A bus travels 2/5 of a total journey at its usual speed. The remaining distance is covered at 6/7 of its usual speed. Because of the reduced speed on the remaining part, it reaches its destination 50 minutes late. If the total distance is 200 km, what is the usual speed of the bus (in km/h)?

Introduction / Context:
This problem combines average speed concepts with a change in speed over different segments of a fixed journey. The bus travels part of the distance at its usual speed and the remaining part at a reduced speed, causing it to arrive late. We must compute the usual speed using the information about the delay and the distances covered at each speed. This type of question is a staple in time and distance sections of aptitude exams.

Given Data / Assumptions:
- Total distance = 200 km.
- Usual speed of the bus = v km/h (unknown).
- First 2/5 of the journey is covered at usual speed v.
- Remaining 3/5 of the journey is covered at reduced speed (6/7) * v.
- The bus reaches 50 minutes late due to the slower speed on the last part.
- 50 minutes = 50/60 = 5/6 hours.
- If the bus had travelled the whole journey at usual speed v, it would arrive on time.

Concept / Approach:
We first calculate the time it would have taken to cover the 200 km at the usual speed v. Then we express the actual time with mixed speeds: one time for the first 2/5 of the distance at v and another time for the remaining 3/5 at (6/7)v. The difference between the actual time and the ideal time is given as 5/6 hours. Setting up this equation allows us to solve for v. The key formula throughout is time = distance / speed.

Step-by-Step Solution:
Step 1: Compute the “on-time” travel time if the entire 200 km is covered at usual speed v.Ideal time T_ideal = 200 / v hours.Step 2: Compute segment distances.First part (2/5 of 200) = 80 km.Second part (3/5 of 200) = 120 km.Step 3: Time for first part at speed v: 80 / v hours.Step 4: Time for second part at speed (6/7)v: 120 / [(6/7)v] = 120 * 7 / (6v) = 140 / v hours.Step 5: Actual time T_actual = 80 / v + 140 / v = 220 / v hours.Step 6: The delay is T_actual - T_ideal = 5/6 hours.So 220 / v - 200 / v = 5/6 ⇒ (20 / v) = 5/6.Step 7: Solve for v: v = 20 * 6 / 5 = 120 / 5 = 24 km/h.

Verification / Alternative check:
Check the times using v = 24 km/h. Ideal time = 200 / 24 ≈ 8.333 hours (8 hours 20 minutes). Actual time: first part 80 / 24 ≈ 3.333 hours, second part at reduced speed (6/7 * 24 = 144/7 ≈ 20.571 km/h) gives 120 / (144/7) = 120 * 7 / 144 = 5.833 hours. Total actual time ≈ 3.333 + 5.833 = 9.166 hours. Difference ≈ 9.166 - 8.333 = 0.833 hours, which is exactly 5/6 hours or 50 minutes, matching the problem statement.

Why Other Options Are Wrong:
- The speeds 20.57 km/h, 28 km/h, and 26.52 km/h, when used in the time calculations, do not produce a delay of 50 minutes. The delay is either more or less than 5/6 hours.
- Only 24 km/h leads to the exact delay specified in the question while covering 200 km.

Common Pitfalls:
Typical mistakes include mixing up the fractional distances (2/5 and 3/5) or incorrectly computing the reduced speed segment. Some students also forget to convert 50 minutes into hours or mistakenly treat 50 minutes as 0.5 hours instead of 5/6. Accuracy with fractions and step-by-step calculation of times makes the solution straightforward.

Final Answer:
The usual speed of the bus is 24 km/h.