Escalators!

Algebra Level 4

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?


The answer is 100.

Prateek Mehra by Prateek Mehra , 22, India

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3 solutions

Prateek Mehra
Dec 12, 2014

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 the following equation: 

L steps - 50 seconds × v steps/second = 50 steps.

Downwards (against the direction of the escalator), you walk 5 steps per second. You need 125 steps, so that takes 25 seconds. This gives the following equation: 

L steps + 25 seconds × v steps/second = 125 steps.

From the two equations follows: L = 100, v = 1. When the escalator stands still, you need 100 steps.

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Can you edit the question for clarity?
What is the final question?
Is the walking and running speed identical?

Calvin Lin Staff - 6 years, 6 months ago

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Let the speed of the escalator be ’a’ steps per sec. Going up the effective speed is (1+a), and takes 50 1 = 50 s e c . For going down effective speed is (5-a), and takes 125 5 = 25 s e c . But steps covered up and down are the same. ⟹ ( 1 + a ) ∗ 50 = s t e p s c o v e r e d = ( 5 − a ) ∗ 25. ⟹ a = 1 s t e p / s e c . ∴ s t e p s c o v e r e d = ( 1 + 1 ) ∗ 50 = 100. \text{Let the speed of the escalator be 'a' steps per sec.}\\\text{Going up the effective speed is (1+a), and takes } \dfrac{50}{1}=50~sec.\\\text{For going down effective speed is (5-a), and takes }\dfrac{125}{5}=25~sec.\\\color{#3D99F6}{\text{But steps covered up and down are the same.} } \\ \implies ~(1+a)*50=steps~covered~=(5-a)*25.\\\implies~a=1~step/sec.~~\therefore~steps ~covered=~~(1+1)*50=~~~~100.

Niranjan Khanderia - 6 years, 1 month ago

What is the question dude?

Pranjal Jain - 6 years, 6 months ago

Let the speed of the escalator be ’a’ steps per second. G o i n g u p t a k e s 50 1 = 50 s e c . G o i n g d o w n t a k e s 125 5 = 25 s e c . For up the effective speed is =1+a. For down it is 5 - a. But the steps covered are the same. ⟹ ( 1 + a ) ∗ 50 = s t e p s c o v e r e d = ( 5 − a ) ∗ 25. ∴ a = 1 s t e p / s e c . a n d s t e p s c o v e r e d = ( 1 + 1 ) ∗ 50 = 100. \text{Let the speed of the escalator be 'a' steps per second.} \\Going ~up ~takes ~\dfrac{50}{1} =50~ sec.~ Going~ down~ takes~\dfrac{125}{5}=25~sec.\\\text{For up the effective speed is =1+a. For down it is 5 - a. }\\\color{#D61F06}{\text{But the steps covered are the same.} }\\\implies (1+a)*50=steps~covered=(5-a)*25.\\\therefore a=1~ step/sec. ~~and~~ steps ~covered = (1+1)*50=100.

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Joseph Yiin
Jan 3, 2015

This question is unclear in whether it is asking total step in going up and down(200) or just the steps going down(100), if the elevator is still

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