Question

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 kmh. 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 10 km before
the railway crossing. Find the speed of the train.​

Answer 1

Let's assume that the man and the train normally meet at the crossing at 8 A.M.,

then the usual time of the cyclist at the bend is 8 A.M.

and he is 6 miles behind at 7.30 A.M.

But when the cyclist is late.

he arrives at the bend at 8.25 A.M.

and therefore he is six miles behind at 7.55 A.M

.Since the train takes 5 minutes to travel the six mile run, the speed of the train is 72 m.p.h.

Answer 2

SOLUTION

Time taken by the cyclist to cover the distance of 10 km.

$t1 = \frac{ \frac{10 \times 1000}{25 \times 1000} }{60} = 24minutes$

It means cyclist was 24 minutes away from the crossing the moment he met the train. It means on a usual day, he will reach the crossing in just 25-24 =1 minutes. It also means the train had to move 10km in 1 minutes to meet the cyclist on a usual day at the crossing.

Train speed = 10km/1minutes

=) 600 km per hour

hope it helps ☺️