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:

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.

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 \(\frac{2x^2-1}{x}<0\), the expression \(2x^2-1\) 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 \(\frac{2x^2-1}{x}>0\), the expression \(2x^2-1\) 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.

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 \(\frac{2x^2-1}{x}<0\), the expression \(2x^2-1\) 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 \(\frac{2x^2-1}{x}>0\), the expression \(2x^2-1\) 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.

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^2-2x < 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.

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 \(x-es\)). 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.

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.
_________________

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.

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 \(\frac{2x^2-1}{x}<0\), the expression \(2x^2-1\) 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 \(\frac{2x^2-1}{x}>0\), the expression \(2x^2-1\) can be negative or positive for \(0<x<1\).

Great explanation Bunuel, but for curiosity purpose, as I understand \(2x^2-1\) should be negative for this equation \(\frac{2x^2-1}{x}<0\) to be true. However if I algebraically find the values for which \(2x^2-1\) 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 \(2x^2-1\) is positive, that I find the end quation satifies .... why is that ?
_________________

Please give me kudos, if you like the above post. Thanks.

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 \(\frac{2x^2-1}{x}<0\), the expression \(2x^2-1\) 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 \(\frac{2x^2-1}{x}>0\), the expression \(2x^2-1\) can be negative or positive for \(0<x<1\).

Great explanation Bunuel, but for curiosity purpose, as I understand \(2x^2-1\) should be negative for this equation \(\frac{2x^2-1}{x}<0\) to be true. However if I algebraically find the values for which \(2x^2-1\) 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 \(2x^2-1\) 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 \(2x^2-1\) 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 \(x-es\)). 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.

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 x-axis and the y-axis 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.

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???

(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.

Re: If x is positive, which of the following could be correct [#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 \(\frac{2x^2-1}{x}<0\), the expression \(2x^2-1\) 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 \(\frac{2x^2-1}{x}>0\), the expression \(2x^2-1\) 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
_________________

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.

Re: If x is positive, which of the following could be correct [#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.
_________________

Re: If x is positive, which of the following could be correct [#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..

gmatclubot

Re: If x is positive, which of the following could be correct
[#permalink]
27 Feb 2012, 08:23

There’s something in Pacific North West that you cannot find anywhere else. The atmosphere and scenic nature are next to none, with mountains on one side and ocean on...

Joe Navarro is an ex FBI agent who was a founding member of the FBI’s Behavioural Analysis Program. He was a body language expert who he used his ability to successfully...