A person walks up a stalled escalator in $90 s$ . when standing on the same escalator, now moving, he is carried in $60 s$ . The time he would take to walk up the moving escalator will be

27 s

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72 s

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18 s

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36 s

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Solution

The correct option is B $36 s$

$t_{1} = 90 s \Rightarrow 90 = \frac{d}{v_{m}}$

speed of man $= \frac{d}{90}$

$t_{2} = 60 \Rightarrow 60 = \frac{d}{v_{s}}$

speed of escalator $v_{s} = \frac{d}{60}$

$t = \frac{d}{v_{m} + v_{s}} = \frac{d}{d / 90 + d / 60} = \frac{60 \times 90}{150} = 36 s$

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A person walks up a stalled escalator in 90s. when standing on the same escalator, now moving, he is carried in 60s. The time he would take to walk up the moving escalator will be

The correct option is B 36s t1=90s⇒90=dvm speed of man =d90 t2=60⇒60=dvs speed of escalator vs=d60 t=dvm+vs=dd/90+d/60=60×90150=36s

A person walks up a stalled escalator in $90 s$ . when standing on the same escalator, now moving, he is carried in $60 s$ . The time he would take to walk up the moving escalator will be

The correct option is B $36 s$

$t_{1} = 90 s \Rightarrow 90 = \frac{d}{v_{m}}$

speed of man $= \frac{d}{90}$

$t_{2} = 60 \Rightarrow 60 = \frac{d}{v_{s}}$

speed of escalator $v_{s} = \frac{d}{60}$

$t = \frac{d}{v_{m} + v_{s}} = \frac{d}{d / 90 + d / 60} = \frac{60 \times 90}{150} = 36 s$