Find the number of escalator steps from the number of steps made by people walking on it

Question

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.

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Answer 1

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.

Answer 2

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):

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