A train travels \(\frac{1}{3}\) of the total distance at \(40\ \text{km/h}\), \(\frac{1}{4}\) of the total distance at \(60\ \text{km/h}\), and the remaining distance at \(50\ \text{km/h}\). Find the average speed of the train.
48 km/hr
Average speed = total distance ÷ total time. Take the total distance as 12 km (LCM of 3 and 4) so that each part is a whole number.
First part: \(\frac{1}{3} \times 12 = 4\) km at 40 km/h, time = \(\frac{4}{40} = \frac{1}{10}\) h.
Second part: \(\frac{1}{4} \times 12 = 3\) km at 60 km/h, time = \(\frac{3}{60} = \frac{1}{20}\) h.
Remaining distance: 12 − 4 − 3 = 5 km at 50 km/h, time = \(\frac{5}{50} = \frac{1}{10}\) h.
Total time = \(\frac{1}{10} + \frac{1}{20} + \frac{1}{10} = \frac{5}{20} = \frac{1}{4}\) h.
Average speed = \(\frac{12}{1/4} = 48\) km/h.
Hence, 48 km/hr.
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