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If x and y are integers, is 2xy<x2+y2?

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Math Expert
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If x and y are integers, is 2xy<x2+y2?  [#permalink]

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New post 29 Apr 2016, 02:38
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  65% (hard)

Question Stats:

54% (01:51) correct 46% (01:40) wrong based on 83 sessions

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Re: If x and y are integers, is 2xy<x2+y2?  [#permalink]

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New post 29 Apr 2016, 11:34
1
st 1: we know (x-y)^2 < 0

thus x^2+y^2 - 2xy<0
given xy<0
then x^2+y^2+2xy<0
x^2+y^2>2xy

Sufficient

St 2:x+y = 5
following can be values of x and y
x = 1, y = 4 or vice versa
for given expression: x^2+y^2 = 1+16 = 17
2xy = 2*1*4 = 8
thus x^2+y^2 > 2xy

another value can be :x = 6 y = -1
then x^2+y^2 =36+1 = 37
2xy = 2*-1*6 = -12
thus x^2+y^2 > 2xy
Sufficient

Ans: d
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Re: If x and y are integers, is 2xy<x2+y2?  [#permalink]

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New post 29 Apr 2016, 12:48
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Becoz (x-y)^2 >0 is possible only when x not equal to y. In 2 we can not say x not equal to y but in 1 we can say.

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Re: If x and y are integers, is 2xy<x2+y2?  [#permalink]

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New post 29 Apr 2016, 12:54
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subha107 wrote:
A
Becoz (x-y)^2 >0 is possible only when x not equal to y. In 2 we can not say x not equal to y but in 1 we can say.

Posted from my mobile device


in S2, if xy<0 ---> means that x and y have OPPOSITE signs and are NON ZERO. Whatever be the values of x,y, you will always have x \(\neq\) y

Hope this helps.
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If x and y are integers, is 2xy<x2+y2?  [#permalink]

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New post 27 Jun 2017, 11:32
I found this question nice so i would like to bring my two cents :)

First lets simplify the inequality: 2xy<x^2+y^2 ----> 0<x^2 +y^2-2xy ----> 0<(x-y)^2. So in fact the question asks whether (x-y) are not equal to 0?

statement 1: xy<0. This statement tells us that x and y have different signs. Thus (x-y) necessarily couldn't equal to 0 because positive number - negative number = positive number and negative number - positive number = negative number. Sufficient.

statement 2: x+y=5. (x-y) could be zero just if both x and y equal to each other. Since x and y are integers we can conclude that x and y are not equal to each other an thus that (x - y) couldn't be equal to 0. Sufficient.

Answer: D
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Re: If x and y are integers, is 2xy<x2+y2?  [#permalink]

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New post 06 Aug 2019, 10:27
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Re: If x and y are integers, is 2xy<x2+y2?   [#permalink] 06 Aug 2019, 10:27
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