Product XY

[GmatKiss]

If x and y are positive integers, is the product xy even?

(1) 5x - 4y is even

(2) 6x + 7y is even

[kmittal82]

(1)
5x - 4y = 2k
5x = 2(k + 2y)

this means 5x is even, which in turn means x is even. If x is even, it would mean xy is even

Sufficient

(2)
6x + 7y = 2k
7y = 2(k - 3x)

7y is even, which in turn means y is even, and if so, xy ought to be even

Sufficient

So, either statement is sufficient, i.e. (D)

OA please?

[neelgandham]

If x and y are positive integers, is the product xy even?

(1) 5x - 4y is even => (5x - Even) is Even => 5x is Even => x is Even => xy is Even - Sufficient

(2) 6x + 7y is even => (7y + Even) is Even => 7y is Even => y is Even => xy is Even - Sufficient

IMO Option D

Concept recap :
Even+Even = Even
Even*Even = Even
Even* Odd = Even

[Brent@GMATPrepNow]

If x and y are positive integers, is the product xy even?
(1) 5x - 4y is even
(2) 6x + 7y is even

Statement 1:
First, we know that 4y will be even for any integer y.
So we get . . . 5x - even is even
This means that 5x must be even.
For 5x to be even, x must be even.
Since x is even, xy must be even (regardless of the even/oddness of y)
SUFFICIENT

Statement 2:
First, we know that 6x will be even for any integer x.
So we get . . . even + 7y is even
This means that 7y must be even.
For 7y to be even, y must be even.
Since y is even, xy must be even (regardless of the even/oddness of x)
SUFFICIENT

So, the answer is D

Cheers,
Brent

[masoom j negi]

Statement 1. 5x - 4y is even i.e. 5x - 4y = 2k
Difference of two numbers is even if either both of them are odd or both of them are even.
Here, 4y is always even. So, 5x is also even.
Since 5x is even, we can say that x is even.
Now, xy will be even because the product of an even no with any positive integer is always even. Hence, Sufficient.
Statement 2. 6x + 7y is even.
6x is eve n because it is divisible by 2. So, 7y has to be even.
For 7y to be even, y has to be even.
We know the product of an even no with a positive integer is even.
Hence, xy is even. Hence, Sufficient.