A man walks up a hill at 2mph, and back down the same way at 4mph. Upon returning to where he started, what is his average speed?


closed as off-topic by manshu, ABcDexter, Carl Löndahl, AeJey, Hugh Meyers Mar 4 '16 at 12:14

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  • "This question is off-topic as it appears to be a mathematics problem, as opposed to a mathematical puzzle. For more info, see "Are math-textbook-style problems on topic?" on meta." – manshu, ABcDexter, Carl Löndahl, AeJey, Hugh Meyers
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  • 2
    $\begingroup$ I vote to close this question as math problem $\endgroup$ – manshu Mar 4 '16 at 11:14
  • $\begingroup$ And the answer is 2.67 mph...if someone is wondering.. $\endgroup$ – manshu Mar 4 '16 at 11:19
  • $\begingroup$ Answered, and voted to close. $\endgroup$ – ABcDexter Mar 4 '16 at 11:22
  • $\begingroup$ So should this be posted on the Mathematics StackExchange? Seems a bit trivial for that site, given the complexity of some of the topics covered there. $\endgroup$ – Nuclear Wang Mar 4 '16 at 12:45
  • 1
    $\begingroup$ It is not an intuitive result, so I would agree that this is still a puzzle. $\endgroup$ – Trenin Mar 4 '16 at 12:52

The answer is :

$\frac{8}{3}$ mph

Let's take the distance up (and down) the hill as $d$.


Time taken in reaching top of the hill is : $\frac{d}{2}$


Time taken in coming down the hill is : $\frac{d}{4}$

Thus, average speed is: $\frac{total \space distance}{total \space time \space taken}$.

which is:


On simplifying,

it gives $\frac{8}{3}$ or $2.67$ mph (approx).


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