Two inlet pipes can fill a cistern in 10 and 12 hours respectively and an outlet pipe can empty 80 gallons of water per hour. All the three pipes working together can fill the empty cistern in 20 hours. What is the capacity (in gallons) of the tank?

Options

A. 360
B. 300
C. 600
D. 900

Answer

To solve the problem, we will follow these steps:

### Step 1: Determine the rates of the inlet and outlet pipes.
- **Inlet Pipe A** fills the tank in 10 hours. Therefore, its rate is:
\[ \text{Rate of A} = \frac{1 \text{ tank}}{10 \text{ hours}} = \frac{1}{10} \text{ tanks per hour} \]

- **Inlet Pipe B** fills the tank in 12 hours. Therefore, its rate is:
\[ \text{Rate of B} = \frac{1 \text{ tank}}{12 \text{ hours}} = \frac{1}{12} \text{ tanks per hour} \]

- **Outlet Pipe C** empties 80 gallons per hour. To find its rate in terms of tanks, we need to express the tank capacity in gallons first. We will denote the capacity of the tank as \( C \) gallons. Therefore, the rate of C in terms of tanks is:
\[ \text{Rate of C} = -\frac{80}{C} \text{ tanks per hour} \]

### Step 2: Set up the equation for the combined rate of the pipes.
When all three pipes are working together, they can fill the tank in 20 hours. Hence, their combined rate is:
\[ \text{Combined Rate} = \frac{1 \text{ tank}}{20 \text{ hours}} = \frac{1}{20} \text{ tanks per hour} \]

### Step 3: Write the equation for the combined rates.
Combining the rates of the three pipes, we have:
\[ \frac{1}{10} + \frac{1}{12} - \frac{80}{C} = \frac{1}{20} \]

### Step 4: Find a common denominator and solve for \( C \).
The common denominator for 10, 12, and 20 is 60. Rewriting the equation:
\[ \frac{6}{60} + \frac{5}{60} - \frac{80}{C} = \frac{3}{60} \]

Combining the fractions:
\[ \frac{6 + 5 - 3}{60} = \frac{80}{C} \]
\[ \frac{8}{60} = \frac{80}{C} \]

Cross-multiplying gives:
\[ 8C = 60 \times 80 \]
\[ 8C = 4800 \]
\[ C = \frac{4800}{8} = 600 \]

### Conclusion
The capacity of the tank is \( \boxed{600} \) gallons.

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