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A boat traveled upstream a distance of 90 miles at an

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A boat traveled upstream a distance of 90 miles at an [#permalink] New post 23 Jul 2006, 08:02
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A boat traveled upstream 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 a half hour longer than the trip downstream, then 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

OPEN DISCUSSION OF THIS QUESTION IS HERE: a-boat-traveled-upstream-90-miles-at-an-average-speed-of-100767.html
[Reveal] Spoiler: OA

Last edited by Bunuel on 10 Feb 2014, 12:34, edited 1 time in total.
Edited the question and added the OA.
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 [#permalink] New post 23 Jul 2006, 08:41
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90/(V-3) = 90/(V+3) + 0.5 hours
=> 90(V+3) = (V-3) [90 + 0.5 V + 1.5]

90V + 270 = (V-3)(183/2 + V/2)
180V + 540 = (V-3)(183+V)= 183V- 549+ V^2 -3V

V^2 - 1089= 0 => V= 33 miles/hr

=> Time downstream = 90/36 = 30/12= 5/2= 2.5 hrs
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 [#permalink] New post 23 Jul 2006, 13:11
90/(v-3)=0.5+90/(v+3)

solve for v=33
Then, Td=90/36=2.5

A
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 [#permalink] New post 26 Jul 2006, 01:39
A

Time taken upstream = 90/v-3
Time taken downstream = 90/v+3

Hence

90/v-3 + 90/v+3 = 1/2

v = 33

Time taken downstream = 90/36 = 5/2 = 2.5 hrs
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Re: GMAT PREP PS Boat (Upstream vs Downstream) [#permalink] New post 24 Jan 2010, 02:51
it took me a while, but i finally got it :)
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Re: GMAT PREP PS Boat (Upstream vs Downstream) [#permalink] New post 24 Jan 2010, 08:50
I forgot to take account for both directions. I add the current on one but forgot to subtract it from the other.
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Re: GMAT PREP PS Boat (Upstream vs Downstream) [#permalink] New post 21 Oct 2011, 10:00
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http://www.beatthegmat.com/boat-traveled-t18353.html
Beautiful solution
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Re: A boat traveled upstream a distance of 90 miles at an [#permalink] New post 10 Feb 2014, 09:07
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Re: A boat traveled upstream a distance of 90 miles at an [#permalink] New post 10 Feb 2014, 12:34
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A boat traveled upstream 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 a half hour longer than the trip downstream, then 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

Trip upstream took \frac{90}{v-3} hours and trip downstream took \frac{90}{v+3} hours. Also given that the difference in times was \frac{1}{2} hours --> \frac{90}{v-3}-\frac{90}{v+3}=\frac{1}{2};

\frac{90}{v-3}-\frac{90}{v+3}=\frac{1}{2} --> \frac{90(v+3)-90(v-3)}{v^2-9}=\frac{1}{2} --> \frac{90*6}{v^2-9}=\frac{1}{2} --> v^2=90*6*2+9 --> v^2=9*(10*6*2+1) --> v^2=9*121 --> v=3*11=33;

Trip downstream took \frac{90}{v+3}=\frac{90}{33+3}=2.5 hours.

Answer: A.

OPEN DISCUSSION OF THIS QUESTION IS HERE: a-boat-traveled-upstream-90-miles-at-an-average-speed-of-100767.html
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Re: A boat traveled upstream a distance of 90 miles at an [#permalink] New post 10 Feb 2014, 21:12
Expert's post
CfaMiami wrote:
A boat traveled upstream 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 a half hour longer than the trip downstream, then 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

OPEN DISCUSSION OF THIS QUESTION IS HERE: a-boat-traveled-upstream-90-miles-at-an-average-speed-of-100767.html


Here is a post discussing how to solve such equations easily: http://www.veritasprep.com/blog/2013/03 ... culations/
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Re: A boat traveled upstream a distance of 90 miles at an   [#permalink] 10 Feb 2014, 21:12
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