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Re: DS: Set-21, Q11 [#permalink]
09 Feb 2008, 13:18
Is the answer A.
(1) 3a + 4b is even ASsuming 3a + 4b is even. 3a and 4b could both be odd, or both be even. Now, 4b cannot be odd, since any number multiplied by 4 is even. Hence, b is even. (2) 3a + 5b is even Applying the above logic to b, b could be even or odd and it cannot be determined.