A train travels at a certain average speed for a distance of 54 km and then travels a distance of 63 km at an average speed of 6 km/hr more than the first speed. If it takes 3 hours to complete the total journey, what is its first speed?

Let the first speed of the train be x km/h

Time taken to cover 54 km =

New speed of train = x + 6 km/h

Time taken to cover 63 km =

According to given question

taking LCM

117x + 324 = 3(x^{2} + 6x)

117x + 324 = 3x^{2} + 18x

3x^{2} – 99x – 324 = 0

x^{2} – 33x – 108 = 0

Using the splitting middle term – the middle term of the general equation is divided in two such values that:

Product = a.c

For the given equation a = 1 b = – 33 c = – 108

= 1. – 108 = – 108

And either of their sum or difference = b

= – 33

Thus the two terms are – 36 and 3

Difference = – 36 + 3 = – 33

Product = – 36.3 = – 108

x^{2} – 36x + 3x – 108 = 0

x (x – 36) + 3(x – 36) = 0

(x – 36) (x + 3) = 0

x = 36 or x = – 3

x = 36 (speed cannot be negative)

Hence the first speed of the train is 36 km/hr

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