1. A pedestrian left point $A$ for point $B$, and after some delay, a second pedestrian followed. When the first pedestrian had walked half the distance, the second had walked 15 km, and when the second pedestrian had walked half the distance, the first had walked 24 km. Both pedestrians arrived at point $B$ simultaneously. What is the distance between points $A$ and $B$?
# Solution:

Let $s$ be the distance between points $A$ and $B$, and $v_{1}, v_{2}$ be the speeds of the pedestrians. Then $\frac{s}{2 v_{1}}=\frac{s-15}{v_{2}}$ and $\frac{s-24}{v_{1}}=\frac{s}{2 v_{2}}$. From this, $\frac{s}{2(s-24)}=\frac{(s-15) \cdot 2}{s} ; s^{2}=4 s^{2}-4 \cdot 39 s+60 \cdot 24 ;$ $s^{2}-52 s+480=0 ; s_{1,2}=26 \pm 14 . s_{1}=40, s_{2}=12$ does not satisfy the conditions $s>15, s>24$.

Answer: 40 km.