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Tricky race and distance question

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Retired Moderator
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Tricky race and distance question [#permalink]

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New post 17 Aug 2006, 09:58
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Hi everyone i wonder if any one can help me with this one????

Three runners A, B and C run a race, with runner A finishing 12 meters ahead of runner B and 18 meters ahead of runner C, while runner B finishes 8 meters ahead of runner C. Each runner travels the entire distance at a constant speed.

What was the length of the race?

(1) 36 meters (2) 48 meters
(3) 60 meters (4) 72 meters

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New post 17 Aug 2006, 11:03
Is there any other information? It seems as though time or rates are missing.

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Schools: Completed at SAID BUSINESS SCHOOL, OXFORD - Class of 2008
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New post 17 Aug 2006, 11:26
48 meters.

Let x = total distance
In the time A finished the race, B covered x-12 and C covered x-18

x/A = (x-12)/B = (x-18)/C

B/C = (x-12)/(x-18)....EQ1

In the time B finished the race, C covered x-8

x/B = (x-8)/C
B/C = x/(x-8).........EQ2

From EQ1 and EQ2 we have
(x-12)/(x-18) = x/(x-8)

x^2 +96 - 20x = x^2 - 18x
x = 48
_________________

SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008

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New post 17 Aug 2006, 12:01
I'm surely not ready for QS.

Good one PS!

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 [#permalink]

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New post 18 Aug 2006, 02:27
OA is 48.........Great Dahiya .... i did it as back solving from the answer choices. but your answer is wut i was exactly looking for but i couldn,t put into action.

Thanks

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  [#permalink] 18 Aug 2006, 02:27
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