4・16 A student attempts to find the average of three numbers $x, y$, and $z$ by first finding the average of $x$ and $y$, and then finding the average of that result and $z$. When $x<y<z$, the student's final result is
(A) correct.
(B) always less than $A$.
(C) always greater than $A$.
(D) sometimes less than $A$ and sometimes equal to $A$.
(E) sometimes greater than $A$ and sometimes equal to $A$.
(37th American High School Mathematics Examination, 1986)
[Solution] By algebraic transformation, we get
$$
\begin{aligned}
& \frac{\frac{x+y}{2}+z}{2}-\frac{x+y+z}{3}=\frac{x+y+2 z}{4}-\frac{x+y+z}{3} \\
= & \frac{x+y+z+z}{4}-\frac{x+y+z}{3}=\frac{3 A+z}{4}-A \\
= & \frac{z}{4}-\frac{A}{4}, \\
\because \quad & x\frac{x+y+z}{3}=A, \\
\therefore \quad & \frac{z}{4}-\frac{A}{4}>0 .
\end{aligned}
$$

Therefore, the answer is $(C)$.