Answer: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 trainick to let others know, how helpful is it

Answer:2

→ Original speed of the train = 45 km/hr .

Step-by-step EXPLANATION:

Given:-

→ Distance = 90 km .

Let the original speed of the train = x km/h .

→ Time taken to travel = 90/x hr .

∴ Then, new speed  = ( x + 15 ) km/hr .

∵ Time taken to travel = 90/( x + 15 ) hr .

Now, A/Q,

∵ 90/x = 90/( x + 15 ) + 1/2 .

⇒ 90/x - 90/( x + 15 ) = 1/2 .

⇒ 90[ 1/x - 1/( x + 15 )] = 1/2 .

⇒ x + 15 - x/x( x + 15 ) = 1/2 × 1/90 . 

⇒ 15/x( x + 15 ) = 180 .

⇒ x( x + 15 ) = 15 × 180 = 2700 .

⇒ x² + 15x - 2700 = 0 .

⇒ x² + 60x - 45x + 2700 = 0 .

⇒ ( x + 60 )( x - 45 ) = 0 .

⇒ x + 60 = 0 or x - 45 = 0 .

⇒ x = 45 or - 60 .

[ ∵ Speed can't be negative. ]

∴ Original speed = 45km/hr .

Hence, it is solved .

THANKS .