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?

For the first part of the journey:

Distance = 63 km

Speed = x km/h

Time taken = $$\frac{63}{x}$$​ hours

For the second part of the journey:

Distance = 72 km

Speed = (x+6 )km/h

Time taken = $$\frac{72}{x + 6}$$ hours  

​$$\frac{63}{x} + \frac{72}{x + 6} = 3\\ \ \\63(x + 6) + 72x = 3x(x + 6)\\ \ \\\text{Simplifying:}\\ \ \\ 63x + 378 + 72x = 3x^2 + 18x\\ \ \\135x + 378 = 3x^2 + 18x\\ \ \\3x^2 - 117x - 378 = 0\\ \ \\ x^2 - 39x - 126 = 0\\ \ \\\text{Factoring:}(x - 42)(x + 3) = 0\\ \ \\ x = 42 \text{ km/h.} \\ \ \\ \text{and x} = -3\text{ km/h.}$$

(Speed cannot be in negative) 

Therefore, Original speed of the train is 42 km/hr.​