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:

X/|X| < X . Which of the following must be true for all ?

a. X > 1 b. X is an element in (-1,0) U (1,inf) c. |X| < 1 d. |X| = 1 e. |X|^2 > 1

Can some one explain how X can be zero for the above condition?

x is in the denominator so it can not equal to zero as division be zero is undefined.

Correct form of this question is below (m09 q22, discussed here: m09-q22-69937.html):

If \(\frac{x}{|x|} \lt x\), which of the following must be true about \(x\)? (\(x \ne 0\)) A. \(x\gt 2\) B. \(x \in (-1,0) \cup (1,\infty)\) C. \(|x| \lt 1\) D. \(|x| = 1\) E. \(|x|^2 \gt 1\)

\(\frac{x}{|x|}< x\) Two cases: A. \(x<0\) --> \(\frac{x}{-x}<x\) --> \(-1<x\). But as we consider the range \(x<0\) then \(-1<x<0\)

B. \(x>0\) --> \(\frac{x}{x}<x\) --> \(1<x\).

So the given inequality holds true in two ranges \(-1<x<0\) and \(x>1\).

Brunel, Can you please explain why option E is not feasible?

Dear AverageuGuy123

As Bunuel explained above,

Either -1 < x < 0 Or x > 1

Now, |x| as you know, represents the magnitude of x. Option E says that |x|^2 must be greater than 1.

Let's first consider the case when -1 < x < 0

A possible value of x in this case is -0.5 So, what is the value of |x|^2? It is equal to 0.25

Is it greater than 1? NO

Let's now consider the case when x > 1

A possible value of x in this case is 2. So, what is the value of |x|^2? It's 4.

Is it greater than 1? YES

So, as we see, that |x|^2 CAN BE greater than 1. But can we say that |x|^2 MUST BE greater than 1? NO, because |x|^2 is not greater than 1 for all possible values of x.

So, the key takeawayfrom this discussion is that:

we need to be careful whether the question is asking about MUST BE TRUE statements or about CAN BE TRUE statements.

Brunel, Can you please explain why option E is not feasible?

You can plug in numbers to eliminate options.

"which of the following must be true about x" means that every acceptable value of x must lie in the range given in the correct option. The acceptable values of x are the values for which x/|x| < x.

A. x>2 Must x be greater than 2?

This should make you check for 2. 2/|2| < 2 1 < 2 (True) So 2 is an acceptable value of x. But 2 is not greater than 2. So this option is not correct. This also makes you eliminate options (C) and (D).

E. |x|^2>1 Must x be greater than 1 or less than -1?

Check for 1/2 (1/2)/|1/2| < 1/2 1 < 1/2 (False)

Check for -1/2 (-1/2)/|-1/2| < -1/2 -1 < -1/2 (True)

So x = -1/2 is an acceptable value but it does not lie in this range. Hence option (E) is also incorrect.

This question can be dealt with in a variety of ways. It's actually really susceptible to TESTing VALUES, which we can use to determine possibilities and eliminate answers.

We're told that X/|X| < X. The question asks what must be TRUE about X.

While this inequality looks complicated, you can quickly prove some things about X....

IF.... X = 1 1/|1| is NOT < 1 So X CANNOT be 1 Eliminate D.

IF..... X = 2 2/|2| IS < 2 So X CAN be 2 Eliminate A and C.

IF.... X = -2 -2/|-2| is NOT < -2 So X CANNOT be -2 Eliminate E.

Re: If x/|x|, which of the following must be true for all [#permalink]

Show Tags

08 Dec 2017, 07:56

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________