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svrkpally
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Is x>3 1) 0 < x^2-9x+18 2) 10-|x|=9 24 May 2005, 06:11
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Is x>3

1) 0 < x^2-9x+18
2) 10-|x|=9



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Is x>3 1) 0 < x^2-9x+18 2) 10-|x|=9 24 May 2005, 06:52
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hm ? 1) and 2) seem to give a different answer of x. thats not possible in gmat-questions.

D)...will explain if correct
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christoph
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Is x>3 1) 0 < x^2-9x+18 2) 10-|x|=9 24 May 2005, 07:46
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sparky
B.
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why is A) insufficient ?

0 < x^2-9x+18 => 0 < (x-3) (x-6) => 0 < x-3 or 0 < x-6 => 3 < x or 6 < x => in both cases x is greater than x. pls correct if i am wrong. thx.
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Is x>3 1) 0 < x^2-9x+18 2) 10-|x|=9 24 May 2005, 08:01
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sparky
B.

why is A) insufficient ?

0 < x^2-9x+18 => 0 < (x-3) (x-6) => 0 < x-3 or 0 < x-6 => 3 < x or 6 < x => in both cases x is greater than x. pls correct if i am wrong. thx.
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Because in 1) 0 < x^2-9x+18 iff x doesn't lie between 3 and 6.
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Dan
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Is x>3 1) 0 < x^2-9x+18 2) 10-|x|=9 24 May 2005, 09:54
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(x-3)(x-6) > 0

either both + or both -

so x could be > 3; take 7 for example
or x could be < 3; take 2 for example
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Is x>3 1) 0 < x^2-9x+18 2) 10-|x|=9 24 May 2005, 10:32
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it is B

from 1) we have (x-3)(x-6) > 0

which means (x-3) and (x-6) are both positive or both negetive . SO A is ruled out

From 2) we have 10 -|x| = 9.

We get an answer for that.....

So B it is
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Is x>3 1) 0 < x^2-9x+18 2) 10-|x|=9 25 May 2005, 01:53
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sparky
B.

why is A) insufficient ?

0 0 0 3 in both cases x is greater than x. pls correct if i am wrong. thx.
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:shock: ok i made a stupid mistake. here is my repair:

i concluded above that 0 that is obviously wrong. it is either 0 x-3 and 0 > x-6 => together it means that x > 6 or x insufficient



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Hi there,
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