st(1) y=-x-1 slope is -1 and we conclude that the point (x,y) may potentially be in any quadrant (except for I) Not Sufficient
st(2) xy>0
case a) x>0, y>0 => y=x slope +1 OR II quadrant
case b) x<0, y<0 => y=x slope -1 OR IV quadrant
Not Sufficient
combined st(1&2): When y is +ve, x is +ve too, BUT we have "-" sign in front of x turns it -ve, hence our(x,y) point is II quadrant. This is not possible as y must be negative at all times and (-ve)+(-ve)=(-ve) and not (+ve) as it's required by st(2) y>0. When y is negative, x is -ve too, BUT "-" sign in front of x turn it +ve, hence our (x,y) point is in III quadrant. Sufficient to answer (x,y) can be ONLY in III quadrant.
c
colakumarfanta wrote:Q. (x, y)lies in which quadrant?
(1) x + y =- 1 (2) xy> 0
OA C
Last edited by
pemdas on Fri Nov 11, 2011 5:27 am, edited 1 time in total.
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