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Of the 10 kids in a chess club, 5 are left-handed and 5 are right-handed. The club holds a round-robin tournament in which every player plays against every other player exactly once.

1. Of all the matches, how many of them have a left-handed player competing against a right-handed player?

2. Of all the matches, how many of them have two left-handed players?

3. What fraction of the games have two right-handed players? Enter your answer as a fraction in simplified form.

4. If Anita is one of the club members, what fraction of all games at the tournament does Anita play in? Enter your answer as a fraction in simplified form.

Jun 13, 2021
edited by Guest  Jun 13, 2021

#1
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We know there are 55 total games in the tournament so keep that in mind.

1. There are 5 left-handed players and 5 right-handed players so each left-handed player will play 5 right-handed players so there are 5*5 matches in there so it is 25.

2. Well, that would just be 4+3+2+1 left-hand only matches which will add up to 10 rounds.

3.  It would be the same as the left-handed players which is 10/55 which simplifies to 2/11.

4. Anita will play 9 games so it is 9/55 which cannot be simplified.

Jun 13, 2021
#2
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Not sure if Ninchux is correct....or if my answer is correct either...but here it is

10 players must each play 9 others  = 90 games BUT two players are in each game   90 / 2 = 45 games total

5 players L  must each play the other 5 players R =   25 games with   L vs R

5 L must each play the other 4 lefties   = 20  but two players in each game means  10 games with L vs L

Two right handed players would also be 10 games out of total 45        10/45 = 2/9

Anita must play 9 other players in the tournament    9 /45 =   1/5 of the games

Jun 13, 2021