4.17 $m$ is the arithmetic mean of $a, b, c, d, e$, $k$ is the arithmetic mean of $a$ and $b$, $l$ is the arithmetic mean of $c, d$, and $e$, $p$ is the arithmetic mean of $k$ and $l$, regardless of how $a, b, c, d, e$ are chosen, then it is always true that
(A) $m=p$.
(B) $m \geqslant p$.
(C) $m>p$.
(D) $m<p$.
(E) None of the above.
(16th American High School Mathematics Examination, 1965)
[Solution] Given $2 k=a+b, 3 l=c+d+e$, thus $m=\frac{a+b+c+d+e}{5}=\frac{2 k+3 l}{5}=\frac{4 k+6 l}{10}$.
And $p=\frac{k+l}{2}=\frac{5 k+5 l}{10}$,

then $4 k+6 l\left\{\begin{array}{ll}>5 k+5 l, & \text { if } k<l .\end{array}\right.$
Since $k$ depends on $a, b$; $l$ depends on $c, d, e$. Therefore, both $k<l$ and $k>l$ are possible.
Hence, the answer is $(E)$.