If set S consists of the numbers n, -2, and 4, is the mean of set S gr

Testing numbers smartly:

For (1), if n = 2 (it can't be technically, but we are figuring out the boundary), then mean = 4/3 and median = 2, so the median > mean. The reason why I tested 2 is because we can imagine our number to actually be 2 and an infinity number of 0s after the decimal point, then 1 at the very end. The result of plugging in such a number (which we can't) would be close enough to plugging a 2 in for this type of question specifically, so we can test 2. So anyways, let's try a much larger n. Much, much larger. Say 100. In that case, the mean is (-2 + 4 + 100)/3 = 102/3, or 34. In this case, our median is 4 (the moment n is greater than or equal to 4, the median will always be 4). Thus the mean here is greater than the median. Hence not sufficient.

(2) would be tested in a similar way. What if we use 3? Then we'd get mean = 5/3, and median = 3. So median > mean. What if n = -2? Mean = (-2 + -2 +4)/3 = 0. Median is -2, so mean > median. So not sufficient.

Together, we know that values between 2 and 3 would result in a mean < median. So, both (1) and (2) together are sufficient. So C is the answer.

I think the proper math calculations for it involves assuming n < -2, -2 < n < 4, and n > 4 or something, and so combining (1) and (2) gives you 2 < n < 3 which is between -2 and 4 so it can be figured out, but I a.) I don't remember all the proper math stuff because it's too long ago, and b.) am doing all this off the top of my head.