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I think the ans should be E, as y =0 and x as odd integer case is not considered.
Guys input your explanations pls.
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In case \(x=odd\) and \(y=0\) --> \(2*x*5*y=0=even\)
Question: \(2*x*5*y=even\). As there is 2 as a multiple, then this expression will be even if \(5xy=integer\). Basically we are asked is \(5xy=integer\) true?
Note that \(x\) and \(y\) may not be integers for \(2*x*5*y\) to be even (example \(x=\frac{7}{9}\) and \(y=\frac{9}{7}\)) BUT if they are integers then \(2*x*5*y\) is even.
(1) \(2+x+5+y=even\) --> \(7+x+y=even\) --> \(x+y=odd\). Not sufficient. (x=1 and y=2 answer YES BUT x=1.3 and y=1.7 answer NO)
(2) \(x-y=odd\). Not sufficient. (x=1 and y=2 answer YES BUT x=1.3 and y=0.3 answer NO)
(1)+(2) Sum (1) and (2) \((x+y)+(x-y)=odd_1+odd_2\) --> \(2x=even\) --> \(x=integer\) --> \(y=integer\) --> Both \(x\) and \(y\) are integers. Hence sufficient.
Answer: C.
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