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

答案：A

知识点：圆周长 inverse proportion percent change

难度评级：1550

小提示：

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

大提示：

比较原半径 $12.5$ 与新半径 $12.25$

Compare the original radius $12.5$ with the new radius $12.25$

解答：

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.

1956 AMC 12 第 45 题

答案：A

难度评级：1550

小提示：

大提示：

解答：

其他年份的第 45 题