If pipe A alone and Pipe B alone can fill a tank in 20 min and 30 min respectively and pipe C alone can empty it in 10 min. If the tank is completely filled, then find the time taken to empty the tank if all the three pipes are opened simultaneously?

Options

A. A)45 min
B. B)50 min
C. C)60 min
D. D)40 min

Answer

To solve the problem step by step, we will calculate the efficiencies of pipes A, B, and C, and then determine the time taken to empty the tank when all three pipes are opened simultaneously.

### Step 1: Determine the efficiencies of the pipes
- **Pipe A** can fill the tank in 20 minutes. Therefore, its efficiency is:
\[ \text{Efficiency of A} = \frac{1 \text{ tank}}{20 \text{ min}} = \frac{1}{20} \text{ tanks/min} \]

- **Pipe B** can fill the tank in 30 minutes. Therefore, its efficiency is:
\[ \text{Efficiency of B} = \frac{1 \text{ tank}}{30 \text{ min}} = \frac{1}{30} \text{ tanks/min} \]

- **Pipe C** can empty the tank in 10 minutes. Therefore, its efficiency (as a negative value since it empties) is:
\[ \text{Efficiency of C} = -\frac{1 \text{ tank}}{10 \text{ min}} = -\frac{1}{10} \text{ tanks/min} \]

### Step 2: Calculate the combined efficiency of all three pipes
To find the total efficiency when all three pipes are opened together, we add the efficiencies:
\[ \text{Total Efficiency} = \text{Efficiency of A} + \text{Efficiency of B} + \text{Efficiency of C} \]
Substituting the values:
\[ \text{Total Efficiency} = \frac{1}{20} + \frac{1}{30} - \frac{1}{10} \]

### Step 3: Find a common denominator and simplify
The least common multiple (LCM) of 20, 30, and 10 is 60. We will convert each fraction to have a denominator of 60:
\[ \frac{1}{20} = \frac{3}{60}, \quad \frac{1}{30} = \frac{2}{60}, \quad -\frac{1}{10} = -\frac{6}{60} \]
Now, substituting these values:
\[ \text{Total Efficiency} = \frac{3}{60} + \frac{2}{60} - \frac{6}{60} = \frac{3 + 2 - 6}{60} = \frac{-1}{60} \text{ tanks/min} \]

### Step 4: Calculate the time to empty the tank
The time taken to empty the tank can be calculated using the formula:
\[ \text{Time} = \frac{\text{Work}}{\text{Efficiency}} \]
Here, the work is 1 tank (or 60 liters, as we assumed) and the efficiency is \(-\frac{1}{60}\) tanks/min:
\[ \text{Time} = \frac{1 \text{ tank}}{-\frac{1}{60} \text{ tanks/min}} = -60 \text{ minutes} \]

The negative sign indicates that the tank is being emptied. Thus, the time taken to empty the tank is:
\[ \text{Time} = 60 \text{ minutes} \]

### Final Answer
The time taken to empty the tank when all three pipes are opened simultaneously is **60 minutes**.

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