It is given that
Distance = 360 km
Consider x as the speed, then the time taken
t = $\dfrac{360}{x}$
If the speed is increased by 5 km/h speed will be (x + 5) km/h
Distance will be the same
t = $\dfrac{360}{x+5}$
We know that
Time with original speed – Time with increased speed = 1
$\dfrac{360}{x}$ – $\dfrac{360}{x+5}$= 1
LCM = x (x + 5)
$\dfrac{[360 (x + 5) – 360x]}{x(x + 5)}$ = 1 360 x + 1800 – 360x = x (x + 5)
$x^{2} + 5x $ = 1800
$x^{2} + 5x – 1800 $ = 0
$x^{2}+ 45x – 40x – 1800 $= 0
x (x + 45) – 40 (x + 45) = 0
(x – 40) (x + 45) = 0
x = 40 km/hr
As we know, the value of speed cannot be negative.
Therefore, the speed of the train is 40 km/h.