Flight A and Flight B both travel the same long route of 7200 kilometres. Under normal conditions, Flight A takes 1 hour more than Flight B to cover this distance. Due to engine trouble, the speed of Flight B decreases by a factor of 1/6 (so it becomes five sixths of its original speed), and now Flight B takes 36 minutes more than Flight A to complete the journey. What is the speed of Flight A in km/h?

Introduction / Context:
This question explores comparative speed and time for two flights covering the same long distance. It involves setting up equations based on the relationship between distance, speed, and time and then adjusting one flight speed due to an engine problem. The problem requires forming and solving linear equations rather than just direct substitution, so it tests algebraic manipulation in the context of time and distance.

Given Data / Assumptions:

Distance for both flights A and B = 7200 kilometres.
Under normal conditions, time taken by A is 1 hour more than time taken by B.
Speed of B under normal conditions is v_B km/h and speed of A is v_A km/h.
Due to engine trouble, new speed of B becomes (5 / 6) * v_B.
With this reduced speed, time taken by B is 36 minutes (0.6 hour) more than time taken by A.
We need to find v_A, the speed of Flight A in km/h.

Concept / Approach:
If distance is D and speed is v, then time = D / v. Using this relation, we translate the two time comparisons into equations. First, we relate the original times of A and B. Second, we relate the time of B after the speed reduction to the time of A. This produces a system of two equations in two unknowns (the times or the speeds). Solving these equations yields the required speed of Flight A.

Step-by-Step Solution:
Step 1: Let the normal speeds be v_A for Flight A and v_B for Flight B. Then time taken by A is 7200 / v_A hours and time taken by B is 7200 / v_B hours. Step 2: From the first condition, A takes 1 hour more than B: 7200 / v_A = 7200 / v_B + 1. Step 3: When B's speed reduces to (5 / 6) * v_B, its new time is 7200 / [(5 / 6) * v_B] = (6 / 5) * 7200 / v_B. Step 4: Second condition states new time of B is 0.6 hour more than time of A: (6 / 5) * 7200 / v_B = 7200 / v_A + 0.6. Step 5: Solve these equations. From the first equation, express 7200 / v_A in terms of 7200 / v_B. Substitute into the second equation and solve to get time of A = 9 hours and time of B = 8 hours. Step 6: Speed of Flight A = distance / time = 7200 / 9 = 800 km/h.

Verification / Alternative check:
At 800 km/h, Flight A takes 9 hours. Since Flight B takes 1 hour less normally, its time is 8 hours, giving speed v_B = 7200 / 8 = 900 km/h. After the engine trouble, speed of B becomes (5 / 6) * 900 = 750 km/h. Now time of B becomes 7200 / 750 = 9.6 hours. Time of A is 9 hours, so B now takes 0.6 hour (36 minutes) more than A. Both conditions are satisfied, confirming that 800 km/h for Flight A is correct.

Why Other Options Are Wrong:
If Flight A had a speed of 900 km/h or 750 km/h or 720 km/h, the derived times and adjusted times would not satisfy both given time differences of 1 hour and 36 minutes. The numerical checks would fail one or both of the conditions. Only 800 km/h leads to consistent times before and after the speed change of Flight B.

Common Pitfalls:
Some learners attempt to compare speeds directly without forming proper equations or forget to convert 36 minutes into 0.6 hour. Others may mistakenly think the speed of Flight A also changes when only Flight B is affected. A systematic approach, carefully setting up equations from the language of the problem, avoids these mistakes and leads to the correct result.

Final Answer:
The speed of Flight A is 800 km/h.