If y>=0, What is the value of x? (1) |x-3|>=y (2) |x-3|<=-y : GMAT Data Sufficiency (DS)
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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]

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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]

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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 $$0$$.

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Re: Gmat Prep DS value question ! [#permalink]

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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]

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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: value of x? [#permalink]

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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: Simple inequality [#permalink]

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18 Jun 2010, 23:37
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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 $$|x-3|={0}=y$$ --> $$x=3$$.

Refer for the full solution to my previous post.

Hope it's clear.
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Re: Simple inequality [#permalink]

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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]

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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

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Re: what is the value of x? [#permalink]

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14 Mar 2011, 02:57
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]

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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]

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14 Mar 2011, 03:16
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]

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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]

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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]

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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

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Re: what is the value of x? [#permalink]

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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]

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

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

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

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16 Oct 2016, 18:49
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 $$0$$.

Bunuel

For A. u removed euqlity sign from the equestion.

equation results in x>=3 or x<=3, Hence x could be 3. Hence A sufficies, the condition.
Please tell me, where i understood it wrong.
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Re: If y>=0, What is the value of x? (1) |x-3|>=y (2) |x-3|<=-y [#permalink]

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17 Oct 2016, 02:10
rahul202 wrote:
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 $$0$$.

Bunuel

For A. u removed euqlity sign from the equestion.

equation results in x>=3 or x<=3, Hence x could be 3. Hence A sufficies, the condition.
Please tell me, where i understood it wrong.

Yes, x can be 3 but how does this help to get the exact value of x? Cannot x be any other value?
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Re: If y>=0, What is the value of x? (1) |x-3|>=y (2) |x-3|<=-y   [#permalink] 17 Oct 2016, 02:10

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