The distance between places U and V is 1008 km. An express train leaves place U at 9:00 a.m., runs at a constant speed of 126 km/hr, and also stops on the way for a total of 20 minutes. At what time in the afternoon (p.m.) will the train reach place V?

Introduction / Context:
This question combines uniform speed motion with a fixed stoppage time. The express train travels a known distance at a constant speed and also halts for a specific duration. The task is to compute the overall journey time including stops and then convert this into an actual clock time of arrival.

Given Data / Assumptions:
- Distance between U and V = 1008 km.
- Speed of the express train while moving = 126 km/hr.
- Total stoppage time during the journey = 20 minutes.
- Departure time from U = 9:00 a.m.
- Train speed is constant whenever it is moving, and stoppage time is simply added to the running time.

Concept / Approach:
The main relationship is time = distance / speed for the running part of the journey. Once we compute the running time in hours, we add the additional stoppage time converted into hours. Finally, we add the total journey time to the 9:00 a.m. departure time to find the arrival time at V. Careful conversion of minutes to hours and correct handling of clock time are the key steps.

Step-by-Step Solution:
Step 1: Compute the pure running time (without stops) using time = distance / speed.Step 2: Running time = 1008 / 126 hours.Step 3: Since 126 * 8 = 1008, the running time is exactly 8 hours.Step 4: Convert the stoppage time of 20 minutes into hours: 20 minutes = 20 / 60 = 1/3 hour.Step 5: Total journey time = running time + stoppage time = 8 + 1/3 hours = 8 hours and 20 minutes.Step 6: Add this total journey time to the departure time of 9:00 a.m. Adding 8 hours gives 5:00 p.m., and adding 20 minutes gives 5:20 p.m.

Verification / Alternative check:
We can check reasonableness. A train at 126 km/hr covers roughly 120–130 km each hour. For a 1008 km journey, 1008 / 126 = 8 hours makes sense (about 8 * 125 = 1000 km). Adding a small 20-minute stop on top naturally leads to an arrival time slightly after 5:00 p.m. There are no other pauses or speed changes mentioned, so 5:20 p.m. is fully consistent.

Why Other Options Are Wrong:
3:50 p.m. and 4:20 p.m. are much earlier than the minimum 8 hours required from 9:00 a.m. even without any stops. 6:50 p.m. would correspond to more than 9 hours and 30 minutes of total travel, which is too long for this distance and speed. 4:50 p.m. would correspond to a shorter travel time than 8 hours, again impossible at the given speed.

Common Pitfalls:
Some learners forget to add the 20-minute stoppage, using only the running time and ending at 5:00 p.m. Others incorrectly convert 20 minutes into hours or accidentally subtract instead of add the stoppage time. Another common mistake is to mis-handle the a.m. to p.m. transition when adding hours to 9:00 a.m.

Final Answer:
The express train will reach place V at 5:20 p.m.