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# Value of X

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Value of X [#permalink]  03 Aug 2009, 15:00
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Re: Value of X [#permalink]  06 Aug 2009, 19:59
because you have no information to figure out y maybe?
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Re: Value of X [#permalink]  07 Aug 2009, 02:29
I choose C .

according to 1 and 2: y<=absolute value" x-3 "<=-y

and y>=0 (y is 0 or positive); that is to say -y is negative

==> only when y=0, the inequation is available. so when y=0, x=3.
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Re: Value of X [#permalink]  09 Sep 2009, 11:59
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what is the OA? Please post it. Think it is D...
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Re: Value of X [#permalink]  09 Sep 2009, 15:52
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barakhaiev wrote:
what is the OA? Please post it. Think it is D...

y>=0
(1) |x-3|>= y, x can be any number >=3, x<=3 Not suff
2) |x-3|<=-y, y>=0, equation says that |x-3|less or equals to zero, but |x-3| never negative (|x-3|>=0), so only solution is if |x-3|=0=y --> x-3=0 --> x=3 Sufficinet

B it is.
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Last edited by Bunuel on 09 Sep 2009, 17:25, edited 1 time in total.
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Re: Value of X [#permalink]  09 Sep 2009, 16:50
1
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y >= 0

1.) |x-3| >= y

=> -y >= (x-3) >= y

=> 3-y >= x >= 3+y

Now, if y = 0 , then, 3 >= x >= 3 => x can only be 3
but, if y = 1 , then, 2 >= x >= 4 => which is not possible for any value of x.

=> x can only be 3 when y = 0 ..sufficient..

2.) |x-3| <= -y

=> y <= (x-3) <= -y

=> 3+y <= x <= 3-y

when y = 0 ; then, 3 <= x <= 3 => x can only be 3..
when y = 1 ; then 4 <= x <= 2, which is not possible for any value of x.

=> x can only be 3 when y = 0 ,, sufficient..

Hence, Ans is D.
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Re: Value of X [#permalink]  09 Sep 2009, 17:24
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gmate2010 wrote:
y >= 0

1.) |x-3| >= y

=> -y >= (x-3) >= y

=> 3-y >= x >= 3+y

Now, if y = 0 , then, 3 >= x >= 3 => x can only be 3
but, if y = 1 , then, 2 >= x >= 4 => which is not possible for any value of x.

=> x can only be 3 when y = 0 ..sufficient..

2.) |x-3| <= -y

=> y <= (x-3) <= -y

=> 3+y <= x <= 3-y

when y = 0 ; then, 3 <= x <= 3 => x can only be 3..
when y = 1 ; then 4 <= x <= 2, which is not possible for any value of x.

=> x can only be 3 when y = 0 ,, sufficient..

Hence, Ans is D.

Just plug numbers for statement (1) if 6=3 y=1 |x-3|=3 > 1 or x=-2 y=2 |x-3|=5>2 not sufficient.

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Re: Value of X [#permalink]  09 Sep 2009, 18:43
If y>=0 specified in the stem. Further mod of anything is positive.
So, 0>=y according to 1.
both y>=0 or y<=0, means y=0. So mod(x-3)>=0. Then x-3>=0 or x-3<=0
So, x>=3 or x<=3. Thus, x=3.
Same thing for 2.
So, D.
gmate2010 wrote:
y >= 0

1.) |x-3| >= y

=> -y >= (x-3) >= y

=> 3-y >= x >= 3+y

Now, if y = 0 , then, 3 >= x >= 3 => x can only be 3
but, if y = 1 , then, 2 >= x >= 4 => which is not possible for any value of x.

=> x can only be 3 when y = 0 ..sufficient..

2.) |x-3| <= -y

=> y <= (x-3) <= -y

=> 3+y <= x <= 3-y

when y = 0 ; then, 3 <= x <= 3 => x can only be 3..
when y = 1 ; then 4 <= x <= 2, which is not possible for any value of x.

=> x can only be 3 when y = 0 ,, sufficient..

Hence, Ans is D.

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Re: Value of X [#permalink]  09 Sep 2009, 20:03
I will go with B.
x can be anything and value of x can not be determined from A
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Re: Value of X [#permalink]  11 Sep 2009, 19:07
Can someone please post the OA?
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Re: Value of X [#permalink]  12 Sep 2009, 08:50
My 2 cents on answer D
Rules for inequalities
1)|a| <b if - b<a<b
2)|a|>b if a<-b or a>b

apply rule 2 and derive the equation just as gmate2010 and tusharvk have done.
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Re: Value of X [#permalink]  12 Sep 2009, 10:07
2
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I looked through the logic of people who think answer should be D and here is were they are making mistake:

(1) First of all:
To check |x - 3|>=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|>=0, y>=0. So ABSOLUTELY NO WAY (1) CAN BE SUFFICIENT.

|x - 3|>=y means that:
x - 3>=y>=0 when x-3>0 --> x>3
OR (not and)
-x+3>=y>=0 when x-3<0 --> x<3
Generally speaking |x - 3|>=y>=0 means that |x - 3|, an absolute value, is at least some positive number. So, there's no way you'll get a unique value for x. INSUFFICIENT.

(2) |x-3|<=-y, y>=0, equation says that |x-3|less or equals to zero, but |x-3| never negative (|x-3|>=0), so only solution is if |x-3|=0=y --> x-3=0 --> x=3 SUFFICIENT

In other words:
(-y) is 0 or less, and the absolute value (|x-3|) must be at (0) 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: Value of X [#permalink]  12 Sep 2009, 11:31
Bunuel wrote:

I looked through the logic of people who think answer should be D and here is were they are making mistake:

(1) First of all:
To check |x - 3|>=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|>=0, y>=0. So ABSOLUTELY NO WAY (1) CAN BE SUFFICIENT.

|x - 3|>=y means that:
x - 3>=y>=0 when x-3>0 --> x>3
OR (not and)
-x+3>=y>=0 when x-3<0 --> x<3
Generally speaking |x - 3|>=y>=0 means that |x - 3|, an absolute value, is at least some positive number. So, there's no way you'll get a unique value for x. INSUFFICIENT.

(2) |x-3|<=-y, y>=0, equation says that |x-3|less or equals to zero, but |x-3| never negative (|x-3|>=0), so only solution is if |x-3|=0=y --> x-3=0 --> x=3 SUFFICIENT

In other words:
(-y) is 0 or less, and the absolute value (|x-3|) must be at (0) or below this value. But absolute value (in this case |x-3|) can not be less than zero, so it must be 0.

Got it finally! Thank you for clear explanation, Bunuel. B it is.
Of course, A is not suff...
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Re: Value of X   [#permalink] 12 Sep 2009, 11:31
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