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 500,000 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 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

(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

(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 \(x>1\).

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

Show Tags

09 Nov 2012, 08:34

I just cannot get the answer, Bunuel can you give a more detailed explanation about how you conjugated the problem. When I solve the equation in the question I get this --> |x| < x^2 -->

-|x| < x^2 --> -x < x^2 --> divide by x --> -1 < x |x| < x^2 --> x < x^2 --> divide by x --> 1 < x

which makes the first equation redundant. How do you get x < -1 ?????

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

Show Tags

05 Dec 2012, 03:04

|x|<x^2

Determine thw value of x first before testing agains I,II and III. Let us find out x by testing against checkpoints -1, 0 and 1. Let x < -1: |-2| < 4 Yes! Let x =-1: 1 < 1 No! Let -1 < x < 0: x=-1/4 : 1/4 < 1/16 No! Let 0 < x < 1 : x =1/4 : 1/4 < 1/16 No! Let x = 1: 1 < 1 No! Let x > 1: x = 2 : 2 < 4 Yes!

Answer: -1 < x < 1 or |x| > 1

I. x^2>1 Remember, x^2 > 1 is the same as |x| > 1 Yes! II. x>0 This doesn't cover x < -1. III. x<-1 This doesn't cover x > 1.

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.

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

Show Tags

07 Mar 2013, 09:50

Here are my two cents.

You cannot divide both the sides by only x since you do not know whether x is positive or not. Granted that the absolute value will always be positive but if x=-1, absolute value would be 1 which would be denoted as -x. Hence, the absolute value, even though positive, could be +/-x.

Now take the negative aspect into the picture.

|x|<x^2 -x<x^2 -1>x Hence, for values such as x <-1, the equation stands. You can test it with values like x=-2,-3 etc.

Hence, c does not stand as it does not necessarily have to be true. Of course, that was evident from you calculations.

Hope this helps!

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]

Show Tags

09 Mar 2013, 22:24

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

PLEASE READ THE WHOLE THREAD AND FOLLOW THE LINKS PROVIDED IN ORDER TO UNDERSTAND THE QUESTION/SOLUTION CORRECTLY.

|x| - x^2 < 0

If x>0, x ( 1 - x) < 0--> x < 0 (can't be true as condition states x>0or x>1 If x<0, -x - x^2 <0-----> x(1+x) >0 --> x>0 or x> -1.

x>1 & x>-1 only option A is valid. x can't be equal to 0. as 0>0 is not true. Is my solution correct?? Why it is different then Bunuel's solution
_________________

|x| will be less than x^2 in the region where the graph of |x| is below the graph of x^2. Graph of y = |x| is a V at (0, 0) and graph of y = x^2 is an upward facing parabola at (0, 0)

Attachment:

Ques3.jpg [ 7.72 KiB | Viewed 1275 times ]

Just looking at the graph should tell you that answer must be (A). You don't need to solve. x will lie in one of two ranges x > a or x < -a. II and III will not work in any case. I must work since one of them has to be true according to the options.
_________________

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

Show Tags

01 Jun 2013, 09:08

The trick is in question, it doesn't ask you to choose possible answers, but it asks for an answer which holds true in all possibilities. Option (i) satisfies this always, but option (iii), not always..
_________________

And many strokes, though with a little axe, hew down and fell the hardest-timbered oak. - William Shakespeare

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

Show Tags

09 Jul 2013, 21:52

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

|x|<x^2

I. x^2>1 If |x|<x^2 then that means x^2 is greater than positive value x. x cannot be between -1 and 1 inclusive because then |x| would be greater than or equal to x^2, not less than it as the stem stipulates. Therefore, for the stem to be true x^2 must be greater than 1.

II. x>0 If x>0 then x could = 1/2. |1/2|<(1/2)^2 1/2<1/4 which is invalid

III. x<-1 While any x value less than -1 will provide a correct answer, valid values for x don't have to be less than -1. They can be greater than one as well.

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

Show Tags

21 Jul 2013, 17:47

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

PLEASE READ THE WHOLE THREAD AND FOLLOW THE LINKS PROVIDED IN ORDER TO UNDERSTAND THE QUESTION/SOLUTION CORRECTLY.

Having seen such a big red quote here, I was wary of this question before starting. For questions like these I have created a method which has helped me and would like to share it.

Bunnel, Please correct me if this is wrong or has limited scope.

If |x|<x^2 , which of the following must be true? this means x>1 and x<-1.

The question stem set must be equal to or a subset of the answer choices i.e. The answer choice must contain at-least all of points of the question

I - x^2>1 => x>1 and x<-1 this choice consists of exactly the same data points as in the question stem, subset and hence, Yes.

II - X>0 this choice does not contain all of the points of the question stem, not a subset and hence, No.

III - x<-1 this choice does not contain all of the points of the question stem, not a subset and hence, No.

Try similar question: If |x| > 3, which of the following must be true?

1) x > 3 2) x^2 > 9 3) |x - 1| > 2

I only II only I and II only II and III only I, II, and III

(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.

Having seen such a big red quote here, I was wary of this question before starting. For questions like these I have created a method which has helped me and would like to share it.

Bunnel, Please correct me if this is wrong or has limited scope.

If |x|<x^2 , which of the following must be true? this means x>1 and x<-1.

The question stem set must be equal to or a subset of the answer choices i.e. The answer choice must contain at-least all of points of the question

I - x^2>1 => x>1 and x<-1 this choice consists of exactly the same data points as in the question stem, subset and hence, Yes.

II - X>0 this choice does not contain all of the points of the question stem, not a subset and hence, No.

III - x<-1 this choice does not contain all of the points of the question stem, not a subset and hence, No.

Try similar question: If |x| > 3, which of the following must be true?

1) x > 3 2) x^2 > 9 3) |x - 1| > 2

I only II only I and II only II and III only I, II, and III

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

Show Tags

21 Jan 2014, 22:49

This is very tricky. If this is a 600-700 level, I am in trouble. I did get it, however, I don't think I would get it if didn't read the explanation for several times. And still a little confused. Thank you.

|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.

Hi bunnel,

Can we write this as x<0 or x>0 instead of x<-1 or x>1

Campus visits play a crucial role in the MBA application process. It’s one thing to be passionate about one school but another to actually visit the campus, talk...

Its been long time coming. I have always been passionate about poetry. It’s my way of expressing my feelings and emotions. And i feel a person can convey...

Marty Cagan is founding partner of the Silicon Valley Product Group, a consulting firm that helps companies with their product strategy. Prior to that he held product roles at...

Written by Scottish historian Niall Ferguson , the book is subtitled “A Financial History of the World”. There is also a long documentary of the same name that the...