Usual time from speed drop and delay A train runs at 5/6 of its usual speed and is late by 10 minutes. What is its usual time for the trip?

Introduction / Context:
With distance fixed, time is inversely proportional to speed. Reducing speed by a factor 5/6 increases time by 6/5. The difference between new and usual times is the observed delay.

Given Data / Assumptions:

• New speed = (5/6) of usual.
• Delay = 10 minutes.
• Let usual time be T minutes.

Concept / Approach:
New time = (6/5)T. Delay = (6/5)T − T = (1/5)T = 10.

Step-by-Step Solution:

Verification / Alternative check:
At usual time 50 min, reduced speed time = (6/5)*50 = 60 min, which is 10 minutes late.

Why Other Options Are Wrong:
40, 45, 55 min won’t produce exactly a 10-minute increase under a 6/5 factor.

Common Pitfalls:
Multiplying the time by 5/6 instead of dividing by 5/6 when speed is reduced.

Final Answer:
50 min