Late arrival due to reduced speed A train runs at 6/7 of its usual speed and arrives 10 minutes late. What is its usual time for the journey?

Introduction / Context:
For a fixed distance, time is inversely proportional to speed. If the speed is scaled by a factor, the time is scaled by the reciprocal factor. The difference between new and usual time gives the delay.

Given Data / Assumptions:

• New speed = (6/7) of usual speed.
• Delay = 10 minutes = 10/60 h.
• Let usual time be T hours.

Concept / Approach:
New time T' = T / (6/7) = (7/6)T. Delay = T' − T = (1/6)T equals 10 minutes.

Step-by-Step Solution:

Verification / Alternative check:
Usual time 60 min; at 6/7 speed time becomes (7/6)*60 = 70 min, exactly 10 min late.

Why Other Options Are Wrong:
25, 15, 35 min cannot produce a 10-min increase when multiplied by 7/6.

Common Pitfalls:
Multiplying by 6/7 instead of dividing by it when converting time.

Final Answer:
60 min