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:

in fact, st2 is sufficient as well so answer is D.

st1 is sufficient as noted above:
if x<0 then |x|=-x
so y=|x|+x = -x+x = 0
suff.

st2 is sufficient as well:
from stem only we can deduce that y is non-negative:
if x<0 then y=0 as we showed above,
if x=0 then obviously y=0.
if x>0 then |x|=x and y = 2x > 0

also note that y is an integer (!!!!!!!)

st2 says y<1. the only non-negative integer less than 1 is 0.
so y must be 0.

Given: y is integer, y = |x| + x
asked : does y = 0 ?

from the question, 0=< y =< infinity; that is, y can range between zero and any positive integer.

because |x| is always positive.
if x is positive then y = 2x
if x is negative then y = 0

(1) x<0
--------------
x is negative then y = x - x = 0 .. YES y = 0
statement 1 is sufficient (2) y<1
--------------
the only possible value for y < 1 is zero because y can never be negative and y is an integer so it can't be some fraction less than 1 like 1/2 or so.

So y in this case has to equal zero [ y = 0 ]
statement 2 is sufficient