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:

the given equation is |x-y| = |x-z| which when squared gives

=> x^2+y^2-2xy = x^2+z^2-2zy => y^2 - z^2 + 2zy - 2xy = 0 => (y-z)[(y+z)-2x] = 0 => (y+z)-2x = 0 From statement 2 it is 10 - 2x = 0 => x = 5

Let me know if mised something

humans - what I'm not sure is, in a condition where y = z so that y = z = 5, we have infinite solutions for the equation (x is any integer). How can it be B then? (your solution seems OK, so I can't make out what I'm missing) _________________

II. insufficient; if y and z are not 5, then x=5; however if y and z are equal to 5 then x can be any value. Thus, I and II combined allows you to solve for x.