A student attempted to compute the average $A$ of $x, y$ and $z$ by computing the average of $x$ and $y$, 
and then computing the average of the result and $z$. Whenever $x < y < z$, the student's final result is
$\textbf{(A)}\ \text{correct}\quad \textbf{(B)}\ \text{always less than A}\quad \textbf{(C)}\ \text{always greater than A}\quad\\ \textbf{(D)}\ \text{sometimes less than A and sometimes equal to A}\quad\\ \textbf{(E)}\ \text{sometimes greater than A and sometimes equal to A}  \quad$
The true average is $A=\frac{x+y+z}{3}$, and the student computed $B=\frac{\frac{x+y}{2}+z}{2}=\frac{x+y+2z}{4}$, so $B-A = \frac{2z-x-y}{12} = \frac{(z-x)+(z-y)}{12}$, which is always positive as $z>x$ and $z>y$. Thus $B$ is always greater than $A$, i.e. $\boxed{C}$.