Mr. and Mrs. Anderson and Mr. and Mrs. Barton competed in a chess tournament. Of the three games played:
(i) In only the first game were the two players married to each other.
(ii) The men won two games and the women won one game.
(iii) The Andersons won more games than the Bartons.
(iv) Anyone who lost a game did not play the subsequent game.
Who did not lose a game?
Since the first game was the only game in which two players married to each other faced one another, the winner of that game moved on to play in the second game against one of the remaining players, and the winner of that game played against the final player.
Also, the winner of the first game must have also won the second game, because if that player lost, then the remaining players in the final game would have again been members of the same couple.
Since the winner of the first game also won the second, that person would have to have been an Anderson since they won more games than the Bartons. It also means that it was Mr. Anderson who beat his wife in the first game, and then beat one of the Bartons in game two.
Since a woman won at least one game, this would mean that Mrs. Barton would be that woman since Mrs. Anderson lost game one. So Mrs. Barton did not lose a game.
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Game 1 - Mr. Anderson beats Mrs. Anderson
Game 2 - Mr. Anderson beats Mr. Barton
Game 3 - Mrs. Barton beats Mr. Anderson
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Posted by hoodat
on 2012-02-04 14:03:13 |