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Problem 318

AMC 12 late, AIME early

Algebra Difficulty 4.6 Multiple choice

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

Pick one

Official solution

［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 $(D)$ .

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.