Data Sufficiency

1. x-3y>0
2. 4y-3x>0

1) x>3y -> plug values, x=4,y=1, x+y>0 or x=1,y=-1/3 x+y=0 so 1 is not suff
2) x<4y/3 ->plug valuesx=y=1, x+y>0 or x=-1,y=1, x+y=0 not suff
taken togetehr, we get
3y<x<4/3y
the above is true only when y is negative (y cannot be equal to 0 as it implies xis both greater than and less than zero-not possible)
so y is negative, x has to be negative too
when x and y both are negative, x+y cannot be > 0
so C is the answer.

ineuqalities in DS are very tricky, i hope im not missing anythgn here.

neerajkumar1_1 wrote:1. x-3y>0
2. 4y-3x>0

1)x-3y>0; now consider x=10 y=3; 10-3*3>0; and also x+y>0; now consider x=-5 y=-10; we have x-3y>0; but x+y<0; also x=10; y=-2 we have x-3y>0; also x+y>0 put x=1, y=-2; we have x-3y>0 but x+y<0; hence 1 alone is not sufficient to answer the question..!!

2) 4y-3x>0; now consider x=1; y=2; 4y-3x>0 ans also x+y>0; now consider x=-3 , y=1; 4y-3x>0; but x+y<0; x=-4, y=-1;we have 4y-3x>0 but x+y<0
hence 2 alone is not sufficient to answer the question..!!!

combining 1 and 2;
multiply x-3y>0 with 3; we have 3x-9y>0; also 4y-3x>0; we know that when we add two positive quantities together their sum would always be positive for example 4>0; 5>0; also 4+5>0;

now add two inequalities, we have 3x-9y+4y-3x>0; -5y>0; i.e. y<0;

and the only case possible that satisfy both the condition is x<0; y<0;
therefore x+y<0;
hence C