Data Sufficiency

If a and b are integers, is b even?
(1) 3a + 4b is even.
(2) 3a + 5b is even.

zagcollins wrote:If a and b are integers, is b even?
(1) 3a + 4b is even.
(2) 3a + 5b is even.

(1) we know that 4b will always be even, regardless of the evenness/oddness of b. So, statement (1) tells us nothing about b: insufficient.

(2) 3a + 5b is even. To get an even sum, you need to add two evens or two odds. So, we could pick odd values for both a and b OR even values for both a and b. Therefore, b could be odd or even: insufficient.

As always, we could have picked numbers if we weren't sure about the rules. Remember, there are two steps to picking numbers in DS:

1) Pick permissible numbers. In other words, we MUST follow all the rules that we're given.

2) Plug your numbers back into the original question, seeing if you can get more than 1 answer.

For (2), we could pick:

a = 3 and b = 3 (permissible because 3*3 + 5*3 = 9 + 15 = 24, which is even as mandated).

Now we ask: is 3 even? NO

We could also pick:

a = 2 and b = 2 (permissible because 3*2 + 5*2 = 6 + 10 = 16, which is even as mandated).

Now we ask: is 2 even? YES

Since we got both an "NO" and a "YES", (2) is also insufficient.

Well, we've eliminated (a), (b) and (d) - now it's combination time!

Let's look at the statements together:

we know from (2) that a and b are both odd or both even. Now that we need to do so, let's examine (1) in more depth:

(1) 3a + 4b is even

As we noted earlier, 4 * any integer will be even, so we can rewrite (1) as:

3a + even = even

which means that 3a must also be even. Of course, the only way that 3*a could be even is if a itself is even.

So, after all that work, we now know that a MUST be even.

Combined with (2), we can eliminate the possibility that both a and b are odd. Therefore, the only possibility that remains is that both a and b are even, which of course means that b MUST be even: together the statements are sufficient, choose (C).

But Stuart, my point is that the question stem only says that they are integers. So why don't we apply negative integer scenario to pick number?
If we do so the picture will be more complicated. Please help me out from this loophole of my thinking.

@saeed,

doesnt really matter if its positive or negative as the problem at hand is dealing with even and odd...so the numbers could be on any side of the number line....

zagcollins wrote:@saeed,

doesnt really matter if its positive or negative as the problem at hand is dealing with even and odd...so the numbers could be on any side of the number line....

100% correct: in odd/even questions, we only care about "oddness" or "evenness", the actual values (and signs) are irrelevant.

Can't we just combine both the statements 1 and 2, so we get 6a + 9b = even. Since 6a is always even, 9b should be even for the sum to be even. So, b should be even. Is may approach OK to add the 2 linear equations from stataement 1 and statement 2 in DS questions?