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# Hard GMATPREP question - boat going upstream

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Senior Manager
Joined: 04 Aug 2008
Posts: 369
Hard GMATPREP question - boat going upstream [#permalink]

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19 Oct 2008, 17:42
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I encountered this insanely hard question while i was doing gmatprep this morning. Spend 10 freakin minutes trying to figure it out and ended up guessing.

When i reviewed this later on, i found a very complicated algebraic solution to this problem which does not pay off in 2 minutes. (better to guess then kill the clock)

My question is - can anyone think of an easy way of solving this in less then 2 mins

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Senior Manager
Joined: 21 Apr 2008
Posts: 265
Location: Motortown
Re: Hard GMATPREP question - boat going upstream [#permalink]

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19 Oct 2008, 18:06
A

90/(v-3) = 90/(v+3)+1/2
180v+180*3 = 180v - 180*3 +v^2 - 9
v^2 = 1089
v = 33
down stream : 90/36 = 10/4 = 2.5
Intern
Joined: 14 Jul 2008
Posts: 12
Re: Hard GMATPREP question - boat going upstream [#permalink]

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19 Oct 2008, 19:42
Agree-- A!

(V-3)(T+1/2)=(V+3)T=90
Current Student
Joined: 28 Dec 2004
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Location: New York City
Schools: Wharton'11 HBS'12
Re: Hard GMATPREP question - boat going upstream [#permalink]

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19 Oct 2008, 20:49
spiridon wrote:
I encountered this insanely hard question while i was doing gmatprep this morning. Spend 10 freakin minutes trying to figure it out and ended up guessing.

When i reviewed this later on, i found a very complicated algebraic solution to this problem which does not pay off in 2 minutes. (better to guess then kill the clock)

My question is - can anyone think of an easy way of solving this in less then 2 mins

easiest way is to plug numbers and i actually did it under 2 mins...with plugging numbers.now.. if you want to do the math way..here is how..

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

90[6/(v^2-9)]=1/2

90*6*2=v^2-9

90*6*2+9=v^2

9(10*12+1)=v^2

9(121)=v^2; notice 121=11^2

3^2 * 11^2=v^2 ; v=33

now that you now v, 90/36 =2.5 hrs..
Senior Manager
Joined: 18 Jun 2007
Posts: 279
Re: Hard GMATPREP question - boat going upstream [#permalink]

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19 Oct 2008, 20:54
A it is.

90/(v-3) - 90/(v+3) = .5
=> 90v+270-90v+270/(v+3)(v-3)= .5
=>540/ v^2-3^2 = .5
=>540/.5 = v^2 - 9
=>1080 = v^2 - 9
=> v^2 = 1089
=> v = 33

now you can see that downstream the speed will be 36
and total time for down stream will be 90/36 = 2.5

I made a calculation mistake and this question sent me to a spin, took 3.5 min. eventually
but as I recognise that silly mistake I knew that this question is qorth only 1 min.
I hope in real test I wont do such mistakes
Senior Manager
Joined: 04 Aug 2008
Posts: 369
Re: Hard GMATPREP question - boat going upstream [#permalink]

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19 Oct 2008, 22:02
yeah i made some silly mistake too and was circling around

i was hoping theres some shortcut to this prob but i guess not

also i wont be able to find sqrt of eg 1089 quickly on the test
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Senior Manager
Joined: 18 Jun 2007
Posts: 279
Re: Hard GMATPREP question - boat going upstream [#permalink]

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19 Oct 2008, 22:40
spiridon wrote:
yeah i made some silly mistake too and was circling around

i was hoping theres some shortcut to this prob but i guess not

also i wont be able to find sqrt of eg 1089 quickly on the test

well the problem in my case wasnt the square.
Anyway finding the square is not that difficult. The best approach is to find a simple perfect square near that number. and then take the last digit into consideration.
take the example of 1089:
the simple perfect square near 1000 is 900(30^2)
so you know that your figure is in theproximity of 30.
Now you see that the last digit is 9 so if 1089 is a perfect square, it has to be 33.
Now all you need is a simple calculation to check.
VP
Joined: 30 Jun 2008
Posts: 1018
Re: Hard GMATPREP question - boat going upstream [#permalink]

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20 Oct 2008, 09:56
fresinha12 wrote:
spiridon wrote:
I encountered this insanely hard question while i was doing gmatprep this morning. Spend 10 freakin minutes trying to figure it out and ended up guessing.

When i reviewed this later on, i found a very complicated algebraic solution to this problem which does not pay off in 2 minutes. (better to guess then kill the clock)

My question is - can anyone think of an easy way of solving this in less then 2 mins

easiest way is to plug numbers and i actually did it under 2 mins...with plugging numbers.now.. if you want to do the math way..here is how..

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

90[6/(v^2-9)]=1/2

90*6*2=v^2-9

90*6*2+9=v^2

9(10*12+1)=v^2

9(121)=v^2; notice 121=11^2

3^2 * 11^2=v^2 ; v=33

now that you now v, 90/36 =2.5 hrs..

Can you show how you plugged numbers for this problem

Thanks
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VP
Joined: 17 Jun 2008
Posts: 1322
Re: Hard GMATPREP question - boat going upstream [#permalink]

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20 Oct 2008, 16:18
LiveStronger wrote:
A

90/(v-3) = 90/(v+3)+1/2
180v+180*3 = 180v - 180*3 +v^2 - 9
v^2 = 1089
v = 33
down stream : 90/36 = 10/4 = 2.5

I agree with this soln simplest and fastest ... i used the same !!!!

--== Message from GMAT Club Team ==--

This is not a quality discussion. It has been retired.

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.

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Re: Hard GMATPREP question - boat going upstream   [#permalink] 20 Oct 2008, 16:18
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