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:

if y is positive then the above condition is not possible as sum of two positives cant be less than difference of two positives. the other possibility is that y is zero and x = 3. hence sufficient?

2) |x-3| <= -y

x -3 <= -y x <=3-y

or

3-x <=-y x>=y+3

so 3-y >= x >= y+3, again since y >=0 , only possibility is that y = 0. hence x = 3. sufficient.

if y >= 0, What is the value of x? 1) |x-3| >= y 2) |x-3| <= -y

I solved this way -

1) |x-3| >= y

x -3 >=y x >= y + 3

or

3 - x >=y x <= 3 - y

so y + 3 <= x <= 3 - y

if y is positive then the above condition is not possible as sum of two positives cant be less than difference of two positives. the other possibility is that y is zero and x = 3. hence sufficient?

2) |x-3| <= -y

x -3 <= -y x <=3-y

or

3-x <=-y x>=y+3

so 3-y >= x >= y+3, again since y >=0 , only possibility is that y = 0. hence x = 3. sufficient.

But the OA is different.

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

(1) \(|x - 3|\geq{y}\). As given that \(y\) is non negative value then \(|x - 3|\) is more than (or equal to) 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) \(|x-3|\leq{-y}\). Now, as \(|x-3|\) is never negative (\(0\leq{|x-3|}\)) then \(0\leq{-y}\) --> \(y\leq{0}\) BUT stem says that \(y\geq{0}\) thus \(y=0\). \(|x-3|\leq{0}\) --> \(|x-3|=0=y\) (as absolute value, in our case |x-3|, can not be less than zero) --> \(x-3=0\) --> \(x=3\). 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 \(0\).

Answer: B.

There is a following problem with your solution: If \(x<3\) --> \(-(x-3)\geq{y}\) --> \(3-y\geq{x}\); OR: If \(x\geq{3}\) --> \((x-3)\geq{y}\) --> \(x\geq{3+y}\);

But you can not combine these inequalities and write: \(3+y\leq{x}\leq{3-y}\) as they are OR scenarios not AND scenarios (meaning that depending on the value of x we'll have either the first one or the second one).

But you can not combine these inequalities and write: \(3+y\leq{x}\leq{3-y}\) as they are OR scenarios not AND scenarios (meaning that depending on the value of x we'll have either the first one or the second one).

Thanks Bunuel..realize my mistake now...great explanation as always...

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

Show Tags

08 Jul 2013, 16:14

If y >= 0, What is the value of x?

(1) |x-3| >= y if y>=0 then |x-3| must be greater than y. |x-3| >= y x>=3: |x-3| >= y x-3 >=y x>=y+3 There is no need to test for a negative case here as we are told that |x-3| is greater than or equal to a positive number. Still, we don't know anything about x except that it must be greater than or equal to 3. INSUFFICIENT

(2) |x-3| <= -y This means that x-3 is negative, so: -(x-3)<=-y 3-x<=-y -(3-x)<=y

Bunuel, I get your explanation, but I have to be honest, I don't think I would have any idea on the test to employ the explanation you got as opposed to the one I started with. Can you shed some light on to how you utilized the particular solution you used (i.e. what signs should I look for to know to solve this problem this way?)

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

Show Tags

08 Jul 2013, 17:38

WholeLottaLove wrote:

If y >= 0, What is the value of x?

(1) |x-3| >= y if y>=0 then |x-3| must be greater than y. |x-3| >= y x>=3: |x-3| >= y x-3 >=y x>=y+3 There is no need to test for a negative case here as we are told that |x-3| is greater than or equal to a positive number. Still, we don't know anything about x except that it must be greater than or equal to 3. INSUFFICIENT

We're not told |x-3| is greater than or equal to a positive number. We're told that y is either 0 or a positive number and that |x-3| >= y. So |x-3| can equal 0. But since |anything|>=0 this doesn't tell us anything more than we already know. In other words, the absolute value of any number is not negative and 1 simply states that |x-3| is not negative.

Quote:

(2) |x-3| <= -y This means that x-3 is negative, so: -(x-3)<=-y 3-x<=-y -(3-x)<=y

|x-3|<=-y does not mean that x-3 is negative. The key to the problem is to remember that the absolute value of anything is not negative. The difference in 2 is that, since we know y>=0, we know that -y <=0. But since 2 states that |x-3|<=-y and it is a fact that the absolute value of anything is not negative, -y=0 and so y=0. Then it's clear that |x-3|=0 and so x=3. Anytime absolute value shows up the trick may likely be remembering the basic fact that absolute value is not negative but that not negative is not the same as positive. There is a possibility of 0.

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

Show Tags

08 Jul 2013, 19:31

Hmmmm...I am still quite lost on your explanation for 2)

Observer wrote:

WholeLottaLove wrote:

If y >= 0, What is the value of x?

(1) |x-3| >= y if y>=0 then |x-3| must be greater than y. |x-3| >= y x>=3: |x-3| >= y x-3 >=y x>=y+3 There is no need to test for a negative case here as we are told that |x-3| is greater than or equal to a positive number. Still, we don't know anything about x except that it must be greater than or equal to 3. INSUFFICIENT

We're not told |x-3| is greater than or equal to a positive number. We're told that y is either 0 or a positive number and that |x-3| >= y. So |x-3| can equal 0. But since |anything|>=0 this doesn't tell us anything more than we already know. In other words, the absolute value of any number is not negative and 1 simply states that |x-3| is not negative.

Quote:

(2) |x-3| <= -y This means that x-3 is negative, so: -(x-3)<=-y 3-x<=-y -(3-x)<=y

|x-3|<=-y does not mean that x-3 is negative. The key to the problem is to remember that the absolute value of anything is not negative. The difference in 2 is that, since we know y>=0, we know that -y <=0. But since 2 states that |x-3|<=-y and it is a fact that the absolute value of anything is not negative, -y=0 and so y=0. Then it's clear that |x-3|=0 and so x=3. Anytime absolute value shows up the trick may likely be remembering the basic fact that absolute value is not negative but that not negative is not the same as positive. There is a possibility of 0.

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

Show Tags

09 Jul 2013, 01:03

WholeLottaLove wrote:

(2) |x-3| <= -y This means that x-3 is negative, so: -(x-3)<=-y 3-x<=-y -(3-x)<=y

Dear WholeLottaLove

Your point "This means that x-3 is negative" is incorrect.

General theory is: \(|X| <= a\), ==> \(-a <= X <= a\) So, X does not have to be negative. X is between -a and a. For example: \(|x -3| <= 9\) ==> \(-9 <= (x -3) <=9\)

The KEY point is absolute value cannot be negative, thus the smallest value of \(|x - 3|\) is 0. Because \(-y <= 0\) ==> \(|x-3|\) must be 0 ==> x = 3.

Hope it helps.
_________________

Please +1 KUDO if my post helps. Thank you.

"Designing cars consumes you; it has a hold on your spirit which is incredibly powerful. It's not something you can do part time, you have do it with all your heart and soul or you're going to get it wrong."

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

Show Tags

09 Jul 2013, 13:05

First number line: y>=0 Second: From properties of absolute value, |x-3|>=0 is true no matter what x is. Third: (2) tells us that |x-3|<=-y. We know nothing about y other than y>=0. Therefore |x-3|<=-y only tells us that |x-3|<=0. Fourth: Since |x-3|<=0 AND |x-3|>=0, we now know the exact value of |x-3|. In the diagram it is the point on the number line that is both <=0 and >=0. The only value for which that is true is 0. So |x-3|=0. From there you know that x-3=0 and so x=3.

WholeLottaLove wrote:

Hmmmm...I am still quite lost on your explanation for 2)

Observer wrote:

WholeLottaLove wrote:

If y >= 0, What is the value of x?

(1) |x-3| >= y if y>=0 then |x-3| must be greater than y. |x-3| >= y x>=3: |x-3| >= y x-3 >=y x>=y+3 There is no need to test for a negative case here as we are told that |x-3| is greater than or equal to a positive number. Still, we don't know anything about x except that it must be greater than or equal to 3. INSUFFICIENT

We're not told |x-3| is greater than or equal to a positive number. We're told that y is either 0 or a positive number and that |x-3| >= y. So |x-3| can equal 0. But since |anything|>=0 this doesn't tell us anything more than we already know. In other words, the absolute value of any number is not negative and 1 simply states that |x-3| is not negative.

Quote:

(2) |x-3| <= -y This means that x-3 is negative, so: -(x-3)<=-y 3-x<=-y -(3-x)<=y

|x-3|<=-y does not mean that x-3 is negative. The key to the problem is to remember that the absolute value of anything is not negative. The difference in 2 is that, since we know y>=0, we know that -y <=0. But since 2 states that |x-3|<=-y and it is a fact that the absolute value of anything is not negative, -y=0 and so y=0. Then it's clear that |x-3|=0 and so x=3. Anytime absolute value shows up the trick may likely be remembering the basic fact that absolute value is not negative but that not negative is not the same as positive. There is a possibility of 0.

(1) |x-3| >= y We are told that |x-3| ≥ y but all we know is that y is a positive # greater than or equal to zero. Therefore, all we know is that |x-3| is greater than or equal to zero and x could be an infinite number of possibilities.

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

(I am having real difficulty understanding the rationale for (2) could someone explain it to me? (preferably like they would to a 5th grader )

(1) |x-3| >= y We are told that |x-3| ≥ y but all we know is that y is a positive # greater than or equal to zero. Therefore, all we know is that |x-3| is greater than or equal to zero and x could be an infinite number of possibilities.

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

(I am having real difficulty understanding the rationale for (2) could someone explain it to me? (preferably like they would to a 5th grader )

Thanks!

Hi WholeLottaLove

The important property concerning absolute value inequalities is: |a| <= b <--> -b <= a <=b [a is in the middle of -b and b, inclusive]

We know y is 0 or positive, -y is 0 or negative. Because there is not any number that is both negative and positive.Thus, there is ONLY one number that is both >= y AND <= -y. That is zero. Therefore, (x-3) must be zero. ==> x = 3

Hope it's clear.
_________________

Please +1 KUDO if my post helps. Thank you.

"Designing cars consumes you; it has a hold on your spirit which is incredibly powerful. It's not something you can do part time, you have do it with all your heart and soul or you're going to get it wrong."

(1) |x-3| >= y We are told that |x-3| ≥ y but all we know is that y is a positive # greater than or equal to zero. Therefore, all we know is that |x-3| is greater than or equal to zero and x could be an infinite number of possibilities.

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

(I am having real difficulty understanding the rationale for (2) could someone explain it to me? (preferably like they would to a 5th grader )

Thanks!

Hi WholeLottaLove

The important property concerning absolute value inequalities is: |a| <= b <--> -b <= a <=b [a is in the middle of -b and b, inclusive]

We know y is 0 or positive, -y is 0 or negative. Because there is not any number that is both negative and positive.Thus, there is ONLY one number that is both >= y AND <= -y. That is zero. Therefore, (x-3) must be zero. ==> x = 3

Hope it's clear.

Just to be clear (because sometimes I can be a bit dense!) we know that y is zero or negative because it is less than or equal to (x-3) which is less than or equal to -y, correct? In other words, y has to be less than or equal to zero because it is less than or equal to a negative number. The intersection of -y and y>= 0 is zero, right?

Also, do you know of any similar problems that I could practice on?

(1) |x-3| >= y We are told that |x-3| ≥ y but all we know is that y is a positive # greater than or equal to zero. Therefore, all we know is that |x-3| is greater than or equal to zero and x could be an infinite number of possibilities.

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

(I am having real difficulty understanding the rationale for (2) could someone explain it to me? (preferably like they would to a 5th grader )

Thanks!

Hi WholeLottaLove

The important property concerning absolute value inequalities is: |a| <= b <--> -b <= a <=b [a is in the middle of -b and b, inclusive]

We know y is 0 or positive, -y is 0 or negative. Because there is not any number that is both negative and positive.Thus, there is ONLY one number that is both >= y AND <= -y. That is zero. Therefore, (x-3) must be zero. ==> x = 3

Hope it's clear.

Just to be clear (because sometimes I can be a bit dense!) we know that y is zero or negative because it is less than or equal to (x-3) which is less than or equal to -y, correct? In other words, y has to be less than or equal to zero because it is less than or equal to a negative number. The intersection of -y and y>= 0 is zero, right?

Also, do you know of any similar problems that I could practice on?

Thanks!

Hi WholeLottaLove

Yes, you're correct. The key point is -y <= 0, and according to the inequality: y <= (x-3) <= -y So y <= -y. But y >= 0. Thus, y must be zero so -y = y = 0.

I would like to clarify a bit. My post above is to explain why the intersection must be zero in the case |A| <= zero or negative number. The KEY of this question is -y <= 0 (a zero or negative number), and the smallest value |x-3| (absolute value) is zero, so (x-3) must be zero. If you see a "regular" question like: |x-3| <= 9 (a positive number). The range of (x-3) is from -9 to 9.

Hope it's more clear.

PS: If I know more similar questions, I will pm you.
_________________

Please +1 KUDO if my post helps. Thank you.

"Designing cars consumes you; it has a hold on your spirit which is incredibly powerful. It's not something you can do part time, you have do it with all your heart and soul or you're going to get it wrong."

Yes, you're correct. The key point is -y <= 0, and according to the inequality: y <= (x-3) <= -y So y <= -y. But y >= 0. Thus, y must be zero so -y = y = 0.

I would like to clarify a bit. My post above is to explain why the intersection must be zero in the case |A| <= zero or negative number. The KEY of this question is -y <= 0 (a zero or negative number), and the smallest value |x-3| (absolute value) is zero, so (x-3) must be zero. If you see a "regular" question like: |x-3| <= 9 (a positive number). The range of (x-3) is from -9 to 9.

Hope it's more clear.

PS: If I know more similar questions, I will pm you.[/quote]

|x-3| can be at the very least, 0 (as it is an absolute value) so if we are told that |x-3| is less than negative y, then y can be at the very most zero (because obviously -y cannot be a positive number) so if y is a negative number and it is greater than something that can be at least zero, the only option is for y to be zero (|0| <= -0)

If y>=0 then |x-3| >= to any positive number. X cannot be determined. INSUFFICIENT

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

So, |x-3| can never be negative which means that it cannot be less than a negative number. For |x-3| <= -y to be true, y must be negative: |x-3| <= -y |x-3| <= -(-y) |x-3| <= (y)

It is possible for |x-3| to be less than or equal to y.

So, y is less than or equal to zero but we are also told that y is greater than or equal to zero, therefore the intersection is at 0.

If y>=0 then |x-3| >= to any positive number. X cannot be determined. INSUFFICIENT

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

So, |x-3| can never be negative which means that it cannot be less than a negative number. For |x-3| <= -y to be true, y must be negative: YOU"RE CORRECT. THAT'S THE KEY.

|x-3| <= -y |x-3| <= -(-y) |x-3| <= (y)

It is possible for |x-3| to be less than or equal to y.

So, y is less than or equal to zero but we are also told that y is greater than or equal to zero, therefore the intersection is at 0.

I think I understand!!

I'm supposed to reply your question, but you got it already Cheer!
_________________

Please +1 KUDO if my post helps. Thank you.

"Designing cars consumes you; it has a hold on your spirit which is incredibly powerful. It's not something you can do part time, you have do it with all your heart and soul or you're going to get it wrong."

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.
_________________

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.
_________________

Happy New Year everyone! Before I get started on this post, and well, restarted on this blog in general, I wanted to mention something. For the past several months...

It’s quickly approaching two years since I last wrote anything on this blog. A lot has happened since then. When I last posted, I had just gotten back from...

Happy 2017! Here is another update, 7 months later. With this pace I might add only one more post before the end of the GSB! However, I promised that...