A railway track runs parallel to a road until a turn brings the road to railway crossing. A cyclist rides along the road everyday at a constant speed 20 km/hr. He normally meets a train that travels in same direction at the crossing. One day he was late by 25 minutes and met the train 10 km before the railway crossing.Find the speed of the train.
My Solution:
I suppose the cyclist starts from a fixed point everyday markes by the red dot. If the situation was A, and if the cyclist wakes up late then train would have crossed the crossing and went straight and there seems no possibility that cyclist would be able to catch it 10 km before the crossing. So situation B would be correct as even if he goes late, train comes to him on his self, so he can catch the train 10 km before.
[contradiction: "a train that travels in same direction"]
As he is 25 minutes late, the train gets 25 minutes extra to cover some distance and cyclist 25 minutes less so he will cover less distance.
[contradiction: if cyclist gets 25 minutes less he would cover $25/60\;.\;20=8.\bar 3$ km before the track but where does 10 km come from]
[contradiction2: if we suppose cyclist takes t min daily to reach crossing, then taking $v_t$ the speed of train $v_t(t+25/60)+v(t-25/60)=v_tt+vt\implies v_t=v=20??$]
Best Answer
Suppose train daily starts $v_tt$ distance before the crossing and the cyclist starts from his house and covers $20t$ distance daily, the difference in starting points thus is $v_tt-20t$. Now since he is late the train travels in time $t_2$: $v_tt_2+10=v_tt$ and cyclist travels in time $t_2$: $20t_2+10=20t$ a distance $v_t(25/60)+v_tt_2=v_tt-10$ where $t_2$ is the time cyclist travels after waking up. The cylist would travel a distance $20t_2=20t-10$, this on slving gives $v_t=120$!!