Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 25 Jun 2016, 02:25

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# m09 q22

Author Message
Intern
Joined: 21 Jun 2008
Posts: 31
Schools: Harvard
Followers: 2

Kudos [?]: 7 [0], given: 0

### Show Tags

07 Sep 2008, 13:18
2
This post was
BOOKMARKED
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

Source: GMAT Club Tests - hardest GMAT questions

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.

___________________________

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.
Intern
Joined: 10 Jan 2008
Posts: 39
Followers: 0

Kudos [?]: 3 [0], given: 0

### Show Tags

08 Sep 2008, 17:10
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
Intern
Joined: 13 Oct 2008
Posts: 16
Followers: 0

Kudos [?]: 11 [0], given: 0

### Show Tags

19 Nov 2008, 07:00
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 ??
Director
Joined: 04 Jan 2008
Posts: 914
Followers: 64

Kudos [?]: 421 [0], given: 17

### Show Tags

19 Nov 2008, 08:15
OA-B
just plug in -0.5 and you will get result
_________________
Intern
Joined: 21 Nov 2008
Posts: 13
Followers: 0

Kudos [?]: 13 [3] , given: 0

### Show Tags

02 Dec 2008, 16:07
3
KUDOS
What if X=0.5?

In this case X>-1, but the inquality is violated.
Attachments

1.JPG [ 34.01 KiB | Viewed 7464 times ]

SVP
Joined: 17 Jun 2008
Posts: 1570
Followers: 11

Kudos [?]: 226 [0], given: 0

### Show Tags

03 Dec 2008, 01:52
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.
Intern
Joined: 21 Nov 2008
Posts: 13
Followers: 0

Kudos [?]: 13 [1] , given: 0

### Show Tags

03 Dec 2008, 14:16
1
KUDOS
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.
CIO
Joined: 02 Oct 2007
Posts: 1218
Followers: 94

Kudos [?]: 861 [1] , given: 334

### Show Tags

26 Feb 2009, 05:58
1
KUDOS
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.
_________________

Welcome to GMAT Club!

Want to solve GMAT questions on the go? GMAT Club iPhone app will help.
Result correlation between real GMAT and GMAT Club Tests
Are GMAT Club Test sets ordered in any way?

Take 15 free tests with questions from GMAT Club, Knewton, Manhattan GMAT, and Veritas.

GMAT Club Premium Membership - big benefits and savings

Manager
Joined: 12 Aug 2008
Posts: 62
Followers: 2

Kudos [?]: 3 [0], given: 2

### Show Tags

26 Sep 2009, 07:32
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?
CIO
Joined: 02 Oct 2007
Posts: 1218
Followers: 94

Kudos [?]: 861 [0], given: 334

### Show Tags

28 Sep 2009, 02:17
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?
_________________

Welcome to GMAT Club!

Want to solve GMAT questions on the go? GMAT Club iPhone app will help.
Result correlation between real GMAT and GMAT Club Tests
Are GMAT Club Test sets ordered in any way?

Take 15 free tests with questions from GMAT Club, Knewton, Manhattan GMAT, and Veritas.

GMAT Club Premium Membership - big benefits and savings

Senior Manager
Joined: 29 Jul 2009
Posts: 314
Followers: 4

Kudos [?]: 273 [1] , given: 9

### Show Tags

05 Oct 2009, 13:05
1
KUDOS
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?
CIO
Joined: 02 Oct 2007
Posts: 1218
Followers: 94

Kudos [?]: 861 [0], given: 334

### Show Tags

06 Oct 2009, 01:08
The change should be visible now. Sorry.
_________________

Welcome to GMAT Club!

Want to solve GMAT questions on the go? GMAT Club iPhone app will help.
Result correlation between real GMAT and GMAT Club Tests
Are GMAT Club Test sets ordered in any way?

Take 15 free tests with questions from GMAT Club, Knewton, Manhattan GMAT, and Veritas.

GMAT Club Premium Membership - big benefits and savings

Senior Manager
Joined: 29 Jul 2009
Posts: 314
Followers: 4

Kudos [?]: 273 [0], given: 9

### Show Tags

07 Oct 2009, 07:50
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?
CIO
Joined: 02 Oct 2007
Posts: 1218
Followers: 94

Kudos [?]: 861 [0], given: 334

### Show Tags

07 Oct 2009, 08:29
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.
_________________

Welcome to GMAT Club!

Want to solve GMAT questions on the go? GMAT Club iPhone app will help.
Result correlation between real GMAT and GMAT Club Tests
Are GMAT Club Test sets ordered in any way?

Take 15 free tests with questions from GMAT Club, Knewton, Manhattan GMAT, and Veritas.

GMAT Club Premium Membership - big benefits and savings

Senior Manager
Joined: 01 Mar 2009
Posts: 372
Location: PDX
Followers: 6

Kudos [?]: 78 [0], given: 24

### Show Tags

15 Oct 2009, 21:27
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.
_________________

In the land of the night, the chariot of the sun is drawn by the grateful dead

CIO
Joined: 02 Oct 2007
Posts: 1218
Followers: 94

Kudos [?]: 861 [1] , given: 334

### Show Tags

15 Oct 2009, 23:36
1
KUDOS
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.

_________________

Welcome to GMAT Club!

Want to solve GMAT questions on the go? GMAT Club iPhone app will help.
Result correlation between real GMAT and GMAT Club Tests
Are GMAT Club Test sets ordered in any way?

Take 15 free tests with questions from GMAT Club, Knewton, Manhattan GMAT, and Veritas.

GMAT Club Premium Membership - big benefits and savings

Senior Manager
Joined: 29 Jul 2009
Posts: 314
Followers: 4

Kudos [?]: 273 [1] , given: 9

### Show Tags

15 Oct 2009, 23:55
1
KUDOS
Hi dzyubam, I think that's the right move. I think that will make option B correct.
CIO
Joined: 02 Oct 2007
Posts: 1218
Followers: 94

Kudos [?]: 861 [0], given: 334

### Show Tags

16 Oct 2009, 00:32
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.

_________________

Welcome to GMAT Club!

Want to solve GMAT questions on the go? GMAT Club iPhone app will help.
Result correlation between real GMAT and GMAT Club Tests
Are GMAT Club Test sets ordered in any way?

Take 15 free tests with questions from GMAT Club, Knewton, Manhattan GMAT, and Veritas.

GMAT Club Premium Membership - big benefits and savings

Senior Manager
Joined: 01 Mar 2009
Posts: 372
Location: PDX
Followers: 6

Kudos [?]: 78 [0], given: 24

### Show Tags

16 Oct 2009, 10:28
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.
_________________

In the land of the night, the chariot of the sun is drawn by the grateful dead

Senior Manager
Joined: 18 Aug 2009
Posts: 303
Followers: 3

Kudos [?]: 215 [1] , given: 9

### Show Tags

14 Nov 2009, 03:10
1
KUDOS
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, 03:10

Go to page    1   2   3    Next  [ 42 posts ]

Similar topics Replies Last post
Similar
Topics:
1 M09 #23 6 19 Nov 2008, 10:17
9 M09 Q16 18 19 Nov 2008, 07:07
14 M09 Q11 18 19 Nov 2008, 07:03
m09 #34 6 10 Nov 2008, 04:10
19 M09 #20 19 10 Nov 2008, 03:24
Display posts from previous: Sort by

# m09 q22

Moderator: Bunuel

 Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.