Just pick numbers (or use number line analogy)with these cases:
case1: x-> positive y->positive
case 2: x->positive y->negative
case 3: x->negative y->negative
case 4: x->negative y->positive
See if u can prove insufficiency. usually case 1 and case 2 would tell u the same thing as case 3 and case 4
Stmt I
y<x
y=2 x=3
|x-y| > |x|- |y| -> no
y= -2 x=3
|x-y| > |x|- |y| -> yes
INSUFF
Stmt II
xy <0
x-> positive y->negative or x->negative and y positive
Always |x-y| > |x| - |y|
SUFF
Choose [spoiler]B[/spoiler]
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