19. Given two fair dice, each with at least six faces, each face has a different number, and is labeled with 1, $2, \cdots, n$ (where $n$ is the number of faces on the die). When the two dice are rolled simultaneously, the probability that their sum is 7 is $\frac{3}{4}$ of the probability that their sum is 10, and is $\frac{1}{12}$ of the probability that their sum is 12. The minimum sum of the number of faces on the two dice is ( ).
(A) 16
(B) 17
(C) 18
(D) 19
(E) 20
19. B.

Assume the number of faces on the two dice are $a$ and $b$ ($a > b$).
Since each die has at least six faces, the number of ways to get a sum of 7 is 6, so the number of ways to get a sum of 10 is $6 \div \frac{3}{4} = 8$.

When $a > b \geq 9$, the number of ways to get a sum of 10 is at least 9.
Since the number of ways to get a sum of 10 is 8, then,
$$
b=8, a \geq 9 \text{. }
$$

Let $N$ be the number of ways to get a sum of 12. Then
$$
\frac{N}{8a} = \frac{1}{12} \Rightarrow N = \frac{2}{3}a \text{. }
$$

Given $b=8, N \leq 8, 3 \mid a$, we know $a=9,12$.
Therefore, the minimum value sought is $8+9=17$.