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Is x < y?

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Is x < y?  [#permalink]

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Updated on: 01 Nov 2013, 01:01
1
5
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Difficulty:

55% (hard)

Question Stats:

62% (02:03) correct 38% (02:03) wrong based on 189 sessions

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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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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.

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Re: Is x < y?  [#permalink]

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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.

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Re: Is x < y?  [#permalink]

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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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19 Apr 2014, 03:25
gauravkaushik8591 wrote:
x^2 - y^2 < 0 --> x^2>y^2

is that a typo or did you mean that?

Yes, it was a typo. Edited. Thank you.
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Re: Is x < y?  [#permalink]

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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.

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Re: Is x < y?  [#permalink]

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11 Dec 2018, 13:42
Veritas Prep OFFICIAL EXPLANATION

Explanation: (1) The values 2 and 3 will give us a YES answer to the question stem, but the values 2 and -3 will give us a response of NO. Another way to approach this is to realize that you can factor the left hand side of the inequality to get (x - y)(x + y) < 0. This means that either the first term, (x - y), is negative or the second term, (x + y), is negative, but not both. If the first term is negative, then YES is answer to the question. However, if the second term is negative, the answer could be YES or NO. Accordingly, this statement is insufficient.

(2) You can simplify this by multiplying both sides by -1, which means you need to flip the direction of the inequality. You'll now have x + y > 0. This, however, doesn't tell us which value is larger, and this statement is insufficient.

Together, we know that (x - y)(x + y) < 0 and x + y > 0. This means that x - y < 0 . Remember, for the product of two values to be negative, one of them and only one of them must be negative. Because we know that (x + y) must be positive, then (x - y) < 0. Simply adding to each side of that inequality, we find that x < y, and know that the statements together are sufficient. Accordingly, the answer is C.
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Re: Is x < y?   [#permalink] 11 Dec 2018, 13:42
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