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?

Solution:

100 Steps.

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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.

## 8 comments:

100 steps

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

thnx for this answer can anyobody tell me

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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

Thanks Barathan. This is clear.

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