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 45th 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?

• we will solve in simplest way
  let its usual time =ut
  1.
  5/4ut=ut+45(as time is inversaly proportional to speed)
  ut=180minutes
  let total distance =d..hence..d-100=180*x.....(i)
  2.5/4ut=ut+30...ut=120 minutes
  d-150=120*x...(ii)
  d-100/180=d-150/120
  d=250(ans)
  
• 8 years agoHelpfull: Yes(6) No(0)
• Let, initial speed of the train = 5 km/h
  Then, speed after the accident = 45 x 5 = 4 km/h
  Time taken to cover 50 km @ 5 km/h = 10 hours
  Time taken to cover 50 km @ 4 km/h = 1212 hours
  Difference between times taken = 1212 - 10 = 212 hours = 150 minutes
  But, actual difference = (45 - 30) minutes = 15 minutes
  = 110 of 150 minutes
  Therefore, Speed is 10 times of assumed speed.
  Therefore, Speed before accident = 10 x 5 = 50 km/h
  And, Speed after accident = 10 x 4 = 40 km/h
  Distance after accident is covered @ 40 km/h instead of 50 km/h
  And, time difference = 45 minutes = 34 hour
  Therefore, Distance between place of accident and B = 50×4050−40×34 = 150 km
  Therefore, Distance between A and B = 100 + 150 = 250 km.
• 8 years agoHelpfull: Yes(0) No(3)