A railway track runs parallel to a road until a turn brings the road to railway crossing. A cyclist rides along the road every day 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.

Options

A. `9 km//hr`
B. `90 km//hr`
C. `120 km//hr`
D. `40 km//hr`

Answer

C

Explanation

`t_("cycle") = (10 km)/(20 kmh^-1) = (1)/(2) h = 30 min`
`Delta t = 5 min = (5)/(60) hr`
Train running as per schedule.
So `v_("train") = (10)/((50//60)) = (10 xx 60)/(5) = 120 kmh^-1`.