DS again!!

1. x-y>-2
=> x>y-2
For y=1 we get x=-1 thus xy<0
For y=3 we get x=1 thus xy>0

Insufficient

2. x-2y<-6
=> x<2y-6
For y=1 we get x=-4 thus xy<0
For y=4 we get x=2 thus xy>0

Insufficient

Combining 1 and 2 gives

x>y-2 and x<2y-6
For y=1 we get -1<x and x<-4 thus for -ve value of x, xy<0
For y=4 we get x>2 and x<2 thus for +ve value of x, xy>0

Hence the solution is E.
Please correct me if i am wrong!
Thank you.

gmatters2vj wrote:1. x-y>-2
=> x>y-2
For y=1 we get x=-1 thus xy<0
For y=3 we get x=1 thus xy>0

Insufficient

2. x-2y<-6
=> x<2y-6
For y=1 we get x=-4 thus xy<0
For y=4 we get x=2 thus xy>0

Insufficient

Combining 1 and 2 gives

x>y-2 and x<2y-6
For y=1 we get -1<x and x<-4 thus for -ve value of x, xy<0
For y=4 we get x>2 and x<2 thus for +ve value of x, xy>0

Hence the solution is E.
Please correct me if i am wrong!
Thank you.

But the official answer is C..that is both the statements together are sufficient..

Sorry Warlock,

I did a mistake while combining the two conditions.
When we combine the 2 conditions we get
y-2<x<2y-6
Please note that it can only be true when y>4

Also for every value of y>4, x>2
Thus xy>0 always.

So C should be the answer.

Apologize for the mistake.

thanks gmatters2vj..nice logic..