A boy reaches school 20 minutes late at 15 km/h. When he walks at 20 km/h, he is still 5 minutes late. What is the distance from home to school?

Introduction / Context:
Another schedule-variance setup: with two speeds and known lateness at each, we deduce the fixed distance by eliminating the (unknown) scheduled time.

Given Data / Assumptions:

• d/15 = T_s + 20/60 h.
• d/20 = T_s + 5/60 h.
• Uniform route and speeds.

Concept / Approach:
Subtract the equations to remove the scheduled time T_s, leaving one equation in d. Convert minutes to hours to keep units consistent.

Step-by-Step Solution:
d/15 − d/20 = (20 − 5)/60 = 15/60 = 1/4 h.d * (1/15 − 1/20) = 1/4.d * (4 − 3)/60 = 1/4 → d/60 = 1/4.d = 15 km.

Verification / Alternative check:
At 15 km/h: 15/15 = 1 h = T_s + 0.333.. → T_s = 0.666.. h. At 20 km/h: 15/20 = 0.75 h = T_s + 0.0833.. → T_s = 0.666.. h; consistent.

Why Other Options Are Wrong:
5 km, 10 km, 20 km violate one or both timing equations.

Common Pitfalls:
Accidentally treating 5 minutes as 0.5 h; forgetting to convert both lateness values to hours.

Final Answer:
15 km