Walking at 3/4 of his usual speed, Sachin covers a fixed distance in 2 hours more than at his usual speed. Find the time (in hours) Sachin takes to cover the distance at his usual speed.

Introduction / Context:
Time varies inversely with speed for a fixed distance. Reducing speed to 3/4 of usual increases time by a factor of 4/3. The difference between the stretched time and the original time is given as 2 hours.

Given Data / Assumptions:

• Usual time = t hours.
• Reduced speed = 3/4 of usual ⇒ time becomes (4/3)t.
• Time increase = (4/3)t − t = 2 hours.

Concept / Approach:
Set (4/3)t − t = 2 and solve for t to find the usual-time value directly.

Step-by-Step Solution:

Verification / Alternative check:
At reduced speed, time = (4/3)*6 = 8 hours, which is 2 hours more than 6 hours—matches the statement.

Why Other Options Are Wrong:
4, 5, 7, 9 hours would not make the time difference exactly 2 hours under a 3/4 speed reduction.

Common Pitfalls:
Using 3/4 on time instead of on speed; forgetting that time scales as the reciprocal of speed.

Final Answer:
6 h