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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Answered by
69
Let speed be 's ', time be 't ' and distance be 'd ' t, t = d / s t = 300 / s ,
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.
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.
dweejareddy:
pls write step-by-step
Answered by
143
total distance=300km
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
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
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