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# What is the value of x given that |x - y| = |x - z| (1). y

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What is the value of x given that |x - y| = |x - z| (1). y [#permalink]  07 Jun 2006, 21:20
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What is the value of x given that |x - y| = |x - z|

(1). y is not equal to z
(2). The sum of y and z is 10.
Director
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[#permalink]  07 Jun 2006, 22:14
If both sides are squared then absolute value sign is eliminated and stem is 2XZ-2XY=Z^2-Y^2 => 2X*(Z-Y)=(Z-Y)*(Z+Y)or 2X=Y+Z
now A) is useless IMO
B) IS SUFF X=5
Manager
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Re: DS: Absolute values [#permalink]  07 Jun 2006, 22:31
Professor wrote:
What is the value of x given that |x - y| = |x - z|

(1). y is not equal to z
(2). The sum of y and z is 10.

C according to me

Statement 1 - does not give any 'value' only expressions

Statement 2 - if y=z, x could be any number

Together x=5
Manager
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[#permalink]  07 Jun 2006, 23:05
It should be B

There are 4 possible conditions from question

1) Both |x - y| and |x - z| > 0 gives y = z which is contradicting to 1st statement.

2) Both |x - y| and |x - z| < 0 also gives y = z not possible.

3) |x - y| > 0 and |x - z| < 0 gives 2x = y + z which is answerable with 2nd statement

4) |x - y| < 0 and |x - z| > 0 also gives 2x = y + z same as previous.

Hence B as 2nd statement is able to answer the question.
Intern
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[#permalink]  08 Jun 2006, 00:23
What is the value of x?
We have abs (x-y) = abs (x-z)
this equation yields 2 equations:
a) x-y=x-z => y=z
b) x-y=-x+z => 2x=z+y

From 1): y is not equal to z, we cannot get the exact value of x => insuf
From 2): y+Z=10 we can find x=5 => sufficient !

What is the OA?
Manager
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[#permalink]  08 Jun 2006, 05:44
It's C.
The same logic as humans. But we need 1 to eliminate the possibility of any other value for x.[/quote]
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[#permalink]  08 Jun 2006, 06:30
I think it has to be B.

statement 1 stops us from being able to solve for x if either both x-y and x-z are > or < 0. this is because we know from 2 that y+z =10

so if we have x-y and x-z >0 then we have

x- y - z = x - 2*z

x - 10 = x - 2*z

z = 5, which means that y =5. So if we choose c then we can not solve for x in two of the 4 instances.
SVP
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[#permalink]  08 Jun 2006, 14:01
I get (C)

|x-y| = |x-z|

x-y=x-z <=> y=z
or
x-y=-x+z <=> 2x=y+z

1) y is not equal to z
thus, equation, 2x=y+z
INSUFF

2) y+z=10
Still have the 2 equations with OR, especially the first one x-y=x-z with x which can be any real number.
INSUFF

Conbining (1) and (2) :
2x=x+y=10 => x=5
Manager
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[#permalink]  08 Jun 2006, 21:28
late answer but B

samel logic as hidalgo01....

i think this is a tricky one...
SVP
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[#permalink]  08 Jun 2006, 22:25
Will go withC.

B looks appealing but has one fault.
Say y and z both be 5 that satisfies B y+z = 10
The equation becomes.......
|x-5| = |x-5|
With this x can be any number.
hence we cannot get the answer with B.
But if we know that y not= z
we will get eh answer
Senior Manager
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[#permalink]  09 Jun 2006, 07:21
I think itâ€™s A.
Stmt 1 tells us that y≠z
Stem tells us that y=z.
Therefore, for the stem to be true, X must equal zero. SUFFICIENT

Stmt 2 tells us that
Y=5
Z=5
Because the stem says y=z, and they must sum to 10.
However, X can be anything for the stem to be true. INSUFFICIENT

Manager
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[#permalink]  09 Jun 2006, 10:49
Another one for C.

We must have both St. 1 and 2 to determine that X = 5. With only St. 2, y could equal Z (say both equals 5) and x could be any number.
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Re: DS: Absolute values [#permalink]  10 Jun 2006, 17:32
Professor wrote:
What is the value of x given that |x - y| = |x - z|

(1). y is not equal to z
(2). The sum of y and z is 10.

|x - y| = |x - z| means
x-y=+/-(x-z)
If x-y=x-z then y=z, and x can be anything
If x-y=-(x-z) then 2x=y+z

(1) y<>z we know then 2x=y+z. However we don't know the value of y+z therefore value of x is not determinable.
(2) y+z=10, we don't know if x-y=x-z or -(x-z) so the value of x is indeterminable.

Combined it's obvious x=5. Therefore C.

Something that is worth of notice is that we often have what's given in (1) in mind when checking (2) and thought that (2) itself is sufficient without realizing that we are actually assuming (1). I've made this mistake before. Be careful with this when you do your test.
_________________

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keep on seeking, and you will find;
keep on knocking, and it will be opened to you.

VP
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[#permalink]  10 Jun 2006, 22:34
No OA for this one too. C looks good as explained by Honghu and others.

but honghu whats wrong with the approach used by BG.

BG wrote:
If both sides are squared then absolute value sign is eliminated and stem is 2XZ-2XY=Z^2-Y^2 => 2X*(Z-Y)=(Z-Y)*(Z+Y)or 2X=Y+Z

now A) is useless IMO

B) IS SUFF X=5
VP
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[#permalink]  10 Jun 2006, 23:07
Professor wrote:
No OA for this one too. C looks good as explained by Honghu and others.

but honghu whats wrong with the approach used by BG.

BG wrote:
If both sides are squared then absolute value sign is eliminated and stem is 2XZ-2XY=Z^2-Y^2 => 2X*(Z-Y)=(Z-Y)*(Z+Y)or 2X=Y+Z

now A) is useless IMO

B) IS SUFF X=5

Prof,

In BG's approach, Z-Y cannot be cancelled from both sides unless we know that Z-Y <> 0. If Z-Y = 0, the expression becomes 0/0 or indeterminate one.
2XZ-2XY=Z^2-Y^2
=> 2X*(Z-Y)=(Z-Y)*(Z+Y)
=> (Z-Y) (2X-Z-Y) =0
=> Z-Y = 0 or X = (Z+Y)/2
1. Z <> Y, but we don't know the value of Z+Y
2. Z+Y = 10, but we don't know if Z<>Y

Combine. SUFF to say X = 10/2 = 5.
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- Bernard Edmonds

[#permalink] 10 Jun 2006, 23:07
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