Find the greatest possible value for the difference in ratings. Seven experts evaluate the picture. Each of them makes an assessment - an integer number of points from 0 to 10 inclusive. It is known that all experts have put different points. According to the old rating system, the rating of the  picture is the arithmetic mean of all expert points. According to the new rating system, the rating of the picture is estimated as follows: the smallest and largest points are discarded and the arithmetic mean of the remaining points is calculated. Find the greatest possible value for the difference in ratings calculated from the old and new assessment systems.
I find seven points 0, 1, 2, 3, 4, 5, 10 for which this difference is $\frac{4}{7}$ but I can't prove that it is maximum value.
 A: Let the seven ratings be $a,b,c,d,e,f,g$ where, WLOG, $a<b<c<d<e<f<g$.
The old system would give $\frac{a+b+c+d+e+f+g}{7}$, whereas the new one would give $\frac{b+c+d+e+f}{5}$
We would like to maximize the absolute difference between the two, which is given by
\begin{align}
\ & |\frac{a+b+c+d+e+f+g}{7}-\frac{b+c+d+e+f}{5}| \\
\ = & |\frac{a+g}{7}-\frac{2(b+c+d+e+f)}{35}| \\
\end{align}
WLOG, let's say we make $\frac{a+g}{7}$ as large as possible, and $\frac{2(b+c+d+e+f)}{35}$ as small as possible.
Then we would want $g=10$. And whatever value that $a$ takes, the smallest possible value for $\frac{2(b+c+d+e+f)}{35}$ would occur when $b=a+1, c=a+2, d=a+3, e=a+4, f=a+5$. And now it is just a matter of directly computing the cases for when $a=0,1,2,3,4$.
It turns out that,
when a=0, the difference is $\frac47$
when a=1, the difference is $\frac37$
when a=2, the difference is $\frac27$
when a=3, the difference is $\frac17$
when a=4, the difference is $0$.
So the answer is indeed $\frac47$
Of course, a more elegant way of doing this would be to say that $$\frac{a+(a+1)+(a+2)+(a+3)+(a+4)+(a+5)+10}{7}=\frac{6a+25}{7},$$ and that
$$\frac{(a+1)+(a+2)+(a+3)+(a+4)+(a+5)}{5}=\frac{5a+15}{5},$$ where $\frac{6a+25}{7} - \frac{5a+15}{5} = \frac{4-a}{7}$ so that the smallest value of $a$, namely $a=0$, gives the largest difference.
A: The highest score $x_{(7)}$ must be at least $6$ more than the lowest $x_{(1)}$ if all the scores are to be distinct integers.
The absolute change from the original average to the truncated average is 
$$\left| \dfrac{x_{(1)}+x_{(2)}+x_{(3)}+x_{(4)}+x_{(5)}+x_{(6)}+x_{(7)}}{7} 
 - \dfrac{x_{(2)}+x_{(3)}+x_{(4)}+x_{(5)}+x_{(6)}}{5}  \right|$$ 
$$= \frac17\left| x_{(1)}+x_{(7)} 
 - 2 \left(\dfrac{x_{(2)}+x_{(3)}+x_{(4)}+x_{(5)}+x_{(6)}}{5}\right)  \right|$$
but the average of the five remaining points must be at least $x_{(1)}+3$ and can be no more than $x_{(7)}-3$, since the values are distinct integers.  
This means that absolute change is bounded above by $\frac17\left( x_{(7)} - x_{(1)}-6  \right)$.  Since $0 \le x_{(1)}$ and $x_{(7)}\le 10$, this in turn is bounded above by $\frac47$.  
This is the change in your example, so that gives the largest possible change.  
$0,5,6,7,8,9,10$ would also achieve the bound.   
