A person walks up a stalled escalator in 90 s. When standing on the same escalator, now moving up, he is carried up in 60 s. How much time (in s) would it take him to walk up the moving up escalator?

Let speed of person be u and length of escalator be $l.$ 

$\therefore$ Time taken by person to walk up on stalled escalator  $=\frac{\mathcal{l}}{u}=90 ....\left(1\right)$

When escalator is moving let its speed be $v,t=\frac{\mathcal{l}}{v}=60 ....\left(2\right)$

$\frac{\left(1\right)}{\left(2\right)}\Rightarrow u=2v, \frac{v}{u}=\frac{3}{2}\Rightarrow 2u=3v$

So time taken by man in walking up the moving escalator $=t\text{'}=\frac{\mathcal{l}}{u+v}.$

$t\text{'}=\frac{\mathcal{l}}{u+\frac{3u}{2}}=\frac{2\mathcal{l}}{5u}=\frac{2}{5}\left(\frac{\mathcal{l}}{u}\right)=\frac{2}{5}\times 90=36\sec .$