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

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If y>=0, What is the value of x? (1) |x-3|>=y (2) |x-3|<=-y [#permalink] New post 09 Jan 2010, 13:17
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If y\geq{0}, what is the value of x?

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

(2) |x - 3|\leq{-y}
[Reveal] Spoiler: OA
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Re: what is the value of x? [#permalink] New post 09 Jan 2010, 13:45
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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.

Answer: B.
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Re: value of x? [#permalink] New post 08 May 2010, 09:35
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neoreaves wrote:
If y > = 0, what is the value of x?
1. |x - 3| >= y
2. |x - 3| <= - y


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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Re: what is the value of x? [#permalink] New post 19 Jun 2011, 08:15
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Ahhhhh i hate absolute value questions. Gets me every time!
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Re: If y>=0, What is the value of x? (1) |x-3|>=y (2) |x-3|<=-y [#permalink] New post 12 Jan 2014, 06:38
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Re: Gmat Prep DS value question ! [#permalink] New post 20 Mar 2010, 23:54
nsp007 wrote:
If y >= 0 , what is the value of x ?

1. |x-3| >= y
2. |x-3| <= -y


Can anyone pls. explain how to approach such a problem?



B

stmnt1 - |x-3| >= y >=0

let y = 0 then we have |x-3| >= 0 .... x can be 0,1,2,3. hence insuff

stmnt2 - |x-3| <= -y

we know that |x-3| will always be +ve,so only value of y can be 0. we have |x-3| = 0 or x = 3. hence suff
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Re: Gmat Prep DS value question ! [#permalink] New post 21 Mar 2010, 22:11
nsp007 wrote:
If y >= 0 , what is the value of x ?

1. |x-3| >= y
2. |x-3| <= -y


Can anyone pls. explain how to approach such a problem?

OA
[Reveal] Spoiler:
B


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
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Re: Simple inequality [#permalink] New post 18 Jun 2010, 23:37
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gmatbull wrote:
If y >= 0, what is the value of x ?

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

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.
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Re: Simple inequality [#permalink] New post 19 Jun 2010, 00:17
Bunuel wrote:
gmatbull wrote:
If y >= 0, what is the value of x ?

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

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
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Re: what is the value of x? [#permalink] New post 08 Oct 2010, 15:00
Graphical Solution

We know y>=0. We need to know x.

(1) y<=|x-3|

Image

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|

Image

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

Answer is (B)
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Re: what is the value of x? [#permalink] New post 14 Mar 2011, 02:57
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vjsharma25 wrote:
Bunuel wrote:
(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.
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Re: what is the value of x? [#permalink] New post 14 Mar 2011, 03:05
Bunuel wrote:
vjsharma25 wrote:
Bunuel wrote:
(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.
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Re: what is the value of x? [#permalink] New post 14 Mar 2011, 03:16
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vjsharma25 wrote:
Does this symbol ---> means "implies that"?. Because I interpreted it that way in the statement.
Rest of the explanation is clear to me.


As it does. But 0\leq{-y} is from |x-3|\leq{-y} not from y\geq{0}.
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Re: what is the value of x? [#permalink] New post 14 Mar 2011, 03:42
If y>=0, what is the value of x?

(1) |x-3| >= y

x-3>=y
x>=y+3
or
x-3<=-y
x<=3-y
N.S.

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

A positive y can never be less than a -ve y. It can be equal only when y=0
so; +y=x-3=-y=0
x-3=0
x=3
Sufficient.

Ans: "B"
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Re: what is the value of x? [#permalink] New post 19 May 2011, 23:56
absolute value can never give negative results.
B is possible for x = 3
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Re: what is the value of x? [#permalink] New post 20 May 2011, 07:38
(1)

x - 3 >= 0 as y is >= 0

=> x >= 3

InSufficient

(2)

|x-3| <= 0 as -y <= 0

but |x-3| can't be negative

=> x-3 = 0

=> x = 3

Sufficient

Answer - B
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Re: what is the value of x?   [#permalink] 20 May 2011, 07:38
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