An oil tank has two pipes connected to it. If the tank is empty, Pipe A can fill it in 2 hours. If the tank is full, Pipe B can empty it in 3 hours. If both pipes are activated at the same time. When the tank is empty, how many hours will it take for the tank to be filled to 60% of its capacity?

Options

A. `2.4`
B. `3.6`
C. `6`
D. `60`

Explanation

To solve the problem step by step, we will analyze the rates of the two pipes and how they affect the filling of the tank.

### Step 1: Determine the rates of Pipe A and Pipe B
- Pipe A can fill the tank in 2 hours. Therefore, the rate of Pipe A (RA) is:
\[ R_A = \frac{D}{2} \]
- Pipe B can empty the tank in 3 hours. Therefore, the rate of Pipe B (RB) is:
\[ R_B = \frac{D}{3} \]

### Step 2: Calculate the effective rate when both pipes are open
When both pipes are activated, the effective rate of filling the tank is:
\[ R_{effective} = R_A - R_B = \frac{D}{2} - \frac{D}{3} \]
To subtract these fractions, we need a common denominator, which is 6:
\[ R_{effective} = \frac{3D}{6} - \frac{2D}{6} = \frac{1D}{6} \]

### Step 3: Calculate the volume to be filled to reach 60% capacity
To find out how much volume corresponds to 60% of the tank's capacity (D):
\[ \text{Volume to fill} = 0.6D \]

### Step 4: Set up the equation to find the time taken to fill 60% of the tank
Using the formula:
\[ \text{Time} = \frac{\text{Volume}}{\text{Rate}} \]
We can substitute the values:
\[ T = \frac{0.6D}{\frac{D}{6}} = 0.6D \times \frac{6}{D} \]
The D cancels out:
\[ T = 0.6 \times 6 = 3.6 \text{ hours} \]

### Final Answer
It will take 3.6 hours to fill the tank to 60% of its capacity when both pipes are activated.

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