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Is xy > 1 ?

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Bunuel
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Is xy > 1 ? [#permalink] New post   21 Mar 2017, 05:39
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E

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57% (01:47) correct 43% (01:45) wrong based on 76 sessions
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Is xy > 1 ?

(1) x^2 + y^2 = 5
(2) x + y > 3
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C
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Diwakar003
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Is xy > 1 ? [#permalink] New post   21 Mar 2017, 07:00
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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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stne
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Re: Is xy > 1 ? [#permalink] New post   16 Jul 2018, 03:43
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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.
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SaahithiV
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Re: Is xy > 1 ? [#permalink] New post   02 Jul 2019, 23:30
Bunuel wrote:
Is xy > 1 ?

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


For 1st option, we do nt know whether x and are positive or negative as they are squared.
For the second option, it can be two cases, both positive-- 1+3> 3 or 10-5>3
Combine both we get,
(x+y)^2> 3^2
x^2+y^2+2xy >9
5+2xy>9
2xy>4
xy>2------- sufficient
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sdilley
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Re: Is xy > 1 ? [#permalink] New post   02 Jul 2019, 23:44
SaahithiV wrote:
Bunuel wrote:
Is xy > 1 ?

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


For 1st option, we do nt know whether x and are positive or negative as they are squared.
For the second option, it can be two cases, both positive-- 1+3> 3 or 10-5>3
Combine both we get,
(x+y)^2> 3^2
x^2+y^2+2xy >9
5+2xy>9
2xy>4
xy>2------- sufficient



I’m still a bit confused how you did the combination here. Can you help walk me through that?

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Is xy > 1 ? [#permalink] New post   03 Jul 2019, 00:41
Is xy > 1? Yes/No
Now no constraints on x and y so they can be non-integers

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

x^2+y^2+2xy = 5+2xy
(x+ y)^2 = 5+2xy
(x+y)^2-5)/2 =xy ,we don’t know what (x+y) is so (Not sufficient)

(2) x+ y > 3

when x=5 ,y=-1 then xy<1 (Nope!)
When x=5 ,y=1 then xy > 1 (yes)
(Not sufficient)

(St 1+ 2) xy= ((x+y)^2-5)/2
and we know (x+y) from (st2) is greater than 3 and can be 3.1, 3.01, 4 , etc.
Now let’s pick 3.01 and other number will still satisfy
((3.01)^2-5)/2 = xy
2.03 = xy , and (2.03) > 1 ,
so YES xy>1 (sufficient)

Answer C
Hope it helps :)
sdilley wrote:
SaahithiV wrote:
Bunuel wrote:
Is xy > 1 ?

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


For 1st option, we do nt know whether x and are positive or negative as they are squared.
For the second option, it can be two cases, both positive-- 1+3> 3 or 10-5>3
Combine both we get,
(x+y)^2> 3^2
x^2+y^2+2xy >9
5+2xy>9
2xy>4
xy>2------- sufficient



I’m still a bit confused how you did the combination here. Can you help walk me through that?

Posted from my mobile device
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kitipriyanka
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Re: Is xy > 1 ? [#permalink] New post   03 Jul 2019, 02:43
Is xy > 1 ?

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


We know, (x+y)^2=x^2+y^2 + 2xy
1. (x+y)^2 = 5+2xy
We don't know what is the value of x+y, hence we can't say anything about xy.
Therefore, Insufficient

2. x+y>3
Insufficient

Together, In 1 we know by 2 that x+y>3, so its square is >9, so xy >1 always. Sufficient
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SaahithiV
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Re: Is xy > 1 ? [#permalink] New post   03 Jul 2019, 02:57
sdilley wrote:
SaahithiV wrote:
Bunuel wrote:
Is xy > 1 ?

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


For 1st option, we do nt know whether x and are positive or negative as they are squared.
For the second option, it can be two cases, both positive-- 1+3> 3 or 10-5>3
Combine both we get,
(x+y)^2> 3^2
x^2+y^2+2xy >9
5+2xy>9
2xy>4
xy>2------- sufficient



I’m still a bit confused how you did the combination here. Can you help walk me through that?

Posted from my mobile device


Taking the second option, I squared both sides -- x+y>3
So (x+y)^2> 3^2
x^2+y^2+2xy >9
Now x^2 +y^2 is given in the first option as 5
Substituting the value we get,
5+2xy > 9
SUbtract 5 from both sides
we get, 2xy> 4
Divide both sides by 2
we get xy>2
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Re: Is xy > 1 ? [#permalink]
03 Jul 2019, 02:57
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Bunuel
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