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

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Manager
Joined: 13 Sep 2006
Posts: 198
Schools: Olin Business School - Washington University
Followers: 1

Kudos [?]: 16 [0], given: 0

A boat traveled upstream a distance of 90 miles at an [#permalink]  23 Sep 2006, 22:01
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0% (00:00) correct 0% (00:00) wrong based on 0 sessions
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

Director
Joined: 06 May 2006
Posts: 781
Followers: 3

Kudos [?]: 16 [1] , given: 0

1
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

Intern
Joined: 01 Jan 2006
Posts: 28
Followers: 0

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

how did u get idea to subtract it ??

I tried average speed, eq etc.. but this is like best..
Intern
Joined: 27 Aug 2006
Posts: 32
Followers: 0

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

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
Joined: 13 Sep 2006
Posts: 198
Schools: Olin Business School - Washington University
Followers: 1

Kudos [?]: 16 [0], given: 0

This is a damn tough question to me...nice job guys. OA is A...
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