By walking at only three-fourths (3/4) of his usual speed, a person reaches his office 20 minutes later than he normally does. What is his usual (normal) time to cover the same route?

Introduction / Context:
This problem tests proportional reasoning between speed and time. When speed changes by a factor, the travel time changes inversely by the same factor. Understanding this inverse relation allows us to translate a given delay into the original time.

Given Data / Assumptions:

• Reduced speed = 3/4 of usual speed.
• Extra time (delay) due to reduced speed = 20 minutes.
• Route distance is the same in both cases.

Concept / Approach:
Time is inversely proportional to speed for a fixed distance. If speed scales by k, then time scales by 1/k. Here, reduced speed factor is 3/4, so the new time factor is 4/3 of the usual time. The difference between new and usual time equals the given delay.

Step-by-Step Solution:
Let T = usual time (minutes). Reduced speed time = (4/3) * T. Delay = ((4/3) * T) - T = (1/3) * T. Given delay = 20 ⇒ (1/3) * T = 20 ⇒ T = 60 minutes.

Verification / Alternative check:
If usual time is 60 minutes, reduced-speed time is 60 * 4/3 = 80 minutes, which is indeed 20 minutes more. Checks out.

Why Other Options Are Wrong:
30 min: Would give delay 10 min, not 20. 75 min: Would give delay 25 min. 1 hr 30 min: Implies a much larger delay. 45 min: Would give delay 15 min. None match the 20-minute delay.

Common Pitfalls:
Mixing up direct versus inverse proportionality and adding 25% to time instead of multiplying by 4/3 are common mistakes.

Final Answer:
60 minutes