Pipes A and B can fill a tank in 18 minutes and `22 1/2` minutes, respectively while pipe C can empty the full tank in 12 minutes. A and B are opened together for 6 minutes and then closed. Now C is opened. C alone will empty the tank in ..........

Options

A. 5 minutes
B. `8 2/5` minutes
C. `7 1/5` minutes
D. 6 minutes

Answer

To solve the problem step by step, we will first determine the filling and emptying rates of the pipes involved, and then calculate how long it will take for pipe C to empty the tank after pipes A and B have filled it for a certain amount of time.

### Step 1: Determine the rates of pipes A, B, and C

1. **Pipe A** can fill the tank in 18 minutes.
- Rate of A = 1 tank / 18 minutes = \( \frac{1}{18} \) tanks per minute.

2. **Pipe B** can fill the tank in \( 22 \frac{1}{2} \) minutes, which is equivalent to \( \frac{45}{2} \) minutes.
- Rate of B = 1 tank / \( \frac{45}{2} \) minutes = \( \frac{2}{45} \) tanks per minute.

3. **Pipe C** can empty the tank in 12 minutes.
- Rate of C = -1 tank / 12 minutes = \( -\frac{1}{12} \) tanks per minute (negative because it is emptying).

### Step 2: Calculate the combined rate of pipes A and B

- Combined rate of A and B:
\[ \text{Rate of A + Rate of B} = \frac{1}{18} + \frac{2}{45} \]

To add these fractions, we need a common denominator. The least common multiple (LCM) of 18 and 45 is 90.

- Convert \( \frac{1}{18} \) to have a denominator of 90:
\[ \frac{1}{18} = \frac{5}{90} \]

- Convert \( \frac{2}{45} \) to have a denominator of 90:
\[ \frac{2}{45} = \frac{4}{90} \]

Now we can add them:
\[ \frac{5}{90} + \frac{4}{90} = \frac{9}{90} = \frac{1}{10} \text{ tanks per minute} \]

### Step 3: Calculate the amount of tank filled by A and B in 6 minutes

- Amount filled in 6 minutes:
\[ \text{Amount filled} = \text{Rate} \times \text{Time} = \frac{1}{10} \times 6 = \frac{6}{10} = \frac{3}{5} \text{ of the tank} \]

### Step 4: Determine how much of the tank is left to be emptied

- Remaining part of the tank to be emptied:
\[ 1 - \frac{3}{5} = \frac{2}{5} \text{ of the tank} \]

### Step 5: Calculate how long it will take for pipe C to empty the remaining part of the tank

- Rate of pipe C is \( -\frac{1}{12} \) tanks per minute.
- To find the time taken to empty \( \frac{2}{5} \) of the tank:
\[ \text{Time} = \frac{\text{Amount}}{\text{Rate}} = \frac{\frac{2}{5}}{-\frac{1}{12}} = \frac{2}{5} \times -12 = -\frac{24}{5} \text{ minutes} \]

Since we are interested in the positive time:
\[ \text{Time} = \frac{24}{5} = 4.8 \text{ minutes} \]

### Step 6: Convert the time into minutes and seconds

- \( \frac{24}{5} \) minutes is equivalent to \( 4 \) minutes and \( 48 \) seconds.

### Final Answer

Pipe C alone will empty the tank in \( 4 \) minutes and \( 48 \) seconds, which can be expressed as \( 4 \frac{4}{5} \) minutes.