gmatnub wrote:

goalsnr wrote:

18. Is x > y ?

(1) ax > ay

(2) a^2x > a^2 y?

Pls explain

B

Because in statement 1, if a = a positive number, then x > y

But if a = a negative number, when you divide by a negative, you switch sign,

if a = negative, then x < y

GmatNub,can you please explain what is wrong about this approach

Stat 1:

(1) ax > ay

a(x-y) >0

a>0 or (x-y)>0

insuff

stat 2:

(2) a^2x > a^2 y?

a^2(x-y) >0

here too a^2 >0 or (x-y) >0

How can we assume a^2 >0 ?