Author 
Message 
TAGS:

Hide Tags

Senior Manager
Joined: 07 Feb 2008
Posts: 311

If x is positive, which of the following could be correct ordering of [#permalink]
Show Tags
03 Oct 2008, 14:37
11
This post received KUDOS
88
This post was BOOKMARKED
Question Stats:
38% (02:06) correct
62% (01:05) wrong based on 3189 sessions
HideShow timer Statistics
If x is positive, which of the following could be correct ordering of \(\frac{1}{x}\), \(2x\), and \(x^2\)? I. \(x^2 < 2x < \frac{1}{x}\) II. \(x^2 < \frac{1}{x} < 2x\) III. \(2x < x^2 < \frac{1}{x}\) (A) none (B) I only (C) III only (D) I and II (E) I, II and III
Official Answer and Stats are available only to registered users. Register/ Login.
Last edited by Bunuel on 18 Jun 2017, 02:09, edited 5 times in total.
Renamed the topic and edited the question.



Manager
Joined: 30 Sep 2008
Posts: 111

Re: If x is positive, which of the following could be correct ordering of [#permalink]
Show Tags
04 Oct 2008, 23:18
6
This post received KUDOS
4
This post was BOOKMARKED
Vavali wrote: If x is positive, which of the following could be correct ordering of 1/x, 2x, and x^2?
(I) X^2 < 2x < 1/x (II) x^2 < 1/x < 2x (III) 2x < x^2 < 1/x
(a) none (b) I only (c) III only (d) I and II (e) I, II and III could be correct orderingSo if we can find any example that satisfy the inequation, that statement will be correct (I) x = 0.1 => 0.01 < 0.2 < 10 (II) x= 1/2 => 1/4 < 1/2 < 1 (III) 2x < x^2 <=> x ( 2 x) < 0, x > 0 then x > 2with x > 2 ==> x^2 < 1/x <=> x^3 < 1 <=> x < 1 So (III) can't happen The answer is D



SVP
Joined: 17 Jun 2008
Posts: 1546

Re: If x is positive, which of the following could be correct ordering of [#permalink]
Show Tags
05 Oct 2008, 13:22
6
This post received KUDOS
2
This post was BOOKMARKED
Vavali wrote: I still dont get why D is the answer. There is a flaw in your reasoning for statement II
if x = 1/2, then 1/1/2 = 2 which is greater than 2(1/2)
can someone please use figures (fractions or decimals) to demonstrate this. Thanks x^2 < 1/x < 2x....If I translate this into two parts, it tell me that x^3 < 1 and 2x^2 > 1...that means, x < 1 and x > 1/1.414 That means approximately, x should be between 0.7 and 1. Take a value of x = 0.9. Here, x^2 = 0.81, 1/x = 10/9 and 2x = 1.8 and these values satisfy the inequality.



Math Expert
Joined: 02 Sep 2009
Posts: 40432

If x is positive, which of the following could be correct ordering of [#permalink]
Show Tags
24 Jan 2010, 01:47
23
This post received KUDOS
Expert's post
42
This post was BOOKMARKED
gautamsubrahmanyam wrote: I am not sure how the OA is D
when x=1/2 then 1/x is 2 ,2x is 1 and x^2 is 1/2 ,this satisfies (I) x^2<2x<1/x
when x =3 then 1/x is 1/3, 2x is 6 and x^2 is 9 ,this does not satisfy (II)
when x = 1/10 then 1/x is 10 , 2x is 1/5 and x^2 is 1/100,this again does not satify (II)
Even ve numbers dont seem to work
when x=3 then 1/x is 1/3 ,2x is 6 and X^2=9,this does not satisfy (II) x=1/3 then 1/x is 3 ,2x is 2/3 and x^2=1/9,this does not satisfy (II)
Can any one give an example which satisfies option (II) If x is positive, which of the following could be the correct ordering of 1/x, 2x and x^2 ? I. \(x^2<2x<\frac{1}{x}\) II. \(x^2<\frac{1}{x}<2x\) III. \(2x<x^2<\frac{1}{x}\) (A) None (B) I only (C) III only (D) I and II only (E) I II and III First note that we are asked "which of the following COULD be the correct ordering" not MUST be. Basically we should determine relationship between \(x\), \(\frac{1}{x}\) and \(x^2\) in three areas: \(0<1<2<\). \(x>2\) \(1<x<2\) \(0<x<1\) When \(x>2\) > \(x^2\) is the greatest and no option is offering this, so we know that x<2. If \(1<x<2\) > \(2x\) is greatest then comes \(x^2\) and no option is offering this. So, we are left with \(0<x<1\): In this case \(x^2\) is least value, so we are left with: I. \(x^2<2x<\frac{1}{x}\) > can \(2x<\frac{1}{x}\)? Can \(\frac{2x^21}{x}<0\), the expression \(2x^21\) can be negative or positive for \(0<x<1\). (You can check it either algebraically or by picking numbers) II. \(x^2<\frac{1}{x}<2x\) > can \(\frac{1}{x}<2x\)? The same here \(\frac{2x^21}{x}>0\), the expression \(2x^21\) can be negative or positive for \(0<x<1\). (You can check it either algebraically or by picking numbers) Answer: D. Second condition: \(x^2<\frac{1}{x}<2x\) The question is which of the following COULD be the correct ordering not MUST be. Put \(0.9\) > \(x^2=0.81\), \(\frac{1}{x}=1.11\), \(2x=1.8\) > \(0.81<1.11<1.8\). Hence this COULD be the correct ordering. Hope it's clear.
_________________
New to the Math Forum? Please read this: All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Manager
Joined: 26 May 2005
Posts: 207

Re: If x is positive, which of the following could be correct ordering of [#permalink]
Show Tags
04 Feb 2010, 18:47
1
This post received KUDOS
zaarathelab wrote: Still not very clear.
I marked I only in Gmatprep by figuring out that x lies between 0 and 1. But how does one use mathematic logic to try out the value 0.9 since plugging 0.9 and 0.5 each gives different results 7. If x is positive, which of the following could be the correct ordering of 1/x,2x and x^2 ? I. x^2<2x<1/x II. x^2<1/x<2x III. 2x<x^2<1/x In all the expression the common expression is x^2 < 1/x which is possible when 0<x<1 Now take expression x^2<2x x^22x < 0 which is possible if x< 0 & x > 2 OR x >0 and x<2 .. so using the above expression 0<x<1 this expression also could be true. now take expression x^2>2x which is possible if x<0 & x<2 OR x>0 & x>2 .. this doesnt fall under the condition of 0<x<1 .. so this expression could not be true.. III is out for this case alone now take expression 2x<1/x x^2 < 1/2 .. the range of values of x is a subset of the range 0<x<1 .. so this expression could be true for some values of x. I can be the correct ordering now take expression 1/x<2x x^2>1/2 .. the range of values of x has some values in the range 0<x<1 .. so this expression could be true for some vaues of x. II can be the correct ordering. D



Math Expert
Joined: 02 Sep 2009
Posts: 40432

Re: If x is positive, which of the following could be correct ordering of [#permalink]
Show Tags
18 Apr 2010, 10:45
9
This post received KUDOS
Expert's post
5
This post was BOOKMARKED
ykaiim wrote: IMO B. for 0<x<1, only statement I holds. Brunuel, if u put x=1/2: II. II. x^2<1/x<2x >>>>> will not hold true. x^2 = 1/4, 1/x=2 and 2x=1 then this expression will not hold. 1/4<2<1 [Incorrect]
If x=1/9 then: x^2=1/81, 1/x=9 and 2x=2/9 1/81<9<2/9 [Incorrect]
Let's check the III option for above values: III. 2x<x^2<1/x For x=1/2: 1<1/4<2 [Incorrect] For x=1/9: 2/9<1/81<9 [Incorrect]
So, B should be the correct answer. Please check. OA IS D.Algebraic approach is given in my solution. Here is number picking: I. \(x^2<2x<\frac{1}{x}\) > \(x=\frac{1}{2}\) > \(x^2=\frac{1}{4}\), \(2x=1\), \(\frac{1}{x}=2\) > \(\frac{1}{4}<1<2\). Hence this COULD be the correct ordering. II. \(x^2<\frac{1}{x}<2x\) > \(x=0.9\) > \(x^2=0.81\), \(\frac{1}{x}=1.11\), \(2x=1.8\) > \(0.81<1.11<1.8\). Hence this COULD be the correct ordering. III. \(2x<x^2<\frac{1}{x}\) > \(x^2\) to be more than \(2x\), \(x\) must be more than 2 (for positive \(xes\)). But if \(x>2\), then \(\frac{1}{x}\) is the least value from these three and can not be more than \(2x\) and \(x^2\). So III can not be true. Thus I and II could be correct ordering and III can not. Answer: D.
_________________
New to the Math Forum? Please read this: All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Manager
Joined: 23 Apr 2010
Posts: 127
Location: Tx
Schools: NYU,UCLA,BOOTH,STANFORD

Re: If x is positive, which of the following could be correct ordering of [#permalink]
Show Tags
03 Sep 2010, 07:28
For the 1 Pick any number 0<x<1 it is correct For the 2 pick one number like 1/2 you will see that it is not correcy but dont rush pick another number which close to 1 but smaller then one like 9/10 and tried you will see it is right. so the answer is D with that kind of guestions pick one less then 1/2 and greater then 1/2 less then 1.
_________________
This is not finished here...Watch me.....



Intern
Joined: 17 Dec 2009
Posts: 21

Re: If x is positive, which of the following could be correct ordering of [#permalink]
Show Tags
03 Sep 2010, 07:44
As we can see that question used term "COULD", so even if any of the choice satisfy the requirement it will be an answer. Now let look at the choices:
Choice I: Just put x=0.5, we can see that it satisfy the condition Choice II: Put x=2, we can see that it also satisfy the condition Choice III: For any value of x, it does not satisfy the condition.
So we can easily figure out that the answer is D.
Lemme me know if any one have any question on this.



Manager
Joined: 16 Jun 2010
Posts: 183

Re: If x is positive, which of the following could be correct ordering of [#permalink]
Show Tags
03 Oct 2010, 14:28
Bunuel wrote: So, we are left with \(0<x<1\): In this case \(x^2\) is least value, so we are left with:
I. \(x^2<2x<\frac{1}{x}\) > can \(2x<\frac{1}{x}\)? Can \(\frac{2x^21}{x}<0\), the expression \(2x^21\) can be negative or positive for \(0<x<1\). (You can check it either algebraically or by picking numbers)
II. \(x^2<\frac{1}{x}<2x\) > can \(\frac{1}{x}<2x\)? The same here \(\frac{2x^21}{x}>0\), the expression \(2x^21\) can be negative or positive for \(0<x<1\).
Great explanation Bunuel, but for curiosity purpose, as I understand \(2x^21\) should be negative for this equation \(\frac{2x^21}{x}<0\) to be true. However if I algebraically find the values for which \(2x^21\) is negative, then on plugging those values in \(x^2<\frac{1}{x}<2x\) I do not find that the equation satisfies. Instead it is the value for which \(2x^21\) is positive, that I find the end quation satifies .... why is that ?
_________________
Please give me kudos, if you like the above post. Thanks.



Math Expert
Joined: 02 Sep 2009
Posts: 40432

Re: If x is positive, which of the following could be correct ordering of [#permalink]
Show Tags
03 Oct 2010, 14:47
devashish wrote: Bunuel wrote: So, we are left with \(0<x<1\): In this case \(x^2\) is least value, so we are left with:
I. \(x^2<2x<\frac{1}{x}\) > can \(2x<\frac{1}{x}\)? Can \(\frac{2x^21}{x}<0\), the expression \(2x^21\) can be negative or positive for \(0<x<1\). (You can check it either algebraically or by picking numbers)
II. \(x^2<\frac{1}{x}<2x\) > can \(\frac{1}{x}<2x\)? The same here \(\frac{2x^21}{x}>0\), the expression \(2x^21\) can be negative or positive for \(0<x<1\).
Great explanation Bunuel, but for curiosity purpose, as I understand \(2x^21\) should be negative for this equation \(\frac{2x^21}{x}<0\) to be true. However if I algebraically find the values for which \(2x^21\) is negative, then on plugging those values in \(x^2<\frac{1}{x}<2x\) I do not find that the equation satisfies. Instead it is the value for which \(2x^21\) is positive, that I find the end quation satifies .... why is that ? You are mixing I and II. If you find the values of \(x\) from the range \(0<x<1\) for which \(2x^21\) is negative then \(x^2<2x<\frac{1}{x}\) will hold true (not \(x^2<\frac{1}{x}<2x\)). Below is number plugging method: I. \(x^2<2x<\frac{1}{x}\) > \(x=\frac{1}{2}\) > \(x^2=\frac{1}{4}\), \(2x=1\), \(\frac{1}{x}=2\) > \(\frac{1}{4}<1<2\). Hence this COULD be the correct ordering. II. \(x^2<\frac{1}{x}<2x\) > \(x=0.9\) > \(x^2=0.81\), \(\frac{1}{x}=1.11\), \(2x=1.8\) > \(0.81<1.11<1.8\). Hence this COULD be the correct ordering. III. \(2x<x^2<\frac{1}{x}\) > \(x^2\) to be more than \(2x\), \(x\) must be more than 2 (for positive \(xes\)). But if \(x>2\), then \(\frac{1}{x}\) is the least value from these three and can not be more than \(2x\) and \(x^2\). So III can not be true. Thus I and II could be correct ordering and III can not. Answer: D.
_________________
New to the Math Forum? Please read this: All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Manager
Joined: 20 Aug 2011
Posts: 146

Re: If x is positive, which of the following could be correct ordering of [#permalink]
Show Tags
24 Oct 2011, 18:15
ammulujnt wrote: A is correct ans check substituting x=3 and x=0.1 if x=3 then x2>2x and if x=0.1 then x2<2x so its no way possible to decide this inequality If x = 0.9, then x^2 < 1/x < 2x (so 2 is possible) Answer is D. With "could" wording, look only for scenarios that work in at least one circumstance rather than in all circumstances.



Intern
Status: single
Joined: 16 Oct 2011
Posts: 6
Location: France, Metropolitan
Concentration: Marketing, Strategy
GMAT 1: 650 Q45 V34 GMAT 2: 660 Q43 V38
GPA: 3.5
WE: Supply Chain Management (Energy and Utilities)

Re: If x is positive, which of the following could be correct ordering of [#permalink]
Show Tags
26 Dec 2011, 11:03
3
This post received KUDOS
IMO the clearest way to solve the problem is to plot the three curves / line:  x^2 is a parabola tangent to the origin and passes through point (1,1) and (2,4)  1/x is the hyperbola tangent to the xaxis and the yaxis that passes through (1,1)  2x is the line that passes through the origin and (1,2) The intersections among the three curves define the 4 possible cases: for x<1/\sqrt{2}: x^2 < 2x < 1/x for 1/\sqrt{2}<x<1: x^2 < 1/x < 2x for 1<x<2: 1/x < x^2 < 2x for x>2: 1/x < 2x < x^2 It's much easier when you draw the three curves and notice the intersection points (but don't know at this point how to include an image). It did not occur to me to draw them during the test and mistankenly chose answer B. With retrospect, the only way I could have figured out the four zones was by drawing. Hope this helps



Intern
Joined: 03 Jan 2011
Posts: 44
Schools: McCombs, ISB, Insead, LBS
WE 1: 1.5
WE 2: 5.5

Re: If x is positive, which of the following could be correct ordering of [#permalink]
Show Tags
07 Feb 2012, 13:51
16
This post received KUDOS
7
This post was BOOKMARKED
Copied from an other forum. Thought it might help someone. This is a great explanation...  Each one of these gives you two inequalities. You know that x is positive, so you don't need to worry about the sign changing direction. (1) x^2<2x<1/x This means that x^2<2x so divide by x to get x<2. The second one tells you that 2x<1/x which simplifies to x < 1/sqrt(2). These can obviously both be satisfied at the same time, so (1) works. (2) x^2<1/x<2x This means that x^2<1/x which gives x^3<1, or x<1. The second half gives you 1/x<2x or 1<2(x^2) or x>1/sqrt(2). So any number that satisfies 1/sqrt(2)<x<1 will work. (3) 2x<x^2<1/x. The first part gives 2x<x^2 or x>2. The second half gives x^2<1/x or x^3<1 or x<1. Since the regions x>2 and x<1 do not overlap, (3) can not be satisfied. The Answer choice is (4), 1 and 2 only.
_________________
You can see from the number of "Kudos Given" on my profile that I am generous while giving Kudos. Is it too much to ask for a similar favor???



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7520
Location: Pune, India

Re: If x is positive, which of the following could be correct ordering of [#permalink]
Show Tags
07 Feb 2012, 23:49
10
This post received KUDOS
Expert's post
5
This post was BOOKMARKED
Vavali wrote: If x is positive, which of the following could be correct ordering of \(\frac{1}{x}\), \(2x\), and \(x^2\)?
(I) \(x^2 < 2x < \frac{1}{x}\) (II) \(x^2 < \frac{1}{x} < 2x\) (III) \(2x < x^2 < \frac{1}{x}\)
(a) none (b) I only (c) III only (d) I and II (e) I, II and III Let's look at this question logically. There will be some key takeaways here so don't focus on the question and the (long) solution. Focus on the logic. First of all, we are just dealing with positives so life is simpler. To compare two terms e.g. \(x^2\) and \(2x\), we should focus on the points where they are equal. \(x^2 = 2x\) holds when \(x = 2\). When \(x < 2, x^2 < 2x\) When \(x > 2, x^2 > 2x\) Similarly \(1/x = x^2\) when \(x = 1\) When \(x < 1, 1/x > x^2\). When \(x > 1, 1/x < x^2\) Going on, \(1/x = 2x\) when \(x = 1/\sqrt{2}\) When \(x < 1/\sqrt{2}, 1/x > 2x\) When \(x > 1/\sqrt{2}, 1/x < 2x\) So now you know that: If \(x < 1/\sqrt{2}\), \(1/x > 2x, 1/x > x^2\) and \(x^2 < 2x\) So \(x^2 < 2x < 1/x\) is possible. If \(1/\sqrt{2} < x < 1\) \(1/x < 2x, 1/x > x^2\) So \(x^2 < 1/x < 2x\) is possible. If \(1 < x < 2\) \(1/x < 2x, 1/x < x^2, x^2 < 2x\) So \(1/x < x^2 < 2x\) is possible. If \(x > 2\) \(1/x < 2x, 1/x < x^2, x^2 > 2x\) So \(1/x < 2x < x^2\) is possible. For no positive values of x is the third relation possible. *Edited
_________________
Karishma Veritas Prep  GMAT Instructor My Blog
Get started with Veritas Prep GMAT On Demand for $199
Veritas Prep Reviews



Senior Manager
Joined: 07 Sep 2010
Posts: 332

Re: If x is positive, which of the following could be correct ordering of [#permalink]
Show Tags
27 Feb 2012, 02:52
Hi Bunnel, Could you please provide a reasoning to the below text... how did you find the range...Pls help Bunuel wrote: gautamsubrahmanyam wrote: I am not sure how the OA is D
when x=1/2 then 1/x is 2 ,2x is 1 and x^2 is 1/2 ,this satisfies (I) x^2<2x<1/x
when x =3 then 1/x is 1/3, 2x is 6 and x^2 is 9 ,this does not satisfy (II)
when x = 1/10 then 1/x is 10 , 2x is 1/5 and x^2 is 1/100,this again does not satify (II)
Even ve numbers dont seem to work
when x=3 then 1/x is 1/3 ,2x is 6 and X^2=9,this does not satisfy (II) x=1/3 then 1/x is 3 ,2x is 2/3 and x^2=1/9,this does not satisfy (II)
Can any one give an example which satisfies option (II) 7. If x is positive, which of the following could be the correct ordering of 1/x,2x and x^2 ? I. x^2<2x<1/x II. x^2<1/x<2x III. 2x<x^2<1/x (A) None (B) I only (C) III only (D) I and II only (E) I II and III First note that we are asked "which of the following COULD be the correct ordering" not MUST be. Basically we should determine relationship between \(x\), \(\frac{1}{x}\) and \(x^2\) in three areas: \(0<1<2<\).
\(x>2\)
\(1<x<2\)
\(0<x<1\)When \(x>2\) > \(x^2\) is the greatest and no option is offering this, so we know that x<2. If \(1<x<2\) > \(2x\) is greatest then comes \(x^2\) and no option is offering this. So, we are left with \(0<x<1\): In this case \(x^2\) is least value, so we are left with: I. \(x^2<2x<\frac{1}{x}\) > can \(2x<\frac{1}{x}\)? Can \(\frac{2x^21}{x}<0\), the expression \(2x^21\) can be negative or positive for \(0<x<1\). (You can check it either algebraically or by picking numbers) II. \(x^2<\frac{1}{x}<2x\) > can \(\frac{1}{x}<2x\)? The same here \(\frac{2x^21}{x}>0\), the expression \(2x^21\) can be negative or positive for \(0<x<1\). (You can check it either algebraically or by picking numbers) Answer: D. Second condition: \(x^2<\frac{1}{x}<2x\) The question is which of the following COULD be the correct ordering not MUST be. Put \(0.9\) > \(x^2=0.81\), \(\frac{1}{x}=1.11\), \(2x=1.8\) > \(0.81<1.11<1.8\). Hence this COULD be the correct ordering. Hope it's clear. [quote="Bunuel"][quote="gautamsubrahmanyam"]I am not sure how the OA is D
_________________
+1 Kudos me, Help me unlocking GMAT Club Tests



Math Expert
Joined: 02 Sep 2009
Posts: 40432

Re: If x is positive, which of the following could be correct ordering of [#permalink]
Show Tags
27 Feb 2012, 03:07
imhimanshu wrote: Hi Bunnel,
Could you please provide a reasoning to the below text... how did you find the range...Pls help The reasoning is that in these ranges x (2x), 1/x and x^2 are ordered differently: For \(x>2\) > \(x^2\) has the largest value. Since no option offers this we know that \(x\) cannot be more that 2; For \(1<x<2\) > \(2x\) has the largest value, then comes \(x^2\). Since no option offers this we know that \(x\) cannot be from this range either; So, we are left with last range: \(0<x<1\). In this case \(x^2\) has the least value. Options, I and II offer this, so we should concentrate on them and test the values of x from 0 to 1. Hope it's clear.
_________________
New to the Math Forum? Please read this: All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Senior Manager
Joined: 25 Nov 2011
Posts: 250
Location: India
Concentration: Technology, General Management
GPA: 3.95
WE: Information Technology (Computer Software)

Re: If x is positive, which of the following could be correct ordering of [#permalink]
Show Tags
27 Feb 2012, 03:23
2
This post received KUDOS
@imhimanshu In this kind of questions, it is best to check with numbers. Also, when fractions are involved in the question, test with 3 values: < 0.5; = 0.5 ; > 0.5 This distribution definitely helps you.
_________________
 Aravind Chembeti



Manager
Joined: 10 Feb 2012
Posts: 71
Location: India
Concentration: Marketing, Strategy
GPA: 3.45
WE: Marketing (Pharmaceuticals and Biotech)

Re: If x is positive, which of the following could be correct ordering of [#permalink]
Show Tags
27 Feb 2012, 08:23
1
This post received KUDOS
The answer would be D
(I) this is clear and is easily deciphered ...
but for (II) x^2 < 1/x < 2x
This means that x^2<1/x which gives x^3<1, or x<1. The second half gives you 1/x<2x or 1<2(x^2) or x>1/sqrt(2). So any number that satisfies 1/sqrt(2)<x<1.. hence (II) COULD also be true for some values..



Retired Moderator
Status: 2000 posts! I don't know whether I should feel great or sad about it! LOL
Joined: 04 Oct 2009
Posts: 1655
Location: Peru
Schools: Harvard, Stanford, Wharton, MIT & HKS (Government)
WE 1: Economic research
WE 2: Banking
WE 3: Government: Foreign Trade and SMEs

Re: If x is positive, which of the following could be correct ordering of [#permalink]
Show Tags
27 Jun 2012, 14:11
1
This post received KUDOS
Bunuel, please check my reasoning about statement III: Statement III says \(2x < x^2 < \frac{1}{x}\) So, we analyze part by part of this compound inequality: a) \(2x < x^2\) then, \(x>2\) b) \(x^2 < \frac{1}{x}\) then,\(x^3<1\) ==> \(x < 1\) So, we have \(x>2\) and \(x<1\). That's impossible! if \(x>2\), x cannot be less than 1 at the same time. Statement III could not be correct! Please, confirm.
_________________
"Life’s battle doesn’t always go to stronger or faster men; but sooner or later the man who wins is the one who thinks he can."
My Integrated Reasoning Logbook / Diary: http://gmatclub.com/forum/myirlogbookdiary133264.html
GMAT Club Premium Membership  big benefits and savings



Math Expert
Joined: 02 Sep 2009
Posts: 40432

Re: If x is positive, which of the following could be correct ordering of [#permalink]
Show Tags
28 Jun 2012, 02:15
metallicafan wrote: Bunuel, please check my reasoning about statement III:
Statement III says \(2x < x^2 < \frac{1}{x}\)
So, we analyze part by part of this compound inequality:
a) \(2x < x^2\) then, \(x>2\)
b) \(x^2 < \frac{1}{x}\)
then,\(x^3<1\) ==> \(x < 1\)
So, we have \(x>2\) and \(x<1\). That's impossible! if \(x>2\), x cannot be less than 1 at the same time. Statement III could not be correct!
Please, confirm. Yes, your reasoning for option III is correct.
_________________
New to the Math Forum? Please read this: All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics




Re: If x is positive, which of the following could be correct ordering of
[#permalink]
28 Jun 2012, 02:15



Go to page
1 2 3
Next
[ 50 posts ]




