A cistern can be filled by two pipes ,A and B in 12 minutes and 14 minutes epectively and can be empited by a third pipe C in 8 minutes.If all the taps be turned on at the same moment,what part of cistern will remain unfilled at the end of 7 minutes

Options

A. 5/24
B. 19/24
C. 7/24
D. 17/24

Answer

To solve the problem, we need to find out how much of the cistern remains unfilled after 7 minutes when pipes A, B, and C are all turned on simultaneously.

### Step-by-Step Solution:

1. **Determine the filling and emptying rates of the pipes:**
- Pipe A fills the cistern in 12 minutes. Therefore, in 1 minute, it fills \( \frac{1}{12} \) of the cistern.
- Pipe B fills the cistern in 14 minutes. Therefore, in 1 minute, it fills \( \frac{1}{14} \) of the cistern.
- Pipe C empties the cistern in 8 minutes. Therefore, in 1 minute, it empties \( \frac{1}{8} \) of the cistern.

2. **Calculate the combined rate of filling when all pipes are open:**
- The filling rate of A in 1 minute: \( \frac{1}{12} \)
- The filling rate of B in 1 minute: \( \frac{1}{14} \)
- The emptying rate of C in 1 minute: \( -\frac{1}{8} \)

To find the total rate of filling when all pipes are open, we add the filling rates of A and B and subtract the emptying rate of C:
\[ \text{Total rate} = \frac{1}{12} + \frac{1}{14} - \frac{1}{8} \]

3. **Find a common denominator to add the fractions:**
- The least common multiple (LCM) of 12, 14, and 8 is 168.
- Convert each fraction:
- \( \frac{1}{12} = \frac{14}{168} \)
- \( \frac{1}{14} = \frac{12}{168} \)
- \( \frac{1}{8} = \frac{21}{168} \)

Now substitute back into the total rate equation:
\[ \text{Total rate} = \frac{14}{168} + \frac{12}{168} - \frac{21}{168} = \frac{14 + 12 - 21}{168} = \frac{5}{168} \]

4. **Calculate the amount filled in 7 minutes:**
- If the total rate of filling is \( \frac{5}{168} \) of the cistern per minute, then in 7 minutes:
\[ \text{Amount filled in 7 minutes} = 7 \times \frac{5}{168} = \frac{35}{168} \]

5. **Determine the remaining part of the cistern:**
- The total capacity of the cistern is 1 (or \( \frac{168}{168} \)).
- The part of the cistern that remains unfilled is:
\[ \text{Remaining part} = 1 - \frac{35}{168} = \frac{168 - 35}{168} = \frac{133}{168} \]

6. **Simplify the remaining part:**
- To simplify \( \frac{133}{168} \), we check for common factors. The fraction is already in its simplest form.
- Therefore, the remaining part of the cistern that is unfilled is \( \frac{133}{168} \).

### Final Answer:
The part of the cistern that will remain unfilled at the end of 7 minutes is \( \frac{133}{168} \).