Let original speed of train = x km/h

We know,

Time = distance/speed

We have

Time taken by train = `360/x` hour

And, Time taken by train its speed increase 5 km/h = `360/( x + 5)`

It is given that,

Time taken by train in first – time taken by train in 2^{nd} case = 48 min = `48/60` hour

`360/x - 360/(x + 5) = 48/60 = 4/5`

3`60(1/x - 1/(x +5)) = 4/5`

`360 xx 5/4 (5/(x^2 + 5x))` = 1

450 × 5 = x^{2} + 5x

x^{2} + 5x – 2250 = 0

`x = (-5 +- sqrt(25 + 9000))/2`

= `(-5 +- sqrt(9025))/2`

= `(-5 +- 95)/2`

= – 50, 45

But x ≠ – 50 because speed cannot be negative

So, x = 45 km/h

Hence, original speed of train = 45 km/h