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:

(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.

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

|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.

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;

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

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.

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):

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.

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.
_________________

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

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 \(x>1\).

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):

Happy New Year everyone! Before I get started on this post, and well, restarted on this blog in general, I wanted to mention something. For the past several months...

It’s quickly approaching two years since I last wrote anything on this blog. A lot has happened since then. When I last posted, I had just gotten back from...

Post-MBA I became very intrigued by how senior leaders navigated their career progression. It was also at this time that I realized I learned nothing about this during my...