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Mathematics 9 Online
OpenStudy (anonymous):

Twice as many girls as boys were participating in a tennis tournament. Each pair of players played only one match and there were no draws. The ratio of girl winnings to boy winnings was 5:7.How many players took part in the tournament

OpenStudy (texaschic101):

twice as many girls as boys.....5 : 7....does that mean there is 12 games ?

OpenStudy (loser66):

I don't think so, it's the ratio of winning girls and boys.

OpenStudy (loser66):

because if there is 12 games, and 5 winning girls that means there is 10 girls, 7 winning boys means 14 boys, and 14 is not twice as 10

OpenStudy (anonymous):

what do we have here lads

OpenStudy (anonymous):

each pair is this doubles?

OpenStudy (anonymous):

need the context haha

OpenStudy (texaschic101):

well...for every 12 games, girls won 5 and boys won 7.....5/7 of x...but I don't know what x is

OpenStudy (texaschic101):

each pair of players....so I would say it is doubles

OpenStudy (anonymous):

nah i don't think it can be 12 games

OpenStudy (anonymous):

would it be affected if it were mixed doubles?

OpenStudy (anonymous):

just putting ideas out there

OpenStudy (texaschic101):

twice as many girls as boys.....2g = 1b.....I have no idea where I am going with this

OpenStudy (anonymous):

loool

OpenStudy (texaschic101):

amistre will get it....just wait

OpenStudy (anonymous):

you can just feel it

OpenStudy (texaschic101):

lol

OpenStudy (texaschic101):

awkward silence...lol

OpenStudy (loser66):

the bad net, I think so

OpenStudy (anonymous):

this sentence is quite ambiguous.. Twice as many girls as boys were participating in a tennis tournament. did twice as many girls actually PLAY though, or they participated by watching the matches

OpenStudy (amistre64):

b g g b b g g g g b b b g g g g g g 2:4 is 3 teams 4:8 is 6 teams 6:12 is 9 teams 2n:4n boys to girls is 3n teams to work with at best

OpenStudy (amistre64):

bg bg bg bg bg gg gg gg gg gg ^ ^ ^ ^ ^ ^ 5 boys win: 5 girls win + 1 more team seems like a bad combination to me ----------------------------- bb bg bg bg gg gg gg gg gg gg ^ ^ ^ ^ ^ ^ 5 boys, 3girl + 4 girls ----------------------------- bb bb bg gg gg gg gg gg gg gg ^ ^ ^ ^ ^ ^ 5 boys, 1girl + 6 girls the setup of by girl teams doesnt seem to matter if we assume all the boys win

OpenStudy (texaschic101):

there is 4 players in each game...right...? so if there was 12 games, then there is 48 players... am I way off on this ?

OpenStudy (rsadhvika):

assume 2 players in each game - singles

OpenStudy (texaschic101):

I thought they were doubles

OpenStudy (anonymous):

thats what i thought with 48..but not 1:2

OpenStudy (amistre64):

bb bb bg bg bg bg gg gg gg gg gg gg ^ ^ ^ ^ ^ ^ +6 5 boys 1girl, need 6 more girls this seems like a solution

OpenStudy (texaschic101):

are we sure that boys and girls are mixed

OpenStudy (amistre64):

i ran thru a pairing that indicated that the bg matchup had no effect

OpenStudy (anonymous):

are we sure that about this ! did twice as many girls actually PLAY though, or they participated by watching the matches!

OpenStudy (amistre64):

we shouldnt try to add in extra content ...

OpenStudy (anonymous):

so how many players do you think played?

OpenStudy (amistre64):

i believe that 8 boys and 16 girls fits for one solution

OpenStudy (anonymous):

In a game, If it is boy VS girl then, probability of boy win =1/2 and probability of girl win=1/2 If it is girl VS girl then, probability of girl win =1 If it is boy VS boy then, probability of boy win =1 Since the number of girls participating is more than the number of boys, the ratio of girl winning to boy winning can never be less than or equal to 1. So, either this question makes no sense or I didn't get the question.

OpenStudy (amistre64):

"experimental" probability does not conform to "theoritical" probability.

OpenStudy (rsadhvika):

the matches happened already, what we have is the data, we can have the ratio of winning of girls as 1 also. so...

OpenStudy (anonymous):

bb bb bg bg bg bg gg gg gg gg gg gg ----> match played b b b b b g g g g g g g ---> win

OpenStudy (anonymous):

then that should fit

OpenStudy (amistre64):

i agree :) but im a little biased

OpenStudy (texaschic101):

does it matter if there is doubles playing ?

OpenStudy (anonymous):

well, if 8 boys and 16 girls fits then 8n boys and 16n girls must also fit where n can be any natural number

OpenStudy (amistre64):

it doesnt matter for doubles; i assumed doubles when i worked it up

OpenStudy (texaschic101):

because I figure if there is doubles playing, then there is 4 playersin each game

OpenStudy (anonymous):

question says a pair of player plays a game

OpenStudy (amistre64):

bb bb bg gg gg gg : win , 5b and 7g gg gg gg bg bg bg : lose

OpenStudy (texaschic101):

ohhhhh...I see

OpenStudy (rsadhvika):

Each pair of players played only one match should we interpret it as "each player played with rest of the players once ?"

OpenStudy (texaschic101):

so is there 24 people ?

OpenStudy (rsadhvika):

"each player played with rest of the players only once ?"

OpenStudy (rsadhvika):

if so, we need to work it from scratch i would guess

OpenStudy (texaschic101):

does that mean there is 24 players

OpenStudy (amistre64):

i would say that there is a way to solve it no matter what your approach is; just as long as you are consistent in your reasoning/interpretation of the information provided.

OpenStudy (anonymous):

the answer is the 9

OpenStudy (rsadhvika):

what you've taken is a simple case which can be worked easily - 12 matches, 24 players but if we let all combinations with all players it gets nasty

OpenStudy (texaschic101):

my brain hurts

OpenStudy (anonymous):

if n is the number of boys then 2 n is the number of girls Let N be the total number of matches

OpenStudy (anonymous):

using combinatrics we see that \[N=\left(\begin{matrix}3n \\ 2\end{matrix}\right)=\frac{ 3n(3n-1) }{ 2 }\]

OpenStudy (rsadhvika):

N = 3n(3n-1) / 2 or N = 3n/2 ? which interpretation is correct ?

OpenStudy (anonymous):

yes 1st one

OpenStudy (anonymous):

\[\text{ Let} N_{ gw} \text{be the number of matches won by a girl}\]

OpenStudy (anonymous):

Since the ratio between the number of matches won by girls and the number of matches won by boys is 5:7 \[\frac{ N_{gw} }{ N-N_{gw} }=\frac{ 5 }{ 7 } \text{ hence } N_{gw}=\frac{ 5 }{ 12 }N=\frac{ 5 }{ 8 }n(3n-1)\]

OpenStudy (anonymous):

The number of matches played between two girls is \[\left(\begin{matrix}2n \\ n\end{matrix}\right)=\frac{ 2n(2n-1) }{ 2 }=n(2n-1)\]

OpenStudy (anonymous):

Since in these matches always girls win, we have \[n(2n-1)\le \frac{5}{8}n(3n-1)\] Solving yields \[n\le 3\] Further, 8 divides \[5n(3n−1)\], therefore n= 3. Thus, there were 9 players.

OpenStudy (anonymous):

do you get the solution guys ,cos i kinda get it but not fully ...can any1 expand it?

OpenStudy (anonymous):

@amistre64 mention 9 tems somehow and experiment attemted the desired solution ,,,,

OpenStudy (anonymous):

qoute of the day @texaschic101 "my brain hurts" lol

OpenStudy (texaschic101):

lol.....

OpenStudy (amistre64):

without knowing how the author of the question is defining the win-lose-rematch structure ... and without having the necessary presupposed rules for such a scenario ... i wouldnt know how to verify the results of "9" people in the tournament

OpenStudy (anonymous):

but theres no rematch,theres win-loose- and no draws so only win or lose

OpenStudy (experimentx):

perhaps you are finding the min num of playes

OpenStudy (amistre64):

9 players do not make even match ups, there will be one person waiting to play how do we determine who the 9th player competes against?

OpenStudy (anonymous):

but one ammusing thing is that girls can play against girls

OpenStudy (experimentx):

the num of boys should be even so should be twice of it. if you multiply it by 2, i think it is also solution of the problem.

OpenStudy (amistre64):

theres some intricate win-lose-rematch structure if we are to assume a 9 member competition

OpenStudy (amistre64):

and by rematch i dont mean 2 people play against each other; but rather there is a re-matching of the players

OpenStudy (anonymous):

do you consider the solution above validated and reputable

OpenStudy (rsadhvika):

everyone is playing with everyone else, so how does even or odd matter here

OpenStudy (anonymous):

OpenStudy (amistre64):

i cant verify the process without knowing how the matching of pairs works in the win-lose structure p1 <-> p2 p3 <-> p4 p5 <-> p6 p7 <-> p8 p9 <-> [p9?]

OpenStudy (experimentx):

seems like one pair meant one player can play more than one matches, i misunderstood it.

OpenStudy (anonymous):

okay so i will check how tunnis is played first,but we are not supposed to be tested on games this is a contest question not english and life skills...??? i guess this is not a good question

OpenStudy (anonymous):

thanx 4 de effort

OpenStudy (experimentx):

No no ... the question is okay, i seem to have presumption abt the question which was not valid. the solution seems fine.

OpenStudy (anonymous):

okay so its just the logic...and understanding game structure...wich is not mentioned in the solution!!

OpenStudy (amistre64):

is the 5 boys to 7 girls ratio a ratio of different people? or is the same person being counted as a gender win for multiple wins?

OpenStudy (experimentx):

except the typo number of matches played between girls is 2nC2

OpenStudy (anonymous):

yes i so that

OpenStudy (anonymous):

thats why i typed a different solution,above

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