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:

A kind of rule to remember is to search the roots of the expression in each absolute value. It's similar to what I said in my response to u on another topic

As we have |x-3|, so we search the roots of x-3. Hence,
x-3 = 0 <=> x = 3

Thus, the domains to consider are x > 3 and x < 3.

Each domain implies a simplified equation that removes the absolute.

We do not consider x > 0 or x < 0 because nothing in the equation asks for it

WE have [x-3]=3-x [] is absolute value if x>= 3 we have x-3 = 3-x. Solve x-3 = 3-x by itself we have x = 3. Put togethere with x>= 3 we have x= 3

if x=<3 we have 3-x = 3-x. it if right with all x =<3 ==> the solution without any additional condition is x =< 3

Let use the condition 1 only we can have x> 3 doesn't belong to the area x=<3==> insufficient Let use the condition 2 only we have x< 0 belong to area x=< 3 ==> sufficient

i. Equality in question is TRUE if x is negative and FALSE if x is positive. Dual solution, hence NOT SUFFICIENT ii. |x| always positive. For -x to be greater than 0, x has to be negative. Single solution, hence SUFFICIENT.

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. _________________

Last year when I attended a session of Chicago’s Booth Live , I felt pretty out of place. I was surrounded by professionals from all over the world from major...

I recently returned from attending the London Business School Admits Weekend held last week. Let me just say upfront - for those who are planning to apply for the...