Pipes A and B can fill a tank in 30 minutes and `37(1)/(2)` minutes, respectively. C is an outlet pipe when all the three pipes are opened together, then the tank is full in 25 minutes. In how much time (in minutes) can C alone empty `(2)/(5)`th part of the tank ?

Options

A. 24
B. 30
C. 20
D. 25

Answer

To solve the problem step by step, we will first determine the rates at which pipes A, B, and C work, and then find out how long it takes for pipe C to empty \( \frac{2}{5} \) of the tank.

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

- Pipe A can fill the tank in 30 minutes.
- Rate of A = \( \frac{1}{30} \) tank per minute.

- Pipe B can fill the tank in \( 37 \frac{1}{2} \) minutes, which is \( 37.5 \) minutes.
- Rate of B = \( \frac{1}{37.5} = \frac{2}{75} \) tank per minute.

### Step 2: Determine the combined rate of pipes A, B, and C

When all three pipes are opened together, they fill the tank in 25 minutes.
- Combined rate of A, B, and C = \( \frac{1}{25} \) tank per minute.

### Step 3: Set up the equation for the combined rate

The combined rate can be expressed as:
\[ \text{Rate of A} + \text{Rate of B} - \text{Rate of C} = \text{Combined Rate} \]
Substituting the rates we found:
\[ \frac{1}{30} + \frac{2}{75} - \text{Rate of C} = \frac{1}{25} \]

### Step 4: Find a common denominator and solve for Rate of C

The common denominator for 30, 75, and 25 is 150. We will convert each rate to have this common denominator:
- Rate of A: \( \frac{1}{30} = \frac{5}{150} \)
- Rate of B: \( \frac{2}{75} = \frac{4}{150} \)
- Combined Rate: \( \frac{1}{25} = \frac{6}{150} \)

Now substituting these into the equation:
\[ \frac{5}{150} + \frac{4}{150} - \text{Rate of C} = \frac{6}{150} \]
Combining the rates:
\[ \frac{9}{150} - \text{Rate of C} = \frac{6}{150} \]
Now, isolate Rate of C:
\[ \text{Rate of C} = \frac{9}{150} - \frac{6}{150} = \frac{3}{150} = \frac{1}{50} \]
This means pipe C can empty \( \frac{1}{50} \) of the tank per minute.

### Step 5: Determine the time taken by pipe C to empty \( \frac{2}{5} \) of the tank

To find out how long it takes for pipe C to empty \( \frac{2}{5} \) of the tank, we can set up the equation:
\[ \text{Time} = \frac{\text{Amount of tank}}{\text{Rate of C}} = \frac{\frac{2}{5}}{\frac{1}{50}} \]
Calculating this gives:
\[ \text{Time} = \frac{2}{5} \times 50 = 20 \text{ minutes} \]

### Final Answer
Pipe C can empty \( \frac{2}{5} \) of the tank in **20 minutes**.