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
How many right angled triangles with integral sides hav one side 15
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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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