A train reaches on time when it travels at an average speed of 40 km/h. If it travels at an average speed of 35 km/h, it reaches 15 minutes late. What is the total journey distance (in km)?

Introduction:
This average speed question tests the relationship between distance, speed, and time. If a journey is on time at one speed but late at a lower speed, the difference in travel times for the same distance equals the lateness. By translating the 15-minute delay into hours and equating the two time expressions, we can solve for the journey time and distance.

Given Data / Assumptions:

• On-time average speed = 40 km/h
• Reduced average speed = 35 km/h
• Lateness at 35 km/h = 15 minutes = 0.25 hours
• Distance is constant in both cases

Concept / Approach:
Let the scheduled travel time be T hours and distance be D km. Then D = 40 * T for the on-time case, and D = 35 * (T + 0.25) for the delayed case. Equate these two expressions for D and solve for T. Finally compute D using D = 40 * T.

Step-by-Step Solution:

Verification / Alternative check:
Time at 35 km/h: 70 / 35 = 2.00 h. Scheduled time: 1.75 h. Difference = 0.25 h = 15 minutes, which matches the given delay.

Why Other Options Are Wrong:
80, 60, 50, and 30 km do not satisfy the 15-minute difference when checked against the two speeds; only 70 km yields exactly 0.25 h extra at 35 km/h.

Common Pitfalls:
Using average of speeds or subtracting speeds directly. The key is equating times for the same distance.

Final Answer:
70 km