$1 \cdot 120 \mathrm{Al}$'s age is 16 years more than the sum of $\mathrm{Bob}$ and $\mathrm{Carl}$'s ages (hereinafter referred to as $\mathrm{A}, \mathrm{B}, \mathrm{C}$). $\mathrm{A}$'s age squared is 1632 more than the square of the sum of $\mathrm{B}$ and $\mathrm{C}$'s ages. The sum of $\mathrm{A}, \mathrm{B}$, and $\mathrm{C}$'s ages is 
(A) 64.
(B) 94.
(C) 96.
(D) 102.
(E) 140.
(28th American High School Mathematics Examination, 1977)
[Solution] Let $a, b$ and $c$ represent the ages of $A l, B o b$ and $C a r l$ respectively. Then
$$
\begin{aligned}
a+b+c & =\frac{(a+b+c)[a-(b+c)]}{a-(b+c)} \\
& =\frac{a^{2}-(b+c)^{2}}{a-(b+c)}=\frac{1632}{16}=102 .
\end{aligned}
$$

Therefore, the answer is $(D)$.