471. A motorcyclist left A for B and at the same time a pedestrian set off from B to A. Upon meeting the pedestrian, the motorcyclist gave him a ride, brought him to A, and immediately set off again for B. As a result, the pedestrian reached A 4 times faster than he would have if he had walked the entire way. How many times faster would the motorcyclist have arrived in B if he had not had to return?
$\triangle$ Let's take the path AB as a unit. Denote the speeds of the pedestrian and the motorcyclist as $v_{1}$ and $v_{2}$ (in fractions of this unit per hour).

Suppose the pedestrian and the motorcyclist met at point C. Since they moved for $\frac{1}{v_{1}+v_{2}}$ hours before meeting, we have

$$
A C=v_{2} \cdot \frac{1}{v_{1}+v_{2}}=\frac{v_{2}}{v_{1}+v_{2}}, \quad B C=\frac{v_{1}}{v_{1}+v_{2}}
$$

Let's calculate the time the pedestrian spent on the path BA - path BC on foot, and path CA - on the motorcycle:

$$
\frac{v_{1}}{v_{1}+v_{2}}: v_{1}+\frac{v_{2}}{v_{1}+v_{2}}: v_{2}=\frac{1}{v_{1}+v_{2}}+\frac{1}{v_{1}+v_{2}}=\frac{2}{v_{1}+v_{2}} \text {. }
$$

Since this time is 4 times less than the time $\frac{1}{v_{1}}$ it would take him to walk the entire path AB, we have

$$
\frac{1}{v_{1}}=\frac{8}{v_{1}+v_{2}}
$$

from which

$$
v_{1}+v_{2}=8 v_{1}, \quad v_{2}=7 v_{1}
$$

Let's find the time the motorcyclist spent on his entire path $2 \mathrm{AC}+\mathrm{AB}$ :

$$
\left(2 \frac{v_{2}}{v_{1}+v_{2}}+1\right): v_{2}=\frac{v_{1}+3 v_{2}}{v_{2}\left(v_{1}+v_{2}\right)}
$$

Now let's calculate how many times this time is greater than the time $\frac{1}{v_{2}}$ it would take the motorcyclist to travel the path AB if he did not return:

$$
\frac{v_{1}+3 v_{2}}{v_{2}\left(v_{1}+v_{2}\right)}: \frac{1}{v_{2}}=\frac{v_{1}+3 v_{2}}{v_{1}+v_{2}}=\frac{v_{1}+21 v_{1}}{v_{1}+7 v_{1}}=\frac{22}{8}=\frac{11}{4}=2.75
$$

Answer: 2.75 times.