A train travels at a certain average speed for a distance of 63 km and then travels a distance of 72 km at an average speed of 6 km/hr more than its original speed. If it takes 3 hours to complete the total journey, what is the original speed of the train in km/hr?

Let the original speed of the train be 's' km/hr. The journey is divided into two parts: Part 1: Distance = 63 km Speed = s km/hr Time taken (T1) = Distance / Speed = 63/s hours Part 2: Distance = 72 km Speed = (s + 6) km/hr (6 km/hr more than original speed) Time taken (T2) = Distance / Speed = 72/(s+6) hours The total journey time is given as 3 hours. So, T1 + T2 = 3 63/s + 72/(s+6) = 3 To find the original speed 's', we can test the given options. Let's test Option C) 42 km/hr: If s = 42 km/hr, then: Time for Part 1 (T1) = 63/42 = 3/2 = 1.5 hours Speed for Part 2 = 42 + 6 = 48 km/hr Time for Part 2 (T2) = 72/48 = 3/2 = 1.5 hours Total time = T1 + T2 = 1.5 + 1.5 = 3 hours. This matches the given total journey time of 3 hours. Therefore, the original speed of the train is 42 km/hr. Analysis of other options: A) If s = 24 km/hr, T1 = 63/24 = 2.625 hours. T2 = 72/(24+6) = 72/30 = 2.4 hours. Total time = 2.625 + 2.4 = 5.025 hours (Incorrect). B) If s = 33 km/hr, T1 = 63/33 approx 1.9 hours. T2 = 72/(33+6) = 72/39 approx 1.8 hours. Total time approx 3.7 hours (Incorrect). D) If s = 66 km/hr, T1 = 63/66 approx 0.95 hours. T2 = 72/(66+6) = 72/72 = 1 hour. Total time approx 1.95 hours (Incorrect). The final answer is C.