A jogger wants to complete a fixed course in one-quarter less time than usual (i.e., reduce time to 75% of normal). By what percentage must the jogger increase average speed to achieve this?

Introduction / Context:
For a fixed distance, speed and time are inversely proportional: speed * time = constant. Reducing time requires increasing speed by the reciprocal factor.

Given Data / Assumptions:

• Target time = 75% of usual time (reduce by 1/4).
• Distance is fixed; conditions unchanged.

Concept / Approach:
If original time is T and original speed is V, then V * T = constant. New time T′ = 0.75T ⇒ new speed V′ must satisfy V′ * 0.75T = V * T ⇒ V′ = (1 / 0.75) * V = (4/3) * V. Increase% = (V′ − V)/V * 100% = (1/3)*100% = 33 1/3%.

Step-by-Step Solution:

Verification / Alternative check:
Example: If usual speed is 12 km/h, new must be 16 km/h to cut time from 1 h to 45 min; increase is 4/12 = 33 1/3%.

Why Other Options Are Wrong:
25% corresponds to reducing time to 80%; 50% is for halving time; 20% underestimates.

Common Pitfalls:
Subtracting percentages instead of using the inverse relationship for fixed distance.

Final Answer:
33 1/3%