On a working escalator, one man walking down it took 50 steps. Another man, walking three times faster, took 75 steps.

How many steps would each man take going down the same escalator during a power outage?

(In reply to posting solution by Sarah)

Sarah, I like yr thinking so am attepting to see how this works out... copying Tristans solution to start, we have;

Let v = the velocity of the escalator (down)
m = the velocity of the first man (down)
s = the number of steps on the escalator

The velocity of the second man is 3m
The amount of time for each man to go down is 50/m and 75/(3m) respectively

50 + v(50/m) = s

But now things start to change due to the change of direction...

75 - v(75/3m) = s
v/m = s/50 - 1
75 - 25(s/50 - 1) = s
100 = 3s/2
s = 66.6 steps

Hope I got the math right?

Posted by Percy on 2006-04-01 20:08:24