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Bunuel
Is xy positive?

(1) (x-y)^2 = 8
(2) (x+y)^2 = 10

From both statement 1 & 2 independently we can't deduce the sign of x and y.

Let's take 1+2 together.

(x-y)^2 - (x+y)^2 = 64-100=-36
(2x)(-2y) = -36
xy= 9. C
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Bunuel
Is xy positive?

(1) (x-y)^2 = 8
(2) (x+y)^2 = 10

#1
(x-y)=√8; 2√2
so
x=6√2 and y=4√2
or x=√2 and y=-√2
insufficient
#2
x+y=√10
x=2√10 & y = -√10
or x=0 & y=√10
insufficient
from 1 & 1
open brackets
we get
x^2+y^2-2xy=8
and x^2+y^2+2xy=10
subtract
-4xy=(-)2
xy=1/2
IMO C
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Bunuel
Is xy positive?

(1) (x-y)^2 = 8
(2) (x+y)^2 = 10



Statement 1 : x - y = underoot 8 . This value could either be positive OR negative . Therefore , this statement is insufficient.

Statement 2 : x + y = underoot 10. This value could either be positive OR negative . Therefore , this statement is insufficient.

Combined , Now either we can solve the entire thing or you can see that since we have 2 variables and 2 equations in which x square and y square can be eliminated ,they can be solved for the value of xy. Hence ANSWER : C
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Bunuel
Is xy positive?

(1) (x-y)^2 = 8
(2) (x+y)^2 = 10

st 1 (x2-2xy+y2) = 8
XY can be positive or negative

st 2 x2+2xy+y2 = 10
XY can be positive or negative
subtract 1 and 2
-4xy = -2
xy = 1/2
ans C
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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 and xy<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 and xy<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
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