1956 AMC 12 第 45 题

Question

一个装有橡胶轮胎的车轮外径为

25

英寸。当半径减少四分之一英寸后，行驶一英里所转的圈数将：

A wheel with a rubber tire has an outside diameter of

25

in. When the radius has been decreased a quarter of an inch, the number of revolutions in one mile will:

增加约

2\%

be increased about

2\%

增加约

1\%

be increased about

1\%

增加约

20\%

be increased about

20\%

增加

\dfrac12\%

be increased

\dfrac12\%

保持不变

remain the same

距离固定时，转数与车轮半径成反比

For a fixed distance, the revolution count is inversely proportional to the wheel’s radius

比较原半径

12.5

与新半径

12.25

Compare the original radius

12.5

with the new radius

12.25

原半径为

12.5

英寸，新半径为

12.25

英寸。对于固定距离，转数与半径成反比，所以相对增幅为

\begin{aligned} \frac{12.5}{12.25}-1 &=\frac{50}{49}-1\\ &=\frac1{49}\\ &\approx2.04\% \end{aligned}\text{。}

这约为

2\%

。

因此，正确答案是 A。

The original radius is

12.5

inches and the new radius is

12.25

inches. For a fixed distance, the number of revolutions varies inversely with radius, so the relative increase is

\begin{aligned} \frac{12.5}{12.25}-1 &=\frac{50}{49}-1\\ &=\frac1{49}\\ &\approx2.04\%. \end{aligned}

This is about

2\%.

Thus, the correct answer is A.

其他年份的第 45 题