Escalator

Math Puzzle:

You walk upwards on an escalator, with a speed of 1 step per second. After 50 steps you are at the end. You turn around and run downwards with a speed of 5 steps per second. After 125 steps you are back at the beginning of the escalator. How many steps do you need if the escalator stands still?

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Solution:

100 Steps.

Let v be the speed of the escalator, in steps per second. Let L be the number of steps that you need to take when the escalator stands still. Upwards (along with the escalator), you walk 1 step per second. You need 50 steps, so that takes 50 seconds. This gives: L - 50 × v = 50. Downwards (against the direction of the escalator), you walk 5 steps per second. You need 125 steps, so that takes 25 seconds. This gives: L + 25 × v = 125. From the two equations follows: L = 100, v = 1. When the escalator stands still, you need 100 steps.

9 comments:

  1. say escalator speed x steps/sec.

    so total steps = 50+50x.(from upward condition).
    Total time to reach up is 50 sec.
    total time to reach down = 25 sec.(125 steps, 5 steps/sec)
    total steps = 125- 25x
    50+50x = 125 - 25x
    75 x = 75=>x =1,
    so total steps = 50 + 50*1= 100

    ReplyDelete
  2. thnx for this answer can anyobody tell me
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  5. dis=speed*time
    let as assume the steps as distance ...there fore we need to find the distance
    (also consider the boat and stream concept here)
    speed of escalator is v
    upwards: d=50(1+v).......(1=speed of man,50 is the time)
    downwards:d=25(5-v)......(5 is the speed of man,25 is the time)
    on solving d=100

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