A cistem can be filled by two pipes A and B 4 hours and 6 hours respectively. When full , the tank can emptied by a third pipe C in 8 hours. If all taps are turned on at the same time the cistern will be full in ?

Options

A. `40/7` hours
B. `24/7` hours
C. `16/7` hours
D. `31/7` hours

Answer

To solve the problem, we will follow these steps:

### Step 1: Determine the work done by each pipe in one hour.
- **Pipe A** fills the tank in 4 hours. Therefore, in one hour, it fills:
\[ \text{Work done by A in 1 hour} = \frac{1}{4} \text{ of the tank} \]

- **Pipe B** fills the tank in 6 hours. Therefore, in one hour, it fills:
\[ \text{Work done by B in 1 hour} = \frac{1}{6} \text{ of the tank} \]

- **Pipe C** empties the tank in 8 hours. Therefore, in one hour, it empties:
\[ \text{Work done by C in 1 hour} = -\frac{1}{8} \text{ of the tank} \]

### Step 2: Calculate the total work done by all pipes in one hour.
Now, we will sum the work done by pipes A, B, and C in one hour:
\[ \text{Total work in 1 hour} = \left(\frac{1}{4} + \frac{1}{6} - \frac{1}{8}\right) \]

### Step 3: Find a common denominator.
The least common multiple (LCM) of 4, 6, and 8 is 24. We will convert each fraction to have a denominator of 24:
- For \(\frac{1}{4}\):
\[ \frac{1}{4} = \frac{6}{24} \]
- For \(\frac{1}{6}\):
\[ \frac{1}{6} = \frac{4}{24} \]
- For \(-\frac{1}{8}\):
\[ -\frac{1}{8} = -\frac{3}{24} \]

### Step 4: Combine the fractions.
Now we can combine the fractions:
\[ \text{Total work in 1 hour} = \frac{6}{24} + \frac{4}{24} - \frac{3}{24} = \frac{6 + 4 - 3}{24} = \frac{7}{24} \]

### Step 5: Calculate the time taken to fill the tank.
If \(\frac{7}{24}\) of the tank is filled in one hour, then to fill the entire tank (1 tank), we can set up the equation:
\[ \text{Time} = \frac{1 \text{ tank}}{\frac{7}{24} \text{ tank/hour}} = \frac{24}{7} \text{ hours} \]

### Final Answer:
The cistern will be full in \(\frac{24}{7}\) hours, which is approximately 3 hours and 24 minutes.

---