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.

Solution: 1 ninth grader participated.  10 tenth graders participated.  The ninth grader won 10 times (lost 0) and the tenth graders won 45 times (lost 55)

N = number of ninth graders
T = number of tenth graders

T = 10N (given)
T + N = total participants

Calculating total wins for all participants (W):
W = (T + N - 1) + (T + N - 2) + (T + N - 3) + . . .
  or
W = (T + N)(T + N - 1) / 2 

Substituting T = 10N:

W = (10N + N) (10N +N - 1) / 2
W = 11N (11N - 1) / 2

Calculating minimum wins for tenth graders (M). This is the total wins accumulated by the 10th graders when they play each other.

M = T (T - 1) / 2 = 10N (10N -1) / 2

M/W = [10N (10N - 1) / 2] / [11N(11N -1) / 2]

simplifying

M/W = 10/11 [(10N - 1) / (11N - 1)]

Given in the problem is that M/W = 4.5 = 9/11 (maximum)
9/11 >= M/W

Substitute
9/11 >= 10/11 [(10N - 1) / (11N - 1)]

9/10 >= (10N - 1) / (11N -1)

9/10 (11N - 1) >= 10N - 1

99/10N - 9/10 >= 100/10N - 10/10

1/10 >= 1/10N

1 >= N  Thus only one 9th grader participated, and won every time.

Edited on September 5, 2008, 12:34 pm

Posted by Leming on 2008-09-05 12:30:48