20.10. a) In a chess tournament, two students from grade 7 and a certain number of students from grade 8 participated. The two seventh-graders scored 8 points, and each of the eighth-graders scored the same number of points. How many eighth-graders participated in the tournament? Find all possible solutions.

b) In a chess tournament, students from grades 9 and 10 participated. There were 10 times more tenth-graders than ninth-graders, and they scored 4.5 times more points in total than all the ninth-graders. How many points did the ninth-graders score?
20.10. a) Answer: 7 or 14. Let $x$ be the number of eighth-graders, and $y$ be the number of points scored by each eighth-grader. By calculating the total points scored by all participants in the tournament in two ways, we arrive at the equation

$$
x y + 8 = \frac{(x+2)(x+1)}{2}
$$

i.e.,

$$
2 y = \frac{(x+2)(x+1) - 16}{x} = x + 3 - \frac{14}{x}
$$

Therefore, $x$ takes one of the values $1, 2, 7, 14$. The values 1 and 2 are eliminated, as in these cases the number $y$ would be negative. The problem has two solutions: $x = 7$ and $x = 14$.

b) Answer: 10. Let $x$ be the number of ninth-graders in the tournament. Then there were $11x$ participants in total, and they scored $\frac{11x(11x-1)}{2}$ points. According to the problem, the ratio of the number of points scored by ninth-graders to the number of points scored by tenth-graders is $1:4.5$. Therefore, the ninth-graders scored $x(11x-1)$ points, which means each ninth-grader won all $11x-1$ games they played. However, if there were two ninth-graders among the participants, they would have had to both win their game against each other, which is impossible. Therefore, only one ninth-grader participated in the tournament; he scored 10 points.