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:

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

Show Tags

28 Jul 2012, 20:24

The question asked "what could be the correct ordering" means it asked for the possibilities. What is question asked "what must be the correct ordering" ? In that case would we be required to choose an option which is true for all scenarios ?

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

Show Tags

29 Jul 2012, 01:37

5

This post received KUDOS

Please, refer to the attached drawing, in which the three graphs \(y=1/x,\) \(y=2x,\) and \(y=x^2\) are depicted for \(x>0\). The exact values for A, B, and C can be worked out, but they are not important to establish the order of the three algebraic expressions.

So, the correct orderings are: If \(x\) between 0 and A: \(x^2<2x<1/x\) If \(x\) between A and B: \(x^2<1/x<2x\) If \(x\) between B and C: \(1/x<x^2<2x\) If \(x\) greater than C: \(1/x<2x<x^2\)

We can see that only the first two of the above options are listed as answers (I and II).

Answer: D.

Attachments

3Graphs.jpg [ 15.15 KiB | Viewed 7121 times ]

_________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

The question asked "what could be the correct ordering" means it asked for the possibilities. What is question asked "what must be the correct ordering" ? In that case would we be required to choose an option which is true for all scenarios ?

The ordering will be different for different values of x so the question cannot ask for a single correct ordering.
_________________

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

Show Tags

08 Aug 2012, 00:50

given that x>0 1 ) x^2 < 2X<1/X lets just check x^2 < 2X => 0 <x<2 ; 2X<1/X => 0<x<1/\sqrt{2} these 2 inequities do not conradict each other. so, 1) is ok

2) x^2 <1/X< 2X check them - x^2 <1/X => 0<x<1 ; 1/X< 2X => x> 1/\sqrt{2} these 2 inequities do not conradict each other. so, 2) is ok

3) 2X< x^2 <1/X check them - 2X< x^2 => x>2 ; x^2 <1/X => x<1 these 2 inequities conradict each other. so, 3) is not ok

p.s. dont know why such symbols as sqroot , fraction ets dont work.
_________________

Happy are those who dream dreams and are ready to pay the price to make them come true

I am still on all gmat forums. msg me if you want to ask me smth

(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 \(x > 1, 1/x > x^2\)Are you sure this is correct?I think.. we can use x=4 here, then 1/4>16 .. which is not correct.

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 \(x > 1\) \(1/x < 2x, 1/x > x^2\) So \(x^2 < 1/x < 2x\) is possible. (Same as above)

For no positive values of x is the third relation possible.

(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 \(x > 1, 1/x > x^2\)Are you sure this is correct?I think.. we can use x=4 here, then 1/4>16 .. which is not correct.

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 \(x > 1\) \(1/x < 2x, 1/x > x^2\) So \(x^2 < 1/x < 2x\) is possible. (Same as above)

For no positive values of x is the third relation possible.

That is a typo. If you notice, for every case, the relation is opposite on the opposite sides of the equality value. So the relation that holds in x < 1 will be opposite to the relation that holds when x > 1. That did mess up the entire explanation. Good you pointed it out. I have edited the original post.
_________________

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

Show Tags

02 Oct 2013, 09:40

Hello Bunuel, Can you show us graphical approach to this question. I was able to draw the graph for all three equations and intersection points, however, I was not able to negate the third ordering. Would you please help me out.

Thanks imhimanshu P.S - How can I post graphs here.
_________________

Hello Bunuel, Can you show us graphical approach to this question. I was able to draw the graph for all three equations and intersection points, however, I was not able to negate the third ordering. Would you please help me out.

Thanks imhimanshu P.S - How can I post graphs here.

Here is the graph:

Attachment:

Ques3.jpg [ 11.4 KiB | Viewed 5305 times ]

III. 2x < x^2 < 1/x

For 2x to be less than x^2, the graph of 2x should lie below the graph of x^2. This happens when the graph of 2x is the red line. For x^2 to be less than 1/x at the same time, the graph of x^2 should lie below the graph of 1/x in the region of the red line. But in the region of the red line, the graph of x^2 is never below the graph of 1/x. It will never be because graph of 1/x is going down toward y = 0 while graph of x^2 is going up toward y = infinity. Hence this inequality will not hold for any region.
_________________

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

Show Tags

22 Feb 2014, 08:48

I picked numbers: 1/2, 1, 3/2, 2, 3

However, it didn occur to me that I must look something like 0.9. Request experts to help me understand the logic behind picking such numbers. Have my actual GMAT in 10 days, any help would be immensely valuable!

However, it didn occur to me that I must look something like 0.9. Request experts to help me understand the logic behind picking such numbers. Have my actual GMAT in 10 days, any help would be immensely valuable!

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

Show Tags

30 Jul 2014, 20:40

What a horrific problem?! I think a fast and simple way is to graph all functions and compare all vertically. Doing this, it can be easy to see that III is impossible.

(a) none (b) I only (c) III only (d) I and II (e) I, II and III

first, let's get rid of at least one x in the given expressions, as x is positive just multiply the expressions by x and we'll get (I) \(x^3 < 2x^2 < 1\) as \(2x^2\) < 1 we must pick a value which is < 1, let's pick 1/2 and it works. COULD BE

(II) \(x^3 < 1< 2x^2\) Here we can see that \(x^3\)<1 so we must pick a value <1 BUT which will make \(2x^2\) >1 if possible. Let's pick 0.9 and it works also here. COULD BE

(III) \(2x^2 < x^3 < 1\) We must pick a value < 1 BUT as we've already seen, if we pick a fraction < 1 we cannot make \(2x^2 < x^3\), in the above cases it \(2x^2 was > x^3\) each time we picked a fraction < 1

Hope it helps.
_________________

When you’re up, your friends know who you are. When you’re down, you know who your friends are.

Share some Kudos, if my posts help you. Thank you !

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

Show Tags

03 Jan 2016, 03:02

Hi Experts, In solving x^2<2x=>x^2-2x<0=>x(x-2)<0=>x<0 or x<2

why did we ignore x<0 is it bcz we are told in question that x is positive number...it is so then why can't it be ignored in statement 3...btw I understood the number picking methodology, just little bit curious...Thanks in advance.