AMC12 2007 A

AMC12 2007 A · Q9

AMC12 2007 A · Q9. It mainly tests Linear equations, Rates (speed).

Yan 在家和体育场之间的某处。要去体育场，他可以直接步行到体育场，或者先步行回家再骑自行车去体育场。他骑车的速度是他步行速度的 7 倍，并且两种选择所用时间相同。Yan 离家的距离与他离体育场的距离之比是多少？

(A) \frac{2}{3} \frac{2}{3}

(B) \frac{3}{4} \frac{3}{4}

(D) \frac{5}{6} \frac{5}{6}

(E) \frac{6}{7} \frac{6}{7}

Answer

Correct choice: (B)

正确答案：（B）

Solution

Let the distance from Yan's initial position to the stadium be $a$ and the distance from Yan's initial position to home be $b$. We are trying to find $b/a$, and we have the following identity given by the problem: \begin{align*}a &= b + \frac{a+b}{7}\\ \frac{6a}{7} &= \frac{8b}{7} \Longrightarrow 6a = 8b\end{align*} Thus $b/a = 6/8 = 3/4$ and the answer is $\mathrm{(B)}\ \frac{3}{4}$ Note: The identity can be given from an application of $t=\frac dr$.

设 Yan 初始位置到体育场的距离为 $a$，到家的距离为 $b$。我们要求 $b/a$，并且由题意有： \begin{align*}a &= b + \frac{a+b}{7}\\ \frac{6a}{7} &= \frac{8b}{7} \Longrightarrow 6a = 8b\end{align*} 因此 $b/a = 6/8 = 3/4$，答案为 $\mathrm{(B)}\ \frac{3}{4}$ 注：该等式可由 $t=\frac dr$ 得到。

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