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Is x < y?

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Is x < y?  [#permalink]

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New post Updated on: 01 Nov 2013, 01:01
1
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A
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C
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Is x < y?

(1) x^2 - y^2 < 0
(2) -x - y < 0

Originally posted by Puneethrao on 31 Oct 2013, 20:02.
Last edited by Bunuel on 01 Nov 2013, 01:01, edited 2 times in total.
Edited the question.
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Re: Is x < y?  [#permalink]

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New post 31 Oct 2013, 22:15
2
Puneethrao wrote:
Is x < y?

x2 - y2 < 0
-x - y <0


St.1- can be rearranged as x^2 If x=1, y=2, the answer would be YES.
If x=1, y=-2, Its a NO. So 1 is insufficient

St.2- Again it can be (2, -1) or (-1, 2) or such values. Insufficient

(1)+(2)--> (x^2-y^2) and -(x+y) are both negative so their division must be positive.

(x^2-y^2)/-(x+y)> 0
(x+y)(x-y)/-(x+y)> 0
-(x-y)> 0
x-y <0
So x Therefore (1)+(2) is sufficient.

Answer is C :)

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Re: Is x < y?  [#permalink]

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New post 01 Nov 2013, 01:01
Puneethrao wrote:
Is x < y?

(1) x^2 - y^2 < 0
(2) -x - y <0


Is x < y?

Is x < y? --> is x - y < 0?

(1) x^2 - y^2 < 0 --> x^2<y^2 --> |x|<|y|. Not sufficient. Consider: (x, y)=(1, -2) and (x, y)=(1, 2).

(2) -x - y < 0 --> x+y>0. The sum of two numbers is greater than 0, from this we cannot say which one is greater. Consider (x, y)=(1, 2) and (x, y)=(2, 1). Not sufficient.

(1)+(2) From (1) we have that (x-y)(x+y)<0 and from (2) we have that x+y>0, thus x-y<0. Sufficient.

Answer: C.
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Re: Is x < y?  [#permalink]

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New post 18 Apr 2014, 23:14
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x^2 - y^2 < 0 --> x^2>y^2

is that a typo or did you mean that?
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Re: Is x < y?  [#permalink]

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New post 19 Apr 2014, 03:25
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Re: Is x < y?  [#permalink]

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New post 23 Sep 2014, 07:42
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This question is also very easily solved algebraically.

1) \(x^2 - y^2 <0. (x+y)(x-y) <0\). One of these is negative, so either\(x+y<0\), which means that \(x<-y\), or \(x<y\). Two options; not sufficient.

2) \(-x-y<0\). Not sufficient.

Using number 2), we can multiply by a negative to see that \(x+y>0\). Now using option 1, this means that x-y must be less than 0. Therefore, \(x-y<0\) and \(x<y\). Sufficient.

Answer: C
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Re: Is x < y?  [#permalink]

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New post 02 Apr 2018, 05:54
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Re: Is x < y? &nbs [#permalink] 02 Apr 2018, 05:54
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