Is xy positive?
Basically, the question is whether x and y have same sign.
(1) (x-y)^2 = 8
As we have square, we cannot say anything about sign.
\(x=\sqrt{2},y=-\sqrt{2}\) will satisfy equation andxy<0
\(x=3\sqrt{2},y=\sqrt{2}\) will satisfy equation and xy>0
Insufficient
(2) (x+y)^2 = 10
As we have square, we cannot say anything about sign.
\(x=\sqrt{10},y=-2\sqrt{10}\) will satisfy equation andxy<0
\(x=\sqrt{10}/2,y=\sqrt{10}/2\) will satisfy equation and xy>0
Insufficient
Combined
\((x-y)^2=8……x^2+y^2-2xy=8\)…(1)
\((x+y)^2=8……x^2+y^2+2xy=10\) ..(2)
Subtract (1) from (2) =>\(4xy=2…..xy=\frac{1}{2}\)
Sufficient
C
...
Basically, the question is whether x and y have same sign.
(1) (x-y)^2 = 8
As we have square, we cannot say anything about sign.
\(x=\sqrt{2},y=-\sqrt{2}\) will satisfy equation andxy<0
\(x=3\sqrt{2},y=\sqrt{2}\) will satisfy equation and xy>0
Insufficient
(2) (x+y)^2 = 10
As we have square, we cannot say anything about sign.
\(x=\sqrt{10},y=-2\sqrt{10}\) will satisfy equation andxy<0
\(x=\sqrt{10}/2,y=\sqrt{10}/2\) will satisfy equation and xy>0
Insufficient
Combined
\((x-y)^2=8……x^2+y^2-2xy=8\)…(1)
\((x+y)^2=8……x^2+y^2+2xy=10\) ..(2)
Subtract (1) from (2) =>\(4xy=2…..xy=\frac{1}{2}\)
Sufficient
C
...
Statistics : Posted by chetan2u • on 26 May 2019, 04:22 • Replies 7 • Views 3678



