This problem involves calculating the capacity of a tank based on the rates at which different pipes fill or empty it. We are given the time taken by two pipes (A and B) to fill the tank individually, the rate at which a third pipe (C) empties it, and the combined time taken by all three pipes working together to fill the tank.

Understanding Pipe and Tank Problems

In problems involving pipes and tanks, we often think of the work done by each pipe in terms of a rate per unit of time. If a pipe can fill a tank in 't' minutes, its filling rate is \(1/t\) of the tank per minute. Similarly, if a pipe can empty a tank in 't' minutes, its emptying rate is \(1/t\) of the tank per minute.

Setting Up the Rates

Let the total capacity of the tank be \(V\) gallons.

• Pipe A fills the tank in 25 minutes. So, in one minute, Pipe A fills \(1/25\) of the tank. In terms of volume, its filling rate is \(V/25\) gallons per minute.
• Pipe B fills the tank in half an hour, which is 30 minutes. So, in one minute, Pipe B fills \(1/30\) of the tank. In terms of volume, its filling rate is \(V/30\) gallons per minute.
• Pipe C empties the tank at a rate of 3 gallons per minute. This is a direct volume rate, not a fraction of the tank per minute unless the capacity is known.

Combined Work Rate

When all three pipes A, B, and C work together, the tank is filled in 15 minutes. The net rate at which the tank is filled is the sum of the filling rates minus the emptying rate.

• Rate of A (filling) + Rate of B (filling) - Rate of C (emptying) = Combined rate (filling)

The combined rate of filling is the total capacity \(V\) divided by the time taken (15 minutes).

Combined filling rate = \(V/15\) gallons per minute.

Formulating the Equation

We can write the equation based on the rates:

$$ \text{Rate of A} + \text{Rate of B} - \text{Rate of C} = \text{Combined Rate} $$

Substituting the values we have:

$$ \frac{V}{25} + \frac{V}{30} - 3 = \frac{V}{15} $$

Solving for the Tank Capacity (V)

Now, we need to solve this equation for \(V\). To eliminate the fractions, we can multiply the entire equation by the least common multiple (LCM) of the denominators (25, 30, and 15).

The prime factorisation of the denominators are:

• 25 = \(5^2\)
• 30 = \(2 \times 3 \times 5\)
• 15 = \(3 \times 5\)

The LCM is \(2 \times 3 \times 5^2 = 6 \times 25 = 150\).

Multiply the equation by 150:

$$ 150 \left( \frac{V}{25} \right) + 150 \left( \frac{V}{30} \right) - 150(3) = 150 \left( \frac{V}{15} \right) $$

Simplify the terms:

$$ 6V + 5V - 450 = 10V $$

Combine the \(V\) terms on the left side:

$$ 11V - 450 = 10V $$

Subtract \(10V\) from both sides:

$$ 11V - 10V = 450 $$

$$ V = 450 $$

So, the capacity of the tank is 450 gallons.

Verification

Let's check if this capacity works with the given times:

• Rate of A = 450 gallons / 25 minutes = 18 gallons/minute
• Rate of B = 450 gallons / 30 minutes = 15 gallons/minute
• Rate of C = 3 gallons/minute (emptying)

Combined rate = Rate of A + Rate of B - Rate of C

Combined rate = 18 gallons/minute + 15 gallons/minute - 3 gallons/minute

Combined rate = 33 gallons/minute - 3 gallons/minute = 30 gallons/minute

Time taken together = Total Capacity / Combined Rate

Time taken together = 450 gallons / 30 gallons/minute = 15 minutes

This matches the information given in the problem, confirming that the capacity of the tank is 450 gallons.

Conclusion

Based on the calculations, the capacity of the tank is 450 gallons.

Revision Table: Tank Capacity Calculation

Additional Information: Pipe and Cistern Concepts

Problems involving pipes filling or emptying tanks (often called cisterns) are a common type of time and work problem. The core idea is to calculate the amount of work done (fraction of the tank filled or emptied) in a unit of time (usually a minute or an hour).

• Efficiency: The rate at which a pipe fills or empties is its efficiency. A pipe that takes less time is more efficient.
• Positive vs. Negative Work: Filling pipes do 'positive work' (adding volume), while emptying pipes do 'negative work' (removing volume).
• Combined Work: When multiple pipes work together, their individual rates are added or subtracted to find the net combined rate. If the net rate is positive, the tank fills; if it's negative, the tank empties.
• Total Work = Rate \(\times\) Time: This fundamental formula is applied. If the total work is filling the entire tank (represented as 1 or the total volume V), then Time = Total Work / Rate.

These principles allow us to solve various problems involving different scenarios, such as pipes working simultaneously, pipes working alternately, or leaks in the tank.