Two pipes can fill a tank in 20 and 24 min, respectively and a waste pipe can empty 6 gallon per min. All the three pipes working together can fill the tank in 15 min. Find the capacity of the tank.

Options

A. 210 gallon
B. 50 gallon
C. 150 gallon
D. 240 gallon

Answer

To solve the problem step by step, we can follow these calculations:

### Step 1: Determine the filling rates of the pipes
- Pipe A can fill the tank in 20 minutes.
- Pipe B can fill the tank in 24 minutes.

To find their filling rates:
- The filling rate of Pipe A = Total capacity of the tank / Time taken by Pipe A = C / 20
- The filling rate of Pipe B = Total capacity of the tank / Time taken by Pipe B = C / 24

### Step 2: Determine the emptying rate of the waste pipe
- The waste pipe can empty 6 gallons per minute.

### Step 3: Calculate the combined filling rate of all pipes
- When all three pipes (A, B, and the waste pipe) work together, they can fill the tank in 15 minutes.
- The combined filling rate = Total capacity of the tank / Time taken when all pipes work together = C / 15

### Step 4: Set up the equation
From the above steps, we can express the filling rates in terms of C:
- Filling rate of Pipe A = C / 20
- Filling rate of Pipe B = C / 24
- Emptying rate of the waste pipe = 6 gallons/minute

The equation for the combined rates is:
\[ \frac{C}{20} + \frac{C}{24} - 6 = \frac{C}{15} \]

### Step 5: Find a common denominator and solve for C
The least common multiple (LCM) of 20, 24, and 15 is 120. We can multiply the entire equation by 120 to eliminate the denominators:
\[ 120 \left(\frac{C}{20}\right) + 120 \left(\frac{C}{24}\right) - 120(6) = 120 \left(\frac{C}{15}\right) \]

This simplifies to:
\[ 6C + 5C - 720 = 8C \]

### Step 6: Rearranging the equation
Now, we can rearrange the equation:
\[ 6C + 5C - 8C = 720 \]
\[ 3C = 720 \]

### Step 7: Solve for C
Dividing both sides by 3 gives:
\[ C = 240 \]

### Step 8: Conclusion
The capacity of the tank is 240 gallons.

---