Question

a railway track runs parallel to a road until a turn brings the road to a railway crossing. a cyclist rides along the road every day at a contant speed od 25kh/hr. he normally meets a train that travels in same direction at the crossing. one day he was late by 25 min. and met the train 10km before the railway crossing. find thespeed of the train

Answer 1

$$\begin{gathered} \mathrm{Time}\mathrm{taken}\mathrm{by}\mathrm{the}\mathrm{cyclist}\mathrm{to}\mathrm{cover}\mathrm{the}\mathrm{distance}\mathrm{of}10\mathrm{km} \\ {t}_1=\frac{10\times 1000}{{\displaystyle \frac{25\times 1000}{60}}}=24\mathrm{minutes} \\ \mathrm{It}\mathrm{means}\mathrm{cyclist}\mathrm{was}24\mathrm{minutes}\mathrm{away}\mathrm{from}\mathrm{the}\mathrm{crossing} \\ \mathrm{the}\mathrm{moment}\mathrm{he}\mathrm{met}\mathrm{the}\mathrm{train}.\mathrm{It}\mathrm{means}\mathrm{on}\mathrm{a}\mathrm{usual}\mathrm{day},\mathrm{he} \\ \mathrm{will}\mathrm{reach}\mathrm{the}\mathrm{crossing}\mathrm{in}\mathrm{just}25-24=1\mathrm{minutes}.\mathrm{It}\mathrm{also} \\ \mathrm{means}\mathrm{the}\mathrm{train}\mathrm{had}\mathrm{to}\mathrm{move}10\mathrm{km}\mathrm{in}1\mathrm{minutes}\mathrm{to}\mathrm{meet} \\ \mathrm{the}\mathrm{cyclist}\mathrm{on}\mathrm{a}\mathrm{usual}\mathrm{day}\mathrm{at}\mathrm{the}\mathrm{crossing}. \\ \mathrm{Train}\mathrm{speed}=\frac{10\mathrm{km}}{1\mathrm{minutes}}=600\mathrm{km}\mathrm{per}\mathrm{hour} \end{gathered}$$