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Bunuel
Is \(x > y\)?

(1) \(x + y > 0\)
(2) \(y^2 > x^2\)
Solution:

Pre-Analysis:
  • We are asked if \(x>y\) or if \(x-y> 0\) or not
  • This is a Yes-No question

Statement 1: \(x + y > 0\)
  • This is not enough to say if \(x-y>0\) or not
  • Thus, statement 1 alone is not sufficient and we can eliminate options A and D

Statement 2: \(y^2 > x^2\)
  • According to this question \(x^2-y^2<0\) or \((x+y)(x-y)<0\)
  • For \((x+y)(x-y)\) to be negative, either \((x+y)\) or \((x-y)\) has to be negative and w ecannot say with surety that \(x-y>0\)
  • Thus, statement 2 alone is also not sufficient

Combining:
  • From statement 1, we got that \(x + y > 0\)
  • From statement 2, we got that \((x+y)(x-y)<0\)
  • Since \(x + y > 0\), \(x-y\) has to be negative or \(x-y<0\)


Hence the right answer is Option C
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Official Solution:


Is \(x > y\)?

(1) \(x + y > 0\)

This statement tells us that the sum of two numbers is greater than 0. This is clearly insufficient to determine which one of them is greater. Not sufficient.

(2) \(y^2 > x^2\)

Taking the square root of the inequality gives \(|y| > |x|\), which implies that \(y\) is further from 0 than \(x\). This is not sufficient to determine whether \(x > y\). For instance, consider \(x = 1\) and \(y = 2\), and \(x = 1\) and \(y = -2\).

(1) + (2) The second statement implies that \((y - x)(y + x) > 0\). Since from the first statement we know that \(x + y\) is positive, then \(y - x\) must also be positive, which implies that \(y > x\), giving a NO answer to the question. Sufficient.


Answer: C
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