AMC 8 2026 Problem 5
AMC8 2026 年第 5 题
行程问题★☆☆☆☆
Casey went on a road trip that covered 100 miles, stopping only for a lunch break along the way. The trip took 3 hours in total and her average speed while driving was 40 miles per hour. In minutes, how long was the lunch break?
Casey 进行了一次总路程为 100 英里的公路旅行，途中只停了一次吃午饭。整个旅程耗时 3 小时，她开车时的平均速度为每小时 40 英里。请问她的午餐休息了多长时间（以分钟计）？
- A.15
- B.30
- C.40
- D.45
- E.60
正确答案 · Correct Answer
B
解析 · Solution
解法1
行驶时间 = 路程 ÷ 速度 = 100 ÷ 40 = 2.5 小时。全程共用 3 小时，午餐时间为 3 − 2.5 = 0.5 小时 = 30 分钟。
若无午餐，按 40 mph 行驶 3 小时应走 120 英里。实际只走 100 英里，差的 20 英里就是午餐占用的时间：20 ÷ 40 = 0.5 小时 = 30 分钟。
设午餐时间 t 小时，则行驶时间 = 3 − t。平均速度公式：100/(3−t) = 40。解得 3−t = 2.5，t = 0.5 小时 = 30 分钟。
Solution 1
Driving time = distance ÷ speed = 100 ÷ 40 = 2.5 hours. Total trip took 3 hours, so the lunch break was 3 − 2.5 = 0.5 hours = 30 minutes.
If there were no break, at 40 mph for 3 hours she would travel 120 miles. She only traveled 100 miles, so the 20-mile difference represents the break: 20 ÷ 40 = 0.5 hours = 30 minutes.
Let t be the lunch break in hours. Then driving time = 3 − t. Average speed = 100/(3−t) = 40. Solving: 3−t = , so t = 0.5 hours = 30 minutes.
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