Let the rate at which pipe A fills the tank be $x$ litres/hour. Since pipe B drains the tank at 8 litres/hour, the effective rate at which the tank is being filled is:

$\text{Net rate} = x - 8 \, \text{litres/hour}.$

Given that the tank has a capacity of 500 litres and it takes 80 hours to fill the tank, we can write the equation:

$\text{(Net rate)} \times \text{(Time)} = \text{Total capacity}.$

Substituting the known values:

$(x - 8) \times 80 = 500.$

Simplifying the equation:

$80x - 640 = 500.$

Now, solve for $x$:

$80x = 500 + 640,$ $80x = 1140,$ $x = \frac{1140}{80} = 14.25.$

So, the value of $x$ is 14.25 litres/hour.

Would you like more details on this solution?

Here are 5 related questions:

1. What happens if the draining rate of pipe B increases?
2. How would the total time change if the capacity of the tank doubled?
3. What is the total amount of water drained by pipe B in 80 hours?
4. What would be the equation if the draining rate of pipe B were unknown?
5. How would the solution change if pipe B was closed halfway through?

Tip: Always ensure units are consistent when solving rate problems involving time and volume.