Multiple choice

A train, an hour after starting, meets with an accident which detains it a half hour, after which it proceeds at $\frac{3}{4}$ of its former rate and arrives $3\frac{1}{2}$ hours late. Had the accident happened $90$km farther along the line, it would have arrived only $3$ hours late. The length of the trip in km was:

  1. $400$
  2. $465$
  3. $600$
  4. $640$
  5. $550$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let d be the total distance and v be the speed. The original time is d/v. The accident happens at distance v (1 hour). Delay is 0.5 hours. Remaining distance (d-v) is covered at 3/4v. Total time = 1 + 0.5 + (d-v)/(0.75v) = d/v + 3.5. This simplifies to (d-v)/(0.75v) - (d-v)/v = 3. Solving for d-v gives 9v. Using the second condition (accident at v+90), we find the distance.

AI explanation

Using the late time difference method, compare the two accident scenarios to find the ratio of original speed to reduced speed. Traveling the 90 km stretch at 3/4 of the original speed causes an extra 0.5 hours of delay, yielding the equation 90/(3S/4) minus 90/S equals 0.5. Solving this shows the original speed is 60 km/hr and the reduced speed is 45 km/hr. The total distance is then found by setting up the total time difference equation, calculating to a total trip length of 600 km.

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