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Re: what is the value of x? [#permalink]
09 Jan 2010, 13:45

4

This post received KUDOS

Expert's post

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.

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]
18 Jun 2010, 23:37

Expert's post

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.

Re: Simple inequality [#permalink]
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]
08 Oct 2010, 15:00

Graphical Solution

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

(1) y<=|x-3|

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

Re: what is the value of x? [#permalink]
14 Mar 2011, 02:57

Expert's post

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]
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.

Re: what is the value of x? [#permalink]
19 Jun 2011, 08:15

1

This post received KUDOS

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]
12 Jan 2014, 06:38

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