Two people $A$ and $B$ need to travel from point $M$ to point $N$, which is 15 km away from $M$. On foot, they can move at a speed of 6 km/h. In addition, they have a bicycle that can be ridden at a speed of 15 km/h. $A$ sets out on foot, while $B$ rides the bicycle until meeting pedestrian $C$, who is walking from $N$ to $M$. Then $B$ continues on foot, and $C$ rides the bicycle until meeting $A$ and hands over the bicycle to him, who then rides it to $N$. When should pedestrian $C$ leave from $N$ so that $A$ and $B$ arrive at point $N$ simultaneously (if he walks at the same speed as $A$ and $B$)?
To arrive at point $N$ simultaneously, $A$ and $B$ must walk the same distance $x$ and cycle the same distance $15-x$. Then, $C$ must also walk the distance $x$ before meeting $B$. Therefore, $C$ will cycle $15-2x$ before meeting $A$. Together, $B$ and $C$ will cycle a total of $30-3x$. In the same time, $A$ will walk $x$, but he walks 2.5 times slower, so $30-3x = 2.5x$, which means $x = 60 / 11$. Before meeting $C$, $A$ will cycle $15 - 60 / 11 = 105 / 11$ km, which takes $7 / 11$ hours. $C$ must walk $60 / 11 : 6 = 10 / 11$ hours before meeting $B$, so he must leave $M$ $3 / 11$ hours before $A$ and $B$ start their journey.

## Answer

$3 / 11$ hours before $A$ and $B$ leave from $M$.