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# rate/time/distance problem

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Intern
Joined: 28 Jun 2008
Posts: 45

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04 Jun 2009, 21:20
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Hi guys, was wondering if anyone could offer a good explanation on how to solve this problem:

A boat traveled upstream a distance of 90 miles at an average speed (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 downstream?

A. 2.5
B. 2.4
C. 2.3
D. 2.2
E. 2.1

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Manager
Joined: 12 Apr 2006
Posts: 210
Location: India

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05 Jun 2009, 02:25
1
KUDOS
Here are the steps.

In question we have 2 unknown variables. V in rate and time T. But if looked carefully it says the differance in average speed between upstream and downstream is 6.

I choose downstream speed as x miles/hour
Now upstream speed is x-6 miles/hour

For downstream, use Rate x Time = Distance which is x*t = 90 => x = 90/t ------------(I)

Now for upstream, (x-6)*(t+1/2) = 90 -----------------(II)

Use value of x from (I) in (II)

(90/t-6)*(t+1/2) = 90

6t^2 + 3t - 45 = 0
On solving this gives

t = 2.5 and -3. Time can't be in -ve so 2.5 is the answer.
Senior Manager
Joined: 16 Jan 2009
Posts: 349
Concentration: Technology, Marketing
GMAT 1: 700 Q50 V34
GPA: 3
WE: Sales (Telecommunications)

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05 Jun 2009, 02:26
1
KUDOS
U D
R (v-3) (v+3)
T t+30 t
D 90 90

(v-3)(t+0.5)=(v+3)t
vt+0.5v-3t-1.5=vt+3t
0.5v-3t-1.5=3t
v/2-6t=3/2
v=3+12t

using downstream eqn.
(3+12t+3)t=90
6t+12t^2-90=0
?????
what nextt.....
OA / explanation plase
_________________

Lahoosaher

Senior Manager
Joined: 16 Jan 2009
Posts: 349
Concentration: Technology, Marketing
GMAT 1: 700 Q50 V34
GPA: 3
WE: Sales (Telecommunications)

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05 Jun 2009, 02:43
1
KUDOS
Hey humans,

Thanks for the explanation .....I missed the difference in the avg speed.

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If you would like to discuss this question please re-post it in the respective forum. Thank you!

To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative | Verbal Please note - we may remove posts that do not follow our posting guidelines. Thank you.

_________________

Lahoosaher

Re: rate/time/distance problem   [#permalink] 05 Jun 2009, 02:43
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