Time and Distance – Train at reduced speed vs own speed: A train running at 8/11 of its normal speed reaches a destination in 5 h 30 min. If the train had run at its full normal speed, how much time would have been saved compared to 5 h 30 min?

Introduction / Context:
This is a speed-time-distance comparison. When a vehicle runs at a fractional multiple of its normal speed, the time taken scales inversely by the reciprocal fraction. We compare the slower trip with the normal-speed trip to compute the time saved.

Given Data / Assumptions:

• Reduced speed = 8/11 of normal speed.
• Time at reduced speed = 5 h 30 min = 5.5 h.
• Distance is the same on both trips.

Concept / Approach:
For fixed distance D and normal speed v, time at normal speed is T = D / v. At reduced speed v' = (8/11) v, time is T' = D / v' = (11/8) T. Hence T = (8/11) T'.

Step-by-Step Solution:
T = (8/11) * 5.5 h.Since 5.5 = 11/2, T = (8/11) * (11/2) = 8/2 = 4 h.Time saved = T' − T = 5.5 − 4 = 1.5 h = 3/2 h.

Verification / Alternative check:
If normal time is 4 h, then at 8/11 speed the required time should be (11/8)*4 = 5.5 h, which exactly matches the given slow trip time.

Why Other Options Are Wrong:
2 h and 1 h are over or under the calculated saving; 5/4 h = 1.25 h and 1 h 20 min = 1.333 h do not equal 1.5 h.

Common Pitfalls:
Applying the 8/11 factor to time directly (should invert), or mixing minutes and hours during subtraction.

Final Answer:
3/2 h