Let $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ be non-negative real numbers, such that $x_{1}+x_{2}+x_{3}+x_{4}+x_{5}=100, M$ is the maximum of $x_{1}+x_{2}$, $x_{2}+x_{3}$, $x_{3} \div x_{4}$, and $x_{1}+x_{3}$, then the minimum value $m$ of $M$ satisfies
(A) $0 \leqslant m<32$
(B) $32 \leqslant m<34$
(C) $34 \leqslant m<36$
(D) $36 \leqslant m<38$
5. (B).

From the problem, we know: $M \geqslant x_{1}+x_{2}, M \geqslant x_{2}+x_{1}, M \geqslant x_{3}+x_{4}, M \geqslant x_{4}+x_{5}$, so
$$
\begin{array}{c}
3 M \geqslant\left(x_{1}+x_{2}\right)+\left(x_{2}+x_{3}\right)+\left(x_{4}+x_{5}\right) \\
=100+x_{2} \geqslant 100, \\
M \geqslant \frac{100}{3} .
\end{array}
$$

Furthermore, when $x_{1}=x_{3}=x_{5}=\frac{100}{3}, x_{2}=x_{1}=0$, we have $M=\frac{100}{3}$, hence $m=\frac{100}{3}$.