A boy walks from his house to school at 5 km/h and reaches 3 minutes late. On the next day, he starts at the same time but walks at 9 km/h and reaches 3 minutes early. What is the distance between his house and the school in kilometres?

Introduction / Context:
This question explores the effect of changing walking speed on arrival time when the start time is fixed. The boy arrives late on one day and early on another day by changing his speed. This difference in arrival times leads to an equation that lets us compute the distance to school. It is a standard type of time and distance problem in aptitude examinations.

Given Data / Assumptions:
On the first day, walking speed = 5 km/h and the boy is 3 minutes late.

Given Data / Assumptions:
On the second day, walking speed = 9 km/h and he is 3 minutes early.
The start time on both days is the same.
The school is at a fixed distance from his house.
Three minutes is equal to 3 / 60 = 1 / 20 hours.
We need to find the distance between house and school in kilometres.

Concept / Approach:
Let the distance be D km and the scheduled correct time to reach be T hours. On the slower day, time taken is D / 5, which equals T plus 1 / 20 hour because he is late. On the faster day, time taken is D / 9, which equals T minus 1 / 20 hour because he is early. Equating these two expressions for T allows us to eliminate T and solve for D.

Step-by-Step Solution:
Step 1: Let D be the distance in km and T be the correct on time travel duration in hours.Step 2: On the first day, D / 5 = T + 1 / 20.Step 3: On the second day, D / 9 = T − 1 / 20.Step 4: From the first equation, T = D / 5 − 1 / 20.Step 5: From the second equation, T = D / 9 + 1 / 20.Step 6: Equate the two expressions for T: D / 5 − 1 / 20 = D / 9 + 1 / 20.Step 7: Bring terms involving D to one side and constants to the other: D / 5 − D / 9 = 1 / 20 + 1 / 20.Step 8: Right side becomes 2 * (1 / 20) = 1 / 10.Step 9: Left side: D * (1 / 5 − 1 / 9) = D * ((9 − 5) / 45) = D * (4 / 45).Step 10: So D * 4 / 45 = 1 / 10.Step 11: Solve for D: D = (1 / 10) * (45 / 4) = 45 / 40 = 9 / 8.Step 12: 9 / 8 km equals 1.125 km.

Verification / Alternative check:
Using D = 1.125 km: Time at 5 km/h is 1.125 / 5 = 0.225 hours, which is 0.225 * 60 = 13.5 minutes. Time at 9 km/h is 1.125 / 9 = 0.125 hours, which is 0.125 * 60 = 7.5 minutes. The average or correct on time is the midpoint between these two times: (13.5 + 7.5) / 2 = 10.5 minutes. Thus, with 5 km/h he is 3 minutes late (13.5 − 10.5) and with 9 km/h he is 3 minutes early (10.5 − 7.5). This matches the problem statement exactly.

Why Other Options Are Wrong:
1 km, 1.25 km, and 1.5 km do not produce the correct 3 minute early and 3 minute late pattern when substituted into the speed calculations.
0.75 km is too short and would yield smaller travel times that do not match the described early and late differences.

Common Pitfalls:
Students sometimes forget to convert minutes into hours and write 3 instead of 1 / 20, which breaks the equation. Another mistake is to assume the difference of times equals 3 minutes instead of 6 minutes, forgetting that he is late by 3 minutes one day and early by 3 minutes another day. Setting up the equations correctly and handling units carefully ensures success in such problems.

Final Answer:
The distance between his house and the school is 1.125 km.