8. In Sergio's class, after correcting the last math assignment, to which all students were present, the arithmetic mean of the failures was 4.6, while the arithmetic mean of the passing grades was 7.1. Knowing that the teacher gave only integer grades, what is the minimum number of students in Sergio's class?
(A) 10
(B) 12
(C) 15
(D) 24
(E) 30.
8. The answer is (C). Let $a$ be the sum of the failing grades, $b$ the sum of the passing grades, $m$ the number of failures in the class, and $n$ the number of passes in the class. The average of the failing grades is $\frac{a}{m}$ and that of the passing grades is $\frac{b}{n}$. We can assume $a, b, m, n > 0$. Moreover, the hypothesis ensures that $a, b$ are natural numbers.

$\frac{a}{m}=4.6=\frac{23}{5} \quad \Rightarrow \quad 23 m=5 a ; \quad \frac{b}{n}=7.1=\frac{71}{10} \quad \Rightarrow \quad 71 n=10 b ;$

Since 5 does not divide $23, m$ will be a multiple of 5. Similarly, $n$ will be a multiple of 10. At a minimum, $m=5$ and $n=10$, thus a total of 15 students. It is easy to verify that there can be a situation like this that meets our assumptions, for example the set of grades $4,4,5,5,5,7,7,7,7,7,7,7$, $7,7,8$.