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

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Director
Joined: 30 Nov 2006
Posts: 591
Location: Kuwait
Followers: 12

Kudos [?]: 169 [0], given: 0

A boat traveled upstream a distance of 90 miles at an [#permalink]  01 Jun 2007, 13:04
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.
Director
Joined: 30 Nov 2006
Posts: 591
Location: Kuwait
Followers: 12

Kudos [?]: 169 [0], given: 0

Yes, but how did you arrive to the answer ? help please
Manager
Joined: 22 May 2007
Posts: 121
Followers: 1

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

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
Manager
Joined: 22 May 2007
Posts: 121
Followers: 1

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

*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!
Director
Joined: 12 Jun 2006
Posts: 536
Followers: 1

Kudos [?]: 29 [0], given: 1

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.
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