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26 May 2017, 04:41
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If \(|x| > |2y|\), is \(x > y\)?
(1) \(x > 0\)
(2) \(y > 0\)
(1) \(x > 0\)
(2) \(y > 0\)
26 May 2017, 04:41
Kudos
Bookmarks
Official Solution:
In general, \(|A| = A\) (when \(A > 0\)), \(|0| = 0\), \(|A| = -A\) (when \(A < 0\)).
In the original condition, there are 2 variables \((x, y)\) and 1 equation \((|x| > |2y|)\), and in order to match the number of variables to the number of equations, there must be 1 more equation. Therefore, D is most likely to be the answer.
In the case of con 1), if \(x > 0\), then \(|x| = x\), so \(x > |2y| \ge |y| \ge y\), and then you get \(x > y\), hence yes, it is sufficient.
In the case of con 2), if \(y > 0\), then \(|x| > |2y| = 2|y| =2y\), and then you get \((x, y) = (3, 1)\) yes, but \((-3, 1)\) no, hence it is not sufficient. The answer is A.
Answer: A
In general, \(|A| = A\) (when \(A > 0\)), \(|0| = 0\), \(|A| = -A\) (when \(A < 0\)).
In the original condition, there are 2 variables \((x, y)\) and 1 equation \((|x| > |2y|)\), and in order to match the number of variables to the number of equations, there must be 1 more equation. Therefore, D is most likely to be the answer.
In the case of con 1), if \(x > 0\), then \(|x| = x\), so \(x > |2y| \ge |y| \ge y\), and then you get \(x > y\), hence yes, it is sufficient.
In the case of con 2), if \(y > 0\), then \(|x| > |2y| = 2|y| =2y\), and then you get \((x, y) = (3, 1)\) yes, but \((-3, 1)\) no, hence it is not sufficient. The answer is A.
Answer: A
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