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by manju_ej » Tue Aug 05, 2008 7:22 am
A boat travelled upstream a distance of 90 miles at an average speed of (v-3) miles per hour and then travelled 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?
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by sudhir3127 » Tue Aug 05, 2008 7:47 am
Speed = distance/time

use this and set up the 2 equations

For going up the river :
90 = t * (v - 3)

For going down the river :
90 = (t - 0.5) * (v + 3)

solve for the equations.to get = 2.5

hence 2.5 for downstream and 3 hrs for upstream..

hope it helps..

PS: be careful while solving the equation...
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Re: Gprep PS

by manju_ej » Tue Aug 05, 2008 8:11 am
Thanks , the answer is 2.5.
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by vishubn » Tue Aug 26, 2008 6:07 pm
Hi Sudhir ,

I somehow feel/I coudnt solve the equations ! Could u please complete the solution??

Vishu
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by sudhir3127 » Tue Aug 26, 2008 8:17 pm
vishubn wrote:Hi Sudhir ,

I somehow feel/I coudnt solve the equations ! Could u please complete the solution??

Vishu
Distance = Rate * Time, so...

90 = (v-3) * (t+1/2) - since it took half an hour longer to go upstream, and 90 = (v+3) * t.

Expand these out:

90 = vt - 3t + v/2 - 3/2...........................................1
90 = vt + 3t............................................................2

Subtract the first from the second:

0 = 6t - v/2 + 3/2

Solve for t in terms of v:

6t = v/2 - 3/2
t = v/12 - 1/4

Substitute into the second equation so we can solve for t:

90 = vt+3t = v(v/12-1/4)+3(v/12-1/4)
= v^2/12-v/4+v/4-3/4
= v^2/12 - 3/4

v = sqrt((90+3/4)*12)
= sqrt(363/4*12)
= sqrt(1089) = 33

And since t = v/12-1/4:
t = 33/12-1/4 = 33/12-3/12 = 30/12 = 5/2. It took 2 1/2 hours downstream and 3 hours upstream


This would probably take a lot of time ...

i just realised there a better way of doing it as well.....


try this

T2-T1 = 1/2
So (90/v-3) - (90/v+3) = 1/2
1089 = v^2
v = 33

so t=2.5
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by vishubn » Tue Aug 26, 2008 8:49 pm
Thanks Sudhir

Yes indeed i went to on to solve the equation after i posted the req ! But Yes second approach was coool

Vishu
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by CSASHISHPANDAY » Sat Oct 20, 2012 6:52 am
Time taken upstream = 90/v-3
Time taken downstream = 90/v+3

Hence

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

v = 33

Time taken downstream = 90/36 = 5/2 = 2.5 hrs simple
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by Brent@GMATPrepNow » Sat Oct 20, 2012 7:39 am
manju_ej wrote:A boat travelled upstream a distance of 90 miles at an average speed of (v-3) miles per hour and then travelled 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?
I like to begin with a "word equation." We can write:
travel time upstream = travel time downstream + 1/2

Time = distance/rate

So, we can replace elements in our word equation to get:
90/(v-3) = 90/(v+3) + 1/2

Now solve for v (lots of work here)
.
.
.
v = 33

So, travel time downstream = 90/(v+3)
= 90/(33+3)
= 90/36
= 5/2
= 2 1/2 hours

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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