$3 \cdot 30$ Three people, A, B, and C, are making a 100-mile journey. A and C travel at a speed of 25 miles per hour by car, while B walks at a speed of 5 miles per hour. After a certain distance, C gets out of the car and walks at 5 miles per hour, and A drives back to pick up B, so that A and B arrive at the destination at the same time as C. The time taken for this journey is
(A) 5.
(B) 6.
(C) 7.
(D) 8.
(E) None of the above.
(2nd American High School Mathematics Examination, 1951)
[Solution] Let $t_{1}, t_{2}, t_{3}$ represent the time for the car to move forward, turn back to pick up person B, and then arrive at the destination, respectively. Then,
$$
\left\{\begin{array}{cl}
25 t_{1}-25 t_{2}+25 t_{3}=100 & \text { (for the car) }, \\
5 t_{1}+5 t_{2}+25 t_{3}=100 & \text { (for person B) } \\
25 t_{1}+5 t_{2}+5 t_{3}=100 & \text { (for person C) } .
\end{array}\right.
$$

Solving the system of equations yields $t_{1}=3, t_{2}=2, \quad t_{3}=3$.
From
$$
t_{1}+t_{2}+t_{3}=8 \text {. }
$$

Therefore, the answer is $(D)$.