A bus running at 9/10 of its usual speed reaches a place in 22 hours. How much time could be saved on the same journey if the bus travelled at its full usual speed?

Introduction / Context:
This question examines the inverse relationship between speed and time for a fixed distance. When speed is reduced to a fraction of its usual value, time increases. You are asked to work backward from the longer time to find the time that would have been taken at the full speed and then calculate the saving in time.

Given Data / Assumptions:
The bus travels at 9/10 of its usual speed and takes 22 hours for the journey.
The distance between the two places remains the same in both situations.
We need to find how many hours would be saved if the bus travelled at its full usual speed instead of 9/10 of it.
Speeds and times are constant during each trip without stops or variations.

Concept / Approach:
Let the usual speed be v km/h and the distance be D km. At reduced speed 9v / 10, the time taken is distance divided by speed. The relationship D = speed * time holds. Using the fact that D is the same in both scenarios, we can equate the expressions and solve for the usual time. The time saved is simply the difference between the slower time and the faster time.

Step-by-Step Solution:
Step 1: Let the usual speed be v km/h and the distance be D km.Step 2: At reduced speed, the bus travels at 9v / 10 and takes 22 hours.Step 3: Distance D = (9v / 10) * 22.Step 4: At full usual speed v, let the time taken be T hours.Step 5: Then D = v * T.Step 6: Equate the two expressions for D: v * T = (9v / 10) * 22.Step 7: Cancel v (since v is not zero): T = (9 / 10) * 22.Step 8: Compute T: T = 9 * 22 / 10 = 198 / 10 = 19.8 hours.Step 9: Time saved = slower time − usual time = 22 − 19.8 = 2.2 hours.

Verification / Alternative check:
We can also compute the ratio of times. Since time is inversely proportional to speed for a fixed distance, time at reduced speed divided by time at usual speed = usual speed divided by reduced speed = v / (9v / 10) = 10 / 9. So 22 / T = 10 / 9, giving T = 22 * 9 / 10 = 19.8 hours. Subtracting from 22 gives 2.2 hours saved, which agrees with the earlier calculation.

Why Other Options Are Wrong:
1.5 hours and 1.7 hours underestimate the time saving and would correspond to incorrect ratios or arithmetic errors.
3.5 hours overestimates the saving and would mean a very large difference in speed not supported by the factor of 9/10 given in the question.
2.0 hours seems reasonable but does not match the exact proportional calculation.

Common Pitfalls:
Students sometimes assume that a 10 percent change in speed leads to a 10 percent change in time, which is incorrect because time is inversely proportional to speed. Others attempt to guess a time saving instead of solving it algebraically. Carefully setting up the equations using distance = speed * time and solving step by step avoids such mistakes.

Final Answer:
If the bus ran at its full usual speed, it would save 2.2 hours on the journey.