A puzzle.

A man on a bike pedals from point A to point B - against the wind - and completes the journey in five minutes.

He then turns around and pedals back to point A - this time with a following wind - and returns in three minutes.

How long would it take him to travel from A to B (or B to A) without any wind at all?

8 minutes?

all things being equal, it seems obvious that it would take 4 mins to peddle from point a to point b.

but if you didn't mention that the wind was very light, and had no bearing on his journey time, or that point b is at the top of a hill, or he stopped to get a paper on his way to point b, or..... well, you see where it could go.

No, mate.[edit}(monqui)[edit]

It takes him eight minutes to travel both legs - with and without the wind.

One leg only, no wind.

I agree with fishy that it should be 4 minutes to travel from A to B without wind and assuming there is no incline.

You said 5 mins against the wind, 3 mins with the wind...

Assuming that the trip was on a flat road and the biker didn't stop,
you could assume that its 4 mins without the wind per leg...

Crap. I suck at life.

One leg only, no wind.

I think that if he has lost a leg, it will take 16 minutes to peddle single-footed.

[edit](assuming that the number of limbs that one has will increase a person's biking power exponentially)[/edit]

Agreed, Fishy. Sorry.

Assume that points A and B are on the same level - no hills - the wind is either face on or directly behind and is constant. His paper is delivered. Ignore friction and all that cobblers, it's not a trick question, honest.

Hint: it isn't four.

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<TD style="FONT-SIZE: 11px; COLOR: gold" bgColor=black>? posted by Biological Component</TD>
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One leg only, no wind.
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I think that if he has lost a leg, it will take 16 minutes to peddle single-footed. </TD></TR></TBODY></TABLE></TD></TR></TBODY></TABLE>Nice one Bio.

Well, I know what you are going after... as the wind doesn't speed you
up as fast as it pushes you back... but I don't recall what the exact
answer is...

this isn't about some wind dynamics equation, and how the faster you go, the more resistance there is, is it?

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<TD bgColor=#151515>this isn't about some wind dynamics equation, and how the faster you go, the more resistance there is, is it?</TD></TR></TBODY></TABLE></TD></TR></TBODY></TABLE>No. No fancy maths. It's more of a logical problem, really

3 minutes 45 seconds?

*Edit:

Let AB=distance from A to B, x=man's speed, y=wind speed. We need to find time with no wind at all, i.e. AB/x

Against wind, AB/(x-y)=5
With the wind, AB/(x+y)=3

Therefore,
(x-y)/AB=1/5
(x+y)/AB=1/3

Adding,
2x/AB=1/5+1/3=8/15

Therefore,
AB/x = 30/8 = 3.75 = 3m 45s

if you're saying that it takes 8mins in total, whether there is wind or not, then half the distance=half the time. so unless you factor in something not already mentioned, like slowing down to stop if there's a return journey involved, then i dont see how it isn't four mins.

does he cross any timezones?

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if you're saying that it takes 8mins in total, whether there is wind or not, then half the distance=half the time. so unless you factor in something not already mentioned, like slowing down to stop if there's a return journey involved, then i dont see how it isn't four mins.

does he cross any timezones?

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I'm not saying that, though. There is a wind. The question is: What if there wasn't?

how... puzzling...

/wanders off pondering the meaning of 'wind'...

hehe, good job fraggard.i knew someone had said 3.45, and i looked back to see who it was, and found that you'd dropped your scrabble set.

i realised that if the wind was slowing your speed in mph by as much as it was increasing it, then you wouldn't get the same amount of extra milage on the return journey as there are less h's.

with an random distance of 3000m, the outward journey would be at a speed of 10m/s, and the return would be at 16.666m/s. so the average speed, being 13.333m/s, would cover the distance of 3000m in 225 secs, or 3.45mins.

i got there in the end :razz:

Simple explanation for those who are still confused: with wind, you are travelling slow for a longer period of time than you are going fast.

hehe, good job fraggard.i knew someone had said 3.45, and i
looked back to see who it was, and found that you'd dropped your
scrabble set.

i realised that if the wind was slowing your speed in mph by as much
as it was increasing it, then you wouldn't get the same amount of extra
milage on the return journey as there are less h's.

with an random distance of 3000m, the outward journey would be at a
speed of 10m/s, and the return would be at 16.666m/s. so the average
speed, being 13.333m/s, would cover the distance of 3000m in 225 secs,
or 3.45mins.

i got there in the end :razz:

thats what i said

and I still couldn't remember it

Didn't we have a thread for these sorta things a few months ago?

Here is an example that illustrates how simply halfing it doesnt work:

Pretend that the one-way trip normally takes 1 min. Now imagine that there is a huge wind practically as powerful as your biking so that the trip takes 1000 minutes. It doesn't matter how fast the wind pushes you back, even if you teleported back your net time would still be 1000 minutes - wayyyy longer than the standard 2 minutes with no wind.

All hail Mathmeister Fraggard.

Sorry, mate, I missed your first post. Impressive explanation, too.

Did you do advanced matsh in college/uni fraggard?

$loth said:
Did you do advanced matsh in college/uni fraggard?

I think you can figure this much out with high school math $loth :smile: and a bit of physics if you count the speed=distance/time part.

*Edit: er... d'oh. I just realised I never answered the actual question Anyway, I've done 4 math courses in 4 semesters so far, 6 if you include discrete math and combinatorics. </freak>