【Answer】 40
【Analysis】If $p$ is greater than $c$, $p-c=\frac{a+b+c+d+e}{5}-c=\frac{a+b-4c+d+e}{5}$. To maximize $p-c$, $a+b-4c+d+e$ must be maximized, which means $a, b, d, e$ should be as large as possible. Among them, $d, e$ can be 100, but the upper limit of $a, b$ cannot exceed $c$. $a+b-4c+d+e \leq a+b-4c+200 \leq c+c-4c+200=200-2c$, the maximum value is when $c$ is 0, at this time $a=b=c=0$. The maximum value of $p-c$ is $\frac{0+200}{5}=40$.
Similarly, when $c$ is greater than $p$, the maximum value of $c-p$ is also 40 when $a=b=0, c=d=e=100$. Therefore, the maximum difference between $p$ and $c$ is 40.
10. There are 5 students participating in a test, and each student's score $a, b, c, d, e$ is an integer, with the score range being $0 \sim 100$, including 0 points and 100 points. Given that $a \leq b \leq c \leq d \leq e$, let the average score of the 5 students be $p$, then the maximum value of $p \vdash_{j} c$ is $\qquad$ .