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This question already has an answer here:

On my way to the supermarket, I walk at a certain speed. I want my average speed for the back and forth trip to be twice that speed. At what speed should I go back home to reach that goal?

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marked as duplicate by Gareth McCaughan, Engineer Toast, f'', Kate Gregory, A E Aug 10 '16 at 15:31

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You can't reach that average speed any more.
You already spent all the time you would have for both trips if you were going at double speed.

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  • $\begingroup$ Can you elaborate on that? Is it because to have double average speed I must travel double the distance in the same amount of time? Is there a way of expressing the answer somewhat like that: To have double average speed, you must halve the amount of time .... of what? $\endgroup$ – plalud Aug 10 '16 at 11:51
  • $\begingroup$ "Is it because to have double average speed I must travel double the distance in the same amount of time?" Yes, that's correct. You can try to calculate the the time that you have for a return trip to reach a certain average speed and use that to get the speed for return trip, but in your example the remaining time for return trip is 0, so you end up dividing by 0 when you try to get the required speed, which is not defined. Therefore there is no solution (Amruth did that). I think that my answer is a better intuitive explanation of why. $\endgroup$ – Meiffert Aug 10 '16 at 12:04
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Solution:

Let x be distance between home and market , t1 time for going forward and t2 for coming back.

$$x/t1=1p kmph$$
p is natural number

$$2x/(t1+t2)=2p kmph $$

Divide both eqn => $$ (t1 +t2)/(t1) =1 $$
**but t2 becomes 0 , so not possible solution.

find $$x/t2$$
answer is not possible..

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  • $\begingroup$ Nice proof. Could you elaborate on the intuition of the result if such an intuition you have? $\endgroup$ – plalud Aug 10 '16 at 11:49
  • $\begingroup$ @plalud I think may be memorizing the formula would have helped.. $\endgroup$ – Amruth A Aug 10 '16 at 11:53

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