Two people A and B start from the same place: A starts at 11:00 a.m. and walks at 3.5 km per hour, while B starts at 1:00 p.m. and walks at 1 km per hour for the first hour, 2 km per hour for the next hour, 3 km per hour for the next hour and so on. At what time will B finally catch up with A?

Introduction / Context:
This problem is a relative speed question with a twist: one person walks at a constant speed, while the other starts later and increases speed in steps every hour. Such questions are common in arithmetic reasoning sections of competitive exams because they test the ability to handle nonuniform motion and to set up equations across time intervals.

Given Data / Assumptions:
- Person A starts at 11:00 a.m. with a constant speed of 3.5 km per hour.
- Person B starts from the same point at 1:00 p.m., which is 2 hours after A.
- B walks at 1 km per hour from 1:00 p.m. to 2:00 p.m.
- B walks at 2 km per hour from 2:00 p.m. to 3:00 p.m.
- B walks at 3 km per hour from 3:00 p.m. to 4:00 p.m., and so on, increasing by 1 km per hour each hour.
- Both move along the same straight path in the same direction.

Concept / Approach:
Let t be the number of hours after 11:00 a.m. When t is large enough, B will have walked several hours, each at different speeds. We compute the distance A has covered as a simple linear function of t, and the distance B has covered as the sum of distances from each hourly segment. We then find the time t when both distances are equal, which gives the catch-up time.

Step-by-Step Solution:
Step 1: Let t be hours after 11:00 a.m. Step 2: Distance covered by A is A(t) = 3.5 * t. Step 3: B starts at t = 2 (1:00 p.m.). For t less than 2, B has not started and cannot catch up. Step 4: For t hours after 11:00 a.m., with t at least 2, B has walked t - 2 hours, but at different speeds each hour. From 1:00 p.m. to 2:00 p.m. the speed is 1 km per hour, from 2:00 p.m. to 3:00 p.m. it is 2 km per hour, from 3:00 p.m. to 4:00 p.m. it is 3 km per hour, and so on. Step 5: For t between 9 and 10 hours after 11:00 a.m. (that is, between 8:00 p.m. and 9:00 p.m.), B has already completed whole hours at speeds 1, 2, 3, 4, 5, 6 and 7 km per hour after starting. The total distance from those full hours is 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28 km. Step 6: In the current hour segment between 8:00 p.m. and 9:00 p.m., B walks at 8 km per hour. So for t between 9 and 10, B(t) = 28 + 8 * (t - 9). Step 7: Set A(t) equal to B(t) for t between 9 and 10: 3.5 * t = 28 + 8 * (t - 9). Step 8: Expand and simplify: 3.5 * t = 28 + 8 * t - 72, so 3.5 * t = 8 * t - 44. Step 9: Rearranging gives 8 * t - 3.5 * t = 44, so 4.5 * t = 44. Step 10: Solve for t: t = 44 / 4.5 ≈ 9.7777 hours after 11:00 a.m. Step 11: Convert 0.7777 hours to minutes by multiplying by 60: 0.7777 * 60 ≈ 46.7 minutes. Step 12: Therefore, the catch-up time is approximately 11:00 a.m. + 9 hours 47 minutes = 8:47 p.m.

Verification / Alternative Check:
Check A’s distance at t ≈ 9.78 hours: A(t) ≈ 3.5 * 9.7777 ≈ 34.22 km. For B, the distance is 28 + 8 * (t - 9) ≈ 28 + 8 * 0.7777 ≈ 34.22 km. The distances agree, confirming that B has caught A at approximately 8:47 p.m. Among the options, 8 : 47 pm is closest to this computed time.

Why Other Options Are Wrong:
Options B, C and D correspond to earlier times at which B’s cumulative distance is still less than A’s distance, since A has had a two hour head start and B’s speed increases gradually. Only by about 8:47 p.m. does B accumulate enough distance to match A.

Common Pitfalls:
Students often assume B’s speed increase can be averaged and use a single average speed, which gives an inaccurate answer. Another common error is to ignore the two hour head start of A, or to assume that B’s speed is constant during the whole journey.

Final Answer:
B will catch up with A at approximately 8 : 47 pm.