cramya wrote:If X, Y, and Z are positive integers, is Y>X?

(1) Y^2=XZ
(2) Z-X>0

(1) we could pick y=2, x=1 and z=4 to get a "yes" answer or y=2, x=4 and z=1 to get a "no" answer.. insufficient.

(2) No info about y, insufficient.

Together:

From (2), we know that z>x. So:

y*y = (x)(more than x)

Since all the variables must be positive, the only way that equation can hold true is if y>x. Choose (C).

Another way we could look at it is to say that z = x + k, in which k is a positive integer. So:

y^2 = x(x+k)
y^2 = x^2 + xk

Since y^2 equals x^2 PLUS something else, y^2 must be greater than x^2, and since x and y are both positive, y > x.

Another way we could look at it is to say that z = x + k, in which k is a positive integer.

I wish I could think more like u in quant but that may never happen I was looking for different approaches and got one.

I did it like this:

z > x

y^ 2 =xz

z = y^2/x

y^2/x>x

y^2/x-x>0

y^2-x^2>0
y^2 > x^2

since both are positive y > x


Thanks again Stuart!!!

f X, Y, and Z are positive integers, is Y>X?

(1) Y^2=XZ
(2) Z-X>0


From one Y is a geometric mean of either XYZ or ZYX

2. Says Z> X
Hence

That GP will be XYZ

So IMO C

Thanks Stuart for the amazing approach. As Cramya mentions. I wish I could think like you.

thanks
Vittal

From one Y is a geometric mean of either XYZ or ZYX

2. Says Z> X
Hence

That GP will be XYZ

So IMO C

Ronnie thanks so much for a very detailed solution. Much appreciated

Love ur expanded IMO's.....

cramya wrote:

From one Y is a geometric mean of either XYZ or ZYX

2. Says Z> X
Hence

That GP will be XYZ

So IMO C

Ronnie thanks so much for a very detailed solution. Much appreciated

Love ur expanded IMO's.....