3. A cyclist traveled the path from A to B and back at a constant speed. A pedestrian walked the path from A to B at a speed three times slower than the cyclist's speed, but on the way back, he took a bus, which traveled at a speed five times greater than the cyclist's speed. How much time did the pedestrian spend on the round trip, if one of them was on the way 36 minutes longer than the other.
Solution:

$$
v_{\mathrm{B}}=x, t_{\mathrm{B}}=\frac{S}{x}+\frac{S}{x}=\frac{2 S}{x}
$$

$v_{n} \cdot A B=\frac{x}{3}, v_{n} \cdot \mathrm{BA}=5 x$, then

$$
t_{n}=\frac{S}{x / 3}+\frac{S}{5 x}=\frac{3 S}{x}+\frac{S}{5 x}=\frac{15 S+S}{5 x}=\frac{16 S}{5 x}
$$

We get $\frac{16 S}{5 x}-\frac{2 S}{x}=\frac{36}{60} ; \frac{6 S}{5 x}=\frac{3}{5}$, therefore,

$\frac{s}{x}=\frac{1}{2}$ then $t_{\mathrm{B}}=2 \cdot \frac{s}{x}=2 \cdot \frac{1}{2}=1$ hour

$t_{n}=1$ hour $+\frac{3}{5}=1.6$ hours.

Answer: 1.6 hours.