A student just emailed me this question, so I figured I'd include the group on my response if anyone finds it helpful:
People say the answer to this data sufficiency question is A, what do you think?
If x is a positive number less than 10, is z greater than the average arithmetic mean of x and 10?
1. On the number line z is closer to 10 than to x
DETAILED SOLUTION APPEARS BELOW
space left blank so that you can work on this one on your own and then scroll down
We know that z is less than 10 and greater than 0, so let’s call it 5 to see how the relationship works. The average of x and 10 would then be 7.5, which is equidistant from x (5) and 10, so if the number z is closer to 10 than to x, then it is, indeed, greater than the average. For any number we choose as x, the same will hold – for two numbers, x and 10, the average will be the same distance from each, so if z is closer to 10, it has to be greater than the average. Therefore, statement 1 is sufficient.
I like using the “prove insufficiency” strategy in cases like these, in which I’ll plug in potential values for the variables to get an answer (either “yes” or “no” in this case), and then try new values to get the opposite answer. Let’s start with some easy-to-calculate values:
x = 1, z = 5. Average of x and 10 = 5.5, and z is LESS THAN the average.
Now, if our goal is to get a choice that is greater than the average, let’s try another set of numbers that has a slightly different set of properties. Maybe fractions are the answer…
x = ½, z = 2.5. Average of x and 10 = just over 5, and z is LESS THAN the average
That didn’t work, but may have tipped us off to an important factor here – the average of 10 and a positive number is always going to be greater than 5, so it won’t help to have small values of z in the mix…we need z to be greater than 5. But because the upper end of x and 10 is 10, the average will be less than 10. So let’s try a set of numbers that ensures z will be greater than 10:
x = 9, z = 45. Average of x and 10 = 9.5, so z is GREATER THAN the average. Here, we’ve proven that we can get both answers, so statement 2 is not sufficient, and the correct answer is A.