A person travels a fixed distance at 3 km/h and arrives 15 minutes late; at 4 km/h the person arrives 15 minutes early. What is the distance (in km)?

Introduction / Context:
With the same departure time, different walking speeds give arrival times relative to a schedule. Setting up two equations in distance and scheduled time allows elimination of the schedule variable to solve for distance directly.

Given Data / Assumptions:

• D/3 = T + 0.25 h (15 min late).
• D/4 = T − 0.25 h (15 min early).

Concept / Approach:
Subtract the equations: D/3 − D/4 = 0.5 h. Solve for D using the difference of reciprocals (1/3 − 1/4) = 1/12.

Step-by-Step Solution:

Verification / Alternative check:
Check with T = 1.25 h: at 3 km/h time = 2 h (0.75 h late); adjust T properly (e.g., T = 1.25 h gives 6/3 = 2 h = 45 min late). Alternatively, use the symmetric 30-minute spread: the 0.5 h difference matches (1/3 − 1/4)×D only when D = 6.

Why Other Options Are Wrong:
4.5, 5, 7.2, 12 km do not satisfy D(1/3 − 1/4) = 0.5.

Common Pitfalls:
Adding the two equations or mixing up early and late signs; forgetting to convert minutes to hours.

Final Answer:
6 km