1954 AMC 12 第 36 题

Question

一艘船在静水中的速度为

15

英里每小时。在水流速度为

5

英里每小时的河流中，它顺流航行一段距离后原路返回。往返平均速度与静水速度之比为：

A boat has a speed of

15

mph in still water. In a stream that has a current of

5

mph it travels a certain distance downstream and returns. The ratio of the average speed for the round trip to the speed in still water is:

\dfrac54

\dfrac11

\dfrac89

\dfrac78

\dfrac98

顺流和逆流速度分别为

20

英里每小时和

10

英里每小时

The downstream and upstream speeds are

20

mph and

10

mph

往返距离相等，用总路程的两倍除以

\frac{d}{20}+\frac{d}{10}

For equal distances, divide twice the distance by

\frac{d}{20}+\frac{d}{10}

若单程距离为

d

，则总时间为

\frac d{20}+\frac d{10}=\frac{3d}{20}\text{。}

因此往返平均速度为

\frac{2d}{\frac{3d}{20}}=\frac{40}{3}\text{。}

它与

15

之比为

\frac{\frac{40}{3}}{15}=\frac89\text{。}

因此，正确答案是 C。

For a one-way distance

d,

the total time is

\frac d{20}+\frac d{10}=\frac{3d}{20}.

Thus the round-trip average speed is

\frac{2d}{\frac{3d}{20}}=\frac{40}{3}.

Its ratio to

15

is

\frac{\frac{40}{3}}{15}=\frac89.

Thus, the correct answer is C.

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