A train covers a distance of 90 km at a uniform speed .if the speed had been increased 15km/h more it would have taken 30 min less for the journey.find the original speed of the train.
Answers
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
Question:-
→ A train covers a distance of 90 km at a uniform speed. Had the speed been 15 km/hour more, it would have taken 30 minutes less for a journey. Find the original speed of the train.
Answer:
→ 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. ]