3.53 $X, Y, Z$ are pairwise disjoint sets of people, and the average age of the people in each set is given in the table below:
Then the average age of the people in the set $X \cup Y \cup Z$ is
(A) 33.
(B) 33.5.
(C) 33.6.
(D) 33.83.
(E) 34.
(39th American High School Mathematics Examination, 1988)
[Solution] If $|X|$ represents the number of elements in $X$, let $|X|=x,|Y|=y$, $|Z|=z$. By the problem's assumption, $X, Y, Z$ are pairwise disjoint, so
$$
|X \cup Y|=x+y,|X \cup Z|=x+z,|Y \cup Z|=y+z \text {. }
$$

According to the problem, we have the system of equations
$$
\left\{\begin{array}{l}
\frac{37 x+23 y}{x+y}=29, \\
\frac{37 x+41 z}{x+z}=39.5, \\
\frac{23 y+41 z}{y+z}=33 .
\end{array}\right.
$$

Solving these equations, we get $y=\frac{4}{3} x, \quad z=\frac{5}{3} x$.
Thus, the average age of $X \cup Y \cup Z$ is
$$
\frac{37 x+23 y+41 z}{x+y+z}=\frac{37 x+23\left(\frac{4}{3} x\right)+41\left(\frac{5}{3} x\right)}{x+\left(\frac{4}{3} x\right)+\left(\frac{5}{3} x\right)}=34 \text {. }
$$

Therefore, the answer is $(E)$.