2.79 Two people start from points $R$ and $S$, which are 76 miles apart, and walk towards each other at the same time. The person at $R$ walks at a constant speed of $4 \frac{1}{2}$ miles per hour, while the person at $S$ walks at a speed of $3 \frac{1}{4}$ miles per hour for the first hour, $3 \frac{3}{4}$ miles per hour for the second hour, and so on, following an arithmetic sequence. If the two people meet at a point that is $x$ miles closer to $R$ than to $S$, and the time taken is exactly an integer number of hours, then $x$ is
(A) 10.
(B) 8.
(C) 6.
(D) 4.
(E) 2.
(14th American High School Mathematics Examination, 1963)
［Solution］Let the time from departure to meeting be $k$ hours. According to the problem, we have
$$
\frac{9}{2} k+\frac{1}{2} k\left[2 \cdot \frac{13}{4}+\frac{1}{2}(k-1)\right]=76,
$$

which simplifies to $k^{2}+30 k-304=0$, solving this gives $k=8$ (considering only the integer solution). Therefore, the meeting point is 36 miles from $R$ and 40 miles from $S$. Thus,
$$
x=40-36=4 \text {. }
$$

Hence, the answer is $(D)$.