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X/|X| <X>1 B) X>-1 C) |X|<1> 1 [#permalink]
 07 Aug 2007, 18:45
X/|X| <X>1
B) X>-1
C) |X|<1> 1
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I guess you will need to post again with HTML disabled.
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ywilfred wrote: I guess you will need to post again with HTML disabled.
Thanks ywilfred. Here it is:
X/|X| < X. Which of the following MUST be true?
1) X > 1
2) X > -1
3) |X| < 1
4) |X| = 1
5) |X|^2 > 1
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shoonya wrote: ywilfred wrote: I guess you will need to post again with HTML disabled. Thanks ywilfred. Here it is: X/|X| < X. Which of the following MUST be true? 1) X > 1 2) X > -1 3) |X| < 1 4) |X| = 1 5) |X|^2 > 1 Only A for me.
X/|X| < X
if X<0, then X/ -X < X => X>-1 => -1<X<0
if X>0, then X>1
By definition, X will not equal to zero, so C and B are out. X cannot be equal to -1, so D is out. E cannot be true because X and be less than -1.
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a) If x = 2, then x/|x| < x. If x = 1.5, then x/|x| < x. Looks good.
b) If x = 1, then x/|x| = x. If x = 2, then x/|x| <x> x. If x = -1/2, then x/|x| <x> 1. Then |x| must be an integer. But x can be positve or negative integer. If x = -4, then x/|x| > x. If x = 4, then x/|x| < x. Out.
A is best.
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ywilfred wrote: a) If x = 2, then x/|x| < x. If x = 1.5, then x/|x| < x. Looks good.
b) If x = 1, then x/|x| = x. If x = 2, then x/|x| <x> x. If x = -1/2, then x/|x| <x> 1. Then |x| must be an integer. But x can be positve or negative integer. If x = -4, then x/|x| > x. If x = 4, then x/|x| < x. Out.
A is best.
This question is from GMATCLUB's question collection 2.0 . I also got A, but the author thinks the answer is B.
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-0.5 also satisfies the equation.
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(B) it is 
X/|X| < X (Not here that X must be !=0 as the equation exists)
<=> X/|X| - X < 0
<=> X - X*|X| < 0 as |X| > 0
<=> X * (1-|X|) < 0
Implies 2 cases :
o If X > 0 then
1 - |X| < 0
<=> |X| > 1
=> X > 1 as X > 0.
o If X < 0 then
1 - |X| > 0
<=> |X| < 1
=> 0 > X > - 1 as X < 0.
So, all in all, to be sure that the equation X/|X| < X is always true, we must take an interval in the answer choice that contains both intervals above. Thus, X > -1.
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bkk145 wrote: Fig wrote: (B) it is X/|X| < X (Not here that X must be !=0 as the equation exists) <=> X/|X| - X < 0 <=> X - X*|X| < 0 as |X| > 0 <=> X * (1-|X|) < 0 Implies 2 cases : o If X > 0 then1 - |X| < 0 <=> |X| > 1 => X > 1 as X > 0. o If X < 0 then1 - |X| > 0 <=> |X| < 1 => 0 > X > - 1 as X < 0. So, all in all, to be sure that the equation X/|X| < X is always true, we must take an interval in the answer choice that contains both intervals above. Thus, X > -1. How about 0? X/|X| < X is an inequation that exists so we cannot have x = 0
So, yes.... it's voluntarily that the author asks x > -1.... because it must be true.... even if -1 < x < 0 U X > 1 is the complete solution
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Fig wrote: bkk145 wrote: Fig wrote: (B) it is X/|X| < X (Not here that X must be !=0 as the equation exists) <=> X/|X| - X < 0 <=> X - X*|X| < 0 as |X| > 0 <=> X * (1-|X|) < 0 Implies 2 cases : o If X > 0 then1 - |X| < 0 <=> |X| > 1 => X > 1 as X > 0. o If X < 0 then1 - |X| > 0 <=> |X| < 1 => 0 > X > - 1 as X < 0. So, all in all, to be sure that the equation X/|X| < X is always true, we must take an interval in the answer choice that contains both intervals above. Thus, X > -1. How about 0? X/|X| < X is an inequation that exists so we cannot have x = 0 So, yes.... it's voluntarily that the author asks x > -1.... because it must be true.... even if -1 < x < 0 U X > 1 is the complete solution Fig, we agree that the actual solution is (-1<x<0) U (X>1). But among the given choices, only X>1 always satisfies the inequality.
For X>-1, test X=1/2. The inequality doesn't stand. So, the answer must be A, not B.
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I too dont think B can be the answer...
take x=1/2 you will 1 <1/2 which cannot be possible...
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fresinha12 wrote: I too dont think B can be the answer...
take x=1/2 you will 1 <1/2 which cannot be possible...
Yes.... X cannot be 1/2... but X must be > -1 to be X=2 or X=-1/2...
When we say X > -1, we do not say X = 1/2... no. That's the 'must be'
If X=6, then X > 1 or X > -1 or X > -10 000 or X must be > -1
Last edited by Fig on 08 Aug 2007, 08:23, edited 1 time in total.
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shoonya wrote: Fig wrote: bkk145 wrote: Fig wrote: (B) it is X/|X| < X (Not here that X must be !=0 as the equation exists) <=> X/|X| - X < 0 <=> X - X*|X| < 0 as |X| > 0 <=> X * (1-|X|) < 0 Implies 2 cases : o If X > 0 then1 - |X| < 0 <=> |X| > 1 => X > 1 as X > 0. o If X < 0 then1 - |X| > 0 <=> |X| < 1 => 0 > X > - 1 as X < 0. So, all in all, to be sure that the equation X/|X| < X is always true, we must take an interval in the answer choice that contains both intervals above. Thus, X > -1. How about 0? X/|X| < X is an inequation that exists so we cannot have x = 0 So, yes.... it's voluntarily that the author asks x > -1.... because it must be true.... even if -1 < x < 0 U X > 1 is the complete solution Fig, we agree that the actual solution is (-1<x<0) U (X>1). But among the given choices, only X>1 always satisfies the inequality. For X>-1, test X=1/2. The inequality doesn't stand. So, the answer must be A, not B. 'Always satisfies' is not 'must be'
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Fig wrote: shoonya wrote: Fig wrote: bkk145 wrote: Fig wrote: (B) it is X/|X| < X (Not here that X must be !=0 as the equation exists) <=> X/|X| - X < 0 <=> X - X*|X| <0> 0 <=> X * (1-|X|) <0> 0 then[/b] 1 - |X| < 0 <X> 1 => X > 1 as X > 0. o If X <0> 0
<=> |X| <1> 0 > X > - 1 as X < 0.
So, all in all, to be sure that the equation X/|X| <X> -1.How about 0? X/|X| < X is an inequation that exists so we cannot have x = 0 So, yes.... it's voluntarily that the author asks x > -1.... because it must be true.... even if -1 < x <0> 1 is the complete solution Fig, we agree that the actual solution is (-1<x<0>1). But among the given choices, only[b] X>1 always satisfies the inequality. For X>-1, test X=1/2. The inequality doesn't stand. So, the answer must be A, not B. 'Always satisfies' is not 'must be' Though i also choose A, i think it should be B as pointed by Fig. He/she has a valid point.
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Darden2010 wrote: what is with all this 'high' math stuff, GMAT is about understanding of simple concepts. X/|X| is either +1 or -1 so we have two scenarios +1<X and -1<X put these on a line and you will see that they overlap starting from 1<X, therefore that is the solution I think most of us know that. That's why we are discussing why the OA is other way round?
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