Question

A train travels a distance of 300 km constant speed if the speed of train is increased by 5km an hour the journey would have taken 2 hours find original speed of the train

Answer 1

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.

Answer 2

Answer:

Let the usual speed of the train be y km / hr .

A.T.Q.

$\displaystyle{\frac{300}{y} -\frac{300}{y+5} =2 }$

300 ( y + 5 ) - 300 y = 2 y ( y + 5 )

y² + 5 y - 750 = 0

y² + 30 y - 25 y - 750 = 0

( y + 30 ( y -25 ) = 0

y = - 30 or y = 25

Since , speed of train can't be negative .

Therefore , the usual speed of the train is 25 km / hr .