Maths question for anyone to pass the time doing.

[Bilthehut]

In a recent cycle journey, we had travelled at an average of 10 miles per hour, but when we did the return journey over the same route, we managed an average of 15 miles per hour.  What was our average speed over the two journeys?

[Vulture]

In a recent cycle journey, we had travelled at an average of 10 miles per hour, but when we did the return journey over the same route, we managed an average of 15 miles per hour.  What was our average speed over the two journeys?

10 mph and 15 mph....

[Bilthehut]

 Not quite the answer I'm looking for, Vulture.  The average speed (singular answer) over the two journeys.

[Malc]

You mean the two journeys combined?

Without even resorting to my calcumulator, I estimate it at 12.5 miles per hour.

[Diane CBPFC]

I don't believe you got your butt out of your armchair never mind your leg over a bicycle.

[Diane CBPFC]

Sorry Bill - I thought I was talking to Mince.

[Jack]

On the way there, you travelled a distance x miles in a time t hours. x/t = 10
On the way back, you travelled a distance x miles in a time T hours. x/T = 15

Overall, you travelled a distance 2x in a time t+T hours, at an average speed of a miles per hour.
a = 2x/(t+T)

Rearranging the first two equations, t=x/10 and T=x/15

Therefore t+T = (x/10)+(x/15) = (3x/30)+(2x/30) = 5x/30 = x/6

Therefore a = 2x/(t+T) = 2x/(x/6) = 12x/x = 12. The overall average speed was 12 miles per hour.


Yet common sense says the answer should be 12.5 miles per hour.

Oh, the mathematical drama of it all.

[Bilthehut]

Well done Jack. Good explanation and you included the error that quite a few make, such as Malky

[Bilthehut]

Sorry Bill - I thought I was talking to Mince.

Apology accepted. But then the way I've been harassed this week, what's another comment.

[Tom]

10 mph and 15 mph....