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Well done Shaan from Watford Grammar School
and many others who sent in well-explained correct solutions for
this problem.
I noticed that neither the distance nor the journey time were
given.
Time there was
d / 50
Time back was
d / 70
The Average speed for the round trip was total distance divided by
total time
The d cancels, indicating that the average
speed does not depend on the distance so long as there and back are
equal distances, and the answer is 58.3 (recurring) mph. which
isn't sixty !
If the average speed is to be 60, I
need to find v so that
:
which solves to give a
v value of 75
And so 75 mph must be the return average speed if the overall
average speed for the round trip is to be 60 mph.
Gemma in Suffolk added this thought :
The reason the average of the average speeds isn't the overall
average speed is because the journey time isn't the same there and
back . . . . but the distance is, and that's useful if instead I
use the reciprocal of the rate.
Speed is a rate (the rate at which distance changes as time
passes) and any rate can be considered in it's reciprocal form (the
rate at which time passes per unit distance)
The distance is the same so adding the reciprocal of the speeds
and then dividing by two does give me the reciprocal of the overall
average speed.
and this gives the answer 75 mph
too.