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# Is |x + y| > |x - y| ?

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Intern
Joined: 06 Feb 2013
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Is |x + y| > |x - y| ? [#permalink]

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25 Sep 2013, 14:07
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Question Stats:

62% (01:29) correct 38% (01:36) wrong based on 238 sessions

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Is |x + y| > |x - y| ?

(1) |x| > |y|
(2) |x - y| < |x|
[Reveal] Spoiler: OA

Last edited by Bunuel on 25 Sep 2013, 14:13, edited 1 time in total.
Renamed the topic and edited the question.

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Re: Is |x + y| > |x - y| ? [#permalink]

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25 Sep 2013, 14:18
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Is |x + y| > |x - y| ?

Is $$|x+y| > |x-y|$$? --> square it: is $$x^2+2xy+y^2>x^2-2xy+y^2$$ --> is $$xy>0$$?

(1) |x| > |y|. We cannot get from this whether x and y have the same sign. Not sufficient.

(2) |x - y| < |x| --> square again: $$x^2-2xy+y^2<x^2$$ --> $$xy>\frac{y^2}{2}$$ --> since $$y^2\geq{0}$$ (the square of any number is more than or equal to 0), then we have that $$xy>\frac{y^2}{2}\geq{0}$$. Sufficient.

Hope it's clear.
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Re: Is |x + y| > |x - y| ? [#permalink]

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26 Sep 2013, 09:16
Bunuel wrote:
Is |x + y| > |x - y| ?

Is $$|x+y| > |x-y|$$? --> square it: is $$x^2+2xy+y^2>x^2-2xy+y^2$$ --> is $$xy>0$$?

(1) |x| > |y|. We cannot get from this whether x and y have the same sign. Not sufficient.

(2) |x - y| < |x| --> square again: $$x^2-2xy+y^2<x^2$$ --> $$xy>\frac{y^2}{2}$$ --> since $$y^2\geq{0}$$ (the square of any number is more than or equal to 0), then we have that $$xy>\frac{y^2}{2}\geq{0}$$. Sufficient.

Hope it's clear.

Sorry Bunuel,

but i don't understand the concept behind the solution...

I understand that an absolute value is always positive. But how can you just square it¿? It supposed to be the root square of the square, isn't it¿??

Kudos [?]: 13 [0], given: 6

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Re: Is |x + y| > |x - y| ? [#permalink]

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26 Sep 2013, 09:19
jacg20 wrote:
Bunuel wrote:
Is |x + y| > |x - y| ?

Is $$|x+y| > |x-y|$$? --> square it: is $$x^2+2xy+y^2>x^2-2xy+y^2$$ --> is $$xy>0$$?

(1) |x| > |y|. We cannot get from this whether x and y have the same sign. Not sufficient.

(2) |x - y| < |x| --> square again: $$x^2-2xy+y^2<x^2$$ --> $$xy>\frac{y^2}{2}$$ --> since $$y^2\geq{0}$$ (the square of any number is more than or equal to 0), then we have that $$xy>\frac{y^2}{2}\geq{0}$$. Sufficient.

Hope it's clear.

Sorry Bunuel,

but i don't understand the concept behind the solution...

I understand that an absolute value is always positive. But how can you just square it¿? It supposed to be the root square of the square, isn't it¿??

If both sides of an inequality are non negative, then we can square it.

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Re: Is |x + y| > |x - y| ? [#permalink]

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09 May 2014, 15:25
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I solved in the following way

First square both sides

We get 'Is xy>0" , or in proper english do 'x' and 'y' have the same sign?

Statement 1: Obviously insufficient
Statement 2: Square both sides again ---> x^2 > x^2 - 2xy + y^2

y^2 - 2xy < 0

y (y-2x)<0

So two cases:

If y>0, then y-2x<0 and 2x>y>0. Answer is YES
If y<0, then y-2x>0 and 2x<y<0. Answer is YES again

Sufficient

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Re: Is |x + y| > |x - y| ? [#permalink]

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04 Nov 2014, 11:05
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Basically the question is asking if x and y have the same sign.

from statement 1: no info on signs.---NSF

for statement 2: I did the following:

The distance between x and y on the number line can be less than the distance of 0 and x only when x and y have the same sign.

---------y----x---------0--------------
or
----------0---------x---y-------------

if
------x-------0--------y---- or ---------y--------0------------x----------
then the statement-2 is not true.

Therefore x and y should have the same sign.

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Re: Is |x + y| > |x - y| ? [#permalink]

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27 Jun 2016, 04:33
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Re: Is |x + y| > |x - y| ?   [#permalink] 27 Jun 2016, 04:33
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