A student walks from home to school at 2.5 km/h and arrives 6 minutes late. The next day, increasing speed by 1 km/h (to 3.5 km/h), the student arrives 6 minutes early. What is the distance from home to school (in km)?

Introduction / Context:
Here two commuting scenarios bracket the scheduled time: one is 6 minutes late, the other is 6 minutes early. For a fixed distance, time is inversely proportional to speed. Setting up equations with the same scheduled time lets us solve the distance cleanly.

Given Data / Assumptions:

• Speed day 1 = 2.5 km/h (late by 6 min).
• Speed day 2 = 3.5 km/h (early by 6 min).
• Let scheduled time be T hours, distance be D km.

Concept / Approach:
Equations: D/2.5 = T + 0.1 and D/3.5 = T − 0.1, since 6 minutes = 0.1 h. Subtract the second from the first to eliminate T and solve for D.

Step-by-Step Solution:

Verification / Alternative check:
At 2.5 km/h: time = 1.75/2.5 = 0.7 h (42 min). At 3.5 km/h: 1.75/3.5 = 0.5 h (30 min). Midpoint is 36 min; late/early by ±6 min as stated.

Why Other Options Are Wrong:
5/4, 9/4, 11/4 km do not produce a symmetrical ±6 minute deviation around a single schedule for the given speeds.

Common Pitfalls:
Using minutes without converting to hours; averaging speeds rather than setting up time equations.

Final Answer:
7/4 km