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Is xy positive? (1) (x-y)^2 = 8 (2) (x+y)^2 = 10

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Bunuel
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Is xy positive? (1) (x-y)^2 = 8 (2) (x+y)^2 = 10 [#permalink] New post   26 May 2019, 04:22
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Is xy positive?

(1) (x-y)^2 = 8
(2) (x+y)^2 = 10
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C
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AkshdeepS
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Is xy positive? (1) (x-y)^2 = 8 (2) (x+y)^2 = 10 [#permalink] New post   Updated on: 26 May 2019, 10:01
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Bunuel wrote:
Is xy positive?

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


Question : Is xy > 0?

Is x & y > 0 Or Is x & y < 0 ?

St 1 : \(x - y = \sqrt{8}\)

x - y = 2.8 approx

If 2.8 - 0 = 2.8 , then xy = 0 --- Not positive

If 5.6 - 2.8 = 2.8, then xy > 0 ---- Positive

Not sufficient

St 2 :

The same way statement 2 can be checked and the answer would be ---- Not sufficient

From 1 and 2 we get,

x^2 + y^2 - 2xy = 8 ------------ (I)

x^2 + y^2 + 2xy = 10 ------------ (II)

Multiplying (II) with -1 and then add I & II,

-4xy = -2

xy = 2/4 = 1/2

Hence SUFFICIENT, IMO (C) ------ xy > 0

Originally posted by AkshdeepS on 26 May 2019, 05:04.
Last edited by AkshdeepS on 26 May 2019, 10:01, edited 3 times in total.
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Re: Is xy positive? (1) (x-y)^2 = 8 (2) (x+y)^2 = 10 [#permalink] New post   26 May 2019, 05:44
statement 1 alone mot sufficient


staement 2 alone also not sufficient



now adding 1 and 2 we get ,

2 (x^2+y^2 )= 18
x^2+y^2 = 9

putting this in 1 , we get

2xy = 1 xy= 1/2 >0


so IMO C

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Re: Is xy positive? (1) (x-y)^2 = 8 (2) (x+y)^2 = 10 [#permalink] New post   26 May 2019, 06:31
Bunuel wrote:
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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Is xy positive? (1) (x-y)^2 = 8 (2) (x+y)^2 = 10 [#permalink] New post   26 May 2019, 10:13
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Bunuel wrote:
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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Re: Is xy positive? (1) (x-y)^2 = 8 (2) (x+y)^2 = 10 [#permalink] New post   26 May 2019, 14:44
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Bunuel wrote:
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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Re: Is xy positive? (1) (x-y)^2 = 8 (2) (x+y)^2 = 10 [#permalink] New post   18 Sep 2023, 21:20
Bunuel wrote:
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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Re: Is xy positive? (1) (x-y)^2 = 8 (2) (x+y)^2 = 10 [#permalink] New post   18 Sep 2023, 23:09
Expert Reply
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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Re: Is xy positive? (1) (x-y)^2 = 8 (2) (x+y)^2 = 10 [#permalink]
18 Sep 2023, 23:09
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