If x and y are positive integers, is x + y even?

Take x=2 y=1; x+y=3=odd; statement 1-xy=even correct
Take x=2 y=4; x+y=6=even; statement 1-xy= even correct
hence statement 1 not sufficient
Take x=2 y=1; x+y=3=odd;statement 1-xy=even correct
Take x=4 y=2; x+y=6=even; statement 1-xy= even correct
Hence statement 2 not sufficient
Take x=2 and y=1 ;x+y=3=odd and take x=4 y=2; x+y=6=evenn
Not sufficient hence E.

Statement1: Even x Even = Even => Even + Even = Even
OR
Even x Odd = Even => Even + Odd = Odd
Hence Insufficient.

Statement2: Even / Even = Even => Even + Even = Even
OR
Even / Odd = Even => Even + Odd = Odd
Hence Insufficient.

Statement 1 and 2 combined also will give similar combination. So can't be determined. Hence answer is E.

Question : If x and y are positive integers, is x + y even?

Statement 1: xy is even
If xy = 8,
Valid combination of x and y are , x= 4 and y =2 . x+y = 6 (true)
Another combination of x and y are , x= 8 and y =1 . x+y = 9 (false)

Insufficient

(2) x/y is even
If x/y = 8,
Valid combination of x and y are , x= 16 and y =2 . x+y = 18 (true)
Valid combination of x and y are , x= 8 and y =1 . x + y = 9 (false)

Insufficient

Combining both statements as well, we will not be able to solve the problem. Hence E) is the correct answer.

x,y > 0; x+y = even if x,y = even or x,y = odd.

1) xy = even --- x or y or x & y - even. So x+y = odd or even. Not Sufficient
2) x/y = even --- x - even and y could be even or odd. So x+y = odd or even. Not Sufficient
1+2 ---- xy = even & x/y = even ---- we can eliminate x and y both odd. So x = even for sure. Y = even or odd. -- x+y = Even or Odd. So 1+2 not sufficient.

Answer. E. OA?

statement 1:
If xy is even that means that at least one of them is even. This is not sufficient because if x=4 and y=5 then xy=20 and x+y=9, odd.
But if x=2 and y = 6 then xy=12 and x+y=8, even.
Hence Not Sufficient. Option A & D are out

statement 2:
X=12 and y=3, x/y=4, even. x+y=15, odd.
x=8 and y=4, x/y=2, even. x+y=12, even.
Hence Not Sufficient. Option B is out

If we combine statements, we still do not have enough information. The most we know is that one of them is even but we need to know for both x and y.

Answer E

from statement 1: xy is even
as we know xy can be even in two cases
1) both x & y are even = x+y=EVEN
2) x is odd and y is even or vice versa. =x+y=ODD
SO its conflicting and no definite result hence INSUFFICIENT.

From statement 2: x/y is even
again x/y is even in following two scenarios
1) both x & y are even =x/y cane both odd and even
2) x is even and y is odd=x/y is even =x+y=ODD
This statement also contradicts and INSUFFICIENT.

Even combination of both statements is also not sufficient.

Answer is E

Official Solution:

If x and y are positive integers, is x + y even?

For the sum of \(x\) and \(y\) to be even, the two variables must both be even or both be odd.

Statement (1): INSUFFICIENT. We know that at least one of the variables is even, but we do not know whether they are both even.

Statement (2): INSUFFICIENT. We know that \(x = y \times \text{(some even integer)}\). Since an even integer multiplied by any other integer is also even, we know that \(x\) is even. However, we do not know whether \(y\) is even.

Statements (1) & (2): INSUFFICIENT. Using both statements, we only know that \(x\) is even. Meanwhile, y could be even, but it does not have to be. As a result, we cannot determine whether the sum of \(x\) and \(y\) is even or odd. All of these results can be confirmed by picking numbers.

Answer: E.

There are a lot of ways to this one.
Given info =>
x,y are integers (This is a very important piece of info and should be always checked )
we need the even/odd nature of x+y
Statement 1=>
xy=even
=> (x,y)=> even,even => sum = even
even,odd => sum odd
odd,even => sum = odd

statement 2
x/y=even
hence x=y*even
so x must be even
but y can be even or odd
hence not sufficient

combining them two cases are possible
(x,y)=> even,even => sum=even
even,odd => sum =odd
hence not sufficient

Hence E

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