A train running at 40 km/h arrives 11 minutes late, but at 50 km/h it is only 5 minutes late. What is the correct scheduled time for the journey (without lateness)?

Introduction / Context:
We compare actual travel times at two speeds to deduce the scheduled time. The arrival-lateness values serve as offsets from the schedule.

Given Data / Assumptions:

• d/40 = T_s + 11/60 h.
• d/50 = T_s + 5/60 h.
• Same distance d; constant speeds.

Concept / Approach:
Subtract the equations to eliminate T_s and solve for d. Then compute T_s from either equation.

Step-by-Step Solution:
d*(1/40 − 1/50) = (11 − 5)/60 = 6/60 = 1/10 h.(1/40 − 1/50) = 1/200 → d * (1/200) = 0.1 → d = 20 km.T_s = d/50 − 5/60 = 20/50 − 1/12 = 0.4 − 0.0833.. = 0.3166.. h = 19 min.

Verification / Alternative check:
At 40 km/h: 20/40 = 0.5 h = 30 min → 30 − 19 = 11 min late; at 50 km/h: 24 min → 24 − 19 = 5 min late.

Why Other Options Are Wrong:
13, 15, or 21 min do not satisfy both lateness conditions.

Common Pitfalls:
Mixing minutes and hours or subtracting late times from each other without anchoring to the schedule.

Final Answer:
19 min