Every day a cyclist meets a train at a crossing. The cyclist travels at 10 km/h. One day the cyclist is 25 minutes late and meets the train 5 km before the crossing. What is the speed of the train?

Every day a cyclist meets a train at a crossing. The cyclist travels at 10 km/h. One day the cyclist is 25 minutes late and meets the train 5 km before the crossing. What is the speed of the train?

• A. 66 km/h.
• B. 40 km/h.
• C. 60 km/h.
• D. 76 km/h.

Answer: (C)

The situation hinges on two ideas: (1) on a normal day the cyclist and the train would reach the crossing at the same time, and (2) when the cyclist is 25 minutes late, they meet 5 km before the crossing. Let the distance from the cyclist’s start to the crossing be A, and the distance from the train’s start to the crossing be B. Normally they meet at the crossing, so their travel times are equal: A/10 = B/v, which gives A = 10B/v. On the late day, the meeting occurs 5 km before the crossing. So the cyclist travels A − 5 before meeting, and the train travels B − 5 before meeting. The cyclist starts 25 minutes (L = 25/60 h) later, so the train’s time to the meeting equals the cyclist’s time to the meeting plus the late start: (B − 5)/v = (A − 5)/10 + L. Substitute A = 10B/v into the equation: (B − 5)/v = (B/v) − 0.5 + L. With L = 25/60 = 5/12, this becomes: (B/v) − 5/v = (B/v) − 0.5 + 5/12. The B/v terms cancel, leaving: −5/v = −1/12, so v = 60 km/h. Thus the train’s speed is 60 km/h.

The situation hinges on two ideas: (1) on a normal day the cyclist and the train would reach the crossing at the same time, and (2) when the cyclist is 25 minutes late, they meet 5 km before the crossing.

Let the distance from the cyclist’s start to the crossing be A, and the distance from the train’s start to the crossing be B. Normally they meet at the crossing, so their travel times are equal: A/10 = B/v, which gives A = 10B/v.

On the late day, the meeting occurs 5 km before the crossing. So the cyclist travels A − 5 before meeting, and the train travels B − 5 before meeting. The cyclist starts 25 minutes (L = 25/60 h) later, so the train’s time to the meeting equals the cyclist’s time to the meeting plus the late start: (B − 5)/v = (A − 5)/10 + L.

Substitute A = 10B/v into the equation:

(B − 5)/v = (B/v) − 0.5 + L.

With L = 25/60 = 5/12, this becomes:

(B/v) − 5/v = (B/v) − 0.5 + 5/12.

The B/v terms cancel, leaving:

−5/v = −1/12, so v = 60 km/h.

Thus the train’s speed is 60 km/h.