Number property - even/odd problem

For sum or difference:
Even and even gives even
Odd and odd gives even
Odd and even gives odd

As such

(1) is sufficient. If A-B is even, then A+B is even, vice versa. If (A-B)-C is even, (A+B)+C is even.

(2) I think you mean (A-C)/B is odd

(2) tells us two possible cases

Case (i) (A-C) is odd and B is odd, and Case (ii) (A-C) is even and B is even

In (i) if (A-C) is odd and B is odd

We can also conclude A or C is odd and the other is even

So two possible combination
(a) A (odd), B (odd), C (even)
(b) A (even), B (odd), C (odd)

In both (a) and (b) prove A + B + C is even

See Case (ii) (A-C) is even and B is even
So two possible combination
(a) A (odd), B (even), C (odd)
(b) A (even), B (even), C (even)

This also gives A+B+C even

So (2) alone is also sufficient.

is something missing with statement 2 ? im not a fan of making assumptions

It is a typical GMAT trap named unwarrant assumption. We are not given that A, B, C are integer!

OA is E!

Let's pick up numbers:

(1) Pick 9/2 - 3/2 - 1 for A - B - C respectively --> A - B - C = 2 (even) but A + B + C = 1 (Odd) --> insuff

(2) similarly, just pick up fraction like 2, 1/3, 1 for A, B,C, we got (A-C)/B odd but A+B+C is even.

(1)&(2), just the same approach.

Wondering if there is a better approach?

That's a good question!

There are a few key number properties you must know:

Adding two evens or adding two odds results in an even number
Adding an even and an odd results in an odd number

For more info, see this Number Properties page that I found.