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Re: John is biking for his trip ... [#permalink]
is it an official question ? ... I did not understand the last past, would you plx explain?
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John is biking for his trip ... [#permalink]
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greenmonomer wrote:
is it an official question ? ... I did not understand the last past, would you plx explain?

Hi there
This is made up problem to stimulate thinking and maybe, add on a flavor to the classical one, "what is the average speed of a car driven half of the route with the speed of 60 km/hour and the other half with the speed of 50 km/hour?". Solution would be accomplished either by introducing d (distance) in numerator and denominator and then cancelling of d for the numerator and the denominator or just applying the weighted average shortcut (1/60)*(1/2)+(1/50)*(1/2) to find the reciprocal of this sum (remember, time/rate are inversely related) for the average speed.

Here, a gist of the question's last part is that we want to apply the average for changing speed itself too. Given the speed of 15 miles/hour is uniformly changing, i.e. decreasing (it's equally spaced interval) till 1 mile/hour, we should apply the average rate of (15+1)*1/2 to find the requried speed, and it would be applied a weight of 1/2 for the remaining half-part of the original route.

Hope this helps.
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Re: John is biking for his trip ... [#permalink]
thank u so much ...
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Re: John is biking for his trip ... [#permalink]
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