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:

y>=0 (1) |x-3|>= y, x can be any number >=3, x<=3 Not suff 2) |x-3|<=-y, y>=0, equation says that |x-3|less or equals to zero, but |x-3| never negative (|x-3|>=0), so only solution is if |x-3|=0=y --> x-3=0 --> x=3 Sufficinet

If y>=0 specified in the stem. Further mod of anything is positive. So, 0>=y according to 1. both y>=0 or y<=0, means y=0. So mod(x-3)>=0. Then x-3>=0 or x-3<=0 So, x>=3 or x<=3. Thus, x=3. Same thing for 2. So, D.

gmate2010 wrote:

\(y >= 0\)

1.) \(|x-3| >= y\)

=> \(-y >= (x-3) >= y\)

=> \(3-y >= x >= 3+y\)

Now, if y = 0 , then, \(3 >= x >= 3\) => x can only be 3 but, if y = 1 , then, \(2 >= x >= 4\) => which is not possible for any value of x.

=> x can only be 3 when y = 0 ..sufficient..

2.) \(|x-3| <= -y\)

=> \(y <= (x-3) <= -y\)

=> \(3+y <= x <= 3-y\)

when y = 0 ; then, \(3 <= x <= 3\) => x can only be 3.. when y = 1 ; then \(4 <= x <= 2\), which is not possible for any value of x.

O.K. as I see there has been kind of confusion about this question. I'm pretty sure answer is B.

I looked through the logic of people who think answer should be D and here is were they are making mistake:

(1) First of all: To check |x - 3|>=0 is sufficient or not just plug numbers A. x=5 y=1>0 and B. x=8 y=2>0; you'll see that both fits in |x - 3|>=0, y>=0. So ABSOLUTELY NO WAY (1) CAN BE SUFFICIENT.

|x - 3|>=y means that: x - 3>=y>=0 when x-3>0 --> x>3 OR (not and) -x+3>=y>=0 when x-3<0 --> x<3 Generally speaking |x - 3|>=y>=0 means that |x - 3|, an absolute value, is at least some positive number. So, there's no way you'll get a unique value for x. INSUFFICIENT.

(2) |x-3|<=-y, y>=0, equation says that |x-3|less or equals to zero, but |x-3| never negative (|x-3|>=0), so only solution is if |x-3|=0=y --> x-3=0 --> x=3 SUFFICIENT

In other words: (-y) is 0 or less, and the absolute value (|x-3|) must be at (0) or below this value. But absolute value (in this case |x-3|) can not be less than zero, so it must be 0.

O.K. as I see there has been kind of confusion about this question. I'm pretty sure answer is B.

I looked through the logic of people who think answer should be D and here is were they are making mistake:

(1) First of all: To check |x - 3|>=0 is sufficient or not just plug numbers A. x=5 y=1>0 and B. x=8 y=2>0; you'll see that both fits in |x - 3|>=0, y>=0. So ABSOLUTELY NO WAY (1) CAN BE SUFFICIENT.

|x - 3|>=y means that: x - 3>=y>=0 when x-3>0 --> x>3 OR (not and) -x+3>=y>=0 when x-3<0 --> x<3 Generally speaking |x - 3|>=y>=0 means that |x - 3|, an absolute value, is at least some positive number. So, there's no way you'll get a unique value for x. INSUFFICIENT.

(2) |x-3|<=-y, y>=0, equation says that |x-3|less or equals to zero, but |x-3| never negative (|x-3|>=0), so only solution is if |x-3|=0=y --> x-3=0 --> x=3 SUFFICIENT

In other words: (-y) is 0 or less, and the absolute value (|x-3|) must be at (0) or below this value. But absolute value (in this case |x-3|) can not be less than zero, so it must be 0.

So answer is B.

Got it finally! Thank you for clear explanation, Bunuel. B it is. Of course, A is not suff... _________________

Re: If y >= 0, what is the value of x? [#permalink]
05 Feb 2014, 15:24

1

This post received KUDOS

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. _________________

Re: If y >= 0, what is the value of x? [#permalink]
16 May 2015, 06:55

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. _________________

Re: If y >= 0, what is the value of x? [#permalink]
16 May 2015, 10:43

1

This post received KUDOS

Expert's post

Hi All,

When complex-looking questions show up on Test Day, there's almost always some type of built-in pattern involved. If you can't immediate spot the pattern, then you have to put in a bit of work to prove what the pattern actually is....TESTing VALUES can help you to prove that a pattern exists.....

Here, we're told that Y >= 0. We're asked for the value of X.

Fact 1: |X-3| >= Y

IF.... Y = 0 Then |X-3| >= 0, so X can be ANY number. As Y gets bigger, certain options are eliminated, but given this 'restriction', X has an infinite number of possibilities. Fact 1 is INSUFFICIENT

Fact 2: |X-3| <= -Y

Here, we have to be CAREFUL with the details. Notice how there's a NEGATIVE sign in front of the Y.....

IF.... Y = 0 |X-3| <= 0

Absolute values CANNOT have negative results - the result is ALWAYS 0 or a positive, so this TEST has JUST ONE solution... X = 3

IF.... Y = 1 |X-3| <= - 1 which is NOT POSSIBLE.

From the prompt, we know that Y >= 0, so choosing a positive value for Y will NOT fit the absolute value given in Fact 2. This means that the ONLY possible value for Y is 0. By extension, there is ONLY ONE possible value for X....X = 3. Fact 2 is SUFFICIENT

MBA Acceptance Rate by Country Most top American business schools brag about how internationally diverse they are. Although American business schools try to make sure they have students from...

McCombs Acceptance Rate Analysis McCombs School of Business is a top MBA program and part of University of Texas Austin. The full-time program is small; the class of 2017...

Now that I have got all the important preparations to move to Switzerland ticked off, I am thinking of the actual curriculum now. IMD will be posting a few...

My swiss visa stamping is done and my passport was couriered through Blue dart. It took 5 days to get my passport stamped and couriered to my address. In...