A train covers a distance of 90 km at a uniform seed. Had the speed been 15 km per hour more , it would have taken half an hour less for the journey. Find the original speed of the train.
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Given, distance = 90 km
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
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
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