3. (10 points) Person A and Person B start riding bicycles from place $A$ to place $B$ at the same time. Person A's speed is 1.2 times that of Person B. After riding 5 kilometers, Person B's bicycle breaks down, and the downtime is equivalent to the time it would take to ride $\frac{1}{6}$ of the entire journey. After fixing the bicycle, Person B's speed increases by 60%, and as a result, both Person A and Person B arrive at place $B$ at the same time. The distance between $A$ and $B$ is $\qquad$ kilometers.
【Analysis】According to the problem, the speed ratio of A and B is $1.2: 1=6: 5$, so the time ratio is $5: 6$. Let's assume A takes $5 t$, then B's original time is $6 t$. The delay time due to the fault for B is $\frac{1}{6} \times 6 t=t$, and the total time taken for the entire journey is $5 t$. Therefore, after the fault is resolved, B's increased speed saves $2 t$ of time. The speed ratio after the increase is $1.6: 1=8: 5$, so the time ratio is $5: 8$, saving three parts of time, so each part is $\frac{2}{3} t$. Therefore, the original planned time for this section is $\frac{2}{3} t \times 8=\frac{16}{3} t$, so the original planned time for the initial 5 kilometers is $6 t-\frac{16}{3} t=\frac{2}{3} t$. Therefore, the distance between A and B is $5 \times\left(6 t \div \frac{2}{3} t\right)$, and then we can calculate it.

【Solution】Solution: The speed ratio of A and B is $1.2: 1=6: 5$, so the time ratio is $5: 6$; assume A takes $5 t$, then B's original time is $6 t$;

The delay time due to the fault for B is $\frac{1}{6} \times 6 t=t$, and the total time taken for the entire journey is $5 t$, so after the fault is resolved, B's increased speed saves $2 t$ of time.

The speed ratio after the increase is $1.6: 1=8: 5$, so the time ratio is $5: 8$, saving three parts of time, so each part is $\frac{2}{3} t$,
Therefore, the original planned time for this section is $\frac{2}{3} t \times 8=\frac{16}{3} t$, so the original planned time for the initial 5 kilometers is $6 t-\frac{16}{3} t=\frac{2}{3} t$, so the distance between A and B is:
$$
\begin{array}{l}
5 \times\left(6 t \div \frac{2}{3} t\right), \\
=5 \times 9, \\
=45 \text { (km); }
\end{array}
$$

Therefore, the answer is: 45.