a car travels 50kph from point a to b and travels 40kph back. what is the average

Here's the "proof" of this......

Total Distance / Total Time = Average Rate

Let D be the one-way distance....then, the total distance = 2D

And the total time is given by    D/R₁ + D/R₂   ......so we have.....

[2D] / [ D/R₁ + D/R₂ ]  =

[2D] / [D (R₂ + R₁)/ (R₁R₂)] =

[2(R₁R₂)] / (R₂ + R₁) ........note that this doesn't depend on "D"

Please post the complete question.

This will be the same answer no matter how far it is so to make life easy I am going to say it is 200km

At 50km/h it will take 4 hours

At 40km/hour it will take 5 hours

That is 9 hours altogether to travel 400km

So the average speed it 400/9  km/hour =    $$44\frac{4}{9}\;\;km/hour$$  

It's always easy to find the average speed in these situations......here's the "formula"...

[2* R₁* R₂] / [R₁ + R₂ ]   ......where R₁ and R₂ are the two different rates.......

I didn't know that Chris.  Thank you.   I'll have to think about it more when I have more time!  :)

$$\\(2* R1* R2) / (R1 + R2 ) \\\\
=\dfrac{\frac{2xy\; km^2}{h^2}}{\frac{x\;km}{h}+\frac{y\;km}{h}}\\\\\\
=\dfrac{\frac{2xy\; km^2}{h^2}}{\frac{x+y\;km}{h}}\\\\\\
=\frac{2xy\; km^2}{h^2}\div \frac{x+y\;km}{h}\\\\
=\frac{2xy\; km^2}{h^2}\times \frac{h}{x+y\;km}\\\\
=\frac{2xy\; km}{h}\times \frac{1}{x+y}\\\\
=\frac{2xy\; km}{(x+y)h}\\\\
=\frac{2xy}{(x+y)}\frac{km}{h}\\\\$$

Umm.  Not a great deal of help although at least the units are right.  

I will have to think about it some more.  

Yep got it.  Thanks Chris.