speed of train= `x`km/hr
case 1:time taken `= 360/x` (1)
case 2 : speed of train `= (x+5)`km/hr
`time= 360/(x+5)` (2)
acc to question
`360/(x+5) = 360/x - 1`
`360x = 360x + 1800- x^2 -5x`
`x^2 +5x +1800= 0`
`x^2+45x-40x - 1800=0`
`x(x-40) + 45(x-40)=0 `
`(x-40)(x+45)=0`
`x=-45 & 40`
speed cant be negative so, sped of train is `40`km/hr