Read the following information carefully and answer the given questions. There are ten buses A, B, C, D, E, F, G, H, J, and K which are at the base station. Four platforms P1, P2, P3, and P4 where the first bus can reach at 5 AM, 6 AM, 7 AM, and 8 AM respectively. Platform P1 is the nearest Platform and Platform P4 is the farthest. There must be a gap of one hour between the two buses reaching a particular station. The speed of the buses is considered to be the same. At most three buses can have a platform as its destination. Bus C and D started from the base station and reached the nearest platform in a gap of 1 hour. C started its journey earlier than D. C reached the destination at 6 AM. Bus A stopped its journey at a platform which is two more than Bus D. Bus J reached the last platform. Bus H's destination is not the same as the destination of C, E and K. Destinations of F and G are the same. The destination of F and G is not the same as the destination of D and J. Bus B's destination is the same as the destination of K. There is only one more bus at the platform where Bus J reached its destination. Bus E's final destination is not P1 and P4. Bus B's destination is ahead of Bus E's destination. Which of the following buses have destination as P3?

### Concept & Logical Deduction This is a logical grouping and linear arrangement problem. The objective is to assign exactly 10 variables to 4 categories under strict constraints. ### Step-by-Step Solution **1. Setup the Matrix:** * 4 Platforms: P1 to P4. * Rule: Max 3 buses per platform. **2. Deduce Destinations:** * **P1:** C and D reached the nearest platform. (C, D are at P1). * **P3:** A stopped at a platform two more than D (1 + 2 = 3). (A is at P3). * **P4:** J reached the last platform (P4). There is exactly "one more bus" here, so P4 has a max capacity of 2 buses. * **P2 & P3:** B and K have the same destination. B is ahead of E (higher platform number). E cannot be at P1 or P4. If E is at P3, B/K must be at P4 (which violates P4's max capacity of 2). Therefore, E must be at P2, leaving B and K to be at P3. * At this point, P3 contains buses A, B, and K, hitting its maximum capacity of 3 buses. * **P2:** F and G go together but not to P1 or P4. P3 is full, so F and G must join E at P2. * **P4:** H cannot be at P1, P2, or P3, so H takes the last slot at P4. **Final Arrangement:** * **P1:** C, D * **P2:** E, F, G * **P3:** A, B, K * **P4:** J, H Based on our deduction, the buses at platform P3 are A, B, and K. ### Exam Strategy & Shortcut You can solve this specific question very quickly without finding every bus. D is at P1, so A is at P3. B and K are together. Since P4 only has 2 slots (taken by J and one other), B and K can't fit there. Thus, A, B, and K naturally cluster at P3. ### Common Pitfall Failing to track the maximum capacity limits for each platform, especially the special constraint that P4 only has 2 buses total. ### Final Answer Therefore, the correct answer is **Bus A, Bus B, Bus K.**