Akhil starts a journey from town A to town B. After travelling 50 km he meets Raju, who advises him to reduce his speed. He then continues the remaining distance at three fourths of his original speed and reaches his destination 35 minutes later than his scheduled arrival time. If instead the meeting with Raju had taken place 24 km further along the same route, that is after Akhil had already covered 74 km from A, he would have arrived only 25 minutes late. What is the initial speed of Akhil in km/h?

Introduction / Context:
This question tests time, speed and distance concepts involving a change of speed midway through a journey and how that affects the delay in arrival. We are given two different hypothetical meeting points with Raju and two corresponding delays, and we must determine Akhil's original speed in km/h.

Given Data / Assumptions:

• Akhil travels from A to B with an original constant speed.
• He meets Raju after 50 km, then reduces his speed to three fourths of the original speed.
• With this slowdown after 50 km, he reaches B 35 minutes late.
• If the meeting and slowdown instead occurred after 74 km from A, he would be 25 minutes late.
• All speeds are constant on each segment and measured in km/h.
• Let the original speed be v km/h and the total distance be D km.

Concept / Approach:
We compare planned travel time at the original constant speed with actual travel times when Akhil slows down partway. By writing equations for both scenarios and using the given delays, we can form two equations in the unknowns D and v. Solving these simultaneously will give the value of the initial speed. The key idea is that delay equals actual time minus scheduled time.

Step-by-Step Solution:
Let the planned travel time (no slowdown) be T_plan = D / v hours. Case 1: Slowdown after 50 km. Time taken = 50 / v + (D - 50) / (3v/4) = 50 / v + (4 * (D - 50)) / (3v). This actual time is 35 minutes later, so 50 / v + (4 * (D - 50)) / (3v) = D / v + 35 / 60. Multiply both sides by v: 50 + (4/3)(D - 50) = D + (35v / 60). Case 2: Slowdown after 74 km. Time taken = 74 / v + (D - 74) / (3v/4) = 74 / v + (4 * (D - 74)) / (3v). This actual time is 25 minutes later, so 74 / v + (4 * (D - 74)) / (3v) = D / v + 25 / 60. Multiply both sides by v: 74 + (4/3)(D - 74) = D + (25v / 60). Now solve the system of two linear equations in D and v. Solving gives D = 134 km and v = 48 km/h. Therefore, Akhil's initial speed is 48 km/h.

Verification / Alternative check:
If v = 48 km/h and D = 134 km, then T_plan = 134 / 48 = 2.7917 hours (about 2 hours 47.5 minutes). In the first case, time = 50 / 48 + 84 / 36 = 1.0417 + 2.3333 = 3.375 hours, which is 0.5833 hours or 35 minutes more than T_plan. In the second case, time = 74 / 48 + 60 / 36 = 1.5417 + 1.6667 = 3.2083 hours, which is 0.4167 hours or 25 minutes more than T_plan. Both match the problem statement.

Why Other Options Are Wrong:
Speeds of 36 km/h, 24 km/h, 20 km/h and 40 km/h do not simultaneously satisfy the delay conditions of 35 minutes for the first meeting point and 25 minutes for the second. Substituting any of these speeds into the equations leads to inconsistent delays, so they cannot be correct.

Common Pitfalls:
A common mistake is to mix minutes and hours without proper conversion, or to assume that the delay depends only on one scenario. Another error is forgetting that after the meeting Akhil travels the remaining distance at the reduced speed only, not switching back. Keeping consistent units and carefully setting up both equations is essential.

Final Answer:
The initial speed of Akhil is 48 km/h.