3.17 A man sold two cigarette holders, each for $1.20, and on one he gained $20 \%$, while on the other he lost $20 \%$. On the whole sale, he
(A) broke even.
(B) lost 4 cents.
(C) gained 4 cents.
(D) lost 10 cents.
(E) gained 10 cents.
(7th American High School Mathematics Examination, 1956)
[Solution] Let the cost prices of the two cigarette holders be $C_{1}, C_{2}$; the selling prices be $S_{1}, S_{2}$. According to the problem,
$$
\begin{array}{l}
S_{1}=C_{1}+\frac{1}{5} C_{1}, \quad S_{2}=C_{2}-\frac{1}{5} C_{2}, \quad \text { and } S_{1}=S_{2} . \\
\therefore \quad C_{1}=\frac{5}{6} S_{1}, \quad C_{2}=\frac{5}{4} S_{2} .
\end{array}
$$

Thus $C_{1}+C_{2}=\frac{5}{6} S_{1}+\frac{5}{4} S_{2}=\left(\frac{5}{6}+\frac{5}{4}\right) \cdot 1.20=2.50$.
And
$$
S_{1}+S_{2}=2.40 \text {. }
$$

Therefore, the answer is $(D)$.