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

(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.

Hope it helps..
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IMO B

Statement 1). |x - 3| >= y >=0
|x - 3| >= 0 , for different values of x, this is true.

Statement 2). |x - 3| <= - y since |x - 3| is always >=0 , and y>=0

|x - 3| <= - y will hold true only when y is 0

=> |x - 3| = 0 , only solution is x=3 hence sufficient.

Thus B
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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.

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

(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]

Apply to statement (2)
|x - 3| <= -y
<--> -(-y) <= (x-3) <= -y,
<--> y <= (x-3) <= -y

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.

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

Final Answer:

GMAT assassins aren't born, they're made,
Rich
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Hi Bunuel,

I understand the OR scenario you mentioned above. However, what if the statement reads \(|x-3|\leq{y}\)? Then we can have \(3-y\leq{x}\leq{3+y}\), can't we? So here we have the AND scenario. Is that right?

And if I am correct, how can we solve the above inequalities, given that \(y\geq{0}\)?

Thank you very much!

Hi,
this too is OR scenario,
because \(3-y\leq{x}\) and \({x}\leq{3+y}\) are dependent on different set of values of x..
\(3-y\leq{x}\) is when \(x<3\)..
and \({x}\leq{3+y}\) is when \(x\geq{3}\)..
Example of AND is
when we have two eqs in x, and not dependent on each other..
say x<3.. and x+2>1..
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Hi chetan2u,

Thank you for the prompt reply. But I still have a doubt.

Take the inequation \(x^2<4\) for example. We then have \(|x|<2\), which means \(-2<x<2\). What I understand is that x must be greater than -2 AND less than 2 for the inequalitiy to hold. So I think this is an AND scenario.

If, however, \(x^2>4\), then \(x<-2\) OR \(x>2\). This is clearly an OR scenario. Here OR makes sense to me.

My inequality skill is pretty rusty. Thank you for bearing with me. I very appreciate your help!

Hi truongynhi,

I am happy to help you and clear a few doubts you have..

WHAt does OR and AND mean..

1) Take the inequation \(x^2<4\) for example. We then have \(|x|<2\), which means \(-2<x<2\)
So here too you had 2 inequalities, x<2 and x>-2..
x<2 can mean x is -3 so this is a solution when we are using OR since we are not looking at both together..
But here x<2 and x>-2 has a range which OVERLAPS, so this is the combined solution for two inequalities..
when you are choosing a value in this range, you are using AND, since taht value will satisfy both the inequalities..

2)If, however, \(x^2>4\), then \(x<-2\) OR \(x>2\). This is clearly an OR scenario.
YES, this is OR situation, because there is no overlap and hence there is no possiblity of a combined solution..
any solution will satisfy just one inequality..

Now when you have two variable as in the case of Q mentioned here..
3+y≤x≤3−y.. you cannot take this as a solution..
WHY..
because the two inequalities you have combined have come in two different scenarios of OR while taking value of x..
there is no OVERLAP in values of x.. x<3 for one inequality and x>= 3 for other..
so don't combine the two..
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Statement 1 is not sufficient, since if y=0, x can be anything.

In Statement 2, the left side is 0 or greater, since it is an absolute value. The right side is 0 or smaller, since it is the negative of y, which is at least zero. If the left side (which is > 0) is no bigger than the right side (which is < 0), the only possibility is that they are both exactly equal to zero. And that only happens if x=3.
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If \(y\geq{0}\), what is the value of x?

(1) \(|x - 3|\geq{y}\)
\(|x - 3|\geq{y}\geq{0}\)
\(|x-3| \geq 0\)
NOT SUFFICIENT

(2) \(|x - 3|\leq{-y}\)
\(|x - 3|\leq{-y} \leq 0\)
\(|x-3| \leq 0\)
|x-3| = 0
x = 3
SUFFICIENT

IMO B
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Hello Buenel,
Could you please let me know how "x-3>=0 became x-3>0"

Or another way:
|x−3|≥y|x−3|≥y means that:

x−3≥y≥0x−3≥y≥0 when x−3>0x−3>0 --> x>3x>3

OR (not and)
−x+3≥y≥0−x+3≥y≥0 when x−3<0x−3<0 --> x<3x<3

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