A man cycles from his house to office at 10 km/h and reaches 6 minutes late. At 12 km/h, he reaches 6 minutes early. How far is his office from his house (in km)?

Introduction / Context:
Two speeds around a schedule produce symmetric early/late differences. As with similar problems, build two equations in distance and scheduled time, subtract to eliminate the schedule, and solve for distance directly.

Given Data / Assumptions:

• D/10 = T + 0.1 h (6 minutes late).
• D/12 = T − 0.1 h (6 minutes early).

Concept / Approach:
Subtract equations: D/10 − D/12 = 0.2 h (12 minutes total spread). Use 1/10 − 1/12 = 1/60 to compute D.

Step-by-Step Solution:

Verification / Alternative check:
At 10 km/h, time = 1.2 h = 72 min; if schedule is 66 min, that is 6 min late. At 12 km/h, time = 60 min, which is 6 min early. Both conditions match.

Why Other Options Are Wrong:
6, 7, 9, 16 km do not satisfy the 12-minute spread with speeds 10 and 12 km/h.

Common Pitfalls:
Using arithmetic mean of speeds; forgetting to convert 6 minutes to 0.1 hour.

Final Answer:
12 km