A student walks from home towards school. If the student walks at 4 meters per minute, arrival is 4 minutes late. If the student walks at 6 meters per minute, arrival is 30 minutes early. How far is the school from the student's home, in meters?

Introduction / Context:
This is a classic punctuality problem involving two different walking speeds and early or late arrival times. The question tests the ability to set up equations using distance, speed, and time, while carefully interpreting what early and late mean in terms of time differences relative to a fixed schedule. Such problems appear frequently in quantitative aptitude tests.

Given Data / Assumptions:

• Walking speed in the first case = 4 meters per minute.
• Walking speed in the second case = 6 meters per minute.
• At 4 m/min, the student is 4 minutes late for school.
• At 6 m/min, the student is 30 minutes early.
• There is a fixed scheduled arrival time, and we assume a fixed starting time from home in both cases.
• We assume straight line travel and constant speeds in both scenarios.

Concept / Approach:
Let D be the distance to school in meters and T be the ideal time (in minutes) that would result in exactly on time arrival at school. At 4 m/min, the actual travel time is T + 4 minutes. At 6 m/min, the travel time is T - 30 minutes. We can express both travel times as distance divided by speed and then solve the resulting system of two equations for D and T.

Step-by-Step Solution:
Let D be the distance in meters and T be the ideal on time travel duration in minutes. Case 1: At 4 m/min, student is 4 minutes late, so travel time = T + 4. Thus, D = 4 * (T + 4). Case 2: At 6 m/min, student is 30 minutes early, so travel time = T - 30. Thus, D = 6 * (T - 30). Equate distances: 4 * (T + 4) = 6 * (T - 30). Expand: 4T + 16 = 6T - 180. Rearrange: 16 + 180 = 6T - 4T, so 196 = 2T. Therefore, T = 98 minutes. Substitute back: D = 4 * (T + 4) = 4 * (98 + 4) = 4 * 102 = 408 meters.

Verification / Alternative check:
Check both scenarios with D = 408 meters. At 4 m/min, time = 408 / 4 = 102 minutes, which is 4 minutes more than T = 98 minutes, so the student is 4 minutes late. At 6 m/min, time = 408 / 6 = 68 minutes, which is 30 minutes less than 98 minutes, meaning the student arrives 30 minutes early. Both conditions are satisfied, confirming that the distance is correct.

Why Other Options Are Wrong:
Distances such as 360 m, 392 m, 420 m, and 450 m do not satisfy both the early and late conditions simultaneously. Substituting any of these values leads to times that are not separated by exactly 4 minutes late and 30 minutes early with respect to the same ideal time T. Only 408 m fits the equations for both speeds and timing conditions.

Common Pitfalls:
Students may confuse early and late conditions, treating both as positive time differences. Another frequent error is mixing minutes and seconds or forgetting that both journeys start at the same clock time. Properly defining an ideal on time duration and writing separate equations for each speed scenario is the most reliable way to solve these problems.

Final Answer:
The distance from the student's home to the school is 408 m.