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?

They live 2 miles apart.

A walks distance W until he is 1 mile from B's house. Then A and B walk toward each other, B walking twice as fast. So they meet when A has traveled an additional 1/3 mile and B has traveled 2/3 mile. A is (W + 1/3) miles from his house, B is 2/3 mile from his house.

A returns to his house -- (W + 1/3) miles.
B returns to his house, then to A's -- (2/3 + 1 + W) miles.

Their times are equal, and B's rate is twice A's. Distance equals rate times time, so time equals distance divided by rate (D/R).

A's D/R is (W + 1/3) / 1.
B's D/R is (W + 5/3) / 2.

Solving, W = 1 mile.
So they live two miles apart.

Howard

Posted by Howard Bandy on 2003-11-01 10:31:48