Every day a cyclist meets a train at a particular crossing. The road is…
2026
Every day a cyclist meets a train at a particular crossing. The road is straight before the crossing and both are traveling in the same direction. The cyclist travels with a speed of 10 Kmph. One day the cyclist comes late by 25 min. and meets the train 5km before the crossing. What is the speed of the train?
Answer: A. 60km/hr — Given: cyclist speed = 10 km/h; cyclist is 25 minutes late; on that day they meet the train 5 km before the crossing. Let: distance from cyclist's start to…
- A.
60km/hr
- B.
72km/hr
- C.
45km/hr
- D.
54km/hr
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Correct answer: A
Given: cyclist speed = 10 km/h; cyclist is 25 minutes late; on that day they meet the train 5 km before the crossing.
Let: distance from cyclist's start to crossing = D, and let v be the train speed (km/h). The normal meeting time at the crossing is t0 = D/10 hours.
On the late day the cyclist departs 25 minutes (25/60 h) late, so he reaches the meeting point (5 km before crossing) at time (D - 5)/10 + 25/60 hours.
The train reaches the crossing at t0, so meeting 5 km before the crossing occurs 5/v hours before t0, i.e. at time t0 - 5/v.
Equate the two expressions for the meeting time: (D - 5)/10 + 25/60 = t0 - 5/v. Substitute t0 = D/10 to get D/10 - 1/12 = D/10 - 5/v.
Cancel D/10 from both sides: -1/12 = -5/v ⇒ 5/v = 1/12 ⇒ v = 5 ÷ (1/12) = 60 km/h.
Answer: 60 km/hr