Avh86 wrote:
AshForGMAT wrote:
B it is ...
by squaring 'a' you are making it +ve ....
What if a=sqrt(-1)?
a^2 will then be negative. I've seen another gmat question where it's tricky like that.
Posted from GMAT ToolKitWelcome to GMAT Club. Below is an answer to your question.
First of all, the GMAT is dealing only with Real Numbers, so
even roots from negative numbers are not defined. So, \(a^{even}\geq{0}\).
Is x>y?Question: is \(x-y> 0\)
(1) \(ax>ay\) --> \(a(x-y)>0\), two cases:
A. \(a>0\) and \(x-y>0\);
OR
B. \(a<0\) and \(x-y<0\);
Depending on \(a\), \(x-y\) may or may not be greater than zero. Not sufficient.
(2) \(a^2x > a^2y\) --> \(a^2(x-y)>0\) --> since \(a^2>0\) (\(a^2\) can not be zero, since the product is more than zero) we can safely reduce by it: \(x-y>0\). Sufficient.
Answer: B.
Hope it's clear.