AMC10 2007 A

AMC10 2007 A · Q13

AMC10 2007 A · Q13. It mainly tests Linear equations, Rates (speed).

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?

Yan 在家和体育场之间某处。为了去体育场，他可以直接走路去体育场，或者先走回家然后骑自行车去体育场。他骑车速度是他走路速度的 $7$ 倍，且两种选择所需时间相同。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

Answer (B): Let $w$ be Yan’s walking speed, and let $x$ and $y$ be the distances from Yan to his home and to the stadium, respectively. The time required for Yan to walk to the stadium is $y/w$, and the time required for him to walk home is $x/w$. Because he rides his bicycle at a speed of $7w$, the time required for him to ride his bicycle from his home to the stadium is $(x+y)/(7w)$. Thus \[ \frac{y}{w}=\frac{x}{w}+\frac{x+y}{7w}=\frac{8x+y}{7w}. \] As a consequence, $7y=8x+y$, so $8x=6y$. The required ratio is $x/y=6/8=3/4$.

答案（B）：设 $w$ 为 Yan 的步行速度，设 $x$ 和 $y$ 分别为 Yan 到家以及到体育场的距离。Yan 步行到体育场所需时间为 $y/w$，步行回家所需时间为 $x/w$。由于他骑自行车的速度为 $7w$，因此他从家骑车到体育场所需时间为 $(x+y)/(7w)$。因此 \[ \frac{y}{w}=\frac{x}{w}+\frac{x+y}{7w}=\frac{8x+y}{7w}. \] 由此可得 $7y=8x+y$，所以 $8x=6y$。所求比值为 $x/y=6/8=3/4$。

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