Question

A train covers a distance of 90 km at a uniform speed. In case of starting late due to some reason, the speed can be increased by 15 km/h and it would take half an hour less for the journey. Find the original speed

Answer 1

1)distance covered = 90 km

let the speed of the train = x km/hr

time (t1 ) = distance /speed

t1 = 90/x -----(1)

2) distance = 90 km

speed = (x+15) km/hr

t2 = 90 /(x+15) ----(2)

given 

t1-t2 = 30 minutes

90/x -  90 / (x+15)  = 1/2 hr

[90(x+15) -90x]/x(x+15) = 1/2

[90x + 1350 -90x]/ (x^2+15x) =1/2

1350 *2= x^2+15x

2700 = x^2+15x

x^2+15x-2700=0

x*x +60x -45x - 45*60=0

x(x+60) - 45(x+60)=0

(x+60)(x-45)=0

x+60=0 or x-45=0

x= -60  0r x= 45

x should not be negative

x= 45

therfore speed of the train = 45km/hr

Answer 2

$- 90 \div x + 5 + 90 \div x = 1 \div 2 \\ 90x + 450 - 90x \div {x}^{2} + 5x = 1 \div 2 \\ 900 = {x }^{2} + 5 x \\ factoriseit$