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?

For Solution SCROLL DOWN...


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.


pg... said...

100 steps

pg... said...

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

class vii a said...

thnx for this answer can anyobody tell me
How many right angled triangles with integral sides hav one side 15

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bharathan sbs said...

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

Mano Ekambaram said...

Thanks Barathan. This is clear.