Last visit was: 18 Nov 2025, 17:36 It is currently 18 Nov 2025, 17:36
Home
GMAT
Messages
Tests
Schools
Events
Rewards
Chat
My Profile
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
705-805 Level|   Inequalities|         
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 18 Nov 2025
Posts: 105,355
Own Kudos:
778,072
 [4]
Given Kudos: 99,964
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,355
Kudos: 778,072
 [4]
12 Days of Christmas GMAT Competition - Day 12: Which of the following 25 Dec 2021, 07:15
1
Kudos
Add Kudos
3
Bookmarks
Bookmark this Post
00:00
Start Timer
Pause Timer
Resume Timer
Show Answer
Hide
Show
History
C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

75% (hard)

Question Stats:

59% (02:19) correct 41% (02:26) wrong based on 367 sessions

History

Date
Time
Result
Not Attempted Yet
12 Days of Christmas GMAT Competition with Lots of Fun

Which of the following can never be correct?

I: \(x < x^2 < 1-x\)

II: \(x < x^2 < \frac{1}{x} \)

III: \(x < \frac{1}{x} < x^2\)

IV: \(x < \frac{1}{x} < x^2 < x^3\)

A. II only
B. II and III only
C. II and IV only
D. I, II, and IV only
E. I, II, III, and IV only


 


This question was provided by Experts'Global
for the 12 Days of Christmas Competition

Win $25,000 in prizes: Courses, Tests & more

 

ShowHide Answer
Official Answer
C
Most Helpful Reply
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 18 Nov 2025
Posts: 105,355
Own Kudos:
778,072
 [1]
Given Kudos: 99,964
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,355
Kudos: 778,072
 [1]
12 Days of Christmas GMAT Competition - Day 12: Which of the following 26 Dec 2021, 08:00
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
12 Days of Christmas GMAT Competition with Lots of Fun

Which of the following can never be correct?

I: \(x < x^2 < 1-x\)

II: \(x < x^2 < \frac{1}{x} \)

III: \(x < \frac{1}{x} < x^2\)

IV: \(x < \frac{1}{x} < x^2 < x^3\)

A. II only
B. II and III only
C. II and IV only
D. I, II, and IV only
E. I, II, III, and IV only


Show more

Expert's Global Official Video Explanation

General Discussion
User avatar
ElninoEffect
User avatar
Joined: 07 Apr 2020
Last visit: 02 Sep 2022
Posts: 319
Own Kudos:
491
 [1]
Given Kudos: 72
Location: India
Concentration: Strategy, Technology
Schools: IIMA PGPX'23 ISB '23 XLRI
WE:Engineering (Computer Software)
Schools: IIMA PGPX'23 ISB '23 XLRI
Posts: 319
Kudos: 491
 [1]
12 Days of Christmas GMAT Competition - Day 12: Which of the following 25 Dec 2021, 07:50
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
12 Days of Christmas GMAT Competition with Lots of Fun

Which of the following can never be correct?

I: \(x < x^2 < 1-x\)

II: \(x < x^2 < \frac{1}{x} \)

III: \(x < \frac{1}{x} < x^2\)

IV: \(x < \frac{1}{x} < x^2 < x^3\)

A. II only
B. II and III only
C. II and IV only
D. I, II, and IV only
E. I, II, III, and IV only


 


This question was provided by Experts'Global
for the 12 Days of Christmas Competition

Win $25,000 in prizes: Courses, Tests & more

 

Show more

We just need to prove it right for only one case, if we find even one value that satisfies the condition that is out.

I: \(x < x^2 < 1-x\)
x = -0.5 satisfies the equation so its out.

II: \(x < x^2 < \frac{1}{x} \)
check for conditions:
x<0
-0.5 No.
-2 No
x>0
0.5 No
2 No
Let's keep it.

III: \(x < \frac{1}{x} < x^2\)
x = -2
Satisfies the equation. Out.

IV: \(x < \frac{1}{x} < x^2 < x^3\)
Putting all the values used in II we will find there are no values which satisfies the condition.
Keep it.


So our answer is II and IV.

C
User avatar
gmatophobia
User avatar
Quant Chat Moderator
Joined: 22 Dec 2016
Last visit: 18 Nov 2025
Posts: 3,170
Own Kudos:
10,413
 [1]
Given Kudos: 1,861
Location: India
Concentration: Strategy, Leadership
Posts: 3,170
Kudos: 10,413
 [1]
12 Days of Christmas GMAT Competition - Day 12: Which of the following 25 Dec 2021, 07:58
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
I: \(x<x^2<1−x\)

x < \(x ^2\) when

1 < x < + infinity

- infinity > x > -1

-1 < x < 0

When x lies between -1 and 0, 1 - x will be positive and greater than 1, while 0 < \(x^2\) < 1

Hence this can be correct.

II: \(x<x^2<\frac{1}{x}\)

x < \(x ^2\) when

1 < x < + infinity

- infinity > x > -1

-1 < x < 0

In this case \(1 / x\) will be negative when x is negative, while \(x^2\) will be positive.

When x is positive and greater than 1, \(1 / x\) will not be less greater than \(x^2\)

Hence this statement is incorrect.

III: \(x<\frac{1}{x}<x^2\)

\(x<\frac{1}{x} \)when

- infinity < x < -1

0 < x < 1

When - infinity < x < -1, \(x^2\) is positive, while x and \(\frac{1}{x}\) is negative.

Hence this can be correct.

IV: \(x<\frac{1}{x}<x^2<x^3\)

\(x^2<x^3\) when

1 < x < + infinity

In this case, \(\frac{1}{x}\) will be less than x.

Hence this statement is incorrect.

IMO - C
User avatar
happypuppy
User avatar
Joined: 29 Mar 2020
Last visit: 03 Dec 2023
Posts: 207
Own Kudos:
437
 [1]
Given Kudos: 238
Location: India
Concentration: Entrepreneurship, Technology
Schools: Stanford '25 (S) Booth '25 (S) Wharton '25 (S) ISB '24 (A) HBS '25 (D)
GMAT 1: 720 Q50 V38 (Online)
GPA: 3.5
Products:
My Reviews
Schools: Stanford '25 (S) Booth '25 (S) Wharton '25 (S) ISB '24 (A) HBS '25 (D)
GMAT 1: 720 Q50 V38 (Online)
Posts: 207
Kudos: 437
 [1]
12 Days of Christmas GMAT Competition - Day 12: Which of the following 26 Dec 2021, 05:01
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
12 Days of Christmas GMAT Competition with Lots of Fun

Which of the following can never be correct?

I: \(x < x^2 < 1-x\)

II: \(x < x^2 < \frac{1}{x} \)

III: \(x < \frac{1}{x} < x^2\)

IV: \(x < \frac{1}{x} < x^2 < x^3\)

A. II only
B. II and III only
C. II and IV only
D. I, II, and IV only
E. I, II, III, and IV only

 


This question was provided by Experts'Global
for the 12 Days of Christmas Competition

Win $25,000 in prizes: Courses, Tests & more

 

Show more


Nothing given for x, so x can assume any value.

For 'Can NEVER be CORRECT' , we need to find one correct value to decide:

For such questions, below is the list of values of x where values change for x, 1/x, x^2 or x^n
x = 10
x = 1
x = 0.9
x = 0.6
x = 0.5
x = 0.4
x = 0.1
x = 0
x = -0.1
x = -0.4
x = -0.5
x = -0.6
x = -0.9
x = -1
x = -10

(Actually, we need not compute for all but it helps us to judge/guess by looking the numbers)
Can never be correct OR ALWAYS incorrect.
So, if we get JUST ONE Correct case, then the statement qualifies.

1. x < x^2 < 1 - x
Works of x = -0.1
-0.1 < 0.01 < 0.9 ------------------ Can be correct for at least one case

2. x < x^2 < 1/x
Doesn't work for any x

3. x < 1/x < x^2
Works for x=-10
-10 < -0.1 < 100 ------------------ Can be correct for at least one case

4. x < 1/x < x^2 < x^3
Doesn't work for any values of x

So, II and IV can never be correct as we got I and III correct for at least one case.

Answer: (C)
User avatar
Snehaaaaa
User avatar
Joined: 09 Mar 2021
Last visit: 13 Feb 2025
Posts: 135
Own Kudos:
429
 [1]
Given Kudos: 161
Location: India
Schools: Booth '24 ISB '23 Ross '24 Sloan '24 Wharton '24 Haas '25 Kellogg '24 Anderson '25
GMAT 1: 640 Q44 V34
GPA: 3.68
Schools: Booth '24 ISB '23 Ross '24 Sloan '24 Wharton '24 Haas '25 Kellogg '24 Anderson '25
GMAT 1: 640 Q44 V34
Posts: 135
Kudos: 429
 [1]
12 Days of Christmas GMAT Competition - Day 12: Which of the following 26 Dec 2021, 05:22
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
12 Days of Christmas GMAT Competition with Lots of Fun

Which of the following can never be correct?

I: \(x < x^2 < 1-x\)

II: \(x < x^2 < \frac{1}{x} \)

III: \(x < \frac{1}{x} < x^2\)

IV: \(x < \frac{1}{x} < x^2 < x^3\)

A. II only
B. II and III only
C. II and IV only
D. I, II, and IV only
E. I, II, III, and IV only

 


This question was provided by Experts'Global
for the 12 Days of Christmas Competition

Win $25,000 in prizes: Courses, Tests & more

 

Show more

Explanation-
In the question statement, there is no limitations define for choosing the value of x
So, x can be any value ............, 1, 10, 0.4, 0.1, 0.5, 0, -0.1, -0.4, -5, -10, .............

We will check the given conditions by substituting any values for x. If the condition holds true for even one value of x, then that condition can be correct.
In this type of question, if we get even a single value correct for a particular condition, the condition will hold true

I: \(x < x^2 < 1-x\)
Case 1: If x = 1,
1 < 1^2 < 1-1
The condition does not holds true

Case 2: If x= -0.1,
-0.1 < 0.01 < 1.1
The condition holds true
So, even though the condition doesn't hold true for 1 value of x, it holds true for other value of x. Therefore, this condition does not fall under 'can never be correct'


II: \(x < x^2 < \frac{1}{x} \)
Case 1: x= -0.1
-0.1 < 0.01 < -10
The condition does not holds true
Checked for multiple values of x, but the condition does not hold true for any value of x


III: \(x < \frac{1}{x} < x^2\)
Case 1: x= -10
-10 < -0.1 < 100
The condition holds true


IV: \(x < \frac{1}{x} < x^2 < x^3\)
Case 1: x = 0.5
0.5 < 2 < 0.25 < 0.025
The condition does not holds true

Case 2: x = -0.1
-0.1 < -10 < 0.01 < -0.001
The condition does not holds true
Checked for multiple values of x, but the condition does not hold true for any value of x


Correct answer is C
User avatar
venkycs98
User avatar
Joined: 07 May 2023
Last visit: 16 Nov 2025
Posts: 317
Own Kudos:
17
Given Kudos: 388
Location: India
Schools: Anderson '26 (S) McCombs '26 Darden '26 McDonough  '26 (S) Mashall '26 (S)
GMAT Focus 1: 685 Q85 V86 DI82
GPA: 3.5
Products:
My Reviews
Schools: Anderson '26 (S) McCombs '26 Darden '26 McDonough  '26 (S) Mashall '26 (S)
GMAT Focus 1: 685 Q85 V86 DI82
Posts: 317
Kudos: 17
12 Days of Christmas GMAT Competition - Day 12: Which of the following 11 Oct 2024, 09:57
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
12 Days of Christmas GMAT Competition with Lots of Fun

Which of the following can never be correct?

I: \(x < x^2 < 1-x\)

II: \(x < x^2 < \frac{1}{x} \)

III: \(x < \frac{1}{x} < x^2\)

IV: \(x < \frac{1}{x} < x^2 < x^3\)

A. II only
B. II and III only
C. II and IV only
D. I, II, and IV only
E. I, II, III, and IV only


 


This question was provided by Experts'Global
for the 12 Days of Christmas Competition

Win $25,000 in prizes: Courses, Tests & more

 

Show more
I solved it in 3 steps:

1. Check if statement I is correct:

x < x^2
=> (i) x is either a positive value greater than 1 or (ii) -ve number
x^2 < 1-x is true only when (ii) is true
=> x = -ve number
take any number say x = -1
-1 < 1 < 1 - (-1)

So clearly statement 1 can be true, so eliminate all answer choices with statement I
=> answer choices D and E are out

Now, notice that the other answers contain statement II, so we don't have to check whether statement II is correct
We only have to check if statement III or IV are correct,

I chose to check if statement III is correct first
IV: x < 1/x < x^2

x < 1/x
=> We can only have 2 types of values for x -> (i) x = +ve decimal value less than 1 (ii) -ve values lesser than -1
1/x < x^2
=> Only (ii) is possible

Lets say x = -5
-5 < -0.2 < 25
Fits, so eliminate all options with statement III

We are left with C and A

now, Check if D is correct
IV:
[ltr]x<1/x<x^2<x^3

x< 1/x
=> There are only 2 types of numbers that satisfy this
(i) x = +ve decimal < 1 and (ii) -ve number lesser than -1

1/x < x^2
We can eliminate (i)

x^2 < x^3
Notice that this statement is not compatible with the possible types of numbers
Since x^2 is always +ve and x^3 is always -ve
So statement 4 can never be true

We can conclude that the answer is C

There's a more optimal approach:
You can start with statement IV
and move on to statement III, with just these 2, you can find the answer :)[/ltr]
Moderators:
Bunuel
Math Expert
105355 posts
Krunaal
Tuck School Moderator
805 posts