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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 350,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Hi, If i solve option 1. i get x^2-1>0 or (x+1)(x-1)>0 this implies : x>-1 or x>1.. however the question in the stem can be rephrased as x<-1 or X>1.. How is option 1 true then?am i missing something?

(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).

Bunuel |x|.|x|=x^2 so when we divide x^2 by |x| we get |x|

(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).

Bunuel |x|.|x|=x^2 so when we divide x^2 by |x| we get |x|

so |x|.|x|= x^2 is this a formula

Yes, |x|*|x|=x^2. Consider x=-2 then |x|*|x|=2*2=4=(-2)^2. _________________

Re: If |x|<x^2 , which of the following must be true? [#permalink]
23 May 2012, 04:10

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.

Re: If |x|<x^2 , which of the following must be true? [#permalink]
23 May 2012, 11:09

picked A in 46 secs. for |X| to be less than x^2 just means that it is x is not a decimal. it can be less than -1 and greater than 1. you can only say this in totality with certainty. And square of nay no less than -1 and greater than 1 will be greater than 1. so A

Re: If |x|<x^2 , which of the following must be true? [#permalink]
23 May 2012, 11:54

Bunuel: I have solved the problem in the following manner: Absolute value of x is positive. So, we can divide both side by x. that leaves us, 1<x or x>1. or x^2>1 Though i have the OA, am i making any fundamental error?

Re: If |x|<x^2 , which of the following must be true? [#permalink]
23 May 2012, 23:45

Expert's post

BDSunDevil wrote:

Bunuel: I have solved the problem in the following manner: Absolute value of x is positive. So, we can divide both side by x. that leaves us, 1<x or x>1. or x^2>1 Though i have the OA, am i making any fundamental error?

Re: If |x|<x^2 , which of the following must be true? [#permalink]
24 May 2012, 17:09

Bunuel, Doesn't the question mean - which of the values of x must hold for |X|<X^2 all values of x<-1 satisfy , although there are other solutions(x>1) but E does not say those solution do not exist. In fact if a superset of values are true then subset of values must be true...

Re: If |x|<x^2 , which of the following must be true? [#permalink]
24 May 2012, 20:16

vikram 4689 what x <-1 says is that x can ONLY hold values of less than -1. Which actually means that x cannot have values of 2,3.. etc. this we all know is not true as x can also takes values of >1. Hence this statement cannot hold true. makes sense?

Re: If |x|<x^2 , which of the following must be true? [#permalink]
25 May 2012, 01:01

Expert's post

vikram4689 wrote:

Bunuel, Doesn't the question mean - which of the values of x must hold for |X|<X^2 all values of x<-1 satisfy , although there are other solutions(x>1) but E does not say those solution do not exist. In fact if a superset of values are true then subset of values must be true...

where am mis-interpreting

No, that's not true.

Again: \(|x|<x^2\) is given as a fact and then we asked to determine which of the following statements MUST be true.

\(|x|<x^2\) means that \(x<-1\) or \(x>1\). Now, evaluate each option:

I. x^2>1. Since \(x<-1\) or \(x>1\) then this option is always true; II. x>0. Since \(x<-1\) or \(x>1\) then this option may or may not be true; III. x<-1. Since \(x<-1\) or \(x>1\) then this option may or may not be true.

Re: If |x|<x^2 , which of the following must be true? [#permalink]
25 May 2012, 01:22

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 ? hope this might help u .. x^2 >|x|---> (implies) x^2>x (since x^2 can not be less than ..it will always by +ve) x^2-x>0 x(x-1)>0 this would give two values x <0 and x>1 but x can not be <0 because values lying between -1 to +1 would nt satisfy the given condition |x|<x^2 ...therefore... x>1 therefore x^2 is also greater than 1

Re: If |x|<x^2 , which of the following must be true? [#permalink]
25 May 2012, 01:25

Expert's post

deepti1206 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 ? hope this might help u .. x^2 >|x|---> (implies) x^2>x (since x^2 can not be less than ..it will always by +ve) x^2-x>0 x(x-1)>0 this would give two values x <0 and x>1 but x can not be <0 because values lying between -1 to +1 would nt satisfy the given condition |x|<x^2 ...therefore... x>1 therefore x^2 is also greater than 1

Please read the thread.

\(|x|<x^2\) means that \(x<-1\) or \(x>1\). _________________

Re: If |x|<x^2 , which of the following must be true? [#permalink]
25 May 2012, 01:36

Sorry i did a mishtake...i would like to revise it,.....

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 ? hope this might help u .. x^2 >|x|---> (implies) i)x^2>x x^2-x>0 x(x-1)>0 this would give two values x <0 and x>1

ii) x^2 <-x x^2-x<0 x(x+1)<0 giving u ..x>0 and x <-1

giving u the common condition that is x >1 therefore x^2 is also greater than 1

Re: If |x|<x^2 , which of the following must be true? [#permalink]
25 May 2012, 01:40

Expert's post

deepti1206 wrote:

Sorry i did a mishtake...i would like to revise it,.....

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 ? hope this might help u .. x^2 >|x|---> (implies) i)x^2>x x^2-x>0 x(x-1)>0 this would give two values x <0 and x>1

ii) x^2 <-x x^2-x<0 x(x+1)<0 giving u ..x>0 and x <-1

giving u the common condition that is x >1 therefore x^2 is also greater than 1

Re: If |x|<x^2 , which of the following must be true? [#permalink]
25 May 2012, 20:41

Bunuel wrote:

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.

Hello Bunuel

I didnot understand this part x^2>|x| holds true for x<-1 and x>1.

if you have x<-1 then why ans "E" is not true?

i also didnot understand folloiwng part "If x>0 then x2-x>0 --> x(x-1)>0 -->[highlight]x<0[/highlight] and x>1" how can x<0 ?

I solved this problem and voted for "A"

for x>1 / 1>x>0 / x<0 (I considered these 3 ranged because if 1>x>0 then the square of number is less then number)

Re: If |x|<x^2 , which of the following must be true? [#permalink]
26 May 2012, 01:36

Expert's post

kuttingchai wrote:

Bunuel wrote:

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.

Hello Bunuel

I didnot understand this part x^2>|x| holds true for x<-1 and x>1.

if you have x<-1 then why ans "E" is not true?

i also didnot understand folloiwng part "If x>0 then x2-x>0 --> x(x-1)>0 -->[highlight]x<0[/highlight] and x>1" how can x<0 ?

E cannot be correct since

I solved this problem and voted for "A"

for x>1 / 1>x>0 / x<0 (I considered these 3 ranged because if 1>x>0 then the square of number is less then number)

Re: If |x|<x^2 , which of the following must be true? [#permalink]
14 Jun 2012, 23:53

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.

Re: If |x|<x^2 , which of the following must be true? [#permalink]
20 Jun 2012, 05:41

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

Hey, Last week I started a few new things in my life. That includes shifting from daily targets to weekly targets, 45 minutes of exercise including 15 minutes of yoga, making...

This week went in reviewing all the topics that I have covered in my previous study session. I reviewed all the notes that I have made and started reviewing the Quant...

I started running as a cross country team member since highshcool and what’s really awesome about running is that...you never get bored of it! I participated in...