2015 AMC 8 第 17 题

先试着解答 2015 AMC 8 第 17 题,然后核对你的答案与视频与精心整理的解答,解答来自 LIVE by Po-Shen Loh。你也可以参加完整限时模拟考试、查看全部 2015 AMC 8 解答,或核对答案

所有题目均经美国数学协会(MAA)官方合法授权使用。

17.

Jeremy 的父亲在交通高峰期开车送他上学要二十分钟。某天没有交通拥堵,所以父亲能以快十八英里每小时的速度开车,并在十二分钟内把他送到学校。到学校有多少英里?

Jeremy's father drives him to school in rush hour traffic in 20 minutes. One day there is no traffic, so his father can drive him 18 miles per hour faster and gets him to school in 12 minutes. How far in miles is it to school?

44

66

88

99

1212

答案:D
知识点:路程、速度与时间一次方程
难度评级:1280
视频讲解:
解答视频缩略图
Play video

Click to load, then click again to play

文字解答:

设高峰期速度为每小时 ss 英里。2020 分钟是 13\dfrac13 小时,所以距离为 s3\dfrac{s}{3}

没有交通拥堵时,速度为每小时 s+18s+18 英里,路程用时 1212 分钟,即 15\dfrac15 小时。同一距离为 s+185\dfrac{s+18}{5}

令距离相等: 得 5s=3s+545s=3s+54,所以 s=27s=27。距离为 27/3=927/3=9 英里。 s3=s+185.\dfrac{s}{3}=\dfrac{s+18}{5}.

所以正确答案是 D

Let the rush-hour speed be ss miles per hour. The 2020-minute rush-hour trip takes 13\dfrac13 hour, so the distance is s3\dfrac{s}{3}.

Without traffic, the speed is s+18s+18 miles per hour and the trip takes 1212 minutes, or 15\dfrac15 hour. The same distance is s+185\dfrac{s+18}{5}.

Set the distances equal: s3=s+185.\dfrac{s}{3}=\dfrac{s+18}{5}. Then 5s=3s+545s=3s+54, so s=27s=27. The distance is 27/3=927/3=9 miles.

Thus, D is the correct answer.

← 第 16 题#16
完整试卷

其他年份的第 17 题

1985 AMC 8 · 1986 AMC 8 · 1987 AMC 8 · 1988 AMC 8 · 1989 AMC 8 · 1990 AMC 8 · 1991 AMC 8 · 1992 AMC 8 · 1993 AMC 8 · 1994 AMC 8 · 1995 AMC 8 · 1996 AMC 8 · 1997 AMC 8 · 1998 AMC 8 · 1999 AMC 8 · 2000 AMC 8 · 2001 AMC 8 · 2002 AMC 8 · 2003 AMC 8 · 2004 AMC 8 · 2005 AMC 8 · 2006 AMC 8 · 2007 AMC 8 · 2008 AMC 8 · 2009 AMC 8 · 2010 AMC 8 · 2011 AMC 8 · 2012 AMC 8 · 2013 AMC 8 · 2014 AMC 8 · 2016 AMC 8 · 2017 AMC 8 · 2018 AMC 8 · 2019 AMC 8 · 2020 AMC 8 · 2022 AMC 8 · 2023 AMC 8 · 2024 AMC 8 · 2025 AMC 8 · 2026 AMC 8