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# Is x>y?

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Current Student
Joined: 19 Dec 2011
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Concentration: Healthcare, Operations
Schools: Olin '17 (M)
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20 Aug 2013, 22:03
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Is x>y?

(1) x+y>0
(2) x^2-y^2>0

[Reveal] Spoiler:
For 1) x+y>0
x> -y Insufficient

For 2) x^2-y^2>0
x^2>y^2
--> x<-y OR x>y

1) and 2)
x> -y
and x<-y OR x>y

So the answer should be E.
However, the answer key says C. What concept am I missing?

Thanks.

[Reveal] Spoiler: OA

Last edited by Bunuel on 21 Aug 2013, 01:20, edited 2 times in total.
Edited the question.
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21 Aug 2013, 01:27
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Is x>y?

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

(1) 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.

(2) x^2-y^2>0 --> x^2>y^2 --> |x|>|y|. This implies that x is further from 0, then y is. Not sufficient. Consider: (x, y)=(-2, 1) and (x, y)=(2, 1).

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

Hope it's clear.
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21 Aug 2013, 17:05
Bunuel:

Your answer totally makes sense - I'm curious, what led you to immediately jump to X>Y --> X-Y>0? To be honest, if I saw the question in the latter form, my first instinct would be to assume that the real question to be is X>Y?. Appreciate the help!
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22 Aug 2013, 02:41
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mid14 wrote:
Bunuel:

Your answer totally makes sense - I'm curious, what led you to immediately jump to X>Y --> X-Y>0? To be honest, if I saw the question in the latter form, my first instinct would be to assume that the real question to be is X>Y?. Appreciate the help!

Hard to say. When I see x>y, I automatically see x-y>0. Guess it comes with practice.
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10 Nov 2015, 01:50
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15 Nov 2015, 01:47
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

Is x>y?

(1) x+y>0
(2) x^2-y^2>0

There are 2 variables (x,y) and 2 equations are given from the 2 conditions, making it likely that (C) will be our answer.
Looking at them together,
x+y>0 and (x-y)(x+y)>0, x+y>0, so x-y>0. This answers the question 'yes' and it is therefore sufficient; the answer is (C).

For cases where we need 2 more equation, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E.
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Re: Is x>y?   [#permalink] 15 Nov 2015, 01:47
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