A train leaves Delhi at 10:00 a.m. and reaches Jaipur at 4:00 p.m. on the same day. Another train leaves Jaipur at 12:00 noon and reaches Delhi at 5:00 p.m. the same day. Approximately at what time of day will the two trains meet?

Introduction / Context:
This problem involves two trains travelling between the same two cities, Delhi and Jaipur, at different departure times and with different total journey durations. We need to determine the clock time at which they meet on the way. It is a typical relative speed and time-line question frequently found in aptitude tests and requires careful handling of departure times and travel durations.

Given Data / Assumptions:
- Train 1 (Delhi to Jaipur) leaves Delhi at 10:00 a.m. and reaches Jaipur at 4:00 p.m., so its journey time is 6 hours.
- Train 2 (Jaipur to Delhi) leaves Jaipur at 12:00 noon and reaches Delhi at 5:00 p.m., so its journey time is 5 hours.
- Let the distance between Delhi and Jaipur be D km.
- Both trains travel along the same straight route and at constant speeds throughout their journeys.

Concept / Approach:
We first find each train’s speed in terms of the unknown distance D. Then we consider a variable time t (in hours) after 10:00 a.m. when the trains meet. By that time, Train 1 will have travelled for t hours, while Train 2, which departs later at 12:00 noon, will have travelled for (t - 2) hours, as long as t is greater than 2 hours. The sum of the distances they have covered by the meeting time must equal the total distance D. This leads to an equation in t, from which we can find the meeting time and convert it to a clock time.

Step-by-Step Solution:
Step 1: Compute speeds in terms of D.Train 1 speed = D / 6 km/h.Train 2 speed = D / 5 km/h.Step 2: Let t be the time in hours after 10:00 a.m. when the trains meet.Then Train 1 has travelled t hours; Train 2, which starts at 12:00 noon, has travelled t - 2 hours.Step 3: Distances covered by meeting time:Train 1: (D / 6) * t.Train 2: (D / 5) * (t - 2).Step 4: Their combined distance equals D:(D / 6) * t + (D / 5) * (t - 2) = D.Divide by D (assuming D > 0): t / 6 + (t - 2) / 5 = 1.Step 5: Solve t / 6 + (t - 2) / 5 = 1.Multiply by 30: 5t + 6(t - 2) = 30 ⇒ 5t + 6t - 12 = 30 ⇒ 11t = 42 ⇒ t = 42 / 11 ≈ 3.818 hours.Step 6: Convert 0.818 hours to minutes: 0.818 * 60 ≈ 49 minutes.Thus, meeting time ≈ 10:00 a.m. + 3 hours 49 minutes = 1:49 p.m.

Verification / Alternative check:
At 1:49 p.m., Train 1 has been travelling for approximately 3.818 hours and Train 2 for 1.818 hours. Train 1 covers (D / 6) * 3.818 ≈ 0.636D, while Train 2 covers (D / 5) * 1.818 ≈ 0.364D. The sum is very close to D (allowing for rounding), showing consistency. The approximate nature of the options also matches the approximate nature of the calculation.

Why Other Options Are Wrong:
- 1:42 p.m., 1:27 p.m., and 2:04 p.m. correspond to different values of t which, when plugged into the equation, do not satisfy t / 6 + (t - 2) / 5 = 1. They either make the combined travelled distance less than D (meeting earlier) or greater than D (meeting later than 2:00 p.m.).

Common Pitfalls:
Common mistakes include ignoring the different start times and treating both trains as if they started at 10:00 a.m., or miscomputing the speeds by assuming a specific distance instead of leaving it symbolic as D. Another issue is working with approximate decimals too early, which can introduce rounding errors before the final step. Keeping t as a fraction until the end helps maintain accuracy.

Final Answer:
The two trains will meet at approximately 1:49 p.m.