# 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.

2
by dweejareddy

2014-10-07T11:07:03+05:30
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 - 2s+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.
pls write step-by-step
2014-10-07T11:11:36+05:30
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