If x and y are integers, is x even?

1.y(x+1)=odd this expression is odd if y=odd and x+1=odd, x=even
2.3(2x-y)=odd, 3 is odd, 2x-y must be also odd -> 2x=odd+y, y is odd (bcs. odd+odd is even), BUT x can be even or odd NOT SUFFICIENT
Answer (A)

IMO a.

Statement 1:for a number to be odd all its factors have to be odd. As such as x=1 is odd, x is even.

Statement 2: no trend can be established.

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

If x and y are integers, is x even?

(1) xy + y is odd

(2) 6x - 3y is odd

There are 2 variables (x,y) and 2 equations, so there is high chance (C) will be the answer. When looking at the conditions together, 6x=even, 6x-3y=odd, so 3y=odd. Also, in (x+1)y=odd, x+1=odd, x=even, so the conditions seem to be sufficient and the answer looks like (C), but this is a commonly made mistake according to 4(A), we should look at the conditions separately. For condition 1, (x+1)y=odd, x+1=odd, y=odd, x=odd-1=even. So this condition is sufficient, but for condition 2, 6x-3y=odd, 6x is always even without regard to the value of x, so this condition is insufficient. The answer becomes (A).
This is a most common type of question in GMAT math DS section.

For cases where we need 2 more equation, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E.

Here's the video solution for that question - https://www.gmatprepnow.com/module/gmat- ... /video/838

Cheers,
Brent

Numerous ways to solve this one =>
Statement 1 =>Here xy+y is odd => y(x+1) is odd so x+1=odd => x=odd-odd=even = suff
Statement 2 => here changing the sides we get that y is odd. not clue of x => no suff

Smash that A

cracked it in 75 seconds...
1. xy + y is odd.
we can get such a result only if:
even+odd
odd+even

if xy is even, then y must be odd. and then x must be even
if xy is odd, then y must be odd. in this case xy+y is even and not odd.
the only way it works is if xy is even, and x is even.
sufficient.
B, C, and E are out.

2. factor out 3
3(2x-y) = odd
since the result of the product is odd, 2x-y must be odd.
2x is 100% even, so y is odd for sure.
but x can be even or odd...since we can't definitely give an answer, D is eliminated.

The answer is A.

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