deagez wrote:This type of problem is why I hate data sufficiency like I hate North Korea.
if x is positive, is x>3?
- (x-1)^2 > 4
- (x-2)^2 > 9
OA is D (each statement alone is sufficient)
Ok so first one makes since: if x >3 (say 4) it will make since 9>4, but if x is 3 then 4 is not > 4
However, the second makes is not so clear cut to me. if X is 4, then 4 is not > 9,, so to make sense here X has to be 6 or higher to make the equation equal 16 > 9
I mean how on earth is the second one sufficient?
If anyone can help me with how I need to properly think about this I would appreciate it, because I am apparently not taking the correct approach here
You're making a very common misinterpretation error. DS is often more about understanding how the questions work than doing math.
As always, let's start with Step 1 of the Kaplan Method for DS: focus on the question stem.
If x is postive, is x > 3?
As soon as you see "is" in the question, you know there are two possible answers: Yes and No. If we always get a "yes" answer, we have sufficiency; if we always get a "no" answer, we have sufficiency; if we get a "maybe yes or maybe no" or "sometimes" or "not sure", we have insufficiency.
So, if x is never greater than 3, that's sufficient; if x is always greater than 3, that's sufficient.
Let's just focus on statement (2):
(2) (x-2)^2 > 9
You approached it by picking numbers, a great way to go in DS. However, when we pick numbers we have to follow TWO key steps:
1) pick permissible numbers; and then
2) plug those numbers back into the original question.
A lot of people mess up step 1. We have to remember that the rules we're given (both in the question stem and the statements) are immutable laws of the universe. All numbers we choose MUST follow those laws.
The first number you chose is 4. As you correctly noted, 4 does not follow the law. Therefore,
we completely ignore 4 for the rest of the exercise. 4 is dead to us, we rip our clothing and move on.
You then determined that x has to be at least 6 - which isn't actually true. 6 is the smallest possible integer value of x, but nowhere does it say that x must be an integer. In fact, x can be any number greater than 5.
Regardless (that particular slip us is irrelevant on this question, but won't be on others), if you see that x must be at least 6, you should conclude that (2) is sufficient. After all, if x>=6, x is ALWAYS bigger than 3. Since we get an "ALWAYS", (2) is sufficient.
You actually interpreted the question as "could x be every number bigger than 3?". You saw that x could not equal 4 and you took that as a "no" answer to the question. It was that misinterpretation that cost you the point.
Thanks for the correction 
