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# Is x > 0? (1) (x + y)^2 < (x - y)^2 (2) x + y < x - y

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Intern
Joined: 05 Jul 2016
Posts: 16
Location: Brazil
Concentration: Finance, Entrepreneurship
WE: Analyst (Investment Banking)
Is x > 0? (1) (x + y)^2 < (x - y)^2 (2) x + y < x - y [#permalink]

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29 Aug 2017, 16:37
00:00

Difficulty:

45% (medium)

Question Stats:

62% (00:57) correct 38% (01:24) wrong based on 79 sessions

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Is x > 0?

(1) (x + y)^2 < (x - y)^2
(2) x + y < x - y
[Reveal] Spoiler: OA
Math Expert
Joined: 02 Aug 2009
Posts: 5660
Re: Is x > 0? (1) (x + y)^2 < (x - y)^2 (2) x + y < x - y [#permalink]

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29 Aug 2017, 18:51
4
KUDOS
Expert's post
guireif wrote:
Is x > 0?
(1) (x + y)^2 < (x - y)^2
(2) x + y < x - y

Hi...
70% going wrong as per timer, so this may help few..
Open the equation, it may give you something to work on.

1) $$(x+y)^2<(x-y)^2$$..
$$x^2+y^2+2xy<x^2+y^2-2xy......4xy<0....xy<0$$..
So we can say, none of the two is 0 and x and y are of OPPOSITE sign
Insufficient

2)$$x+y<x-y$$
2y<0......y<0
Insufficient

Combined..
Both x and y are of OPPOSITE sign and y<0..
So x >0
Sufficient

C
_________________

Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

BANGALORE/-

Intern
Joined: 05 Jul 2016
Posts: 16
Location: Brazil
Concentration: Finance, Entrepreneurship
WE: Analyst (Investment Banking)
Re: Is x > 0? (1) (x + y)^2 < (x - y)^2 (2) x + y < x - y [#permalink]

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29 Aug 2017, 19:04
chetan2u wrote:
guireif wrote:
Is x > 0?
(1) (x + y)^2 < (x - y)^2
(2) x + y < x - y

Hi...
70% going wrong as per timer, so this may help few..
Open the equation, it may give you something to work on.

1) $$(x+y)^2<(x-y)^2$$..
$$x^2+y^2+2xy<x^2+y^2-2xy......4xy<0....xy<0$$..
So we can say, none of the two is 0 and x and y are of OPPOSITE sign
Insufficient

2)$$x+y<x-y$$
2y<0......y<0
Insufficient

Combined..
Both x and y are of OPPOSITE sign and y<0..
So x >0
Sufficient

C

Thanks. I just don't get why in this question you can simplify the inequalities to something like xy<0 or 2y<0. Can I always simplify inequalities that way?
Math Expert
Joined: 02 Aug 2009
Posts: 5660
Re: Is x > 0? (1) (x + y)^2 < (x - y)^2 (2) x + y < x - y [#permalink]

### Show Tags

29 Aug 2017, 19:11
1
KUDOS
Expert's post
1
This post was
BOOKMARKED
guireif wrote:
chetan2u wrote:
guireif wrote:
Is x > 0?
(1) (x + y)^2 < (x - y)^2
(2) x + y < x - y

Hi...
70% going wrong as per timer, so this may help few..
Open the equation, it may give you something to work on.

1) $$(x+y)^2<(x-y)^2$$..
$$x^2+y^2+2xy<x^2+y^2-2xy......4xy<0....xy<0$$..
So we can say, none of the two is 0 and x and y are of OPPOSITE sign
Insufficient

2)$$x+y<x-y$$
2y<0......y<0
Insufficient

Combined..
Both x and y are of OPPOSITE sign and y<0..
So x >0
Sufficient

C

Thanks. I just don't get why in this question you can simplify the inequalities to something like xy<0 or 2y<0. Can I always simplify inequalities that way?

Yes, you can cancel out 4 in 4xy<0 without changing the inequalities because 4 is POSITIVE..
1) If say I knew that x is POSITIVE, I can cancel out 4x in 4xy<0 and get y<0..
2) if say x is NEGATIVE, then you can cancel out the 4x but change the INEQUALITY sign.
That is 4xy<0...x<0 so y>0..

But if you don't know the signs, don't touch the INEQUALITY sign

Hope it helps
_________________

Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

BANGALORE/-

Re: Is x > 0? (1) (x + y)^2 < (x - y)^2 (2) x + y < x - y   [#permalink] 29 Aug 2017, 19:11
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