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Students in a tournament (Posted on 2008-09-05) Difficulty: 2 of 5
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.

See The Solution Submitted by pcbouhid    
Rating: 2.3333 (3 votes)

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Solution Solution | Comment 3 of 7 |

Let the respective numbers of Ninth- and tenth-grade students be x and 10x.

Then, by the problem, the total number matches and therefore the total number of accumulated by all the students
= comb(10x,2) + comb (x,2) + 10*x^2
= (11/2)*(11*x^2 - x)
= N(say) ...... (i)

Let the total score by the ninth grade students be m. Then, it follows that the total numer of points accumulated by tenth grade students is N -m. Accordingly, by the given conditions, we must have:

N-m = 9m/2
or N = 11m/2
or, (11/2)*(11*x^2 - x) = 11m/2
or, m = 11*x^2 - x ........(ii)

Now, total number of points accumulated by ninth grade students
= total number of points -  total number of points accumulated by tenth grade students
= (11/2)*(11*x^2 - x) - comb(10x,2)
= (21/2)*x^2 - x/2

Accordingly, we must have:

m< = (21/2)*x^2 - x/2
or, 11*x^2 - x  < = (21/2)*x^2 - x/2
or, x^2 <=x.
or, x<=1, since x must be a positive integer

The only positive integer satisfying x<=1 occurs at x=1, so that:
m = 11*x^2 - x = 10

Consequently, there was only one ninth grade student who participated and he collected a total of 10 points

Edited on September 5, 2008, 4:31 pm
  Posted by K Sengupta on 2008-09-05 12:58:44

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