sunita123 wrote:
Is \(x + y > 0\)?
(1) \(x - y > 0\)
(2) \(x^{2} - y^{2} > 0\)
We can rephrase the question by subtracting y from both sides of the inequality: Is \(x > -y\) ?
(1)
INSUFFICIENT: If we add y to both sides, we see that x is greater than y. We can use numbers here to show that this does not necessarily mean that \(x > -y\). If \(x = 4\) and \(y = 3\), then it is true that \(x\) is also greater than \(-y\). However if \(x = 4\) and \(y = -5\), \(x\) is greater than \(y\) but it is
NOT greater than \(-y\).
(2)
INSUFFICIENT: If we factor this inequality, we come up \((x + y)(x – y) > 0\).
For the product of \((x + y)\) and \((x – y)\) to be greater than zero, the must have the same sign, i.e. both negative or both positive.
This does not help settle the issue of the sign of \(x + y\).
(1) AND (2)
SUFFICIENT: From statement 2 we know that \((x + y)\) and \((x – y)\) must have the same sign, and from statement 1 we know that \((x – y)\) is positive, so it follows that \((x + y)\) must be positive as well.
The correct answer is
C.
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