Question

After covering a distance of 30 km with a speed there is some defect in a train engine and therefore, its speed is reduced to 4/5 of its original speed. Consequently, the train reaches its destination late by 45 minutes. Had it happened after covering 18 kilometers more, the train would have reached 9 minutes earlier. Find the speed of the train and the distance of journey.

Answer 1

Let the speed of train is x km/h and distance of journey is y km

so, original time, t = y/x hrs

case 1 : speed of train = 4x/5 km/h

distance travelled by train with speed 4x/5 = (y - 30)km

and time taken = (t + 45/60) = (t + 3/4) hrs

we know, time = distance/speed

or, t + 3/4 = 30/x + (y - 30)/(4x/5)

or, y/x + 3/4 = 30/x + 5(y - 30)/4x

or, 4y + 3x = 120 + 5y - 150

or, y - 3x = 30 ......(1)

case 2 : speed of train = 4x/5

distance travelled by train with speed 4x/5 = (y - 48)

time taken = (t + 45/60 -9/60) = (t + 36/60) hrs = (t + 3/5) hrs

so, (t + 3/5) = 48/x + (y - 48)/(4x/5)

or, y/x + 3/5 = 48/x + 5(y - 48)/4x

or, 4y + 12x/5 = 192 + 5y - 240

or, - y + 12x/5 = -48

or, y - 12x/5 = 48.......(2)

from equations (1) and (2),

-12x/5 + 3x = 48 - 30

or, 3x/5 = 18

or, x = 30 and y = 120

so, speed of train = 30 km/h

so, speed of train = 30 km/hand distance of journey = 120km