A train travels at a certain average speed of a distance of 63 km and then travels at 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 average speed?

Let, the original average speed be x km/hr.
Time taken at a speed of x km/hr (t1) = 63/x hrs
Time taken at a speed of x+6 km/hr (t2) = 72/(x+6) hrs
t1 + t2 = 3
63/x + 72/(x+6) = 3
=> 63(x+6) + 72x = 3x(x+6)
=> 63x + 378 + 72x = 3x² + 18x
=> 3x² + 18x - 135x - 378 = 0
=> 3x² - 117x - 378 = 0
=> 3(x² - 39x - 126) = 0
=> x² - 39x - 126 = 0
=> x² - 42x + 3x - 126 = 0
=> x(x - 42) + 3(x - 42) = 0
=> (x - 42)(x + 3) = 0
Either x = 42 or x = -3
Since speed cannot be negative, x = 42 is considered.
Therefore, the original average speed is 42 km/hr