a train covers a distance of 90 km at a uniform speed. Had the speed been 15 km/hr it would have taken 30 minutes less for the journey. find the oroginal speed of the train
Answers
let speed (original) = x km/h
time taken to travel = 90/x hr
∴ New speed = (x+15) km/hr
Time taken to travel = 90/(x+15) hr
∴ 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 [Quadratic Equation]
⇒(x+60)(x-45) = 0
⇒x = 45 or - 60
∵ Speed can't be negative..
∴ Original speed = 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. ]