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A boat traveled upstream a distance of 90 miles at an

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Director
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New post 01 Jun 2007, 13:04
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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 an hour longer than the trip downstream, how many hours did it take the boat to travel downstram ?

a. 2.5
b. 2.4
c. 2.3
d. 2.2
e. 2.1

I will provide the OA soon.
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New post 01 Jun 2007, 13:31
Yes, but how did you arrive to the answer ? help please
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New post 01 Jun 2007, 13:45
First eq. (v-3)t=90 t-time up the stream
Second eq. (v+3)(t-0.5)=90 - downstream

t=90/(v-3) from first eq.

Plug into the second equation and get v=33

Then t=90/(33-3)=3

time downstream t-30 min or 2 and 1/2 hours
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New post 01 Jun 2007, 13:55
*Note

On the GMAT, I wouldn't try to solve this problem - it would be too time-consuming.

Just start plugging in each answer. If it comes down to solving quadratic equations and stuff like that - you must have taken the wrong approach.

Just my opinion!
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New post 01 Jun 2007, 15:14
yeah, can/will someone work thru this one?

U
(v-3)(3t/2) = 90

D
(v+3)t=90 => t=90/v+3

When I try to substitute t into (v-3)(3t/2) = 90 it becomes messy.
  [#permalink] 01 Jun 2007, 15:14
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