The Drunkard Riddle

Puzzle of the day 2038 – The drunkard problem

Puzzle 2038 - The drunkard Puzzle

A drunkard walking in a narrow lane takes 5 steps forward and 3 steps backward, followed again by 5 steps forward and 3 steps backward, and so on. Each step is 1 m long and requires 1 s.

How long the drunkard takes to fall in a pit 13 m away from the start?

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7 Responses

  1. Anonymous says:

    49 secs.

  2. Anonymous says:

    I got 48

  3. Anonymous says:

    Each step takes 1 second, and he takes 5 steps forward = 5 seconds. Then he takes 3 steps back, which takes 3 seconds. So in 8 seconds (5+3), he’s moved forward 2 meters (each step = 1 m, 5 m forward, 3 backwards leaves him 2 m from where he started). So, 8 seconds, 2 meters; 16 seconds, 4 meters; 24 seconds, 6 meters; 32 seconds, 8 meters.

    Now, this is where it gets tricky. If we assume that once he reaches the edge of the hole, he immediately falls in, then at the end of the next 5 steps forward (8 meters + 5 meters = 13 meters), he falls in and we’re done. Which would give us a time of 32+5 seconds, for 37 seconds total.

    If however, he has to step past the edge of the hole at 13 meters to fall in, then we add 1 round of 5 forward, 3 back, in which he gains 2 meters (40 seconds, 10 meters), followed by the four steps forward to put him in the hole. So, 44 seconds and he ends at 14 meters/inside the hole.

    Of course, this is all assuming that he walks in a straight line towards the hole the entire time.

  4. Anonymous says:

    It doesnt ask how long it takes him to get to the pit, it asks how long it takes to fall in it…