A tank is connected with 8 pipes. Some of them are inlet pipes and rest work as outlet pipes. Each of the inlet pipe can fill the tank in 8 hours, individually, while each of those that empty the tank i.e outlet pipe, can empty it in 6 hours individually. If all the pipes are kept open when the tank is full, it will take exactly 6 hours for the tank to empty. How many of these are inlet pipes?

Options

A. 2
B. 4
C. 5
D. 6

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the problem
We have a tank connected with 8 pipes, some of which are inlet pipes and some are outlet pipes. Each inlet pipe can fill the tank in 8 hours, and each outlet pipe can empty the tank in 6 hours. We need to find out how many of the pipes are inlet pipes.

### Step 2: Determine the rates of filling and emptying
- **Inlet Pipe Rate**: Each inlet pipe can fill the tank in 8 hours. Therefore, the rate of one inlet pipe is:
\[ \text{Rate of one inlet pipe} = \frac{1 \text{ tank}}{8 \text{ hours}} = \frac{1}{8} \text{ tanks per hour} \]
If there are \( x \) inlet pipes, the total rate of filling by all inlet pipes is:
\[ \text{Total rate of inlet pipes} = x \times \frac{1}{8} = \frac{x}{8} \text{ tanks per hour} \]

- **Outlet Pipe Rate**: Each outlet pipe can empty the tank in 6 hours. Therefore, the rate of one outlet pipe is:
\[ \text{Rate of one outlet pipe} = \frac{1 \text{ tank}}{6 \text{ hours}} = \frac{1}{6} \text{ tanks per hour} \]
If there are \( y \) outlet pipes, the total rate of emptying by all outlet pipes is:
\[ \text{Total rate of outlet pipes} = y \times \frac{1}{6} = \frac{y}{6} \text{ tanks per hour} \]

### Step 3: Set up the equation
We know that the total number of pipes is 8, so we have:
\[ x + y = 8 \]
When all pipes are open, the tank empties in 6 hours. This means the net rate of change in the tank's volume is:
\[ \text{Net rate} = \text{Total rate of inlet pipes} - \text{Total rate of outlet pipes} \]
Given that the tank empties in 6 hours, the net rate is:
\[ \text{Net rate} = -\frac{1}{6} \text{ tanks per hour} \]
Thus, we can write:
\[ \frac{x}{8} - \frac{y}{6} = -\frac{1}{6} \]

### Step 4: Solve the equations
Now we have two equations:
1. \( x + y = 8 \)
2. \( \frac{x}{8} - \frac{y}{6} = -\frac{1}{6} \)

From the first equation, we can express \( y \) in terms of \( x \):
\[ y = 8 - x \]

Substituting \( y \) into the second equation:
\[ \frac{x}{8} - \frac{8 - x}{6} = -\frac{1}{6} \]
To eliminate the fractions, we can multiply through by 24 (the least common multiple of 8 and 6):
\[ 3x - 4(8 - x) = -4 \]
Expanding and simplifying:
\[ 3x - 32 + 4x = -4 \]
\[ 7x - 32 = -4 \]
\[ 7x = 28 \]
\[ x = 4 \]

### Step 5: Find the number of outlet pipes
Using \( x = 4 \) in the first equation:
\[ y = 8 - x = 8 - 4 = 4 \]

### Conclusion
Thus, the number of inlet pipes is \( \boxed{4} \).