xy <> 0, x>1?
xy <> 0 implies x can be -ve or +ve, and y can be -ve or +ve.
st(1) x>y, Obviously not sufficient as x=-2 and y=-3, x>y BUT x=3 and y=2 again x>y
st(2) x>-y, This is Not Sufficient, as y can be any number -ve or +ve, y=-2 OR y=2 then x=3 or -3 --> x>-y, 3>2 OR -3<-2, -3<2
Combined st(1&2): as x is greater of both +ve and -ve y we can write down x>|y| which means that y is always +ve and x>y And if y>0 y could be 1/2 or 1/100 still x>y could be less than 1 BUT when y=1 or 2 x>y, x>1 Not Sufficient

IOM E

mariah wrote:1. xy not equal to 0 is x >1?

1. x>y
2. x>-y

For the second drill the answer would be C, as x>0 is our condition given with the combined statements (1&2)

mariah wrote: 2. xy not equal to 0 is x>0?

1. x>y
2. x>-y

xy!=0 => x!=0, y!=0 is x>1

1. x>y , since only thing we knew is neither x/y =0, any possibility there, x=1, y=2, or x=0, y=-1
2. x>-y => x+y>0, still any possible soln, as x=1/2, y=1/3 or x=2, y=1

combine: only thing we knew is x>y and x+y>0 so x,y both are greater than 0, but still it doesn't mean x>1, as x=1, y=1/2, x=2, y=1.

E.

Using the above observationyes x>0, so C for 2.

mariah wrote:1. xy not equal to 0 is x >1?

1. x>y
2. x>-y

2. xy not equal to 0 is x>0?

1. x>y
2. x>-y

Did you mean that each statement consists of 2 further statements, or are there two questions in your post?

xy not equal to 0 is x > 0?

1. x>y
2. x>-y

I dont think alone these statements can stand so please opt out A,B,D
1) We know x,y are not equal to zero
2) Combined the two options we have x > |y|
---- we dont know the range of y to determin x... solution is x > 0 if y > 0 else x < 0...

IMO E ...what berserk problem???

HSPA wrote:xy not equal to 0 is x > 0?

1. x>y
2. x>-y

I dont think alone these statements can stand so please opt out A,B,D
1) We know x,y are not equal to zero
2) Combined the two options we have x > |y|
---- we dont know the range of y to determin x... solution is x > 0 if y > 0 else x < 0...

IMO E ...what berserk problem???

an abosulte value is always non-negative. If x is greater than an absoulte value, then x cannot be negative regardless of the value of y: even if y is negative (e.g. y=-5), then x is greater than |-5|=5, so x is positive.