Q. x^2> y^2?
this can be rewritten as : x^2-y^2>0 => (x+y) (x-y)>0?? i.e do (x+y) and (x-y) have same signs.
Going to the stmts:
Stmt1: x<y => (x-y)<0; this does not tells us anything about the sign of x+y. Therefore insufficient.
stmt2: -y>x => (x+y)<0; this does not tells us anything about the sign of x-y. Therefore insufficient.
Combining, we know signs of both. Hence sufficient. Answer =C.
this can be rewritten as : x^2-y^2>0 => (x+y) (x-y)>0?? i.e do (x+y) and (x-y) have same signs.
Going to the stmts:
Stmt1: x<y => (x-y)<0; this does not tells us anything about the sign of x+y. Therefore insufficient.
stmt2: -y>x => (x+y)<0; this does not tells us anything about the sign of x-y. Therefore insufficient.
Combining, we know signs of both. Hence sufficient. Answer =C.
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