13. (15 points) A car travels from location A to location B. If the speed is increased by $20 \%$, it can arrive 25 minutes earlier; if it travels 100 kilometers at the original speed and then increases the speed by $25 \%$, it can arrive 10 minutes earlier. Find the distance between A and B.
【Solution】Solution: The original time is:
$$
\begin{array}{l}
25 \div[1-1 \div(1+20 \%)], \\
=25 \div\left[1-\frac{5}{6}\right], \\
=25 \div \frac{1}{6}, \\
=25 \times 6=150 \text { (minutes); }
\end{array}
$$

The time after traveling 100 kilometers is:
$$
\begin{array}{l}
10 \div[1-1 \div(1+25 \%)], \\
=10 \div\left[1-\frac{4}{5}\right], \\
=10 \div \frac{1}{5} \\
=10 \times 5, \\
=50 \text { (minutes); }
\end{array}
$$

The time required to travel 100 kilometers at the original speed is:
$$
150-50=100 \text { (minutes), }
$$

The distance is:
$$
150 \div 100 \times 100=150 \text { (kilometers). }
$$

Answer: The distance between A and B is 150 kilometers.