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Updated on: 06 Apr 2013, 04:57
Originally posted by shreerajp99 on 06 Apr 2013, 04:47.
Last edited by Bunuel on 06 Apr 2013, 04:57, edited 2 times in total.
Last edited by Bunuel on 06 Apr 2013, 04:57, edited 2 times in total.
Renamed the topic and edited the question.
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
5%
(low)
Question Stats:
88% (00:52) correct
12%
(01:11)
wrong
based on 190
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If a and b are both even integers, which of the following must be true?
A. The product of a and b is odd
B. The sum of a and b is odd
C. a - b > 0
D. a raised to the b power is even
E. (a + b)/4 is even
I know the answer but my question is why? Since zero is an even integer,if we consider 2^0->1 and thus D cant be always true.
A. The product of a and b is odd
B. The sum of a and b is odd
C. a - b > 0
D. a raised to the b power is even
E. (a + b)/4 is even
I know the answer but my question is why? Since zero is an even integer,if we consider 2^0->1 and thus D cant be always true.
06 Apr 2013, 05:02
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shreerajp99
If a and b are both even integers, which of the following must be true?
A. The product of a and b is odd
B. The sum of a and b is odd
C. a - b > 0
D. a raised to the b power is even
E. (a + b)/4 is even
I know the answer but my question is why? Since zero is an even integer,if we consider 2^0->1 and thus D cant be always true.
A. The product of a and b is odd
B. The sum of a and b is odd
C. a - b > 0
D. a raised to the b power is even
E. (a + b)/4 is even
I know the answer but my question is why? Since zero is an even integer,if we consider 2^0->1 and thus D cant be always true.
You are right, D cannot be the correct answer: 2^0=1=odd. So, none of the options must be true.
I guess the question should read "if a and b are both positive even integers..."
06 Apr 2013, 10:00
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Bunuel,does this mean 800score tests questions are not upto the mark?I took one of their quant tests today and got 31/37.Do u think their questions are a fair indicator of my actual level?
Thanks,
Shreeraj
Thanks,
Shreeraj
