¶ 1995 AMC8 Problem 11
Problem:
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 -block area as shown. When they meet for the first time, they will be closest to
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Answer Choices:
A.
B.
C.
D.
E.
Solution:
Jane covers twice as much distance as Hector in the same time. So Jane goes to then , as Hector goes to . Jane continues to then as Hector goes to .
Jane covers twice as much distance as Hector in the same time. When they meet, they will have covered blocks. Since and is twice , they must meet blocks counterclockwise from the start, and this is point .
Answer: .
The problems on this page are the property of the MAA's American Mathematics Competitions