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m09 q22

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m09 q22 [#permalink] New post 07 Sep 2008, 12:18
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If \frac{X}{|X|} \lt X , which of the following must be true about X ? ( X \ne 0 )

(A) X \gt 2
(B) X \in (-1,0) \cup (1,\infty)
(C) |X| \lt 1
(D) |X| = 1
(E) |X|^2 \gt 1

[Reveal] Spoiler: OA
B

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Solution:

If X > 0, the inequality turns into X > 1,

If X < 0 , the inequality turns into X > -1

In each of the cases X > -1, therefore X > -1 is true for all possible X . Note that -10 and -0.5 can serve as counter-examples for other options.
The correct answer is B.

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Isn't the correct answer -1<x<0 and x>1?
Using formula B with x=0.9 gives you 1<0.9 which doesn't work.
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Re: m09 q22 [#permalink] New post 08 Sep 2008, 16:10
Hi Crush:

I agree that the solution is -1 < X < 0 OR X > 1.

I think the trick here is wording. Think about all the possible values of X.

They are all greater than -1. So, B doesn't define the set of possible values of X, but all the possible values of X are greater than -1. So it is ture.

HTH
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Re: m09 q22 [#permalink] New post 19 Nov 2008, 06:00
Can anybody explain this in detail.

I did not understand this. guess I am dumbo :(

X/|X| < X

if x is say -2 then

isn't it -2/2 < -2 ==> -1 < -2 hence cannot satisfy ?? :cry:
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Re: m09 q22 [#permalink] New post 19 Nov 2008, 07:15
OA-B
just plug in -0.5 and you will get result
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M09, Q22 [#permalink] New post 02 Dec 2008, 15:07
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What if X=0.5?

In this case X>-1, but the inquality is violated.
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Re: M9, Q22 [#permalink] New post 03 Dec 2008, 00:52
The question is expecting the value of X and not necessarily for all values of X.

The answers are x > 1 and -1 < x < 0.

Hence, for any value of x within these ranges, x > -1.
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Re: M9, Q22 [#permalink] New post 03 Dec 2008, 13:16
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scthakur wrote:
The question is expecting the value of X and not necessarily for all values of X.

The answers are x > 1 and -1 < x < 0.

Hence, for any value of x within these ranges, x > -1.


Actually the question stem states: which of the following MUST be true about X..

MUST means for any values, right? There is no limitations on the Value of X, so if one can find one case that doesn't satisfy our inequality, then the answer is wrong. And X>-1 includes values 0<x<1
For example if the stem had a note: X is an integer that would make the question clear.

IMHO the answer choices or the question stem should be reviewed.
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Re: M9, Q22 [#permalink] New post 26 Feb 2009, 04:58
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The questions stem is changed to:

\frac{X}{|X|} < X . Which of the following must be true about integer X ? ( X \ne 0 )

The first option changed to:

X > 2

Is the problem resolved now? Thanks for bringing up the issue. +1.
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Re: m09 q22 [#permalink] New post 26 Sep 2009, 06:32
Can someone explain why |X|^2 > 1 is wrong?

If X is a non-zero integer and X has to be > -1, X has to be an integer greater than 1, as the other constraint says X>1, correct?

Am I missing something?

The explanation says, "Note that -10 and -0.5 can serve as counter-examples for other options." How can X be 0.5 or -10?
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Re: m09 q22 [#permalink] New post 28 Sep 2009, 01:17
You're right, guys. The question still needs revision. Here's what I've come up with. Is there any problem if we change the question stem and options to the following? I've removed the "integer" from the stem:

If \frac{X}{|X|} < X , which of the following must be true about X ? ( X \ne 0 )

(C) 2008 GMAT Club - m09#22

* X > 2
* X > -1
* |X| < 1
* |X| = 1
* |X|^2 > 1

The explanation stays the same. -0.5 can serve as a counter example for A, D and E. If you plug -0.5 into the inequality from the stem, you will see that it holds true. -10 could be probably removed from the OE.
In order to answer the question, you have to solve the inequality from the question stem. The OE solves the modulus inequality. Is there anything wrong with the way the OE solves the inequality that I don't see?

What do you all think about the changes?
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Re: m09 q22 [#permalink] New post 05 Oct 2009, 12:05
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I've come across this question and the stimulus still contains "about integer". This is the question I saw

If \frac{X}{|X|} < X , which of the following must be true about integer X ? ( X \ne 0 )

(C) 2008 GMAT Club - m09#22

* X > 2
* X > -1
* |X| < 1
* |X| = 1
* |X|^2 > 1


According to my understanding B cannot be the correct answer choice. As other members pointed out the solution is

The answers are x > 1 and -1 < x < 0. Since X is an integer x cannot take any values from -1 < x < 0 so the inequality only makes sense when x > 1 --> x >=2

IMO if you change option A for X >=2, I think it should be the correct answer choice. What do you think?
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Re: m09 q22 [#permalink] New post 06 Oct 2009, 00:08
The change should be visible now. Sorry.
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Re: m09 q22 [#permalink] New post 07 Oct 2009, 06:50
I see the change but I still think that the correct answer cannot be B.

According to answer B, X can be 1 which is not a solution for the problem. Could someone explain me why I'm wrong?
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Re: m09 q22 [#permalink] New post 07 Oct 2009, 07:29
I see. The question is really tricky. What if we change the B option to X \in (-1, 1) \cup (1, \infty)? It's stated in the stem that X \not= 0, so this corrected B option should work.
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Re: m09 q22 [#permalink] New post 15 Oct 2009, 20:27
dzyubam wrote:
I see. The question is really tricky. What if we change the B option to X \in (-1, 1) \cup (1, \infty)? It's stated in the stem that X \not= 0, so this corrected B option should work.


IMHO this should work but just wanted to clarify one basic question. Is the x x/|x| different from the X on the right side. I spent nearly 30 minutes trying to understand this.
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Re: m09 q22 [#permalink] New post 15 Oct 2009, 22:36
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I'm not sure if I understand your question correctly. In the inequality from the stem, all three X should be the same values when plugging in any number instead of X. All three are the same X.
I'll change the B option as stated above when some other people confirm that is a right move. Please tell me if it is the right move :).
pleonasm wrote:
dzyubam wrote:
I see. The question is really tricky. What if we change the B option to X \in (-1, 1) \cup (1, \infty)? It's stated in the stem that X \not= 0, so this corrected B option should work.


IMHO this should work but just wanted to clarify one basic question. Is the x x/|x| different from the X on the right side. I spent nearly 30 minutes trying to understand this.

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Re: m09 q22 [#permalink] New post 15 Oct 2009, 22:55
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Hi dzyubam, I think that's the right move. I think that will make option B correct.
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Re: m09 q22 [#permalink] New post 15 Oct 2009, 23:32
Thank you. +1. I've edited the B option. Hope this question is OK now.
mikeCoolBoy wrote:
Hi dzyubam, I think that's the right move. I think that will make option B correct.

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Re: m09 q22 [#permalink] New post 16 Oct 2009, 09:28
dzyubam wrote:
I'm not sure if I understand your question correctly. In the inequality from the stem, all three X should be the same values when plugging in any number instead of X. All three are the same X.
I'll change the B option as stated above when some other people confirm that is a right move. Please tell me if it is the right move :).
pleonasm wrote:
dzyubam wrote:
I see. The question is really tricky. What if we change the B option to X \in (-1, 1) \cup (1, \infty)? It's stated in the stem that X \not= 0, so this corrected B option should work.


IMHO this should work but just wanted to clarify one basic question. Is the x x/|x| different from the X on the right side. I spent nearly 30 minutes trying to understand this.


Yeah ok that's what I thought. Thanks for the clarification. It's the modulus tag that makes the x look a bit different.
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Re: m09 q22 [#permalink] New post 14 Nov 2009, 02:10
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dzyubam wrote:
I'm not sure if I understand your question correctly. In the inequality from the stem, all three X should be the same values when plugging in any number instead of X. All three are the same X.
I'll change the B option as stated above when some other people confirm that is a right move. Please tell me if it is the right move :).
pleonasm wrote:
dzyubam wrote:
I see. The question is really tricky. What if we change the B option to X \in (-1, 1) \cup (1, \infty)? It's stated in the stem that X \not= 0, so this corrected B option should work.


IMHO this should work but just wanted to clarify one basic question. Is the x x/|x| different from the X on the right side. I spent nearly 30 minutes trying to understand this.



Shouldn't the answer option B be:
X \in (-1,0) \cup (1,\infty)

Even the solution describes the same. If X falls in the range (0,1) the statement is not true.
Re: m09 q22   [#permalink] 14 Nov 2009, 02:10
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