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 platforms have more than two buses?

### Concept & Logical Deduction This is a constraint satisfaction and linear arrangement problem. The key is tracking the maximum capacity of each platform and anchoring the known destinations first before solving the conditional clues. ### Step-by-Step Solution **1. Understand the Basic Rules:** * 10 Buses: A, B, C, D, E, F, G, H, J, K. * 4 Platforms: P1 (nearest) to P4 (farthest). * Max capacity per platform: 3 buses. **2. Place the Direct Anchors:** * Bus C and D reached the nearest platform, so **C and D are at P1**. * Bus A is at a platform that is two more than D's platform (P1 + 2). Thus, **A is at P3**. * Bus J reached the last platform, so **J is at P4**. **3. Deduce the Capacities & Relative Positions:** * P4 has exactly "one more bus" besides J. So, **P4 has exactly 2 buses**. * Bus B and K have the same destination. If they were at P4, P4 would have 3 buses (J, B, K), violating the 2-bus limit. So, B and K are NOT at P4. * Bus B is "ahead" of E (meaning B is at a higher-numbered platform). E is not at P1 or P4. If E were at P3, B would have to be at P4 (impossible). Thus, **E is at P2**, and **B and K are at P3**. (P3 now has A, B, K and is at its max capacity of 3). **4. Place the Remaining Buses (F, G, H):** * F and G share the same destination. They cannot be at P1 (with D) or P4 (with J). P3 is already full. Therefore, **F and G must be at P2**. (P2 now has E, F, G and is full). * H is not at P1 (with C), P2 (with E), or P3 (with K). Therefore, **H must be at P4**. **Final Platform Distribution:** * **P1:** C, D (2 buses) * **P2:** E, F, G (3 buses) * **P3:** A, B, K (3 buses) * **P4:** J, H (2 buses) Based on this complete arrangement, the platforms that have more than two buses are P2 and P3. ### Exam Strategy & Shortcut Always map out a hard matrix of Platforms (P1 to P4) with their known limits (max 3). Fill in the definitive clues first (C, D at P1; J at P4) and cross off impossibilities to force the remaining groups into the only available slots. ### Common Pitfall Assuming "ahead" refers to the nearest platform (lower number) instead of the direction of travel (higher number). Contextual clues usually frame P1 as the start and P4 as the furthest point "ahead". ### Final Answer Therefore, the correct answer is **P2 and P3**.