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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

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

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. _________________

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

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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