Question

在全程为 $d$ 码的匀速赛跑中， $A$ 可领先 $B$ $20$ 码， $B$ 可领先 $C$ $10$ 码，且 $A$ 可领先 $C$ $28$ 码。则以码为单位的 $d$ 等于：

In racing over a distance $d$ at uniform speed, $A$ can beat $B$ by $20$ yards, $B$ can beat $C$ by $10$ yards, and $A$ can beat $C$ by $28$ yards. Then $d,$ in yards, equals:

无法由已知信息确定

not determined by the given information

$58$

$100$

$116$

$120$

答案：C

知识点：路程、速度与时间比与比例分式方程

难度评级：1550

小提示：

将每个领先距离转化为速度比

Translate each winning margin into a ratio of speeds

大提示：

将 $\frac{v_B}{v_A}$ 与 $\frac{v_C}{v_B}$ 相乘得到 $\frac{v_C}{v_A}$

Multiply the ratios $\frac{v_B}{v_A}$ and $\frac{v_C}{v_B}$ to get $\frac{v_C}{v_A}$

解答：

由领先距离得 $\begin{aligned} \frac{v_B}{v_A}&=\frac{d-20}{d},\\ \frac{v_C}{v_B}&=\frac{d-10}{d},\\ \frac{v_C}{v_A}&=\frac{d-28}{d} \end{aligned}\text{。}$ 因此 $\frac{(d-20)(d-10)}{d^2}=\frac{d-28}{d}\text{。}$ 展开并化简得 $2d=200$ ，所以 $d=100$ 。

所以，正确答案是 C。

The margins give $\begin{aligned} \frac{v_B}{v_A}&=\frac{d-20}{d},\\ \frac{v_C}{v_B}&=\frac{d-10}{d},\\ \frac{v_C}{v_A}&=\frac{d-28}{d}. \end{aligned}$ Therefore $\frac{(d-20)(d-10)}{d^2}=\frac{d-28}{d}.$ Expanding and simplifying yields $2d=200,$ so $d=100.$

Thus, the correct answer is C.

1961 AMC 12 第 37 题

答案：C

难度评级：1550

小提示：

大提示：

解答：

其他年份的第 37 题