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| Change of lunch plans (Posted on 2003-10-29) |
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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?
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Submitted by Gamer
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Rating: 2.8571 (7 votes)
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Solution:
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They live two miles away from each other.
If you call the distance from A's house to where they met x, and the distance from where they met to B's house, then the distance covered by A after they meet is x, and the distance covered by B after they meet is x+2y.
Since A walks twice as slowly, 2x=x+2y, so x=2y. So they met at 2/3 of the way to B's house. (closer to B's house)If you mark off the distance in sixths, A and B met at 4/6 the distance to B's house. Since A covers half of what B does, 1 mile back (where A was when B started out to meet A) was 4/6 minus half of 1-(4/6), or 1/2. Since half the way there is 1 mile, the distance between the two must be two miles. |
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