$3 \cdot 24$ A train departed and after 1 hour it stopped for half an hour due to an incident, then proceeded at $\frac{3}{4}$ of its original speed, resulting in a delay of $3 \frac{1}{2}$ hours. If the incident had occurred 90 miles further ahead, the train would have been only 3 hours late. The total distance of the journey in miles is
(A) 400.
(B) 465.
(C) 600.
(D) 640.
(E) 550.
(5th American High School Mathematics Examination, 1954)
［Solution］Let the total distance be $x$ miles, and the original speed be $v$, then we have
$$
\left\{\begin{array}{l}
\frac{x-v}{\frac{3}{4} v}+1+\frac{1}{2}-\frac{x}{v}=3 \frac{1}{2}, \\
\frac{x-v-90}{\frac{3}{4} v}+1+\frac{1}{2}+\frac{90}{v}-\frac{x}{v}=3,
\end{array}\right.
$$

Subtracting the two equations yields $\frac{\frac{90}{3}}{\frac{3}{4} v}-\frac{90}{v}=\frac{1}{2}$, solving for $v$ gives $v=60$.
Substituting $v$ into the first equation gives
$$
x=600 \text {. }
$$

Therefore, the answer is $(C)$.