Time and Distance – Early or late arrival with two walking speeds: A student walks from home to school at 3 km/h and arrives 5 minutes late relative to the scheduled time. The next day, walking at 4 km/h, the student arrives 3 minutes early. What is the distance from home to school?

Introduction / Context:
Problems with early and late arrivals hinge on equating the same distance under different speeds and different arrival offsets. Translating minutes into hours keeps units consistent and allows a clean equation for distance.

Given Data / Assumptions:

• Speed case 1: 3 km/h and arrival is 5 min late.
• Speed case 2: 4 km/h and arrival is 3 min early.
• Scheduled travel time is the same reference in both cases.

Concept / Approach:
Let D be distance and T be the scheduled travel time (in hours). Then D/3 = T + 5/60 and D/4 = T − 3/60. Subtracting eliminates T and solves D directly.

Step-by-Step Solution:
D/3 − D/4 = (5/60) + (3/60) = 8/60 = 2/15.Left side = D*(1/12) ⇒ D/12 = 2/15.Therefore D = 24/15 = 8/5 = 1.6 km.

Verification / Alternative check:
Check quickly: At 3 km/h, time = 1.6/3 ≈ 0.533 h = 32 min; at 4 km/h, time = 1.6/4 = 0.4 h = 24 min. The 8-minute gap matches the 5 min late vs 3 min early difference.

Why Other Options Are Wrong:
1.24, 1.36, 1.5, and 1.8 km do not satisfy the pair of equations built from the arrival offsets.

Common Pitfalls:
Using 5 − 3 = 2 minutes rather than summing offsets to 8 minutes, or forgetting to convert minutes to hours.

Final Answer:
1.6 km