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If z = x-2, y = x-1, then
|z-x| = |z-y| + |y-z|
|x-2-x| = |x-2-x+1| + |x-1-x+2|
|-2| = |-1| + |-1|

So can be x > y > z, or x < y <z>x. Nothing else about y. x could be = y, or <y> y. Insufficient.

St1 & St2:
z = x+2, y = x+1, then |z-x| = |z-y| + |y-z| ( x < y < z)
Only x<y<z yields a solution that fit st1 and st2. The rest do not yield such a solution.

Re: x y z are Three integers is x>y>z? [#permalink]

Show Tags

16 Sep 2007, 22:17

Fistail wrote:

b14kumar wrote:

Fistail wrote:

studentnow wrote:

x y z are Three integers is x>y>z?

1) |z-x|=|z-y|+|y-z| 2) z>x

should be B. if z > x, then definitely x>y>z.

How?

If Z > X then how would X > Y >Z ?

- Brajesh

if z>x, then x cannot be grater than z. so suff...

do not need st. 1.

You said:

if z > x, then definitely x>y>z

I thought you were saying that if z > x then x>y>z would be true.
However, I think you meant to say that x>y>z would not be true....ST2 is sufficient to answer the question.....That is what you meant...Right?

Re: x y z are Three integers is x>y>z? [#permalink]

Show Tags

17 Sep 2007, 03:08

[quote="studentnow"]x y z are Three integers is x>y>z?

1) |z-x|=|z-y|+|y-z|

2) z>x[/quote]

I think it is E

1) |z-x|=|z-y|+|y-z| => |z-x|=|z-y|+|z-y|=2|z-y|=>there are many combinations=>insufficient
2) z>x=>insufficient
1&2) z-x=2|z-y|=> a) z-x=2z-2y=>x=2y-z or b) x-z=2z-2y=>x=z-2y we have at least two combinations=> so insufficient

Changed my opinion to B

Last edited by Vlad77 on 17 Sep 2007, 04:50, edited 1 time in total.

Re: x y z are Three integers is x>y>z? [#permalink]

Show Tags

17 Sep 2007, 10:43

b14kumar wrote:

Fistail wrote:

b14kumar wrote:

Fistail wrote:

studentnow wrote:

x y z are Three integers is x>y>z?

1) |z-x|=|z-y|+|y-z| 2) z>x

should be B. if z > x, then definitely x>y>z.

How?

If Z > X then how would X > Y >Z ?

- Brajesh

if z>x, then x cannot be grater than z. so suff...

do not need st. 1.

You said:

if z > x, then definitely x>y>z

I thought you were saying that if z > x then x>y>z would be true. However, I think you meant to say that x>y>z would not be true....ST2 is sufficient to answer the question.....That is what you meant...Right?