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Updated on: 28 May 2023, 11:29
Originally posted by cyberjadugar on 24 May 2012, 03:38.
Last edited by Bunuel on 28 May 2023, 11:29, edited 3 times in total.
Last edited by Bunuel on 28 May 2023, 11:29, edited 3 times in total.
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
5%
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Question Stats:
87% (00:36) correct
13%
(00:40)
wrong
based on 458
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24 May 2012, 04:16
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cyberjadugar
No, in this case the answer would be C.
Is a even?
(1) 2a is even --> \(2a=even\) --> \(a=\frac{even}{2}\), so \(a\) is either even or odd (in any case \(a\) must be an integer). Not sufficient.
(2) \(a^2\) is even --> \(a^2=even\) --> \(a=\sqrt{even}\), so \(a\) is either even (for example if \(a^2=4\)) or an irrational number (for example if \(a^2=6\)). Not sufficient. Generally \(a^2\) to be an integer \(a\) must be either an integer or an irrational number (for example: \(\sqrt{3}\)), (notice that \(a\) can not be reduced fraction, for example \(\frac{2}{3}\) or \(\frac{11}{3}\) as in this case \(a^2\) won't be an integer).
(1)+(2) Since from (1) \(a\) is an integer, then in order \(a^2\) to be even \(a\) must be even too. Sufficient.
Answer: C.
Hope it's clear.
General Discussion
24 May 2012, 03:44
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Is a even?
(1) 2a is even --> a can be even as well as odd. Not sufficient.
(2) \(\sqrt{a}\) is even --> \(\sqrt{a}=even\) --> \(a=even^2=even\). Sufficient.
Answer: B.
(1) 2a is even --> a can be even as well as odd. Not sufficient.
(2) \(\sqrt{a}\) is even --> \(\sqrt{a}=even\) --> \(a=even^2=even\). Sufficient.
Answer: B.
