1972 AMC 12 第 24 题

Question

一个人以恒定速度步行了一段路。若他每小时快走

\frac12

英里，所需时间将为原来的五分之四；若他每小时慢走

\frac12

英里，则会多走

2\frac12

小时。他步行的距离（英里）为：

A man walked a certain distance at a constant rate. If he had gone

\frac12

mile per hour faster, he would have walked the distance in four-fifths of the time; if he had gone

\frac12

mile per hour slower, he would have been

2\frac12

hours longer on the road. The distance in miles he walked was:

13\frac12

15

17\frac12

20

25

设实际速度和时间分别为

v

和

t

Let the actual speed and time be

v

and

t

先用

vt=(v+\frac{1}{2})(\frac{4t}{5})

求出

v

First use

vt=(v+\frac{1}{2})(\frac{4t}{5})

to determine

v

设实际速度和时间为

v

和

t

。较快的情形给出

vt=\left(v+\frac12\right)\frac{4t}{5}\text{，}

所以

v=2

。较慢的情形于是给出

2t=\frac32\left(t+\frac52\right)\text{，}

从而

t=\frac{15}{2}

。步行距离为

vt=2\cdot\frac{15}{2}=15\text{。}

因此，正确答案是 B。

Let the actual speed and time be

v

and

t.

The faster case gives

vt=\left(v+\frac12\right)\frac{4t}{5},

so

v=2.

The slower case then gives

2t=\frac32\left(t+\frac52\right),

whence

t=\frac{15}{2}.

The distance was

vt=2\cdot\frac{15}{2}=15.

Therefore, the correct answer is B.

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