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# Efficient approaches are welcome

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Director
Joined: 29 Aug 2005
Posts: 854

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21 Jul 2009, 12:01
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Efficient approaches are welcome
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Manager
Joined: 03 Jul 2009
Posts: 103
Location: Brazil

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21 Jul 2009, 13:42
Rearranging the expression we have
$$p < (r^2 + 2) /r$$

1) Use "picking numbers":
For r=10, the expression is 10,2. It is major
For r=-10, the expression is -10,2. It is minor.
Therefore NOT SUFFICIENT

NOT SUFFICIENT

Together, it seems to work, but if you pick the number 1/10...
For r=1/10, the expression is 20.1. It is major
Therefore with the two expressions it is possible to verify the validity. C

PS.: When using picking numbers, remember to use positives, negatives and fractions between 0 and 1.

Last edited by coelholds on 22 Jul 2009, 05:31, edited 2 times in total.
GMAT Tutor
Joined: 24 Jun 2008
Posts: 1346

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21 Jul 2009, 16:57
Using Statement 1, we can rephrase the question (replacing p with r). Notice that we can multiply on both sides by r^2 + 2 since we know r^2 + 2 must be positive. We can't multiply by r, since we don't know if r is positive or negative (unless we use Statement 2):

Is 1/r > r/(r^2 + 2) ?
Is (r^2 + 2)/r > r ?
Is (r^2/r) + (2/r) > r ?
Is r + (2/r) > r ?
Is 2/r > 0 ?
Is r > 0 ?

So using Statement 1 alone, the question is equivalent to "Is r > 0?", something we only know for certain if we use Statement 2. So the answer is C - both statements are required.
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Manager
Joined: 15 Apr 2008
Posts: 50
Location: Moscow

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21 Jul 2009, 22:34
Agree with C:
Knowing, that p=r and r>0, we can rewrite the equation: $$\frac{1}{p}>\frac{r}{r^2+2}?$$ -> $${1}>\frac{r*p}{r^2+2}?$$ -> $${1}>\frac{r^2}{r^2+2}?$$; $$r^2+2$$ is always bigger than $$r^2$$, then $$\frac{r^2}{r^2+2}$$ will always be less than 1.
Manager
Joined: 03 Jul 2009
Posts: 103
Location: Brazil

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22 Jul 2009, 05:32
I had did a mistake, but after seeing the explanation of IanStewart, I corrected.
Re: GMATPrep2: number prop   [#permalink] 22 Jul 2009, 05:32
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