A person walks at only 6/7 of his usual walking speed and therefore reaches his destination 30 minutes late. What is his usual time to cover the same distance?

Introduction / Context:
This is a classical speed and time problem where a change in speed causes a change in time taken. The key relationship is that for a fixed distance, speed and time are inversely proportional. When the person walks slower than usual, the time taken increases, and we are given this increase as 30 minutes. Using the ratio of speeds, we can find the ratio of times and then determine the usual time for the journey.

Given Data / Assumptions:

• The person walks at 6/7 of his usual speed.
• Because of this reduced speed, he is 30 minutes late.
• The distance remains the same in both scenarios.
• We must find his usual time to cover the distance.

Concept / Approach:
For a fixed distance, speed * time is constant, so time is inversely proportional to speed. If the usual speed is S and usual time is T, then distance = S * T. When speed changes to 6S/7, the new time T' must satisfy S * T = (6S/7) * T', so T' = (7/6) * T. The extra time taken is T' − T, which is given as 30 minutes. Setting (7/6) * T − T equal to 30 minutes allows us to solve for T and find the usual travel time.

Step-by-Step Solution:
Step 1: Let the usual speed be S and the usual time be T hours. Step 2: The distance is D = S * T. Step 3: At the reduced speed, the speed becomes 6S / 7. Step 4: Let the new time taken be T' hours. Then D = (6S / 7) * T'. Step 5: Since D is the same, S * T = (6S / 7) * T'. Step 6: Cancel S on both sides to get T = (6 / 7) * T'. Step 7: Rearranging gives T' = (7 / 6) * T. Step 8: The person is 30 minutes late, so T' − T = 30 minutes. Step 9: Express 30 minutes in hours: 30 minutes = 0.5 hours. Step 10: Substitute T' = (7 / 6) * T: (7 / 6) * T − T = 0.5. Step 11: Simplify the left side: (7T / 6 − 6T / 6) = T / 6. Step 12: So T / 6 = 0.5, giving T = 0.5 * 6 = 3 hours.

Verification / Alternative check:
With a usual time of 3 hours, at reduced speed the time becomes T' = (7 / 6) * 3 = 3.5 hours, that is 3 hours and 30 minutes. The difference is 30 minutes, exactly matching the given delay. This confirms that the usual time is 3 hours.

Why Other Options Are Wrong:
Option (a) 2 hrs 50 min: This value would not produce an exact 30 minute delay under the given speed ratio.
Option (b) 3 hrs 25 min: Using this as T would yield a different delay than 30 minutes.
Option (d) 2 hrs 5 min: This is significantly smaller and inconsistent with the specified ratio and delay.

Common Pitfalls:
A common error is to treat speed and time as directly proportional instead of inversely proportional, leading to wrong equations. Some learners also forget to convert 30 minutes into hours or mis-handle the fractions involved in (7 / 6). Carefully setting up the proportional relationship and using consistent units helps avoid mistakes.

Final Answer:
The person's usual time to cover the distance is 3 hours.