The distance between cities $X$ and $Y$ is equal to $21$ kilometers. A pedestrian leaves $X$ and goes to $Y$ at a constant speed of $5$ kilometers per hour. At the same moment, a cyclist leaves $Y$ and goes towards $X$, the speed of the latter can vary between $10$ and $13$ km/h throughout the journey. After meeting each other, the motorcyclist goes $26$ minutes more towards city $X$, and then turns back and returns to $Y$. What is the maximum difference in time the pedestrian and the cyclist arrive to $Y$? Express the answer in minutes.
1 hour (h) = 60 minutes (min); 1 kilometer (km) = 1000 meters (m)
Clearly the cyclist will arrive first, as in the worst case scenario when he goes $10$ km/h, then $13$ km/h, he will be less then $20$ km $\left(14+\frac{26\cdot13}{60}\right)$away while the pedestrian will be more than $10$ km $\left(14-\frac{26\cdot5}{60}\right)$ away. Thus, the cyclist should go as fast as possible, meet the pedestrian, go as slow as possible, turn back and go as fast as possible.