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If y is greater than or equal to 0, what is the value of x?

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Manager
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If y is greater than or equal to 0, what is the value of x? [#permalink] New post 08 Sep 2007, 11:58
If y is greater than or equal to 0, what is the value of x?

(1) |x-3| >= y
(2) |x-3| <= -y
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Re: absolute value [#permalink] New post 08 Sep 2007, 13:16
A.Haung wrote:
If y is greater than or equal to 0, what is the value of x?

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


I get B

Given: y>=0

(1) |x-3| >= y
Use the given: 0 >= -y
Add inequalities to obtain:
|x-3| >= 0
This means that x can be pretty much anything.
INSUFFICIENT

(2) |x-3| <= -y
Use the given: 0 <= y
Do the same thing...get
|x-3| <= 0
Since |x-3| is always positive, the only possible solution for this is:
|x-3| = 0
x=3
SUFFICIENT
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Re: absolute value [#permalink] New post 27 Sep 2007, 21:26
A.Haung wrote:
If y is greater than or equal to 0, what is the value of x?

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


B. since Y cannot be -ve, so lx-3l must be 0.
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Re: absolute value [#permalink] New post 27 Sep 2007, 22:19
[quote="A.Haung"]If y is greater than or equal to 0, what is the value of x?

(1) |x-3| >= y
(2) |x-3| <y>=0 or -x+3>=0

What am I missing I get D???

S1:

x-3>=0 or -x+3>=0

x>=3 3>=x So 3=<x<=3


S2:

x-3<=0 or -x+3<=0

x<=3 or 3<=x 3<=x<=3
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 [#permalink] New post 28 Sep 2007, 07:22
I get E cause we dont know the exact value of y...

suppose if Y=1 ?

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

(1)
if x>3

x-3 >0; x>3 but x could be 4 if y=1..x could be 5 if y=2

if x<3

-(x-3) >0
-x+3>0 -x>-3 which means x<3 if y=0 x<3 if y=1 x<2 if y=2 x<1


(2)
x>3
x-3 <-0 x< 3 if y=0, x<2 if y=1 x<1

if x<3
-(x-3) <-0
-x+3 <-0
-x<-3 x>3 if y=0 if y=1 x>4 if y=2 x>5

Insuff..

I get E ..what am i doing wrong?
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 [#permalink] New post 28 Sep 2007, 07:48
I get D

[1] |x-3| >= y --> from this we can deivse two equations and can solve each for x:
1) x-3 >= y --> x >= y+3
2) 3-x >= y --> -x >= y-3 --> x <= 3-y

Putting both together we get: y+3 <= x <3>= 0 we can start plugging in numbers.
For y=0 we get 3<=x<=3 and therefore x = 3
For y=1 we get 4<=x<=2 and obviously this cannot work
Therefore, the only way the above inequality works is if y=0 and x=3
SUFFICIENT

[2] |x-3|<y> we can solve this the same way as [1] and get the same inequality: y+3 <= x <= 3-y
Therefore, SUFFICIENT
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 [#permalink] New post 28 Sep 2007, 12:33
I get D.

1)
X>=Y+3
X<=-Y+3
Y>=0
=> Only one solution : (X;Y) = (3;0)
=> SUFF

2)
X<=-Y+3
X>=Y+3
Y>=0
=> Only one solution : (X;Y) = (3;0)
=> SUFF

Therefore ans D !
What's the OA ?
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Re: absolute value [#permalink] New post 28 Sep 2007, 13:38
A.Haung wrote:
If y is greater than or equal to 0, what is the value of x?

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


ST1:

|x-3| >= y

Given that , y>=0

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

So x can have many values hene NOT SUFF.

ST2:

|x-3| <= -y

Given that y>=0
Multiple both sides by (-1), we get,

-y<=0

Not put the above value in ST2, we get,

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

=> |x-3| <=0
Absolute value can not be -ve so x has to 3 to make '|x-3|' zero.

Hence B is SUFF.

- Brajesh
Re: absolute value   [#permalink] 28 Sep 2007, 13:38
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