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 find out how long it takes for the outlet pipe C to empty a full cistern. We will follow these steps:

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

### Step 2: Calculate the combined rate of pipes A and B
- The combined rate of pipes A and B when both are working together is:
\[ \text{Combined Rate of A and B} = \frac{1}{10} + \frac{1}{15} \]
- To add these fractions, we need a common denominator. The least common multiple of 10 and 15 is 30:
\[ \text{Combined Rate} = \frac{3}{30} + \frac{2}{30} = \frac{5}{30} = \frac{1}{6} \text{ cisterns per hour} \]

### Step 3: Determine the effective rate when pipe C is also open
- When pipe C is open, the cistern can be filled in 18 hours. Therefore, the effective rate when all pipes are working together is:
\[ \text{Effective Rate} = \frac{1}{18} \text{ cisterns per hour} \]

### Step 4: Set up the equation to find the rate of pipe C
- The effective rate can be expressed as the combined rate of A and B minus the rate of C:
\[ \frac{1}{6} - \text{Rate of C} = \frac{1}{18} \]

### Step 5: Solve for the rate of pipe C
- Rearranging the equation gives:
\[ \text{Rate of C} = \frac{1}{6} - \frac{1}{18} \]
- To subtract these fractions, we need a common denominator. The least common multiple of 6 and 18 is 18:
\[ \text{Rate of C} = \frac{3}{18} - \frac{1}{18} = \frac{2}{18} = \frac{1}{9} \text{ cisterns per hour} \]

### Step 6: Calculate the time taken by pipe C to empty the cistern
- If pipe C can empty \(\frac{1}{9}\) of the cistern in one hour, then the time taken by pipe C to empty a full cistern is the reciprocal of its rate:
\[ \text{Time taken by C} = \frac{1}{\frac{1}{9}} = 9 \text{ hours} \]

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

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