AMC 8 1995 Problem 11
AMC8 1995 年第 11 题
行程问题★★★☆☆
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
Jane 走任何距离所花的时间是 Hector 走同样距离所花时间的一半。他们从如图所示的 18 个街区区域的外围向相反方向出发。当他们第一次相遇时，他们最接近哪里？
- A.A
- B.B
- C.C
- D.D
- E.E
正确答案 · Correct Answer
D
解析 · Solution
Solution 1
Counting around, when Jane walks 12 steps, she will be at . When Hector walks 6 steps, he will also be at . Since Jane has walked twice as many steps as Hector, they will reach this spot at the same time. Thus, the answer is .
We count to find that the perimeter of the rectangle is 18. We divide by 3 to see how far Hector has walked when he meets Jane, this distance is 6. We t(en move 6 steps counterclockwise to find the point they met or .
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