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# A boat traveled upstream a distance of 90 miles at an

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Manager
Joined: 14 Sep 2006
Posts: 197

Kudos [?]: 58 [0], given: 0

Schools: Olin Business School - Washington University
A boat traveled upstream a distance of 90 miles at an [#permalink]

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23 Sep 2006, 23:01
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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

Kudos [?]: 58 [0], given: 0

Director
Joined: 06 May 2006
Posts: 790

Kudos [?]: 39 [1], given: 2

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24 Sep 2006, 02:51
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KUDOS
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

_________________

Uh uh. I know what you're thinking. "Is the answer A, B, C, D or E?" Well to tell you the truth in all this excitement I kinda lost track myself. But you've gotta ask yourself one question: "Do I feel lucky?" Well, do ya, punk?

Kudos [?]: 39 [1], given: 2

Intern
Joined: 01 Jan 2006
Posts: 28

Kudos [?]: 5 [0], given: 0

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24 Sep 2006, 08:25
how did u get idea to subtract it ??

I tried average speed, eq etc.. but this is like best..

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Intern
Joined: 27 Aug 2006
Posts: 32

Kudos [?]: 1 [0], given: 0

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25 Sep 2006, 02:24
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

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Manager
Joined: 14 Sep 2006
Posts: 197

Kudos [?]: 58 [0], given: 0

Schools: Olin Business School - Washington University

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25 Sep 2006, 10:05
This is a damn tough question to me...nice job guys. OA is A...

Kudos [?]: 58 [0], given: 0

25 Sep 2006, 10:05
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