$1 \cdot 35$ If $\frac{m}{n}=\frac{4}{3}$ and $\frac{r}{t}=\frac{9}{14}, \frac{3 m r-n t}{4 n t-7 m r}$ is
(A) $-5 \frac{1}{2}$.
(B) $-\frac{11}{14}$.
(C) $-1 \frac{1}{4}$.
(D) $\frac{11}{14}$.
(E) $-\frac{2}{3}$.
(5th American High School Mathematics Examination, 1954)
[Solution]
$$
\begin{aligned}
\text { Original expression } & =\frac{\frac{3 m r}{n t}-\frac{n t}{n t}}{\frac{4 n t}{n t}-\frac{7 m r}{n t}}=\frac{3 \frac{m r}{n t}-1}{4-7 \frac{m r}{n t}} \\
& =\frac{3 \cdot \frac{4}{3} \cdot \frac{9}{14}-1}{4-7 \cdot \frac{4}{3} \cdot \frac{9}{14}} \\
& =-\frac{11}{14} .
\end{aligned}
$$

Therefore, the answer is $(B)$.