I use to use this question in my drill sets...lol

Statement (1): Consider x and y have the same signs

x = 2, y= 2
4 + 4 + 4 +4 > 0 True

Consider x is neg y is pos

x = -2 y = 2

4 + (-4) + 4 + (-4) >0 is false

consider x is pos y is neg

x = 2 y = -2

4 + 4 + (-4) + (-4) > 0 is false

consider both x and y are negative

x = -2 y = -2

4 + (-4) + (-4) + 4 > 0 is false


Therefore, that statement is only true when x and y are positive. That means that x and y lie in Quandrant 1. Statement(1) sufficient.

Do the same thing with statement (2)

x = 2 y = 1

-2 < -1 < 1 is true

x = -2 y =1

- (-2) < -1 < 1 is false

x = 2 y - -1

-2 < 1< 1 is false


x = -2 y =-1

- (-2) < -(-1) < 1 is false

Once again the only combination that worked was when x and y were both positive. Therefore choose D

IMO D

s1) only for x>0 and y >0 the statement holds true -----> x,y are in 1st quardant
suff
s2)-X<-Y<+OR-Y
--->-X<-Y<Y OR -X<-Y<-Y

-X<- Y<Y--->X>Y>-Y---->ADDING Y TO THE INEQ---->X+Y>2Y>0--->X>Y>0---->1ST QDT

-X<-Y<-Y----->X>Y>Y---->X-Y>0----->X>Y SINCE Y>0 PROVED ABOVE ----> 1ST QDT
SUFF


oa PLS

thephoenix wrote:IMO D

s1) only for x>0 and y >0 the statement holds true -----> x,y are in 1st quardant
suff
s2)-X<-Y<+OR-Y
--->-X<-Y<Y OR -X<-Y<-Y

-X<- Y<Y--->X>Y>-Y---->ADDING Y TO THE INEQ---->X+Y>2Y>0--->X>Y>0---->1ST QDT

-X<-Y<-Y----->X>Y>Y---->X-Y>0----->X>Y SINCE Y>0 PROVED ABOVE ----> 1ST QDT
SUFF


oa PLS

Its D. Its from MGMAT CAT 2

OA : D
Its from MGMAT CAT 5

sumanr84 wrote:OA : D
Its from MGMAT CAT 5

Oh wow...I guess there is no order, I haven't even taken CAT 5...lol