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Bunuel
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I don't think that \(a\) can be a negative integer.

a = |x| absolute value of x is always positive

a < 1 means tht a can only be a fraction < 1.

Please check again.
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vitaliyGMAT
I don't think that \(a\) can be a negative integer.

a = |x| absolute value of x is always positive

a < 1 means tht a can only be a fraction < 1.

Please check again.

ya , you are correct.

My bad :(
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From 1

A is +ve , y is -ve and x is non zero
Question becomes is
Ax-y> 0

Since we know nothing about x except it ain't zero

insuff

From 2

Clearly insuff

Both


C - edited from E.





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Bunuel
If y is not equal to zero, is ax > y?

(1) a = -y = |x|
(2) a < 1


Hi
We are looking at ax>y..


(1) a = -y = |x|
This tells us the value of each discarding SIGN is same..
a is +, y is - and x can be any thing..

Substitute y=-a in equation ax>y....
So ax>-a.......ax+a>0.......a(1+x)>0...
We know a is positive, so if 1+x is positive ans is YES otherwise NO..
SIGN or value of a/y/x is not known ..
Insuff

(2) a < 1
Nothing about x and y ..
Clearly insuff

Let's see combined..
Is a(1+x)>0....
The answer depends on value of x..
Now a is between 0 and 1..
So x will be between -1 and 1, both exclusive and 0 not included..
Now 1+x will always be POSITIVE, when x is between -1 & 1..
So answer is YES
Suff

C
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