Inequalities Problem

IMO C.

Separately:

1. X could be smaller or bigger - we dont know.

2. No Y in there, so we cant say anything either.

Together:

Solve the 1st equation for Z = Y^2/X.
Switch in the 2. Z-X>0 to Z > X.

Put the 1st equation into the second --> Z = Y^2/X>X.
--> Y^2>X^2. Therefore we know that Y > X. --> Solution C.

schakiiieee wrote:IMO C.

Separately:

1. X could be smaller or bigger - we dont know.

2. No Y in there, so we cant say anything either.

Together:

Solve the 1st equation for Z = Y^2/X.
Switch in the 2. Z-X=0 to Z = X.

Put the 1st equation into the second --> Z = Y^2/X=X.
--> Y^2=X^2. Therefore we know that Y and X are equal. Y is NOT bigger than X --> Solution C.

Only problem is in option B, Z-X>0 (and not equal to 0)
Therefore, Z>X and not equal to X

Hmm... true. So did I mess up things or is the answer E?


EDIT: (was a bit confused when i wrote this )

schakiiieee wrote:Hmm... true. So did I mess up things or is the answer E?

OA is C

gmat009 wrote:If X, Y, and Z are positive integers, is Y>X?
(1) Y^2=XZ
(2) Z-X>0

Another way to think about it:

(1) Y*Y = X*Z
as (2) states that Z>X we know that Y*Y < Z*Z. So X<Y enables Y*Y = X*Z.

x, y, and z are postivie INTEGERS is y >x?

1. Y^2= xz
For example, if y=3, y^2 =9. The only intergers that work for x and z are either 3x3 or 1x9 or 9x1. Therefore Insufficient

2. z-x>0
from this we know z>X but we have no information about Y so Insuffient

From 1+2 we know z is greater than x and to satisfy statement 1 the only values that work is x=1 and z=9. Therefore, Y>X (3>1) suffient