To determine whether x + y is even or odd

Statement 1x + 2y is odd

=> 2y is always even irrespective of even/odd nature of y (any integer multiplied by even results in an even integer)

=> x + 2y = odd

=> x = odd - 2y = odd - even = odd

So, x is odd and y can either be odd or even

if x is odd and y is odd => x + y = odd + odd = even

if x is odd and y is even => x + y = odd + even = odd

As we have two possible cases, statement 1 is insufficient

Statement 2xy is odd

=> xy is odd if and only if x is odd and y is odd

=> x + y = odd + odd = even

Statement 2 is sufficientHence option B
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