$1 \cdot 29$ If $\frac{x}{3}=\frac{y}{-4}=\frac{z}{7}$, then the value of $\frac{3 x+y+z}{y}$ is
(A) 0.
(B) -1.
(C) -2.
(D) -3.
(E) 3.
(China Junior High School Mathematics Correspondence Competition, 1985)
[Solution] $\because \quad \frac{x}{3}=\frac{y}{-4}=\frac{z}{7}, \quad \therefore \quad \frac{3 x}{9}=\frac{y}{-4}=\frac{z}{7}$.
$\therefore$ By the property of equal ratios, $\frac{3 x+y+z}{9-4+7}=\frac{y}{-4}$,
$\therefore \quad \frac{3 x+y+z}{y}=\frac{12}{-4}=-3$.
Therefore, the answer is $(D)$.