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Director
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If x > 0, then x = A. 3x B. 1 C. x2 D. x E. 1+x2 [#permalink]
 11 Jun 2007, 07:22
Question Stats:
100% (01:16) correct
0% (00:00) wrong based on 0 sessions
If x > 0, then x =
A. 3x
B. 1
C. x2
D. x
E. 1+x2
Last edited by nick_sun on 28 Jun 2008, 02:30, edited 1 time in total.
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Manager
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A, unless there is some subtlety that I'm missing.
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Director
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sludge wrote: A, unless there is some subtlety that I'm missing.
It is not "A". The unofficial answer is D. Could anybody please explain why it is D.
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SVP
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(D) it is 
1/[sqrt(2x)+sqrt(x)]
= 1/[sqrt(2)*sqrt(x)+sqrt(x)]
= 1/[sqrt(x)*(sqrt(2)+1)]
= [1/sqrt(x)] * [1/(sqrt(2)+1)]
= 1/sqrt(x) * (sqrt(2)-1)/[(sqrt(2)-1)*(sqrt(2)+1)]
= 1/sqrt(x) * (sqrt(2)-1)/[ 2 -1 ]
= (sqrt(2)-1) / sqrt(x)
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Director
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Guys, what happens if 'x > 0' not provided. what 'x > 0' means here?
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SVP
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asaf wrote: Guys, what happens if 'x > 0' not provided. what 'x > 0' means here?
We need it because sqrt(x) is in the expression : x has to be positive or nul to make srqt() existing. Also, if x=0, we have 1/0, impossible for the GMAT context
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GMAT Club Legend
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For a moment, I got A because that square root notation appears like a letter 'v' on my screen. (With a variable like 'v', it works out to be A)
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Manager
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Fig,
You solved the problem as if it were an Theorem and you knew where to arrive at the end.
This is an ambiguous question. Thanks for the info
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Director
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LOL yes, the sqrt appears as V which left me blinking !!
its a simple conjucate problem !
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SVP
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vijay2001 wrote: Fig,
You solved the problem as if it were an Theorem and you knew where to arrive at the end.
This is an ambiguous question. Thanks for the info
On such question, I try to simplify the expression by:
> factorizing what I can (ie: 1/sqrt(x))
> pulling up from the denominator to the numerator the constants containing sqrt() (Note that it could be to move a sqrt(x) up also)
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Manager
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these kind of questions (i have forgot their real name) were there in my 9th grade and loved to do them back then. Now when i have seen one such question, i dont know what like after 15 years or so, it took me a few moments to figure out what to do.
anyway, the rules that my teacher told us were, 1)always try to get rid of sqrt from the denominators and 2) use a^2-b^2=(a+b)(a-b) formula.
SO by multiplying and dividing the fraction with sqrt2x+sqrtx, we can easily solve the problem.
and yes, the answer is D.
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Director
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Using x is positive, 1/[v(2x)+vx] = 1/v[2x+x] = 1/v(3x)
Answer: A
This question is too easy to be on the GMAT.
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SVP
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Mishari wrote: Using x is positive, 1/[v(2x)+vx] = 1/v[2x+x] = 1/v(3x)
Answer: A
This question is too easy to be on the GMAT.
v is actually sqrt()... Hardening the question  & making it a GMAT question?
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Manager
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Director
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Now this is a GMAT question, a good one actually:
sqrt(2x) = sqrt(2) * sqrt(x)
1/[sqrt(2x)+sqrt(x)] = 1/[(sqrt(2)*sqrt(x)) + sqrt(x)]
= 1/[sqrt(x)*(sqrt(2)+1)]
= 1/[sqrt(x)*(sqrt(2)+sqrt(1)]
= 1/sqrt(x)*sqrt(3)
= 1/sqrt(3x)
ANSWER: A
Thanks Fig
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SVP
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Mishari wrote: Now this is a GMAT question, a good one actually: sqrt(2x) = sqrt(2) * sqrt(x) 1/[sqrt(2x)+sqrt(x)] = 1/[(sqrt(2)*sqrt(x)) + sqrt(x)] = 1/[sqrt(x)*(sqrt(2)+1)] = 1/[sqrt(x)*(sqrt(2)+sqrt(1)]
= 1/sqrt(x)*sqrt(3)= 1/sqrt(3x) ANSWER: AThanks Fig U are weclome
Also, sqrt(2)+sqrt(1) != sqrt(3)... Similarly sqrt(4) = 2 != sqrt(2) + sqrt(2) = 1,7 + 1,7 = 3,4
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Director
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Okay then .. I dunno how to solve .. I give up 
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