Question

Jane 走任意距离所用时间是 Hector 走同样距离所用时间的一半。他们从图中所示 $18$ 个街区区域外侧出发，沿相反方向行走。他们第一次相遇时，最接近哪个点？

Jane can walk any distance in half the time it takes Hector to walk the same distance. They set off in opposite directions around the outside of the $18$ -block area as shown. When they meet for the first time, they will be closest to

$A$

$B$

$C$

$D$

$E$

答案：D

知识点：路程、速度与时间周长相对速度

难度评级：1030

解答：

该区域的周长为 $18$ 个街区，所以 Jane 和 Hector 相遇时合计走了 $18$ 个街区。因为 Jane 速度是 Hector 的两倍，所以 Jane 走了 $12$ 个街区，Hector 走了 $6$ 个街区。

从底边中点出发，Hector 走 $6$ 个街区，先到 $E$ ，再向上到 $D$ ；Jane 走 $12$ 个街区，先到 $A$ ，再向上到 $B$ ，最后沿上边到 $D$ 。所以他们在 $D$ 相遇。

所以正确答案是 D。

The perimeter of the region is $18$ blocks, so when Jane and Hector meet they have together walked $18$ blocks. Since Jane walks twice as fast, she covers $12$ blocks and Hector covers $6.$

Starting from the middle of the bottom edge, Hector walks $6$ blocks (to $E,$ then up to $D$ ), and Jane walks $12$ blocks (to $A,$ up to $B,$ then across the top to $D$ ). They meet at $D.$

Thus, the correct answer is D .

1995 AMC 8 第 11 题

答案：D

难度评级：1030

解答：

其他年份的第 11 题