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

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Intern
Joined: 20 Jul 2007
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A boat traveled upstream a distance of 90 miles at an avg [#permalink]  20 Jul 2007, 08:31
A boat traveled upstream a distance of 90 miles at an avg speed of (v-3) mph and then traveled the same distance downstream at an avg speed of (v+3) miles per hour. If the upstream took 1/2 an hour longer than the trip downstream, how many hours did it take the boat to travel downstream?

2.5
2.4
2.3
2.2
2.1
Manager
Joined: 22 May 2007
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[#permalink]  20 Jul 2007, 08:51
You could try solving it, but there is no point. By substituting you can easily get 2.5.

Any other strategies?

ps. The first number that I would plug in would be A (2.5) because when you add 1/2 hour to it, you would get 3 hours - all good numbers to work with 90.
Senior Manager
Joined: 29 Nov 2006
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[#permalink]  20 Jul 2007, 09:23
Same as Hayabusa's way, but showing the setup:

Set up a table:
Upstream Downstream
R V-3 V+3
T T+0.5 T
D 90 90

Simplify:
Upstream Downstream
R V-6 V
T T+0.5 T
D 90 90

Start plugging in numbers for T to find V. Verify that they match the downstream calculations.
Intern
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[#permalink]  20 Jul 2007, 10:07
i did it algebraically, was wondering if there was a quicker way.

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

solving first for v and plugging into the second

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

cleaning up the fractional eqtn:

90(x - 1/2) +6(x)(x - 1/2) = 90(x)

6x2 - 3x +45 = 0
(2x + 5)(x - 3) = 0
x = 3
x - 1/2 = 2.5

was looking for a simpler way that maybe utilized the difference in the avg speed maybe and the 1/2 hour difference.
Manager
Joined: 22 May 2007
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[#permalink]  20 Jul 2007, 10:15
On the actual exam, you don't have all the time in the world. These problems can be solved by looking at the possible choices.

If you can set up and solve a bunch of quadrtic equations in less than a min. - more power to you
[#permalink] 20 Jul 2007, 10:15
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# A boat traveled upstream a distance of 90 miles at an avg

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