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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]

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24 May 2012, 18:09

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]

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24 May 2012, 21:16

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?

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]

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25 May 2012, 02:22

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

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]

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25 May 2012, 02:36

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

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]

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25 May 2012, 21:41

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)

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]

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15 Jun 2012, 00:53

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.

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.

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.

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

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20 Jun 2012, 06:41

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

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.

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.
_________________

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

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21 Jun 2012, 19:15

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

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

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04 Oct 2012, 11:09

A quicker way of solving take the square of both sides x^2 < x^4 since both sides ought to be positive we can simplify by x^2 w/t changing the direction of the inequality 1 < x^2 or x^2 >1 Brother Karamazov