Last visit was: 06 Sep 2026, 19:58 It is currently 06 Sep 2026, 19:58
Home
Messages
Tests
Schools
Events
Close
GMAT Club Daily Prep
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.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel
Back to Forum
Create Topic
   1   2   3 
User avatar
ramiav
User avatar
Joined: 31 Oct 2022
Last visit: 19 Jun 2025
Posts: 27
Own Kudos:
60
Given Kudos: 16
Location: Spain
Concentration: Entrepreneurship, Technology
Schools: MIT '27 (WL) HBS '27 Stanford '27 (I) Columbia '27 (D) Haas '27 (S)
GMAT Focus 1: 705 Q86 V85 DI84
Schools: MIT '27 (WL) HBS '27 Stanford '27 (I) Columbia '27 (D) Haas '27 (S)
GMAT Focus 1: 705 Q86 V85 DI84
Posts: 27
Kudos: 60
If x is positive, which of the following could be correct ordering of 14 Mar 2024, 02:53
Kudos
Add Kudos
Bookmarks
Bookmark this Post
­My answer for visual thinkers, similar to the Bunuel with the critical points

( I find useful to learn how to draw 1/x - with vertical and horizontal offsets - the same as known line equations and parabolas )

You can see the 4 different zones A,B, P, Q - sorry for the letters
the limits are (0,1) (1, something between 1 and 2)((something between 1 and 2, 2 )(2, inf)


Show Spoiler
Attachment:
GMAT-Club-Forum-167dfckl.png
GMAT-Club-Forum-167dfckl.png [ 267.01 KiB | Viewed 725 times ]
User avatar
de1234
User avatar
Joined: 04 Dec 2024
Last visit: 02 Aug 2026
Posts: 36
Own Kudos:
2
Given Kudos: 2
GMAT Focus 1: 685 Q83 V83 DI86
GMAT Focus 2: 695 Q88 V82 DI83
Products:
My Reviews
GMAT Focus 2: 695 Q88 V82 DI83
Posts: 36
Kudos: 2
If x is positive, which of the following could be correct ordering of 15 Jul 2025, 06:11
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Vavali
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
Show more
Here, Since we are asked a possible ordering, we should check each equation for certain value. If none of the iterations are true for the equality, we can say that the equality is wrong and if even one is correct we will say that the equality is correct.

Here selecting numbers to test the equality is the tricky part.

let's say that we have the numbers 2x,1/x and x^2 to compare. obviously the equality will change depending on the value of x. Observe each individual equality and make a decision. ( i assumed 1/2,0.9,1.5,3)
User avatar
wArrior902
User avatar
Joined: 19 Jan 2023
Last visit: 08 Aug 2026
Posts: 20
Own Kudos:
7
Given Kudos: 48
Posts: 20
Kudos: 7
If x is positive, which of the following could be correct ordering of 18 Aug 2025, 09:56
Kudos
Add Kudos
Bookmarks
Bookmark this Post
This is very easily solvable using graphs. Create graphs of y = x^2, 2x and 1/x.

Graph of 1/x would be close to infinite at x = 0 and a smooth curve which goes to 0 as x goes to infinite. Make sure to intersect y = 1/x and x^2 at x = 1 and y = 1/x and 2x at x = 1/sqrt(2). Now you can see option 1 and 2 are correct from graphs directly.
User avatar
Dream009
User avatar
Joined: 05 Nov 2024
Last visit: 13 Mar 2026
Posts: 276
Own Kudos:
119
Given Kudos: 60
Location: India
Concentration: Strategy, Leadership
Schools: NYU Stern ESSEC IIM
GMAT Focus 1: 615 Q82 V79 DI80
GRE 1: Q30 V50
GPA: 84
WE:General Management (Consulting)
Products:
My Reviews
Schools: NYU Stern ESSEC IIM
GMAT Focus 1: 615 Q82 V79 DI80
GRE 1: Q30 V50
Posts: 276
Kudos: 119
If x is positive, which of the following could be correct ordering of 20 Dec 2025, 21:28
Kudos
Add Kudos
Bookmarks
Bookmark this Post
easy way alert x is positive and that's the biggest saving grace here.
Let me show for statement 2
x^2<1/x
=> x^3 < 1
=> x < 1
So x^# is b/w 0 & 1
Also 1< 2x^2
0.5 < x^2
So X^2 can be 0.9 and whatever is that square root multiple with that again and it will surely be less than 1
Bingo this is possible

Apply same to Statement 3
2x<x^2
2<x
Also
X^2 <1/x
x^3<1
=> x <1
X has to be greater than 2 but other side says x is less than 1 - not possible

Similarly solve 1 and you will get the ans
User avatar
Rickooreoisb
User avatar
Joined: 18 Jul 2025
Last visit: 02 Sep 2026
Posts: 153
Own Kudos:
46
Given Kudos: 575
Location: India
Concentration: Finance, Marketing
GPA: 9
WE:Investment Banking (Finance: Investment Banking)
Posts: 153
Kudos: 46
If x is positive, which of the following could be correct ordering of 12 Jan 2026, 11:18
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Whenever you are given such cases to such, universe of values are

-2, -1.1, -1, -0.9, -0.5, 0, 0.5, 0.9, 1, 1.1 and 2

Since here the value of x is positive, try only 0.5, 0.9, 1, 1.1 and 2 and for 0.5 use it in fraction form
User avatar
Kinshook
User avatar
Major Poster
Joined: 03 Jun 2019
Last visit: 06 Sep 2026
Posts: 6,129
Own Kudos:
6,089
Given Kudos: 164
Location: India
Schools: HBS '26 CBS '26 Wharton '26
GMAT 1: 690 Q50 V34
WE:Engineering (Transportation)
Products:
My Reviews
Schools: HBS '26 CBS '26 Wharton '26
GMAT 1: 690 Q50 V34
Posts: 6,129
Kudos: 6,089
If x is positive, which of the following could be correct ordering of 13 Jan 2026, 03:49
Kudos
Add Kudos
Bookmarks
Bookmark this Post
If x is positive, which of the following could be correct ordering of \(\frac{1}{x}\), \(2x\), and \(x^2\)?

I. \(x^2 < 2x < \frac{1}{x}\)
x^2 - 2x < 0; x(x-2)<0; 0<x<2
2x - 1/x < 0; (2x^2 - 1)/x < 0; Since x>0; 2x^2 - 1< 0; x^2 < 1/2; \(x < 1/\sqrt{2}\)
For\( 0<x<\frac{1}{\sqrt{2}}\), Possible

II. \(x^2 < \frac{1}{x} < 2x\)
x^2 - 1/x < 0; (x^3 - 1)/x < 0; x^3 - 1 < 0; (x-1)(x^2 + x + 1) < 0; x < 1 since (x+1/2)^2 + 3/4 > 0
1/x - 2x < 0; (1 - 2x^2)/x < 0; x^2 > 1/2; \(x > \frac{1}{\sqrt{2}}\)
\(\frac{1}{\sqrt{2}} < x < 1\); Possible

III. \(2x < x^2 < \frac{1}{x}\)
2x - x^2 < 0; x(x-2) > 0; x > 2
x^2 - 1/x < 0
(x^3 -1)/x < 0; (x-1)(x^2 + x + 1) < 0; x < 1 since (x+1/2)^2 + 3/4 > 0
Not possible since x < 1 & x> 2 is not possible simultaneously

(A) none
(B) I only
(C) III only
(D) I and II
(E) I, II and III

IMO D
User avatar
messi16
User avatar
Joined: 13 Aug 2025
Last visit: 03 May 2026
Posts: 44
Own Kudos:
4
Given Kudos: 3
Posts: 44
Kudos: 4
If x is positive, which of the following could be correct ordering of 01 Mar 2026, 17:57
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Honestly, this one is just one where I recommend solving by trying different numbers. If you try 1, 2, 3, 4, etc. you will notice a pattern that is not one of the options. HOWEVER, if you use decimals (I used .5 and .75) you will notice that I and II are the correct answers, thus answer choice D.
User avatar
kaustubhkohli12
User avatar
Joined: 25 Jul 2023
Last visit: 05 Sep 2026
Posts: 78
Own Kudos:
11
Given Kudos: 4
Products:
GMAT Club Tests
Posts: 78
Kudos: 11
If x is positive, which of the following could be correct ordering of 26 Jun 2026, 09:52
Kudos
Add Kudos
Bookmarks
Bookmark this Post
In this question, if x can be negative then all three can be possible right ?

Bunuel
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


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.

Answer: D.
Show more
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 06 Sep 2026
Posts: 113,167
Own Kudos:
839,601
Given Kudos: 111,337
Products:
GMAT Club Tests
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 113,167
Kudos: 839,601
If x is positive, which of the following could be correct ordering of 26 Jun 2026, 10:35
Kudos
Add Kudos
Bookmarks
Bookmark this Post
kaustubhkohli12
In this question, if x can be negative then all three can be possible right ?


Show more

No. If x is negative, then 1/x and 2x are both negative, while x^2 is positive.

So x^2 cannot be less than either 1/x or 2x. Therefore, I and II are impossible for negative x.

Also, III is impossible because it requires x^2 < 1/x, but x^2 is positive and 1/x is negative.

So if x is negative, none of the three orderings is possible.
   1   2   3 
Moderator:
Bunuel
Math Expert
113167 posts