Question

Yan 位于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。他骑车的速度是步行速度的 $7$ 倍，并且两种选择所需时间相同。Yan 到家的距离与到体育场的距离之比是多少？

Yan is somewhere between his home and the stadium. To get to the stadium he can walk directly to the stadium, or else he can walk home and then ride his bicycle to the stadium. He rides $7$ times as fast as he walks, and both choices require the same amount of time. What is the ratio of Yan’s distance from his home to his distance from the stadium?

$\dfrac{2}{3}$

$\dfrac{3}{4}$

$\dfrac{4}{5}$

$\dfrac{5}{6}$

$\dfrac{6}{7}$

答案：B

知识点：路程、速度与时间比与比例

难度评级：1440

小提示：

设步行速度为 $w$ ，到家的距离为 $x$ ，到体育场的距离为 $y$ 。

Let the walking speed be $w,$ and let $x$ and $y$ be the distances to home and to the stadium

大提示：

令两种时间相等： $\dfrac{y}{w}=\dfrac{x}{w}+\dfrac{x+y}{7w}$ 。

Equate the two times: $\dfrac{y}{w}=\dfrac{x}{w}+\dfrac{x+y}{7w}$

解答：

设 $w$ 为步行速度， $x$ 和 $y$ 分别为 Yan 到家和到体育场的距离。

步行到体育场需要 $\dfrac{y}{w}$ 。先走回家再骑车需要 $\dfrac{x}{w}+\dfrac{x+y}{7w}=\dfrac{8x+y}{7w}$ 。

令两者相等，得 $7y=8x+y$ ，所以 $8x=6y$ ， $\dfrac{x}{y}=\dfrac{3}{4}$ 。

因此，正确答案是 B。

Let $w$ be the walking speed and let $x$ and $y$ be Yan’s distances from home and from the stadium.

Walking to the stadium takes $\dfrac{y}{w}.$ Walking home then biking takes $\dfrac{x}{w}+\dfrac{x+y}{7w}=\dfrac{8x+y}{7w}.$

Setting these equal gives $7y=8x+y,$ so $8x=6y$ and $\dfrac{x}{y}=\dfrac{3}{4}.$

Thus, the correct answer is B.

2007 AMC 12A 第 9 题

答案：B

难度评级：1440

小提示：

大提示：

解答：

其他年份的第 9 题