Is x+y an even number?

We are not given that x and y are integers, so for (2) it's possible say x = 1.1 and y = 0.1. In this case x + y = 1.2, which is not even.

st 1) 5x + 4 is odd
no information on y
and only says 5x is odd
not sufficient.

st 2) x - y is odd
x and y could be even/odd or decimals/fractions.
not sufficient.

combining 1 and 2
x could be odd or any fraction/decimal value.
hence, y could be even or fraction/decimal value.
but, by plugging in values, x + y turns out to be either odd or fraction/decimal value.
in either case we can say that x + y is not even, which should be sufficient to answer the question.

shouldn't the answer be C and not E?
what am I missing here? Bunuel

Is \(x + y\) an even number?

First of all notice that we are not given that \(x\) and \(y\) are integers

(1) \(5x + 4\) is an odd number

\(5x + 4 = odd\);

\(5x = odd - 4 = odd - even = odd\);

\(x = \frac{odd}{5}\). So, \(x\) is either a fraction of the form \(\frac{odd}{5}\) (..., -3/5, -1/5, 1/5, 3/5, ...) or an odd integer (..., -3, -1, 1, 3, ...).

If \(x = 1\) and \(y = 2\), then \(x + y = 3\), which is NOT even.
If \(x = 1\) and \(y = 1\), then \(x + y = 2\), which IS even.

Not sufficient.


(2) \(x - y\) is an odd number.

If \(x = 1.1\) and \(y = 0.1\), then \(x + y = 1.2\), which is NOT even.
If \(x = 1.5\) and \(y = 0.5\), then \(x + y = 2\), which IS even.

Not sufficient.


(1)+(2) From (1) we have that \(x = \frac{odd}{5}\), so from (2):

\(x - y = odd\);

\(\frac{odd}{5} - y = odd\);

\(y=\frac{odd}{5}-odd=\frac{odd-5*odd}{5}=\frac{even}{5}\). So, \(y\) is either a fraction of the form \(\frac{even}{5}\) (..., -4/5, -2/5, 2/5, 4/5, ...) or an even integer (..., -4, -2, -0, 2, 4, ...).


We got that \(x = \frac{odd}{5}\) and \(y=\frac{even}{5}\), thus \(x+y=\frac{odd}{5}+\frac{even}{5}=\frac{odd+even}{5}=\frac{odd}{5}\). An odd number divvied by 5 is either an odd number or not an integer, so \(x + y\) cannot be even. So, we have a definite NO answer to the question. Sufficient.


Answer: C.

Yes, the OA should be C, not E. Edited. Thank you!

Here is a solution: https://gmatclub.com/forum/is-x-y-an-ev ... l#p2945090

Hope it helps.