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Is xy > 1 ?

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Math Expert
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V
Joined: 02 Sep 2009
Posts: 47983
Is xy > 1 ?  [#permalink]

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New post 21 Mar 2017, 05:39
00:00
A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

58% (01:12) correct 42% (00:59) wrong based on 33 sessions

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Manager
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S
Joined: 02 Aug 2015
Posts: 83
Is xy > 1 ?  [#permalink]

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New post 21 Mar 2017, 07:00
Bunuel wrote:
Is xy > 1 ?

(1) x^2 + y^2 = 5
(2) x + y > 3


Please correct me if I'm missing something.

(1) x^2 + y^2 = 5 - Equation of a circle with radius root(5). Since the circle passes through all 4 quadrants, xy will be negative if we choose a point on circumference of the circle from 2nd and 4th quadrant. xy will be positive if we choose a point on circumference from 1st and 3rd quadrant. - Not sufficient.

(2) x+y > 3. Can be any line passing through quadrant 1, 2, and 4. Again xy can be negative or positive. - Not sufficient.

(1) + (2) The line and the circle doesn't meet. So not sufficient.

Hence E.

Cheers!
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Re: Is xy > 1 ?  [#permalink]

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New post 16 Jul 2018, 03:43
Bunuel wrote:
Is xy > 1 ?

(1) x^2 + y^2 = 5
(2) x + y > 3


We can clearly see that statement 1 and 2 by themselves are insuff. After plugging some numbers I couldn't get that xy<1 hence choose C.

but can anyone explain how x+y can be greater than 3 and \(x^2 +y^2\) = 5
Please can some one give me values of X and Y which satisfy both the statements
as of now I can only come up with x= -2, or 2 and y = -1 or 1 and vice versa in both cases X+ Y \(\leq\) 3
I understand that X and Y need not be integers then lets take \(y^2\)= 5 and x= 0 ( for \(X^2\)+\(Y^2\)= 5) then y ~2.23 here also x+y < 3
so it is possible that \(x^2\) + \(y^2\)= 5 and x + y > 3

Thank you.
_________________

- Stne

Re: Is xy > 1 ? &nbs [#permalink] 16 Jul 2018, 03:43
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