A water tank is `2/5` th full. Inlet A can fill the tank in 12 minutes while an outlet pipe B can empty it in 6 minutes. If both the pipes are open, how long will it take to empty or fill the tank completely?

Options

A. It fills in 4.8 min
B. It empties in 4.8 min
C. It fills in 5.6 min
D. It empties in 5.6 min

Answer

To solve the problem step by step, let's break it down clearly:

### Step 1: Understand the problem
We have a water tank that is \( \frac{2}{5} \) full. There are two pipes:
- Inlet A fills the tank in 12 minutes.
- Outlet B empties the tank in 6 minutes.

### Step 2: Determine the rates of filling and emptying
- The rate of Inlet A (filling) can be calculated as:
\[ \text{Rate of A} = \frac{1 \text{ tank}}{12 \text{ minutes}} = \frac{1}{12} \text{ tanks per minute} \]

- The rate of Outlet B (emptying) can be calculated as:
\[ \text{Rate of B} = \frac{1 \text{ tank}}{6 \text{ minutes}} = \frac{1}{6} \text{ tanks per minute} \]

### Step 3: Calculate the net rate when both pipes are open
When both pipes are open, the net rate is:
\[ \text{Net Rate} = \text{Rate of A} - \text{Rate of B} = \frac{1}{12} - \frac{1}{6} \]

To perform the subtraction, we need a common denominator:
\[ \frac{1}{12} - \frac{1}{6} = \frac{1}{12} - \frac{2}{12} = -\frac{1}{12} \text{ tanks per minute} \]
This negative sign indicates that the tank is being emptied.

### Step 4: Determine how long it takes to empty the tank
Since the tank is \( \frac{2}{5} \) full, we need to find out how long it will take to empty this amount at the rate of \( -\frac{1}{12} \) tanks per minute.

Let \( t \) be the time in minutes to empty \( \frac{2}{5} \) of the tank:
\[ \text{Amount emptied} = \text{Net Rate} \times t \]
\[ \frac{2}{5} = -\frac{1}{12} \times t \]

### Step 5: Solve for \( t \)
Rearranging the equation gives:
\[ t = \frac{2/5}{1/12} = \frac{2}{5} \times \frac{12}{1} = \frac{24}{5} = 4.8 \text{ minutes} \]

### Conclusion
It will take 4.8 minutes to empty the tank when both pipes are open.