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 2nd 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 = x2 + 5x
x2 + 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