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Renata walks down an escalator that moves up and counts $150$ steps. Her sister Fernanda climbs the same escalator and counts $75$ steps. If the speed of Renata (in steps per time unit) is three times the speed of Fernanda, determine how many steps are visible on the escalator at any time.

The answer is 120 steps.

This is my first post ever on stackexchange, hello!

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    $\begingroup$ Welcome to SE and Math SE! What do you mean by visible? $\endgroup$
    – BCLC
    Aug 23, 2014 at 19:43
  • $\begingroup$ Is this question even possible? I don't even know where to start... $\endgroup$
    – Shahar
    Aug 23, 2014 at 19:43
  • $\begingroup$ When you say has 150 steps do you mean that the escalator has 150 steps or that Renata counted 150 steps? $\endgroup$
    – Shahar
    Aug 23, 2014 at 19:45
  • $\begingroup$ @Shahar From the context, it has to be the latter; otherwise the question would not make sense. The visible steps are those that are exposed on the escalator at any time. $\endgroup$
    – user147263
    Aug 23, 2014 at 20:09
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    $\begingroup$ @BCLC, imagine you can see the escalator: at any time, the quantity of steps you can see is constant, that's what I mean! And thank you for your welcome! :) $\endgroup$
    – Ashitaka
    Aug 23, 2014 at 22:28

2 Answers 2

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Let:

  • $x$ be the number of visible steps.
  • $y$ be the speed of the escalator (in steps per time unit).
  • $z$ be the speed of Fernanda (in steps per time unit), relative to the escalator.

Consider Renata's trip. Notice that in the time that Renata takes to finish her trip, she has travelled $150$ steps at a speed of $3z$, while the escalator has travelled $150 - x$ steps at a speed of $y$. This yields: $$ \frac{150}{3z} = \frac{150 - x}{y} $$ Now consider Fernanda's trip. Notice that in the time that Fernanda takes to finish her trip, she has travelled $75$ steps at a speed of $z$, while the escalator has travelled $x - 75$ steps at a speed of $y$. This yields: $$ \frac{75}{z} = \frac{x - 75}{y} $$ Dividing the first equation by the second yields: \begin{align*} \frac{\frac{150}{75}}{\frac{3z}{z}} &= \frac{\frac{150 - x}{x - 75}}{\frac{y}{y}} \\ \frac{2}{3} &= \frac{150 - x}{x - 75} \\ 2(x - 75) &= 3(150 - x) \\ 2x - 150 &= 450 - 3x \\ 5x &= 600 \\ x &= 120 \end{align*} as desired.

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  • $\begingroup$ Thank you, Adriano! Good solution, thanks! $\endgroup$
    – Ashitaka
    Aug 23, 2014 at 22:46
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Let ${ t }_{ 1 }$ and ${ t }_{ 2 }$ be the time that Renata and Fernanda, respectively, need to complete the trip. And $x$ the number of visible steps.

Considering Renata's trip and knowing that ${ v }_{ r }=3{ v }_{ f }$

$3v_{ f }-v_{ e }=\frac { x }{ t_{ 1 } } \\ v_{ f }=\frac { 150 }{ { t }_{ 1 } }$

Dividing the first equation by the second, we find:

$x=\frac { 50(3v_{ f }-v_{ e }) }{ v_{ f } } (I)$

Now, considering Fernanda's trip:

$v_{ f }+v_{ e }=\frac { x }{ { t }_{ 2 } } \\ v_{ f }=\frac { 75 }{ t_{ 2 } }$

Dividing the first equation by the second, we find:

$x=\frac { 75(v_{ f }+v_{ e }) }{ v_{ f } } (II)$

And we have:

$(I)=(II)→50(3v_{ f }-v_{ e })=75(v_{ f }+v_{ e })→\frac { v_{ e } }{ v_{ f } } =\frac { 3 }{ 5 } (III)$

Now, using (III) in (II):

$x=75\left( 1+\frac { v_{ e } }{ v_{ f } } \right) =75\left( 1+\frac { 3 }{ 5 } \right) =120$

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