Two people (A and B) want to meet each other to have lunch, and plan to meet at B's house party. A starts out for B's house, but when A is a mile away from B's house, B realizes that his house is a mess and A's isn't. So B takes off to meet A.

Once they meet, they talk for a while, and decide to meet at A's house instead. A goes back to her house to wait for B, and B goes to his house to pick up his food, then goes to A's house. They arrive at A's house at exactly the same time. If B walks twice as fast as A, how far apart do the two people live?

Let X be the distance in miles between the houses of A and B. After A walks X-1 miles, B starts out. At this time they are a mile apart. Since B walks twice as fast as A, they meet when they are 2/3 mile from B's house. A now begins a journey of X-2/3 miles back to her house, and B walks 2/3 mile to his house, picks up his food very quickly, and commences to walk X miles to A's house, a total distancce of X+2/3 miles. (Let's say the time required for A to "pick up his food" is negligible). Since B walks twice as fast as A, and A and B arrive at A's house simultaneously, it must be true that B was required to walk twice as far as A on this last leg.
2*(X-2/3)=X+2/3
X=2 miles
They live two miles apart.

Posted by Dan on 2003-11-05 04:00:37