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.
Let n be the number of students in ninth grade. The number of students
in tenth grade is equal to 10*n. In total there are 11n students. These
students play a total of 11n(11n-1)/2 games, and that's the total
number of points, too.
The students in ninth grade play n(n-1)/2 games between each other, these points they have automatically.
The students in tenth grade have 10n(10n-1)/2 points in the same way.
There are 10*n*n games between a ninth- and a tenth grade student.
Let k be the number of points the ninth-grade students get in total.
This number must be n(n-1)/2 + [a number between 0 and 10*n*n].
The number of wins by tenth grade is 11n(11n-1)/2 - k, and this is 4.5 * k.
11n(11n-1)/2 - k = 4.5 k
11n(11n-1)/2 = 5.5 k
11n(11n-1) = 11k
n(11n-1) = k
11*n*n - n = k
Since we know this must be between n(n-1)/2 and n(n-1)/2 + 10*n*n, we know that
0 ≤ 10.5*n*n - n/2 ≤ 10*n*n
We also know that n is an integer, and larger than zero (otherwise there would have been no tournament, or a trivial solution).
We get from the rightmost inequality:
0.5*n*n ≤ 0.5*n
n ≤ 1
This gives us the only solution (excluding the one with no students at all):
1 student of ninth grade and 10 students from tenth grade participated.
The student from ninth grade won all games, which gave him 10 points.
The students of tenth grade, amongst each other, won 45 points.
Solution
