Walking at a speed of 10 km/h, Madhu reaches her university 15 minutes late. Next time she increases her walking speed to 12 km/h but is still late by 5 minutes. What is the distance of the university from her house (in km)?

Introduction / Context:
This problem links time, speed, and punctuality. Madhu walks to her university at two different speeds on two different days, and we are told how late she is in each case. From this, we must deduce the distance between her home and the university. The question tests your ability to model lateness or earliness as a difference between actual travel time and scheduled (on-time) travel time, and then use algebra to solve for distance.

Given Data / Assumptions:
- Speed on the first day = 10 km/h; she arrives 15 minutes late.
- Speed on the second day = 12 km/h; she arrives 5 minutes late.
- Let the scheduled (on-time) travel time be T hours.
- Let the distance between her home and the university be d km.
- 15 minutes is 15/60 = 0.25 hours; 5 minutes is 5/60 = 1/12 hours.
- Madhu walks at constant speed on a fixed, straight route.

Concept / Approach:
We express the actual travel times using time = distance / speed. On each day, this actual time is equal to the scheduled time plus a certain lateness. This gives us two equations linking d and T. Subtracting one equation from the other eliminates T and allows us to solve for d directly. Once we have the distance, we can check that both lateness conditions are satisfied.

Step-by-Step Solution:
Step 1: Let the exact on-time journey time be T hours.On the first day: d / 10 = T + 0.25.On the second day: d / 12 = T + 1/12.Step 2: Subtract the second equation from the first:d / 10 - d / 12 = (T + 0.25) - (T + 1/12).Left side: (1/10 - 1/12) * d = (6/60 - 5/60) * d = (1/60) * d.Right side: 0.25 - 1/12 = 1/4 - 1/12 = 3/12 - 1/12 = 2/12 = 1/6.So d / 60 = 1/6 ⇒ d = 60 * 1/6 = 10 km.

Verification / Alternative check:
With d = 10 km, first day time at 10 km/h is 10 / 10 = 1 hour. If she should have arrived in T hours, then T = 1 - 0.25 = 0.75 hours (45 minutes). Second day time at 12 km/h is 10 / 12 ≈ 0.8333 hours (50 minutes). Compared to the on-time 0.75 hours, she is now 0.0833 hours late, which is 5 minutes. Both lateness values match the problem statement perfectly.

Why Other Options Are Wrong:
- 15 km, 20 km, or 30 km, when substituted into the two speed conditions, do not yield lateness values of exactly 15 minutes and 5 minutes. Either one or both conditions fail, so these distances cannot be correct.

Common Pitfalls:
Some students mistakenly assume the scheduled time is the same as one of the actual times, or they attempt to work directly with differences between speeds without explicitly writing equations. Others misconvert minutes to hours. Expressing everything in hours and setting up two clear equations is the safest path to the correct answer. Always remember that “late” means the actual travel time is longer than the scheduled time by the given amount.

Final Answer:
The distance of the university from Madhu’s house is 10 km.