Question

A train after travelling 150 km meets with an accident and then proceed at 3/5 of its former speed and arrived at its destination 8 hrs late. Had the accident occured 360 km futher it would have reached the destination 4hrs late. What is the total distance travelled by train

Answer 1

Answer:

let s = the normal speed of the train

then

0.6s = the speed after the accident (3/5 the normal speed)

let d = total distance

Then

(d-150)km = distance traveled at the slower speed, first scenario

and

(d - (150+360)) = distance traveled at the slower speed (360 km further)

(d - 510) km

Write a time equation for the 1st scenario

slower time - normal time = 8 hrs

(150%2Fs + %28d-150%29%2F.6s) - d%2Fs = 8

Multiply by 0.6s to clear the denominators

(.6(150) + (d-150)) - 0.6d = 0.6s(8)

90 + d - 150 - 0.6d = 4.8s

0.4d - 60 = 4.8s

0.4d - 4.8s = 60

Same with the 2nd scenario, (accident 360 km further down the road)

(510%2Fs + %28d-510%29%2F.6s) - d%2Fs = 4

Multiply by 0.6s to clear the denominators

(0.6(510) + (d-510)) - 0.6d = 0.6s(4)

306 + d - 510 - 0.6d = 2.4s

0.4d - 204 = 2.4s

0.4d - 2.4s = 204

Find the speed using elimination with these two equations

0.4d - 2.4s = 204

0.4d - 4.8s = 60

------------------subtraction eliminates d find s

0 + 2.4s = 144

s = 144/2.4

s = 60 km/hr is the normal speed

then

0.6(60) = 36 km/hr is the "after accident speed"

Find the distance using the 1st scenario equation, replace s with 60

0.4d - 4.8(60) = 60

0.4d - 288 = 60

0.4d = 60 + 288

0.4d = 348

d = 348/.4

d = 870 km is the total distance

:

:

Check solution using the 2nd scenario

510%2F60 + %28870-510%29%2F36 - 870%2F60 =

8.5 + 10 - 14.5 = 4 hrs,

The speed of train is=870km