Bunuel
If x and z are positive numbers less than 16, is z greater than the average (arithmetic mean) of x and 16 ?
(1) z > x, and both x and z are perfect squares.
(2) x ≠ 1
Average of x and 16 will lie exactly in the middle of them (eg, average of 2 and 16 is 9, and 9 is 7 steps away from both 2 and 16. Average of 3 and 16 is 9.5, and 9.5 is 6.5 steps away from both 3 and 16). So we have to determine whether z is greater than the number which lies in the middle of x and 16.
(1) Both x, z are perfect squares and z>x but both less than 16. So positive perfect squares less than 16 are: 1, 4, 9. If x=1, then average of x and 16 = 8.5. z could be 4 and thus less than 8.5 or z could be 9 and thus greater than 8.5. So Insufficient.
(2) x is not 1, but nothing given about z. Clearly Insufficient.
Combining the two statements, if x is not 1, then x and z can be perfect squares among 4 & 9 only. x < z so only possibility is x=4, z=9. Since we have unique values we can find mean of x and 16 and determine whether z is greater than that or not. Sufficient.
Hence
C answer