Question

A train left station a for station b at a certain speed. After travelling for 100 km, the train meets with an accident and could travel at 4 5 th of the original speed and reaches 45 minutes late at station

b. Had the accident taken place 50 km further on, it would have reached 30 minutes late at station

b. What is the distance between station a and b

Answer 1

Answer:

50 km/hr

250 km

Step-by-step explanation:

Let say Distance  = D  km

Speed  = 5S  km/Hr

Time Required for Journey = D/5S  hr

Time Taken for 100 km = 100/5S  hr

4/5 of original Speed = (4/5)5S = 4S

Time taken for D - 100 km = (D-100)/4S hr

100/5S + (D-100)/4S  = D/5S   + 3/4     (45 minutes = 3/4 hrs)

Multiplying by 20S

=> 400 + (D -100)5  = 4D  + 15S

=> D = 15S + 100

Had the accident taken place 50 km further on

=> 150/5S + (D-150)/4S  = D/5S   + 1/2

=> 600 + (D - 150)5 = 4D + 10S

=> D = 150 + 10S

=> 15S + 100 = 150 + 10S

=> 5S = 50

=> S = 10

Speed = 5S = 50 km /hr

D = 15*10 + 100 = 250 km