a common denominator (which is 18):
\[ R_C = \frac{3}{18} - \frac{1}{18} = \frac{2}{18} = \frac{1}{9} \text{ cistern per hour} \]
- Since this is the rate at which C empties the cistern, it means that pipe C can empty the cistern in 9 hours.

### Final Answer:
The outlet pipe C can empty a full cistern in **9 hours**.