Solution:

The answer is (C). Let $a$ be the sum of the failing grades, $b$ the sum of the passing grades, $m$ the number of failing grades in the class, $n$ the number of passing grades in the class. The mean of the failing grades is $\frac{a}{m}$ and that of the passing grades $\frac{b}{n}$. We can take $a, b, m, n > 0$. Moreover the hypothesis assures us that $a, b$ are two natural numbers.

$\frac{a}{m} = 4,6 = \frac{23}{5} \quad \Rightarrow \quad 23m = 5a ; \quad \frac{b}{n} = 7,1 = \frac{71}{10} \quad \Rightarrow \quad 71n = 10b ;$

Since 5 does not divide $23$, $m$ will be a multiple of 5. Likewise $n$ will be a multiple of 10. At minimum, $m = 5$ and $n = 10$, giving a total of 15 students. It is easy to verify that such a situation respecting our hypotheses can occur, for example the set of grades $4, 4, 5, 5, 5, 7, 7, 7, 7, 7, 7, 7, 7, 7, 8$.