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:

(1) Given \(y\) is non negative value and \(|x - 3|\geq{y}\), so \(|x - 3|\) is more than some non negative value, (we could say the same ourselves as absolute value in our case (\(|x - 3|\)) is never negative). So we can not determine single numerical value of \(x\). Not sufficient.

Or another way: to check \(|x - 3|\geq{y}\geq{0}\) 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 \(|x - 3|>=y\), \(y\geq{0}\).

Or another way: \(|x - 3|\geq{y}\) means that:

\(x - 3\geq{y}\geq{0}\) when \(x-3>0\) --> \(x>3\)

OR (not and) \(-x+3\geq{y}\geq{0}\) when \(x-3<0\) --> \(x<3\)

Generally speaking \(|x - 3|\geq{y}\geq{0}\) means that \(|x - 3|\), an absolute value, is not negative. So, there's no way you'll get a unique value for \(x\). INSUFFICIENT.

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

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

stmt1: |x-3| >= y in case y>= 0 if we consider y = 0, |x-3| >= 0 but y can be 0, 1,2,3 anything so insufficient.

stmt2: |x-3| <= -y since y >= 0 => -y <= 0 so we can say |x-3| <= -y <= 0 but |x-3| has to be positive which is only possible if y = 0 and |x-3| = 0 only one condition suffice this if x=3, hence sufficient. so B is the answer
_________________

I know that the value of an abs expression cannot be less than 0. so does that imply that x is 0 units from 3?

As \(y\) is some non-negative value (0, 7, 1.4, ...) then \(-y\) is some non-positive value (0, -9, -5.6, ...). Statement 2 says that \(|x-3|\leq{-y}\) (\(|x-3|\leq{non-positive}\)) - absolute value is less than or equal to some non-positive value, BUT absolute value can not be negative, least value of it is zero, thus \(|x-3|={0}=y\) --> \(x=3\).

I know that the value of an abs expression cannot be less than 0. so does that imply that x is 0 units from 3?

As \(y\) is some non-negative value (0, 7, 1.4, ...) then \(-y\) is some non-positive value (0, -9, -5.6, ...). Statement 2 says that \(|x-3|\leq{-y}\) (\(|x-3|\leq{non-positive}\)) - absolute value is less than or equal to some non-positive value, BUT absolute value can not be negative, least value of it is zero, thus \(|x-3|={0}=y\) --> \(x=3\).

Refer for the full solution to my previous post.

Hope it's clear.

I trust your explanations - very clear. Thanks
_________________

KUDOS me if you feel my contribution has helped you.

The inequality represents the regino between the x-axis and the blue line. Knowing y>=0 is not enough to know x

(2)-y>=|x-3| OR y<=-|x-3|

The inequality represents the region below the blue line. We know y>=0, this can only represent the point where, the blue line meets the x-axis or x=3. Sufficient

(2) \(|x-3|\leq{-y}\), \(y\geq{0}\) --> [highlight]\(0\leq{-y}\)[/highlight], equation says that \(|x-3|\) less or equals to zero, but \(|x-3|\) never negative (\(|x-3|\geq{0}\)), so only solution is if \(|x-3|=0=y\) --> \(x-3=0\) --> \(x=3\). SUFFICIENT It should be [highlight]\(0\geq{-y}\)[/highlight] in my opinion

No, it should be as it is.

(2) \(|x-3|\leq{-y}\) --> now, as LHS is absolute value, which is always non-negative, then we have that \(-y\) is more than or equal to some non-negative value: \(0\leq{-y}\) --> \(y\leq{0}\). As also is given that \(y\geq{0}\) then \(y=0\) --> \(|x-3|\leq{0}\) --> absolute value can not be negative, so \(|x-3|=0\) --> \(x=3\).
_________________

(2) \(|x-3|\leq{-y}\), \(y\geq{0}\)-->[highlight]\(0\leq{-y}\)[/highlight], equation says that \(|x-3|\) less or equals to zero, but \(|x-3|\) never negative (\(|x-3|\geq{0}\)), so only solution is if \(|x-3|=0=y\) --> \(x-3=0\) --> \(x=3\). SUFFICIENT It should be [highlight]\(0\geq{-y}\)[/highlight] in my opinion

No, it should be as it is.

(2) \(|x-3|\leq{-y}\) --> now, as LHS is absolute value, which is always non-negative, then we have that \(-y\) is more than or equal to some non-negative value: \(0\leq{-y}\) --> \(y\leq{0}\). As also is given that \(y\geq{0}\) then \(y=0\) --> \(|x-3|\leq{0}\) --> absolute value can not be negative, so \(|x-3|=0\) --> \(x=3\).

Does this symbol ---> means "implies that"?. Because I interpreted it that way in the statement. Rest of the explanation is clear to me.

Ahhhhh i hate absolute value questions. Gets me every time!
_________________

******************************************************************* ~ PickyTooth - Eat Like a Local Foodie // http://www.pickytooth.com ~ *******************************************************************

Re: If y>=0, What is the value of x? (1) |x-3|>=y (2) |x-3|<=-y [#permalink]

Show Tags

12 Jan 2014, 07:38

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? (1) |x-3|>=y (2) |x-3|<=-y [#permalink]

Show Tags

29 Jun 2015, 05:13

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? (1) |x-3|>=y (2) |x-3|<=-y [#permalink]

Show Tags

17 Sep 2016, 23:08

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? (1) |x-3|>=y (2) |x-3|<=-y [#permalink]

Show Tags

16 Oct 2016, 19:49

Bunuel wrote:

If \(y\geq{0}\), what is the value of x?

(1) \(|x - 3|\geq{y}\)

(2) \(|x - 3|\leq{-y}\)

(1) Given \(y\) is non negative value and \(|x - 3|\geq{y}\), so \(|x - 3|\) is more than some non negative value, (we could say the same ourselves as absolute value in our case (\(|x - 3|\)) is never negative). So we can not determine single numerical value of \(x\). Not sufficient.

Or another way: to check \(|x - 3|\geq{y}\geq{0}\) 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 \(|x - 3|>=y\), \(y\geq{0}\).

Or another way: \(|x - 3|\geq{y}\) means that:

\(x - 3\geq{y}\geq{0}\) when \(x-3>0\) --> \(x>3\)

OR (not and) \(-x+3\geq{y}\geq{0}\) when \(x-3<0\) --> \(x<3\)

Generally speaking \(|x - 3|\geq{y}\geq{0}\) means that \(|x - 3|\), an absolute value, is not negative. So, there's no way you'll get a unique value for \(x\). INSUFFICIENT.

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

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

(1) Given \(y\) is non negative value and \(|x - 3|\geq{y}\), so \(|x - 3|\) is more than some non negative value, (we could say the same ourselves as absolute value in our case (\(|x - 3|\)) is never negative). So we can not determine single numerical value of \(x\). Not sufficient.

Or another way: to check \(|x - 3|\geq{y}\geq{0}\) 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 \(|x - 3|>=y\), \(y\geq{0}\).

Or another way: \(|x - 3|\geq{y}\) means that:

\(x - 3\geq{y}\geq{0}\) when \(x-3>0\) --> \(x>3\)

OR (not and) \(-x+3\geq{y}\geq{0}\) when \(x-3<0\) --> \(x<3\)

Generally speaking \(|x - 3|\geq{y}\geq{0}\) means that \(|x - 3|\), an absolute value, is not negative. So, there's no way you'll get a unique value for \(x\). INSUFFICIENT.

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

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

There’s something in Pacific North West that you cannot find anywhere else. The atmosphere and scenic nature are next to none, with mountains on one side and ocean on...

This month I got selected by Stanford GSB to be included in “Best & Brightest, Class of 2017” by Poets & Quants. Besides feeling honored for being part of...

Joe Navarro is an ex FBI agent who was a founding member of the FBI’s Behavioural Analysis Program. He was a body language expert who he used his ability to successfully...