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studentnow
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If z = x+2, y = x+1, then

St1:
|z-x| = |z-y| + |y-z|
|x+2-x| = |x+2-x-1| - |x+1-x-2|
|2| = |1| + |1|

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.

Ans C
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studentnow
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.
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studentnow
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?

- Brajesh
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studentnow
x y z are Three integers is x>y>z?


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


2) z>x


I get D.

Start w/ S2. S1 looks ugly.

z>x, well then x>y>z can't be true.


S1.
z-x=z-y+y-z---> z-x=0. z has to equal x.

-z+x=-z+y -y+z. -z+x=0---> x-z=0 x=z.
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[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
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Should be 'B'

stmt1: INSUFF
take x =5 , y =3, z =1 satisfies stmt1
x=-3 , y =3, z=1 also satisfies stmt1

Stmt2; suff
as Z>x, this means that x>y>z can never be true.
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I will go with B as well. What is the OA.
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Should be 'B'

stmt1: INSUFF
take x =5 , y =3, z =1 satisfies stmt1
x=-3 , y =3, z=1 also satisfies stmt1

Stmt2; suff
as Z>x, this means that x>y>z can never be true.



oops ya change to B from D
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b14kumar
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b14kumar
[quote="Fistail"][quote="studentnow"]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.[/quote]

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?

- Brajesh[/quote]

thats absolutely right.



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