Escalator Problems!!

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Okay.. These are one of the more trickier problems in Quants... *Problem #1* A person walking takes 26 steps to come down on a escalator and it takes 30 seconds for him for walking. The same person while running takes 18 second and 34 steps. H...
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Thanx man.

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You are Da man! :)
Thanx for everything! Now I know tooo.. 😃

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hi,
here r the answers

Problem #1
A person walking takes 26 steps to come down on a escalator and it takes 30 seconds for him for walking. The same person while running takes 18 second and 34 steps. How many steps are there ??
Sol:
Let's suppose that escalator moves n steps/sec.
It is given that if he walks he takes 30 sec and covers 26 steps.
So in that 30 sec escalator would have covered 30n steps.
Hence the total number of steps on the escalator is 26 + 30n----(1)

Similarly when he runs he takes 18 sec and covers 34 steps.
So in 18 sec escalator covers 18n steps.
Hence total steps on the escalator must be 34 + 18n-------(2)

Equating (1) & (2) 26 +30n = 34 + 18n we get n= 2/3
Hence no. steps is 26+30(2/3) = 46.

Problem #2
An escalator is descending at constant speed. A walks down and
takes 50 steps to reach the bottom. B runs down and takes 90 steps
in the same time as A takes 10 steps. How many steps are visible
when the escalator is not operating?

Sol: Lets suppose that A walks down 1 step / min and
escalator moves n steps/ min
It is given that A takes 50 steps to reach the bottom
In the same time escalator would have covered 50n steps
So total steps on escalator is 50+50n.

Again it is given that B takes 90 steps to reach the bottom and time
taken by him for this is equal to time taken by A to cover 10 steps i.e
10 minutes. So in this 10 min escalator would have covered 10n steps.
So total steps on escalatro is 90 + 10n

Again equating 50 + 50n = 90 +10n we get n = 1
Hence total no. of steps on escalator is 100.

Problem #3
There is a escalator and 2 persons move down it. A takes 50 steps and B takes 75 steps while the escalator is moving down. Given that the time taken by A to take 1 step is equal to time taken by B to take 3 steps, find the no. of steps in the escalator while it is staionary.

Sol: Let A take 1 step/min.
Hence B takes 3 steps/min. Let the escalator take n steps/min
Given that A takes 50 steps and hence in the same time escalator will
take 50n steps. So total no. of steps must be 50 + 50n

It is given that B takes 75 steps (which means he takes 25 min)
So in the same time escalator will cover 25n steps.
Hence no. of steps must be 75 +25n

Equating 50+50n = 75+25n we get n = 1.
Hence total no. of steps must be 100.
I hope my answers are correct and my explanation is lucid

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Okay.. These are one of the more trickier problems in Quants...
Problem #1
A person walking takes 26 steps to come down on a escalator and it takes 30 seconds for him for walking. The same person while running takes 18 second and 34 steps. How many steps are there ??

Problem #2
An escalator is descending at constant speed. A walks down and
takes 50 steps to reach the bottom. B runs down and takes 90 steps
in the same time as A takes 10 steps. How many steps are visible
when the escalator is not operating?

Problem #3
There is a escalator and 2 persons move down it. A takes 50 steps and B takes 75 steps while the escalator is moving down. Given that the time taken by A to take 1 step is equal to time taken by B to take 3 steps, find the no. of steps in the escalator while it is staionary.

I need detailed answers with the solutions..
Thanx for al ya help! and Happy solving!

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