8. Every day, Xiaoming has to pass through a flat section $A B$, an uphill section $B C$, and a downhill section $C D$ (as shown in the right figure). It is known that $A B: B C: C D=1: 2: 1$, and the speed ratio of Xiaoming on the flat, uphill, and downhill sections is $3: 2: 4$. If the time ratio for Xiaoming to go to school and return home is $\frac{n}{m}$ (where $\mathrm{m}$ and $\mathrm{n}$ are coprime natural numbers), then the value of $\mathrm{m}+\mathrm{n}$ is $\qquad$
Answer: 35.
The simple assignment method can calculate the result. Let $\mathrm{AB}$ be 100 meters, then the distance of $\mathrm{BC}$ is 200 meters, and the distance of $\mathrm{CD}$ is 100 meters; assume Xiao Ming's speed on flat ground is 3 meters/second, then his speed on uphill and downhill sections is 2 meters/second and 4 meters/second, respectively. According to the assignment, 
the time for Xiao Ming to go to school is: $\frac{100}{3}+\frac{200}{2}+\frac{100}{4}=\frac{475}{3}$
the time for Xiao Ming to return home from school is: $\frac{100}{3}+\frac{200}{4}+\frac{100}{2}=\frac{400}{3}$
Therefore, the time ratio is $\frac{475}{3}: \frac{400}{3}=\frac{19}{16}!$
If you feel that the assignment method is not rigorous enough, then you can set $A B=s$, then $B C=2 s, C D=s$; set $v_{\text {flat }}=3 a$, then $v_{\text {up }}=2 a, v_{\text {down }}=4 a$. You can calculate these expressions to get:
$$
\begin{array}{l}
t_{\text {to school }}=\frac{s}{3 a}+\frac{2 s}{2 a}+\frac{s}{4 a}=\frac{4+12+3}{12} \times \frac{s}{a}=\frac{19}{12} \times \frac{s}{a}, \\
t_{\text {return home }}=\frac{s}{3 a}+\frac{2 s}{4 a}+\frac{s}{2 a}=\frac{4+6+6}{12} \times \frac{s}{a}=\frac{16}{12} \times \frac{s}{a},
\end{array}
$$

Therefore,
$$
\frac{t_{\text {to school }}}{t_{\text {return home }}}=\frac{19}{16}=\frac{n}{m}, \mathrm{~m}+\mathrm{n}=35 \text { . }
$$