A train, an hour after starting, meets with an accident which detains it for a…

2024

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

  1. A.

    400

  2. B.

    465

  3. C.

    600

  4. D.

    640

Attempted by 5 students.

Show answer & explanation

Correct answer: C

Explanation:

  1. Let the original speed be v km/h and the total distance be D km.

  2. First scenario (accident after 1 hour): Time before accident = 1 h. Detention = 0.5 h. Remaining distance = D − v, travelled at reduced speed 3v/4, so time after = 4(D − v)/(3v). Actual total time = 1 + 0.5 + 4(D − v)/(3v). Scheduled time = D/v. Given actual = scheduled + 3.5, so:

    1.5 + 4(D − v)/(3v) = D/v + 3.5

    Solve: 4(D − v)/(3v) = D/v + 2 ⇒ multiply by 3v ⇒ 4(D − v) = 3D + 6v ⇒ 4D − 4v = 3D + 6v ⇒ D = 10v.

  3. Second scenario (accident 90 km farther): Time before accident = (v + 90)/v = 1 + 90/v. Detention = 0.5 h. Remaining distance = D − v − 90, time after = 4(D − v − 90)/(3v). Actual total time = 1 + 90/v + 0.5 + 4(D − v − 90)/(3v). Given actual = scheduled + 3, so:

    1.5 + 90/v + 4(D − v − 90)/(3v) = D/v + 3

    Multiply by 3v: 270 + 4(D − v − 90) = 3D + 4.5v ⇒ 4D − 4v − 90 = 3D + 4.5v ⇒ D = 8.5v + 90.

  4. Use D = 10v from the first result: 10v = 8.5v + 90 ⇒ 1.5v = 90 ⇒ v = 60 km/h. Then D = 10v = 600 km.

  5. Therefore the length of the trip is 600 km.

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