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# Is xy > 1 ?

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Math Expert
Joined: 02 Sep 2009
Posts: 50627
Is xy > 1 ?  [#permalink]

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21 Mar 2017, 04:39
00:00

Difficulty:

45% (medium)

Question Stats:

63% (01:46) correct 38% (01:32) wrong based on 34 sessions

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Is xy > 1 ?

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

_________________
Manager
Joined: 02 Aug 2015
Posts: 115
Is xy > 1 ?  [#permalink]

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21 Mar 2017, 06:00
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!
Director
Joined: 27 May 2012
Posts: 611
Re: Is xy > 1 ?  [#permalink]

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16 Jul 2018, 02:43
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, 02:43
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