3.26 An airplane flying straight between two towns against the wind took 84 minutes, and the return trip with the wind took 9 minutes less than it would have in still air. The number of minutes for the return trip is (two answers)
(A) 54 or 18.
(B) 60 or 15.
(C) 63 or 12.
(D) 72 or 36.
(E) 75 or 20.
(24th American High School Mathematics Examination, 1973)
[Solution] Let $d, v, w$ represent the distance between the two towns, the speed against the wind, and the speed with the wind, respectively. Then, from $\frac{d}{v}=84$, we have $d=84 v$, so
$$
\frac{d}{w}=\frac{d}{\frac{1}{2}(w+v)}-9, \text{ or } 84 \frac{v}{w}=\frac{84 v}{\frac{1}{2}(w+v)}-9.
$$

Let $x=\frac{v}{w}$, the above equation can be simplified to $28 x=\frac{56}{\frac{1}{x}+1}-3$,
which is $28 x^{2}-25 x+3=0$, or $(4 x-3)(7 x-1)=0$,
solving for $x$ gives
$$
x=\frac{3}{4} \text { or } \quad \frac{1}{7}.
$$

The time in minutes for the distance $d$ to be flown at the speed with the wind $w$ is
$\frac{d}{w}=84 \frac{v}{w}=84 \cdot \frac{3}{4}$ or $84 \cdot \frac{1}{7}$, which is 63 or 12.
Therefore, the answer is $(C)$.