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(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.
Re: If |x|<x^2 , which of the following must be true? [#permalink]
23 May 2012, 05:09
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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;
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):
Re: If |x|<x^2 , which of the following must be true? [#permalink]
15 Jun 2012, 01:59
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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.
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]
20 Jun 2012, 05:44
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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.
Re: If |x|<x^2 , which of the following must be true? [#permalink]
20 Jun 2012, 21:12
1
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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. _________________
Re: If |x|<x^2 , which of the following must be true? [#permalink]
04 Oct 2012, 20:21
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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
(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 \(x>1\).
If |x|<x^2 , which of the following must be true? [#permalink]
04 Oct 2015, 04:14
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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
Classroom Centre Address: GMATinsight 107, 1st Floor, Krishna Mall, Sector-12 (Main market), Dwarka, New Delhi-110075 ______________________________________________________ Please press the if you appreciate this post !!
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):
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