Question

A train met with an accident 150 km from station A.
It completed the remaining journey at of the
previous speed and reached 15 min late at station B.
Had the accident taken place 30 km further, it would
have been only 7 min late. Find the speed of the train
and the distance between the two stations A and B.​

Answer 1

Answer:

Disatance=206.25km , Speed=1125km/hr

Step-by-step explanation:

Heya there is some mistake in question according to sources speed gets reduced by 5/6 of previous speed note this it is missing in question.

Let the initial speed= x & distance= y

original time taken be= y/x

After the accident speed reduces by 5/6 of previous speed,so speed after accident= x - (x(5/6)) = x/6.

1st condition:-

(150/x) + ((6(y-150))/x) =(y/x) +15,

or, 150 + 6y -900 = y + 15x,

15x - 5y = -750 ---------- (1)

2nd condition:-

(180/x) + ((6(y-180))/x) = (y/x) + 7,

180 + 6y - 1080 = y + 7x,

7x - 5y = -900 -----------(2)

Solving (1) & (2),

we get x= 75/4 & y= 206.25 

Therefore x = (75/4)km/minute = 1125 km/hr

and

y = 206.25km

Answer 2

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