Question

Aya 每天早晨都会步行 $9$ 千米，然后在咖啡店停留。若她以每小时 $s$ 千米的恒定速度行走，包括在咖啡店停留的 $t$ 分钟在内，全程共用 $4$ 小时。若她以每小时 $s + 2$ 千米的速度行走，包括同样的 $t$ 分钟在内，全程共用 $2$ 小时 $24$ 分钟。假设 Aya 以每小时 $s + \frac{1}{2}$ 千米的速度行走。求包括在咖啡店停留的 $t$ 分钟在内，全程需要的分钟数。

Every morning Aya goes for a $9$ -kilometer-long walk and stops at a coffee shop afterwards. When she walks at a constant speed of $s$ kilometers per hour, the walk takes her $4$ hours, including $t$ minutes spent in the coffee shop. When she walks $s + 2$ kilometers per hour, the walk takes her $2$ hours and $24$ minutes, including $t$ minutes spent in the coffee shop. Suppose Aya walks at $s + \frac{1}{2}$ kilometers per hour. Find the number of minutes the walk takes her, including the $t$ minutes spent in the coffee shop.

答案

答案：204

知识点：路程、速度与时间二次方程

难度评级：1890

解答：

以小时为单位，两种情形给出 $\frac{9}{s} + \frac{t}{60} = 4$ $\frac{9}{s+2} + \frac{t}{60} = \frac{12}{5}.$ 和相减得 $\frac{9}{s} - \frac{9}{s+2} = \frac{8}{5}$ ，所以 $\frac{18}{s(s+2)} = \frac{8}{5}$ ，从而 $s(s+2) = \frac{45}{4}$ 。方程 $s^2 + 2s - \frac{45}{4} = 0$ 的正根为 $s = \frac{5}{2}$ 。

因此 $\frac{t}{60} = 4 - \frac{9}{5/2} = \frac{2}{5}$ ，所以 $t = 24$ 分钟。以 $s + \frac{1}{2} = 3$ 千米每小时行走时，步行本身需要 $\frac{9}{3} = 3$ 小时，所以总时间为 $180 + 24 = 204$ 分钟。

Measuring time in hours, the two scenarios say $\frac{9}{s} + \frac{t}{60} = 4$ and $\frac{9}{s+2} + \frac{t}{60} = \frac{12}{5}.$ Subtracting, $\frac{9}{s} - \frac{9}{s+2} = \frac{8}{5},$ so $\frac{18}{s(s+2)} = \frac{8}{5},$ giving $s(s+2) = \frac{45}{4}.$ The positive root of $s^2 + 2s - \frac{45}{4} = 0$ is $s = \frac{5}{2}.$

Then $\frac{t}{60} = 4 - \frac{9}{5/2} = \frac{2}{5},$ so $t = 24$ minutes. Walking at $s + \frac{1}{2} = 3$ kilometers per hour takes $\frac{9}{3} = 3$ hours, so the total is $180 + 24 = 204$ minutes.

2024 AIME I 第 1 题

答案：204

难度评级：1890

解答：

其他年份的第 1 题