$3 \cdot 28$ Two people, A and B, start from point $M$ to point $N$ at the same time. A walks at a speed of $a$ kilometers per hour for half the time and at a speed of $b$ kilometers per hour for the other half. B walks at a speed of $a$ kilometers per hour for half the distance and at a speed of $b$ kilometers per hour for the other half. If $a \neq b$, then the order of their arrival at point $N$ is
(A) Both arrive at the same time.
(B) A arrives first.
(C) B arrives first.
(D) If $a > b$, A arrives first; if $a < b$, B arrives first.
(2nd "Hope Cup" National Mathematics Invitational Competition, 1991)
[Solution] Let the distance between $M$ and $N$ be $S$, the time it takes for A to reach $N$ be $t_{1}$, and the time it takes for B be $t_{2}$. According to the problem,
$$
\left\{\begin{array}{l}
\frac{a t_{1}}{2} + \frac{b t_{1}}{2} = S, \\
t_{2} = \frac{\frac{S}{2}}{a} + \frac{\frac{S}{2}}{b},
\end{array} \text { solving gives } \quad \left\{\begin{array}{l}
t_{1} = \frac{2 S}{a+b}, \\
t_{2} = \frac{(a+b) S}{2 a b} .
\end{array}\right.\right.
$$

Then $t_{1} - t_{2} = \frac{2 S}{a+b} - \frac{(a+b) S}{2 a b} = \frac{S\left[4 a b - (a+b)^{2}\right]}{2 a b(a+b)}$
$$
= \frac{-S(a-b)^{2}}{2 a b(a+b)} < 0 \text{. }
$$
$\therefore t_{1} < t_{2}$, which means A is ahead.
Therefore, the answer is $(B)$.