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There are  5 girls and  5 boys in a chess club. The club holds a round-robin tournament in which every player plays against every other player exactly once.

What fraction of the games are boy-versus-boy? Enter your answer as a fraction in simplified form

 Jan 13, 2019
 #1
avatar+533 
+1

there are 10c2 or 45 games total

 

and there are 5c2 or 10 games where it is boy-v-boy

 

so it is 10/45 or 2/9

 

HOPE THIS HELPED!

 Jan 13, 2019
 #2
avatar+18138 
+3

We found from Chris' previous answer that girl vs girl  (or boy vs boy) is 10 games

 

Total games would be   10C2 = 45

10/45 = 2/9

 Jan 13, 2019
 #3
avatar+99736 
+2

As in your other question....the girls play 10 games against each other

And the boys play the same number

Each boy (or girl)  plays the opposite counterpart  5 * 5  = 25 times

 

So......the number of total games =  10 + 10 + 25  =  45

 

So....the fractional number of boy-boy games is   10 / 45   =  2 / 9

 

 

cool cool cool

 Jan 13, 2019
 #4
avatar+99736 
+1

OK...by unanimous consensus.....2 / 9   it is.....!!!!!

 

cool cool cool

 Jan 13, 2019
edited by CPhill  Jan 13, 2019
 #5
avatar+533 
+1

yep!!

 Jan 13, 2019
 #6
avatar+99736 
+1

LOL!!!!!

 

cool cool cool

CPhill  Jan 13, 2019
 #7
avatar+533 
0

LOLOLOLOLOLOLOLOL

 Jan 13, 2019
 #8
avatar+533 
0

i am going to post my own question; i can probably solve it on my own but i will post it anyway

 Jan 13, 2019
 #9
avatar+99736 
+1

OK......post away, asdf......!!!!!

 

 

 

cool cool cool

CPhill  Jan 13, 2019
 #10
avatar+533 
+1

i have posted the question cphill you can solve it if you want :P

 Jan 13, 2019

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