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. At what time Bus D reached the platform P1?

### Concept & Logical Deduction This question focuses entirely on time sequence tracking based on defined intervals. It requires parsing the specific rules for bus arrival times at individual platforms. ### Step-by-Step Solution **1. Identify Relevant Constraints:** * The first bus reaches platform P1 at exactly 5 AM. * There is a mandatory gap of 1 hour between any two buses reaching the exact same station. * Bus C and D both reached the nearest platform (P1). **2. Map the Timeline for P1:** * Slot 1 (1st Bus): 5 AM * Slot 2 (2nd Bus): 6 AM * Slot 3 (3rd Bus): 7 AM **3. Apply the Clues:** * The problem explicitly states: "C reached the destination at 6 AM." * This means C took the 2nd slot at P1. * The problem also states: "Bus C and D reached the nearest platform in a gap of 1 hour. C started its journey earlier than D." * Since C arrived earlier at 6 AM, and D arrived with a 1-hour gap, D must have arrived exactly 1 hour after C. * 6 AM + 1 hour = 7 AM. ### Exam Strategy & Shortcut You do not need to solve the entire 10-bus platform matrix for this particular question. Scan the text strictly for clues related to C, D, and their timings. The math is a simple addition of the 1-hour gap to C's explicitly given arrival time. ### Common Pitfall Confusing the arrival times of different platforms (e.g., mixing up the starting time of P3 with P1) instead of using the uniform 1-hour gap rule for buses arriving at the *same* platform. ### Final Answer Therefore, the correct answer is **7AM**.