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28 Jul 2018, 21:33
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If x and y are integers, is x + y even?
(1) x + 2y is odd.
(2) xy is odd.
(1) x + 2y is odd.
(2) xy is odd.
28 Jul 2018, 21:53
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To determine whether x + y is even or odd
Statement 1
x + 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 2
xy is odd
=> xy is odd if and only if x is odd and y is odd
=> x + y = odd + odd = even
Statement 2 is sufficient
Hence option B
Statement 1
x + 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 2
xy is odd
=> xy is odd if and only if x is odd and y is odd
=> x + y = odd + odd = even
Statement 2 is sufficient
Hence option B
General Discussion
28 Jul 2018, 21:38
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This one (from GMATPrep Exam 5) has me pulling my hair out. According to the OA, only (2) is sufficient. I think (1) is sufficient alone too, however.
My logic for (1):
- x + 2y -> O
- 2y must be even (any integer times 2 is even)
- thus x must be odd
- "x + y" is only odd if either x or y is odd
- thus (1) indicates that "x + y" is odd
My logic for (1):
- x + 2y -> O
- 2y must be even (any integer times 2 is even)
- thus x must be odd
- "x + y" is only odd if either x or y is odd
- thus (1) indicates that "x + y" is odd
