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 350,000 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:

GMATBLACKBELT wrote:
If x is not equal to 0, is |x| less than 1?

x > x/|x|
|x| > x

i get C please confirm if i am correct.. i still struggle with these

from the stem
x =! 0
-1 < x < 1

st 1: x > x/|x|

positive: x > x/x => x*x > x
x^2 - x > 0
x(x-1) > 0
x > 0
x > 1
3 > 3/3 = 3 > 1
really just x > 1 valid

negative: x < - x/x => x*x < -x
x^2 + x < 0
x(x+1) < 0
x < 0 valid
x < -1 not valid
-1/2 > -1/2 / |-1/2| = -1/2 > -1 okay
-2 > -2/|-2| = -2 > -1 no good
so -1< x < 0
not sufficient

st 2: |x| > x
means x can only be any negative # with the exception of -1
x = -1/2 would satisfy the condition in the stem
x = -3 would not
not sufficient

putting 1 and 2 together, we know that x > 1 cannot be true, so it would have to be the x < 0, sufficient

EDIT: sorry I messed up the first time and messed up/ didnt complete my work

EDIT #2 and #3 and #4: forgot to disable HTML again... whats wrong with this feature??

Last edited by beckee529 on 04 Nov 2007, 10:45, edited 4 times in total.

if x = -1, -1 > -1 Not possible
if x = -2, -2 > -1 Not possible
if x = -0.5, -0.5 > -1 True
So -1 < x <0> 1 Not possible
if x = 2, 2 > 1 True
if x = 0.5, 0.5 > 1 Not possible
So x > 1
Therefore x can take values from o to -1 of greater than 1
1 not suff

2. lxl > x

if x = -1, 1 > -1 true
if x = -2, 2 > -2 true
here x > 0 is not possible
So x < 0.
Therefore x can take values less than 0
2 not suff

Together we get -1 < x < 0
which is sufficient...So C

I m still not sure ....may be I have made some mistake...Can anyone please check

1: x > x/|x| implies 0 > x > -1 therefore x is any negative fraction between 0 and 1 that gives us 1> |x| > 0 : is |x| less than 1 - YES
or x > 1 that gives |x| > 1 : is |x| less than 1 - NO
IN SUFFICIENT

2: |x| > x implies x < 0 therefore |x| > 0 : is |x| less than 1 - YES & NO
INSUFFICIENT

1: x > x/|x| implies 0 > x > -1 therefore x is any negative fraction between 0 and 1 that gives us 1> |x| > 0 : is |x| less than 1 - YES or x > 1 that gives |x| > 1 : is |x| less than 1 - NO IN SUFFICIENT

2: |x| > x implies x <0> 0 : is |x| less than 1 - YES & NO INSUFFICIENT

Answer: E

But 1 also true for all x > 1
So 1 is not sufficient