It is currently 11 Dec 2017, 02:33

### 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
Your Progress

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

# If |x|<x^2 , which of the following must be true?

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

### Hide Tags

Intern
Joined: 22 Jun 2010
Posts: 41

Kudos [?]: 159 [15], given: 1

If |x|<x^2 , which of the following must be true? [#permalink]

### Show Tags

20 Aug 2010, 12:59
15
This post received
KUDOS
71
This post was
BOOKMARKED
00:00

Difficulty:

75% (hard)

Question Stats:

49% (00:56) correct 51% (01:04) wrong based on 1988 sessions

### HideShow timer Statistics

If |x|<x^2 , which of the following must be true?

I. x^2>1
II. x>0
III. x<-1

(A) I only
(B) II only
(C) III only
(D) I and II only
(E) I and III only

PLEASE READ THE WHOLE THREAD AND FOLLOW THE LINKS PROVIDED IN ORDER TO UNDERSTAND THE QUESTION/SOLUTION CORRECTLY.
[Reveal] Spoiler: OA

Last edited by Bunuel on 20 Jun 2012, 06:46, edited 2 times in total.
Edited the question and added the OA

Kudos [?]: 159 [15], given: 1

Math Expert
Joined: 02 Sep 2009
Posts: 42536

Kudos [?]: 135188 [8], given: 12671

Re: not that hard [#permalink]

### Show Tags

20 Aug 2010, 13:18
8
This post received
KUDOS
Expert's post
20
This post was
BOOKMARKED
mehdiov wrote:
If |x|<x^2 , which of the following must be true?

I. x^2>1
II. x>0
III. x<-1

(A) I only
(B) II only
(C) III only
(D) I and II only
(E) I and III only

what is the fastest way to solbe it ?

Given: $$|x|<x^2$$ --> reduce by $$|x|$$ (side note: we can safely do this as absolute value is non-negative and in this case we know it's not zero too, as if x would be zero inequality wouldn't hold true, so $$|x|>0$$) --> $$1<|x|$$ --> $$x<-1$$ or $$x>1$$.

So we have that $$x<-1$$ or $$x>1$$.

I. x^2>1 --> always true;
II. x>0 --> may or may not be true;
III. x<-1 --> --> may or may not be true.

Answer: A (I only).
_________________

Kudos [?]: 135188 [8], given: 12671

Intern
Joined: 15 Aug 2010
Posts: 23

Kudos [?]: 29 [6], given: 0

Location: Mumbai
Schools: Class of 2008, IIM Ahmedabad
Re: not that hard [#permalink]

### Show Tags

20 Aug 2010, 13:22
6
This post received
KUDOS
Hi,

The X-Y plane of looking at it is the fastest way - I would guess.

Mod(x) looks like a V

x^2 is a parabola which just touches the x-axis (tangent) at the origin.

When x < -1; Mod(x) < x^2
When x = -1; Mod (x) = x^2
When -1<x<0; Mod(x) > x^2
When x = 0; Mod (x) = x^2
When 0<x<1; Mod(x) > x^2
When x = 1; Mod (x) = x^2
When x > 1; Mod(x) < x^2

Hope this helps. Thanks.

Wrong forum ?!?!?
_________________

Naveenan Ramachandran
4GMAT - Mumbai

Kudos [?]: 29 [6], given: 0

Math Expert
Joined: 02 Sep 2009
Posts: 42536

Kudos [?]: 135188 [3], given: 12671

Re: not that hard [#permalink]

### Show Tags

21 Aug 2010, 02:24
3
This post received
KUDOS
Expert's post
1
This post was
BOOKMARKED
qweert wrote:
I'm confused, because if x= -2 ==> |x|< x^2

|x|< x^2, X^2 should always be > 1
Can eliminate II, because if x=1/2, |x|> x^2

Could you give an example Bunuel

Question is: "which of the following statements MUST be true", not COULD be true.

We are GIVEN that $$|x|<x^2$$, which means that either $$x<-1$$ OR $$x>1$$.

So GIVEN that: $$x<-1$$ OR $$x>1$$.

Statement II. is $$x<-1$$ always true? NO. As $$x$$ could be more than 1, eg. 2, 3, 5.7, ... and in this case $$x<-1$$ is not true. So statement II which says that $$x<-1$$ is not always true.

Hope it's clear.
_________________

Kudos [?]: 135188 [3], given: 12671

Math Expert
Joined: 02 Sep 2009
Posts: 42536

Kudos [?]: 135188 [3], given: 12671

Re: If |x|<x^2 , which of the following must be true? [#permalink]

### Show Tags

23 May 2012, 06:09
3
This post received
KUDOS
Expert's post
1
This post was
BOOKMARKED
pavanpuneet wrote:
Hi Bunuel, I understood your logic that we can safely divide by |x|, however, I tried to solve it this way, just wanted to confirm is this is also right...

x2>|x| => if x is positive then x2-x>0 => x(x-1)>0 => x<0 and x>1................case 1

and if x is negative, then x2+x>0 => x(x+1)>0 =>x<-1 and x>0...case2

From case 1 and case2....x<-1 and x>1..[Same conclusion but lengthier method].. I want to especially confirm because I tried to apply modulus and inequalities theory.. so want to be sure.

x^2>|x|

If x>0 then x2-x>0 --> x(x-1)>0 --> x<0 and x>1. Since we are considering x>0 range then x>1;
If x<0 then x2-x>0 --> x(x+1)>0 --> x<-1 and x>0. Since we are considering x<0 range then x<-1;

So, x^2>|x| holds true for x<-1 and x>1.

Hope it's clear.
_________________

Kudos [?]: 135188 [3], given: 12671

Math Expert
Joined: 02 Sep 2009
Posts: 42536

Kudos [?]: 135188 [2], given: 12671

Re: not that hard [#permalink]

### Show Tags

20 Aug 2010, 13:33
2
This post received
KUDOS
Expert's post
4gmatmumbai wrote:
Bunuel wrote:
III. x<-1 --> --> may or may not be true.

Bunuel, Isn't this always true when |x|<x^2 ... Isn't the answer E.

Thanks.

$$|x|<x^2$$ is given as fact and then we asked to determine which of the following statements MUST be true.

$$|x|<x^2$$ means that either $$x<-1$$ or $$x>1$$, $$x$$ can be ANY value from these two ranges, (I think in your own solution you've reached this conclusion: when $$x<-1$$ the graph of $$|x|$$ is below (less than) the graph of $$x^2$$ and when $$x>$$1 again the graph of $$|x|$$ is below the graph of $$x^2$$).

Now, III says $$x<-1$$ this statement is not always true as $$x$$ can be for example 3 and in this case $$x<-1$$ doesn't hold true.

Hope it's clear.
_________________

Kudos [?]: 135188 [2], given: 12671

Director
Joined: 17 Dec 2012
Posts: 623

Kudos [?]: 547 [2], given: 16

Location: India
Re: If |x|<x^2 , which of the following must be true? [#permalink]

### Show Tags

21 Jul 2013, 20:25
2
This post received
KUDOS
Expert's post
1. |x|<x^2 =>

(i) x < x^2 or
(ii) -x < x^2

2. (1) => x>1 or x<-1 . Only this satisfies both (i) and (ii)

Is x^2>1 always true? Yes follows from (2)
Is x>0 always true, No because x could be <-1
Is x<-1 always true?, No because x could be >1

So answer is choice A.
_________________

Srinivasan Vaidyaraman
Sravna
http://www.sravnatestprep.com/regularcourse.php

Premium Material
Standardized Approaches

Kudos [?]: 547 [2], given: 16

SVP
Joined: 08 Jul 2010
Posts: 1856

Kudos [?]: 2392 [2], given: 51

Location: India
GMAT: INSIGHT
WE: Education (Education)
If |x|<x^2 , which of the following must be true? [#permalink]

### Show Tags

04 Oct 2015, 05:14
2
This post received
KUDOS
Expert's post
longfellow wrote:
Thanks GMATInsight. I would still like practice on more questions like this one? Could you point out a few?

Question 1: What are the values of n that satisfy the condition 1/|n| > n?
(A) 0<n<1 (and) - infinity < n < 0
(B) 0 < n < infinity (or) -infinity < n < -1
(C) 0 < n < 1 (and) -1 < n < 0
(D) - infinity < n < 0 (or) 0 < n < 1
(E) 0<n<1 (or) - infinity < n < 0

Answer - Option A
Link: what-are-the-values-of-n-that-satisfy-the-condition-1-n-n-174746.html

Question 2: If 0 < x < 1 < y, which of the following must be true?

A. 1 < 1/x < 1/y
B. 1/x < 1 < 1/y
C. 1/x < 1/y < 1
D. 1/y < 1 < 1/x
E. 1/y < 1/x < 1

Answer: Option D
Link: if-0-x-1-y-which-of-the-following-must-be-true-198235.html

Question 3: If x is positive, which of the following could be correct ordering of 1/x, 2x, and x^2?

(I) X^2 < 2x < 1/x
(II) x^2 < 1/x < 2x
(III) 2x < x^2 < 1/x

(a) none
(b) I only
(c) III only
(d) I and II
(e) I, II and III

Answer: Option D
Link: if-x-is-positive-which-of-the-following-could-be-correct-71070.html

I hope this helps!
_________________

Prosper!!!
GMATinsight
Bhoopendra Singh and Dr.Sushma Jha
e-mail: info@GMATinsight.com I Call us : +91-9999687183 / 9891333772
Online One-on-One Skype based classes and Classroom Coaching in South and West Delhi
http://www.GMATinsight.com/testimonials.html

22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION

Kudos [?]: 2392 [2], given: 51

Intern
Joined: 15 Aug 2010
Posts: 23

Kudos [?]: 29 [1], given: 0

Location: Mumbai
Schools: Class of 2008, IIM Ahmedabad
Re: not that hard [#permalink]

### Show Tags

20 Aug 2010, 13:25
1
This post received
KUDOS
Bunuel wrote:
III. x<-1 --> --> may or may not be true.

Bunuel, Isn't this always true when |x|<x^2 ... Isn't the answer E.

Thanks.
_________________

Naveenan Ramachandran
4GMAT - Mumbai

Kudos [?]: 29 [1], given: 0

Math Expert
Joined: 02 Sep 2009
Posts: 42536

Kudos [?]: 135188 [1], given: 12671

Re: not that hard [#permalink]

### Show Tags

07 Sep 2010, 04:03
1
This post received
KUDOS
Expert's post
nonameee wrote:
mehdiov, what's the OA?

Bunuel, I understand your reasoning but I don't think you are right. IMO the answer is E. Should we follow your logic:

Quote:
Statement II. is x<-1 always true? NO. As x could be more than 1, eg. 2, 3, 5.7, ... and in this case x<-1 is not true. So statement II which says that x<-1 is not always true.

Then we can't even accept I, since x doesn't always have to be greater than 1 (it can be -10, -50 etc.).

The question is similar to (but I still think that it is different):

ps-inequality-13943.html

Maybe I don't understand something.

Thanks.

OA is A.

$$|x|<x^2$$ means that either $$x<-1$$ or $$x>1$$. For example $$x$$ could be -5, -3, 2, 3, ...

Basically the question is: If $$x<-1$$ or $$x>1$$ which of the following must be true?

You are saying that answer is: E (I and III only). Is III: $$x<-1$$ ALWAYS true? Is it true for $$x=3$$? NO. So E is not correct.
_________________

Kudos [?]: 135188 [1], given: 12671

Math Expert
Joined: 02 Sep 2009
Posts: 42536

Kudos [?]: 135188 [1], given: 12671

Re: If |x|<x^2 , which of the following must be true? [#permalink]

### Show Tags

15 Jun 2012, 02:59
1
This post received
KUDOS
Expert's post
pavanpuneet wrote:
Hi Bunuel, I tried to solve it this way, please let me know if this is correct as I am getting confused with the final wording of MUST be true. I am kind of stuck at the final stages to make the decision.

The problem mentions: |x|<x2

Case 1: if x is +ve; then x<x2, given that we do not know if x=0, then x-x2<0 ==> x(1-x)<0 ==> multiplying by negative sign; x(x-1)>0...which means x<0 and X>1

Case 2: if x is -ve; then -x<x2, then -x-x2<0 ===>-(x+x2) <0...==> x(1+x)>0 ===> x<-1 and x>0

From C1 AND C2..on the number x<-1 and x>1; after this I am a little lost, how to make the final call.

First of all please study these posts:
if-x-x-2-which-of-the-following-must-be-true-99506.html#p767256
if-x-x-2-which-of-the-following-must-be-true-99506.html#p767264
if-x-x-2-which-of-the-following-must-be-true-99506.html#p767437
if-x-x-2-which-of-the-following-must-be-true-99506.html#p776341
if-x-x-2-which-of-the-following-must-be-true-99506.html#p1088622
if-x-x-2-which-of-the-following-must-be-true-99506-20.html#p1089273

Because the question you ask was answered before.

Anyway, given that $$|x|<x^2$$, which means that $$x<-1$$ or $$x>1$$. Next we are given three options and are asked which of the following MUST be true.

I. x^2>1. Since $$x<-1$$ or $$x>1$$ then this option is always true (ANY number less than -1 or more than 1 when squared will be more than 1);
II. x>0. This option is NOT always true, for example x could be -2 and in this case x>0 won't not true;
III. x<-1. This option is also NOT always true, for example x could be 2 and in this case x<-1 won't not true.

So, we have that only option I is always true.

Answer: A (I only).

Check our Must or Could be True Questions to practice more: search.php?search_id=tag&tag_id=193

Hope it helps.
_________________

Kudos [?]: 135188 [1], given: 12671

Math Expert
Joined: 02 Sep 2009
Posts: 42536

Kudos [?]: 135188 [1], given: 12671

Re: If |x|<x^2 , which of the following must be true? [#permalink]

### Show Tags

20 Jun 2012, 06:44
1
This post received
KUDOS
Expert's post
idreesma wrote:
Hi,
I am having issues with a fundanmental concept. In the below explanation x(x+1)>0 , isnt x>0 and x>-1 (from x+1>0) rather than x<-1 and x>0. clearly by inspection my answer is wrong, but was confused over how you go correct answer (is it possible to show the steps)

If x>0 then x2-x>0 --> x(x-1)>0 --> x<0 and x>1. Since we are considering x>0 range then x>1;
If x<0 then x2-x>0 --> x(x+1)>0 --> x<-1 and x>0. Since we are considering x<0 range then x<-1;
appreciate your input
Thanks

Welcome to GMAT Club. Below links might help you to understand the concept.

x2-4x-94661.html#p731476 (check this one first)
inequalities-trick-91482.html
everything-is-less-than-zero-108884.html?hilit=extreme#p868863
xy-plane-71492.html?hilit=solving%20quadratic#p841486
data-suff-inequalities-109078.html
range-for-variable-x-in-a-given-inequality-109468.html?hilit=extreme#p873535

Hope it helps.

Hope it helps.
_________________

Kudos [?]: 135188 [1], given: 12671

Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7787

Kudos [?]: 18094 [1], given: 236

Location: Pune, India
Re: If |x|<x^2 , which of the following must be true? [#permalink]

### Show Tags

20 Jun 2012, 22:12
1
This post received
KUDOS
Expert's post
idreesma wrote:
Hi,
I am having issues with a fundanmental concept. In the below explanation x(x+1)>0 , isnt x>0 and x>-1 (from x+1>0) rather than x<-1 and x>0. clearly by inspection my answer is wrong, but was confused over how you go correct answer (is it possible to show the steps)

If x>0 then x2-x>0 --> x(x-1)>0 --> x<0 and x>1. Since we are considering x>0 range then x>1;
If x<0 then x2-x>0 --> x(x+1)>0 --> x<-1 and x>0. Since we are considering x<0 range then x<-1;
appreciate your input
Thanks

Responding to a pm:

The problem you are facing is that you do not know how to handle inequalities.

How do you get the range for which this inequality holds? x(x+1) > 0

Think of it this way: Product of x and (x+1) should be positive. When will that happen? When either both the terms are positive or both are negative.

Case 1: When both are positive
x > 0
x + 1 > 0 i.e. x > -1

For both to be positive, x must be greater than 0. Hence this inequality will hold when x > 0.

Case 2: When both are negative
x < 0
x + 1 < 0 i.e. x < -1

For both to be negative, x must be less than -1. Hence this inequality will hold when x < -1.

So we get two ranges in which this inequality holds: x > 0 or x < -1.

The fastest way to solve it is using the number line.
Check this post for the explanation of this method: inequalities-trick-91482.html

Also, your Veritas book discusses this concept too.
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Kudos [?]: 18094 [1], given: 236 Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7787 Kudos [?]: 18094 [1], given: 236 Location: Pune, India Re: If |x|<x^2 , which of the following must be true? [#permalink] ### Show Tags 04 Oct 2012, 21:21 1 This post received KUDOS Expert's post idreesma wrote: Hi Krishma, the link that you posted is probably best for summary. It ends with the comment "Anyone wants to add ABSOLUTE VALUES....That will be a value add to this post ". Can you suggest any such thread where absolute value side is summarized like this as well. Appreciate your input Thanks I have discussed mods here: http://www.veritasprep.com/blog/2011/01 ... edore-did/ http://www.veritasprep.com/blog/2011/01 ... h-to-mods/ http://www.veritasprep.com/blog/2011/01 ... s-part-ii/ _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

Veritas Prep Reviews

Kudos [?]: 18094 [1], given: 236

Math Expert
Joined: 02 Sep 2009
Posts: 42536

Kudos [?]: 135188 [1], given: 12671

Re: not that hard [#permalink]

### Show Tags

29 Oct 2012, 02:51
1
This post received
KUDOS
Expert's post
mario1987 wrote:
Bunuel wrote:
mehdiov wrote:
If |x|<x^2 , which of the following must be true?

I. x^2>1
II. x>0
III. x<-1

(A) I only
(B) II only
(C) III only
(D) I and II only
(E) I and III only

what is the fastest way to solbe it ?

Given: $$|x|<x^2$$ --> reduce by $$|x|$$ (side note: we can safely do this as absolute value is non-negative and in this case we know it's not zero too, as if x would be zero inequality wouldn't hold true, so $$|x|>0$$) --> $$1<|x|$$ --> $$x<-1$$ or $$x>1$$.

So we have that $$x<-1$$ or $$x>1$$.

I. x^2>1 --> always true;
II. x>0 --> may or may not be true;
III. x<-1 --> --> may or may not be true.

Answer: A (I only).

Hi bunuel,
sorry , but I didn't undestrand how you reduce |x|<x^2 by |x| to get the solution...x<-1 and x>1
please, could you explain me in more details?
Thanks

Express $$x^2$$ as $$|x|*|x|$$, so we have that: $$|x|<|x|*|x|$$ --> reduce by $$|x|$$ (notice that $$|x|$$ is positive, thus we can safely divide both parts of the inequality by it) --> $$1<|x|$$ --> $$x<-1$$ or $$x>1$$.

Hope it's clear.
_________________

Kudos [?]: 135188 [1], given: 12671

Intern
Joined: 13 Aug 2012
Posts: 6

Kudos [?]: 6 [1], given: 4

Location: United States
Concentration: Entrepreneurship, Marketing
GMAT 1: 650 Q49 V28
GPA: 2.9
WE: Project Management (Computer Software)
Re: If |x|<x^2 , which of the following must be true? [#permalink]

### Show Tags

10 Mar 2013, 02:01
1
This post received
KUDOS
|X| < x^2 means X is not a decimal. means X < -1 or x > 1. Square of anything will be positive. Hence. A
Hope it helps.

Kudos [?]: 6 [1], given: 4

Manager
Joined: 27 May 2010
Posts: 97

Kudos [?]: 6 [0], given: 13

Re: not that hard [#permalink]

### Show Tags

21 Aug 2010, 00:10
thanks bunuel. great explanation.

Kudos [?]: 6 [0], given: 13

Manager
Joined: 27 May 2010
Posts: 197

Kudos [?]: 72 [0], given: 3

Re: not that hard [#permalink]

### Show Tags

21 Aug 2010, 01:01
I'm confused, because if x= -2 ==> |x|< x^2

|x|< x^2, X^2 should always be > 1
Can eliminate II, because if x=1/2, |x|> x^2

Could you give an example Bunuel

Kudos [?]: 72 [0], given: 3

Manager
Joined: 27 May 2010
Posts: 197

Kudos [?]: 72 [0], given: 3

Re: not that hard [#permalink]

### Show Tags

21 Aug 2010, 08:09
Thank you.

I get it now...

Kudos [?]: 72 [0], given: 3

Director
Joined: 23 Apr 2010
Posts: 573

Kudos [?]: 100 [0], given: 7

Re: not that hard [#permalink]

### Show Tags

07 Sep 2010, 01:40
mehdiov, what's the OA?

Bunuel, I understand your reasoning but I don't think you are right. IMO the answer is E. Should we follow your logic:

Quote:
Statement II. is x<-1 always true? NO. As x could be more than 1, eg. 2, 3, 5.7, ... and in this case x<-1 is not true. So statement II which says that x<-1 is not always true.

Then we can't even accept I, since x doesn't always have to be greater than 1 (it can be -10, -50 etc.).

The question is similar to (but I still think that it is different):

ps-inequality-13943.html

Maybe I don't understand something.

Thanks.

Kudos [?]: 100 [0], given: 7

Re: not that hard   [#permalink] 07 Sep 2010, 01:40

Go to page    1   2   3   4   5    Next  [ 90 posts ]

Display posts from previous: Sort by

# If |x|<x^2 , which of the following must be true?

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne 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®.