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At a certain race, two athlets run around a stadium. The

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stolyar
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At a certain race, two athlets run around a stadium. The [#permalink] New post   29 Jun 2004, 06:32
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At a certain race, two athlets run around a stadium. The first one runs a lap in 6 minutes; the second one—in 10 minutes. If they start running simultaneously, then how long will it take the first athlet to be exacly two laps ahead of the second one?



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Re: At a certain race, two athlets run around a stadium. The [#permalink] New post   29 Jun 2004, 06:54
30 minutes.
First athlete will cover 5 laps in 30 minutes.
Second athlete will cover 3 laps in 30 minutes.
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Re: At a certain race, two athlets run around a stadium. The [#permalink] New post   29 Jun 2004, 20:57
This answer wworks for non integers too!

Speed 1 x time = distance 1
Speed 2 x time = distance 2
n = number of laps

distance 1 - distance 2 = 2 laps

so,

n x laps x 10 - n x laps x 6 = 2 laps (dividing by laps)

10n-6n = 2

4n = 2

n = 0.5 hours = 30 minutes
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Re: At a certain race, two athlets run around a stadium. The [#permalink] New post   29 Jun 2004, 21:16
Alfred wrote:
This answer wworks for non integers too!

Speed 1 x time = distance 1
Speed 2 x time = distance 2
n = number of laps

distance 1 - distance 2 = 2 laps

so,

n x laps x 10 - n x laps x 6 = 2 laps (dividing by laps)

10n-6n = 2

4n = 2

n = 0.5 hours = 30 minutes



Your answer is a little confusing ( though I am not saying its wrong )
You define n = number of laps

and later say n = 30 mins which is the required time !!!

I think you might have missed something here :)

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Re: At a certain race, two athlets run around a stadium. The [#permalink] New post   29 Jun 2004, 21:30
Here's my attempt.

let slower guy make n laps his vel = 1/10 lap/min

then,faster guy makes n+2 laps his vel = 1/6 lap/min

time taken for this is same for both guys.

n/ (1/10) = (n+2)/(1/6)

10n =6n+12

n=3 , slower guy makes 3 laps, takes 3*10 = 30 mins.

The LCM method is much quicker.
LCM of 6 and 10 gives the time when both guys meet at a point and complete an Integer number of laps.

The edit function has come back ......thank god...( or actually the programmer)

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Re: At a certain race, two athlets run around a stadium. The [#permalink] New post   29 Jun 2004, 21:44
30 min.
difference b/w 1/6 and 1/10 is 1/15
1/15*X = 2
X = 30
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Re: At a certain race, two athlets run around a stadium. The [#permalink] New post   29 Jun 2004, 21:59
I was wondering why I got a wrong answer using relative speed.

Initially,I was using relative speed as 10-6=4 ;-)
I declare myself a birdbrain !

Thanks Paul.
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Re: At a certain race, two athlets run around a stadium. The [#permalink] New post   29 Jun 2004, 23:16
My solution:

D1=D2+2 laps

T*1/6=T*1/10 + 2 (T is the time)

T=30



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Re: At a certain race, two athlets run around a stadium. The [#permalink]
29 Jun 2004, 23:16
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