AMC8 2025

AMC8 2025 · Q19

AMC8 2025 · Q19. It mainly tests Linear equations, Rates (speed).

Two towns, $A$ and $B$, are connected by a straight road that is $15$ miles long. Travelling from city $A$ to town $B$, the speed limit changes every $5$ miles: from $25$ to $40$ to $20$ miles per hour (mph). Two cars, one at town $A$ and one at town $B$, start moving toward each other at the same time. They drive at exactly the speed limit in each portion of the road. How far from town $A$, in miles, will the two cars meet?

两个城镇$A$和$B$由一条15英里长的直路连接。从$A$到$B$，每5英里限速变化：25、40、20英里每小时（mph）。两辆车，一辆在$A$，一辆在$B$，同时开始向对方行驶。它们在每段路严格按限速行驶。两车将在距离$A$镇多少英里处相遇？

(A) \ 7.75 \ 7.75

(B) \ 8 \ 8

(D) \ 8.5 \ 8.5

(E) \ 8.75 \ 8.75

Answer

Correct choice: (D)

正确答案：（D）

Solution

The first car, moving from town $A$ at $25$ miles per hour, takes $\frac{5}{25} = \frac{1}{5} \text{hours} = 12$ minutes. The second car, traveling another $5$ miles from town $B$, takes $\frac{5}{20} = \frac{1}{4} \text{hours} = 15$ minutes. The first car has traveled for 3 minutes or $\frac{1}{20}$th of an hour at $40$ miles per hour when the second car has traveled 5 miles. The first car has traveled $40 \cdot \frac{1}{20} = 2$ miles from the previous $5$ miles it traveled at $25$ miles per hour. They have $3$ miles left, and they travel at the same speed, so they meet $1.5$ miles through, so they are $5 + 2 + 1.5 = \boxed{\textbf{(D) }8.5}$ miles from town $A$.

第一辆车从$A$以25英里每小时行驶，行驶5英里用$\frac{5}{25} = \frac{1}{5}$小时$=12$分钟。第二辆车从$B$再行驶5英里，用$\frac{5}{20} = \frac{1}{4}$小时$=15$分钟。当第二辆车行驶完5英里时，第一辆车已行驶12分钟，然后再行驶3分钟即$\frac{1}{20}$小时以40英里每小时，行驶$40 \cdot \frac{1}{20} = 2$英里。此时第一辆车总计行驶$5+2=7$英里，还剩3英里，它们速度相同，各走1.5英里，因此距离$A$为$5 + 2 + 1.5 = \boxed{\textbf{(D) }8.5}$英里。

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