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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 Jun 2007, 03:04
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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 (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?
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 [#permalink] New post 23 Jun 2007, 04:42
Upstream:
D = 90 m
S= (v-3) m/hr
Let T = t

Downstream:
D=90
S=(v+3) m/hr [let this be statement 1]
T = t-1/2 = (2t-1)/2 [let this be statement 2]

D=ST
Since D is the same upstream & downstream:
=> t(v-3) = (v+3)(2t-1)/2
=> 2tv-6t = 2tv+6t-v-3
=> v+3=12t or t=(v+3)/12 [let this be statement 3]

Using the value of t from statement 3 in statement 2 we get T=(V-3)/12
D= S * T
=> 90= (v+3)(v-3)/12
=>v^2=1089
=>v=33 [let this be statement 4]

Therefore, when going downstream:
D= 90
S = v+3 = 33+3 = 36 m/hr
T = 90/36 = 2.5 hrs.

Not sure if the above is correct. Even if it is, I'm sure there's a much simpler way to solve this.
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 [#permalink] New post 23 Jun 2007, 06:14
GK_Gmat wrote:
Upstream:
D = 90 m
S= (v-3) m/hr
Let T = t

Downstream:
D=90
S=(v+3) m/hr [let this be statement 1]
T = t-1/2 = (2t-1)/2 [let this be statement 2]

D=ST
Since D is the same upstream & downstream:
=> t(v-3) = (v+3)(2t-1)/2
=> 2tv-6t = 2tv+6t-v-3
=> v+3=12t or t=(v+3)/12 [let this be statement 3]

Using the value of t from statement 3 in statement 2 we get T=(V-3)/12
D= S * T
=> 90= (v+3)(v-3)/12
=>v^2=1089
=>v=33 [let this be statement 4]

Therefore, when going downstream:
D= 90
S = v+3 = 33+3 = 36 m/hr
T = 90/36 = 2.5 hrs.

Not sure if the above is correct. Even if it is, I'm sure there's a much simpler way to solve this.


Your approach is perfect and answer is absolutely correct. But what "I" or "we" need is a fast approach to be able to solve it within 2 minutes!

Anyone with some fast approach?
  [#permalink] 23 Jun 2007, 06:14
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