Question

三名跑步者同时从一条 $500$ 米环形跑道的同一点出发，沿顺时针方向跑。他们的速度分别为每秒 $4.4, 4.8$ 和 $5.0$ 米。三人第一次再次同时相遇时，他们已经跑了多少秒？

Three runners start running simultaneously from the same point on a $500$ -meter circular track. They each run clockwise around the course maintaining constant speeds of $4.4, 4.8,$ and $5.0$ meters per second. The runners stop once they are all together again somewhere on the circular course. How many seconds do the runners run?

$1,000$

$1,250$

$2,500$

$5,000$

$10,000$

答案：C

知识点：相对速度最小公倍数路程、速度与时间

难度评级：1600

解答：

先求速度为 $4.8$ 米/秒的跑者追上最慢跑者所需的时间。

必须有其中 $x$ 是较快跑者追上另一人的时间。 $4.8x - 4.4x = 500$ $x = 1250,$

注意 $4.4 \cdot 1250 = 5500$ ，所以这两名跑者每次相遇都在起点。

现在要求最小的 $t$ ，使 $t$ 是 $1250$ 的倍数，并且最快的跑者也回到起点。

每 $1250$ 秒，最快的跑者跑 $1250 \cdot 5 = 6250$ 米；到 $2500$ 秒时，他跑了 $12500$ 米，正好是整数圈，其余条件也满足。

所以正确答案是 C。

Let us find the amount of time that it takes for the runner running at $4.8$ meters per second to lap the slowest person.

We must have that $4.8x - 4.4x = 500$ $x = 1250,$ where $x$ is the amount of time it takes for the faster runner to lap the other.

Note that $4.4 \cdot 1250 = 5500,$ which means that these two runners always intersect at the starting line.

We now have to find the least time, $t,$ such that $t$ is a multiple of $1250$ and the fastest runner ends up at the starting line.

Every $1250$ seconds, the fastest runner runs $1250 \cdot 5 = 6250$ meters. Then in $2500$ seconds, the fastest runner runs $12500$ meters, which is a whole number of laps.

Thus, C is the correct answer.

2012 AMC 10A 第 16 题

答案：C

难度评级：1600

解答：

其他年份的第 16 题