# A train travels a distance of 300 km at a constant speed.If the speed of the train is increased by 5 km hour, the journey would have taken 2 hours less. Find the original speed of the train.

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by dweejareddy

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by dweejareddy

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d = (s + 5)(t -2) => d = (s+5)(300/s-2) {t=300/s} => 300s = (s+5)s(300/s-2) [multiplyning both sides with s]we get = > 300s = (s+5)(300 - 2s)=> 300s = 300s - 2s2 +1500 - 10s => 300s = 300s - 2s2 + 1500 - 10s=> 0 = -2s2 +1500 - 10s=> 0 = 2s2 +10s - 1500 [multiplying both sides -1 ]=> 0 = 2s2 +60s - 50s - 1500=> 0 = 2s(s + 30) -50(s + 30)=> 0 = (2s - 50)(s + 30)=> s = 25, -30 Since, speed cannot be negative, we take speed 25Km/h.

let speed=xkm/hr

therefore time = d/s = 300/x hr---------------1

new speed=x+5 km/hr

therefore time=d/s = 300/x+5 hr-------------2

given difference in times = 2 hrs

therefore from 1 and 2

300/x - 300/x+5 = 2

300x+1500-300x/x(x+5) = 2

1500 = 2xsquare + 10x

2x square + 10x - 1500 = 0

x2 + 5x - 750 = 0

x2 + 30x - 25x -750 =0

x(x+30) - 25(x+30)

(x+30)(x-25)=0

x+30=0 ! x-25=0

x= -30 ! x = 25

speed cannot be measured in negetive

therefore original speed of train is 25 km/hr