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