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A boat traveled upstream a distance of 90 miles at an

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Manager
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Joined: 13 Sep 2006
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Schools: Olin Business School - Washington University
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A boat traveled upstream a distance of 90 miles at an [#permalink] New post 23 Sep 2006, 22:01
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A boat traveled upstream a distance of 90 miles at an average speed of (V-3) miles per hour and then traveled the same distance downstream at an average speed of (v+3) miles per hour. If the trip upstream took half hour longer than the trip downstream, how many hours did it take the boat to travel downstream?

A) 2.5
B) 2.4
C) 2.3
D) 2.2
E) 2.1

Can someone please explain?
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Director
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 [#permalink] New post 24 Sep 2006, 01:51
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Hmm.

90/(v-3) - 90/(v+3) = 1/2
=> 180* (v + 3 - v + 3) = v^2 - 9
=> 1089 = v^2
=> v = 33 mph

To travel downstream, time = 90/(v+3) = 90/36 = 2.5 hr

Answer: A
Intern
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 [#permalink] New post 24 Sep 2006, 07:25
aewsome paddy,
how did u get idea to subtract it ??

I tried average speed, eq etc.. but this is like best..
Intern
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 [#permalink] New post 25 Sep 2006, 01:24
The question says that it took additional half an hour for Boat to go upstream ,hence T(Upstream)=90/(v-3) must be subtracted from T2(DS)=90/(V+3)

hence as result ,the equation equates to 1/2

hope this helps
Manager
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Joined: 13 Sep 2006
Posts: 198
Schools: Olin Business School - Washington University
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 [#permalink] New post 25 Sep 2006, 09:05
This is a damn tough question to me...nice job guys. OA is A...
  [#permalink] 25 Sep 2006, 09:05
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