Last visit was: 21 Apr 2026, 19:15

It is currently 21 Apr 2026, 19:15

Toolkit

Practice Tests

Quantitative Questions

Problem Solving (PS)

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

Request Expert Reply

Confirm Cancel

Apr 16
Sneak Peek: $200 Off GMAT Prep
12:00 PM EDT
-
11:59 PM EDT
You’re first to know: Save $200 on GMAT prep starting tomorrow, 4/13. Get 6 official practice tests and study with real questions. Be ready to save!
Apr 26
Score High on the GMAT with Target Test Prep
10:00 AM EDT
-
11:00 AM EDT
The Target Test Prep course represents a quantum leap forward in GMAT preparation, a radical reinterpretation of the way that students should study. Try before you buy with a 5-day, full-access trial of the course for FREE!
Apr 27
Try OnDemand GMAT Masterclass
11:00 AM EDT
-
12:00 PM EDT
Prefer video-based learning? The Target Test Prep OnDemand course is a one-of-a-kind video masterclass featuring 400 hours of lecture-style teaching by Scott Woodbury-Stewart, founder of Target Test Prep and one of the most accomplished GMAT instructors
Apr 28
Join - GMAT Day!
08:00 AM PDT
-
11:00 AM PDT
Whether you are just beginning your MBA journey or fine-tuning your path to a 700+ score, GMAT Day is designed to give you a competitive edge.

Back to Forum

Create Topic

If x/|x|<x which of the following must be true about x?

1 2 3

christoph

Joined: 30 Sep 2004

Last visit: 23 Nov 2011

Posts: 684

Own Kudos:

1,587

[228]

Location: Germany

Posts: 684

Kudos: 1,587

[228]

Updated on: 19 Sep 2025, 12:56

Originally posted by christoph on 06 Feb 2005, 09:16.
Last edited by Bunuel on 19 Sep 2025, 12:56, edited 8 times in total.

Kudos

214

Bookmarks

00:00

Start Timer

Pause Timer

Resume Timer

Show Answer

Hide

Show

History

Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

95% (hard)

Question Stats:

33% (01:37) correct

67% (01:50) wrong

based on 4369 sessions

History

Date

Time

Result

Not Attempted Yet

If x ≠ 0 and x/|x|<x. Which of the following must be true about x ?

A. x > 1

B. x > -1

C. |x| < 1

D. |x| = 1

E. |x|^2 > 1

Note: The question and the OA are 100% correct. To fully understand the reasoning, please study the discussion carefully. Read the expert explanations and check the linked similar questions for additional practice.

M09-22

Experience GMAT Club Test Questions

Yes, you've landed on a GMAT Club Tests question

Craving more? Unlock our full suite of GMAT Club Tests here

Want to experience more? Get a taste of our tests with our free trial today

Rise to the challenge with GMAT Club Tests. Happy practicing!

GMAT Club Tests Check Our Products Try Free Tests

Show Answer

Official Answer

Most Helpful Reply

Bunuel

Math Expert

Joined: 02 Sep 2009

Last visit: 21 Apr 2026

Posts: 109,728

Own Kudos:

810,476

[154]

Given Kudos: 105,800

Products:

Expert

Expert reply

Posts: 109,728

Kudos: 810,476

[154]

18 Jan 2010, 04:11

Kudos

Bookmarks

nmohindru

If $\frac{x}{|x|}<x$ which of the following must be true about $x$?

(A) $x>1$

(B) $x>-1$

(C) $|x|<1$

(D) $|x|=1$

(E) $|x|^2>1$

This question was well explained by Durgesh and Ian Stewart, but since there are still some doubts, I'll try to add my 2 cents.

First of all let's solve this inequality step by step and see what is the solution for it, or in other words let's see in which ranges this inequality holds true.

Two cases for $\frac{x}{|x|}<x$:

A. $x<0$ --> $|x|=-x$ --> $\frac{x}{-x}<x$ --> $-1<x$ --> $-1<x<0$;

B. $x>0$ --> $|x|=x$ --> $\frac{x}{x}<x$ --> $1<x$.

So given inequality holds true in the ranges: $-1<x<0$ and $x>1$. Which means that $x$ can take values only from these ranges.

------{-1}xxxx{0}----{1}xxxxxx

Now, we are asked which of the following must be true about $x$. Option A cannot be ALWAYS true because $x$ can be from the range $-1<x<0$, eg $-\frac{1}{2}$ and $x=-\frac{1}{2}<1$.

Only option which is ALWAYS true is B. ANY $x$ from the ranges $-1<x<0$ and $x>1$ will definitely be more the $-1$, all "red", possible x-es are to the right of -1, which means that all possible x-es are more than -1.

Answer: B.

Bunuel

Math Expert

Joined: 02 Sep 2009

Last visit: 21 Apr 2026

Posts: 109,728

Own Kudos:

810,476

[66]

Given Kudos: 105,800

Products:

Expert

Expert reply

Posts: 109,728

Kudos: 810,476

[66]

01 Sep 2010, 07:32

Kudos

Bookmarks

x/|x|<x, which of the following must be true about x ?

A. x>1
B. x>-1
C. |x|<1
D. |x|=1
E. |x|^2>1

We are given that $\frac{x}{|x|}<x$ (this is a true inequality), so first of all we should find the ranges of $x$ for which this inequality holds true.

$\frac{x}{|x|}< x$ multiply both sides of inequality by $|x|$ (side note: we can safely do this as absolute value is non-negative and in this case we know it's not zero too) --> $x<x|x|$ --> $x(|x|-1)>0$:
Either $x>0$ and $|x|-1>0$, so $x>1$ or $x<-1$ --> $x>1$;
Or $x<0$ and $|x|-1<0$, so $-1<x<1$ --> $-1<x<0$.

So we have that: $-1<x<0$ or $x>1$. Note $x$ is ONLY from these ranges.

Option B says: $x>-1$ --> ANY $x$ from above two ranges would be more than -1, so B is always true.

Answer: B.

IanStewart

GMAT Tutor

Joined: 24 Jun 2008

Last visit: 17 Apr 2026

Posts: 4,143

Own Kudos:

11,267

[47]

Given Kudos: 99

Expert

Expert reply

Posts: 4,143

Kudos: 11,267

[47]

17 Aug 2008, 08:29

Kudos

Bookmarks

I'm pretty sure Durgesh has already explained his answer convincingly, but just in case, I'll present a different problem:

If x is positive, what must be true?

I) x > 10
II) x > -10
III) x > 0

Of course, III) says exactly 'x is positive', so III) must be true. But if x is positive, x is obviously larger than -10. II) must also be true. I) doesn't need to be true; x could be 4, or 7, for example.

The same situation applies with the question that began this thread. We know that either -1 < x < 0, or x > 1. What must be true? Well, "x > 1" does not need to be true. We know that x might be -0.5, for example. On the other hand, "x > -1" absolutely must be true: if x satisfies the inequality x/|x| < x, then x is certainly larger than -1. If we had been asked whether "x > -1,000,000" must be true, the answer would also be yes. The question did not ask "What is the solution set of "x/|x| < x"; nor did it ask "For which values of x is x/|x| < x true?".

KarishmaB

Tutor

Joined: 16 Oct 2010

Last visit: 21 Apr 2026

Posts: 16,438

Own Kudos:

79,375

[29]

Given Kudos: 484

Location: Pune, India

Expert

Expert reply

Posts: 16,438

Kudos: 79,375

[29]

13 Jan 2012, 10:46

Kudos

Bookmarks

gautammalik

I think answer should be A and not B because x cannot be equal to zero. If x equals to zero then the equation will lead to infinity.

Think again.
Every value greater than -1 need not satisfy the inequality but every value satisfying the inequality must be greater than -1.

x/|x| can take only 2 values: 1 or -1
If x is positive, x/|x| = 1
If x is negative, x/|x| = -1
x cannot be 0.

Now let's look at the question.
x > x/|x| holds for x > 1 (x is positive) or -1 < x < 0 (x is negative)

x can take many values e.g. -1/3, -4/5, 2, 5, 10 etc

Which of the following holds for every value that x can take?
(A) X > 1
(B) X > -1

I hope that you agree that X > 1 doesn't hold for every possible value of X whereas X > -1 holds for every possible value of X.

Mind you, every value greater than -1 need not be a possible value of x.

durgesh79

Joined: 27 May 2008

Last visit: 14 Dec 2021

Posts: 229

Own Kudos:

647

[23]

Posts: 229

Kudos: 647

[23]

15 Aug 2008, 08:37

Kudos

Bookmarks

i think OA is correct....

the question is If x / |x| < x. so we have to consider only those values of x for which this inequalty is true and what are those values
1. when x is between -1 and 0
2. when x is more than 1

lets call these conditions our universe.

Now the question is for all values of x (in our universe) which of the following is true
option B, x > -1, has both conditions 1 and 2

now you may say what about x = 1/2 .... that wasnt even part of our universe... so even if x = 1/2 is satisfying option B and not the question stem, we dont have to worry... becuase we are not supposed to take it as an example ...

for all values of x in our universe, option B is ALWAYS true...
Option A is not always true...

General Discussion

HongHu

Joined: 03 Jan 2005

Last visit: 25 Apr 2011

Posts: 962

Own Kudos:

798

[12]

Posts: 962

Kudos: 798

[12]

06 Feb 2005, 22:32

Kudos

Bookmarks

x/|x|<x
if x>0, then 1/x<1 ie, x>1
if x<0, then 1/(-x)>1, ie, x>-1

So combine these two, we know that x must be greater than -1.

baner is right, (B).

nonameee

Joined: 23 Apr 2010

Last visit: 30 Sep 2013

Posts: 475

Own Kudos:

412

[2]

Given Kudos: 7

Posts: 475

Kudos: 412

[2]

02 Sep 2010, 00:20

Kudos

Bookmarks

Bunuel, thanks for your reply. But I have to disagree with you because:

The answer can't be B, since let x=1/2. We get:
(1/2)/(1/2) < (1/2)
1< 1/2
Contradiction.

What do you think about that?

Bunuel

Math Expert

Joined: 02 Sep 2009

Last visit: 21 Apr 2026

Posts: 109,728

Own Kudos:

810,476

[8]

Given Kudos: 105,800

Products:

Expert

Expert reply

Posts: 109,728

Kudos: 810,476

[8]

02 Sep 2010, 03:30

Kudos

Bookmarks

nonameee

Bunuel, thanks for your reply. But I have to disagree with you because:

The answer can't be B, since let x=1/2. We get:
(1/2)/(1/2) < (1/2)
1< 1/2
Contradiction.

What do you think about that?

OA for this question is B and it's not wrong.

Consider following:
If $x=5$, then which of the following must be true about $x$:
A. x=3
B. x^2=10
C. x<4
D. |x|=1
E. x>-10

Answer is E (x>-10), because as x=5 then it's more than -10.

Or:
If $-1<x<10$, then which of the following must be true about $x$:
A. x=3
B. x^2=10
C. x<4
D. |x|=1
E. x<120

Again answer is E, because ANY $x$ from $-1<x<10$ will be less than 120 so it's always true about the number from this range to say that it's less than 120.

The same with original question:

If $-1<x<0$ or $x>1$, then which of the following must be true about $x$:
A. x>1
B. x>-1
C. |x|<1
D. |x|=1
E. |x|^2>1

As $-1<x<0$ or $x>1$ then ANY $x$ from these ranges would satisfy $x>-1$. So B is always true.

$x$ could be for example -1/2, -3/4, or 10 but no matter what $x$ actually is it's IN ANY CASE more than -1. So we can say about $x$ that it's more than -1.

On the other hand A says that $x>1$, which is not always true as $x$ could be -1/2 and -1/2 is not more than 1.

Hope it's clear.

nades09

Joined: 18 Feb 2010

Last visit: 24 Feb 2014

Posts: 21

Own Kudos:

616

Given Kudos: 5

Posts: 21

Kudos: 616

26 Oct 2010, 03:12

Kudos

Bookmarks

Hi Bunnuel,
Can it be solved as-

x/|x|<x

Hence 1/|x|<1

1/x<1 OR -1/x<1
when 1/x<1
then 1<x = x>1

when -1/x<1
then -1<x = x>-1

The two possible outcomes are x>1 or x>-1
The more restrictive is x>-1
Hence B

Thanks
Neelam

Bunuel

Math Expert

Joined: 02 Sep 2009

Last visit: 21 Apr 2026

Posts: 109,728

Own Kudos:

810,476

[3]

Given Kudos: 105,800

Products:

Expert

Expert reply

Posts: 109,728

Kudos: 810,476

[3]

26 Oct 2010, 09:44

Kudos

Bookmarks

nades09

No, that't not correct.

First of all, when you are writing 1/|x|<1 from x/|x|<x you are reducing (dividing) inequality by x:

Never multiply or reduce inequality by an unknown (a variable) unless you are sure of its sign since you do not know whether you must flip the sign of the inequality.

Consider a simple inequality $4>3$ and some variable $x$.

Now, you can't multiply (or divide) both parts of this inequality by $x$ and write: $4x>3x$, because if $x=1>0$ then yes $4*1>3*1$ but if $x=-1$ then $4*(-1)=-4<3*(-1)=-3$. Similarly, you can not divide an inequality by $x$ not knowing its sign.

Next, inequality $\frac{1}{|x|}<1$ holds true in the following ranges: $x<-1$ and $x>1$ (and not: x>1 or x>-1, which by the way simply means x>1).

Hope it's clear.

myfish

Joined: 15 Sep 2011

Last visit: 03 Oct 2012

Posts: 7

Own Kudos:

[5]

Given Kudos: 6

Posts: 7

Kudos: 18

[5]

19 Dec 2011, 15:20

Kudos

Bookmarks

Typical GMAT nonsense. A lot of people seem to talk themselves into a solution but in mathematics, there are no GMAT-truths. x>-1 must not be true, since it ignores the fact that 0 does not fulfill the requirement.

IanStewart

GMAT Tutor

Joined: 24 Jun 2008

Last visit: 17 Apr 2026

Posts: 4,143

Own Kudos:

11,267

[10]

Given Kudos: 99

Expert

Expert reply

Posts: 4,143

Kudos: 11,267

[10]

20 Dec 2011, 03:08

Kudos

Bookmarks

myfish

No one is "talking themselves into a solution" here, and there's nothing wrong with the mathematics. I explained why earlier, but I can use a simpler example. If a question reads

If x = 5, what must be true?

I) x > 0

then clearly I) must be true; if x is 5, then x is certainly positive. It makes no difference that x cannot be equal to 12, or to 1000.

The same thing is happening in this question. We know that either -1 < x < 0, or that 1 < x. If x is in either of those ranges, then certainly x must be greater than -1. It makes no difference that x cannot be equal to 1/2, or to 0.

This is an important logical point on the GMAT (even though the question in the original post is not a real GMAT question), since it comes up all the time in Data Sufficiency. If a question asks

Is x > 0?

1) x = 5

that is exactly the same question as the one I asked above, but now it's phrased as a DS question. This question is really asking, when we use Statement 1, "If x = 5, must it be true that x > 0?" Clearly the answer is yes. If you misinterpret this question, and think it's asking "can x have any positive value at all", you would make a mistake on this question and on most GMAT DS algebra questions.

myfish

Joined: 15 Sep 2011

Last visit: 03 Oct 2012

Posts: 7

Own Kudos:

[3]

Given Kudos: 6

Posts: 7

Kudos: 18

[3]

20 Dec 2011, 07:29

Kudos

Bookmarks

IanStewart

myfish

Dear Ian,
I truly appreciate your efforts on here. 'Must be true' is a condition without exceptions. And when i plug in 0, the inequality is NOT true. That five apples are more than 0 apples is clear to me. However, the question asks for what 'Must be true'. Several ranges that make the inequality true makes this question inaccurate. Unless, if the GMAT translates 'Must be true' into 'may or may not be true' then explanation with the ranges make sense. For me, these type of questions make the GMAT into a lottery and I am not alone since many test takers have trouble with a logic that ignores exceptions. I have another example, fresh from Kaplan.

If it is true to -6<= n <= 10, which of the following must be true?

n<8
n=-6
n>-8
-10<n<7
none of the above

Same case. Official solution is n>-8, however n= -7 does not fulfill the first requirement, it therefore CAN BE TRUE - but not MUST BE TRUE

Again, I am no stranger to logic but, I am sure many will agree, these kind of questions are nonsense, especially when one considers the official (you and others) translation of the question into "Are 3 apples more than 2?" - what kind of a question is that?

IanStewart

GMAT Tutor

Joined: 24 Jun 2008

Last visit: 17 Apr 2026

Posts: 4,143

Own Kudos:

11,267

[1]

Given Kudos: 99

Expert

Expert reply

Posts: 4,143

Kudos: 11,267

[1]

20 Dec 2011, 19:07

Kudos

Bookmarks

myfish

I've tried to explain the logic behind this question twice, so I won't try again, but I can assure you that every mathematician in the world would agree with the answer to this question - this has nothing to do with some kind of logic exclusive to the GMAT. The same is true of the Kaplan question you quote; if n is greater than -6, it is surely true that n is greater than -8. You seem to be looking at these problems backwards: you're assuming n > -8 is true, and asking if it needs to be true that -6 < n < 10. That's the opposite of what the question is asking you to do.

gautammalik

Joined: 20 Jul 2010

Last visit: 29 Jun 2012

Posts: 87

Own Kudos:

[1]

Given Kudos: 12

Status:Target MBA

Location: Singapore

Posts: 87

Kudos: 64

[1]

13 Jan 2012, 09:40

Kudos

Bookmarks

I think answer should be A and not B because x cannot be equal to zero. If x equals to zero then the equation will lead to infinity.

koro12

Joined: 26 May 2012

Last visit: 27 Jan 2013

Posts: 2

Own Kudos:

Posts: 2

Kudos: 11

26 May 2012, 11:50

Kudos

Bookmarks

Bunuel

nmohindru

If $\frac{x}{|x|}<x$ which of the following must be true about $x$?

(A) $x>1$

(B) $x>-1$

(C) $|x|<1$

(D) $|x|=1$

(E) $|x|^2>1$

This question was well explained by Durgesh and Ian Stewart, but since there are still some doubts, I'll try to add my 2 cents.

First of all let's solve this inequality step by step and see what is the solution for it, or in other words let's see in which ranges this inequality holds true.

Two cases for $\frac{x}{|x|}<x$:

A. $x<0$ --> $|x|=-x$ --> $\frac{x}{-x}<x$ --> $-1<x$ --> $-1<x<0$;

B. $x>0$ --> $|x|=x$ --> $\frac{x}{x}<x$ --> $1<x$.

So given inequality holds true in the ranges: $-1<x<0$ and $x>1$. Which means that $x$ can take values only from these ranges.

------{-1}xxxx{0}----{1}xxxxxx

Now, we are asked which of the following must be true about $x$. Option A can not be ALWAYS true because $x$ can be from the range $-1<x<0$, eg $-\frac{1}{2}$ and $x=-\frac{1}{2}<1$.

Only option which is ALWAYS true is B. ANY $x$ from the ranges $-1<x<0$ and $x>1$ will definitely be more the $-1$, all "red", possible x-es are to the right of -1, which means that all possible x-es are more than -1.

Answer: B.

isnt E true ? from the above 1<mod (x) which implies 1< mod(x) squared

Bunuel

Math Expert

Joined: 02 Sep 2009

Last visit: 21 Apr 2026

Posts: 109,728

Own Kudos:

810,476

[2]

Given Kudos: 105,800

Products:

Expert

Expert reply

Posts: 109,728

Kudos: 810,476

[2]

27 May 2012, 02:37

Kudos

Bookmarks

koro12

Bunuel

nmohindru

If $\frac{x}{|x|}<x$ which of the following must be true about $x$?

(A) $x>1$

(B) $x>-1$

(C) $|x|<1$

(D) $|x|=1$

(E) $|x|^2>1$

This question was well explained by Durgesh and Ian Stewart, but since there are still some doubts, I'll try to add my 2 cents.

First of all let's solve this inequality step by step and see what is the solution for it, or in other words let's see in which ranges this inequality holds true.

Two cases for $\frac{x}{|x|}<x$:

A. $x<0$ --> $|x|=-x$ --> $\frac{x}{-x}<x$ --> $-1<x$ --> $-1<x<0$;

B. $x>0$ --> $|x|=x$ --> $\frac{x}{x}<x$ --> $1<x$.

So given inequality holds true in the ranges: $-1<x<0$ and $x>1$. Which means that $x$ can take values only from these ranges.

------{-1}xxxx{0}----{1}xxxxxx

Now, we are asked which of the following must be true about $x$. Option A can not be ALWAYS true because $x$ can be from the range $-1<x<0$, eg $-\frac{1}{2}$ and $x=-\frac{1}{2}<1$.

Only option which is ALWAYS true is B. ANY $x$ from the ranges $-1<x<0$ and $x>1$ will definitely be more the $-1$, all "red", possible x-es are to the right of -1, which means that all possible x-es are more than -1.

Answer: B.

isnt E true ? from the above 1<mod (x) which implies 1< mod(x) squared

We are told that $x$ can take values only from these ranges.

------{-1}xxxx{0}----{1}xxxxxx

Now, $|x|^2>1$ means that $x^2>1$ --> $x<-1$ or $x>1$. Since $x<-1$ is not true about $x$ (we know that $-1<x<0$ and $x>1$), then this option is not ALWAYS true.

Hope it's clear.

vibhav

Joined: 28 Dec 2010

Last visit: 13 Oct 2014

Posts: 211

Own Kudos:

816

Given Kudos: 33

Location: India

Posts: 211

Kudos: 816

27 May 2012, 05:05

Kudos

Bookmarks

Bunuel, the question doesn't mention anywhere that x is an integer. so why have we not considered the values of 0<x<1 which also doesnt satisfy the equation? If it has been considered then the solution should be A and not B. x>1 will always hold true aka must be true! while x>-1 is sometimes true.. aka can be true.

Bunuel

Math Expert

Joined: 02 Sep 2009

Last visit: 21 Apr 2026

Posts: 109,728

Own Kudos:

810,476

[2]

Given Kudos: 105,800

Products:

Expert

Expert reply

Posts: 109,728

Kudos: 810,476

[2]

27 May 2012, 05:16

Kudos

Bookmarks

vibhav

I think you don't understand the question.

Given: $-1<x<0$ and $x>1$. Question: which of the following must be true?

A. $x>1$. This opinion is not always true since $x$ can be $-\frac{1}{2}$ which is not more than 1.
B. $x>-1$. This option is always true since any $x$ from $-1<x<0$ and $x>1$ is more than -1.