Tournament's problem Suppose that we have boys and girls who competes in a tournament where each of the person will play against each of the person exactly one time and if he wins then he gets $1$ point if he loses then he gets $0$ point and if the result of game is draw then both of players get $0.5$ point.
We know that quantity of boys is three times bigger than quantity of girls.
And we know that after the end of tournament overall points of boys are $20\%$ bigger than girls overall points.
We need to find how many people could participate in this tournament.
Let $x$ is quantity of girls and then $3x$ is quantity of boys, $4x$ is quantity of people.
I know that overall points of all people is $\frac{4x(4x-1)}{2} = S$
Boys got $20\%$ more points so if $y$ are points which were got by girls then we have
$y+1.2y=S \Rightarrow y = \frac{10x(4x-1)}{11}$.
At most girls can get $x\cdot3x + \frac{x(x-1)}{2}$ points, because they can win all boys (each girl wins against each boy) and they need to distribute other points beetwen them.
So we get inequaltiy: $\frac{10x(4x-1)}{11} \leq x\cdot3x + \frac{x(x-1)}{2}$ and we get from it that $x = 1$ and overall quantity of people is $4$.
I think that there is an error (or errors) in my solution.
Can somebody help me find them?
UPDATE.
From inequality $\frac{10x(4x-1)}{11} \leq x\cdot3x + \frac{x(x-1)}{2}$ we get that $0 \leq x \leq 3$.
So quantity of people can be $4$, $8$ or $12$.
Am i right now?
 A: Addendum added to respond to the comment/question of
Arty.

Thanks to Ross Millikan for pointing out:

*

*The obvious mistake of my confusing the total number of points scored with the total number of points scored by all of the girls.


*The more subtle mistake that the average score of all of the girls must not be greater than what their score would be, if every girl beat every boy.

If $x$ girls scored $P$ points, then $3x$ boys scored $1.2P$ points, so $2.2P$ points were scored in total.
Let $~Q = 2.2P = ~$ be the total number of points scored.
Then, the total score of all the boys is $~\dfrac{6Q}{11}~$ and the total score of all the girls is $~\dfrac{5Q}{11},~$ so $~Q~$ must be a multiple of $~11.~$
Also, each game generated $1$ point, distributed as either [1,0] or [1/2,1/2].  Therefore $~Q~ = ~$ the number of games played.
Since there were $~x~$ girls and $~3x~$ boys, you have that
$$Q = \binom{4x}{2} = (2x) \times (4x - 1).$$
So, $~Q~$ must be chosen so that there
exists a positive integer $x$ such that 
$(2x) \times (4x-1) = Q.$
So, either $2x$ is a multiple of $11$ or $(4x - 1)$ is a multiple of $11$.
The smallest positive integer $x$ that can serve as a $\color{red}{\text{candidate value}}$ is $x = 3.$
Exploring this $\color{red}{\text{candidate value}}$:

*

*The total number of games played is 
$\displaystyle \binom{12}{2}~ = 66 = Q.$


*The total number of girls is $~3~$ and the total number of boys is $~9.$


*The total number of points collectively scored by all of the girls is $~\dfrac{5Q}{11} = 30.$ 
So, the average score of each girl was $~10.$


*The total number of points collectively scored by all of the boys is $~\dfrac{6Q}{11} = 36.$ 
So, the average score of each boy was $~4.$
$\color{red}{\text{Does this work?}}$
Suppose that every girl beat all 9 boys.
Further suppose that every girl:girl game and every boy:boy game ended in a draw.
Then each girl would score 
$\displaystyle 9  + \left[2 \times \frac{1}{2}\right] = 10,$
and each boy would score 
$\displaystyle 0  + \left[8 \times \frac{1}{2}\right] = 4.$
So, the $\color{red}{\text{candidate value}}$ of $x = 3$ works okay.

Addendum 
Responding to the comment/question of Arty.
First of all, your overall approach was $\color{red}{\text{better}}$ than my $\color{red}{\text{eventual}}$ approach, in one way, and flawed in another way.
That is:

*

*You elegantly confronted the issue that the total score of the girls, assuming that each girl beat each boy, had to be at least $~\dfrac{5}{11}~$ of the total games played. 
Initially, I totally overlooked that issue.  Then, after Ross Millikan commented, I edited my answer after the fact.  However, I did not try for the elegant (and better) approach of establishing an inequality involving $~x.~$ 
Instead, I (inelegantly) used the notion of a candidate value to verify that my solution worked.


*You totally overlooked the constraint that the number of games played, $~\dfrac{4x(4x-1)}{2}~$ has to be a multiple of $~11.~$  This is because the total number of points scored by the girls is 
$\displaystyle y = \frac{5}{11} \times \frac{4x(4x-1)}{2}.$

$\displaystyle \frac{10x(4x - 1)}{11} \leq [x \times 3x] + \frac{x(x-1)}{2}.$

I agree with your updated interpretation of the above inequality.  That is you have that:
$\displaystyle \frac{40x^2 - 10x}{11} \leq 3x^2 + \frac{x^2 - x}{2} = \frac{7x^2 - x}{2}.$
Cross multiplying, this implies that
$\displaystyle 80x^2 - 20x \leq 77x^2 - 11x.$
Since $x$ must be positive, (so $x$ is non-zero), you can divide the above inequality through by $x$ to give
$80x - 20 \leq 77x - 11 \implies 3x \leq 9 \implies x \leq 3.$
This begs the question, which value of $x$ is satisfactory.
You need $~\displaystyle \frac{4x(4x-1)}{2} = (2x) \times (4x-1)~$ to be a multiple of $11$.
Of the candidate values given by $~x \in \{1,2,3\}~$, only $~x=3~$ satisfies this constraint.
