3-4. 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 travel 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 \(M\). After the meeting, \(B\) continues on foot, and \(C\) rides the bicycle until meeting \(A\) and hands over the bicycle to him, on which \(A\) then arrives at \(N\).

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 solve 4. For \(A\) and \(B\) to arrive at point \(N\), they 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 \(A\), and after that, \(A\) will cycle \(15-x\). In total, \(A\) and \(B\) together will cycle \(30-3x\). In the same time, \(B\) will walk \(x\), so \(\frac{30-3x}{15}=\frac{x}{6}\), i.e., \(x=60/11\). \(B\) travels \(\frac{15-x}{15}=\frac{1}{15} \cdot \frac{105}{11}=\frac{7}{11}\) hours before meeting \(C\), and \(C\) walks \(\frac{x}{6}=\frac{1}{6} \cdot \frac{60}{11}=\frac{10}{11}\) hours before meeting \(B\), so \(C\) should leave \(M\) \(\frac{3}{11}\) hours before \(A\) and \(B\) start their journey from \(M\).

\section*{8th Grade}