Pipes P and Q can fill a storage tank in full with water in 10 and 6 minutes, respectively. Pipe R draws the water out from the storage tank at a rate of 34 litres per minute. P, Q and R operate at a constant rate.

If it takes one hour to completely empty a full storage tank with all the pipes operating simultaneously, what is the capacity of the storage tank (in litres)?

Calculate Storage Tank Capacity: Pipes P, Q, R Rates

This problem involves calculating the capacity of a storage tank based on the rates at which different pipes fill and drain it. We are given the time taken by pipes P and Q to fill the tank individually, the rate at which pipe R drains the tank, and the time taken to empty the tank when all three pipes operate simultaneously.

Understanding Pipe Rates

First, let's determine the rate at which each pipe works:

- Pipe P fills the tank in 10 minutes. Its filling rate is $\frac{1}{10}$ of the tank per minute.
- Pipe Q fills the tank in 6 minutes. Its filling rate is $\frac{1}{6}$ of the tank per minute.
- Pipe R draws water out at a rate of 34 litres per minute.

Combined Filling Rate

When pipes P and Q operate together, their filling rates add up. The combined filling rate is:

Rate${}_{P+Q}$ = Rate${}_P$ + Rate${}_Q$

Rate${}_{P+Q}$ = $\frac{1}{10}+\frac{1}{6}$ tank/minute

To add these fractions, we find a common denominator, which is 30:

Rate${}_{P+Q}$ = $\frac{1\times 3}{10\times 3}+\frac{1\times 5}{6\times 5}$ = $\frac{3}{30}+\frac{5}{30}$ = $\frac{8}{30}$ tank/minute

Simplifying the fraction:

Rate${}_{P+Q}$ = $\frac{4}{15}$ tank/minute

Net Rate of Change

Let the total capacity of the storage tank be $C$ litres.

We can express the rates in litres per minute:

- Rate${}_P$ = $\frac{C}{10}$ litres/minute
- Rate${}_Q$ = $\frac{C}{6}$ litres/minute
- Combined filling rate (P+Q) = $\frac{4C}{15}$ litres/minute
- Rate${}_R$ = $34$ litres/minute

When all three pipes operate simultaneously, the net rate is the combined filling rate minus the draining rate:

Net Rate = Rate${}_{P+Q}$ - Rate${}_R$

Net Rate = $(\frac{C}{10}+\frac{C}{6})-34$ litres/minute

Net Rate = $\frac{4C}{15}-34$ litres/minute

Calculating Tank Capacity

The problem states that it takes 1 hour (which is 60 minutes) to completely empty a full storage tank when all pipes are operating. This implies that the draining rate is greater than the combined filling rate, and the net effect is the tank emptying.

The rate at which the tank empties is the total capacity divided by the time taken to empty it:

Emptying Rate = $\frac{\text{Capacity}}{\text{Time to Empty}}$

Emptying Rate = $\frac{C}{60}$ litres/minute

This emptying rate is equal to the net rate of change when the tank is emptying. Therefore, we can set up the equation:

Emptying Rate = Rate${}_R$ - Rate${}_{P+Q}$

$\frac{C}{60}=34-(\frac{C}{10}+\frac{C}{6})$

We already calculated the combined filling rate $\frac{C}{10}+\frac{C}{6}$ as $\frac{4C}{15}$. Substituting this into the equation:

$\frac{C}{60}=34-\frac{4C}{15}$

Now, we need to solve this equation for $C$. First, let's move the term with $C$ to the left side:

$\frac{C}{60}+\frac{4C}{15}=34$

To add the fractions on the left side, we find a common denominator, which is 60:

$\frac{C}{60}+\frac{4C\times 4}{15\times 4}=34$

$\frac{C}{60}+\frac{16C}{60}=34$

Combine the terms on the left:

$\frac{C+16C}{60}=34$

$\frac{17C}{60}=34$

To find $C$, multiply both sides by 60 and divide by 17:

$C=\frac{34\times 60}{17}$

Simplify the expression:

$C=2\times 60$

$C=120$

Conclusion

The capacity of the storage tank is 120 litres.