Multiple choice

A train met with an accident $60$ km away from Anantpur station. It completed the remaining journey at $\dfrac {5}{6}^{th}$ of the previous speed and reached Barmula station $1$ hour $12$ in late. Had accident taken place $60$ km further, it would have been only $1$ hour late.

  1. $60$ km/hr, $420$ km
  2. $80$ km/hr, $800$ km
  3. $8$ km/hr, $80$ km

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let speed be v and distance be d. 60/v + (d-60)/(5/6)v = d/v + 1.2. (d-60)/(5/6)v - (d-60)/v = 1.2 - 60/v. This leads to the correct speed and distance values.

AI explanation

The time difference between the two accident scenarios is 12 minutes for the 60 km stretch. This means the train loses 12 minutes at the reduced speed for 60 km. By calculating the difference between the normal time (60/v) and the delayed time (60 divided by 5v/6), we find the normal speed v is 60 km/hr. Using this speed to calculate the total delay of 1 hour and 12 minutes for the rest of the journey from the first accident point gives a remaining distance of 360 km, making the total distance 420 km.

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