Ninth- and tenth-grade students participated in a tournament. Each contestant played each other contestant once. There were ten times as many tenth-grade students, but they were able to win only four-and-a-half times as many points as ninth graders.

How many ninth-grade students participated, and how many points did they collect?

Note: one point for every win.

Say there are n ninth graders and t tenth graders.

The total number of points earned is (n+t)(n+t-1)/2.

But t=10n, so the number of points is 11 n(11n-1)/2.

And 9/11 of the points were gained by tenth graders and 2/11 by ninth graders, or, in numbers: 2 n(11n-1)/2 were scored by ninth graders and 9 n(11n-1)/2 by tenth graders.

The number scored by ninth graders simplifies to n(11n-1), so the average ninth grader scored 11n-1, which, in other words, was the total number of players other than himself. So every ninth grader must have defeated every other player, including any other ninth graders.  But if there were more than one ninth grader, they couldn't all have defeated each other, as they play only once against each opponent.

Therefore there was only one ninth grader and ten tenth graders. The lone ninth grader gained 10 points by defeating each of the tenth graders. The tenth graders, playing among themselves, won the other 45 points.

Posted by Charlie on 2008-09-05 13:23:20